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https://mathoverflow.net/questions/426409 | 0 | Let $X$ be a compact metric space, and fix an arbitrary point $x\_\ast \in X$. By the Kantorovich-Rubinstein duality theorem, the $1$-Wasserstein metric $W\_1$ on the set of Borel probability measures on $X$ can be expressed as
$$ W\_1(\mu,\nu) = \max\left\{ \int f \, d\mu - \int f \, d\nu \, : \, f \in \mathcal{K}\_0\right\}\text{,} $$
where $\mathcal{K}\_0$ is the compact set (in the topology of uniform convergence) consisting of all $1$-Lipschitz functions $f \colon X \to \mathbb{R}$ with $f(x\_\ast)=0$.
I now want to move from the purely topological setting to a "smooth" setting.
---
Suppose $X$ is the closure of a bounded convex open subset of $\mathbb{R}^N$, with the boundary being an $(N-1)$-dimensional $C^\infty$ embedded submanifold of $\mathbb{R}^N$. For each $r \in \mathbb{N}$, let $C^r(X,\mathbb{R})$ be the set of functions $f \colon X \to \mathbb{R}$ admitting a $C^r$ extension to an $\mathbb{R}^N$-neighbourhood of $X$.
>
> For each $r \in \mathbb{N}$, does there exist a set $\mathcal{K}\_r \subset C^r(X,\mathbb{R})$ that is compact in the $C^0$ topology on $X$ (i.e. topology of uniform convergence) such that $W\_1$ can be expressed as
>
>
> $$ W\_1(\mu,\nu) = \max\left\{ \int f \, d\mu - \int f \, d\nu \, : \, f \in \mathcal{K}\_r\right\} \text{?} $$
>
>
>
| https://mathoverflow.net/users/15570 | Is there a "smooth Kantorovich-Rubinstein duality" for Wasserstein distances on smooth/Euclidean space? | $\newcommand{\K}{\mathcal K}\newcommand{\R}{\mathbb R}\newcommand{\de}{\delta}$No, the formula
\begin{equation\*}
W\_1(\mu,\nu) = \max\Big\{ \int f \, d\mu - \int f \, d\nu \, : \, f \in \K\_r\Big\}, \tag{1}\label{1}
\end{equation\*}
with $\max$ rather than $\sup$, does not hold in general, for whatever choices of an integer $r\ge1$ and a set $\K\_r \subset C^r(X,\R)$
(the $\sup$ version of \eqref{1} holds by approximation).
Indeed, let $N=1$ and $X:=[-1,1]$.
Note that all functions in $\K\_r$ must be $1$-Lipschitz. Otherwise, there would exist a function $f\in\K\_r$ and points $x,y$ in $X$ such that $f(x)-f(y)>|x-y|$. But then we would have
\begin{equation\*}
|x-y|=W\_1(\de\_x,\de\_y)\ge\int f \, d\de\_x - \int f \, d\de\_y=f(x)-f(y)>|x-y|,
\end{equation\*}
a contradiction. (Here, of course, $\de\_z$ is the Dirac probability measure supported on $\{z\}$.)
Suppose now that \eqref{1} holds for
\begin{equation\*}
\mu:=\tfrac12\,(\de\_{-1}+\de\_1)\quad\text{and}\quad\nu:=\de\_0.
\end{equation\*}
Then there exists a $1$-Lipschitz function $f\colon[-1,1]\to\R$ in $C^1(X,\R)$ such that
\begin{equation\*}
\begin{aligned}
1&= W\_1(\mu,\nu)=\int f \, d\mu- \int f \, d\nu=\tfrac12\,(f(-1)+f(1))-f(0) \\
&=\tfrac12\,(f(-1)-f(0)+f(1)-f(0))\le\tfrac12\,(|-1-0|+|1-0|)=1.
\end{aligned}
\end{equation\*}
It follows that
\begin{equation\*}
f(-1)-f(0)=f(1)-f(0)=1.
\end{equation\*}
Since $f$ is $1$-Lipschitz, it now follows that $f(x)-f(0)=|x|$ for all $x\in[-1,1]$. So, $f\notin C^1(X,\R)$, a contradiction. $\quad\Box$
| 3 | https://mathoverflow.net/users/36721 | 426415 | 173,066 |
https://mathoverflow.net/questions/426420 | 1 | A connected sum decomposition of a closed $n$-manifold $M^n$:
$M^n = M\_1^n \# M\_2^n$, is to view $M^n$ as two closed $M\_1^n$ and $M\_2^n$, joined
by a neck $I\times S^{n-1}$.
Similarly, a $k$-connected sum decomposition of a closed $n$-manifold $M^n$:
$M^n = M\_1^n \#\_k M\_2^n$, is to view $M^n$ as two closed $M\_1^n$ and $M\_2^n$, joined
by $k$ necks, each has a form $I\times S^{n-1}$.
I like to know the reference is this "$k$-connected sum decomposition". Is there a classification of $n$-manifolds via the $k$-connected sum decomposition? Is there a classification of 3-manifolds via the $k$-connected sum decomposition?
| https://mathoverflow.net/users/17787 | Multi-connected sum decomposition of $n$-manifolds | I don't have a reference, but I think it's not too hard to see that $M\_1 \#\_k M\_2 \approx M\_1 \# X\_k \# M\_2$, where $X\_k = (S^1 \times S^{n-1})^{\# (k-1)}$. (Connected sum depends on choices of embedded disks, and so does $k$-connected sums; I'm assuming you want all $k$ pairs of disks to be isotopic.) So decomposing using $k$-connected sum is not so different from decomposing using usual connected sum, except for some bookkeeping about extra summands of $S^1 \times S^{n-1}$.
The idea of this formula $M\_1 \#\_k M\_2 \approx M\_1 \# X\_k \# M\_2$ is that gluing the first neck gives $M\_1 \# M\_2 \approx M\_1 \# S^n \# M\_2$; the remaining $k-1$ necks may then be attached to the middle $S^n$ instead of between $M\_1$ and $M\_2$, each of which creates an $(S^1 \times S^{n-1})$-summand.
| 5 | https://mathoverflow.net/users/171227 | 426421 | 173,068 |
https://mathoverflow.net/questions/426423 | 3 | As is known to all, the Lie algebra $\frak{sl}\_2$ admits a very nice representation on
$$
\mathbb{K}[X,Y]
$$
the polynomials in two variables, given by
$$
E \mapsto X\frac{\partial }{\partial Y}, ~~ F \mapsto Y\frac{\partial }{\partial X}, ~~ H \mapsto X\frac{\partial }{\partial X} - Y\frac{\partial }{\partial Y}.
$$
Does an analogous representation exist for general $\frak{sl}\_n$?
| https://mathoverflow.net/users/153228 | A representation of $\frak{sl}_n$ as partial derivatives on polynomials | The group $GL(V)$ acts on a vector space $V$ by linear automorphisms, and this induces the action of its Lie algebra $\mathfrak{gl}(V)$, i.e. a homomorphism to the Lie algebra of vector fields (differentiations). Explicitly
$$e\_{ij} \mapsto X\_i \partial\_j.$$
The composition of two differentiations is a second order differential operator, and the commutator is another differentiation. All these polynomial differential operators naturally act on the ring $\mathbb{K}[V]$ of polynomial functions.
For $\mathfrak{sl}(V) \subset \mathfrak{gl}(V)$ just restrict to a subalgebra. For explicit operators like $E,F,H$ choose a basis of $\mathfrak{sl}(V)$, e.g. $e\_{ij}$ for $i\neq j$ and $e\_{ii} - e\_{i+1,i+1}$.
| 5 | https://mathoverflow.net/users/6468 | 426424 | 173,070 |
https://mathoverflow.net/questions/424857 | 1 | I have a set of points $X=\{x\_1,\ldots,x\_N\}$ on the unit sphere in $\mathbb{R}^d$ ($N>d\ge 3$). What is an algorithm to find any facet of the polyhedron whose vertices are $X$?
The set $X$ has some features that may be helpful:
* It is highly symmetric in the sense that for each pair $x,x'\in X$ there is an isometry that maps $X$ to itself while mapping $x\to x'$. ($X$ "looks the same" from the vantage of any point in the set.) For this reason it is sufficient for me to find any facet; by symmetry the others will look the same.
* I can efficiently enumerate the points closest to any given $x\in X$.
Note, I am *not* looking for a software tool. The set $X$ is a theoretical construct and I wish to use my knowledge about $X$ and some facet-finding procedure TBD to determine the facet geometry analytically. An example of the *kind* of answer I am looking for might be something like "1. Pick any vertex. 2. Find one of its nearest neighbors. 3. Find one of that vertex's nearest neighbors ... "
| https://mathoverflow.net/users/484293 | Algorithm to find a facet of a polyhedron given the vertices? | Inspired by Matt F's response, I figured out a solution (I think). Intuitively, a procedure that will find *some* facet starting from a given vertex $x\_1$ is the following: Start with a hyperplane containing the point $x\_1$ and orthogonal to the vector $x\_1$. Then tilt the hyperplane in some direction until it intersects another vertex. Using those two vertices as pivot points, tilt the hyperplane in some other direction until it intersects a third vertex, and so on until you have $d$ linearly independent vertices.
More formally, let $x\_i$ denote the vertex selected in iteration $i$ and let $u\_i$ denote a vector in the span of $x\_1,...,x\_i$ and orthogonal to the face defined by these vertices. "Tilting the hyperplane" is achieved by adding to $u\_i$ any vector orthogonal to $x\_1,\ldots,x\_i$.
Given an initial vertex $x\_1$, start by taking $u\_1 = x\_1$. Then to find $x\_{i+1}$ and $u\_{i+1}$,
1. Define
$$
\lambda\_i(x) = \frac{1 - x \cdot u\_i}{||\Pi\_i x||}
$$
where $\Pi\_i$ is the projector onto the subspace orthogonal to $x\_1,\ldots,x\_i$.
2. Take
$$
x\_{i+1} = \arg \min\_{x\in X} \lambda\_i(x) \\
u\_{i+1} = u\_i + \lambda\_i(x\_i) \frac{\Pi\_i x\_i}{||\Pi\_i x\_i||}
$$
After $d$ iterations, one obtains $d$ linearly independent vertices $F = \{x\_1,\ldots,x\_d\}$. It can be shown inductively that $u\_d \cdot x\_1 = \cdots = u\_d \cdot x\_d = 1$ and $u\_d \cdot x < 1$ for all $x\in X - F$. Thus $F$ defines a facet.
| 1 | https://mathoverflow.net/users/484293 | 426428 | 173,072 |
https://mathoverflow.net/questions/254953 | 5 | This question is closely connected to the following paper and to a prior question posted to Mathoverflow titled "Can Cantor's theorem be proved in Paraemter Free Zermelo",
<https://wwwmath.uni-muenster.de/u/rds/ZFC_without_parameters.pdf>
That paper provided an answer to a question about whether $\text{ZFC}^o$ is equivalent to $\text{ZFC}$, where $\text{ZFC}^o$ is the theory axiomatized by axioms of Extensionality, Foundation, Pairing, Union, Power, Infinity and a Parameter free version of Replacement and Separation denoted by $\text{Repl}^o$, $\text{Aus}^o$ respectively; where the axioms of Pairing, Union and Power are written in full, i.e. one can prove existence the sets $\{a,b\}$; $\bigcup a$; $\mathcal P(a)$ for any sets $a,b$ using those three axioms and Extensionality without using an instance of schema of specification.
In the Wikipedia, axioms of $\text{ZFC}$ are written in a different manner, see the following link:
<https://en.wikipedia.org/wiki/Zermelo%E2%80%93Fraenkel_set_theory>
The axioms of Pairing, Union, Power and Replacement only asserts existence of a non specific set that contains sets $a,b$ ; elements of elements of $a$; subsets of $a$, replacements of elements of $a$ respectively; one needs to use Specification in order to prove existence of the sets $\{a,b\}$; $\bigcup a$; $\mathcal P (a)$, $F(a)$
Now if by $\text{eZFC}^o$, denoting $\text{"extreme Parameter Free ZFC"}$, it is meant $\text{ZFC}$ axiomatically presented as how it is mentioned in the Wikipeida but with Specification and Replacement axioms replaced by parameter free versions of them, i.e. by $\text{Repl}^o$ and $\text{Aus}^o$ with these two schemata being formalized exactly as in the above-mentioned article (see: Postscript) for full exposition of $\text {eZFC}^o$. Then apparantly the proof presented in the above article fails, because it clearly depends on specifying pairs, unions, powers and replacment sets without the need to use Specification, and thus in some manner it allows the use of Parameters external to Specification scheme. Along the same line $\text{eZC}^o$ denotes $\text{"extreme Parameter Free Zermelo"}$, and this is $\text{eZFC}^o$ minus Replacment.
Given that, the questions are:
>
> Is $\text{eZFC}^o$ equivalent to $\text{ZFC}$?
>
>
>
>
> Is $\text{eZC}^o$ equivalent to $\text{ZC}$?
>
>
>
If both are false then is Cantor's theorem provable in $\text{ZFC}^o$ or $\text{ZC}^o$?
---
PS: $\text{eZFC}$ is formulated in the first order language of set theory with
the following axioms:
Extensionality: $\forall A,B (\forall X (X \in A \iff X \in B) \iff \forall Y (A \in Y \iff B \in Y))$
Foundation: $\forall A (\exists X \in A \implies \exists B \in A (\not \exists C \in B (C \in A)))$
Specification: if $\phi$ is a formula in which only symbol $X$ occurs free, then
$\forall A \exists B \forall X (X \in B \iff X\in A \wedge \phi)$
is an axiom.
Pairing: $\forall A,B \exists C (A \in C \wedge B\in C)$
Union: $\forall A \exists B \forall X,Y (X \in Y \wedge Y \in A \implies X \in B)$
Power: $\forall A \exists B \forall X (\forall Y \in X (Y \in A) \implies X \in B)$
Replacement: If $\phi(Y,Z)$ is a formula in which only symbols $Y,Z$ occur free, then:
$\forall A \exists B \forall X (\exists Y \in A (\forall Z (\phi(Y,Z) \iff Z=X)) \implies X \in B)$
is an axiom.
Infinity: $\exists X (\varnothing \in X \wedge \forall Y \in X (\{Y\} \in X))$
$\text {+\-}$
$AC$
| https://mathoverflow.net/users/95347 | Is Extreme Parameter Free ZFC equivalent to ZFC? | The answer is to the negative for both questions. A result proved by Levy. This can be seen in Kanamori's article: [Levy and set theory](http://math.bu.edu/people/aki/11.pdf). Annals of Pure and Applied Logic 140 (2006) 233–252
In page 247, he wrote:
>
> Levy’s main, negative results addressed another issue of self-refinement in axiomatics and showed that his aforementioned positive result is reasonably sharp. In the presence of Separation the generative axioms are sometimes given parsimoniously in a weaker, conditional form, e.g. $\forall u \exists y
> \forall x (x \subseteq u \to x \in y) $ for the Power Set Axiom. With his positive result, the set $\sf T $ consisting of the usual $\sf ZF$ axioms, but with the Separation schema replaced by $S\_0$ and the Replacement schema replaced by the conditional version, is an axiomatization of $\sf ZF$. **Levy established that full Separation is not a consequence of $\sf T$ if the Power Set axiom is weakened to the conditional form. He also established the analogous results for the conditional version of Union and the conditional version of Pairing.**
>
>
>
[Note: $S\_0$ is the same as $\text{Aus}^o$ here]
| 5 | https://mathoverflow.net/users/95347 | 426433 | 173,074 |
https://mathoverflow.net/questions/426435 | 4 | Let $F: \mathcal{C} \to \mathcal{D}$ be a left Quillen functor between model categories. In Definition 2.16 of [Goerss–Schemmerhorn - Model Categories and Simplicial Methods](https://arxiv.org/abs/math/0609537), the left derived functor $LF: \mathcal{C} \to \operatorname{Ho}(\mathcal{D})$ is defined by $X \mapsto F(P)$. However, the article does not define the functor on maps.
In practice, this is often done under the assumption that $\mathcal{C}$ has functorial factorisation (or at least functorial cofibrant replacement), but Goerss–Schemmerhorn seem to imply that this assumption is not needed. Could someone help me understand a construction that works in general, without requiring functorial factorisation?
My best guess is to fix an arbitrary cofibrant replacement $P\_X \to X$ for each $X \in \mathcal{C}$, and then try to construct a well-defined map $F(P\_X) \to F(P\_Y)$ in the homotopy category of $\mathcal{D}$, given a morphism $f:X\to Y$. We get morphisms
$$
F(P\_X) \rightarrow F(X) \overset{f}{\to} F(Y) \leftarrow F(P\_Y).
$$
If the right-hand arrow were a weak equivalence, then we would be done, since it would be invertible in the homotopy category. This is the case if $Y$ is cofibrant, since left Quillen functors preserve weak equivalences between cofibrant objects. They do not preserve weak equivalences in general though, so the approach doesn't seem to work.
| https://mathoverflow.net/users/169035 | Can we define derived functors in model categories without functorial factorisations? | This depends on whether one insists on derived functors landing in the original model category (as is the case with modern approaches of Hinich and Dwyer–Hirschhorn–Kan–Smith), or in its homotopy category (as is the case with the classical approaches of Quillen, Grothendieck, etc.).
In the latter case it is indeed possible to define the left derived functor of a left Quillen functor in the original setting of (closed) model categories of Quillen, i.e., without the existence of functorial factorizations and small (co)limits, added later by Kan.
Given a left Quillen functor $$F\colon C→D,$$
pick a cofibrant replacement (not necessarily functorial) $Q\_X$
and an acyclic fibration $q\_X\colon Q\_X→X$
for every object $X∈C$.
Now construct the left derived functor $$\def\ldf{{\bf L}}\def\Ho{{\rm Ho}} \ldf F\colon C→\Ho(D)$$ as follows.
Send an object $X∈C$ to $Q\_X∈\Ho(D)$.
Given a morphism $f\colon X→Y$ in $C$,
factor the composition $f∘q\_X\colon Q\_X→Y$ through the acyclic fibration $q\_Y\colon Q\_Y→Y$ obtaining a map $g\colon Q\_X→Q\_Y$. Send $f$ to the morphism $$F(g)$$ in $\Ho(D)$.
To see that this construction yields a functor, consider composable morphisms $f\colon X→Y$, $g\colon Y→Z$ together with their chosen lifts $Q\_f\colon Q\_X→Q\_Y$, $Q\_g\colon Q\_Y→Q\_Z$, as well as the lift of the composition $Q\_{gf}\colon Q\_X→Q\_Z$.
Now $Q\_g Q\_f\colon Q\_X→Q\_Z$ is another lift for $gf\colon X→Z$.
Any two lifts of the same morphism through an acyclic fibration are left homotopic (see, for example, Proposition 1.2.5(iv) in Hovey's book).
Thus, the two lifts map to the same morphism in $\Ho(D)$.
An elementary argument then shows that a different choice of $q$ and $Q$ produces the same morphism in $\Ho(D)$, and therefore the same functor $C→\Ho(D)$.
**Added**: This is essentially the construction given by Quillen in Proposition I.4.1 of *Homotopical Algebra*, except that Quillen's construction starts by defining a functor $C→\Ho(C\_c)$, where $C\_c$ denotes the full subcategory of $C$ on cofibrant objects. This functor is then composed with $F$, using Ken Brown's lemma.
| 8 | https://mathoverflow.net/users/402 | 426439 | 173,076 |
https://mathoverflow.net/questions/426440 | 7 | Let $\mathcal A$ be an abelian category. In [this lecture](https://youtu.be/7RNh_cxDYmk), Thomas Nikolaus
1. Defines the unbounded derived category $\mathcal D(\mathcal A)$ as $\mathcal K(\mathcal A)[W^{-1}]$, where $\mathcal K(\mathcal A)=N\_{\mathrm{dg}}(\operatorname{Ch}(\mathcal A))$ and $W$ is the set of quasi-isomorphisms of chain complexes; and
2. Defines the notion of left-derived functor from an $\infty$-category with a set of weak equivalences to an arbitrary $\infty$-category, and shows that the left derived functor of a functor from $\mathcal K(\mathcal A)$ to an arbitrary $\infty$-category can be computed as a limit, provided it exists.
Point (1) is identical to the construction of the unbounded derived category of an abelian category as a triangulated category, before you know anything about K-injective complexes (Lurie’s approach in Higher Algebra is to require the existence of K-injectives to define the $\infty$-categorical unbounded derived category).
Point (2) is identical to Deligne’s approach to defining derived functors ([05S7](https://stacks.math.columbia.edu/tag/05S7)).
My question is, where is (2), or more generally an $\infty$-categorical notion of derived functor, written down? I wasn’t even able to find in HTT, HA, or SAG a definition of derived functor from an $\infty$-category with weak equivalences to another $\infty$-category, but I may have overlooked it in the thousands of pages. I’m especially interested in Nikolaus’s computation of derived functors out of $\mathcal K(\mathcal A)$ as a limit/colimit because it’s easy to connect to everything that’s written down in the language of triangulated categories (e.g. you immediately know you can compute the derived tensor product using a K-flat resolution etc.).
| https://mathoverflow.net/users/37110 | Derived functors out of an unbounded derived $\infty$-category | An account of derived functors between ∞-categories equipped with weak equivalences and fibrations can be found in Section 7.5 of Cisinski's *Higher Categories and Homotopical Algebra*. This setting is sufficient to treat the case of the unbounded derived category of an abelian category.
A more general treatment of derived functors in the setting of ∞-categories equipped with a calculus of fractions can be found in Section 7.2. See, in particular, Theorem 7.2.8 and Corollary 7.2.9, as well as Remark 7.2.21, which connects this setting to the setting of fibrations.
| 5 | https://mathoverflow.net/users/402 | 426444 | 173,077 |
https://mathoverflow.net/questions/426445 | 6 | Not thinking about $h$-cobordism, one usually defines a *cobordism* between manifolds, realizes it is an equivalence relation, chooses an appropriate class of *structured manifolds* (framed, unoriented, spin, ...), and dreams about the group of cobordism classes of these manifolds.
Thinking about $h$-cobordism, whose definition I learned today, one seems to not think this way. At least, I've glanced through the Wikipedia page and some notes/articles about $h$-cobordism and there is no explicit mention of 'the set of $h$-cobordism classes of $n$-manifolds (possibly with structure)' or its (group) structure.
(Taking after Will Sawin and Connor Malin's comments)
There is not an interesting group of $h$-cobordism classes of manifolds in general (see comments.) But there is, for instance, interesting groups of $h$-cobordism classes of homotopy spheres.
>
> Is there any interpretation of these groups of a similar spirit (e.g., using spectra) to the relationship between cobordism and Thom spectra?
>
>
>
| https://mathoverflow.net/users/472967 | Do $h$-cobordism groups arise from a 'Thom-like' spectrum? | This topic has a very different flavor from what is usually meant by cobordism. No, the correspondence between cobordism classes and homotopy classes (of a Thom space or Thom spectrum) has no analogue here; when manifolds, or cobordisms between manifolds, are created by transversality, we don't have much control over their homotopy types.
I would say that the main point of considering h-cobordisms is that surgery theory (in the sense of Wall, Browder, Novikov, ...) is pretty good at answering questions like "How many manifolds, if any, have a given homotopy type?" but where "how many" means up to the equivalence relation of h-cobordism. So, to complete the picture we need to know something about the distinction between this equivalence relation and the one that we originally cared about.
| 10 | https://mathoverflow.net/users/6666 | 426446 | 173,078 |
https://mathoverflow.net/questions/426453 | 8 | Let $\Lambda\subset \mathbb{R}^4$ be a lattice. We identify $\mathbb{R}^4$ with the space $M\_2(\mathbb{R})$ of $2\times 2$ matrices over $\mathbb R$. It then is is clear that the set
$$
\det(\Lambda)=\big\{\det(\lambda):\lambda\in\Lambda\big\}
$$
is not necessarily discrete in $\mathbb R$.
But now we additionally insist that the group
$$
\Gamma=\big\{\gamma\in\mathrm{SL}\_2(\mathbb{R}): \gamma\Lambda=\Lambda\big\}
$$
be a cocompact lattice in $\mathrm{SL}\_2(\mathbb{R})$.
Under this condition, is the set $\det(\Lambda)$ discrete in $\mathbb R$?
| https://mathoverflow.net/users/473423 | Are the determinants of a lattice discrete? | This seems to be related to issues of Diophantine approximation. Fix such $\Lambda$ and $\Gamma$. Fix $\lambda \in \Lambda$ so that $\det(\lambda) \neq 0$. After rescaling $\Lambda$, we may assume that $\det(\lambda) = 1$. Fix some $\gamma \in \Gamma$ other than the identity. Conjugating $\Gamma$ (and transforming $\Lambda$ via the conjugating matrix) we may assume that $\gamma$ is of the following form.
\begin{pmatrix}
\alpha & 0 \\
0 & 1/\alpha
\end{pmatrix}
Suppose that $p$ and $q$ are integers.
Then $p\lambda$ and $q\lambda$ lie in $\Lambda$, thus so does $p\gamma\lambda$ and thus so does $p\gamma\lambda - q\lambda = (p\gamma - q I) \lambda$. Here $I$ is the two-by-two identity matrix.
We now compute:
$$
\det(p\gamma\lambda - q\lambda) = \det(p\gamma - qI) \det(\lambda)
= p^2 + q^2 - pq(\alpha + 1/\alpha)
$$
So if $\alpha$ has unusually good Diophantine approximations then this quadratic form should "approximate zero". In this case, $\det(\Lambda)$ is indiscrete.
| 5 | https://mathoverflow.net/users/1650 | 426460 | 173,080 |
https://mathoverflow.net/questions/426172 | 22 | I am able to construct functions $\sin,\cos\colon \mathbb R \to \mathbb R$ satisfying the following properties:
1. $\sin^2 x + \cos^2 x = 1$,
2. $\sin(x+y)=\sin x \cos y + \sin y\cos x$, $\cos(x+y)=\cos x \cos y - \sin x \sin y$,
3. $\sin(0)=0$, $\cos(0)=1$
4. there exists $\tau>0$ such that $\sin$ and $\cos$ are $\tau$-periodic
5. $\sin$ is strictly increasing if restricted to $[-\tau/4,\tau/4]$.
If you prefer, last condition can also be replaced with
6. $\sin$ and $\cos$ are continuous.
How can I prove that $\lim\_{x\to 0} \frac{\sin x}{x}$ exists and is finite?
I'm trying to define the trigonometric functions following the geometric intuition so I cannot assume that the trigonometric function have been defined and cannot use the complex exponential. I also don't want to use integrals or differential equations.
The above functions are obtained by choosing some unit to measure angles and following the geometric definition of trigonometric functions. If we replace $\sin x$ and $\cos x$ with $\sin cx$ and $\cos cx$ for any $c>0$ we obtain another pair of trigonometric functions with the same properties as above and period $\tau/c$ in place of $\tau$.
The above construction can be achieved by using the natural isomorphism between totally ordered, dense, continuous, additive groups in exactly the same way which allows us to define an isomorphism between the additive group of real numbers and the multiplicative group of positive real numbers obtaining the exponential function $a^x$ for any positive $a$. In fact we can restate the trigonometric functions as real and imaginary part of a complex function $\phi\colon \mathbb R \to \mathbb C$ which takes values on the unit circle in $\mathbb C$ and such that $\phi(x+y)=\phi(x)\phi(y)$.
The constant $\pi$ is defined as being $\tau/2$ when choosing such a unit of measure so that
$$
\lim\_{x\to 0} \frac{\sin x}{x} = 1.
$$
This last property of trigonometric functions settles the radian as natural unit for angles in the same way that the corresponding limit
$$
\lim\_{x\to 0} \frac{a^x-1}{x}=1
$$
settles $a=e$ as the natural base of logarithms.
Additional thoughts
* I know that by defining the exponential complex function $\exp(z)= \sum\_{k=0}^{+\infty} \frac{z^k}{k!}$ we can define everything at once: real exponential, trigonometric function, $e$ and $\pi$. However I find interesting to follow the usual elementary definition of such functions and to settle it rigorously.
* the analogous problem for the exponential function is solved by proving that the limit
$$
\lim\_{n\to +\infty}\left(1+\frac 1 n\right)^n
$$
exists and is finite. Once this is settled to be the number $e$ one can sort of extend this result by monotonicity and find the limit
$$
\lim\_{x\to 0} \left(1+x\right)^{\frac 1 x} = e
$$
which then gives
$$
\lim\_{x\to 0} \frac{\log\_e(1+x)}{x} = 1
$$
and finally
$$
\lim\_{x\to 0} \frac{e^x-1}{x} = 1.
$$
Maybe I should find the analogous path for trigonometric functions. In fact using the angle bisection formulas it is easy to show that
$$
\lim\_{n\to \infty}\frac{\sin(2^{-n})}{2^{-n}}
$$
exists because such a sequence is increasing. What I'm missing is to extend such a result to all sequences going to $0$.
| https://mathoverflow.net/users/36826 | Axiomatic construction of trigonometric functions | This is to complete the [answer](https://mathoverflow.net/a/426247) by Emanuele Paolini by showing that
\begin{equation\*}
a\_n:=\frac{\sin \frac xn}{\frac xn}
\end{equation\*}
is increasing in natural $n$ for each $x\in(0,h]$, where $h$ is any positive real number such that $\sin>0$, $\cos>0$, and $\tan<1$ on the interval $(0,h]$.
The desired inequality,
\begin{equation}
a\_n<a\_{n+1}, \tag{1}\label{1}
\end{equation}
can be rewritten as
\begin{equation\*}
(n+1)\sin ny>n\sin(n+1)y, \tag{2}\label{2}
\end{equation\*}
where $y:=\frac x{n(n+1)}\in(0,\frac h{n(n+1)}]$. Note that
\begin{equation\*}
\sin(n+1)y=\sin ny\cos y+\sin y\cos ny<\sin ny+\sin y\cos ny,
\end{equation\*}
so that
\begin{equation\*}
(n+1)\sin ny-n\sin(n+1)y>\sin ny-n\sin y\cos ny \\
=(\tan ny-n\sin y)\cos ny\ge0,
\end{equation\*}
in view of the inequality $n\sin y\le\tan ny$ for $y\in(0,\frac hn]$, proved in the answer by Emanuele Paolini.
This proves \eqref{2} and hence \eqref{1}. $\quad\Box$
| 4 | https://mathoverflow.net/users/36721 | 426488 | 173,088 |
https://mathoverflow.net/questions/426403 | 14 | Let $A$ be a set of $n$ integers and consider the quantity:
$$\int\_{0}^1 \left| \sum\_{a \in A} e^{2\pi i a x} \right|dx. $$
The (now solved) Littlewood conjecture is the claim that this quantity is lower bounded by $c \log n$ for some universal constant $c>0$.
The (still open) strong Littlewood conjecture is the claim that
$$\min\_{A; |A|=n} \int\_{0}^1 \left| \sum\_{a \in A} e^{2\pi i a x} \right|dx = \int\_{0}^1 \left| \sum\_{a=1}^{n} e^{2\pi i a x} \right|dx$$
Or in, other words, the exponential sum above is minimized when the set $A$ is an arithmetic progression. This seems to be a hard open problem not near resolution.
My question is:
>
> Can one prove the "first case" of the strong Littlewood conjecture. Or, in other words, prove that a three-term arithmetic progression has the smallest $L^1$ norm of any three-term idempotent trigonometric polynomial?
>
>
>
| https://mathoverflow.net/users/630 | The first case of the strong Littlewood conjecture | The following human-verifiable proof is in collaboration\* with Fedja.
**Lemma** $1$: We have the following for $0 \le x \le 3$: $$\frac{6204}{6750}x^2-\frac{8429}{60750}x^4+\frac{4475}{546750}x^6\le x.$$
>
> Proof: The above is equivalent to $$\frac{6204}{750}\left(\frac{x}{3}\right)^2-\frac{8429}{750}\left(\frac{x}{3}\right)^4+\frac{4475}{750}\left(\frac{x}3\right)^6\le x,$$ so letting $x=3y$, multiplying through by $750$, and dividing by $y$, we wish to show $$2250-6204y+8429y^3-4475y^5 \ge 0$$ for $0 \le y \le 1$. The above polynomial is divisible by $1-y$, and dividing yields $$4475y^4+4475y^3-3954y^2-3954y+2250.$$ But $$25\cdot (4475y^4+4475y^3-3954y^2-3954y+2250)$$ $$= 179(5y-3)^4 + (15215y+4638)(5y-3)^2 + 15y+9.$$ $\hspace{130mm}$ $\square$
>
>
>
**Lemma** $2$: For any $a,b \in \mathbb{N}$ with $b > a$ and $b \not = 2a$, we have $$\int\_0^1 \left|1+e^{2\pi i a \theta}+e^{2\pi i b \theta}\right|^4 d\theta \le 15.$$
>
> Proof: A quick proof is that when $b=2a$, the integral is $19$ and there are only $4$ nontrivial 4-tuples $(c\_1,c\_2,d\_1,d\_2) \in \{0,a,b\}$ with $c\_1+c\_2 = d\_1+d\_2$, namely when one is $0$, one is $b$, and the other two are $a$. $\square$
>
>
>
**Lemma** $3$: For any $a,b \in \mathbb{N}$ with $b > a$, we have $$\int\_0^1 \left|1+e^{2\pi i a\theta}+e^{2\pi i b\theta}\right|^6d\theta \ge 93.$$
>
> Proof: We wish to lower-bound the number of 6-tuples $(c\_1,c\_2,c\_3,d\_1,d\_2,d\_3) \in \{0,a,b\}$ with $c\_1+c\_2+c\_3 = d\_1+d\_2+d\_3$. Having $|\{c\_1,c\_2,c\_3\}| = 3$ gives $6\cdot 6 = 36$ 6-tuples, having $|\{c\_1,c\_2,c\_3\}| = 2$ gives $6\cdot 3\cdot 3 = 54$ 6-tuples, and having $|\{c\_1,c\_2,c\_3\}| = 1$ gives $3$ 6-tuples. $\square$
>
>
>
**Proposition** $4$: For any $a,b \in \mathbb{N}$ with $b > a$, we have $$\int\_0^1 \left|1+e^{2\pi i a\theta}+e^{2\pi i b\theta}\right|d\theta \ge \int\_0^1 \left|1+e^{2\pi i a\theta}+e^{2\pi i (2a)\theta}\right|d\theta.$$ Consequently, the first (nontrivial) case of the strong Littlewood conjecture is true.
>
> Proof: Suppose $b \not = 2a$. By Lemmas $1$, $2$, and $3$, we have \begin{align\*} \int\_0^1 \left|1+e^{2\pi i a\theta}+e^{2\pi i b\theta}\right|d\theta & \ge \frac{6204}{6750}\cdot 3 - \frac{8429}{60750}\cdot 15 + \frac{4475}{546750}\cdot 93 \\ &= \frac{130972}{91125} \\ &> 1.437 \\ &> \int\_0^1 \left|1+e^{2\pi i a\theta}+e^{2\pi i (2a)\theta}\right|d\theta,\end{align\*} where we used \begin{align\*} \int\_0^1 |1+e^{2\pi i a\theta}+e^{2\pi i (2a)\theta}|d\theta &= \int\_0^1 |1+e^{2\pi i \theta}+e^{2\pi i 2\theta}| d\theta \\ &= \frac{2\sqrt{3}}{\pi}+\frac{1}{3} \\ &< \frac{2\cdot 1.733}{3.141}+\frac{1}{3} \\ &< 1.437.\end{align\*}
>
>
>
---
\*If this were a paper, Fedja would be an author $99$ times and I once (if lucky).
| 8 | https://mathoverflow.net/users/129185 | 426498 | 173,092 |
https://mathoverflow.net/questions/426491 | 1 | Let $f\_n: \mathbb R\_+\to (0,1]$ be continuous and strictly decreasing for every $n\ge 1$. Assume that the pointwise limit of $(f\_n)\_{n\ge 1}$ exists, denoted by $f$, and is also strictly decreasing. Can we prove
$$\lim\_{n\to\infty}f\_n^{-1}(t)=f^{-1}(t), \quad \mbox{for almost every } t\in (0,1)?$$
The definition of $f\_n^{-1}$ is standard and we set $f\_n^{-1}(t):=\infty$ for all $t\in [0, f\_n(\infty)]$. As for $f$, which generalized inverse we should take for the above purpose?
| https://mathoverflow.net/users/nan | Identification of $\lim_{n\to\infty}f_n^{-1}$ with $f_n:\mathbb R_+\to (0,1]$ strictly decreasing and converging pointwise | $\newcommand{\de}{\delta}\newcommand{\ep}{\varepsilon}$Yes, this is true. Moreover, the continuity and the strictness of the decrease of the $f\_n$'s are not needed.
Indeed, for any nonincreasing function $g\colon[0,\infty)\to(0,1]$ and any $t\in(0,1)$, let
\begin{equation\*}
g^{-1}(t):=\sup\{x\in[0,\infty)\colon g(x)\ge t\}, \tag{0}\label{0}
\end{equation\*}
with $g^{-1}(t):=0$ if $\{x\in[0,\infty)\colon g(x)\ge t\}=\emptyset$. Of course, in the particular case when the function $g$ is continuously and strictly decreasing from $1$ to $0$ on $[0,\infty)$, the generalized inverse $g^{-1}$ defined by formula \eqref{1} coincides with the usual inverse of $g$.
Let us show that
\begin{equation\*}
f\_n^{-1}\to f^{-1} \tag{1}\label{1}
\end{equation\*}
pointwise on the set
\begin{equation\*}
C\_\*:=C\cup(0,t\_\infty),
\end{equation\*}
where $t\_\infty:=\lim\_{x\to\infty}f(x)\in[0,1]$ and $C$ is the set of all points $t\in(0,1)$ such that $f^{-1}$ is finite and continuous at $t$. Note that $f^{-1}=\infty$ on $(0,t\_\infty)$.
Since the function $f^{-1}$ is nonincreasing, the set $(0,1)\setminus C\_\*$ is at most countable. So, it will follow that the convergence \eqref{1} is almost everywhere (a.e.), as desired.
To prove \eqref{1}, take first any $t\_0\in C$ and let $x\_0:=f^{-1}(t\_0)$, so that $x\_0\in[0,\infty)$. Take any real $\ep>0$. The condition $t\_0\in C$ implies that for some real $\de=\de\_\ep>0$ we have
\begin{equation\*}
|f^{-1}(t\_0\pm\de)-f^{-1}(t\_0)|<\ep.
\end{equation\*}
By the pointwise convergence $f\_n\to f$, there is some natural $n\_\ep=n\_{\ep,\de\_\ep}$ such that
\begin{equation\*}
n\ge n\_\ep\implies |f\_n(x\_0\pm\ep)-f(x\_0\pm\ep)|<\de.
\end{equation\*}
So, for $n\ge n\_\ep$ we have the following implications:
\begin{equation\*}
\begin{aligned}
&f\_n^{-1}(t\_0)>x\_0+\ep\implies f\_n(x\_0+\ep)\ge t\_0
\implies f(x\_0+\ep)>t\_0-\de \\
& \implies x\_0+\ep\le f^{-1}(t\_0-\de)<f^{-1}(t\_0)+\ep=x\_0+\ep,
\end{aligned}
\end{equation\*}
a contradiction. So, for $n\ge n\_\ep$,
\begin{equation\*}
f\_n^{-1}(t\_0)\le x\_0+\ep=f^{-1}(t\_0)+\ep.
\end{equation\*}
Similarly, for $n\ge n\_\ep$ we have
\begin{equation\*}
\begin{aligned}
& f\_n^{-1}(t\_0)<x\_0-\ep\implies f\_n(x\_0-\ep)<t\_0
\implies f(x\_0-\ep)<t\_0+\de \\
& \implies x\_0-\ep\ge f^{-1}(t\_0+\de)>f^{-1}(t\_0)-\ep=x\_0-\ep,
\end{aligned}
\end{equation\*}
a contradiction. So, for $n\ge n\_\ep$,
\begin{equation\*}
f\_n^{-1}(t\_0)\ge x\_0-\ep=f^{-1}(t\_0)-\ep.
\end{equation\*}
Thus, convergence \eqref{1} holds on $C$.
Take now any $t\_0\in(0,t\_\infty)$, so that $f^{-1}(t\_0)=\infty$. Take any real $x\ge0$. Then $f(x)>t\_0$. So, there is some natural $n\_x$ such that for all $n\ge n\_x$ we have $f\_n(x)>t\_0$ and hence $f\_n^{-1}(t\_0)\ge x$. It follows that $f\_n^{-1}(t\_0)\to\infty=f^{-1}(t\_0)$. So, convergence \eqref{1} holds on $(0,t\_\infty)$ as well, and thus it does hold on $C\_\*$, as claimed. $\quad\Box$
| 1 | https://mathoverflow.net/users/36721 | 426504 | 173,094 |
https://mathoverflow.net/questions/426503 | 5 | I have a feeling the following is true.
Assume that there are $n$ mutually disjoint closed disks $D\_i$ in the complex plane and $n$ complex polynomials $p\_i(z)$ of degree $n - 1$, with both types of objects indexed from $1$ to $n$, such that
condition: the polynomial $p\_i(z)$ has exactly 1 complex root in each disk $D\_j$, where $1 \leq j \leq n$ and $j \neq i$ and no other roots.
Are the $n$ polynomials $p\_1(z), \ldots, p\_n(z)$ necessarily linearly independent over $\mathbb{C}$? I think it is true.
So for example, if $n = 3$, then $p\_1(z)$ has exactly $1$ root in $D\_2$ and exactly $1$ root in $D\_3$ and no other roots, and so on.
If my intuition is correct, this problem should be some kind of "dual" to the problem where you have $n$ polynomials, say $q\_i(z)$, of degree $n - 1$, where all the roots of $q\_i(z)$ are in the closed disk $D\_i$ (we also assume that the closed disks $D\_i$, for $i = 1, \dots, n$, are mutually disjoint). I know how to prove this possibly "dual" problem using the Grace-Walsh-Szego (accents?) coincidence theorem.
I did not run any numerical simulations yet. Any ideas or suggestions? I came across this problem by thinking about the Atiyah problem on configurations of points.
Edit: the conjecture in this post was proved to be false for $n \geq 3$ by Noam D. Elkies (see his answer below). However, for a follow-up question, please see [On well separated circular regions in the Riemann sphere and complex polynomials](https://mathoverflow.net/questions/426655/on-well-separated-circular-regions-in-the-riemann-sphere-and-complex-polynomials)
| https://mathoverflow.net/users/81645 | A statement on complex polynomials | The conjecture is easily seen to be true for $n<3$.
We give a counterexample for $n=3$.
Let $p\_i = z^2 - \omega\_i$ where the $\omega\_i$ are the cube roots of unity.
These are linearly dependent because they're in the $2$-dimensional subspace
spanned by $1$ and $z^2$. But their zeros are the sixth roots of unity,
forming a regular hexagon $H$ each of whose three long diagonals $d\_i$ joins
the roots of one of the $p\_i$. Let $s\_i$ be pairwise disjoint sides of $H$
such that $s\_i$ is parallel to the corresponding $d\_i$. It's easy to find
discs $D\_i$ such that each $D\_i$ contains $s\_i$ and is disjoint from $s\_j$
for each $j \neq i$; for instance start from the discs with diameter $d\_i$
and dilate each one by a factor $1.01$. Then each $p\_i$ has no roots in $D\_i$
and exactly one root in $D\_j$ for each $j \neq i$, **QEF**.
A similar construction works for $n>3$ using
$p\_i = z^{n-1} - \omega\_i$ where the $\omega\_i$ are the $n$-th roots of unity.
Note that these $p\_i$ are not only linearly dependent but span a linear space
of dimension only $2$, the minimum possible; moreover, their roots are all
on the unit circle, so a fractional linear transformation such as
$z = (z'+i) / (z'-i)$ makes all the roots real, which means that
the $p\_i$ can be taken to have real coefficients.
| 16 | https://mathoverflow.net/users/14830 | 426505 | 173,095 |
https://mathoverflow.net/questions/426499 | 6 | Let $X$ be a separable metric space which is *homogeneous*, i.e. for every two points $x,y\in X$ there is a homeomorphism $h$ of $X$ onto itself such that $h(x)=y$.
A compactification of $X$ is a compact metric space which contains a dense homeomorphic copy of $X$.
Does $X$ have a homogeneous compactification?
Examples of homogeneous compactifications include the circle for the real line, the torus for the plane etc.
| https://mathoverflow.net/users/95718 | Do all homogeneous spaces have homogeneous compactifications? | Since you want a connected example:
A surface of infinite genus has no homogeneous compactification.
Indeed first observe a dense locally compact subset has to be open.
So the surface has to be open, and by homogeneity the compactification is a closed surface. But an open subset of a closed surface has (each component of) finite genus.
| 14 | https://mathoverflow.net/users/14094 | 426508 | 173,097 |
https://mathoverflow.net/questions/426511 | 1 | Let $\hat{\pi}^N$ be an AW-consistent estimator of $\pi$ (i.e., $\hat{\pi}^N$ is a strongly consistent estimator of $\pi$ under adapted (or called nested) Wasserstein distance $AW(\pi, \hat{\pi}^N)\to 0 $ a.s.).
How to prove that $W(\hat{\pi}^N)$ is a consistent estimator of $W(\pi)$?
| https://mathoverflow.net/users/168083 | How to prove that is a consistent estimator? | Take $\pi^N$ with $AW(\pi^N, \pi) \leq \frac{1}{N}$, where we denote by $\mu^N$ and $\nu^N$ the marginals of $\pi^N$.
Note that by the backward induction for $AW$ (cf. [here](https://epubs.siam.org/doi/pdf/10.1137/16M1080197)), it holds
$$
AW(\pi, \pi^N) = \inf\_{\kappa\_1 \in \Pi(\mu, \mu^N)} d\_{X\_1}(x\_1, y\_1) + W\_1(\pi\_{x\_1}, \pi^N\_{y\_1}) \kappa\_1(dx\_1, dy\_1),
$$
and thus we can choose $\kappa\_1^N \in \Pi(\mu, \mu^N)$ such that
$$
\int W\_1(\pi\_{x\_1}, \pi^N\_{y\_1}) \kappa\_1^N(dx\_1, dy\_1) \leq \frac{1}{N},
$$
and further clearly the second marginal converges, i.e., $W\_1(\nu^N, \nu) \leq \frac{1}{N}$.
By applying twice the triangle inequality, we get
\begin{align}
\int W\_1(\pi\_{x\_1}, \nu) \mu(dx\_1) - \int W\_1(\pi^N\_{y\_1}, \nu^N) \mu^N(dy\_1) &= \int W\_1(\pi\_{x\_1}, \nu) - W\_1(\pi^N\_{y\_1}, \nu^N) \kappa\_1(dx\_1, dy\_1) \\
&\leq \int W\_1(\nu, \nu^N) + W\_1(\pi\_{x\_1}, \pi\_{y\_1}^N) \kappa\_1^N(dx\_1, dy\_1) \\
&\leq \frac{2}{N}
\end{align}
and vice versa.
Since the denominator of $W(\pi)$ should be strictly larger than zero (and converges as well since $W\_1(\nu^N, \nu)$ goes to zero), we get that the estimator is consistent.
| 1 | https://mathoverflow.net/users/106046 | 426529 | 173,103 |
https://mathoverflow.net/questions/372903 | 5 | Let $C$ be a category with finite hom-sets.
Suppose that $X$ and $Y$ are objects in $C$ such that $C(Z,X)\cong C(Z,Y)$ for any Z (with no naturality condition).
For which categories $C$ does it follow that $X \cong Y$?
(Of course, it is true for posets).
A somewhat related question is the following.
Let $C$ be a symmetric monoidal closed category.
Suppose that $X$ and $Y$ are objects in $C$ such that $[X,Z]\cong [Y,Z]$ for any Z (with no naturality condition).
For which categories $C$ does it follow that $X \cong Y$?
| https://mathoverflow.net/users/166165 | When is an object determined by the number of maps from the other objects? | The first question is the main topic of two recent papers:
* Reggio's [Polyadic Sets and Homomorphism Counting](https://arxiv.org/abs/2110.11061), which is also a good reference for previous results in the literature (including that of Pultr mentioned in the comments). In particular, see Theorems 4.3, 5.10, and 5.12, the latter of which generalise beyond locally-finite categories.
* Fujino–Matsumoto's [Lovàsz's hom-counting theorem by inclusion-exclusion principle](https://arxiv.org/abs/2206.01994). In particular, see Theorems 2.11 and 2.12.
As far as I understand, the theorems give orthogonal sufficient conditions: it's unclear whether there is a general theorem subsuming both.
I don't believe there are answers for enriched categories yet.
| 3 | https://mathoverflow.net/users/152679 | 426540 | 173,105 |
https://mathoverflow.net/questions/426537 | 1 | Let $X$ be an infinite moving average time series, i.e.
$$
X(t) = \sum\_{k = -\infty}^\infty a\_j \varepsilon\_{t-j}, \quad t \in \mathbb{Z},
$$
where $\varepsilon\_{j}$ are uncorrelated zero mean, finite variance and identically distributed random variables.
In my opinion, $\mathbb{E}X(t) = 0$ for all $t \in \mathbb{Z}$ is easy to prove with if $\sum\_j \vert a\_j \vert < \infty$. This is often referred to as short memory of the moving average, cf. Section 4.2.4 in [1].
But what happens if we only assume $\sum\_j \vert a\_j \vert^2 < \infty$? This is a common assumption, e.g. in Wold's decomposition, and can include both long and short memory moving averages. Is there some alternative way to prove $\mathbb{E}X(t) = 0$ for all $t \in \mathbb{Z}$ under these conditions?
References
----------
[1] J. Beran, Y. Feng, S. Ghosh, and R. Kulik. Long-memory processes. Springer, 2016.
| https://mathoverflow.net/users/302666 | Expected value of long memory moving average | This is a slightly more elementary version of Anthony Quas's answer, without appealing to martingales.
By the condition $\sum\_j|a\_j|^2<\infty$, the sequence of random variables $X\_n:=\sum\_{j=-n}^n a\_j\epsilon\_{-j}$ is [Cauchy convergent](https://en.wikipedia.org/wiki/Cauchy%27s_convergence_test) in $L^2$ and hence in $L^1$. So, $X\_n$ converges to a limit $X(0)$ in $L^1$. Since $EX\_n=0$ for all $n$, we have $EX(0)=0$.
(In view of [Kolmogorov's maximal inequality](https://en.wikipedia.org/wiki/Kolmogorov%27s_inequality), $X\_n\to X(0)$ almost surely as well.)
| 1 | https://mathoverflow.net/users/36721 | 426544 | 173,108 |
https://mathoverflow.net/questions/426502 | 4 | We work in weakly predicatively constructive mathematics, in that we accept function sets but do not accept power sets or excluded middle. More specifically, we shall assume a sequential universe hierarchy in our foundations, where the elements of every universe are (codes for) sets, and every universe $\mathcal{V}$ embeds into a successor universe $\mathcal{V}'$. While there is no global 'set of all propositions', for every universe $\mathcal{V}$ there is a 'set of all propositions of a universe' $\mathrm{Prop}\_\mathcal{V}$ which lies in the next universe $\mathcal{V}'$, where 'proposition' is defined as 'set where every element is equal to every other element'.
In the following question, a Dedekind real number is a two-sided, located, open, rounded, bounded Dedekind cut. A generalised Cauchy real number is an equivalence class of Cauchy nets of rational numbers, in the same way that a Cauchy real number is an equivalence class of Cauchy sequences of rational numbers. Unlike the case with Cauchy sequences in constructive mathematics, here it doesn't matter whether the Cauchy nets come with moduli of convergence or not, because the moduli of convergence are given by a net regardless whether the foundations is classical or constructive. In addition, given a set of rational numbers $\mathbb{Q}$ in a universe $\mathcal{V}$, due to the weakly predicative nature of the foundations, both the set of Dedekind real numbers and the set of generalised Cauchy real numbers lie in the next universe $\mathcal{V}'$. This isn't the case in impredicative mathematics, as only the generalised Cauchy real numbers lie in $\mathcal{V}'$, the Dedekind real numbers lie in $\mathcal{V}$.
This paragraph appears on [the nLab's article about Cauchy real numbers](https://ncatlab.org/nlab/show/Cauchy+real+number#relations_between_these_definitions):
>
> Every generalised Cauchy real number becomes a Dedekind real number in the usual way, defining lower and upper sets in terms of the order relation on generalised Cauchy real numbers. Furthermore, any two generalised Cauchy real numbers are equal as generalised Cauchy real numbers if and only if they are equal as Dedekind cuts. Conversely, any Dedekind cut is equal (as a Dedekind cut) to some generalised Cauchy real number.
>
>
>
However the nLab does not provide a proof that these two are equivalent. Is the nLab's statement correct, and if so, is there a proof to show the validity of the statement?
| https://mathoverflow.net/users/483446 | Equivalence of real numbers in terms of Dedekind cuts and Cauchy nets of rational numbers | This seems like something that would be in the literature somewhere, but I couldn't find a reference, so here's a proof outline. I'll sketch how to get multivalued Cauchy real from Dedekind, generalised Cauchy from multivalued Cauchy and Dedekind from generalised Cauchy. It follows that all three are equivalent. In order for this to work, the multivalued Cauchy condition on a multivalued function $C$ from $\mathbb{N}$ to $\mathbb{Q}$ needs to be that there is $N$ with $|x - y| < \epsilon$ for all $x \in C(n)$, $y \in C(m)$ with $n, m > N$ (not some $x, y$ like on the nlab page).
It follows directly from the definition of Dedekind cut that Dedekind real numbers are located and bounded. If $x$ is a Dedekind real number, there are rational numbers $p, q$ such that $p < x < q$. By induction on natural numbers $n > 1$, using locatedness, we can show that for all $n$ there merely exists a rational number $r$ such that $|x - r| < |p - q| 2^{-n}$. Hence if $x$ is a Dedekind real number, then for all $n$ the set $C(n) := \{ q \in \mathbb{Q} \;|\; x - 2^{-n} < q < x + 2^{-n}\}$ is merely inhabited. Note that $C$ defines a multivalued Cauchy sequence.
Now given any multivalued Cauchy sequence $C$ we can define a generalised real number as follows. The coproduct $\sum\_{n \in \mathbb{N}} C(n)$ is a directed set taking $(n, p) \leq (m, q)$ when $n \leq m$. The second projection followed by inclusion $C(n) \hookrightarrow \mathbb{Q}$ defines a net on this directed set. This is a generalised Cauchy real number.
Finally, given any generalised Cauchy number, say $(q\_i)\_{i \in I}$, we define a Dedekind real as follows. We take the left cut $L$ to be those $p \in \mathbb{Q}$ such that there exists $r > p$ and $j \in I$ such that $q\_i \geq r$ for all $i \geq j$. We similarly define the right cut $R$ by reversing the ordering relations in the above.
| 6 | https://mathoverflow.net/users/30790 | 426553 | 173,111 |
https://mathoverflow.net/questions/426552 | 5 | In the paper by L.Clozel [in this book](https://www.jmilne.org/math/Books/AA1988a.pdf) (a French text), there is this conjecture (conjecture 4.5 p139)
**Conjecture**: Given $\pi$ an algebraic cuspidal representation of $Gl(n)$ of weight $w$ and denote by $E$ it's definition field. Then there exists a degree $n$ motive $M$ over a finite extension of $E$ such that the euler factors of the two L-function coïncide up to a translation of $\frac{1-n}{2}$.
**First question**: How can we generalize this conjecture for the group $GSp\_{4}$?
On the over hand, in [this paper](https://webusers.imj-prg.fr/%7Emichael.harris/occult.pdf), M.Harris associate to $\pi\_{f}$ the finite part of a cuspidal representation $\pi$ of $GSp\_{4}$ a motive $M(\pi\_{f})$ as follows
$$M(\pi\_{f})=H^{3}\_{!}(Sh,\tilde{V}\_{\rho})[\pi\_{f}]$$
It is done p.5, and note by the way that he does not tell us that it "is a motive associated to $\pi$", I just believe that it is what he mean because of the name of the section 1 and notation..
**Second question**: Assume $\pi$ is an algebraic representation of $GSp\_{4}$, does the motive $M(\pi\_{f})$ satisfy the conjecture stated in Clozel's paper and why??
**Last question**: Does the compatible system of Galois representation of $M(\pi\_{f})$ coïncide with the representation in the Theorem 1 p.3 of [this paper](http://www.numdam.org/item/AST_2005__302__67_0.pdf) by Weissauer?
**Bonus question**: What about the more general case of $GSp\_{2n}$ (including $Gl\_{2}$!) ????
| https://mathoverflow.net/users/169282 | Motive associated to a cuspidal representation of $GSp_{4}$ | The formula you quote from Harris defines a Galois representation, not a motive. We expect that there is a motive whose etale realisation is Harris' space, but that is not immediate.
The problems are: firstly, the Siegel threefold is not proper; secondly, the projector cutting out M(pi\_f) might not be an idempotent in the Chow ring.
The first problem was [solved by Wildeshaus](https://arxiv.org/abs/1706.02743) using his theory of "interior motives". This gives a Grothendieck motive attached to pi, but sadly not a Chow motive, because of the second problem above. It is not known if Wildeshaus' Grothendieck motive can be upgraded to a Chow motive.
(Sorry for lack of nice formatting, typing on my phone.)
| 3 | https://mathoverflow.net/users/2481 | 426570 | 173,115 |
https://mathoverflow.net/questions/426567 | 4 | Let $ \pi : X \rightarrow B $ be a family of compact complex manifolds parametrized by a connected base $ B $. (By this I mean $ \pi $ is a proper holomorphic submersion.) Let $ L $ be a holomorphic line bundle on $ X $ and $ n $ a positive integer. What can be said about the locus of points $ b \in B $ such that $ L|\_{X\_b} $ on $ X\_b $ has an $n$-th root?
I posted it originally on MSE (without a response) but I guess it's more appropriate here.
| https://mathoverflow.net/users/152391 | Roots of line bundles in a family | The locus you consider is either empty, or equal to $B$.
This can be seen as follows. Line bundles on a fiber $F$ of $\pi $ are parameterized by $H^1(F,\mathscr{O}^\*\_F)$. This group fits into an exact sequence
$$H^1(F,\mathscr{O}\_F) \rightarrow H^1(F,\mathscr{O}^\*\_F) \xrightarrow{\ c\_1\ }H^2(F,\mathbb{Z})\xrightarrow{\ i\ }H^2(F, \mathscr{O}\_F)\ .$$
I claim that a line bundle $M$ on $F$ admits a $n$-th root if and only if $c\_1(M)$ is divisible by $n$ in $H^2(F,\mathbb{Z})$. Indeed, assume that $c\_1(M)=n\alpha $ for some $\alpha $ in $H^2(F,\mathbb{Z})$; we have $i(\alpha )=0$ since $H^2(F,\mathscr{O}\_F)$ is torsion-free, hence $\alpha =c\_1(N)$ for some $N$ in $H^1(F,\mathscr{O}^\*\_F)$. Then $M\otimes N^{-n}$ has $c\_1=0$, hence comes from a class in the vector space $H^1(F,\mathscr{O}\_F) $, which is of course divisible by $n$.
Now assume first that $B$ is simply connected; then $R^2\pi \_\*\mathbb{Z}$ is the constant sheaf $B\times H^2(F,\mathbb{Z})$ on $B$, so if $c\_1(L\_{|F})$ is divisible by $n$, the same holds for $c\_1(L)$ restricted to any fiber. In general, suppose that $L\_{|X\_b}$ admits a $n$-th root for some $b\in B$; for any $c\in B$ we can choose a path from $b$ to $c$ and cover it by simply connected open subsets, so that $L\_{|X\_c}$ admits a $n$-th root.
| 7 | https://mathoverflow.net/users/40297 | 426577 | 173,116 |
https://mathoverflow.net/questions/236323 | 11 | Is there a closed form expression for the determinant of the $n\times n$ Vandermonde-type matrix
$$A = \left(\begin{array}{}
1&g\_1 & x\_1&g\_1 x\_1 & x\_1^2&g\_1 x\_1^2 & \cdots & x\_1^{n/2-1} & g\_1 x\_1^{n/2-1} \\
1&g\_2 & x\_2&g\_2 x\_2 & x\_2^2&g\_2 x\_2^2 & \cdots & x\_2^{n/2-1} & g\_2 x\_2^{n/2-1} \\
\vdots &\vdots &\vdots &\vdots &\vdots & \vdots & \ddots & \vdots & \vdots \\
1&g\_n & x\_n&g\_n x\_n & x\_n^2&g\_n x\_n^2 & \cdots & x\_n^{n/2-1} & g\_n x\_n^{n/2-1}
\end{array}\right), $$
with $n$ even?
The $g\_i$ can be assumed to be functions of $x\_i$: $g\_i\equiv g(x\_i)$.
Furthermore, it might be helpful to assume the $x\_i$ to be the complex roots of a known polynomial of order $n$.
| https://mathoverflow.net/users/90413 | Determinant of a certain Vandermonde matrix | In Appendix B of <https://arxiv.org/abs/2103.10776> (J. Phys. A: Math. Theor. 54, 375201 (2021)), I derived a transformation of the block Vandermonde determinant above to a Hankel matrix, which in my case dramatically simplified the problem. This is a block-matrix generalization of the well known "Vandermonde factorization of a Hankel matrix":
Let $\mathbf V\_{\!\vec{x}}$ be a general Vandermonde matrix
\begin{equation}
\mathbf V\_{\!\vec{x}} \equiv \left[\vphantom{A\_{2}^{2}}x\_{\mu}^{j}\right]\_{\mu=1,\,j=0}^{M\hspace{1.5ex}M-1}\label{eq:V\_x}\tag{1}
\end{equation}
and let
\begin{equation}
P\_{\vec{x}}(x)=\prod\_{\mu=1}^{M}(x-x\_{\mu})=\sum\_{n=0}^{M}b\_{n}x^{n},\label{eq:P\_x(x)}\tag{2}
\end{equation}
be the related characteristic polynomial in $x$,
where $b\_{M}=1$ by construction. If we further let
\begin{align}
\mathbf D\_{\vec{x}} & \equiv \left[\frac{\delta\_{\mu\nu}}{P\_{\vec{x}}'(x\_{\mu})}\right]\_{\mu,\nu=1}^{M}, & \mathbf H\_{\vec{x}}^{-1} & \equiv\left[\vphantom{A\_{2}^{2}}b\_{i+j+1}\right]\_{i,j=0}^{M-1},\tag{3}
\end{align}
then
\begin{align}
\mathbf V^T\_{\!\vec{x}}\,\mathbf D\_{\vec{x}}\mathbf V\_{\!\vec{x}} & = \mathbf H\_{\vec{x}}, & {\det}^{2}\,\mathbf V\_{\!\vec{x}}\det\mathbf D\_{\vec{x}} & =1.\label{eq:Vandermonde\_Hankel\_Identity}\tag{4}
\end{align}
While $\mathbf H\_{\vec{x}}^{-1}$ is upper anti-triangular, $\mathbf H\_{\vec{x}}$
itself is lower anti-triangular, e.g. for $M=4$,
\begin{equation}
\mathbf H\_{\vec{x}}^{-1}=\begin{pmatrix}b\_{1} & b\_{2} & b\_{3} & 1\\
b\_{2} & b\_{3} & 1 & 0\\
b\_{3} & 1 & 0 & 0\\
1 & 0 & 0 & 0
\end{pmatrix}\quad\Rightarrow\quad\mathbf H\_{\vec{x}}=\begin{pmatrix}0 & 0 & 0 & 1\\
0 & 0 & 1 & \tilde{b}\_{5}\\
0 & 1 & \tilde{b}\_{5} & \tilde{b}\_{6}\\
1 & \tilde{b}\_{5} & \tilde{b}\_{6} & \tilde{b}\_{7}
\end{pmatrix},\label{eq:H\_x\_example}\tag{5}
\end{equation}
with $\tilde{b}\_{n}=\sum\_{\mu}x\_{\mu}^{n-1}/P\_{\vec{x}}'(x\_{\mu})$.
As a side note, both $\tilde{b}\_{n}=s\_{(n-M)}(\boldsymbol{x})$ and
$b\_{n}=(-1)^{n}s\_{\boldsymbol{1}\_{M-n}}(\boldsymbol{x})$ are Schur
polynomials.
We now turn to the block matrix version:
Let $\boldsymbol{\mathcal{V}}\_{\vec{g},\vec{x}}$ be the generalized $1\times B$ block Vandermonde matrix with $M\times N$ blocks
\begin{equation}
\boldsymbol{\mathcal{V}}\_{\vec{g},\vec{x}} \equiv \left[\vphantom{A\_{2}^{2}}g\_{\mu}^{b}\,x\_{\mu}^{n}\right]\_{b=0;\,\,\mu=1,\,n=0}^{B-1\,M\hspace{2ex}N-1} = \left[\vphantom{A\_{2}^{2}}\mathbf G^{b}\,\mathbf V\_{\!\vec{x}}\right]\_{b=0}^{B-1},\tag{6}
\end{equation}
with
\begin{equation}
\mathbf V\_{\!\vec{x}} \equiv \left[\vphantom{A\_{2}^{2}}x\_{\mu}^{n}\right]\_{\mu=1,\,n=0}^{M\hspace{1.5ex}N-1},\qquad \mathbf G \equiv \left[\vphantom{A\_{2}^{2}}\delta\_{\mu\nu}\,g\_{\mu}\right]\_{\mu,\nu=1}^{M}.\tag{7}
\end{equation}
As an example, for $M=4$, $B=3$ and $N=3$ we have
\begin{equation}
\boldsymbol{\mathcal{V}}\_{\vec{g},\vec{x}} =
\left[\begin{array}{ccc|ccc|ccc}
1 & x\_{1} & x\_{1}^{2} & g\_{1} & g\_{1}x\_{1} & g\_{1}x\_{1}^{2} & g\_{1}^{2} & g\_{1}^{2}x\_{1} & g\_{1}^{2}x\_{1}^{2}\\
1 & x\_{2} & x\_{2}^{2} & g\_{2} & g\_{2}x\_{2} & g\_{2}x\_{2}^{2} & g\_{2}^{2} & g\_{2}^{2}x\_{2} & g\_{2}^{2}x\_{2}^{2}\\
1 & x\_{3} & x\_{3}^{2} & g\_{3} & g\_{3}x\_{3} & g\_{3}x\_{3}^{2} & g\_{3}^{2} & g\_{3}^{2}x\_{3} & g\_{3}^{2}x\_{3}^{2}\\
1 & x\_{4} & x\_{4}^{2} & g\_{4} & g\_{4}x\_{4} & g\_{4}x\_{4}^{2} & g\_{4}^{2} & g\_{4}^{2}x\_{4} & g\_{4}^{2}x\_{4}^{2}
\end{array}\right].
\end{equation}
Furthermore, let $\mathbf D$ be an arbitrary $M\times M$ diagonal matrix.
Then, the $B\times B$ block Hankel matrix with $N\times N$ blocks
\begin{equation}
\boldsymbol{\mathcal{H}}\_{\vec{g},\vec{x}} \equiv \left[\sum\_{\mu=1}^{M}d\_{\mu}g\_{\mu}^{a+b}\,x\_{\mu}^{m+n}\right]\_{a,b=0;\,m,n=0}^{B-1\;\;N-1} = \left[\mathbf V^T\_{\!\vec{x}}\,\mathbf D\,\mathbf G^{a+b}\,\mathbf V\_{\!\vec{x}}\right]\_{a,b=0}^{B-1}\tag{8}
\end{equation}
trivially fulfills the identity
\begin{equation}
\boldsymbol{\mathcal{H}}\_{\vec{g},\vec{x}} = \boldsymbol{\mathcal{V}}^T\_{\vec{g},\vec{x}} \, \mathbf D \, \boldsymbol{\mathcal{V}}\_{\vec{g},\vec{x}}.\tag{9}
\end{equation}
Note that the upper left block $\hat{\mathbf H}\_{0} \equiv (\boldsymbol{\mathcal{H}}\_{\vec{g},\vec{x}})\_{0,0} = \mathbf V^T\_{\!\vec{x}}\,\mathbf D\,\mathbf V\_{\!\vec{x}}$
is free of $g\_{\mu}$ and is therefore a usual Vandermonde product
in $x\_{\mu}$. Consequently, if the matrix $\mathbf D$ is set to the diagonal matrix $\mathbf D\_{\vec{x}}$ containing the reciprocal first derivatives of the characteristic polynomial $P\_{\vec{x}}(x)=\prod\_{\mu=1}^M(x-x\_\mu)$,
\begin{equation}
\mathbf D\_{\vec{x}} \equiv \left[\frac{\delta\_{\mu\nu}}{P\_{\vec{x}}'(x\_{\mu})}\right]\_{\mu,\nu=1}^{M},\tag{10}
\end{equation}
and if additionally $M\geq2N$, then $\hat{\mathbf H}\_{0}=\mathbf 0$
vanishes identically.
For the problem above, $B=2$ and $N=M/2$, such that
\begin{equation}
\mathbf A^T \mathbf D\_{\vec{x}}\mathbf A =
\boldsymbol{\mathcal{H}}\_{\vec{g},\vec{x}} =
\begin{bmatrix}
\mathbf 0 & \hat{\mathbf H}\_{1} \\
\hat{\mathbf H}\_{1} & \hat{\mathbf H}\_{2}
\end{bmatrix}.\tag{11}
\end{equation}
The desired determinant $\det \mathbf A$ therefore fulfills
\begin{equation}
\det(\mathbf A^T \mathbf D\_{\vec{x}} \mathbf A)
= {\det}^{2}\,\mathbf A\det\mathbf D\_{\vec{x}}
= \det \boldsymbol{\mathcal{H}}\_{\vec{g},\vec{x}}
= {\det}^{2}\,\hat{\mathbf H}\_{1}.\tag{12}
\end{equation}
For further details, see <https://arxiv.org/abs/2103.10776>.
| 3 | https://mathoverflow.net/users/90413 | 426579 | 173,117 |
https://mathoverflow.net/questions/426585 | 3 | I have some doubts about what an abelian covering is, and I'll try my best to articulate them.
In Serre's *Algebraic groups and class fields* Chapter VI.2, he fixed a base field $k$ with algebraic closure $\bar{k}$, and $V$ a normal irreducible algebraic variety over $\bar{k}$ with function field $K = \bar{k}(V)$. For a finite separable extension $L/K$, the *normalization* $W$ of $V$ in $L$ is the variety whose local rings are obtained by decomposing the integral closure in $L$ of the local rings $\mathcal{O}\_P$ of points $P \in V$. The variety $W$ thus defined comes with a projection $\pi: W \rightarrow V$ and we say that the covering $\pi$ is *abelian* if the field extension $L/K$ is.
Now let $G$ be a finite abelian etale group scheme over $k$. Then the classes of $G$-torsors $Y \rightarrow V$ over $V$ are elements of the etale cohomology group $H^1(V,G)$. So an element $[Y] \in H^1(V,G)$ is such that we have $Y \times G \cong Y \times Y$.
I know that the $G$-torsors $Y$ are called *abelian coverings* of $V$, but is it in the same sense as the one given by Serre? In other words, is $Y$ the normalization of $V$ in some finite extension $L$ of $\bar{k}(V)$ such that $\mathrm{Aut}(L/K) \cong G$? Also, how do we check if a $G$-torsor is unramified, are there convenient results?
Any references or ideas are much appreciated.
| https://mathoverflow.net/users/172132 | The notion of abelian covers | (1) $G$-torsors are always unramified, because they are étale-locally trivial, and unramifiedness may be checked etale-locally.
(2) A $G$-torsor is a $G$-covering in the sense of Serre as long as it is irreducible. We simply take $L$ to be the function field $\overline{k}(Y)$. Since the morphism $Y \to V$ is finite, $Y \times\_V \operatorname{Spec} \mathcal O\_P \to \operatorname{Spec} \mathcal O\_P$ is finite, so every function on $Y \times\_V\operatorname{Spec} \mathcal O\_P$ is integral over $\mathcal O\_P$, and because $V$ is normal and $Y \to V$ is etale, $Y$ is normal, so every element of $L$ integral over $\mathcal O\_P$ is a function in $Y \times\_V\operatorname{Spec} \mathcal O\_P$.
Thus the ring of functions on $Y \times\_V\operatorname{Spec} \mathcal O\_P$ is the integral closure, which decomposes into local rings corresponding to the fibers, as desired.
| 8 | https://mathoverflow.net/users/18060 | 426587 | 173,118 |
https://mathoverflow.net/questions/426304 | 1 | Language: Multi-sorted first order logic with equality and membership, where for each natural $n$ we have variables $x\_i^n$ of sort $n$, and for each decidable monotonic strictly increasing sequence of naturals $s$ [with 0 in its domain] ,we have binary relation symbols $=^s, \in^s$ with the following syntatical restrictions: the symbol $\in^s$ can only occur between variables of sort $s\_n$ on the left to variables of sort $s\_{n+1}$ on the right, generally denoted as $x\_i^{s(n)} \in^s x\_j^{s(n+1)}$ [where $s(n)$ is the $n^{th}$ item in sequence $s$]. On the other hand, the symbol $=^s$ can only occur between variables of the same sort, generally denoted as $x\_i^{s(n)} =^s x\_j^{s(n)}$.
Notation: for simplicity we'll only write the type of a variable at quantification.
**Axioms:** [Multi-sorted ID axioms for each sequence $s$] +
**Extensionality:** $ \forall x^{s\_{n+1}} \, \forall y^{s\_{n+1}}: \forall z^{s\_n} \, ( z \in^s x \iff z \in^s y ) \implies x=^sy$
**Comprehension:** $\exists x^{s\_{n+1}} \forall y^{s\_n} (y \in^s x \iff \phi^s(y))$;
where $\phi^s$ only uses $\in^s,=^s$ as predicates, and the sorts of all variables written as items of $s$.
>
> Is this equivalent to Tangled Type Theory "$\sf TTT$" of Holmes [see: [Holmes p:11](https://arxiv.org/pdf/1503.01406.pdf), [Holmes p:4-5](https://randall-holmes.github.io/Papers/tangled.pdf)]?
>
>
>
In the presentation by Holmes there is seeminly one membership and equality relation, unlike here where there is one per type sequence. I was personally thinking of a proof by compactness since every finite fragment of $\sf TTT$ per sequence $s$ is interpretable here, but I'm not that sure?!
| https://mathoverflow.net/users/95347 | Is this theory equivalent to Tangled Type Theory? | It is not. It is very important, very awful, and quite intended that the membership relation between the same two types is the same in different sequences of types.
| 4 | https://mathoverflow.net/users/485780 | 426599 | 173,120 |
https://mathoverflow.net/questions/426591 | 1 | Assume $A$ is an operator on a Hilbert space with discrete spectrum. Assume $B$ is a positive operator on the same Hilbert space also with a discrete spectrum. Also assume $A$ and $B$ commute.
I'm looking at the small $t$ expansion of the expression,
$$W(t) = \mathrm{Tr} \left( A e^{-tB} \right) = \sum\_{n=0}^\infty D\_n A\_n e^{-t B\_n}$$
where $A\_n$ and $B\_n$ are the eigenvalues of $A$ and $B$, $D\_n$ is the dimension of the eigenspace corresponding to $A\_n$ and $B\_n$, and we can further assume that $B\_n$ grows quadratically with $n$, $D\_n$ grows at most polynomially with $n$ and $A\_n$ is such that $W(t)$ is finite for $t>0$, i.e. $A\_n$ does not grow too fast.
I'd like to get the $\log(t)$ term in the small $t$ expansion of $W(t)$ i.e. not the leading polynomial ones. I mean on general grounds I'll have, for $t\to 0$,
$$W(t) = \sum\_j \frac{W\_j}{t^{\alpha\_j}} + w \log(t) + O(t\log(t)) + \ldots$$
with exponents $\alpha\_j > 0$ and what I'd like to have is the coefficient $w$.
In principle I can successively find all polynomial terms, subtract them, and then the leading term will be the $\log(t)$ term, but I suppose there is a way to get directly the $\log(t)$ term, what would it be?
| https://mathoverflow.net/users/41312 | $\log(t)$ term in the small time expansion of $\mathrm{Tr}( A e^{-tB} )$ | The answer by Carlo Beenakker is correct that any contribution to $D\_n A\_n \sim n^p$ with $p\ge 0$ does not generate any logarithmic asymptotic terms. Instead of numerical checks, one can also examine the [Taylor-Maclaurin](https://en.wikipedia.org/wiki/Euler%E2%80%93Maclaurin_formula#Asymptotic_expansion_of_sums) asymptotic series for $\sum\_{n=0}^\infty n^p e^{-t n^2}$ (no $\log(t)$ terms are generated by differentiating the summand w.r.t $n$).
What is left to cover is the case $p=-1$. Then
$$
\sum\_{n=1}^\infty n^{-1} e^{-t n^2} \sim \int\_1^\infty n^{-1} e^{-t n^2} dn
= \frac{1}{2} \int\_t^\infty \frac{e^{-x}}{x} dx
= -\frac{1}{2} \log(t) + O(1) .
$$
The subleading terms from the Euler-Maclaurin formula and in $O(1)$ above will also not be logarithmic (compare with the series expansion of the exponential integral $E\_1(x)$ [DLMF §6.6.2](https://dlmf.nist.gov/6.6.E2)).
So to get the logarithmic contribution to $W(t)$, you can directly look at the $O(n^{-1})$ term in the asymptotic expansion of $D\_n A\_n$ (provided the expansion itself doesn't have funny terms involving $\log(n)$ or something like that).
| 1 | https://mathoverflow.net/users/2622 | 426604 | 173,121 |
https://mathoverflow.net/questions/426605 | 4 | In his Tohoku article, Grothendieck says that the category $\mathbf{Ab}$ of abelian groups satisfies the axiom AB6, namely
"All small colimits exist in $\mathbf{Ab}$. Moreover for any index family $J$ and filtered categories $I\_j, j\in J$, with functors $I\_j\to \mathbf{Ab}, i\mapsto M\_i$, the natural map
$
\underset{\substack{\rightarrow \\ (i\_j\in I\_j)\_j}}{\lim} \prod\_{j\in J} M\_{i\_j} \rightarrow \prod\_{j\in J}\underset{\substack{\rightarrow \\ i\_j\in I\_j}}{\lim}M\_{i\_j} $
is an isomorphism."
It seems that this is saying that filtered limits and arbitrary products commute, but I don't think that this commutation happens.
Let us consider, for example (this is probably not the simplest example, but it is how I recognized the problem), a profinite space $S=\lim\_i S\_i$, projective limit of finite discrete sets $S\_i$, where $i\in I=\{\text{finite quotients of } S\}$. For all $n\in \mathbb{N}$, let us consider the family $I\_j=I$, with functors $I\_j\to \mathbf{Ab}, i\mapsto \mathrm{Map}(S\_i,\mathbb{Z})$ the abelian group of maps of sets from $S\_i$ to $\mathbb{Z}$.
The term on the left is, if I understood correctly, the abelian group
$\underset{\substack{\rightarrow\\(i\in I)\_n}}{\lim} \prod\_{n\in \mathbb{N}} \mathrm{Map}(S\_i,\mathbb{Z})=\underset{\substack{\rightarrow\\(i\in I)\_n}}{\lim} \mathrm{Map}(S\_i,\prod\_{n\in \mathbb{N}} \mathbb{Z})$
i.e. the abelian group of continuous function from $S$ to $\prod\_{n\in\mathbb{N}}\mathbb{Z}$, where the latter has the **discrete topology** (every map from the profinite $S$ to a discrete abelian group factors via some quotient $S\_i$).
On the other hand, the term on the right is
$\prod\_{n\in\mathbb{N}} \underset{\substack{\rightarrow\\ i\in I}}{\lim} \mathrm{Map}(S\_i,\mathbb{Z})=\prod\_{n\in\mathbb{N}}\mathrm{Cont}(S,\mathbb{Z})=\mathrm{Cont}(S,\prod\_{n\in\mathbb{N}}\mathbb{Z})$
which is the abelian group of continuous functions $S\to \prod\_{n\in\mathbb{N}}\mathbb{Z}$, where this time the latter has the **product topology**.
Since the continuous identity morphism from the discrete $\prod \mathbb{Z}$ to itself with the product topology is not a homeomorphism, the two abelian groups $\mathrm{Cont}(S,(\prod\mathbb{Z})^{\delta})$ and $\mathrm{Cont}(S,(\prod\mathbb{Z})$ cannot coincide for all profinite space $S$.
I am sure that in the formulation of Grothendieck there are no errors, so I guess I misunderstood the meaning of the left-hand side (the filtered colimit of the products): can someone help me?
| https://mathoverflow.net/users/485722 | Why does the category of abelian groups satisfy the axiom AB6? | The objects we are taking the limit over in the left side are $(i\_j \in I\_j)\_j$, i.e. tuples of, for each $j\in J$, an element $i\_j$ of $I\_j$. These are the same as elements of $\prod\_{j\in J}$.
Thus, the left side is a limit over $\prod\_{j\in J} I\_j$.
On the other hand, you have defined $I\_j = I$ for all $j$ and then taken the limit over $I$. Of course in this case $I$ embeds into $\prod\_{j\in J} I$ as a "diagonal" subset, but not every element is bounded by an element on the diagonal, unless $I$ is finite.
| 5 | https://mathoverflow.net/users/18060 | 426607 | 173,122 |
https://mathoverflow.net/questions/426600 | 3 | Consider the permutation group $\mathfrak{S}\_n$ on $n$ letters $\{1,2,\dots,n\}$. Let $\iota=(1,2,3,\dots,n)\in\mathfrak{S}\_n$ be the *identity* permutation in a $1$-line notation. Given $\pi, \rho\in\mathfrak{S}\_n$ define the *dot product* $\pi\cdot\rho=\pi\_1\rho\_1+\cdots+\pi\_n\rho\_n$.
The $q$-factorial is given by $[n]!\_q=\prod\_{k=1}^n(1+q+\cdots+q^{k-1})$. Denote $D=\frac{d}{dq}$ to be the derivative evaluated at $q=1$. The following is a fact
$$\sum\_{\pi\in\mathfrak{S}\_n}\pi\cdot\iota=\frac{(n+1)!}2\binom{n+1}2=D\,[n+1]!\_q.$$
Consider the *lexicographic ordering* on $\mathfrak{S}\_n$ for $i=1$ through $n!$. For example, if $n=3$ then write (in order)
$$\mathfrak{S}\_3=\{\pi^{(1)},\pi^{(2)},\pi^{(3)},\pi^{(4)},\pi^{(5)},\pi^{(6)}\}
=\{(1,2,3),(1,3,2),(2,1,3),(2,3,1),(3,1,2),(3,2,1)\}.$$
Denote $\eta=(\frac11,\frac12,\frac13,\cdots,\frac1n)$ and let $R(n)$ be the *number of runs* of length $1$ in all permutations in $\mathfrak{S}\_n$ (see [OEIS A097900](https://oeis.org/A097900)).
>
> **QUESTION 1.** Is this true? For $n\geq2$, we have
> $$\sum\_{i=1}^{n!}(-1)^i\,(\pi^{(i)}\cdot\eta)=\frac{(n-2)!(n+1)}6.$$
>
>
>
>
> **QUESTION 2.** Is this true? For $n\geq5$, we have
> $$\frac1{n-2}\sum\_{i=1}^{n!}(-1)^i\,(\pi^{(i)}\cdot\eta)=R(n-3).$$
>
>
>
| https://mathoverflow.net/users/66131 | Sums over permutations relates to permutations? | **Question 1.** All entries except for the last $2$ positions remain constant in blocks of even size and, therefore, contribute $0$ to the signed sum. Now consider the last $2$ entries. Each pair of values $(a,b)$, where $1\le a<b\le n$, comes in consecutive pairs $\pi^{(2j-1)}=(\dots,a,b)$ and $\pi^{(2j)}=(\dots,b,a)$, for some $1\le j\le n!/2$. The net contribution from (the last $2$ entries of) this pair is, therefore,
$$
-\left(\frac{1}{n-1}\cdot a + \frac{1}{n}\cdot b\right) + \left(\frac{1}{n-1}\cdot b + \frac{1}{n}\cdot a\right)= \frac{1}{n(n-1)}\cdot(b-a).
$$
Each such pair of values $(a,b)$, $a<b$, occurs in the last $2$ positions of $(n-2)!$ permutations. Each difference $k=b-a$ occurs for $n-k$ pairs $(a,b)$, $a<b$. Therefore, the signed sum in question is equal to
$$
\frac{1}{n(n-1)}(n-2)!\sum\_{k=1}^{n-1}{k(n-k)}=\frac{1}{n(n-1)}(n-2)!\binom{n+1}{3}=\frac{(n-2)!(n+1)}{6}
$$
*Update:* In general, by the same argument, given any vector $\eta=(\eta\_1,\dots,\eta\_n)$, the signed sum in question is equal to
$$
\frac{(n+1)!}{6}(\eta\_{n-1}-\eta\_n).
$$
| 5 | https://mathoverflow.net/users/113161 | 426608 | 173,123 |
https://mathoverflow.net/questions/426598 | 3 | Let $(X, T, \mathcal F, \mu)$ be an ergodic measure preserving system with finite measure.
Suppose $T$ is uniformly recurrent, in the following sense:
For every $A \in \mathcal F$, there exists an $M \in \mathbb N$ such that for almost every $x \in A$, we have $T^{n(x)} \, (x) \in A$ for some $n(x) \leq M$.
**Question:** Does it follow that the convergence of the Birkhoff sums is essentially uniform? That is, for every $f \in L^1 (X)$, there exists a measurable set $E$ of full measure such that
$$\frac{1}{N}\sum\_{i = 0}^{N-1} f(T^n (x)) \to \int f \, d\mu$$
uniformly on $E$ as $N \to \infty$.
| https://mathoverflow.net/users/173490 | Does uniform recurrence imply uniform convergence of the Birkhoff sums? | This is just a comment on your definition of "uniformly recurrent". Can you give an example of a system you'd consider uniformly recurrent?
If this notion is standard, the following must be nonsense, but it seems to me that your notion does not make much sense.
>
> Lemma. If $T$ is ergodic and uniformly recurrent, then it is a single finite cycle.
>
>
>
Proof. Suppose first that the measure of points with an eventual period is $0$. Then by the Rokhlin lemma for noninvertible transformations [1], there exists an **$n$-tower** for $f$ of measure at least $1 - \epsilon$ for each $\epsilon > 0$, meaning a set $B\_{(n)}$ such that for some **base** $B$, $B\_{(n)} = \bigcup\_{0 \leq k < n} T^{-k}(B)$ is disjoint and has measure at least $1 - \epsilon$ and the sets $f^{-1}\_k(B)$ are disjoint.
From this it is easy to see that there also exist arbitrarily small such sets, just take small subsets of $B$. From this we obtain that we can find for each $n \geq 1$ pick an $n$-tower $B\_{(n)}$ such that these towers are disjoint for distinct $n$. Now denoting by $B^n$ is the base of $B\_{(n)}$, define $A = \bigcup\_{n > 1} B^n \cup T^{-n+1}(B^n)$.
Note that $B^n$ and $T^{-n+1}(B^n)$ have equal measure, so if you pick a random point $x$ in $A$, it will be in in a set of the form $T^{-n+1}(B^n)$ (for some $n$) with probability $0.5$. But then the minimal $n$ of $T$ such that $T^n(x) \in A$ is $n-1$. It follows that $T$ is not uniformly recurrent.
We conclude that there must exist $n, p$ such that the probability that $T^n(x) = T^{n+p}(x)$ is positive. In particular, the shift-invariant set of points satisfying $T^p(x) = x$ has positive measure, so this measure must be $1$. It is well-known that in this case, we can find a cross-section $S$ such that $X = S \cup T(s) \cup \cdots T^{n-1}(S)$ and the union is disjoint (assuming this is on a standard Lebesgue space $[0, 1]$, I think the usual construction is to pick the minimal point on each orbit).
Ergodicity implies that $S$ must be a singleton, as otherwise we can split it in two. Square.
Of course in this case, the answer to your question is "yes".
[1] *Avila, Artur; Candela, Pablo*, [**Towers for commuting endomorphisms, and combinatorial applications**](http://dx.doi.org/10.5802/aif.3042), Ann. Inst. Fourier 66, No. 4, 1529-1544 (2016). [ZBL1360.28012](https://zbmath.org/?q=an:1360.28012).
| 3 | https://mathoverflow.net/users/123634 | 426610 | 173,125 |
https://mathoverflow.net/questions/426615 | 2 | Let $X$ be a smooth projective algebraic surface over $\mathbb C$, and let $L\_1, L\_2$ be ample line bundles satisfying $H^i(X,L\_j) = 0$ for $i > 0$.
Do we necessarily have $(h^0(L\_1) + h^0(L\_2))^2 \geq h^0(L\_1 \otimes L\_2)$, or are there examples where this inequality is violated?
| https://mathoverflow.net/users/485787 | Inequalities of dimensions of global sections for line bundles on surfaces | Here is a counterexample. Let $X = C \times \mathbb{P}^1$ where $C$ is a curve of genus $g \geq 2$ and let $L\_1 = L\_2 = p\_1^\*L \otimes p\_2^\*\mathcal{O}(1)$ where $L$ s a generic line bundle of degree $g-1$. Then $H^0(L) = 0$ since the map $\mathrm{Sym}^{g-1}(C) \to \mathrm{Pic}^{g-1}(C)$ is not surjective for dimension reasons, and $\chi(L) = 0$ by Riemann-Roch so $L$ is an ample line bundle on $C$ with no sections and no cohomology.
In particular, $L\_1 = L\_2$ is ample as its the sum of pullbacks of amples, and by Künneth formula, we deduce that $h^0 = h^1 = h^2 = 0$. Thus $L\_1$ and $L\_2$ satisfy the conditions and the left hand side of the inequality is $0$. On the other hand,
$$
L\_1 \otimes L\_2 = \pi\_1^\*L^{\otimes 2} \otimes \pi\_2^\*\mathcal{O}(2)
$$
which has sections by Riemann-Roch + Künneth again.
| 3 | https://mathoverflow.net/users/12402 | 426616 | 173,126 |
https://mathoverflow.net/questions/426575 | 1 | I've got a question to the Hoeffding Inequality which states, that for data points $X\_1, \dots, X\_n \in X$, which are i.i.d. according to a probability measure $P$ on $X$, we find an upper bound for:
$$P\left(\left|{\sum\_{I=1}^n}X\_i - \int\_XX\_1dP(x)\right| \ge \varepsilon\right) \le \alpha.$$
In machine learning proofs, this is often used for bounding functions, hence we get for a function $f: X \rightarrow \mathbb{R}$ a bound:
$$P\left(\left|{\sum\_{I=1}^n}f(X\_i) - \int\_Xf(X\_1)dP(x)\right| \ge \varepsilon\right) \le \alpha.$$
I don't understand, why we are able to obtain this bound with the same probability measure $P$. Why don't we need to look at a different / transformed probability measure $P\_f$?
| https://mathoverflow.net/users/485752 | Using Hoeffding inequality for risk / loss function | I think this question is more on the definition side. Often times people don't distinguish $\mathbb{P}$ the probability measure in the underlying probability space, and $P$ the distribution or the probability measure induced by a random variable. Also people tend to use $P$ directly, without making any explicit reference to the original probability space. This can cause confusion.
The first inequality may be restated more precisely as follows. Let $X\_i: \Omega \to \mathcal{X}$ be a real-valued Borel measurable map from the probability space $(\Omega, \mathcal{F}, \mathbb{P})$, for each $i=1,...,n$. Let $X\_i$'s have a common distribution $P$. Then the first inequality should be
$$
\mathbb{P}\Big\{ \omega \in \Omega : \Big| \sum\_{i=1}^n X\_i(\omega) - \int\_{\mathcal{X}} x P(dx) \Big| > \varepsilon \Big\} \leq \alpha,
$$
which can be written as
$$
\mathbb{P}\Big\{ \Big| \sum\_{i=1}^n X\_i - \int\_{\mathcal{X}} x P(dx) \Big| > \varepsilon \Big\} \leq \alpha.
$$
Notice $\mathbb{P}$ should be used instead of the distribution $P$.
In fact, the first inequality in your post doesn't really make sense when you read it rigorously: Since $P$ is a probability measure on $\mathcal{X}$, the argument of function $P$ should be a subset of $\mathcal{X}$.
The second inequality should be of the form
$$
\mathbb{P}\Big\{ \omega \in \Omega : \Big| \sum\_{i=1}^n f(X\_i(\omega)) - \int\_{\mathcal{X}} f(x) P(dx) \Big| > \varepsilon \Big\} \leq \alpha'.
$$
Notice you're essentially taking probability of some collection of outcomes in $\Omega$, which has nothing to do with the function $f$.
| 2 | https://mathoverflow.net/users/483494 | 426619 | 173,128 |
https://mathoverflow.net/questions/426569 | 2 | Let $K$ be a CM-field with totally real subfield $F$. Let $(V\_1, h\_1)$ and $(V\_2, h\_2)$ be two $n$-dimensional $K$-vector spaces with nondegenerate Hermitian forms, where $n\geq 3$.
>
> **Question 1** Does every $F$-group isomorphism $PU(V\_1, h\_1)\cong PU(V\_2, h\_2)$ arise from an isometry of $(V\_1, h\_1)\cong (V\_2, \lambda h\_2)$ for some $\lambda\in F^{\times}$ ?
>
>
>
>
> **Edit. Question 2** Suppose there exists an $F$-group isomorphism $PU(V\_1, h\_1)\cong PU(V\_2, h\_2)$, can we deduce $(V\_1, h\_1)\cong (V\_2, \lambda h\_2)$ for some $\lambda\in F^{\times}$, or any other relations between $h\_1$ and $h\_2$?
>
>
>
A related statement for orthogonal group is given in the comment to the question [$p$-adic orthogonal groups in four variables](https://mathoverflow.net/questions/226214/p-adic-orthogonal-groups-in-four-variables) .
We are interested in the case that $K$ is a cyclotomic field and $h$ has signature $(1,n-1)$ for an embedding $F\to \mathbb{R}$ and is definite for all the other embeddings.
| https://mathoverflow.net/users/108424 | Does the F-unitary group isomorphism arises from a conformal isometry? | The answer to Question 1 is **No.** Indeed, take $V\_1=V\_2=V:=K^n$ and write
$$ h\_1(z)=h\_2(z)=h(z):=\lambda\_1 z\_1 \bar z\_1+\dots+\lambda\_n z\_n\bar z\_n\quad\text{with}\
\lambda\_i\in F^\times.$$
Write $\widetilde G=U(V,h),\ G={\rm PU}(V,h)$.
Define
$$\tilde\sigma\colon\,\widetilde G\to \widetilde G,\quad g\mapsto \bar g\quad\text{for}\ g\in \widetilde G(F)\subset {\rm GL}(n,K).$$
Then the $F$-automorphism $\tilde \sigma$ of $\widetilde G$
induces an $F$-automorphism $\sigma$ of $G$.
Since $n\ge 3$, this autmorphism $\sigma$ is *outer*, and hence it is not a conformal isometry. The difference with the case of [Brian Conrad's comment](https://mathoverflow.net/questions/226214/p-adic-orthogonal-groups-in-four-variables#comment558491_226214) is that in his case the algebraic group has no outer automorphisms.
**Edit.** The answer to Question 2 is **Yes**.
The isomorphism classes of twisted forms $(V',\,F^\times\cdot h')$ of the Hermitian space $(V,\,F^\times\cdot h)$ bijectively correspond to $H^1(F, {\rm GU}(V,h))$,
where ${\rm GU}(V,h)={\rm Aut}(V,\,F^\times\cdot h)$.
The isomorphism classes of twisted forms of the algebraic $F$-group ${\rm PU}(V,h)$ bijectively correspond to $H^1(F,{\rm Aut}({\rm PU}(V,h)))$.
We wish to show that the kernel
$$\ker\big[ H^1(F, {\rm GU}(V,h))\longrightarrow H^1(F,{\rm Aut}({\rm PU}(V,h)))\big]$$
is trivial.
We factor the above arrow as
$$H^1(F, {\rm GU}(V,h))\to H^1(F,{\rm Aut}({\rm PU}(V,h))^0)\to H^1(F,{\rm Aut}({\rm PU}(V,h)))$$
(where $^0$ denotes the identity component),
and show that both arrows have trivial kernels.
For the first arrow we have an exact sequence
$$1=H^1(F,K^\times)\to H^1(F, {\rm GU}(V,h))\to H^1(F,{\rm Aut}({\rm PU}(V,h))^0),$$
which shows that the kernel is trivial. For the second arrow we have a short exact sequence of $F$-groups,
$$1\to {\rm Aut}({\rm PU}(V,h)^0)\to {\rm Aut}({\rm PU}(V,h))\to {\rm Aut}(A\_{n-1})\to 1,$$
where $A\_{n-1}$ is the corresponding Dynkin diagram.
We obtain a cohomology exact sequence
\begin{align\*}
{\rm Aut}({\rm PU}(V,h))(F)\to {\rm Aut}(A\_{n-1})(F)\to H^1(F,&{\rm Aut}({\rm PU}(V,h))^0)\\
&\to H^1(F,{\rm Aut}({\rm PU}(V,h)))
\end{align\*}
where the arrow ${\rm Aut}({\rm PU}(V,h))(F)\to {\rm Aut}(A\_{n-1})(F)$
is surjective by the answer to Question 1.
This shows that the arrow
$$H^1(F,{\rm Aut}({\rm PU}(V,h))^0)\to H^1(F,{\rm Aut}({\rm PU}(V,h)))$$
has trivial kernel, which completes the proof.
| 3 | https://mathoverflow.net/users/4149 | 426620 | 173,129 |
https://mathoverflow.net/questions/426606 | 1 | Let $C^{m,\alpha}\_M([0,1])$ be a Holder ball consisting of real-valued functions $g$ on $[0,1]$ such that
$$ \|g\|\_{C^{m,\alpha}} := \max\_{0\leq j \leq m } \sup\_{x\in [0,1]} |g^{(j)}(x)| + \sup\_{x,y\in [0,1], x\neq y} \frac{ |g^{(m)}(x) -g^{(m)}(y)|}{|x-y|^\alpha} \leq M. $$
Let $f \in C^{m,\alpha}\_M([0,1])$ be fixed.
By Weierstrass approximation theorem, there exists a sequence of polynomials
$$p\_n(x) = \sum\_{j=0}^{J\_n} \alpha\_{n,j} x^j$$
such that $\|f-p\_n\|\_{\infty} \to 0$ as $n \to \infty$, where $\|\cdot\|\_\infty$ denotes the uniform norm. Here $J\_n \to \infty$.
**Question:**
Does there exist approximating polynomials $p\_n \in C^{m,\alpha}\_{M'}([0,1])$ for some $M'>0$, such that $\|f-p\_n\|\_{\infty} \to 0$? If this is not true in general, for what kind of functions $f$ it may hold?
**Motivation of the question:**
I am reading some papers in nonparametric statistics, [Chen (2007)](https://www.sciencedirect.com/science/article/abs/pii/S157344120706076X) and [Chen and Ai (2003)](https://onlinelibrary.wiley.com/doi/abs/10.1111/1468-0262.00470). They rely on approximating an unknown function in some space, i.e. $f\in C\_M^{m,\alpha}$ here, using functions in some approximating spaces increasing with $n$, i.e. space of $J\_n$-th order polynomials here.
To have desirable statistical properties, a key assumption they used is that, the approximating spaces are subsets of the original space $C^{m,\alpha}\_M([0,1])$. Their practice corresponds to the question. I am not sure whether adding such further restriction on the approximating spaces will cause serious issues on the approximating ability. This issue is not discussed in the listed papers, nor in other related papers I have checked.
Any reference or discussions is greatly appreciated!
| https://mathoverflow.net/users/483494 | Approximating a smooth function under some restrictions | The revised question, where you just want uniformly bounded Holder norm on the approximating polynomials, the answer is **yes**.
In fact, you have more.
**Theorem 1** If $f\in C^{m,\alpha}\_M([0,1])$ is such that the function $\tilde{f}(x) = \begin{cases} f(x) & x\in [0,1]\\ 0 & x\not\in [0,1]\end{cases}$ is in $C^{m,\alpha}(\mathbb{R})$, then there exists a polynomial approximation sequence in $C^{m,\alpha}\_M([0,1])$.
This theorem is based on a proof of the Stone-Weierstrass Theorem using a polynomial approximation to identity. See e.g. Theorem 11.7.1 of [Lebl's textbook (volume 2)](https://www.jirka.org/ra/).
A nice property of the convolution argument is that:
*If $\phi$ is a non-negative function with total integral 1, then $\|\phi\*f\|\_{\infty} \leq \|f\|\_\infty$.*
Since $(\phi\*g)^{(j)} = \phi\* (g^{(j)})$, for any function $g$, applying this to $\tilde{f}$ you see that the convolution by the polynomial approximation to identity means that the polynomials constructed have $C^m$ norms bounded by that of $\tilde{f}$.
Similarly, letting $g = \tilde{f}^{(m)}$, observe for each fixed $t\neq 0$, $x\mapsto \frac{1}{|t|^\alpha}({g}(x+t) - {g}(x))$ is a $C^{0,\alpha}$ function, we have that
$$ \frac{1}{|t|^\alpha} \| \phi\*{g}(x+t) - \phi\*{g}(x)\|\_{L^\infty\_x} \leq \frac{1}{|t|^\alpha} \|{g}(x+t) - {g}(x)\|\_{L^\infty\_x} $$
Taking the sup of both sides over $t\neq 0$ this shows that the Holder semi-norm of $\phi\*\tilde{f}$ is bounded by that of $\tilde{f}$. And this proves the Theorem.
---
Now, given $f\in C^{m,\alpha}\_M([0,1])$, there exists a polynomial $q$ such that $f-q$ satisfies the hypotheses of Theorem 1. The polynomial $q$ is determined entirely by $f(0), f'(0), \ldots, f^{(m)}(0), f(1), f'(1), \ldots, f^{(m)}(1)$.) You can select $q$ to be of degree $2m+1$, in which case the $q$ is uniquely determined by those values. The relation between those values and the coefficients of $q$ are given by a linear operator, depending only on $m$. This means that there exists $\tilde{M}$ that uniformly bounds the coefficients of $q$, where $\tilde{M}$ depends only on $M$ and on $m$ (since by assumption $f(0), \ldots f^{(m)}(1)$ are all bounded by $M$). (Note that in the context of Theorem 1, this unique $q$ is the 0 polynomial, and hence $\tilde{M} = 0$.) The mapping from $\mathbb{R}^{2m+2}\times \mathbb{R} \to \mathbb{R}$ where the first factor represents the coefficients of $q$ and the last factor represents $x$ mapping $(q,x)\mapsto q(x)$ is continuous (in fact a polynomial), and hence achieves a maximum on compact regions. Therefore knowing that the coefficients are bounded by $\tilde{M}$ and $x\in [0,1]$ means that there exists $\hat{M}$ depending only on $M$ and $m$ that bounds the $C^{m+1}$ and hence $C^{m,\alpha}$ norm of $q$. Finally set $M' = M + 2 \hat{M}$.
So you conclude:
**Theorem 2** Given $m, \alpha,M$, there exists $M' = M'(m,M,\alpha)$ such that whenever $f\in C^{m,\alpha}\_M([0,1])$, you can approximate $f$ by polynomials $p\_j$ in the uniform norm, with $p\_j$ selected from $C^{m,\alpha}\_{M'}([0,1])$.
| 1 | https://mathoverflow.net/users/3948 | 426621 | 173,130 |
https://mathoverflow.net/questions/426623 | 2 | Under what conditions can an orientable Riemannian 3-manifold $\Sigma$ be defined implicitly?
What I mean by implicitly is that there exists a smooth function $f:\mathbb{R}^n\to \mathbb{R}^m$, such that $\Sigma$ is diffeomorphic to $f^{-1}(0)$, and the Euclidean metric on $\mathbb{R}^n$ pulled back to $f^{-1}(0)$ is equal to the metric on $\Sigma$.
| https://mathoverflow.net/users/485792 | Under what conditions can an orientable Riemannian 3-manifold be defined implicitly? | By the Nash embedding theorem every Riemannian manifold $M$ embeds isometrically into some ${\Bbb R}^n$. You may then take $f(x)=dist(x,M)$ for $x\in{\Bbb R}^n$.
| 9 | https://mathoverflow.net/users/39082 | 426625 | 173,132 |
https://mathoverflow.net/questions/426346 | 1 | Let $F:\mathbb{R}^n\to\mathbb{R}^n$ be a continuous non-linear map, and let $A$ be a connected subset of $\mathbb{R}^n$ with $\text{dim}(A)=d\leq n$. When can we say that the dimension of the image, $\text{dim}(F(A))$, is also $d$? In other words, when does the map $F$ preserve dimension?
A non-example would be some sort of non-linear projection that embeds a set into a lower dimensional space.
I think that if $F$ is injective then this would be sufficient, but it also seems that we could weaken this significantly. For instance, maybe something along the lines of $F^{-1}(y)$ has at most countable cardinality for all $y\in\mathbb{R}^n$, although I am not sure how to guarantee this for a non-linear map.
If $F$ were measure-preserving then I think this would also be sufficient, but again somewhat overkill. In my case it is fine if $F$ changes the measure, so long as it doesn't change the dimension.
| https://mathoverflow.net/users/484618 | Dimension-preserving non-linear map | *Dimension Theory* by Ryszard Engelking provides the answer I was looking for. Specifically, theorem 1.12.8 which I have provided below, although there are other results in this same area that may be useful to someone.
**Alexandroff's Theorem**: *If $f:X\to Y$ is an open mapping of a locally compact seperable metric space $X$ onto a separable metric space $Y$ such that $|f^{-1}(y)|\leq \aleph\_0$ for every $y\in Y$, then $\text{ind}(X)=\text{ind}(Y)$.*
Here, $\text{ind}$ is the small inductive dimension, which for separable metrizable spaces is equal to the Lebesgue covering dimension (i.e. the topological dimension).
| 1 | https://mathoverflow.net/users/484618 | 426652 | 173,137 |
https://mathoverflow.net/questions/426655 | 3 | It started with a conjecture I had, see [A statement on complex polynomials](https://mathoverflow.net/questions/426503/a-statement-on-complex-polynomials?noredirect=1#comment1096888_426503), which was false for $n \geq 3$, as shown by Noam D. Elkies in his answer there. The present post is an attempt to salvage the main idea of the conjecture by adding an extra hypothesis, namely that the $n$ circular disks are well separated. I will try to keep the remainder of this post self-contained, as much as possible.
Let $n \geq 3$ be an integer and let $D\_i$, for $i = 1, \ldots, n$, be $n$ closed circular regions on the Riemann sphere $\widehat{\mathbb{C}}$, which are mutually disjoint. Note that for example a closed half-plane in the complex plane with $\infty$ is to be considered a closed circular region, since a line in the complex plane can be thought of as a circle on $\widehat{\mathbb{C}}$ passing through $\infty$.
The first problem I would like to solve, is how to define a measure, say $\delta$, of how well the $D\_i$ are separated in such a way that $\delta$ is invariant under the group $PSL(2, \mathbb{C})$ of Moebius transformations.
Here is a suggestion. Given $i$, with $1 \leq i \leq n$, choose a Moebius transformation $\phi\_i$ which maps $D\_i$ onto the closed upper hemisphere of $\widehat{\mathbb{C}}$ (or, if you prefer, the complement of the open unit disk). Note that $\phi\_i$ is not unique, as one could post-compose $\phi\_i$ with an element of $PSU(1,1)$ (the group of Moebius transformations which preserve the open unit disk, isomorphic to the group of hyperbolic symmetries of the hyperbolic plane) and obtain another equally valid map.
Consider now the set $\mathcal{D}\_i$ of circular regions $\phi\_i(D\_j)$, for $1 \leq j \leq n$ and $j \neq i$, all lying in the open unit disk.
Using the Poincare model on the open unit disk and the corresponding notion of hyperbolic distance, we let $\delta\_i$ be the smallest pairwise hyperbolic distance between the elements of $\mathcal{D}\_i$.
Finally, we let
$$ \delta = \operatorname{min}\{ \delta\_i \,;\, 1 \leq i \leq n \}. $$
The number $\delta$ can be thought of as measuring how well the $D\_i$ are separated. I believe, unless I am missing something, that $\delta$ is well defined and invariant under $PSL(2, \mathbb{C})$. If there are some issues with $\delta$, please inform me in the comments.
Let $p\_i(z)$, for $1 \leq i \leq n$, be $n$ complex polynomials of degree at most $n - 1$ (so that it is allowed to have $\infty$ as a root, so to speak, when thinking about the Riemann sphere picture). We say that the $p\_i(z)$ are compatible with the closed circular regions $D\_i$ if, for each $i$, with $1 \leq i \leq n$, $p\_i(z)$ has exactly $1$ root in each $D\_j$, with $1 \leq j \leq n$ and $j \neq i$, and no other roots.
Given an integer $n \geq 3$, does there exist a constant $c\_n > 0$, depending on $n$ only, such that if the $D\_i$ ($1 \leq i \leq n$) are mutually disjoint closed circular regions in $\widehat{\mathbb{C}}$ with $\delta > c\_n$ and if the $p\_i(z)$ ($1 \leq i \leq n$) are complex polynomials of degree at most $n - 1$ which are compatible with the $D\_i$, then the $p\_i(z)$, for $1 \leq i \leq n$, are then linearly independent over $\mathbb{C}$?
I am also interested if one could find, assuming the above question has a positive answer, a specific value of $c\_n$ for which the above statement is true (ideally the best value of $c\_n$, though that may prove to be too difficult).
| https://mathoverflow.net/users/81645 | On well separated circular regions in the Riemann sphere and complex polynomials | Let $\{ z\_1,\ldots,z\_n\}$ be any finite set,
and
$$L\_k(z)=\prod\_{j\neq k}(z-z\_j).$$
Then $L\_k,\; 1\leq k\leq n$ are linearly independent since
the set of their linear combinations consists of all polynomials
of degree $\leq n-1$ (Lagrange interpolation formula says that),
and the space of all polynomials of degree $\leq n-1$ is of dimension $n$.
Now we can easily prove (by contradiction) that for every
$\{ z\_1,\ldots,z\_n\}$ there exists $\epsilon>0$ such that
the disks $\{ z:|z-z\_k|<\epsilon\}$ have your desired property.
I understand that this is not what you desire; you want a
conformally invariant "separation parameter".
However this argument shows that if the "separation parameter" of a finite collection of disks has the property that when it tends to infinity, the disks shrink to points, then your statement about linear independence is true.
The separation parameter that you propose does not have this property, and in fact it has another important drawback: it can stay $>$ some positive number while some of your disks are arbitrarily close to the unit circle (which is the boundary of one of the disks of your family). And your separation parameter depends on
which disk you choose to send to the "upper hemisphere". So your
"separation parameter" can be very large, while some disks are very close to each other.
I can propose a separation parameter which will have the desired property: it is the minimum (over $1\leq k\leq n$) of the reciprocal extremal length of the family of curves
which separate one disk $D\_k$ from the rest.
When this tends to $\infty$, this means that all these extremal lengths are small, so the disks are well separated, and must shrink to points.
For the definition and properties of extremal length, see any of the two books by Ahlfors, Conformal invariants, or Lectures on quasiconformal mappings.
| 2 | https://mathoverflow.net/users/25510 | 426667 | 173,139 |
https://mathoverflow.net/questions/426601 | 6 | Let $F$ be a number field ($F=\mathbb Q$ is fine for my purposes) and let $n\geq2$ be an integer. Is it known whether the first motivic cohomology groups
$$\mathrm H^1(\mathrm{Spec}(F),\mathbb Z(n))$$
are finitely generated? We know that they are finite-dimensional after tensoring with $\mathbb Q$, since they become identified with the rational $K$-theory $K\_{2n-1}(F)\_{\mathbb Q}$ whose dimension was computed by Borel. But are they finitely generated integrally? A reference, if it exists, would be appreciated.
| https://mathoverflow.net/users/126183 | Finite generation of motivic cohomology of number fields | Indeed, I believe it is known that these are finitely generated. First, the Gysin sequence shows that the map
$$H^1(\mathcal{O}\_F;\mathbb{Z}(n))\rightarrow H^1(F;\mathbb{Z}(n))$$
is injective with cokernel given by
$$\oplus\_\nu H^0(k\_\nu;\mathbb{Z}(n-1))$$
where $\nu$ runs over all maximal ideals of $\mathcal{O}\_F$ with residue field $k\_\nu$. As we assumed $n\geq 2$, this direct sum vanishes, so the question is equivalent to showing that $H^1(\mathcal{O}\_F;\mathbb{Z}(n))$ is finitely generated. But now that we've replaced $F$ by $\mathcal{O}\_F$, this finite generation actually holds in any degree and weight.
Indeed, the K-groups of $\mathcal{O}\_F$ are finitely generated as shown by Quillen (a "simple" argument is to use homological stability to reduce to showing that the homology groups of general linear groups over $\mathcal{O}\_F$ are finitely generated, which follows from Borel-Serre). Now, the spectral sequence from motivic cohomology to K-theory degenerates rationally by the Adams operations, but in fact more is true, as noted by Kahn: it degenerates "up to isogeny". So K-theory and motivic cohomology can only differ by bounded torsion. Thus, to deduce finite generation of motivic cohomology from that of K-theory, it suffices to see that (mod $p$) motivic cohomology of $\mathcal{O}\_K$ is finitely generated for any prime $p$, in any degree and weight. By another application of Gysin and the fact that mod $p$ motivic cohomology of a finite field of characteristic $p$ is only nontrivial when degree = weight = zero, this reduces to the same claim for $\mathcal{O}\_K[1/p]$. Now we are in the Bloch-Kato regime where we can compare to etale cohomology, but we should take a bit of care because $\mathcal{O}\_K$ is not itself a field. But if you compare Gysin sequences for motivic cohomology and etale cohomology and use Bloch-Kato for the quotient field and residue fields, you do indeed see that the claim reduces to the finiteness of etale cohomology of $\mathcal{O}\_K[1/p]$ with $\mathbb{F}\_p(n)$-coefficients.
| 3 | https://mathoverflow.net/users/480363 | 426674 | 173,141 |
https://mathoverflow.net/questions/426668 | 3 | The [Tracy–Widom distributions](https://en.wikipedia.org/wiki/Tracy%E2%80%93Widom_distribution) admit many interpretations.
[One of them](https://www.ggi.infn.it/sft/SFT_2018/LectureNotes/Majumdar_slides.pdf) is related to quantum mechanics: If we consider $N$ non-interacting fermions confined by the potential $V(x) = x^2$, then in the ground state, the position of the rightmost fermion (approximately) has probability density $P=\sqrt{2N} + \frac{N^{-1/6}}{\sqrt 2} F\_2$ where $F\_2$ is the PDF of the unitary Tracy-Widom distribution.
In the ground state, the wave-function $\Phi(x\_1, \dots , x\_N ) = \frac{1}{\sqrt{N!}} \det[\phi\_i(x\_j)]$ where $\phi \_i$ runs over the first $N$ Hermitian polynomials and $j$ runs over $1 \dots N$. The probability density of observing the particles, $\Phi^2$, can be written as a determinant $\frac{1}{N!}\det[K\_N(x\_i,x\_j)]$ where $K\_N$ is a kernel depending on $N$. When $N\rightarrow \infty$, the kernel $K\_N$ converges to the Airy kernel $K\_A(u,v) = \int\_0^{+\infty} Ai(u+x) Ai(v+x)dx$ at the edge location $\sqrt{2N}$ after scaling. This is why we have the Fredholm determinant representation $F\_2=\det(1-K\_A)\_{L^2([s,+\infty])}$.
When the temperature is above zero, there is no determinantal structure for $\Phi^2$ when $N$ is fixed. However, if we consider the grand canonical potential, then there is a similar edge kernel $K\_b(u,v)=\int\_{-\infty}^{+\infty}\frac{ Ai(u+x) Ai(v+x)}{1+e^{-b x}}dx$ where $b$ is related to the temperature. Thus we call the Fredholm determinant $F\_{2,b}=\det(1-K\_b)\_{L^2([s,+\infty])}$ the **Finite-temperature Tracy-Widom distribution** function.
**Question:** How can we find a similar distribution that degenerates to the orthogonal/symplectic Tracy-Widom distribution when $T\rightarrow 0$?
| https://mathoverflow.net/users/125498 | What is the finite-temperature orthogonal/symplectic Tracy-Widom distribution? | This is not a complete answer, but more of an approach and an invitation to look at the relevant literature. As you write, you would like to insert the so-called Fermi factor into the Fredholm Pfaffian expression for $F\_1$ or $F\_4$ (by analogy with the Fredholm determinant for $F\_2$). Such Fermi factors are also present in exact formulas for solutions of the KPZ equations. On the line, this is due to Amir-Corwin-Quastel and Sasamoto-Spohn (both 2010). For $F\_1$ or $F\_4$, you want to change the geometry. Some particle systems, in particular, the Facilitated/Open TASEP evolving on the half line, have $F\_4$ fluctuations. Therefore, you want to consider exact Pfaffian formulas for the KPZ equation on the half line. There is a growing literature on this topic, both rigorous and physics level. I found [this paper](https://arxiv.org/pdf/1905.05718.pdf) which contains formulas you can look at (start with formula (64) there). Hope this helps.
| 2 | https://mathoverflow.net/users/979 | 426680 | 173,142 |
https://mathoverflow.net/questions/426681 | 3 | Let $(G,+)$ be an abelian group and $A$, $B$ and $C$ be finite subsets of $G$ with $A+B=C$. One may conclude that $A\subset C-B$. However, $A$ need not be equal to $C-B$. What is a necessary and sufficient condition to have $A=C-B$?
| https://mathoverflow.net/users/165074 | Sumsets with the property "$A+B=C$ implies $A=C-B$" | Regardless of whether $A,\ B$, and $G$ are finite or infinite, the necessary and sufficient condition is that $B$ lies in a coset of the *period* (or *stabilizer*) of $A$, defined to be the subgroup of all those group elements $g$ with $A+g=A$.
To see this, notice that $A\subseteq (A+B)-B$ holds true in a trivial way, so that your condition reduces to $(A+B)-B\subseteq A$, which is thus equivalent to $A+(b\_1-b\_2)=A$ for any $b\_1,b\_2\in B$.
| 6 | https://mathoverflow.net/users/9924 | 426686 | 173,144 |
https://mathoverflow.net/questions/426279 | 5 | I have the $N$x$N$ matrix below where $N$ is a power of 2 (usually 64 or 256) and $\omega = 2\pi/N$. What is its largest eigenvalue?
$\begin{bmatrix}
2 & 1 & 0 & 0 & \cdots & 0 & 0 & 1\\
1 & 2\cos(\omega) & 1 & 0 & \cdots & 0 & 0 & 0\\
0 & 1 & 2\cos(2\omega) & 1 & \cdots & 0 & 0 & 0\\
0 & 0 & 1 & 2\cos(3\omega) & \cdots & 0 & 0 & 0\\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots & \vdots \\
0 & 0 & 0 & 0 & \cdots & 2\cos((N-3)\omega) & 1 & 0\\
0 & 0 & 0 & 0 & \cdots & 1 & 2\cos((N-2)\omega) & 1\\
1 & 0 & 0 & 0 & \cdots & 0 & 1 & 2\cos((N-1)\omega)
\end{bmatrix}$
This is exactly the $S$ matrix [here](https://www.cs.princeton.edu/%7Eken/Eigenvectors82.pdf) (which explains how this eigenvector is also the simplest eigenvector of the Discrete Fourier Transform); I am wondering if our analysis of this eigenvector has improved since 1982.
For large $N$, I know this eigenvalue tends to 4 and its eigenvector tends to a Gaussian, and I can numerically find the roots of the characteristic polynomial for more precision (e.g., this eigenvalue is 3.903025 for $N$=64), but is there a faster more-accurate method? Is there a closed-form solution?
To reiterate, the helpful points here are that $N$ is a power of 2 and I am only concerned with the highest eigenvalue.
| https://mathoverflow.net/users/20757 | The discrete Fourier transform's Gaussian-like eigenvector | A quick numerical investigation for $N$ up to 512 leads to a conjectured exact large-$N$ expansion of the largest eigenvalue,
$$
\lambda\_0(N) = 4 - \frac{2\pi}{N} + \frac{\pi^2}{2 N^2} - \frac{\pi^3}{24 N^3} + \frac{\pi^4}{48 N^4}
- c\_5 \frac{\pi^5}{N^5} + \ldots.
$$
The coefficient $c\_5^{-1} = -114.63(1)$ is negative and seems to be a more complicated expression.
**Added 16.07.22, 18:00 CEST**:
Extending the high-precision calculation to $N=1024$, the next two constants seem to be $c\_5=-67/(2^6 5!)$ and $c\_6=653/(2^7 6!)$. These terms should be verified with a precision higher than 30 digits.
**Edited 17.07.22, 15:00 CEST:**
I did this verification with 100 digits precision, verifying $c\_{5,6}$ and extending the series to 10th order.
Therefore,
$$
\lambda\_0(N) = 4 - \frac{2\pi}{N}
+ \frac{\pi^2}{2 N^2}
- \frac{\pi^3}{24 N^3}
+ \frac{\pi^4}{48 N^4}
+ \frac{67\pi^5}{7680 N^5}
+ \frac{653\pi^6}{92160 N^6}
+\frac{32519 \pi ^7}{5160960 N^7}
+\frac{135001 \pi ^8}{20643840 N^8}
+\frac{45750727 \pi ^9}{5945425920 N^9}
+\frac{1198585643 \pi ^{10}}{118908518400 N^{10}}
+\ldots.
$$
Note that the denominators of $c\_n$ are divisible by $n!$.
All constants were successively determined by a least-square fit of a 20th order polynomial in $\pi/N$ to $λ\_0(N)$. I did not find a known function with this series expansion.
For reference, I'll append the 30 digits results:
\begin{array}{ll}
N & λ\_0(N)\\
128 & 3.95121320281088603898478521135 \\
256 & 3.97553152996970070083424756460 \\
384 & 3.98367098180919839228844082991 \\
512 & 3.98774696887736782094115794509 \\
640 & 3.99019456589827273460033342852 \\
768 & 3.99182713285240169577614505117 \\
896 & 3.99299366147114034625800714548 \\
1024 & 3.99386878184121881247367446866 \\
\end{array}
| 6 | https://mathoverflow.net/users/90413 | 426691 | 173,146 |
https://mathoverflow.net/questions/426682 | 3 | Maps between real numbers are often defined by convergent series. For example, to define the exponential map, we can just prove that series
$$
\sum\_{n = 0}^{\infty} \frac{x^n}{n!}
$$
converges, which implies that there is a map to which it converges. Can we use this approach to define (constructively) localic maps between localic reals? If not, how do we define maps like these?
| https://mathoverflow.net/users/62782 | Localic maps given by series | Here is a fairly general methods for this sort of thing :
**Step 1)** We give a constructive proof that for each (Dedekind) real $x$, the serie $\sum \frac{x^n}{n!}$ converge. We define $exp(x)$ as the limit.
**Step 2)** We show that $x \mapsto exp(x)$ is a geometric construction. That is if $f : \mathcal{E} \to \mathcal{S}$ is a geometric morphism and $x$ is a Dedekind real in $\mathcal{S}$ then $exp(f^\*x) = f^\*exp(x)$.
For limits of series this is relatively easy to do : The key point is that quantification over $\mathbb{N}$ is well behaved when it comes to pullback along a geometric morphism: if $P(n)$ denotes some proposition in $\mathbb{S}$ indexed by the NNO (so essentially a subobject of $\mathbb{N}\_S$) then
$$f^\*(\exists n \in \mathbb{N}, P(n)) = \exists n \in \mathbb{N}, f^\*(P(n))$$
and
$$f^\*(\forall n \in \mathbb{N}, P(n)) \Rightarrow \forall n \in \mathbb{N}, f^\*(P(n))$$
and as the definition of limit can be written using only quantification over $\mathbb{N}$ that's enough.
**Step 3)** Just use the Yoneda lemma : the locale $\mathbb{R}$ represent the functor on the category of locale that sends a locale $X$ to the set of Dedekind real in $Sh(X)$, and the two points above shows that $exp$ is a natural transformation of this functor.
I think Step 2 alway work for things defined as limits of a series or sequence (at least I can't imagine a situation where it doesn't), and step 3) is completely formal, so in general, the only things you need to do to define this kind of functor is Step 1) (though you should quickly check step 2).
| 3 | https://mathoverflow.net/users/22131 | 426692 | 173,147 |
https://mathoverflow.net/questions/426684 | 14 | This question is motivated by the superficial observation that [Birkhoff's representation theorem](https://en.wikipedia.org/wiki/Birkhoff%27s_representation_theorem) and the [cryptomorphism between matroids and geometric lattices](https://en.wikipedia.org/wiki/Geometric_lattice#Cryptomorphism) are sort of similar. The former says that for a finite distributive lattice $L$ with set of join-irreducible elements $P$ the subsets $\{p\in P|p\le a\}\subset P, a\in L$ form the set of order ideals of a partial order on $P$ (which uniquely defines the partial order). The latter says that if $L$ is, instead, a geometric lattice, then the same family of subsets is the set of flats for a matroid structure on $P$ (for a geometric lattice join-irreducible elements are precisely its atoms). The former provides a bijection between finite distributive lattices and finite posets, the latter provides a bijection between geometric lattices and finite simple matroids. Both bijections have functorial interpretations.
Broadly speaking, my question is: **are there other interesting correspondences of this form and do these two phenomena have some interesting common generalization?** More specifically, here is an example of what such a result may look like. We may consider an arbitrary (finite?) lattice $L$ with set of join-irreducible elements $P$ and define a family of subsets of $P$ in the same way. Does this family of subsets define some interesting algebraic structure on $P$? Perhaps, for some specific classes of lattices, maybe classes that include both geometric and distributive $L$? (Apparently, in full generality we can obtain any family which is closed under intersection, has least common supersets and for any $p,q\in P$ includes a subset containing exactly one of $p$ and $q$. But I'm not sure where to go from here.)
**Update.** Two further nice examples of such correspondences were given by Richard Stanley and Sam Hopkins.
* For finite join-distributive $L$ this family of subsets is the family of feasible sets of an [antimatroid](https://en.wikipedia.org/wiki/Antimatroid) on $P$. This generalizes Birkhoff's representation theorem. See details [here](https://en.wikipedia.org/wiki/Antimatroid#Join-distributive_lattices).
* In their [2019 paper](https://arxiv.org/abs/1907.08050) Reading, Speyer and Thomas introduce the notion of a *finite two-acyclic factorization system*: a finite set equipped with a specific kind of binary relation. For a finite semidistributive lattice our family of subsets is the set of first components of *maximal orthogonal pairs* of such a binary relation on $P$ and the relation is recovered from this data.
I've accepted the first answer but **I still very much hope to see other examples!**
| https://mathoverflow.net/users/19864 | Birkhoff's representation theorem vs matroid-geometric lattice correspondence | [Antimatroids](https://en.wikipedia.org/wiki/Antimatroid)
are a good example. We have the syllogism "Antimatroids are to
matroids as join-distributive lattices are to geometric lattices."
Two other examples are the characterizations of complemented modular
lattices of finite length $n\geq 4$ and primary modular lattices of
finite length $n\geq 4$. (A modular lattice of finite length is
*primary* if for every join-irreducible $t$, the interval
$[\hat{0},t]$ is a chain, and dually, e.g., the lattice of subgroups of
a finite abelian group.) Their characterizations are analogous to
characterizing finite distributive lattices as a collection of sets
closed under union and intersection. See for instance Theorems 5 and 6
of Alan Day, Geometric applications in modular lattices, in
*Universal Algebra and Lattice Theory (Puebla, 1982)*, Springer
Lecture Notes in Mathematics **1004**, pp. 111-141. I don't know a
common generalization of distributive and geometric lattices with the
type of structure asked for.
| 7 | https://mathoverflow.net/users/2807 | 426695 | 173,148 |
https://mathoverflow.net/questions/426696 | 3 | This might be related to an open problem.
For odd natural $n$ define the Euler quotient:
$$ a(n)=\frac{(2^{\phi(n)}-1) \bmod n^2}{n}=\frac{2^{\phi(n)}-1}{n} \bmod n$$
>
> Q1 Are there infinitely many $n$ with unbounded smallest prime factor
> for which $a(n)$ is non-zero?
>
>
>
>
> Q2 Are there infinitely many $n$ for which $a(n)$ is non-zero?
>
>
>
$a(n)=0$ is [OEIS Wieferich numbers (1)](https://oeis.org/A077816)
$a(n)=1$ for prime $n$ is [OEIS A125854](https://oeis.org/A125854)
$a(n)=3$ for prime $n$ is [OEIS A175866](https://oeis.org/A175866)
Conditional results are welcome, but we believe abc implies positive
answers.
| https://mathoverflow.net/users/12481 | Are there infinitely many nonzero Euler quotients $a(n)=\frac{2^{\phi(n)}-1}{n} \bmod n$? | Let $p$ be any odd prime. Suppose that $2^{p-1}-1=p^ku$ for some integer $u$ with $(p,u)=1$. Then by Lifting The Exponent Lemma for all integer $K\geq 0$ we have
$$
2^{p^K(p-1)}-1=p^{K+k}u\_K
$$
For some integer $u\_K$ such that $(p,u\_K)=1$. Therefore, for all $K>k$ the number $n=p^{K+1}$ satisfies
$$
2^{\varphi(n)}-1\not\equiv 0\pmod {n^2}
$$
| 8 | https://mathoverflow.net/users/101078 | 426700 | 173,150 |
https://mathoverflow.net/questions/426701 | 0 | If $G = (V, E)$ is a simple, undirected graph and $T \subseteq V$, let $$N(T) = \{v \in V: \{v, t\}\in E \text{ for some }t\in T\}.$$
Given $v\in V$ we let $N\_0(v) = \{v\}$ and $N\_{k+1}(v) = N\_k(v) \cup N(N\_k(v))$ for all $k\geq 1$. The *iterated degree sequence* of $v$, denoted by $(\text{deg}\_k(v))\_{k\in\omega}$, is defined by $$\text{deg}\_k(v) = |N\_k(v)|\text{ for every }k\in \omega.$$
To every finite graph $G = (V,E)$ we associate the *iterated degree matrix* $\mathbb{D}(G) \in \mathbb{N}^{n\times n}$ (where $n=|V|$) in the following way: for every $v\in V$, take the first $n$ elements of its iterated degree sequence; order these $n$-element integer vectors [lexicographically](https://en.wikipedia.org/wiki/Lexicographic_order), and put these lexicographically ordered vectors in the matrix.
**Question.** Are there finite $G\_i = (V\_i, E\_i)$ for $i = 1,2$ with $|V\_1| = |V\_2|$, $G\_1\not\cong G\_2$, but $\mathbb{D}(G\_1) = \mathbb{D}(G\_2)$?
| https://mathoverflow.net/users/8628 | Non-isomorphic graphs with identical iterated degree matrix | Yes. Consider all graphs with $V=\{1,\ldots,n\}$ for which the vertex $n$ has degree $n-1$. There are $2^{n^2/2+o(n^2)}$ isomorphism classes of such graphs. But $\deg\_k(v)=n$ for all $v$ and all $k\geqslant 2$, thus there exist only at most $n^{n-1}$ distinct matrices.
| 4 | https://mathoverflow.net/users/4312 | 426702 | 173,151 |
https://mathoverflow.net/questions/426554 | 1 | Let $\rho : \mathbb{R}^n\to \mathfrak{so}(2m)$ be a faithful representation of the commutative Lie algebra $\mathbb{R}^n$ into the Lie algebra of skew-symmetric matrices. There is an orthonormal basis $d\_1,\ldots,d\_{2m}$ of $\mathbb{R}^{2m}$ such that for all $i=1,\ldots,n$ and $j=1,\ldots,m$
$$\rho(e\_i)(d\_{2j-1})=\lambda\_{ij}d\_{2j}\ \text{and}\ \rho(e\_i)(d\_{2j})=-\lambda\_{ij}d\_{2j-1}$$
for some real numbers $\lambda\_{ij}$.
The **degeneracy** property of this representation is as follows
\begin{equation}\label{deg}\tag{1}
\lambda\_{ki}^2=\lambda\_{kj}^2\ \text{for all}\ k=1,\ldots,n\ \text{and for all basis of}\ \mathbb{R}^n.
\end{equation}
If $v\_k=\sum\_{p=1}^na\_{pk}e\_p$ $k=1,\ldots,n$ is another basis of $\mathbb{R}^n$, then
\begin{align\*}
\text{$\eqref{deg}$}& \Longleftrightarrow \left(\sum\_{p=1}^na\_{pk}\lambda\_{pi}\right)^2-\left(\sum\_{p=1}^na\_{pk}\lambda\_{pj}\right)^2=0\ \text{for all}\ k=1,\ldots,n\\
& \Longleftrightarrow \sum\_{p=1}^na\_{pk}^2(\lambda\_{ki}^2-\lambda\_{kj}^2)+2\sum\_{1\leq p<q\leq n}a\_{pk}a\_{qk}\left(\lambda\_{pi}\lambda\_{qi}-\lambda\_{pj}\lambda\_{qj}\right)=0\ \text{for all}\ k=1,\ldots,n\\
& \Longleftrightarrow \sum\_{1\leq p<q\leq n}a\_{pk}a\_{qk}\left(\lambda\_{pi}\lambda\_{qi}-\lambda\_{pj}\lambda\_{qj}\right)=0\ \text{for all}\ k=1,\ldots,n
\end{align\*}
>
> So I am studying the following question:
> $$\sum\_{1\leq i<j\leq n}a\_{ik}a\_{jk}x\_{ij}=0\ \text{for all invertible matrix}\ A=(a\_{ij})$$
>
>
>
>
> 1. Does this imply that $x\_{ij}=0$ for all $1\leq i<j\leq n$?
> 2. Is it possible to write the equality above using some quadratic form?
>
>
>
| https://mathoverflow.net/users/56980 | Degenerate representation | Suppose $V = \mathbb{R}^n$ has a basis $(e\_1,\dots,e\_n)$. Your assumption is that you have a family of linear maps $\lambda\_1,\dots,\lambda\_m \in V^\*$ which are defined such that $\lambda\_r(e\_i) = \lambda\_{ir}$ for any $1 \leqslant r \leqslant m$. Working inside the algebra of all functions $\mathrm{Fun}(V,\mathbb{R})$, with multiplication given by pointwise evaluation, we have a function
$$q\_{rs} = \lambda\_r\lambda\_r - \lambda\_s\lambda\_s$$
for any $1 \leqslant r,s \leqslant m$. This is, indeed, a quadratic form on $V$.
If $(e\_1^\*,\dots,e\_n^\*)$ is the dual basis of $V^\*$ then $\lambda\_r = \sum\_{i=1}^n \lambda\_{ir}e\_i^\*$ and
$$\lambda\_r\lambda\_r = \sum\_{i=1}^n \lambda\_{ir}^2e\_i^\* + 2\sum\_{1 \leqslant i < j \leqslant n} \lambda\_{ir}\lambda\_{jr}e\_i^\*e\_j^\*$$
Your assumption that $\lambda\_{ir}^2 = \lambda\_{is}^2$ for any $1 \leqslant r,s \leqslant m$ means that
$$q\_{rs} = 2\sum\_{1 \leqslant i < j \leqslant n} (\lambda\_{ir}\lambda\_{jr}-\lambda\_{is}\lambda\_{js})e\_i^\*e\_j^\*$$
As you do, we could divide by $2$ to get a quadratic form $q\_{rs}'$ so that $q\_{rs} = 2q\_{rs}'$.
If $g \in \mathrm{End}(V)$ then $ge\_k = \sum\_{i=1} a\_{ki}e\_i$ for some $a\_{ki} \in \mathbb{R}$ and we have $e\_i^\*(ge\_k)e\_j^\*(ge\_k) = a\_{ki}a\_{kj}$ so you have
$$q\_{rs}'(ge\_k) = \sum\_{1 \leqslant i < j \leqslant n} a\_{ki}a\_{kj}(\lambda\_{ir}\lambda\_{jr}-\lambda\_{is}\lambda\_{js})$$
So your condition reads that $q\_{rs}'(ge\_k) = 0$ for all $g \in \mathrm{GL}(V)$.
Now let $q : V \to \mathbb{R}$ be a quadratic form. Your question amounts to asking whether $q = 0$ given that $q(ge\_i) = 0$ for all $g \in \mathrm{GL}(V)$ and $1 \leqslant i \leqslant n$. If we were working over $\mathbb{C}$ then we could show this by viewing $V$ as an affine space and using that $q$ is a polynomial function. So as $\mathrm{GL}(V)$ is dense in $\mathrm{End}(V)$ we must have $q = 0$. Probably there is an equivalent argument over $\mathbb{R}$ but I'll leave this to someone else.
One can instead argue as follows. For any $i \neq j$ we see that the map $\tau\_{ij} \in \mathrm{End}(V)$ given by $\tau\_{ij}e\_k = e\_k + \delta\_{ki}e\_j$ is invertible and $\tau\_{ij}e\_i = e\_i+e\_j$. Hence $q(e\_i+e\_j) = 0$ for any $1 \leqslant i,j \leqslant n$ (if $i=j$ then this is $q(2e\_i) = 4q(e\_i) = 0$).
Now let $\beta(u,v) = q(u+v) - q(u) - q(v)$ be the symmetric bilinear form determined by $q$. Then for any $1 \leqslant i,j \leqslant n$ we have
$$\beta(e\_i,e\_j) = q(e\_i+e\_j)-q(e\_i) - q(e\_j) = 0$$
Using the bilinearity we see that $\beta(u,v) = 0$ for any $u,v \in V$ which shows that $q(v) = \frac{1}{2}\beta(v,v) = 0$ for any $v \in V$. Thus $q = 0$ is identically zero.
| 1 | https://mathoverflow.net/users/22846 | 426718 | 173,157 |
https://mathoverflow.net/questions/426727 | 1 | I am looking for preferably a closed form (or series solution if not possible) for the following integral:
$$\int\_0^a x^{3/2} J\_\nu (bx) dx$$
where $\nu$ is an integer. This 1D integral appears when taking the polar Fourier transform of a separable radially symmetric function in 2D that I would like to propagate using the angular spectrum method. I understand that I can express this as a finite Hankel transform of $\sqrt{x}$ but I was hoping there was an analytic solution for this simple case.
| https://mathoverflow.net/users/483438 | Definite integral of Bessel function of the first kind times $x^{3/2}$ | The integral requires $\nu>-5/2$ for convergence, and then becomes a hypergeometric function:
$$\int\_0^a x^{3/2} J\_\nu (bx) dx=\frac{2^{1-\nu} a^{\nu+\frac{5}{2}} b^{\nu}}{(2 \nu+5) \Gamma (\nu+1)}\, \_1F\_2\left(\frac{\nu}{2}+\frac{5}{4};\frac{\nu}{2}+\frac{9}{4},\nu+1;-\frac{1}{4} a^2 b^2\right).$$
| 4 | https://mathoverflow.net/users/11260 | 426729 | 173,161 |
https://mathoverflow.net/questions/426732 | 2 | Suppose that $X,X\_1,X\_2,X\_3\dots$ is a sequence of $\mathbb{P}$-i.i.d. random variables **supported in the interval $[0,1]$**. Let $F$ be the cumulative distribution of $X$, i.e. $F(x):=\mathbb{P}[X \le x]$ and let $\hat{F}\_n$ be the empirical cumulative distribution given $X\_1,\dots,X\_n$, i.e, $\hat{F}\_n(x) := \frac{1}{n}\sum\_{k=1}^n\mathbb{I}\{X\_k \le x\}$.
It is known from the [DKW inequality](https://en.wikipedia.org/wiki/Dvoretzky%E2%80%93Kiefer%E2%80%93Wolfowitz_inequality#The_DKW_inequality) that for every $n \in \mathbb{N}$ and $\varepsilon >0$
\begin{equation\*}
\mathbb{P}[\|\hat{F}\_n-F\|\_{L^\infty([0,1])} \ge \varepsilon ] =\mathbb{P}\bigg[\sup\_{x \in [0,1]}|\hat{F}\_n(x)-F(x)| \ge \varepsilon\bigg] \le 2 \exp(-2\cdot\varepsilon^2 \cdot n) \;.
\end{equation\*}
I'm wondering if can *strengthen* the DKW inequality if we replace the norm of $L^\infty([0,1])$ with the norm of $L^1([0,1])$. Specifically, can we find a universal constant $\alpha < 2$ and other two universal constants $c\_1>0, c\_2>0$ such that, regardless what is the distribution $F$ of $X$, it holds that for each $n \in \mathbb{N}$ and $\varepsilon >0$:
\begin{equation\*}
\mathbb{P}[\|\hat{F}\_n-F\|\_{L^1([0,1])} \ge \varepsilon ] =\mathbb{P}\bigg[\int\_0^1|\hat{F}\_n(x)-F(x)| \mathrm{d}x \ge \varepsilon\bigg] \le c\_1 \exp(-c\_2\cdot\varepsilon^\alpha \cdot n) \;?
\end{equation\*}
I value even something not exactly like this but similar in spirit. Any pointer to the literature is very welcome.
| https://mathoverflow.net/users/126216 | DKW inequality for $L^1$-norm | The exponent 2 on $\epsilon$ cannot be improved by passing to the $L^1$ norm. Consider $X\_i$ i.i.d. uniform in $[0,1]$. Let $C$ be a constant, and denote $\varepsilon\_n= C/\sqrt{n}$. Let $A\_n $ be the event that $\hat{F}\_n(1/2)<1/2-9\varepsilon\_n$ and let $B\_n$ be the event that $\|\hat{F}\_n-F\|\_{L^1([0,1])} \ge \varepsilon\_n$.
Then $P(A\_n) \to \Phi(-18C)$ as $n \to \infty$ by the central limit theorem, and for large $n$ we can infer from the DKW inequality that
$P(B\_n|A\_n)>1/2$. To do that, first condition on $A\_n$, and then apply DKW separately in $[0,1/2]$ and in $[1/2,1]$.
Thus $\liminf\_n P(B\_n) >0$, but for $\alpha<2$ we have
$\exp(-c\_2\cdot\varepsilon^\alpha \cdot n) \to 0$ as $ n \to \infty$.
| 2 | https://mathoverflow.net/users/7691 | 426741 | 173,164 |
https://mathoverflow.net/questions/426534 | 1 | Let $(W\_t)\_{t\ge 0}$ be a standard Brownian motion and $\tau$ be a stopping time lying in $[1,2]$. For $x, y>0$, can we show
$$\mathbb E\big[{\bf 1}\_{\{x+\inf\_{0\le t\le 2}W\_t>0\}}(W\_{\tau}-y)^+\big]>0?$$
| https://mathoverflow.net/users/261243 | Is this expectation $\mathbb E\big[{\bf 1}_{\{x+\inf_{0\le t\le 2}W_t>0\}}(W_{\tau}-y)^+\big]$ strictly positive? | Let $A= \{W\_1>y+2, \, \inf\_{0 \le t \le 1} W\_t >-x\}.$
By the reflection principle,
$$P(A)=P(W\_1>y+2)-P(W\_1<-2x-y-2)$$ $$=P(W\_1>y+2)-P(W\_1>2x+y+2)>0\,.$$
Let $D= \{ \inf\_{1 \le t \le 2} W\_t >y\}.$ Then by the Markov property and the reflection principle,
$$P(D|A) \ge P(\inf\_{0 \le t \le 1} W\_t >-1)=P(W\_1>-1)-P(W\_1<-1)=P(W\_1 \in (-1,1]\:)>0\,.$$
Thus $P(D)>0$. We deduce that
$$\mathbb E\big[{\bf 1}\_{\{x+\inf\_{0\le t\le 2}W\_t>0\}}(W\_{\tau}-y)^+\big]>0$$
Since the integrand is strictly positive on the event $D$.
| 2 | https://mathoverflow.net/users/7691 | 426744 | 173,165 |
https://mathoverflow.net/questions/426227 | 14 | In their celebrated paper "[A new approach to the representation theory of the symmetric group. II](https://doi.org/10.1007/s10958-005-0421-7)", Okounkov and Vershik prove that $Z(n-1,1)$, the centralizer of $\mathbb{C}[S\_{n-1}]$ in $\mathbb{C}[S\_n]$ is the (surprisingly commutative!) algebra generated by $Z(\mathbb{C}[S\_{n-1}])$ and $X\_n$, the Jucys–Murphy element $(1,n)+\dotsb+(n-1,n)$. In the same paper, they give a new proof of a result that generalizes this: the centralizer $Z(l,k):=\mathbb{C}[S\_{l+k}]^{\mathbb{C}[S\_{l}]}$ is generated by $Z(\mathbb{C}[S\_{l}])$, the group $S\_k$ permuting the elements $l+1,\dotsc,l+k$, and the JM elements $X\_{l+1},\dotsc,X\_{l+k}$.
Both of these are cases of centralizers of group algebras of Young subgroups, namely of $S\_{n-1}\times S\_1$ and $S\_{l}\times S\_1\times \dotsb \times S\_1$ respectively.
>
> Are there similar results about $\mathbb{C}[S\_{a\_1+\dotsb+a\_k}]^{\mathbb{C}[S\_{a\_1}\times\dotsb\times S\_{a\_k}]}$?
>
>
>
| https://mathoverflow.net/users/138150 | What is the centralizer of a Young subgroup of $S_n$? | This answer is largely inspired by the wonderful paper of Samuel Creedon, [The Farahat-Higman Algebra of Centralizers of Symmetric Group Algebras](https://arxiv.org/abs/2206.02939), which studies in detail the case of $\mathbb{C}S\_n^{S\_{n-m}}$, where $m$ is considered fixed and $n$ varies. Let $\lambda = (\lambda\_1, \lambda\_2, \ldots, \lambda\_l)$ be a partition of $n$, and write $S\_\lambda$ for the Young subgroup $S\_{\lambda\_1} \times S\_{\lambda\_2} \times \cdots \times S\_{\lambda\_l}$ of $S\_n$. It seems likely that one can extend some of the work in the paper mentioned above to the case where all parts of $\lambda$ may vary, though there might be less structure to that case.
First of all, let us describe what the centraliser looks like. Just as the centre of a group algebra, $Z(\mathbb{C}G) = \mathbb{C}G^G$, has a basis of conjugacy class sums, we have that $\mathbb{C}S\_n^{S\_\lambda}$ has a basis of sums of elements within an $S\_\lambda$ orbit of $S\_n$ (under the conjugation action).
To understand these $S\_\lambda$-conjugacy classes, recall that if $\sigma, \tau \in S\_n$, and $(a\_1, a\_2, \ldots, a\_k)$ is a cycle of $\sigma$, then $(\tau(a\_1), \tau(a\_2), \ldots, \tau(a\_k))$ is a cycle of $\tau \sigma \tau^{-1}$. If we were allowed to take $\tau \in S\_n$, then we could replace the numbers appearing in the cycle with any other numbers (in the range $1$ to $n$), and only the size of the cycle would matter. However, since we can only conjugate by $\tau \in S\_{\lambda}$, we have some restrictions on what each $a\_i$ can be sent to. Let us say that a number $k$ ($1 \leq k \leq n$) is permuted by $S\_{\lambda\_r}$ if we have
$$
\lambda\_1 + \cdots + \lambda\_{r-1}+ 1 \leq k \leq \lambda\_1 + \cdots + \lambda\_r,
$$
so that in $S\_{\lambda\_1} \times S\_{\lambda\_2} \times \cdots \times S\_{\lambda\_l}$, it is the factor $S\_{\lambda\_r}$ that acts non-trivially on $k$. The upshot of this is that the $S\_\lambda$ conjugation action allows arbitrary renumberings of cycles that preserve which $S\_{\lambda\_r}$ permutes a given entry. So the appropriate data here is to remember which $S\_{\lambda\_r}$ permutes a given element in a cycle. We do this by replacing the number with a label (or "colour") to record this information. The simplest thing to do is to use the number $r$ corresponding to $S\_{\lambda\_r}$. Let's see an example.
Suppose that $\lambda = (2,2)$ and $n=4$. Then some examples of $S\_\lambda$-conjugacy classes are:
$$(1)(2)(3)(4)$$
(the identity), whose label becomes (1)(1)(2)(2) (since the first two 1-cycles are permuted by $S\_{\lambda\_1}$, and the second two 1-cycles are permuted by $S\_{\lambda\_2}$)
$$(13)(2)(4), (23)(1)(4), (14)(2)(3), (24)(1)(3)$$
The label for this class is (12)(1)(2), since the two-cycle contains one element permuted by $S\_{\lambda\_1}$ and one permuted by $S\_{\lambda\_2}$. (Note that the 2-cycles $(12)(3)(4)$ and $(34)(1)(2)$ are in classes by themselves, and in particular not with the remaining 2-cycles above. Their labels are $(1,1)(2)(2)$ and $(2,2)(1)(1)$, respectively.)
So now it is not difficult to check that the $S\_\lambda$-conjugacy classes in $S\_n$ are in bijection with the following combinatorial objects (which are the labels we have constructed:
multisets of [necklaces](https://en.wikipedia.org/wiki/Necklace_(combinatorics)) using the "colours" $1,2,\ldots,l$, where the colour $r$ appears $\lambda\_r$ times among all necklaces in the multiset.
If $M$ is such a multiset of necklaces, let us write $X\_M$ for the sum of elements of the corresponding $S\_\lambda$-conjugacy class. We can check that the centraliser is generated by those $X\_M$ indexed by multisets of necklaces where exactly one necklace has size larger than 1. This is a filtration argument. Let us put $X\_M$ in filtration degree equal to $n$-$(\mbox{number of necklaces of size 1})$. This way, the $S\_\lambda$-conjugacy class of a permutation $\sigma$ is in filtration degree equal to the number of elements of $1,2,\ldots,n$ that are moved by $\sigma$ (not fixed points).
It is not difficult to check that (1) this is indeed a filtration, (2) the associated graded multiplication is computed by multiplying the two elements and assuming that they move disjoint elements in $1,2,\ldots,n$.
Having said all this, things are neater if instead of writing every element of $M$, we only write the necklaces of size greater than one. This is enough information to reconstruct the original multiset of necklaces because the sizes $\lambda\_r$ tell us how many necklaces of size 1 and colour $r$ have neen omitted. So let us adopt this simplified notation. Then $X\_M$ is in filtration degree equal to the sum of sizes of the necklaces in $M$.
In the (non-identity) conjugacy-class example above, where in our new notation $M = (12)$, squaring this element in the associated graded would give us $2X\_N$, where $N = (12)(12)$.
From this associated graded rule, it follows up to some rational scalar, any $X\_M$ is the product of $X\_N$ where $N$ ranges across the constituent necklaces of $M$ (counted with multiplicity), plus lower order terms. In particular $X\_N$ with $N$ having one necklace generates the centraliser.
So we've found a description of the centraliser, and a generating set. Let us see how it generalises the case $\lambda = (n-m, 1^m)$. I would like to take the case $m=1$ for granted, so that the centraliser of $S\_{n-1}$ in $\mathbb{C}S\_n$ is generated by conjugacy-class sums of $S\_{n-1}$ and the JM element $L\_n = (1,n) + (2,n) + \ldots + (n-1,n)$. As an indication of what is to come, note that for the partition $(n-1,1)$, we have $L\_n = X\_{(12)}$ (it would be $X\_{(12)(1)(1)\cdots(1)}$ if we weren't removing necklaces of size 1).
Claim: when $\lambda = (n-m,1^m)$ the centraliser of $S\_{\lambda}$ in $\mathbb{C}S\_n$ is generated by $\mathbb{Z}(\mathbb{C}S\_{n-m})$ and those $X\_M$ where $M$ has one necklace of size 2.
I will sketch the proof (since this post is already quite verbose):
* The base case $m=1$ was mentioned above.
* For general $m$, pick out any one part of size 1 in the partition, and think about the sub-partition $(n-m,1)$ of $\lambda$. This embeds inside $S\_n$ as $H = \mathrm{Sym}(1,2,\ldots,n-m) \times \mathrm{Sym}(k)$, where $k$ depends on which part of size 1 we chose. We refer to this smaller centraliser problem as a subcase.
* By the base case, we can get any element in $\mathbb{C}\mathrm{Sym}(1,2,\ldots,n-m,k)^H$ from $Z(\mathbb{C}S\_{n-m})$ and a "modified" JM element, which will equal $(1,k) + (2,k) + \cdots + (n-m,k)$. If we add $(n-m+1,k) + (n-m+2,k) + \cdots + (k-1,k)$ we will recover the usual JM element $L\_k$. Each of the transpositions $(p,k)$ we added is of the form $X\_M$ where $M$ consists of the 2-element necklace $(p-(n-m-1), k-(n-m-1))$. So up to introducing some terms in our generating set, we recover the usual JM element.
* The base case tells us that using $Z(\mathbb{C}S\_{n-m})$ and the appropriate JM element we can generate the whole centraliser in the subcase we are working. This includes $X\_M$ where $M$ has one necklace $(A,1,1,\ldots,1)$, where $A=k-(n-m-1)$ is the label of the part of size 1 that we are working with in this subcase.
* We can "stitch together" $X\_M$ where $M$ are of the fom $(A,1,1,\ldots,1)$ for varying $A$ as follows. Consider $ X\_{(A,B)} X\_{(B,1,1,\ldots,1)} X\_{(A,1,1,\ldots,1)}$. This turns out to be $X\_{(A,1,1,\ldots,1,B,1,1,\ldots,1)}$ plus lower-order terms (lower-order terms arise because the entries of cycles labelled by 1 can coincide in the two terms, but the leading order term in the product is where they are all distinct). Iterating this we can construct $X\_M$ with an arbitrary necklace $M$.
* So we can construct arbitrary $X\_M$, and in particular the whole centraliser using the following generators: $Z(\mathbb{C}S\_{n-m})$, the modified JM elements, and $X\_{(A,B)}$ where $A,B > 1$. In turn these generators can be expressed in terms of the usual JM elements and $X\_{(A,B)}$.
Thus we have deduced the two (equivalent) statements:
* The centraliser of $S\_{\lambda}$ in $\mathbb{C}S\_n$ is generated by $Z(\mathbb{C}S\_{n-m})$ and the usual JM elements and $S\_{m}$ (which is generated by transpositions).
* The centraliser of $S\_{\lambda}$ in $\mathbb{C}S\_n$ is generated by $Z(\mathbb{C}S\_{n-m})$ and $X\_{M}$ (where we may take $M$ to have a single necklace of size 2).
It would be nice to have similar statements (that we can restrict ourselves to a smaller set of generators) when $\lambda$ has more than one part of size larger than one. I don't currently know of such a statement, but maybe they could be obtained from a good understanding of the case where $\lambda = (n-d,d)$ has two parts ($d=1$ was the base case above).
| 10 | https://mathoverflow.net/users/159272 | 426751 | 173,167 |
https://mathoverflow.net/questions/426664 | 6 | Let me recall the following definition. Let $F: C \to D$ be a functor between homotopical categories. Denote by $\gamma\_C: C \to \mathrm{Ho} C$ the localization and similary for $D$. A **total left derived functor** $\mathbf{L}F: \mathrm{Ho}C \to \mathrm{Ho}D$ for $F$ is a right Kan extension of $\gamma\_D \circ F$ along $\gamma\_C$. This will be called an **absolute** derived functor if it is absolute as a Kan extension.
It is well-know that the following two properties of derived functors:
(a) if $F: C \to D$ and $F': D \to E$ are functors between homotopical categories that admit total left derived functors $\mathbf{L}F$ and $\mathbf{L}F'$, then $\mathbf{L}F' \circ \mathbf{L}F$ is a total left derived functor for $F' \circ F$.
(b) if $F : C \to D$ is left adjoint to $G: D \to C$ and $F$ and $G$ admit left and right total derived functors $\mathbf{L}F $ and $\mathbf{R}G$ respectively, then $\mathbf{L}F $ is left adjoint to $\mathbf{R} G$.
are in fact true if these functors are constructed via deformations (or, in particular, via (co)fibrant replacements in model categories), but are no more true in general, meaning, for derived functors that plainly satisfy the definition above.
Anyway, it is known, by a theorem of Maltsiniotis, that (b) holds true whenever the involved derived functors are not simply Kan extensions, but absolute ones.
I do not know if this is true also for (a), and I can not find any reference in the literature. For instance, Riehl in various places in her books or articles mentions the theorem by Maltsiniotis for adjunctions, but never says a word about composition of absolute derived functors. Cisinski in his book on [*Higher categories and homotopical algebra*](https://cisinski.app.uni-regensburg.de/publikationen.html) seems to suggest (Corollary 2.3.4 and Proposition 2.3.6) that (a) is in fact true for derived functors which are absolute Kan extensions, but does not give a proof.
So the question is: is property (a) true if $\mathbf{L}F $ and $\mathbf{L}F'$ are absolute Kan extensions?
| https://mathoverflow.net/users/482564 | Is the composite of absolute derived functors a derived functor? | Here is a somewhat degenerate example that illustrates what can go wrong.
Let $\textbf{Ab}$ be the category of abelian groups, considered as a homotopical category where the weak equivalences are the isomorphisms, and let $\textbf{Ch}$ be the category of chain complexes of abelian groups, considered as a homotopical category where the weak equivalences are the quasi-isomorphisms.
The obvious functor $\textbf{Ab} \to \textbf{Ch}$ sending each abelian group to the corresponding chain complex concentrated in degree 0 is homotopical.
On the other hand, if $M$ is not a flat abelian group, then $M \otimes\_\mathbb{Z} {-}$ is not a homotopical functor $\textbf{Ch} \to \textbf{Ch}$.
Nonetheless, the obvious functor $\textbf{Ab} \to \textbf{Ch}$ followed by $M \otimes\_\mathbb{Z} {-}$ is homotopical, so is its own (absolute) derived functor.
On the other hand, if $N$ is an abelian group such that $\textrm{Tor}^1\_\mathbb{Z} (M, N)$ is non-zero, then $M \otimes^\textbf{L}\_\mathbb{Z} N$ (i.e. the value of the (absolute) left derived functor of $M \otimes\_\mathbb{Z} {-}$ at $N$) is not quasi-isomorphic to $M \otimes\_\mathbb{Z} N$ (considered as a chain complex concentrated in degree 0).
Thus, we have a homotopical functor $\textbf{Ab} \to \textbf{Ch}$ and a left deformable functor $\textbf{Ch} \to \textbf{Ch}$ such that the composite of the (absolute) left derived functors is different from the (absolute) left derived functor of the composite.
What is true is that the universal property of absolute Kan extensions gives you a comparison between the composite of absolute derived functors and the absolute derived functor of the composite.
| 6 | https://mathoverflow.net/users/11640 | 426752 | 173,168 |
https://mathoverflow.net/questions/426762 | 17 | Are there uncountably many $A\_\alpha $ of subsets of $\mathbb{N}$ with the following two properties:
1. Each $A\_\alpha$ has positive upper natural density
2. $A\_\alpha \cap A\_\beta$ is a finite set for $\alpha \neq \beta$
If the answer is no then the next question:
Are there uncountably many $A\_\alpha $ of subsets of $\mathbb{N}$ with the following two properties:
1. Each $A\_\alpha$ has positive upper natural density
2. $A\_\alpha \cap A\_\beta$ has zero natural density for $\alpha \neq \beta$
**Note:** As indicated [at this MathSE post](https://math.stackexchange.com/questions/162387/), there are uncountably many infinite subsets of $\mathbb{N}$ with pairwise finite intersection. However,
I could not modify the method indicated in the discussion of that post to get an answer to my question.
| https://mathoverflow.net/users/36688 | Uncountably many subsets of the natural numbers with certain natural density condition | Amazingly, the answer to the main question is yes. For each $n$ let $I\_n = [2^{2^n}, 2^{2^{n+1}})$. Then let $\mathcal{C}$ be an uncountable family of infinite subsets of $\mathbb{N}$ any two of which have finite intersection. For each $B\in \mathcal{C}$ let $B' = \bigcup \{I\_n: n \in B\}$. Then $\{B': B \in \mathcal{C}\}$ is the desired family. The point is that if $B$ is any infinite subset of $\mathbb{N}$ then $B'$ has upper density 1.
Edit: or just take $I\_n = [n!, (n+1)!)$, so that $\frac{|I\_n|}{(n+1)!} = \frac{n}{n+1}$.
| 25 | https://mathoverflow.net/users/23141 | 426766 | 173,171 |
https://mathoverflow.net/questions/426772 | 0 | I want to plot the two surfaces which are defined in $ \mathbb{ R }^3 \ni ( x, y, z ) $ via the equations $ 0 = y^2 - x\*(x^2 + 1) $ and $ 0 = z^2 - y\*(y^2 + 1) $, respectively.
Moreover, I want also to highlight the intersection curve of these surfaces.
Any tool is ok for me, but I know that sage has some functions for plotting implicit functions in 3d:
The commands
```
var('x, y, z')
s1 = implicit_plot3d(0 == y^2 - x*(x^2 + 1), (x,-3,3), (y,-3,3), (z,-3,3))
s2 = implicit_plot3d(0 == z^2 - y*(y^2 + 1), (x,-3,3), (y,-3,3), (z,-3,3))
plot(s1 + s2)
```
plot both surfaces.
But I would like to highlight the intersection curve or at least change the transparency-level of the surfaces (changing the alpha-values seems not to do anything)
A bonus would also be if there were a way to directly plot the intersection of these surfaces.
Does anybody know a tool to do this?
| https://mathoverflow.net/users/132492 | Plot two implicit surfaces in 3D and highlight their intersection | Something like this?
Blue curve is the intersection of the two surfaces $ 0 = y^2 - x(x^2 + 1) $ and $ 0 = z^2 - y(y^2 + 1) $.
The 3D plots were generated with Mathematica, as documented [here.](https://reference.wolfram.com/language/example/HighlightTheIntersectionOfTwoSurfaces.html)
| 2 | https://mathoverflow.net/users/11260 | 426774 | 173,173 |
https://mathoverflow.net/questions/426759 | 6 | Let $f \in L^1 (\mathbb R)$. Suppose $g\_n \in L^1 (\mathbb R)$ are a sequence of positive functions.
Define, for each $n$, the function $f\_n$ by
$$f\_n (x) := \frac{1}{2g\_n (x)} \int\_{x - g\_n (x)}^{x + g\_n (x)} f(y) \, dy.$$
**Question:** Is it true that if $g\_n \to 0$ in weak $L^1$ norm, then $f\_n \to f$ in $L^1$ strong?
*Remark:* Implicit here is that one has to prove the $f\_n$ are all eventually in $L^1$.
| https://mathoverflow.net/users/173490 | Convergence of integral averages in $L^1$ | The answer is negative even if $f$ is supported in $[0,1]$ and $g\_n \to 0$ uniformly in $[0,1]$. The reason is that $f \in L^1$ does not suffice for the Hardy-Littlewood maximal function to be in $L^1$, so choosing $g\_n$ to capture that maximal function will yield a counterexample. It is still useful to see an explicit counterexample. I will add that next.
For $x \in (0,1]$, let $$f(x)=\frac1{x \cdot \log^2(2/x)}\,,$$ with $f(x)=0$ for all $x \notin (0,1]$. Then $f$ is in $L^1[0,1]$.
Define $g\_n(x)= \min\{x,1/n\}$ for $x \in (0,1]$, with $g\_n(x)=\frac1{n(x^2+1)}$ for $x \notin [0,1]$. Then
$$f\_n (x) := \frac{1}{2g\_n (x)} \int\_{x - g\_n (x)}^{x + g\_n (x)} f(y) \, dy$$
satisfies
$$f\_n(x)=\frac{1}{2x\log(2/x)} \quad \text{for all} \quad x \in (0,1/n)\,,$$
so $f\_n \notin L^1[0,1]$.
**Remark** If we assume that $f\log\_+(f) \in L^1$ then the answer to the original question is positive, because in that case The Hardy-Littlewood maximal function is in $L^1$, so one can appeal to dominated convergence. See, e.g., Theorem 3.4 in [this paper](https://www.jstor.org/stable/2314249).
| 7 | https://mathoverflow.net/users/7691 | 426784 | 173,178 |
https://mathoverflow.net/questions/426780 | 1 | $S\_5$ is a set consisting of the following 5-length sequences $s$: (1) each digit of $s$ is $a$, $b$, or $c$; (2) $s$ has and only has one digit that is $c$.
$T\_5$ is a set consisting of the following 5-length sequences $t$: (1) each digit of $t$ is $a$, $b$, or $c$; (2) $t$ has two digits that are $c$.
Is there a 3-partition $S\_5= \bigcup\_{j=0,1,2} A\_j$, $A\_j
\cap A\_i=\varnothing$ with the following property: for any $A\_j$ and any $t$ $\in$ $T\_5$, there is a $s$ $\in$ $A\_j$ such that exsits a $n\in\{1,2,3,4,5\}$, $s\_n \neq t\_n$, $t\_n=c$ and $s\_m=t\_m $ for any $m\neq n$, where $s\_n$ (or $t\_n$) is the n-th digit of $s$ (or $t$).
For example, $t=ccaab$ and $s=acaab$ where $n=1$.
And is there a 3-partition of $S\_6$ with $T\_6$ for 6-length sequences?
Thank you!
| https://mathoverflow.net/users/57534 | 3-partition of a special set | Here is one such partition of $S\_5$, obtained via integer linear programming:
\begin{align}
A\_0 &= \{aaaac,aaacb,aabca,aacba,ababc,abbac,abcab,abcba,acaaa,acbbb,baabc,
babac,bacaa,bacbb,bbaca,bbbbc,bbbcb,bbcaa,bcaab,bcbba,caaba,cabab,cabbb,
cbaab,cbabb,cbbaa\} \\
A\_1 &= \{aaaca,aabbc,aacab,abaac,abbca,abbcb,abcbb,acaba,acabb,acbaa,acbab,
baaac,babca,babcb,bacba,bbabc,bbacb,bbbac,bbcab,bbcba,bcaaa,bcbbb,caaab,
caabb,cabaa,cbaaa,cbbba\} \\
A\_2 &= \{aaabc,aabac,aabcb,aacaa,aacbb,abaca,abacb,abbbc,abcaa,acaab,acbba,
baaca,baacb,babbc,bacab,bbaac,bbbca,bbcbb,bcaba,bcabb,bcbaa,bcbab,caaaa,
cabba,cbaba,cbbab,cbbbb\}
\end{align}
And here is one for $S\_6$:
\begin{align}
A\_0 &= \{aaaaac,aaabca,aaacbb,aababc,aabacb,aabbac,aabcba,aacaaa,aacbab,aacbba,
abaaac,abaacb,ababbc,abacaa,abbabc,abbbca,abbcab,abcaba,abcbaa,abcbbb,acaaba
,acaabb,acabab,acbaaa,acbaab,acbbbb,baaabc,baabcb,baacaa,babaac,babaca,
babbbc,babcab,bacabb,bacbab,bacbba,bbaabc,bbaaca,bbabac,bbacbb,bbbaac,bbbbcb
,bbbcba,bbcaab,bbcbaa,bbcbbb,bcaaaa,bcaaab,bcabba,bcbaba,bcbabb,bcbbaa,
caaaab,caaaba,caabaa,caabbb,cabbaa,cabbbb,cbabab,cbabba,cbbaaa,cbbabb,cbbbab
,cbbbba\} \\
A\_1 &= \{aaaaca,aaabbc,aaacab,aabaac,aabbbc,aabbcb,aabcaa,aacaba,aacabb,aacbaa,
abaabc,ababac,ababcb,abacba,abbaca,abbbac,abbcbb,abcaaa,abcaab,abcbba,acaaab
,acabaa,acabbb,acbabb,acbbab,acbbba,baaacb,baabac,baacba,bababc,babbac,
babbca,babcbb,bacaaa,bacaab,bacbbb,bbaaac,bbabbc,bbabca,bbacab,bbbacb,bbbbbc
,bbbcaa,bbcaba,bbcabb,bbcbab,bcaaba,bcabaa,bcabbb,bcbaaa,bcbbab,bcbbba,
caaaaa,caaabb,caabab,caabba,cabaab,cababa,cbaaaa,cbaabb,cbbaab,cbbaba,cbbbaa
,cbbbbb\} \\
A\_2 &= \{aaaabc,aaaacb,aaabac,aaabcb,aaacaa,aaacba,aabaca,aabbca,aabcab,aabcbb,
aacaab,aacbbb,abaaca,ababca,abacab,abacbb,abbaac,abbacb,abbbbc,abbbcb,abbcaa
,abbcba,abcabb,abcbab,acaaaa,acabba,acbaba,acbbaa,baaaac,baaaca,baabbc,
baabca,baacab,baacbb,babacb,babbcb,babcaa,babcba,bacaba,bacbaa,bbaacb,bbabcb
,bbacaa,bbacba,bbbabc,bbbaca,bbbbac,bbbbca,bbbcab,bbbcbb,bbcaaa,bbcbba,
bcaabb,bcabab,bcbaab,bcbbbb,cabaaa,cababb,cabbab,cabbba,cbaaab,cbaaba,cbabaa
,cbabbb\}
\end{align}
| 3 | https://mathoverflow.net/users/141766 | 426788 | 173,180 |
https://mathoverflow.net/questions/426793 | 2 | Define a coupling $\pi\in \Pi(\mu,\nu)$ on the product space $(X\times X,\mathcal{F}\times\mathcal{F})$. let $\pi\_x$ be the disintegration of $\pi$ with respect to the $\mu$, i.e. there exists a Borel measurable function $x\mapsto\pi\_x$ such that $$\pi(dx,dy)=\mu(dx)K(x,dy)$$ where $K(\cdot,\cdot)$ is a probability transition kernel.
If $\pi\_x=\delta\_{f(x)}$ ($\delta$ is the Dirac measure), can we say that there exists a measurable function $f$ such that
$$
\nu=f\_{\#}\mu
$$ where $f\_{\#}\mu$ is the pushforward of measure $\mu$, i.e. $\mu(f^{-1}(A))=\nu(A)$ for any Borel measurable sets $A\subset X$.
| https://mathoverflow.net/users/168083 | Can we say that there exists a measurable function $f$ such that $ \nu=f_{\#}\mu$? | Yes: if $\pi\_x=\delta\_{f(x)}$, then
$$\nu(A)=\pi(X\times A)=\int\_X\mu(dx)\pi\_x(A) \\
=\int\_X\mu(dx)\,1(f(x)\in A)=\int\_X\mu(dx)\,1(x\in f^{-1}(A))
=\mu(f^{-1}(A))$$
for all Borel subsets $A$ of $X$, as desired.
| 2 | https://mathoverflow.net/users/36721 | 426795 | 173,182 |
https://mathoverflow.net/questions/426796 | 3 | It is known that
$$
\cos(\frac{x}{2})\cos(\frac{x}{4})\cos(\frac{x}{8})\dots = \frac{\sin x}{x} = O\_{x \rightarrow \infty}(x^{-1})
$$
Is it true that
$$
f(x) = \cos(\frac{x}{3})\cos(\frac{x}{9})\cos(\frac{x}{27})\dots = o\_{x \rightarrow \infty} (1) ?
$$
If so, what is the rate of convergence?
It seems to me that $f(x)$ converges to zero, but very slowly. For example $f(1081882100) \approx 0.27$. I guess the reason is that $f(x)$ is the Fourier transform of the uniform distribution on the Cantor set $C$ supported in $[-1/2, 1/2]$, which is highly irregular. To see this, let $X \sim \mathcal U (C)$, then by the self-similarity of $C$, we have
$$
X \stackrel{(d)}{=} X/3 + Y
$$
where $Y \sim \mathcal U(\pm 1/3)$ is independent of $X$. So
$$
\mathbb E[e^{itX}] = \mathbb E[e^{itX/3}]\cos(x/3)
$$
From which we obtain
$$
\mathbb E[e^{itX}] = \cos(\frac{x}{3})\cos(\frac{x}{9})\cos(\frac{x}{27})\dots
$$
**Update.** The answer is no by Noam Elkies. However now I want to ask the same question for
$$
f\_a(x) = \prod\_{n \geq 1} \cos(\frac{x}{a^k})
$$
for $a>1$ and $a \neq 2$.
| https://mathoverflow.net/users/97209 | $\cos(\frac{x}{3})\cos(\frac{x}{9})\cos(\frac{x}{27})\dots$ as $x \rightarrow \infty$ | No. If $x = 3^n \pi$ then $|f(x)| = f(\pi) \neq 0$
(numerically it's about $0.466$), and $3^n \pi$ can be arbitrarily large.
Such a construction fails for $\prod\_{k=1}^\infty \cos(x/2^k)$
because if some $x/2^k$ is nearly $\pi$,
or more generally some *odd* multiple of $\pi$, then
$x/2^{k+1}$ is nearly a half-integral multiple of $\pi$
and thus has a very small cosine.
[**added later**] For $f\_a(x) := \prod\_{k=1}^\infty \cos(x/a^k)$
with $a > 2$, the same construction does work if $a$ is an integer
(consider $x = a^n \pi$), and more generally if $a$ is a
Pisot-Vijayaraghavan number (which need not exceed $2$,
e.g. if $a = (1+\sqrt5)/2$ then there are arbitrarily large $x$
such that $f\_a(x)$ remains bounded away from zero).
| 9 | https://mathoverflow.net/users/14830 | 426797 | 173,183 |
https://mathoverflow.net/questions/426709 | -1 | Let $N(T)$ be the number of zeros of Riemann zeta function upto height $T$ in the critical strip and $N\_0(T)$ be the number of zeros on the critical line.
What will be the significance of proving that there is an $H$ such that for $T\geq H$, $$N\_0(T+1)-N\_0(T)\sim \frac{1}{2\pi}\log \frac{T}{2\pi}$$ and $$N(T+1)-N(T)\sim \frac{1}{2\pi}\log \frac{T}{2\pi}$$
Are the above results (especially the first one) known? Please explain the importance of the above results.
**edit** So if the above two results are proved we can conclude that $$\liminf\_{T\to \infty}\frac{N\_0(T+1)-N\_0(T)}{N(T+1)-N(T)}=1$$
Thank you.
| https://mathoverflow.net/users/nan | Significance of $N_0(T+1)-N_0(T)\sim \frac{1}{2\pi}\log \frac{T}{2\pi}$ | The first two displays (together) are in between the Lindelöf hypothesis and the Riemann hypothesis. That is, they imply the Lindelöf hypothesis, while they follow from the Riemann hypothesis. They are not known unconditionally. See Sections 13.5-13.6 in Titchmarsh: The theory of the Riemann zeta-function.
It is straightforward that the first two displays imply the third display in the stronger form
$$\lim\_{T\to \infty}\frac{N\_0(T+1)-N\_0(T)}{N(T+1)-N(T)}=1.$$
To see this, divide both the numerator and the denominator by $\frac{1}{2\pi}\log \frac{T}{2\pi}$, and observe that the new numerator and denominator tend to $1$ by assumption.
| 2 | https://mathoverflow.net/users/11919 | 426800 | 173,184 |
https://mathoverflow.net/questions/426773 | 2 | It is mentioned in multiple occasions [here](https://terrytao.wordpress.com/tag/mobius-function/) that the bound
$$
\mathop{\sum\_{n=1}^{N}}\_{n\equiv a\mod l} \mu(n) = o(N)
$$
is equivalent to the prime number theorem in arithmetic progressions. But I am not able to find a proof of this equivalence (or the proof of the above bound) except for the case when $l=1$ which is the classical prime number theorem. If anyone can point me to a reference, it will be much appreciated.
Also, there is a result in [Edwards' book](https://www.google.co.in/books/edition/_/-66eHCf2Y8kC?hl=en&gbpv=0) where he proves the result
$$
\left|\sum\_{n=1}^{N} \frac{\mu(n)}{n}\right| \leq \frac{K}{\log(N)}.
$$
Is there an analogue of this bound for arithmetic progressions as well?
There are some discussions [here](https://mathoverflow.net/questions/356251/analytic-equivalents-for-primes-in-arithmetic-progressions), but they don't consider the Möbius function
| https://mathoverflow.net/users/148866 | Averages of Möbius function in arithmetic progressions | When $(a,m) = 1$, let $\pi(x;a \bmod m)$ be the number of primes $p\leq x$ such that $p \equiv a \bmod m$.
The prime number theorem for arithmetic progressions
mod $m$ says for all $a \in (\mathbf Z/m\mathbf Z)^\times$ that $\pi(x;a \bmod m) \sim (1/\varphi(m))x/\log x$.
Harold Shapiro, in the paper *Some assertions equivalent to the prime number theorem for arithmetic progressions*, Comm. Pure Appl. Math. 2 (1949), 293-308, showed
that theorem for an integer $m \geq 1$ is equivalent to each of the following conditions for the same $m$:
$$
\sum\_{\substack{n \leq x \\ n \equiv a \bmod m}} \mu(n) = o(x)
$$
for all $a$ with $(a,m) = 1$ and
$$
\sum\_{\substack{n \geq 1 \\ n \equiv a \bmod m}} \frac{\mu(n)}{n} \ \ {\rm converges}
$$
for all $a$ with $(a,m) = 1$. This should address both of your questions.
Concerning the second equivalent condition above, note that the prime number theorem is in many places expressed as being equivalent to the calculation $\sum \mu(n)/n = 0$, but it is also equivalent just to the convergence of $\sum \mu(n)/n$ because it is easy to show that this series must be $0$ if it converges thanks to Abel's theorem for Dirichlet series: convergence of $\sum \mu(n)/n$ implies this series must equal
$\lim\_{s \to 1^+} \sum \mu(n)/n^s = \lim\_{s \to 1^+} 1/\zeta(s) = 0$.
I explained in my answer [here](https://mathoverflow.net/questions/356251/analytic-equivalents-for-primes-in-arithmetic-progressions) how the prime number theorem in arithmetic progressions mod $m$ is equivalent to nonvanishing of $L(s,\chi)$ on the line ${\rm Re}(s) = 1$
for all Dirichlet characters $\chi \bmod m$,
With this information, let's see how to derive the first Moebius analogue above. When $(a,m) = 1$,
$$
\sum\_{\substack{n \leq x \\ n \equiv a \bmod m}} \mu(n) = \sum\_{n \leq x} \frac{1}{\varphi(m)}\left(\sum\_{\chi \bmod m} \overline{\chi}(a)\chi(n)\right)\mu(n),
$$
which is
$$
\frac{1}{\varphi(m)}\sum\_{\chi \bmod m} \overline{\chi}(a)\left(\sum\_{n \leq x} \chi(n)\mu(n)\right).
$$
To show this is $o(x)$, we show each inner sum is $o(x)$.
The inner sum at $\chi$ is the partial sum of the coefficients of the Dirichlet series $\sum \chi(n)\mu(n)/n^s$, which is $1/L(s,\chi)$ for ${\rm Re}(s) > 1$.
The coefficients of this Dirichlet series are bounded in absolute value (by $1$) and $1/L(s,\chi)$ has an analytic continuation to the line ${\rm Re}(s) = 1$ (this is where we use the nonvanishing of all $L(s,\chi)$ on that line, including the pole at $s = 1$ when $\chi$ is the trivial character, so the reciprocal there is analytic at $s=1$ with a zero), so by Newman's Tauberian theorem
$$
\frac{1}{x}\sum\_{n \leq x} \chi(n)\mu(n) \to {\rm Res}\_{s=1} \frac{1}{L(s,\chi)} = 0.
$$
The convergence of $\sum\_{n \equiv a \bmod m} \mu(n)/n$ when $(a,m) = 1$ is proved by a similar approach: reduce the task to showing for all Dirichlet characters $\chi \bmod m$ that $\sum\_{n \leq x, n \equiv a \bmod m} \chi(n)\mu(n)/n$ converges as $x \to \infty$.
| 5 | https://mathoverflow.net/users/3272 | 426811 | 173,185 |
https://mathoverflow.net/questions/426354 | 4 | Let $G$ be a countable abelian group and let $H \le G$ be a subgroup. Let $G \curvearrowright (X,\mu)$ be an ergodic measure preserving action on some probability space $(X,\mu)$. Now we know that the action $H \curvearrowright (X,\mu)$ may not be ergodic, but it has an ergodic decomposition $\mu = \int\_Y\mu\_y d\nu\_y$ where each $\mu\_y$ is ergodic for the action of $H$.
Questions:
1. Are almost all the ergodic components $\mu\_y$ isomorphic to each other (as $H$-systems)?
2. If $G$ is weakly mixing then are almost all $\mu\_y$ weakly mixing?
The answer seems to be yes if $H$ is finite index in $G$. I am interested in the case of $H = \mathbb{Z} \times \{0\} $ and $G = \mathbb{Z}^2$.
| https://mathoverflow.net/users/47704 | Ergodic decomposition of the action of a subgroup | The answer to both questions is "no" - a counterexample (labeled "folklore") appears immediately prior to Question 6.6 in *Austin, Tim*, [**Extensions of probability-preserving systems by measurably-varying homogeneous spaces and applications**](http://dx.doi.org/10.4064/fm210-2-3), Fundam. Math. 210, No. 2, 133-206 (2010). [ZBL1206.28023](https://zbmath.org/?q=an:1206.28023)..
Here it is, essentially verbatim:
Let $G = \mathbb Z\times \mathbb Z$, let $\mathbb T = \mathbb R/\mathbb Z$ with Haar probability measure $m$, and let $X = \mathbb T^2 \times \mathbb T^2$ and $\mu=$ Haar probability measure on $X$. Let $A:\mathbb T^2\to \mathbb T^2$ be any ergodic toral automorphism, and define an action of $\mathbb Z\times \mathbb Z$ on $X$ by $(n,m)\cdot (\mathbf x, \mathbf y) = (A^n\mathbf x, A^n(\mathbf y+m\mathbf x))$. In other words, $G \curvearrowright (X,\mu)$ is the action generated by $A\times A$ and the map $(\mathbf x,\mathbf y)\mapsto (\mathbf x, \mathbf y+\mathbf x)$. The action generated by $A\times A$ is ergodic (and in fact mixing), since the action generated by $A$ is mixing. The entire action is therefore weak mixing.
Taking $H:= \{(0,m):m\in \mathbb Z\}$, the ergodic components are given by rotation $(0,m)\cdot \mathbf y = \mathbf y + m \mathbf x$ on the vertical $Y\_{\mathbf x}:=\{(\mathbf x, \mathbf y): \mathbf y\in \mathbb T^2\}$. Examining the spectrum of these components reveals that they are not (almost all) mutually isomorphic, and no component is weak mixing.
| 4 | https://mathoverflow.net/users/10457 | 426818 | 173,186 |
https://mathoverflow.net/questions/426776 | 4 | A well known result of A. Zuk states that for $\frac{1}{3} < d < \frac{1}{2}$, a random group $\Gamma$ with respect to Gromov's density model with density $d$ has Kazhdan's property (T) with overwhelming probability.
On the other hand, [C. J. Ashcroft](https://arxiv.org/pdf/2206.14616.pdf) has recently proved that that at densities below $\frac{1}{4}$, with overwhelming probabilty, $\Gamma$ acts with unbounded orbits on a finite dimensional
CAT(0) cube complex, and hence does not have Property (T).
It is also [known](https://mathoverflow.net/a/192098/136187) that below density $\frac{1}{6}$, with overwhelming probability, $\Gamma$ will be residually finite (by deep results of Agol, Ollivier-Wise).
**My question is whether there are ranges of densities $d$ where a random group $\Gamma$ sampled according to Gromov's density model at density $d$ is known to be both Kazhdan and residually finite.**
Any reference/solution will be greatly appreciated.
| https://mathoverflow.net/users/136187 | Residual finiteness of random groups with property (T) | This is an open question: there are no densities $1/2>d\geq 1/6$ where a random group is known to be residually finite. Any progress would be a major step forward.
As mentioned in the question, at density $<1/6$, [Ollivier--Wise](https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=&cad=rja&uact=8&ved=2ahUKEwj1vejzl4L5AhXHh1wKHXhqDAwQFnoECAMQAQ&url=http%3A%2F%2Fwww.yann-ollivier.org%2Frech%2Fpubls%2Fsixthcubes.pdf&usg=AOvVaw2Gzerup-rJoFvtRF0WWI4y) showed that a random group is the fundamental group of a compact, non-positively curved cube complex. By [Agol's theorem](https://arxiv.org/abs/1204.2810), such groups are virtually special in the sense of [Haglund--Wise](https://link.springer.com/article/10.1007/s00039-007-0629-4), and in particular residually finite.
At density $\geq 1/2$, random groups are a.a.s. finite.
Ashcroft's paper mentioned in the question, like the previous work of [Mackay--Przytycki](https://arxiv.org/abs/1407.0332) and [Montee](https://arxiv.org/abs/2106.14931) improving the Ollivier--Wise bound, gives a non-trivial action on a cube complex (which is enough to contradict (T)), but not the proper, cocompact action needed to apply Agol's theorem.
The bottom line is that, unless a hyperbolic group with (T) happens to be linear, we have no tools to prove residual finiteness. Away from the context of random groups, a very concrete class of examples is provided by the recent [Caprace--Conder--Kaluba--Witzel census](https://arxiv.org/abs/2011.09276) of generalised triangle groups. Some of these have since been [shown](https://arxiv.org/abs/2012.09019) to be residually finite, again by Ashcroft, but since the methods use cubulation these ones certainly don't have (T).
| 4 | https://mathoverflow.net/users/1463 | 426831 | 173,189 |
https://mathoverflow.net/questions/426638 | -1 |
>
> On the other hand, if $\mu$ and $\nu$ are completely dependent then $\pi\_{x\_1}=\delta\_{f(x\_1)}$ for some function $f$. Then $W(\pi)=1$.
>
>
>
---
Note that
$$
\pi(dx, dy)=\pi\_x(dy)\mu(dx).
$$
If $\pi\_x(dy)=\delta\_{f(x)}(dy)$, then
$$
\pi(dx, dy)=\pi\_x(dy)\delta\_{f(x)}(dy).
$$
| https://mathoverflow.net/users/168083 | What does $\mu$ and $\nu$ "dependent" mean? | It makes no sense to say that two probability measures $\mu$ and $\nu$ are "completely dependent". Dependence (or lack thereof) is a property of random elements, not of probability measures.
Also, a disintegration $(\pi\_x)$ is an attribute, not of probability measures $\mu$ and $\nu$, but of a coupling $\pi$ of $\mu$ and $\nu$. On the other hand, $\mu$ and $(\pi\_x)$ determine $\pi$ and hence $\nu$.
E.g., if $\mu$ and $\nu$ are each the uniform distribution on $[0,1]$, then the formula $\pi\_x=\nu$ for $x\in[0,1]$ determines the "independent" coupling $\pi=\mu\otimes\nu$ of $\mu$ and $\nu$, whereas the formula $\pi\_x=\delta\_{f(x)}$ for $f(x):=x$ and $x\in[0,1]$ determines the "completely dependent" coupling $\pi$ (of $\mu$ and $\nu$) that is the uniform distribution on the diagonal $D:=\{(x,x)\colon x\in[0,1]\}$. Functions $f$ defined by formulas $f(x):=1-x$ and $f(x):=\lfloor2x\rfloor$ for $x\in[0,1]$ provide another two examples of "completely dependent" couplings $\pi$ of $\mu$ and $\nu$, with disintegration $\pi\_x=\delta\_{f(x)}$.
Note also that the author of the linked paper never defines "completely dependent" probability measures, and I have never seen such nonsense elsewhere. Apparently, what the author meant by
>
> On the other hand, if $\mu$ and $\nu$ are completely dependent then $\pi\_{x\_1}=\delta\_{f(x\_1)}$ for some function $f$. Then $W(\pi)=1$.
>
>
>
is just the following:
>
> If the coupling $\pi$ of $\mu$ and $\nu$ is such that $\pi\_{x\_1}=\delta\_{f(x\_1)}$ for some Borel function $f$ and $\mu$-almost all $x\_1\in X$, then $W(\pi)=1$.
>
>
>
| 1 | https://mathoverflow.net/users/36721 | 426837 | 173,191 |
https://mathoverflow.net/questions/426833 | 3 | *Note: This is a refinement of a [previous problem](https://mathoverflow.net/questions/426759/convergence-of-integral-averages-in-l1).*
Let $f \in L^1 (\mathbb R)$. Suppose $g\_n \in L^1 (\mathbb R)$ are a sequence of positive functions.
Define, for each $n$, the function $f\_n$ by
$$f\_n (x) := \frac{1}{2g\_n (x)} \int\_{x - g\_n (x)}^{x + g\_n (x)} f(y) \, dy.$$
**Question:** Is it true that if $g\_n \to 0$ in $L^1$ strong, then $f\_n \to f$ in weak $L^1$ norm?
*Remarks:*
Yuval Peres has shown in the linked post that the implication $g\_n \to 0$ in $L^1$ $\Rightarrow$ $f\_n \to f$ in $L^1$ does not generally hold.
In fact it is possible that none of the $f\_n$ are even in $L^1$. As he points out, this is related to the fact that the Hardy-Littlewood maximal function of $f$ is not in $L^1$, and one can capture the behaviour of the maximal function adequately.
| https://mathoverflow.net/users/173490 | Weak convergence of integral averages | This is true, because we can approximate $f$ in $L^1$ by continuous functions. For them the statement clearly holds, and for the difference we can apply the classical
Hardy-Littlewood maximal inequality. Next, I will add some details.
Define the operator $T\_n$ on functions in $L^1(\mathbb R)$ by $$(T\_nf)(x) := \frac{1}{2g\_n (x)} \int\_{x - g\_n (x)}^{x + g\_n (x)} f(y) \, dy.$$
By [1], $T\_n$ maps $L^1$ to functions with finite weak L^1 norm, see [2] for the definition.
Given $f \in L^1(\mathbb R)$ and $\epsilon>0$, there exists a continuous function $h$ with compact support, $h \in C\_c(\mathbb R)$, such that $\|f-h\|\_1<\epsilon.$ Let $\mu$ be Lebesgue measure. By [1], for all $\lambda>0$,
$$\mu\{x \in \mathbb R: |T\_n(f-h)(x)|>\lambda/2 \} \le \frac{2C\_1 \epsilon}{\lambda} \,, \tag{1}$$ where $C\_1$ is an absolute constant.
Since $h$ is uniformly continuous and $g\_n \to 0$ in probability, the bounded convergence theorem gives that
$\|T\_n h-h\|\_1 \to 0$ as $n \to \infty$. Thus for all $n>n\_\epsilon$, we have
$\|T\_n h-h\|\_1<\epsilon$, so
$$\mu\{x \in \mathbb R: |T\_n(h)(x)|>\lambda/2 \} \le \frac{2\epsilon} {\lambda}\,. \tag{2}$$
By (1) and (2), for all $n>n\_\epsilon$,
$$\mu\{x \in \mathbb R: |T\_n(f)(x)|>\lambda \} \le \frac{(2+2C\_1)\epsilon} {\lambda}\,,\tag{2}$$
as needed.
[1] <https://en.wikipedia.org/wiki/Hardy%E2%80%93Littlewood_maximal_function>
[2] <https://en.wikipedia.org/wiki/Lp_space#Weak_Lp>
| 3 | https://mathoverflow.net/users/7691 | 426843 | 173,193 |
https://mathoverflow.net/questions/426849 | 1 | I am trying a find a reference to/proof of the following result:
>
> Let $(M, g)$ be a compact Riemannian manifold. Then there is $b$ so that the following holds: for any $r>0$, there is a covering $\mathscr U$ of $M$ by open geodesic balls of radius $r$ so that (i) every member of $\mathscr U$ intersects at most $b$ other members of $\mathscr U$, and (ii) the collection
> $$ \mathscr U\_{1/2} = \{ B(a, r/2) : B(a, r) \in \mathscr U\}$$
> still covers $M$.
>
>
>
Without condition (ii), it is a consequence of Besicovitch Covering Theorem. However, (ii) cannot holds in the general form (e.g. $r$ is not constant) as was shown in the MO post [Stronger version of Besicovitch covering theorem](https://mathoverflow.net/questions/334376/stronger-version-of-besicovitch-covering-theorem).
I am guessing the fact that the radius is constant (or bounded by below by some positive constant) would be sufficient. The above theorem is used quite frequently: in p.16, start of proof of Proposition 4.3 in [Sacks and Uhlenbeck - The existence of minimal immersions of 2-spheres](https://math.jhu.edu/%7Ejs/Math748/sacks-uhlenbeck.pdf), the last paragraph in p.390 of [Choi and Schoen - The space of minimal embeddings of a surface into a three-dimensional manifold of positive Ricci curvature](https://doi.org/10.1007/BF01388577), and a more recent occurrence: p.23 of [Chen and Warren - Compactification of the space of Hamiltonian stationary Lagrangian submanifolds with bounded total extrinsic curvature and volume](https://arxiv.org/abs/1901.03316), where they cited the Besicovitch Covering Theorem.
| https://mathoverflow.net/users/41094 | A different version of Besicovitch Covering Theorem involving balls of half radius | Are you allowing $b$ to depend on the manifold (as it appears to me from your statement)? In that case, Besicovitch is overkill, and this statement holds in much more generality than Besicovitch. One only needs to know that a compact Riemannian manifold is a [doubling metric space](https://en.wikipedia.org/wiki/Doubling_space). Doubling has many equivalent definitions, but here you can think of it as: any ball of radius $s$ contains at most $C$ $s/2$-separated points, for some fixed constant $C$ independent of $s$.
With that, let $N$ be a maximal $\frac{r}{2}$-separated set in $M$, i.e., a set of points such that any two have distance $\geq \frac{r}{2}$ and such that every point in $M$ is distance $<\frac{r}{2}$ away from one of them. Let $\mathcal{U}$ be the balls of radius $r$ centered on $N$. By construction $\mathcal{U}$ satisfies (ii) immediately. Property (i) follows from the doubling condition: The centers of all balls in $\mathcal{U}$ that touch a fixed ball from $\mathcal{U}$ form an $\frac{r}{2}$-separated set inside a ball of radius $2r$, and the cardinality of such a set is controlled by the doubling constant.
To see that every *compact* Riemannian manifold $M$ is doubling is easy. Cover $M$ by finitely many charts that are bi-Lipschitz to subsets of $\mathbb{R}^d$, and the doubling constant can then be estimated based on the number of charts, the bi-Lipschitz constants, and the doubling constant of $\mathbb{R}^d$. More refined estimates based on curvature are possible too; it depends what you need.
| 4 | https://mathoverflow.net/users/486022 | 426858 | 173,196 |
https://mathoverflow.net/questions/426850 | 3 | Say that a clone (in the sense of universal algebra) $\mathfrak{C}$ has *almost abelian symmetry groups* (= **aasg**) iff for each function $f(x\_1,...,x\_n)\in\mathfrak{C}$ there is an abelian subgroup $A\subseteq S\_n$ and a set $X\subseteq \{1,...,n\}$ such that $$\langle A\cup \mathsf{Fix}(X)\rangle=\{\sigma: f=\lambda x\_1,...,x\_n. f(x\_{\sigma(1)},...,x\_{\sigma(n)})\}$$ (where "$\mathsf{Fix}(X)$" denotes the subgroup of $S\_n$ consisting of permutations fixing $X$ pointwise and "$\langle U\rangle$" denotes the subgroup of $S\_n$ generated by $U$). Basically, $\mathfrak{C}$ has aasg if every function in $\mathfrak{C}$ has an abelian symmetry group *once* we ignore "dummy variable" issues.
(Note that we're looking at the symmetry groups of the elements of the clone, not the symmetry group of the clone - in any sense - itself.)
Coming from [this old question of mine](https://mathoverflow.net/questions/412563/what-classes-of-groups-can-arise-as-symmetry-groups-of-terms), I'm interested in understanding aasg-ness. It seems natural to start with the case where we look at clones on a $2$-element set, since things are already nontrivial but well-understood there:
>
> **Question 1**: Which clones on a $2$-element set have aasg?
>
>
>
Note that [Post's lattice](https://en.wikipedia.org/wiki/Post%27s_lattice#/media/File:Post-lattice.svg) consists of finitely many elements together with eight infinite families, so - since aasg-ness is inherited by subclones - there are only a few different "shapes" that the set of aasg clones could have. In particular, the set of aasg clones on a two-element set is (in the appropriate sense) computable (thanks to Emil Jerabek for pointing this out!).
A follow-up question is whether, like general clones, things "blow up" once we get to three elements:
>
> **Question 2**: How many aasg clones are there on $\{1,2,3\}$? If there are only countably many, what is the least finite $n$ (if any) such that there are continuum-many aasg clones on $\{1,...,n\}$?
>
>
>
| https://mathoverflow.net/users/8133 | When does a clone on a two-element set have almost abelian symmetry groups? | **Question 1**:
Is the set of clones on $\{1,2\}$ which have aasg computable?
The answer is Yes. The aasg clones are just the essentially unary clones
on $\{1,2\}$, and there are six of them.
**Easy direction:**
Assume that $\mathfrak C$ is essentially unary
and $f(x\_1,\ldots,x\_n)\in \mathfrak C$
does not depend on its first $n-1$
variables. The symmetry group of $f$ is
$\textrm{Sym}(\{1,\ldots,n-1\})$ if
$f$ depends on its last variable and is
$\textrm{Sym}(\{1,\ldots,n\})$ otherwise.
Both cases satisfy the aasg property.
**Less easy direction:**
Assume that $\mathfrak C$ is a clone on $\{0,1\}$
that is not essentially unary.
[I changed my underlying set to $\{0,1\}$ so that I can
write operations as Boolean expressions.]
I will argue that (i) $\mathfrak C$
contains a totally symmetric operation
of arity $\geq 3$ that depends on all variables, and that (ii)
a clone with a totally symmetric operation
of arity $\geq 3$ that depends on all variables
fails the aasg property.
For (i), just examine Post's lattice.
Every clone that is not essentially unary
contains one of the Boolean operations
(a) $x\wedge y\wedge z$ (ternary conjunction),
(b) $x\vee y\vee z$ (ternary disjunction),
(c) $x \oplus y \oplus z$ (ternary addition modulo $2$),
(d) $(x\wedge y)\vee (x\wedge z)\vee (y\wedge z)$ (ternary majority).
For (ii), assume that $\mathfrak C$ contains
a totally symmetric operation $f(x\_1,\ldots,x\_m)$
which depends on all variables and that $m\geq 3$.
For $n\geq 3$, the symmetry group of
$g(x\_1,\ldots,x\_m,x\_{m+1},\ldots,x\_{m+n}):=f(x\_1,\ldots,x\_m)$
is $S\_m\times S\_n$, and both factors
are nonabelian.
**Claim.** The symmetry group of $g$,
which is $S\_m\times S\_n\leq S\_{m+n}$, cannot have the form
$\langle A\cup \textrm{Fix}(X)\rangle$ for any
abelian subgroup $A\subseteq S\_{m+n}$
and any $X\subseteq \{1,\ldots,m+n\}$.
Assume otherwise.
Note that $\textrm{Fix}(X)$
is generated by the transpositions it contains,
and that this set of transpositions is `connected'
in the sense that $(i\;j), (k\;l)\in\textrm{Fix}(X)$
implies $(j\;k)\in \textrm{Fix}(X)$.
The only connected subsets of transpositions
in $S\_m\times S\_n$ lie in $S\_m\times \{1\}$
or $\{1\}\times S\_n$, so $\textrm{Fix}(X)$
lies in either
$S\_m\times \{1\}$
or $\{1\}\times S\_n$. This is enough to see that
if $N$ is the normal subgroup $S\_m\times S\_n$
that is generated by $\textrm{Fix}(X)$, then
$(S\_m\times S\_n)/N$ is nonabelian,
which can't happen if
$S\_m\times S\_n=\langle A\cup \textrm{Fix}(X)\rangle$.
| 4 | https://mathoverflow.net/users/75735 | 426859 | 173,197 |
https://mathoverflow.net/questions/426863 | 3 | Let $X$ be a smooth projective surface and $f:X\to\mathbb{P}^1$ be a $\mathbb{P}^1$-fibration with a singular fiber consisting of a tree with three irreducible rational ($-2$)-curves $D\_1$, $D\_2$, $D\_3$ and a ($-1$)-curve $L$ intersecting one of $D\_i$'s. I want to know is it possible to calculate multiplicity of $L$ in the scheme theoretic fiber of this singular fiber? Is it possible to calculate multiplicity of the singular fiber from above information?
| https://mathoverflow.net/users/211682 | Multiplicity of irreducible component of a singular fiber of a $\mathbb{P}^1$-fibration | If we assume that the rational curves intersect in nodes, then we can compute it from the data of the dual graph. Since $L$ intersects only one of the $D\_i$, then the $D\_i$ necessarily form a chain, say $D\_1 \cup D\_2 \cup D\_3$ and without loss of generality, there are two possibilities: 1) $L$ intersects $D\_1$, or 2) $L$ intersects $D\_2$. With that in mind we can compute the multiplicities of the curves in the fiber using the following basic lemma.
**Lemma:** If $F = f^\*p$ is a scheme theoretic fiber, then $F.E = 0$ for all curves $E$ contained in a fiber of $f$.
Then writing our fiber $F = a\_1D\_1 + a\_2D\_2 + a\_3D\_3 + bL$, we have a system of equations $F.D\_i = F.L = 0$. On the other hand, we know the self intersections $D\_i^2 = -2$ and $L^2 = -1$, we know that $D\_1.D\_2 = D\_2.D\_3 = 1$ and $D\_1.D\_3 = 0$, and finally we know $L.D\_i$ in each of the two cases. Using this we can solve completely for the $a\_i$ and $b$ in each of the two cases.
**Edit:** I think the curves actually have to intersect in nodes under the assumption that $X$ is smooth. This is just a question about the reduced fiber, call it $C$, which is a tree of rational curves on a smooth surface $X$. The idea is to compute the contribution of the $\delta$-invariant of the singularity to the arithmetic genus.
Let's suppose that $p \in C$ is a singular point and let $C' \to C$ be the partial normalization at $p$. If $p$ lies on a unique irreducible component of $C$ then $p\_a(C) - p\_a(C') = \delta > 0$ which is impossible as $p\_a(C) = 0$. Otherwise, $p$ is at the intersection of exactly two components of $C$ (by the assumption that $C$ is a tree). Then $C'$ is a disjoint union of two components $A$ and $B$ so $p\_a(C') = p\_a(A) + p\_a(B) - 1 = -1$ and $p\_a(C) - p\_a(C') = \delta$ so we conclude that $\delta = 1$. Then its not hard to see that the node is the only plane curve with $2$ branches and $\delta = 1$.
**Edit 2:** It turns out that in the first case there are no solutions which means that such a configuration cannot appear as the fiber of a genus $0$ fibration on a smooth surface. We can see this geometrically as follows. Since $L$ is a $(-1)$-curve we can contract it to obtain a smooth surface that the fibration factors through, but then $D\_1$ becomes a $(-1)$-curve which can contract to another smooth surface through which the fibration factors. Now $D\_2$ becomes a $(-1)$ curve so after one more contraction we have a smooth surface with a fibration and a fiber that is set theoretically just $D\_3$, but we also have that $D\_3^2 = -1$ (after these three contractions) and this is impossible.
In the second case solving the linear system just yields that $a\_2 = b = 2a\_3$ and $a\_1 = a\_3$ so we still need one more input. This comes from Tsen's theorem which guarantees (at least over an algebraically closed field) that the fibration $f$ has a section. This means that the fiber $F$ contains at least one reduced component. Since the $a\_i$ and $b$ are all positive integers, this gives us the unique solution $a\_1 = a\_3 = 1$ and $a\_2 = b = 2$. We can achieve such a fiber by starting with a fiber that is just $\mathbb{P}^1$, blowing up once, then blowing up again at the new node we produced which gives us a fiber that looks like $D\_1 + 2D\_2 + D\_3$, and finally blowing up a generic point on $D\_2$ to obtain the $D\_1 + 2D\_2 + D\_3 + 2L$ fiber.
| 7 | https://mathoverflow.net/users/12402 | 426870 | 173,200 |
https://mathoverflow.net/questions/426828 | 2 | Here I am dealing with the difference of triangulated hull and thick hull. Let $\mathcal{D}$ be a triangulated category and $\mathcal{E}\subset\mathcal{D}$ be a collection of objects. The triangulated hull $\mathcal{E}^{\star}$ in $\mathcal{D}$ is the smallest triangulated subcategory of $\mathcal{D}$ containing $\mathcal{E}$, the thick hull $\langle\mathcal{E}\rangle$ in $\mathcal{D}$ is the smallest thick subcategory of $\mathcal{D}$ containing $\mathcal{E}$.
Consider $X$ a complex smooth projective variety of dimension $n$ with a polarization $\mathcal{O}\_X(1)$ or say $\mathcal{O}\_X(1)$ is very ample, then we know that the thick hull of $\mathcal{E}=\{\mathcal{O}\_X,\dots,\mathcal{O}\_X(n)\}$ in $D^b(X)$ is $D^b(X)$.
>
> Is $\mathcal{E}^{\star}=D^b(X)$?
>
>
>
I know it is true in the case of $X=\mathbb{P}^n$.
| https://mathoverflow.net/users/nan | triangulated hull and thick hull of $\mathcal{O}_X,\dots,\mathcal{O}_X(n)$ | No. The simplest counterexample is a smooth quadric surface $X = \mathbb{P}^1 \times \mathbb{P}^1 \subset \mathbb{P}^3$. If $F$ belongs to the triangulated hull of $\mathcal{O}\_X$, $\mathcal{O}\_X(1)$, $\mathcal{O}\_X(2)$, then its first Chen class is proportional to
$$
c\_1(\mathcal{O}\_X(1)) = h\_1 + h\_2,
$$
where $h\_i$ are the classes of the rulings. In particular, the line bundle $\mathcal{O}\_X(h\_1)$ is not in this triangulated hull.
| 3 | https://mathoverflow.net/users/4428 | 426875 | 173,202 |
https://mathoverflow.net/questions/426872 | 4 | Are all Frey elliptic curves semi-stable? If so, where exactly is this needed in the modularity approach, now that we know modularity for all rational elliptic curves?
Thank you!
| https://mathoverflow.net/users/478525 | Are Frey elliptic curves semi-stable? | See Proposition 1 in [http://www.fen.bilkent.edu.tr/~franz/ta/ta-flt.pdf](http://www.fen.bilkent.edu.tr/%7Efranz/ta/ta-flt.pdf) and the paragraph following it. You have to adjust the terms in a potential counterexample to Fermat's Last Theorem for prime exponent $p \geq 5$ to make the Frey curve semistable. Whether or not modularity is *now* known in greater generality, Wiles could not handle the general case in the early 1990s but he could handle the semistable case and that is good enough for FLT.
If you look at Ribet's article [here](https://math.berkeley.edu/%7Eribet/Articles/invent_100.pdf) that first established Frey's idea that FLT should follow from modularity of elliptic curves over $\mathbf Q$, the link appears on the second page as Corollary 1.2. In the statement of this result, Ribet assumes *all* elliptic curves over $\mathbf Q$ are modular. At that time it did not seem worth making a weaker hypothesis of modularity only for semistable elliptic curves over $\mathbf Q$ even though that is the kind of elliptic curve arising in Frey's construction and Ribet is quite explicit about that in his proof since it's essential in the argument (as Wuthrich indicates in the comment above).
| 9 | https://mathoverflow.net/users/3272 | 426876 | 173,203 |
https://mathoverflow.net/questions/426864 | 6 | I am interested in existing algorithms to compute whether a given non-cyclic, non-pure cubic extension $K/\mathbb{Q}$ is monogenic or not, and if so, to give me a defining polynomial for the integral power basis $\mathbb{Z}[\alpha]$. I am also interested in general about the following statement: If we are to fix an integer $\Delta$ and count the number of cubic fields having this discriminant, what proportion of these fields are monogenic/non-monogenic as either $\Delta$ increases in absolute value or the number of number fields having $\Delta$ as a discriminant increases?
| https://mathoverflow.net/users/138669 | Algorithm for computing whether a cubic field is monogenic? | In the paper "[Computing all power integral bases of cubic fields](https://doi.org/10.2307/2008731)" (by Gaal and Schulte, published in Mathematics of Computation in 1989) the authors give an algorithm to determine if a cubic field $K/\mathbb{Q}$ is monogenic or not. It boils down to solving a cubic Thue equation, which the authors effectively solve using Baker's linear forms in logarithms.
Apparently, it is expected that 0% of cubic fields are monogenic, although this has not been proven. In the paper "[A positive proportion of cubic fields are not monogenic yet have no local obstruction to being so](https://arxiv.org/abs/2011.01186)", Alpoge, Bhargava, and Shnidman prove the statement that is the title of the paper. They also show that a positive proportion of cubic fields are locally monogenic (and respectively have no local obstruction to being monogenic, which is a stronger condition). While I don't think this completely answers your second question, I think it's the best that's presently known.
| 9 | https://mathoverflow.net/users/48142 | 426880 | 173,206 |
https://mathoverflow.net/questions/157630 | 8 | This question is [cross-posted](https://math.stackexchange.com/questions/661801/infinite-matrix-leading-eigenvalue-problem) at Math.StackExchange.com.
I'm trying to find the leading eigenvalue and corresponding left and right eigenvectors of the following infinite matrix, for $\lambda>0$:
$$
\mathrm{A}=\left(
\begin{array}{cccccc}
1 &e^{-\lambda} & 0 &0 &0 & \dots\\
1 &e^{-\lambda} & e^{-2\lambda} &0 &0 & \dots\\
1 &e^{-\lambda} & e^{-2\lambda} &e^{-3\lambda} &0 & \dots\\
\vdots & \vdots & \vdots & & \ddots
\end{array}
\right)
$$
Note that there are terms above the main diagonal.
I know that in general infinite matrices aren't really a self-consistent idea. However, this problem arises from the infinite-$n$ limit of $n\times n$ matrices with the same values. That is, if $A^{(n)}$ is an $n\times n$ matrix such that $A^{(n)}\_{ij}=A\_{ij}$ then I'm looking for $\lim\_{n\to\infty} \eta^{(n)}$, $\lim\_{n\to\infty} u^{(n)}$ and $\lim\_{n\to\infty} v^{(n)}$, where $\eta^{(n)}$, $u^{(n)}$ and $v^{(n)}$ are the leading eigenvalue of $A^{(n)}$ and its corresponding left and right eigenvectors.
I hope that the above leads to a consistent definition. I don't know much about the theory of operators on sequence spaces (which I gather is what's required to think about infinite matrix type problems correctly - pointers about how to apply it to my problem would be appreciated), but it seems to me that defining it as the limit of a sequence of finite problems should at least lead to something well-defined.
One thing that might be important is that this arises in a context in where $p\_i = u\_iv\_i$ forms a probability distribution. So it's the per-element product of the left and right eigenvectors that has to be normalisable according to the $L\_1$ norm. (This is why I'm not sure which sequence space I should be asking about.)
Of course these limits might not converge, but from investigating the $n\times n$ case numerically using [power iteration](https://en.wikipedia.org/wiki/Power_iteration), it looks like they do. The convergence is slower for smaller values of $\lambda$, and rounding errors become a problem when it gets too small, but it looks like it probably converges for all $\lambda>0$.
Note that I only care about the leading eigenvalue, i.e. the one with the largest magnitude, which should be real and positive. Its corresponding eigenvectors should have only positive entries, due to the Perron-Frobenius theorem.
Alternatively, if it's easier, a solution for the following matrix will be just as useful to me:
$$
\mathrm{B}=\left(
\begin{array}{cccccc}
1 & 1& 0 &0 &0 & \dots\\
e^{-\lambda} &e^{-\lambda} & e^{-\lambda} &0 &0 & \dots\\
e^{-2\lambda} & e^{-2\lambda} &e^{-2\lambda} &e^{-2\lambda} &0 & \dots\\
\vdots & \vdots & \vdots & & \ddots
\end{array}
\right)
$$
Again note the terms above the diagonal. (The two problems are not equivalent, it's just that either one of them will help me solve a larger problem.)
The problem is, I just don't have much of an idea how to do this. I've tried a variety of naive methods, along the lines of writing the eigenvalue equation $\mathrm{A}\mathbf{x} = \eta \mathbf{x}$ as an system of equations and then trying to find $\{x\_i >0\}$ and $\eta>0$ to satisfy them, but this doesn't seem to lead anywhere nice.
It could be that there is no analytical solution. Or even worse it could be that these matrices have unbounded spectra after all (in which case I'd really like to know!), but if anyone has any insight into how to solve one of these two problems I'd really appreciate it.
| https://mathoverflow.net/users/46551 | Infinite matrix leading eigenvector problem | The characteristic polynomial of $\mathrm A\_n$ can be calculated exactly in the limit $n\to\infty$: As pointed out by @Gottfried, the characteristic polynomial $c\_n(x)=\det(\mathrm A\_n-x 1)$ can be written in terms of the $q$-binomial as
$$
c\_n(x)=\sum\_{k=0}^{n/2+1}q^{k(k-1)}\binom{n-k+1}{k}\_{\!q} \, (-x)^{n-k},
$$
where $q=e^{-\lambda}$. Inserting the definition of the $q$-binomial
$$
\binom{n-k+1}{k}\_{\!q} \,= \frac{(q;q)\_{n-k+1}}{(q;q)\_{n-2k+1}(q;q)\_{k}},
$$
and dropping $n$-dependent prefactors, we can perform the limit $n\to\infty$ to get
$$
c\_\infty(x) = \sum\_{k=0}^{\infty}q^{k(k-1)}
\frac{\left(-x\right)^{-k}}{(q;q)\_{k}}.
$$
This sum exactly matches the definition of the basic hypergeometric series, or [$q$-hypergeometric function](https://reference.wolfram.com/language/ref/QHypergeometricPFQ.html),
$$
{\_r \phi\_s}(a;b;q;z)=\sum\_{k=0}^\infty
\frac{(a\_1;q)\_k \ldots (a\_r;q)\_k}{(b\_1;q)\_k \ldots (b\_s;q)\_k}
\left((-1)^k q^{k(k-1)/2}\right)^{1+s-r} \frac{z^k}{(q;q)\_k}.
$$
The characteristic polynomial of the matrix $\mathrm A\_n$ for $n\to\infty$ then becomes the nice expression
$$
c\_\infty(x) = {\_0 \phi\_1}\!\left(;0;q;-x^{-1}\right).
$$
After I derived this result, a web search revealed that this function is well known.
The case ${\_0 \phi\_1}(;0;q;-q z)$ is also known as Ramanujan function or $q$-Airy function, see page 27 of [these slides](https://web.math.pmf.unizg.hr/najman_conference/3rd/slides/stovicek.pdf) and references therein, where also the zeroes of $c\_\infty(x)$ are discussed.
| 5 | https://mathoverflow.net/users/90413 | 426882 | 173,207 |
https://mathoverflow.net/questions/426881 | 4 | I was reading Brown's Cohomology Theory of Finite groups and was wondering whether there's an example of the following.
Let $n\in \mathbb{N}$. Does there exist a $\mathbb{Z}\_n$ module $M$ which is cohomologically trivial (i.e. $\smash{\hat{H}}^i(\mathbb{Z}\_n,M)$ is trivial for each $i$) but such that there exists some $p,i\in \mathbb{N}$ such that $\smash{\hat{H}}^i(\mathbb{Z}\_n,M/pM)$ is non-trivial?
In the case that $M$ has no $n$ torsion this is not possible by Lemma 8.6 of Brown's Cohomology of groups, I was wondering what goes wrong in the sense of a counterexample when $M$ has $n$ torsion.
| https://mathoverflow.net/users/161454 | Cohomologically trivial module $M$ such that $M/pM$ is not cohomologically trivial for some $p\in\mathbb{N}$ | For $n=p$ an odd prime, $M=\mathbb Z/p^2$, where a generator of $\mathbb Z/p$ acts by multiplication by $1+p$, is an example.
$M/pM$ is $\mathbb Z/p$ with the trivial action of $\mathbb Z/p$, which has cohomology groups $\mathbb Z/p$ in each degree.
But for $M$ itself, the invariants are generated by $p$ and the coinvariants by $1$, and the norm map from invariants to coinvariants sends $1$ to $\sum\_{i=1}^{p-1} (1+p)^i = p + p \binom{p}{2} + \dots$, which is $p$ times a unit since $\binom{p}{2}$ is divisible by $p$, so the norm map is an isomorphism, and thus the Tate cohomology in degrees $0$ and $-1$ vanish. By periodicity, they vanish in all degrees.
| 7 | https://mathoverflow.net/users/18060 | 426883 | 173,208 |
https://mathoverflow.net/questions/426905 | 1 | I was wondering if anything could be said at all about the well-psedness of the following time-inhomogeneous singular diffusion SDE:
\begin{align}d X\_t&=\sigma(X\_t,t ) d W\_t , \qquad t\geq 0, \, X\_0 \in \mathbb R \\
\sigma(x,t )&=t^{-\alpha} b(x) \qquad \, \alpha>0,
\end{align}
with $b$ as nice as required (say bounded, or Lipschitz continuous and sublinearly growing).
I can imagine for existence one needs at least $\alpha <1/2$ so that $\sigma \in L^2(\mathbb R, [0,T])$. Perhaps something could be achieved by time reversal?
The following paper may be relvant, but I am not entirely sure of its applicability
<https://projecteuclid.org/journals/annals-of-probability/volume-46/issue-3/Strong-solutions-to-stochastic-differential-equations-with-rough-coefficients/10.1214/17-AOP1208.pdf>
| https://mathoverflow.net/users/97651 | Solution of SDE with time power law singular diffusion | It seems natural to view the Itô process on a different time scale where it becomes a Itô diffusion, and in turn, invoke existence/uniqueness theory for Itô diffusion.
In particular, under the (deterministic) time change given by $\tau(t) = (t-2 \alpha t )^{\frac{1}{1-2 \alpha}}$ (so that $\tau'(t) = \tau(t)^{2 \alpha}$ and $\tau(0)=0$), the time-changed process is (in law) an Itô diffusion that satisfies $$
d \tilde{X}\_t = b(\tilde{X}\_t) d B\_t
$$ where $\tilde{X}\_t = X\_{\tau(t)}$ and $B\_t=\int\_0^{\tau(t)} s^{-\alpha} dW\_s$ is again a standard Brownian motion. Note that this time change only makes sense for $\alpha \in [0,1/2)$.
Sufficient conditions for existence and uniqueness of the time-changed process can be found in, e.g., Section 1.2 of the monograph
*Cherny, Alexander S.; Engelbert, Hans-Jürgen*, [**Singular stochastic differential equations.**](http://dx.doi.org/10.1007/b104187), Lecture Notes in Mathematics 1858. Berlin: Springer (ISBN 3-540-24007-1/pbk). viii, 128 p. (2005). [ZBL1071.60003](https://zbmath.org/?q=an:1071.60003).
For more about the correspondence between an Itô diffusion and Itô process via time change, see
*Øksendal, Bernt*, [**When is a stochastic integral a time change of a diffusion?**](http://dx.doi.org/10.1007/BF01045159), J. Theor. Probab. 3, No. 2, 207-226 (1990). [ZBL0698.60046](https://zbmath.org/?q=an:0698.60046).
| 3 | https://mathoverflow.net/users/64449 | 426908 | 173,213 |
https://mathoverflow.net/questions/426910 | -2 | Let $E:y^2＝x^3-x/ \Bbb{Q}(i)$ be elliptic curve and $L(E,1)$ be a special value of $L$ function of $E$ at $1$.
Let $L(ψ,1)$ be value at $1$ of Hecke $L$ function with respect to Hecke character $ψ$, It is known that $L(E,1)＝L( \bar{ψ},1)L(ψ,1)$.
In this case, why $L(ψ,1)＝1$ ?
I may forget some trivial fact about Hecke $L$ character.
Thank you for your help.
| https://mathoverflow.net/users/144623 | Special value of Hecke $L$ function | Not clear what you mean by $\varphi$ or $\psi$. At any rate, $E$ is a CM curve, hence $L(s,E)$ equals $L(s,\psi)$ for a suitable Hecke character $\psi$ of the number field $\mathbb{Q}(i)$. You can find the details (e.g. the definition of $\psi$) in Sections 8.3-8.4 of Iwaniec's book "Topics in classical automorphic forms".
**Added.** As David Loeffler pointed out, if $E$ is regarded as a curve over $\mathbb{Q}(i)$, its $L$-function is not $L(s,\psi)$ but $L(s,\psi)L(s,\bar\psi)$. In my answer, $E$ was regarded as a curve over $\mathbb{Q}$, as was in the original question (cf. [revision history](https://mathoverflow.net/posts/426910/revisions)).
| 3 | https://mathoverflow.net/users/11919 | 426911 | 173,214 |
https://mathoverflow.net/questions/426915 | 4 | Let $C$ be a closed convex set of $\mathbb{R}^n$ $(n\geq 1)$, and $u\in\mathbb{R}^n\setminus\{0\}$ such that $u$ does not belong to the asymptotic cone of $C$ and is nowhere tangent to $\partial C$. For $x\in C$, let $p(x)$ be the projection of $x$ on $\partial C$ toward the direction $u$:
$$p(x)=x+\inf\{\lambda\geq 0;\,x+\lambda u\notin C\}\,u$$
Hence $p$ is defined from $C$ to $\partial C$.
**Question:** Is $p$ Lipschitz?
The answers given lead to asking the following close question: [Lipschitz aspect of a projection on the boundary of a convex](https://mathoverflow.net/questions/427057/lipschitz-aspect-of-a-projection-on-the-boundary-of-a-convex)
| https://mathoverflow.net/users/159940 | Is this projection on the boundary of a convex Lipschitz? | No. E.g., let $n=2$ and
$$C=\{(s,t)\in\mathbb R^2\colon t\ge s^2\}.$$
Then the asymptotic cone of $C$ is $K:=\{(0,t)\colon t\ge0\}$. Let $u=(1,0)\notin K$. Then
$$p((0,t))=(\sqrt t,t)$$
for real $t\ge0$. So, $p$ is not Lipschitz.
---
Somehow I missed the condition that the direction $u$ be nowhere tangent to $\partial C$ -- my apologies for that. Yet, for this $C$ still there is a truly "bad" admissible direction $u$, such that the projection along $u$ is not Lipschitz.
Indeed, let $u=(0,-1)$. Then $u\not\in K$ and $u$ is nowhere tangent to $\partial C$. Take any $s$ and $h$ such that $0<h<s<\infty$. Let $x=(s,s^2)$ and $y=(s-h,s^2)$. Let $d$ denote the Euclidean distance. Then $p(x)=(s,s^2)$, $p(y)=(s-h,(s-h)^2)$, $d(x,y)=h$, and $d(p(x),p(y))\ge s^2-(s-h)^2>sh$. So, letting $s\to\infty$, we have $\dfrac{d(p(x),p(y))}{d(x,y)}\to\infty$. So, $p$ is not Lipschitz.
| 4 | https://mathoverflow.net/users/36721 | 426918 | 173,218 |
https://mathoverflow.net/questions/426906 | 3 | While playing around with random matrices and I arrived at a different formula for the mean of the limiting normal distribution for a spectral CLT for sample covariance matrices. More precisely I have the formula
\begin{align\*}
& \sum\limits\_{b=1}^{r-1} c^{b} \left(A(r,b) - {r \choose b}^2 \, b\right)
\end{align\*}
for the expression (1.23) from Bai and Silverstein's famous paper 'CLT for Linear Spectral Statistics of Large-Dimensional Sample Covariance Matrices', where
\begin{align\*}
A(r,b) :=& \sum\limits\_{m=1}^{\min(b,r-b+1)} (2m-1) {r \choose b-m}{r \choose b+m-1}\\
& + \sum\limits\_{m=1}^{\min(b,r-b)} (2m+1) {r \choose b-m}{r \choose b+m} \ .
\end{align\*}
In order to show that my expression is equal to that of Bai and Silverstein, I need to show the equality
$$
A(r,b) = \frac{1}{2} {2r \choose 2b} + \left( b - \frac{1}{2} \right) {r \choose b}^2
$$
for all $r \in \mathbb{N}$ and $b<r$. I am not too well versed with combinatorial identities and so far my efforts have lead nowhere. Does someone have an idea how to prove this?
Testing with the computer shows this to be true for at least all $b<r \leq 1000$.
Technically I could just use the fact that my formula and Bai and Silverstein's formula describe the same object to show this identity, which seems rather crude unless this identity is not jet known.
| https://mathoverflow.net/users/409412 | Is this combinatorial identity known? (of interest for random matrix theory) | Firstly, exploit the finite support to simplify the limits of the sums. Secondly, split the second sum. We get
$$\begin{align\*}A(r,b) =& \sum\limits\_{m=1}^{r} (2m-1) {r \choose b-m}{r \choose b+m-1} \\
& + \sum\limits\_{m=1}^{r} 2m {r \choose b-m}{r \choose b+m} \\
& + \sum\limits\_{m=1}^{r} {r \choose b-m}{r \choose b+m}
\end{align\*}$$
The first two are mechanical using Wilf-Zeilberger, so ask your favourite CAS (e.g. Wolfram Alpha or Sage) to get
$$\begin{align\*}A(r,b) =
& \frac{b(r-b+1)\binom{r}{b-1}\binom{r}{b} + (b-2r-1)(b+r)\binom{r}{b-r-1}\binom{r}{b+r}}{r} \\
& + \frac{(b+1)(r-b+1)\binom{r}{b-1}\binom{r}{b+1} + (b-2r-1)(b+r+1)\binom{r}{b-r-1}\binom{r}{b+r+1}}{r} \\
& + \sum\_{m=1}^r \binom{r}{b-m} \binom{r}{b+m}
\end{align\*}$$
For the third sum, we apply $$\sum\_k \binom{r}{m+k} \binom{s}{n-k} = \binom{r+s}{m+n}$$ (see e.g. identity (5.22) of *Concrete Mathematics* by Graham, Knuth and Patashnik). So $$\sum\_{m=1}^r \binom{r}{b-m} \binom{r}{b+m} = \frac{\sum\_{m=-r}^r \binom{r}{b-m} \binom{r}{b+m} - \binom{r}{b}^2}{2} = \frac{\binom{2r}{2b} - \binom{r}{b}^2}{2}$$
I've left you some cancellation, which I fully expect to be routine.
| 7 | https://mathoverflow.net/users/46140 | 426919 | 173,219 |
https://mathoverflow.net/questions/426903 | 2 | Let $M$ be a closed oriented real manifold with a free smooth circle action. Denote $BS^1$ to be the classifying space of principal circle bundles and $ES^1\rightarrow BS^1$ to be the universal principal circle bundle.
Now we consider the fibre product $M\_{S^1}:=M\times\_{S^1}ES^1$. When the fundamental group $\pi\_1(M)=0$, one can show that $$H^2(M\_{S^1}, \mathbb{Z})\cong H^2(M, \mathbb{Z})\oplus \mathbb{Z}.\ \ \ \ \ \ \ \ \ \ \ \ \ (\star)$$
My question is: Is it possible to loosen the condition "$\pi\_1(M)=0$"? For example, does $(\star)$ still hold true if we only assume $H^1(M, \mathbb{Z})=0$?
| https://mathoverflow.net/users/169860 | on second cohomology of $S^1$-manifold | Yes. This follows from the Leray-Serre spectral sequence of the fibre bundle
$$
M\to M\_{S^1} \to BS^1
$$
which has $E\_2^{p,q}=H^p(BS^1;H^q(M;\mathbb{Z}))$ and converges to (the associated graded of the filtration on) $H^\*(M\_{S^1};\mathbb{Z})$. Note that $BS^1\simeq \mathbb{C}P^\infty$ has cohomology concentrated in even degrees. So the only possible differentials into or out of the diagonal $p+q=2$ are $$d\_2:H^0(BS^1;H^2(M;\mathbb{Z}))\to H^2(BS^1;H^1(M;\mathbb{Z}))$$ and $$d\_2:H^0(BS^1;H^1(M;\mathbb{Z}))\to H^2(BS^1;H^0(M;\mathbb{Z})),$$ both of which must be zero if we assume that $H^1(M;\mathbb{Z})=0$.
| 7 | https://mathoverflow.net/users/8103 | 426920 | 173,220 |
https://mathoverflow.net/questions/426712 | 9 | I have done a course in Sieve Theory from [the notes of Prof. Rudnick](http://www.math.tau.ac.il/%7Erudnick/courses/sieves2015.html). Before this, I did 2 courses in Number Theory from the 2 volumes of Apostol.
I don't have any guidance by professor as I am living in a very poor country in Europe and I study mathematics in spare time though I am serious about it.
>
> While I was searching for papers in Sieve theory( I wanted to read more papers in Sieve Theory), there are not many people working in Sieve Theory. Comparatively, much more work is been going in L-functions ( Low lying zeroes, twists, Multiple L-functions and so on) and Modular forms. Can you please let me know why? Has research in Sieve Theory attained saturation?
>
>
>
Kindly let me know!
Thanks!
| https://mathoverflow.net/users/151209 | Status of current research in Sieve Theory | Sieve theory is not saturated. It is alive and thriving. The ICM just awarded the Fields medal to James Maynard in no small part due to his work in sieve theory (see [here](https://arxiv.org/abs/1311.4600) and [here](https://arxiv.org/abs/1408.5110)). Because of the natural role of multiplicative structure in sieve theory, there are lots of fascinating and important results in the theory of $L$-functions and modular forms that rely decisively on sieve theory. In these instances, it is usually the flexibility of sieve methods that ends up being the key to success. On the other hand, there are lots of things that one can study about $L$-functions and modular forms that have nothing to do with sieve theory.
For a first look at sieve theory, I think that "An Introduction to Sieve Methods and Their Applications" by Cojocaru and Murty is very nice. As Stanley Xiao mentioned above, a more advanced book (but still nice to read) is "Opera de Cribro" by Friedlander and Iwaniec.
Here are some examples where sieve theory, $L$-functions, modular forms, and other ideas combine, resulting in truly splendid results. This list is far from exhaustive.
* [Matomäki and Radziwill's work](https://arxiv.org/abs/1405.7671) on sign changes for Hecke eigenvalues of holomorphic cusp forms;
* [Radziwill and Soundararajan's work](https://arxiv.org/abs/1403.7067) on moments of $L$-functions associated to quadratic twists of elliptic curves (where they, in essence, prove a robust "Brun-Hooley type" sieve for $L$-functions);
* [Holowinsky's work](https://arxiv.org/abs/0809.1640) on shifted convolution sums, which played a key role in the resolution of the holomorphic variant of the Rudnick-Sarnak quantum unique ergodicity conjecture.
| 15 | https://mathoverflow.net/users/111215 | 426932 | 173,223 |
https://mathoverflow.net/questions/426666 | 4 | Consider a $1$-periodic Hamiltonian $H:S^{1}\times M\rightarrow \mathbb{R}$ defined on a compact symplectic manifold $M$. Let's suppose $M$ is nice enough so that we can develop Floer theory on it. Now when defining Floer's equation I have seen people considering time-dependent compatible almost complex structures $J\_t$ and as well just a compatible almost complex structure $J\_0$ to define the Floer equation and hence the differential of the Floer complex.
What I am trying to understand is that why is there this interplay ? Why not just pick an almost complex structure $J\_0$ instead of the family $J\_t$?
Is it because sometimes we want to obtain transversality results for the almost complex structure $J\_t$? That is we want to obtan that there exists a dense set of almost complex strcutures instead of Hamiltonians , and this forces us to take the time-dependency?
However I'm not sure if there is something deeper going on and would like to hear from people that know more about this subject than me.
Thanks in advance.
| https://mathoverflow.net/users/155363 | Choice of a family of almost complex structures when defining Floer Homology | For a lot of things, you can work with a generic time-independent $J\_0$ as you suggest; for instance, Audien & Damien work in this context in their book (so for most of the "fundamental" constructions in Hamiltonian Floer theory, you don't need time-dependence in your almost complex structure).
But for other things, it doesn't seem to be easy to do this; for instance, it's often convenient (see for instance Seidel's paper $\pi\_1$ *of symplectic automorphism groups and invertibles in quantum homology rings* for real development of this idea) to be able to have an action of a Hamiltonian loop $(g\_t)\_{t \in S^1} \subset Ham(M,\omega)$ on Floer complexes $CF\_\*(H,J)$ which is essentially an isomorphism of filtered chain complexes (up to a potential shift in action and degree in non-aspherical manifolds), and the natural way to do this in the aspherical case (see either the Seidel paper or the section on the Seidel representation in the chapter on Floer Homology in McDuff & Salamon's *J-holomorphic curves and symplectic topology* to see how to define the map on capped orbits) is to send $1$-periodic orbits of $\phi\_H:=(\phi\_H^t)\_{t \in [0,1]}$ to $1$-periodic orbits of $g^{-1} \circ \phi\_H = (g^{-1}\_t \circ \phi\_H^t)\_{t \in [0,1]}$ (this is the Hamiltonian isotopy generated by $g^\*H$), and to then note that a cylinder $u$ satisfies the $(H,J)$-Floer equation if and only if the cylinder $v(s,t)=g^{-1}\_t(u(s,t))$ satisfies the $(g^\*H,g^\*J)$-Floer equation. This shows that the aforementioned map gives an isomorphism of chain complexes between $CF\_\*(H,J)$ and $CF\_\*(g^\*H,g^\*J)$, but for this latter complex to even make sense, you'll need to allow time-dependent almost-complex structures, even if your initial $J$ was time-independent!
So you actually gain something by working with time-dependent almost complex structures. Of course, you pay a bit of a cost too, in that when holomorphic spheres bubble off you need to keep track of the $t$-coordinate in the domain at which the bubble occurs, since the resulting sphere will be $J\_t$-holomorphic for that value of $t$ (in the time-independent case you always get a $J\_0$-holomorphic sphere, of course). As a consequence, if you don't want to use virtual techniques, then you're forced to work in the setting of **strong** semipositive symplectic manifolds rather than semipositive symplectic manifolds *tout court* in order to compensate for the extra dimension of freedom you're adding in permitting time-dependence (the proof of the invariance of Floer homology under changes of the almost complex structure requires that $J$-holomorphic spheres with negative Chern number don't appear in generic $3$-parameter families of almost complex structures. Hofer-Salamon's *Floer homology and Novikov rings* is a good place for most of the details on this, and it makes clear why you'd need the strong semipositivity hypothesis in the time-dependent case, although they work in the time-independent setting). Also, it stops being so clear that you can just identify the Floer complex of a pair $(f,J)$, where $f$ is a small Morse function, with the Morse complex (with quantum coefficients) of $(f,g\_J)$, since autonomous almost complex structures aren't generic in the space of $t$-dependent almost complex structures, so you either have to work with the "equivariant action functional" and quotient out by the additional $S^1$-symmetry here (this is what Floer-Hofer-Salamon do in their "Transversality" paper), or you have to develop the PSS-isomorphism in order to instantiate the usual identification of Floer homology with quantum homology. If I had to guess, I'd imagine that it's this last difficulty which motivated Audin & Damien to do things in the time-independent setting.
So, in some sense, there's a real difference between the two settings, and in practice it's often just more flexible to work with $t$-dependent almost-complex structures, but also, a lot of these differences are mainly just technical and all of the "main results" work in both settings --- and most of these with only fairly minor variations in statements and proofs, but the identification of Floer homology with quantum homology becomes more involved in the $t$-dependent case.
| 5 | https://mathoverflow.net/users/134697 | 426943 | 173,226 |
https://mathoverflow.net/questions/425095 | 2 | When developing floer theory for an Hamiltonian $H:M\times S^{1}\rightarrow \mathbb{R}$ we will want $H$ to satisfy a non-degenerancy condition, that is, for every $x\in \mathcal{P}(H)$, a periodic solution of the hamiltonian system, $1$ is not an eigenvalue of $d\phi^{1}\_{H}(x(0))\in GL(T\_{x(0)}T^\*M)$. From all the places I have been reading about Floer homology, it's said that this assumption holds for a generic choice of $H$. However I would like to see the exact statement regarding this result, is it that we have a dense set of such functions in the $C^{\infty}-$topology ? Does anyone know a reference for this ?
Any insight is appreciated, thanks in advance.
| https://mathoverflow.net/users/155363 | Generic choice of non-degenerate Hamiltonians $H$ in Floer theory | You can find a statement (and proof) of such a theorem in Hofer-Salamon's *Floer homology and Novikov rings*, where it appears as Theorem $3.1$. They require also that no holomorphic spheres with first Chern number $\leq 1$ lie on the periodic orbit (which becomes necessary when one deals with bubbling off in the Floer-theoretic setting), but the first part of the proof pertains just to the non-degeneracy condition.
The genericity is in the sense that there is a set $\mathcal{U} \subset C\_{\epsilon}^\infty(S^1 \times M)$ of the second category in the sense of Baire such that every Hamiltonian $H \in \mathcal{U}$ is non-degenerate in your sense. Here $\epsilon=(\epsilon\_k)\_{k=0}^{\infty}$ is a sequence of positive numbers which decrease to $0$ sufficiently rapidly, and $C\_{\epsilon}^\infty(S^1 \times M)$ denotes the Banach space of smooth functions $H : S^1 \times M \rightarrow \mathbb{R}$ such that $\sum\_{k=0}^\infty \epsilon\_k \| H \|\_{C\_k} < \infty$. If the $\epsilon\_k$ are chosen to decrease sufficiently rapidly, then this space is dense in $L^2(S^1 \times M)$ (see Lemma $5.1$ in Floer's *The unregularized gradient flow of the symplectic action*, or the proof of Proposition $8.3.1$ in Audin-Damien's book on Floer homology), and so since $\mathcal{U}$ is in particular dense in $C\_{\epsilon}^\infty(S^1 \times M)$ and since $C^\infty(S^1 \times M) \subset L^2(S^1 \times M)$, $\mathcal{U}$ is also dense in $C^\infty(S^1 \times M)$.
| 1 | https://mathoverflow.net/users/134697 | 426950 | 173,229 |
https://mathoverflow.net/questions/426948 | 0 | Let's take a simple random walk on $\mathbb{Z}$, $(S\_n)\_{n\geq0}$, started at zero. If $\tau^+\_0 = \inf\{n \geq 1: S\_n = 0\}$ is the first time the walk returns on zero, we know that $\mathbb{E}[\tau^+\_0] = +\infty$, since the walk is recurrent null. Now I need to know the tail, $\mathbb{P}[\tau^+\_0> K]$, when $K\to+\infty$. I've understood that it's $\tfrac{C}{\sqrt{K}}$, but does someone have a proof or a reference ? Thanks.
| https://mathoverflow.net/users/486100 | Simple random walk return time | This can be done by the reflection principle. Also, one can use [Theorem 0.6](http://cgm.cs.mcgill.ca/%7Ebreed/MATH671/lecture2corrected.pdf), which implies
$$P(\tau\_0^+>k)=\tfrac1k\,E|S\_k|.$$
By the central limit theorem and uniform integrability, for $k\to\infty$,
$$E|S\_k|\sim\sqrt k\,E|Z|=\sqrt k\,\sqrt{\frac2\pi},$$
where $Z$ is a standard normal random variable.
So,
$$P(\tau\_0^+>k)\sim\sqrt{\frac2\pi}\frac1{\sqrt k}.$$
| 1 | https://mathoverflow.net/users/36721 | 426952 | 173,230 |
https://mathoverflow.net/questions/426947 | 1 | $\DeclareMathOperator\GL{GL}\DeclareMathOperator\Tor{Tor}$Let's suppose $V$ is a $k$-vector space equipped with its standard (left) $\GL (V)$-action. The shuffle algebra is the graded dual of the tensor algebra with the standard Hopf algebra structure, and for ease I'm just going to denote this shuffle algebra by $T^\bullet (V)$.
Notice that the shuffle algebra is equivariant with respect to the induced $\GL (V)$-action, by which I mean $g (v\_1 \cdot v\_2) = (g v\_1) \cdot (g v\_2)$ for any $g \in \GL(V)$, $v\_i \in T^\bullet (V)$. The *decomposables* of $T^\bullet (V)$ are going to be all elements in the image of the multiplication map $T^+ (V) \otimes\_k T^+ (V) \to T^+ (V)$ ($-^+$ means positive degree); this is evidently a $\GL (V)$-representation by the equivariance mentioned above. The *indecomposables* are the elements that are not decomposable, unsurprisingly.
My question is this: is it known what the indecomposable elements of the tensor algebra are as a $\GL (V)$-representation? Or, maybe a better way to say this: is there a "nice" (interpretation up to you) description of the representation theoretic structure of the indecomposables?
I can give one possible description that I've come up with but isn't totally satisfying (and likely hard to read): the shuffle algebra may be viewed as the Tor-algebra $\Tor\_\bullet^R (k,k)$ of the residue field $k$ over the ring $R = S\_\bullet (V) / (V)^2$ (that is, take a polynomial ring and mod out by the irrelevant ideal squared). This is a so-called Golod ring and hence its homotopy Lie algebra is free on (a shift of) $\textrm{Tor}\_\bullet^{S\_\bullet (V)} (R , k)^\*$, which is well-known to be equivariantly isomorphic to a direct sum of Schur modules corresponding to hook shapes $(2,1^i)$. Since the Ext algebra is the universal enveloping algebra of the homotopy Lie algebra in this case, we deduce that the indecomposable elements correspond to a plethysm problem for Schur modules, which I would like to avoid.
Anyway, any help is appreciated.
| https://mathoverflow.net/users/73780 | Equivariant description of indecomposable elements in shuffle algebra | There is a duality on covariant functors from $k$--vector spaces to itself: $DF(V) = F(V^{\vee})^{\vee}$, where $V^{\vee}$ is the dual of the vector space $V$. $D$ is known to preserve irreducible functors, so in the semisimple case (e.g. if $k$ has characteristic 0), $D$ acts as the identity on appropriately finite objects. (Things are more exciting in characteristic $p$.)
Now towards your question ... In any homogeneous degree, the functors of indecomposables in the shuffle algebra is dual to the functor of primitives in tensor algebra. The functors of primitives in $T\_\*(V)$ are known: it is $Lie\_\*(V)$, the free Lie algebra generated by the vector space $V$. (In characteristic $p$, I guess we need the free restricted Lie algebra generated by $V$.)
So one description of the functors of indecomposables in degree $d$ in the shuffle algebra $T^\*(V)$ will be $DLie\_d(V)$, which in char 0, will be just $Lie\_d(V)$.
| 5 | https://mathoverflow.net/users/102519 | 426958 | 173,232 |
https://mathoverflow.net/questions/426896 | 2 | Let $u(x, t)$ be a (non-negative, bounded) function on $\mathbb{R}^{n}\times [0, +\infty)$ and suppose that $u$ satisfies some time-independent PDE, e.g. $\partial\_{t}u=\Delta\_{p}u$. Let us assume that $u$ has the following property: if there exists a ball $B(x\_{0}, r\_{0})$ in $\mathbb{R}^{n}$ such that $$B(x\_{0}, r\_{0})\cap \text{supp}~u(\cdot, 0)=\emptyset,$$ then there is $t\_{0}=t\_{0}(r\_{0})$ so that for all $0\leq t<t\_{0}$, $$B(x\_{0}, r\_{0}/2)\cap \text{supp}~u(\cdot, t)=\emptyset.$$
Does this property imply that, if we suppose that $\text{supp}~u(\cdot, 0)$ is compact, there exists $T=T(r)>0$ and an increasing, non-negative function $$r:[0, T)\to [0, \infty)$$ so that for any $0\leq t<T$, $$\text{supp}~u(\cdot, t)\subset U\_{r(t)},$$ where $U\_{r}=\{x\in \mathbb{R}^{n}:\text{dist}(x, \text{supp}~u(\cdot, 0))\leq r\}$?
In other words, the latter property says that $u$ has finite propagation speed.
Does anyone have a reference for this?
Any help is appreciated!
| https://mathoverflow.net/users/163368 | Property implies finite propagation speed | As noted by the OP in a comment, $u(x, t)$ is assumed to be a (non-negative, bounded) function on $\mathbb{R}^{n}\times [0, +\infty)$ which solves a time-homogenous PDE.
``Time homogenous'' means that if $u(x,t)$ is a solution, then $\tilde{u}(x,t):=u(x,s+t)$ is also a solution for every $s \ge 0$. In particular, this holds for equations of the form $u\_t=F(u,\nabla u, \nabla^2u)$, and the equation is of this form.
Indeed, suppose $u(x,t)$ solves $$u\_t(x,t)=F\bigl(u(x,t),\nabla u(x,t), \nabla^2 u (x,t)\bigr) \,.$$
Then
$$\tilde{u}\_t(x,t)=u\_t(x,s+t)=F\bigl(u(x,t),\nabla u(x,s+t), \nabla^2 u (x,s+t)\bigr)$$ $$=F\bigl(\tilde{u}(x,t),\nabla\tilde{u}(x,t), \nabla^2 \tilde{u}(x,t)\bigr) \,.$$
Thus the hypothesis in the problem can be rewritten in the following form:
For every $s\ge 0$, if there exists a ball $B(x\_{0}, r\_{0})$ in $\mathbb{R}^{n}$ such that $$B(x\_{0}, r\_{0})\cap \text{supp}~u(\cdot, s)=\emptyset,$$ then there is $t\_{0}=t\_{0}(r\_{0})$ so that for all $0\leq t<t\_{0}$, $$B(x\_{0}, r\_{0}/2)\cap \text{supp}~u(\cdot, s+ t)=\emptyset.$$
Since supp $u(\cdot,0)$ is compact, it is contained in a closed ball $\bar{B}(0,R)$.
**Claim:** If for some $s\ge 0$ and $R\_1>0$, we have supp $u(\cdot,s) \subset \bar{B}(0,R\_1)$, then for all $t\in [0,t\_0/2]$ we have supp $u(\cdot,s+t) \subset \bar{B}(0,R\_1+r\_0/2)$.
Proof: It suffices to verify that every $z \notin \bar{B}(0,R\_1+r\_0/2)$ satisfies $z \notin \: \text{supp} \: u(\cdot,s+t)$ for every $t\in [0,t\_0/2]$.
If $|z| \ge R\_1+r\_0$, , let $z\_1=z$.
If $|z|<R\_1+r\_0$, let $z\_1:= \displaystyle(R\_1+r\_0)\frac{z}{|z|},$
and observe that in this case,
$$ z\_1-z =(R\_1+r\_0-|z|)\frac{z}{|z|} \quad \text{so} \quad |z\_1-z|=R\_1+r\_0-|z| \in (0, r\_0/2)\,.$$
In both cases, we have
$$B(z\_1, r\_{0})\cap \text{supp}~u(\cdot, s)=\emptyset\, ,$$
so for all $t\in [0,t\_0/2]$,
$$B(z\_1, r\_{0}/2)\cap \text{supp}~u(\cdot, s+t)=\emptyset\, .$$
In particular, $ z \notin \: \text{supp} \: u(\cdot,s+t)$.
$\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \Box$
The claim implies, by induction on $k$, that
$$\forall k \ge 1, \quad \forall t\in [(k-1)t\_0/2,\, kt\_0/2),
\quad \text{supp} \:u(\cdot,t) \subset \bar{B}(0, R+kr\_0/2) \tag{\*} \,. $$
as required.
| 5 | https://mathoverflow.net/users/7691 | 426959 | 173,233 |
https://mathoverflow.net/questions/426957 | 9 | Let $k$ be a field. By the Quillen-Suslin theorem, all vector bundles on $\mathbb{A}^n\_k$ are trivial for all $n \geq 0$. If $U \subset \mathbb{A}^n\_k$ is an affine open subset, then vector bundles on $U$ (of small rank) need not be trivial in general, but there do exist (non-trivial) such $U$, e.g., (split) algebraic tori, on which all vector bundles are indeed trivial.
>
> Are there simple conditions on $U$ as above which imply that all vector bundles on $U$ are trivial?
>
>
>
My interest is in conditions for arbitrary $n$; I would also be interested in conjectural answers or results for special $k$, e.g., $k$ algebraically closed or $k = \mathbb{C}$.
I am particularly interested in the following special case:
>
> Let $U$ be the complement of a finite union of (affine) hyperplanes in $\mathbb{A}^n\_k$. Is every vector bundle on $U$ trivial? If not, is there a large class of such $U$ for which it is known (or conjectured) that all vector bundles on $U$ are trivial?
>
>
>
| https://mathoverflow.net/users/519 | Triviality of vector bundles on affine open subsets of affine space | For your final question, the answer is that all vector bundles over $U$ are trivial.
Sketch of proof: Let $R=k[x\_1,\ldots,x\_n]$.
Let $L\_1, L\_2,\ldots, L\_m$ be the equations of hyperplanes in $R$. Without loss of generality, we can assume $L\_1=x\_1$.
Set $A=R[L\_1^{-1},\ldots, L\_m^{-1}]$ and set $B=R[L\_2^{-1},\ldots,L\_m^{-1}]$. We want to show that all projective modules over $A$ are trivial. Of course if $m=0$, we are done by Quillen-Suslin. Otherwise, we can assume by induction that all projective modules over $B$ are trivial.
Now consider this diagram:
$$\matrix{B&\rightarrow&A\cr
\downarrow&&\downarrow\cr
B\otimes\_{k[x\_1]}k[x\_1]\_{(x\_1)}&\rightarrow &
B\otimes\_{k[x\_1]}k(x\_1)\cr
}$$
Check that this allows you to construct projective modules on $B$ by patching.
Now by induction all projective modules over the lower right corner are trivial. So any projective module over $A$ can be patched to a trivial module, and therefore lifts to a projective module over $B$, where we've already agreed that the statement is true.
I believe this argument is essentially due to Ofer Gabber.
| 11 | https://mathoverflow.net/users/10503 | 426961 | 173,234 |
https://mathoverflow.net/questions/426954 | 3 | In symbolic method, one often considers two operators on ordinary generating functions, namely
$$
\operatorname{PSET}F(x) = \exp\left(F(x)-\frac{F(x^2)}{2}+\frac{F(x^3)}{3}-\dots\right),
$$
and
$$
\operatorname{MSET}F(x) = \exp\left(F(x)+\frac{F(x^2)}{2}+\frac{F(x^3)}{3}+\dots\right).
$$
These operators allowed to enumerate sets (for $\operatorname{PSET}$) or multisets (for $\operatorname{MSET}$) constructed of *unlabeled* $F$-structures.
One may note that $\operatorname{MSET} F(x)$ is, essentially, unlabeled generating functions for the species $E \circ F$, where $E$ is the species of sets and $\circ$ denotes the composition of species. Correspondingly, the relevant exponential generating function would be $e^{F(x)}$.
Is there any meaningful way to interpret $\operatorname{PSET}$ in the terms of labeled species as well? Ideally, to get some formula for the exponential generating function of the labeled version of $\operatorname{PSET}$.
| https://mathoverflow.net/users/116776 | Representing PSET as species | See Gilbert Labelle, *[On asymmetric structures](https://doi.org/10.1016/0012-365X(92)90371-L)*, Discrete Math. 99 (1992), 141–164.
| 5 | https://mathoverflow.net/users/10744 | 426963 | 173,235 |
https://mathoverflow.net/questions/425847 | 3 | Let $A$ and $B$ be two (non commuting) self-adjoint bounded operator acting on a Hilbert space and let $p,q>1$ such that $\frac1p+\frac1q=1$
Do we have a Young-type inequality such as $ \frac12|AB+BA| \leq \frac{|A|^p}{p}+ \frac{|B|^q}{q}$ in the sense of quadratic form ? It is of course obvious for $p=q=2$.
| https://mathoverflow.net/users/107004 | Young-type inequality for bounded operator | As a general criterion, these kind of inequalities hold at the level of singular values (which implies a version of the inequality where one of the sides is conjugated by a unitary/partial isometry), but they tend to fail at the plain operator level.
Here is a counterexample for $p=3$, $q=3/2$. Take
$$
A=\begin{bmatrix}1&0\\0&3\end{bmatrix},\qquad\qquad B=\begin{bmatrix}1&1\\1&1\end{bmatrix}.
$$
Then
$$
\frac{A^3}3+\frac{2B^{3/2}}3=\frac13\,\Big(\begin{bmatrix}1&0\\0&27\end{bmatrix}
+\begin{bmatrix}2\sqrt2&2\sqrt2\\2\sqrt2&2\sqrt2\end{bmatrix}\Big) =\frac13\,\begin{bmatrix}1+2\sqrt2&2\sqrt2\\2\sqrt2&27+2\sqrt2\end{bmatrix}.
$$
We have
$$
AB+BA=\begin{bmatrix}2&21\\21&40\end{bmatrix},
$$
so
$$
(AB+BA)^2=\begin{bmatrix} 20&32\\32&52\end{bmatrix}
$$
and then [Wolfram Alpha](https://www.wolframalpha.com/input?i=simplify+MatrixPower%5B%7B%7B20%2C32%7D%2C%7B32%2C52%7D%7D%2C1%2F2%5D) tells us that
$$
|AB+BA|=\begin{bmatrix} 20&32\\32&52\end{bmatrix}^{1/2}=\frac2{\sqrt5}\,\begin{bmatrix}3& 4\\ 4& 7
\end{bmatrix}.
$$
Then, [using Wolfram Alpha](https://www.wolframalpha.com/input?i=%281%2F3%29%7B%7B1%2C0%7D%2C%7B0%2C3%5E3%7D%7D%2B%282sqrt%282%29%2F3%29%7B%7B1%2C1%7D%2C%7B1%2C1%7D%7D-%281%2F2%29%7B%7B6%2Fsqrt%285%29%2C+8%2Fsqrt%285%29%7D%2C+%7B8%2Fsqrt%285%29%2C+14%2Fsqrt%285%29%7D%7D) again, we see that
$$
\frac{A^3}3+\frac{2B^{3/2}}3-\frac12\,|AB+BA|
$$
is not positive, as it has a negative eigenvalue.
| 5 | https://mathoverflow.net/users/3698 | 426964 | 173,236 |
https://mathoverflow.net/questions/426942 | 3 | Say we have a power series of two variables, with an associated function $f$ defined as
$$
\begin{split}
f(x, y) =\, & \sum\_{n,m} a\_{n,m}x^ny^m,\\
& a\_{n,m} \geq 0 \quad \forall n, m \in\mathbb{N},
\end{split}
$$ which is known to converge in each point of a compact set $\mathcal{C} \subset \mathbb{R\_+}^2$. We can also assume $f$ to be bounded on $\mathcal{C}$.
Can $f$ have a point of discontinuity (or more) on $\mathcal{C}$?
Edit : As shown by [Tom Goodwillie with this nice counter-example](https://mathoverflow.net/a/426980/102634), it turns out to be possible for $f$ to be discontinuous. Thanks!
Edit : clarified question as per the inputs from Tom Goodwillie, David E Speyer and Daniele Tampieri.
| https://mathoverflow.net/users/102634 | Can a power series of several variables be discontinuous on a compact set if it converges in every point of this set? | The series $f(x,y)=y+xy+x^2y+x^3y+\dots$ converges to $0$ when $y=0$, and converges to $y/(1-x)$ when $|x|<1$. This function is not continuous at $(x,y)=(1,0)$.
| 7 | https://mathoverflow.net/users/6666 | 426980 | 173,240 |
https://mathoverflow.net/questions/426972 | 6 | Let $X$ be a topological space and $\mathscr{F}$ a sheaf on $X$. In the paper [Tropical cycle classes for non-archimedean spaces and weight decomposition of de Rham cohomology sheaves](https://doi.org/10.24033/asens.2423) by Yifeng Liu, the notation $\underline{H}^p(X, \mathscr{F})$ is used. (More generally, $\mathscr{F}$ could be an object of the derived category of sheaves on $X$, e.g. some complex of sheaves.) For example we have the expression $H^0(X, \underline{H}^p(X, i\_{Z!} i\_Z^! \mathscr{K}\_X^p))$ appearing right before Lemma 2.4 or $H^0(U, \underline{H}^p(U, i\_{Z!}i\_Z^! \mathscr{D}\_U^{p,\bullet}))$ in the proof of Theorem 3.9.
It seems to be some kind of sheaf, but I cannot find its definition anywhere. Is it the sheaf associated to the presheaf $U \mapsto H^p(U, \mathscr{F})$? Where can I read some basic properties about this construction? I searched in the Stacks Project, in Hartshorne, Lei Fu's book on etale cohomology, Gelfand-Manin's homological algebra, but I couldn't find it.
| https://mathoverflow.net/users/112369 | Where can I find a definition of $\underline{H}^p(X, \mathscr{F})$? | As indicated in the comments, the notation $\def\HH{\underline{\rm H}}\def\H{{\rm H}}\HH^p(X,F)$ is defined (for example) by Milne in *Étale Cohomology* as the $p$th right derived functor of the inclusion functor from the category of sheaves of abelian groups to the category of presheaves of abelian groups on the same site.
This is also known as *hypercohomology* and is studied under this name by Grothendieck, although I could not locate this notation in his paper.
In modern terms, we can compute $\HH^p(X,F)$ by replacing $F$ with a locally quasi-isomorphic presheaf $G$ of chain complexes that satisfies homotopy descent, e.g., by fibrantly replacing in the local injective model structure on presheaves or sheaves of chain complexes, or the local projective model structure on presheaves of chain complexes.
Then $\HH^p(X,F)$ is simply the $p$th cohomology group of $G$ computed in the abelian category of presheaves of abelian groups.
The keyword to look for in older sources is [hypercohomology](https://ncatlab.org/nlab/show/hypercohomology), the newer sources may simply be talking about the derived category of sheaves, and even newer sources might say something like the “cohomology presheaf of an ∞-sheaf of chain complexes”.
| 6 | https://mathoverflow.net/users/402 | 426989 | 173,243 |
https://mathoverflow.net/questions/426984 | 2 | An integer partition is a sequence $\lambda=(\lambda\_1\geq\lambda\_2\geq\dotsb\geq\lambda\_k)$ of positive integers, for some $k\geq1$. Consider the following two sets of partitions of $n$. Fix a positive integer $s>1$.
Let $\mathcal{O}\_{n,s}=\{\lambda\vdash n: \lambda\_i\in\{1,3,5,\dots,2s-1\}\}$. Parts are odd integers.
Also, let
$\mathcal{A}\_{n,s}=\{a\_1+2a\_2+2a\_3+\cdots+2a\_s=n: a\_1\geq a\_2\geq a\_3\geq\cdots\geq a\_s\geq0\}$.
I would like ask:
>
> **QUESTION.** Is there a combinatorial or bijective proof of the equinumerosity $\#\mathcal{O}\_{n,s}=\#\mathcal{A}\_{n,s}$?
>
>
>
**Postscript.** Fix $\ell>1$. Define
$\mathcal{L}\_{n,s}=\{\lambda\vdash n: \lambda\_i\equiv 1\mod \ell, \,\,i=1,\dots,s\}$ and
$\widetilde{\mathcal{L}}\_{n,s}=\{a\_1+\ell a\_2+\ell a\_3+\cdots+\ell a\_s=n: a\_1\geq a\_2\geq a\_3\geq\cdots\geq a\_s\geq0\}$.
Then, Per Alexandersson's solution below appears to prove that $\#\mathcal{L}\_{n,s}=\#\widetilde{\mathcal{L}}\_{n,s}$.
| https://mathoverflow.net/users/66131 | Seeking a bijective proof enumerating two partition sets: Part I | Take a partition (Young diagram) in $O\_{n,s}$.
Number of elements in first column is $a\_1$.
Removing the first column now gives a partition with only even parts.
Divide all parts by $2$, and take conjugate. This gives $a\_2 \geq a\_3 \geq \dotsc \geq a\_s$, and you now have an element in the second set.
*It is a nice problem, I will most likely steal it!*
| 4 | https://mathoverflow.net/users/1056 | 426990 | 173,244 |
https://mathoverflow.net/questions/426956 | 8 | Since [this question](https://mathoverflow.net/questions/41310/) is on the front page again, a generalization.
>
> Let $p$ be prime, and let $a$ and $b$ be positive integers with $a+b=p-1$. Is it possible to have two loaded dice, one with sides labeled $\{ 0,1,\ldots, a \}$ and the other with sides labeled $\{ 0,1,\ldots, b \}$, such that, when we roll the two dice, each of the totals $\{ 0,1,\ldots, p-1 \}$ occurs with equal probability?
>
>
>
I require that $p$ is prime because, if $p=uv$, we can take $(a,b) = (u-1, u(v-1))$ so that the first die produces the rolls $\{ 0,1,2,\ldots, u-1 \}$ each with equal probability, and the second die produces the rolls $\{ 0,u,2u, \ldots, u(v-1) \}$ each with equal probability, and has probability zero of landing on any other face.
Looking at the answers to the previous question, Jim Propp and Michael Lugo's solutions rule out $a=b=(p-1)/2$, and Michael Lugo's other solution rules out solutions with $a$ and $b$ odd. But I don't see any of the solutions that can be adapted to the general case.
Taking a generating function approach, we want to factor $x^{p-1} + x^{p-2} + \cdots + x+ 1$ into $\alpha(x) \beta(x)$ with each of $\alpha$ and $\beta$ of positive degree and having nonnegative real coefficients. For $p \leq 31$, I have tested the following claim in Mathematica:
>
> If $x^{p-1} + x^{p-2} + \cdots + x+ 1 = \alpha(x) \beta(x)$ with $\alpha$ and $\beta$ both of positive degree with real coefficients, and if $\alpha$ is the factor with $\alpha(e^{(2 \pi i)/p})=0$, then $\alpha$ has a negative coefficient.
>
>
>
Caveat: I was using floating point arithmetic without interval bounds, so this might be the result of numerical error. Jim Propp's argument proves this when $\deg \alpha \leq (p-1)/2$, but that's only half the cases.
I don't see why this should be true, does anyone?
| https://mathoverflow.net/users/297 | Two dice yielding uniform distribution, part 2 | Suppose that you can factor $x^{p-1} + x^{p-2} + \cdots + 1$ as $f(x)g(x)$ with $\deg(f)=a,\deg(g) = b$ and $f,g$ having nonnegative coefficients.
Lemma: It must be the case that $f = \sum\_{i=0}^a c\_i x^i$ and $g = \sum\_{i=0}^b d\_i x^i$, where $c\_i, d\_j > 0$ for all $0 \le i \le a$ and $0 \le j \le b$.
Proof: Suppose that some $c\_i$ is zero; for concreteness say $c\_a = 0$. Now, let's fix the set of $a$ distinct $p$th roots of unity that $f$ vanishes on, say $\omega^{r\_1}, \ldots, \omega^{r\_a}$ with $\omega$ a primitive root. The set of polynomials satisfying these conditions is the subspace $(c\_0, \ldots, c\_{a-1})$ such that $M.(c\_0, \ldots, c\_{a-1}) = (0,\ldots, 0)$, and $M$ is the $a \times a$ matrix whose $i,j$th entry is $\omega^{r\_i \cdot j}$. But this is a submatrix of the DFT matrix for $\mathbb{Z}\_p$, and hence it is invertible; see Lemma 1.3 of <https://arxiv.org/abs/math/0308286> for a reference. Hence $f$ must be identically zero, a contradiction. Note this argument works when we omit any monomial from the support of $f$, not just $x^a$.
Now, note that by rescaling $f$ and $g$ we may assume that $c\_0 = 1$, and therefore also $d\_0 = 1$. Now consider $c\_ad\_b$, the coefficient of $x^{p-1}$ in the product. Note that $c\_a$ must be strictly less than 1, since the coefficient of $x^a$ in the product equals $c\_a \cdot 1$ plus some positive stuff (by the lemma), and this should equal 1. Similarly, $d\_b < 1$. So $c\_ad\_b < 1$, a contradiction.
| 1 | https://mathoverflow.net/users/141277 | 426994 | 173,245 |
https://mathoverflow.net/questions/427000 | 1 | In [the proof that Martin's maximum implies (\*)](https://ivv5hpp.uni-muenster.de/u/rds/MM_implies_star_final_version_March_18_2021.pdf), the introduction gives the following theorem:
>
> Assume there is a proper class of Woodin cardinals. Then, the following are equivalent:
>
>
> * $(^\*)$
> * For any $\Pi\_2$ $\mathcal{L}$-sentence $\sigma$, where $\mathcal{L}$ is the language of the structure $(H\_{\omega\_2}; \in, \mathrm{NS}\_{\omega\_1}, \mathcal{A})\_{\mathcal{A} \in \mathcal{P}(\mathbb{R}) \cap L(\mathbb{R})}$, if $\sigma$ is $\Omega$-consistent, then $\sigma$ is true.
>
>
>
Now, let's define the axiom $(^\*)^+$ to be the conjunction of the following two statements:
* There exists a proper class of Woodin cardinals.
* For any $\mathcal{L}$-sentence $\sigma$, where $\mathcal{L}$ is the language of the structure $(H\_{\omega\_2}; \in, \mathrm{NS}\_{\omega\_1}, \mathcal{A})\_{\mathcal{A} \in \mathcal{P}(\mathbb{R}) \cap L(\mathbb{R})}$, if $\sigma$ is $\Omega$-consistent, then $\sigma$ is true.
Namely, the restriction to $\Pi\_2$-sentences (statements of the form $\forall x \exists y \varphi(x,y)$) is removed. Now, here's my question.
>
> Is $(^\*)^+$ inconsistent? If it is consistent (assuming the consistency of certain large cardinal assumptions), then is it equiconsistent with the regular axiom $(^\*)^+$, or strictly stronger?
>
>
>
| https://mathoverflow.net/users/473200 | An extension of Woodin's star axiom | This principle is inconsistent, even if we just look at $(H\_{\omega\_2};\in)$. This is because - for example - the continuum hypothesis is expressible as a sentence in this structure ("There is an $\omega\_1$-sequence of reals such that every real appears in the sequence"), but both $\mathsf{CH}$ and $\neg\mathsf{CH}$ are $\Omega$-consistent.
Indeed, the example of $\mathsf{CH}$ shows that $\Pi\_2$ is basically the best we can hope for: already at $\Pi\_2\vee\Sigma\_2$ we get inconsistency.
| 3 | https://mathoverflow.net/users/8133 | 427003 | 173,247 |
https://mathoverflow.net/questions/426998 | 3 | Let $T$ be a ring with involution $s:T\rightarrow T$. And let
$$h:T\otimes T^\text{op} \rightarrow T\otimes T^\text{op}$$ be the ring automorphism given by $h(a\otimes b)=s(b)\otimes s(a)$.
I was wondering if the induced homorphism
$$ K\_{\ast}(h): K\_{\ast}(T\otimes T^\text{op})\rightarrow K\_{\ast}(T\otimes T^\text{op}) $$ in K-theory, is the identity map?
Notice that "op" is used for the opposite multiplication.
| https://mathoverflow.net/users/165456 | Involution map, and induced morphism in K-theory | No. Let $R$ be commutative, $T=R\times R$, $s(x,y)=(y,x)$. Then $T\otimes T^{op}=T\otimes T$ is the product of four copies of $R\otimes R$, so that $K(T\otimes T^{op})$ is the product of four copies of $K(R\otimes R)$ and $h$ permutes the copies in some nontrivial way.
| 6 | https://mathoverflow.net/users/6666 | 427006 | 173,250 |
https://mathoverflow.net/questions/426861 | 1 | Let $k$ be an integer with $k>1$ and let $$a\_k=\frac{2 k(k-1)}{\Gamma[k-1]} \int\_0^\infty \frac{e^{-r^2} r^{2k-3}}{2+4 r^2+r^4} dr.$$
How to prove that $a\_k<1$?
| https://mathoverflow.net/users/486023 | A bounded sequence $a_k=\frac{2 k(k-1)}{\Gamma[k-1]} \int_0^\infty \frac{e^{-r^2} r^{2k-3}}{2+4 r^2+r^4} dr$ | $\newcommand\Ga\Gamma$We have
$$a\_k=\frac{2 k(k-1)}{\Ga(k-1)} J\_k,\quad J\_k:=\int\_0^\infty e^{-r^2} r^{2k-3}g(r)\,dr,$$
$$g(r):=\frac1{2+4 r^2+r^4}.$$
Next (for $r>0$),
$$g(r)=h(r)-\frac{8 \left(70 r^2+41\right)}{r^{12} \left(r^4+4 r^2+2\right)}<h(r),$$
where
$$h(r):=\frac{1}{r^4}-\frac{4}{r^6}+\frac{14}{r^8}-\frac{48}{r^{10}}+\frac{164}{r^{12}}.$$
So, for $k\ge8$
$$a\_k<\frac{2 k(k-1)}{\Ga(k-1)}\,\int\_0^\infty e^{-r^2} r^{2k-3}h(r)\,dr
=\frac{(k-1) k (k (k ((k-26) k+265)-1296)+2768)}{(k-7) (k-6) (k-5) (k-4) (k-3) (k-2)}<1$$
if $k\ge14$.
---
It remains to check the inequality $a\_k<1$ for $k=2,\dots,13$. This is straightforward to do. For instance,
$$a\_2=\frac{e^{2-\sqrt{2}} \left(e^{2 \sqrt{2}}
\text{Ei}\left(-2-\sqrt{2}\right)-\text{Ei}\left(-2+\sqrt{2}\right)\right)}{\sqrt{2}}
=0.427\ldots<1.$$
Mathematica can compute values of the built-in special function $\text{Ei}$ with any degree of accuracy.
More generally, one can note that
$$K\_2(t):=\int\_0^\infty e^{-t\,r^2} r^{2\times2-3}g(r)\,dr
=-\frac{e^{-\left(\left(\sqrt{2}-2\right) t\right)} \left(\text{Ei}\left(\left(-2+\sqrt{2}\right) t\right)-e^{2 \sqrt{2} t}
\text{Ei}\left(-\left(\left(2+\sqrt{2}\right) t\right)\right)\right)}{4 \sqrt{2}},$$
which follows by the partial fraction decomposition
$$\frac{2 \sqrt{2}}{2+4 u+u^2}=-\frac{1}{u+\sqrt{2}+2}-\frac{1}{-u+\sqrt{2}-2}.$$
Then one can use the formulas
$$J\_k=(-1)^{k-2}K\_2^{(k-2)}(1)$$
for $k\ge2$ and $\text{Ei}'(u)=e^u/u$ to find $J\_k$ in closed form for $k=2,\dots,13$, to get
$$(a\_2,\dots,a\_{13})=(0.42\ldots,0.66\ldots, 0.78\ldots, 0.85\ldots, 0.89\ldots, 0.91\ldots, \\
0.93\ldots, 0.95\ldots,
0.95\ldots, 0.96\ldots, 0.97\ldots, 0.97\ldots).$$
In particular,
$$a\_{13}=\frac{13}{1663200}\\
\times\left(-8 \left(8119+5741 \sqrt{2}\right) e^{2+\sqrt{2}} \text{Ei}\left(-2-\sqrt{2}\right)+8 \left(5741 \sqrt{2}-8119\right) e^{2-\sqrt{2}}
\text{Ei}\left(-2+\sqrt{2}\right)+94224\right)
=0.97\ldots.$$
| 3 | https://mathoverflow.net/users/36721 | 427010 | 173,251 |
https://mathoverflow.net/questions/426969 | 6 | Consider simple random walks that stop when reaching a given node $x$ in an undirected, unweighted and connected graph on $n$ nodes.
Let
* $H(i,x)$ denote the (expected) hitting time from $i$ to $x$, with $H(x,x)=0$.
* $H\_{\max} = \max\_{i \in V} H(i,x)$ and $H\_{\text{avg}} = \frac{1}{n}\sum\_{i \in V} H(i,x)$.
I am interested in the (asymptotic) largest possible ratio of $\frac{H\_{\max}}{H\_{\text{avg}}}$ in function of $n$.
Similar to [this](https://cs.stackexchange.com/questions/42917/average-vs-worst-case-hitting-time) question, but now considering the single-source variant w.r.t a specified node.
The worst example I was able to construct has ratio $\Omega(n^{3/4})$: Consider an $n$-star graph with center node $x$, attach to $x$ a lollipop graph of $2n^{1/4}$ vertices (i.e., a path of length $n^{1/4}$, attached with a clique of size $n^{1/4}$). So $H\_{\max} \approx n^{3/4}$, and $$H\_{\text{avg}} \approx \frac{n\cdot 1 +n^{1/4}\cdot n^{3/4}}{n} \in O(1),$$
since the average hitting time (over all the lollipop vertices) to $x$ is cubic in the size of the lollipop graph.
Are there any worse examples where $\frac{H\_{\max}}{H\_{\text{avg}}}$ is larger, perhaps even $\Omega(n)$?
| https://mathoverflow.net/users/486114 | Average and max. hitting time to a specific vertex | **Notation:** Let $G=(V,E)$ be an undirected simple graph of $n$ nodes. If $\tau\_x$ is the (random) time it takes the walk to reach the node $x$,
then write $H(v,x)=E\_v(\tau\_x)$. Denote $H\_{\max}(x):=\max\_{u \in V} H(u,x)$ and
$ H\_{\rm avg}(x) =\frac1n\sum\_{v \in V} H(v,x)$.
**Claim:** $ H\_{\max}(x) \le 2n^{3/4} H\_{\rm avg}(x)$ for all $x$.
Proof:
Given $M>0$, define
$$S=\{v \in V: H(v,x) \le M H\_{\rm avg}(x) \},$$
where clearly
$$n H\_{\rm avg}(x) \ge \sum\_{v \in S^c} H(v,x) \ge |S^c|\cdot M H\_{\rm avg}(x) \,,$$
so $|S^c| \le n/M.$
Given a starting node $u \in V$, denote by $\tau\_{S}$ the first time the random walk from $u$ visits $S$, and let $H(u,S):=E\_u[\tau\_{S}]$ be its expectation (which is $0$ if $u \in S$, the interesting case will be $u \in S^c$.)
Consider the graph $G^\*$ on $|S^c|+1$ nodes, obtained from $G$ by contracting all nodes in $S$ to a single supernode $s$. To ensure $G^\*$ is a simple graph, if $w\in S^c$ has several edges connecting it to $S$, we only keep one of these edges in $G^\*$ connecting $w$ to $s$. Then for every $w \in S^c$ we have $P\_G(w,S) \ge P\_{G^\*}(w,s)$. Thus we can couple the walks in $G$ and $G^\*$ so that
$H(u,S)$ in $G$ is at most the expected hitting time $H^\*(u,s)$ in $G^\*$ from $u$ to $s$, which is well known to be at most
$ |S^c|^3$. See, e.g., [1] for the slightly weaker bound $(|S^c|+1)^3$. In fact, one can multiply this upper bound by $4/27+o(1)$, see [2], but we will not worry about optimizing the constants. A related stronger result on exploration time is in [3].
Then the strong Markov property at time $\tau\_S$ implies that
$$H(u,x) \le H(u,S)+\max\_{v \in S} H(v,x) \le (n/M)^3+M \cdot H\_{\rm avg}(x) $$ $$\le
\Bigl((n/M)^3+M\Bigr) \cdot H\_{\rm avg}(x) \,. \tag{1}$$
(We may assume that $ H\_{\rm avg}(x) \ge 1$, otherwise $G$ is a star.)
To minimize the right hand side of $(1)$, without optimizing the constants, choose
$M=n^{3/4}$. We conclude that
$ \max\_{u \in V} H(u,x) \le 2n^{3/4} \cdot H\_{\rm avg}(x) \,. $$
$$ \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \Box$
[1] <https://www.yuval-peres-books.com/markov-chains-and-mixing-times/>
<https://pages.uoregon.edu/dlevin/MARKOV/mcmt2e.pdf> Proposition 10.16 page 134.
[2] Brightwell, Graham, and Peter Winkler. "Maximum hitting time for random walks on graphs." Random Structures & Algorithms 1, no. 3 (1990): 263-276.
[3] Barnes, Greg, and Uriel Feige. "Short random walks on graphs." In Proceedings of the Twenty-Fifth annual ACM symposium on Theory of computing, pp. 728-737. 1993. See Theorem 1.
| 9 | https://mathoverflow.net/users/7691 | 427013 | 173,254 |
https://mathoverflow.net/questions/427007 | 1 | I am currently reading "Smoothing of Multivariate Data" by Klemela. It contains Lemma 11.6, which upper and lower bounds the KL-divergence of two densities in terms of the $L\_{2}$-metric. The Lemma specifically states the following 2 bounds ***without*** proof:
**Upper bound**
>
> Let $f, f\_{0}$ be densities. In particular, if $\inf \_{x \in \mathbf{R}^{d}} f\_{0}(x)>0$, then
> $$
> D\_{K}^{2}\left(f, f\_{0}\right) \leq \frac{\left\|f-f\_{0}\right\|\_{2}^{2}}{\inf \_{x \in \mathbf{R}^{d} f\_{0}(x)}}
> $$
>
>
>
**Lower bound**
>
> Also, if $f$ and $f\_{0}$ are both bounded and bounded away from zero, then
> $$
> D\_{K}^{2}\left(f, f\_{0}\right) \geq \int\_{\left\{x \in \mathbf{R}^{d}: f\_{0}(x)>0\right\}}\left(f-f\_{0}\right)+C\left\|f-f\_{0}\right\|\_{2}^{2}
> $$
> for a positive constant $C$.
>
>
>
I can prove the upper bound rigorously as follows:
\begin{align}
D\_{K}^{2}(f, f\_{0})
&\leq \chi^{2}(f || f\_{0})
\tag{well known upper bound} \\
&=: \int\_{\left\{x \in \mathbf{R}^{d}: f\_{0}(x)>0\right\}} \frac{\left(f-f\_{0}\right)^{2}}{f\_{0}}
\tag{by definition} \\
&\leq \frac{1}{\inf \_{x \in \mathbf{R}^{d} f\_{0}(x)}}
\int\_{\left\{x \in \mathbf{R}^{d}: f\_{0}(x)>0\right\}}
\left(f-f\_{0}\right)^{2}
\tag{since $\inf$ exists.} \\
&=: \frac{\left\|f-f\_{0}\right\|\_{2}^{2}}{\inf \_{x \in \mathbf{R}^{d} f\_{0}(x)}}
\end{align}
Note here that the **definition** of the KL divergence used in the Lemma is as follows:
>
> If $f$ and $g$ are densities of $P$ and $Q$ with respect to the Lebesgue measure, then we may write
> $$
> D\_{K}^{2}(f, g)=\int\_{\mathbf{R}^{d} \cap\{x: g(x)>0\}} f \log \_{e}\left(\frac{f}{g}\right)
> $$
>
>
>
I'm unsure on how to prove the lower bound. Some comments (for the lower bound to hold):
* It appears that the densities here must also be bounded above (else see first counterexample [here](https://mathoverflow.net/a/393477/486152)).
* Since the densities are assumed to be bounded above and positively bounded from below over their common support, the support set in $\mathbb{R}^{d}$ must also be bounded (else see second counterexample [here](https://mathoverflow.net/a/393477/486152)).
Could anyone please show how to prove this **lower bound** rigorously (or provide a citable rigorous proof reference)? Also, if I've made a mistake in my two comments above, please also let me know.
**Aside:** I had originally posted this on math.SE. To respect math.overflow [cross-posting etiquette](https://meta.mathoverflow.net/a/1726/486152), I've deleted that post due to no responses. I realized that it had also been asked [there before](https://math.stackexchange.com/questions/2635261/kullback-leibler-divergence-is-equivalent-to-l2-norm-in-some-cases) without any suitable responses. Since it is a research related question with no well citable proof, I believe it is fair to post here on math.overflow to settle the issue.
| https://mathoverflow.net/users/486152 | Lower bound for KL divergence of bounded densities and $L_{2}$ metric | $\newcommand{\ep}{\varepsilon}$As in your post and comments, suppose that $f$ and $f\_0$ are supported on a compact set $S$, and
\begin{equation\*}
a\le f\le b,\quad a\le f\_0\le b
\end{equation\*}
on $S$ for some real $a,b$ such that $0<a<b$.
Then
\begin{equation\*}
\int\_{\left\{x \in \mathbf{R}^{d}: f\_{0}(x)>0\right\}}\left(f-f\_{0}\right)
=\int\_S\left(f-f\_{0}\right)=0.
\end{equation\*}
So, the inequality in question is simply is
\begin{equation\*}
D\_{K}^{2}\left(f, f\_{0}\right) \ge C\left\|f-f\_{0}\right\|\_{2}^{2}. \tag{1}\label{1}
\end{equation\*}
By definition,
\begin{equation\*}
D\_K^2(f,f\_0)=\int\_S f\ln\frac f{f\_0}=-\int\_S f\ln\frac{f\_0}f.
\end{equation\*}
We have the elementary inequality
\begin{equation\*}
\ln x\le(x-1)-\ep\_M(x-1)^2 \tag{2}\label{2}
\end{equation\*}
for any real $M>1$ and all $x\in(0,M]$, where
\begin{equation\*}
\ep\_M:=\frac{M-1-\ln M}{(M-1)^2}>0.
\end{equation\*}
Note that $0<\frac{f\_0}f\le\frac ba$ on $S$. So, using \eqref{2} with $x=\frac{f\_0}f$ and $M=\frac ba$, we get
\begin{equation\*}
\begin{aligned}
D\_K^2(f,f\_0)&\ge-\int\_S f\Big(\frac{f\_0}f-1-\ep\_{b/a}\Big(\frac{f\_0}f-1\Big)^2\Big) \\
& =\ep\_{b/a}\int\_S f\Big(\frac{f\_0}f-1\Big)^2
=\ep\_{b/a}\int\_S \frac{(f\_0-f)^2}f \\
&\ge\frac{\ep\_{b/a}}b\,\int\_S (f\_0-f)^2
=C\_{a,b}\|f-f\_0\|\_2^2,
\end{aligned}
\end{equation\*}
where
\begin{equation\*}
C\_{a,b}:=\frac{\ep\_{b/a}}b>0,
\end{equation\*}
as desired.
| 1 | https://mathoverflow.net/users/36721 | 427017 | 173,255 |
https://mathoverflow.net/questions/427025 | 1 | I have a problem in understanding the concept of trianguline representation. Maybe someone can enlighten me.
Let $K$ be a finite extension of $\mathbb{Q}\_p$ and $V$ be a $p$-adic representation of $G\_K$ (absolute Galois group of $K$) of dimension $n$. By $p$-adic Hodge theory $V$ corresponds to some étale $(\varphi,\Gamma)$-module $M$ over the Robba ring $\mathcal{R}\_K$. Lets call $M$ *trianguline* if there exists a filtration $$ 0=M\_0 \subset M\_1 \subset \cdots \subset M\_n=M,$$ of sub-$(\varphi,\Gamma)$-modules of $M$ such that the successive quotients $M\_i/M\_{i-1}$ are of rank $1$. Since rank $1$ $(\varphi,\Gamma)$-moduels over $\mathcal{R}\_K$ are of the form $\mathcal{R}\_K(\delta)$ for some unique continuous character $\delta: \mathbb{Q}\_p^{\times} \rightarrow K^{\times}$ we can describe a triangulation by a sequence of such characters $(\delta\_i)\_{1 \leq i \leq n}$.
So lets us say that $M$ is trianguline, is there a *unique* triangulation? What confuses me is that in such a case the generalized Hodge-Tate weights of $V$ correspond to the weights of the $\delta\_i$ defined as $$ w(\delta\_i):= \log\_p \delta\_i(u) / \log\_p u $$ for some $u \in 1 +p \mathbb{Z}\_p$ doesn't this make the triangulation unique? Where is my mistake here...
| https://mathoverflow.net/users/118220 | Trianguline representation | No, triangulations are not in general unique.
A simple way of seeing this is to consider the case when $K = \mathbf{Q}\_p$, $V$ is 2-dimensional and crystalline with distinct Hodge–Tate weights, say $\{0, -r\}$ and the eigenvalues of $\varphi$ on $D\_{cris}(V)$ are distinct, say $\alpha$ and $\beta$. Weak admissibility forces $v\_p(\alpha)$ and $v\_p(\beta)$ to be in the interval $[0, r]$ and sum to $r$. I'm going to suppose $\alpha \ne \beta$ and both $v\_p(\alpha)$ and $v\_p(\beta)$ are strictly positive (so $V$ is irreducible).
If you want to write down a triangulation of $V$, then the character $\delta\_1$ has to be unramified, and $\delta\_2$ has to restrict to $x \mapsto x^r$ on $\mathbf{Z}\_p^\times$ (or maybe $x^{-r}$, I forget.) Moreover, $\delta\_1(p)$ and $\delta\_2(p)$ have to be $\alpha$ and $\beta$ **in some order**. However, we are free to choose which order, and this gives 2 distinct triangulations.
(This is exactly what the Galois reps of non-ordinary modular forms of prime-to-$p$ level and weight $\ge 2$ look like locally at $p$. The two triangulations correspond to the two $p$-stabilisations of $f$.)
Your question "Where is my mistake here..." is hard to answer, because I can't work out why you expect the assertion about HT weights to imply any uniqueness.
| 3 | https://mathoverflow.net/users/2481 | 427031 | 173,258 |
https://mathoverflow.net/questions/427023 | 0 | Consider the scalar conservation law
$$u\_t+f(u)\_x=0, \hspace{0.4 cm} \text{in $\hspace{0.2 cm}$ $\mathbb{R} \times (0,\infty)$}$$
where $f \in C^{2}(\mathbb{R})$ is a strictly convex function ($f''>c>0$).
The solution can be shown to satisfy so-called "Oleinik's entropy condition":
$$ \frac{u(x+a,t)-u(x,t)}{a} \leq \frac{c}{t} \hspace{0.7 cm} a>0,t>0.$$
**Question:** How does this condition imply $u(\cdot,t) \in BV\_{\mathrm{loc}}(\mathbb R)$?
| https://mathoverflow.net/users/nan | Oleinik inequality (one-sided Lipschitz condition) implies $BV_{\mathrm{loc}}$ for solution of conservation law | The function $f(x):=u(x,t)-cx/t$ decreases, thus $u(x,t)=f(x)+cx/t$ is a sum of two monotone functions, so belongs to local BV.
| 3 | https://mathoverflow.net/users/4312 | 427033 | 173,259 |
https://mathoverflow.net/questions/426974 | 6 | Suppose the dynamical system $(X,T)$ has only proper factors (i.e. not $(X,T)$ itself) of zero topological entropy. Does the system $(X,T)$ also have zero entropy?
| https://mathoverflow.net/users/45092 | Topological dynamical systems with only zero-entropy factors | This question is very related to the question of **lowering topological entropy**, introduced in ``Can one always lower topological entropy?'' by Shub and Weiss and then very nearly solved by Lindenstrauss in "Lowering topological entropy" and "Mean Dimension, Small Entropy Factors, and an Embedding Theorem." There's too much to state everything here, but here's a quick overview:
The question is: does every $(X,T)$ of positive entropy possess a nontrivial (i.e. not one point) factor $(Y,S)$ with $h(Y) < h(X)$?
1. Shub and Weiss showed that the answer is "yes" for subshifts and more generally, systems with the so-called **small boundary property**.
2. Lindenstrauss showed that the answer is "yes" for systems $(X,T)$ where $X$ is finite-dimensional. He then showed that the same is true when $(X,T)$ has zero **mean dimension** and a nontrivial minimal factor. Mean dimension is too long to define here, but in particular, $(X,T)$ has zero mean dimension if $(X,T)$ has finite entropy!
3. Germane to your question: Lindenstrauss also gives examples of $(X,T)$ (with infinite entropy and $X$ infinite-zero dimension) for which every nontrivial factor $(Y,S)$ factors onto the original system $(X,T)$! The easiest is $X = [0,1]^{\mathbb{Z}}$ and $T$ the shift $\sigma$, but he also constructs a minimal example.
To summarize: any finite-entropy system (with a nontrivial minimal factor) has a nontrivial factor of smaller entropy, thus not conjugate to the original. So for systems with a nontrivial minimal factor, the only possibilities for your condition are zero or infinite entropy.
For infinite entropy, we can't yet rule out your condition, since technically the example above doesn't satisfy your criteria (the nontrivial factors are not isomorphic to the original system). However, the factors are in a way "bigger" than the original system, which is perhaps even more surprising!
| 8 | https://mathoverflow.net/users/116357 | 427036 | 173,260 |
https://mathoverflow.net/questions/427026 | 1 | Let $W$ be a standard one dimensional Brownian motion, and $\mathcal F\_t$ its natural filtration. Consider the SDE
$$dX\_t = \mu\_X (t, \omega) \, dt + \sigma\_X (t, \omega) \, dW\_t$$
$$dY\_t = \mu\_Y (t, \omega) \, dt + \sigma\_Y (t, \omega) \, dW\_t$$
$$X\_0 = x\_0, Y\_0 = y\_0 \text{ a.s.}$$
where $\mu\_X, \mu\_Y, \sigma\_X, \sigma\_Y \geq 0$ are progressively measurable with respect to $\mathcal F\_t$, and $x\_0, y\_0$ are constants.
Assume the existence of a solution to the above two equations up to a determinstic time $T$.
**Question:** Suppose $\sigma\_X \neq \sigma\_Y$ on a subset of $\Omega \times [0, T]$ of positive measure. Then is it true that
$$\mathbb P(Y\_T > X\_T) > 0?$$
| https://mathoverflow.net/users/173490 | A comparison principle for SDE | Don't think so. Take both to satisfy something like $dZ = \sigma |Z| dW$, start one from -1, and one from 1. regardless of the exact parameters, one stays positive and one stays negative.
| 2 | https://mathoverflow.net/users/143907 | 427038 | 173,261 |
https://mathoverflow.net/questions/427035 | 15 | I feel that the following problem should have a clean and simple solution, but so far I couldn't find one.
>
> Suppose that $p$ is a prime, and that $A\subseteq\mathbb Z/p\mathbb Z$ is a set such that
>
>
> * for any $z\in\mathbb Z/p\mathbb Z$, either $z\notin A$ or $z+1\notin A$;
> * for any $z\in\mathbb Z/p\mathbb Z$, either $z\notin A$ or $2z\notin A$.
>
>
> What is the largest possible density of such a set?
>
>
>
It is easy to construct a set $A$ with these properties of density $1/3+o(1)$, like for instance $\{2z\colon z\in(p/3,2p/3)\pmod p\}$. A more elaborated construction: assuming $p\equiv\pm3\pmod 8$, consider the set of all those quadratic residues $r$ such that if $r'$ is the smallest quadratic non-residue exceeding $r$, then $r'-r$ is odd. Can $A$ have a density exceeding $1/3+\varepsilon$ for infinitely many primes $p$, with fixed $\varepsilon>0$?
I also wonder what the expander properties and the eigenvalues of the corresponding graph are ($x,y\in\mathbb Z/p\mathbb Z$ connected by an edge whenever $y=x+1$, $y=x-1$, $y=2x$, or $y=x/2$).
| https://mathoverflow.net/users/9924 | Sets with both additive and multiplicative gaps | I can give an upper bound of $2p/5$ and a lower bound of $(2/5-o(1))p$ for a conjecturally infinite set of $p$.
For the upper bound, the numbers $z, z+1, 2z+2, 2z+1, 2z$ form a pentagon in your graph so at most $2$ out of the $5$ may be in the set $A$. As $z$ runs over $\mathbb Z/p\mathbb Z$, each of $z, z+1, 2z+2, 2z+1, 2z$ run over $\mathbb Z/p\mathbb Z$, so adding up the number of members of $A$ in each of the $p$ pentagons we obtain $5 |A| \leq 2 p$, giving $|A| \leq 2p /5$.
For the lower bound, take $p=2^n-1$ to be a Mersenne prime. Then for $a \in \mathbb Z/p \mathbb Z$, the binary digits of $a/p$ after the "decimal" point are periodic with period $n$. The length of a string of zeroes in this decimal expansion is bounded, so there must be a longest length. Let $k\_a$ be the place coming just before the first string of zeroes with the longest length, e.g. if the longest length is $3$ and $a/p = .10100010001\dots $ then $k\_a=3$.
Let $c\_a$ be the rational number obtained by taking $2^{k\_a} a $, lifting from $\mathbb Z/p\mathbb Z$ to $\{0,\dots, p-1\}$, and then dividing by $2^{k\_a}$. Note that $c\_a$ is a rational number with denominator a power of $2$ and numerator a positive integer $<p$, and usually much smaller than $p$ since the string of digits is quite large.
Call $a$ good if $c\_{a+1} =c\_a+1$ and and $c\_{2a} = 2 c\_a$, and bad otherwise.
We can take $$A= \{ a \textrm{ good}, c\_a \equiv 1 \textrm{ or } 4 \bmod 5 \}$$ and then we will have $|A|= (2/5- o(1))p$ as long as the number of bad $a$ is $o(p)$ (since that itself implies most good $a$ are in a large interval of good $a$).
We have either $a\notin A$ or $a+1\notin A$ because if $a\in A$ then $a$ is good and $c\_a$ is congruent to $1$ or $4$ mod $5$ so $c\_{a+1} = c\_a + 1$ is congruent to $2$ or $0$ mod $5$, and thus $a+1\notin A$ (regardless of whether $a+1$ is good). For $2a$, it's the same argument, except congruent to $2$ or $3$ mod $5$.
So it suffices to, for each of the finitely many reasons that $a$ can be bad, show that the number of $a$ which are bad for that reason is $o(p)$.
This can be done as follows. Suppose $c\_{a+1} \neq c\_a$. The binary expansion of $c\_{a+1}$ is obtained from $c\_a$ by adding a $1$ in the $n-1$st place (and $2n-1$, and $3n-1, \dots$) and carrying. The only way that this can affect the first longest string of zeroes is if the longest string of zeroes contains the $n-1$st place, or the longest string of zeroes is followed by a string of $1$s that touches the $n-1$st place. These are both clearly events with probability going to $0$ as $n$ goes to $\infty$. If it doesn't do that, the only way it can change $k\_a$ is if it creates a new, longer string of zeroes, which can only happen if $n-1$ ends a string of $1$s followed by the beginning of a string of zeroes whose total length is greater than the length of the largest existing string of zeroes. Since the length of the largest existing string of zeroes will typically be $\log n$, this is a low-probability event.
For multiplication by $2$, the argument is similar. The typical case is that this subtracts $1$ from $k\_a$, leaving $c\_a$ unchanged. Because it just shifts bits to the left, it fixes the length of the longest string of zeroes, so the only way it can do anything but subtract $1$ from $k\_a$ is if it pushes the first bit of the first longest string of zeroes past the decimal point, i.e. if $k\_a=1$. But this is again obviously a low probability event.
---
The adjacency matrix of the graph has a relatively large number of eigenvalues close to $4$. This is clearest to me if we look at the action of this adjacency matrix $M$ on a Fourier basis consisting of the vectors $v\_\alpha$ whose $x$'th coordinate is $e(x\alpha/p)$ for $\alpha\in \mathbb Z/p\mathbb Z$. We have $M v\_\alpha = v\_{\alpha/2} + 2 \cos (2\pi /p) v\_\alpha + v\_{2\alpha}$. If we let $k$ be a small odd integer and $j$ a very small positive integer then $v\_k + v\_{2k} + v\_{4k} + \dots + v\_{2^jk}$ is close to an eigenvector with eigenvalue close to $4$, and these vectors are orthogonal. This can only happen if there are many eigenvectors with eigenvalues close to $4$.
| 23 | https://mathoverflow.net/users/18060 | 427042 | 173,262 |
https://mathoverflow.net/questions/426995 | 0 | An integer partition is a sequence $\lambda=(\lambda\_1\geq\lambda\_2\geq\dotsb\geq\lambda\_k)$ of positive integers, for some $k\geq1$. Consider the following two sets of partitions of $n$. Fix a positive integer $s>1$.
Let $\mathcal{O}\_{n,s}=\{\lambda\vdash n: \lambda\_i\in\{1,3,5,\dotsc,2s-1\}\}$. Parts are odd integers.
Also, let $\mathcal{A}\_{n,s}=\{a\_1+2a\_2+2a\_3+\dotsb+2a\_s=n: a\_1\geq a\_2\geq a\_3\geq\dotsb\geq a\_s\geq0\}\subset\mathbb{Z}^s\_{\geq0}$. In my earlier MO post [Seeking a bijective proof enumerating two partition sets: Part I](https://mathoverflow.net/questions/426984/seeking-a-bijective-proof-enumerating-two-partition-sets-part-i), I proposed a question on
$\#\mathcal{O}\_{n,s}=\#\mathcal{A}\_{n,s}$ which was subsequently [answered](https://mathoverflow.net/a/426990) by Per Alexandersson.
Let's add one more set of partitions
$$\mathcal{B}\_{n,s}=\{a\_1+a\_2+\cdots+a\_s=n: a\_2\geq 2a\_1, 2a\_3\geq 3a\_2, 3a\_4\geq 4a\_3,\dotsc,(s-1)a\_s\geq sa\_{s-1}\}.$$
I would like ask:
>
> **QUESTION.** Is there a combinatorial or bijective proof of the equinumerosity $\#\mathcal{O}\_{n,s}=\#\mathcal{B}\_{n,s}$?
>
>
>
| https://mathoverflow.net/users/66131 | Seeking a bijective proof enumerating two partition sets: Part II | Just to mark this question as answered, let me convert my comments into an answer.
The set of partitions $\mathcal{B}\_{n,s}$ you define are called the "lecture hall partitions" and they were introduced by Mireille Bousquet-Mélou and Kimmo Eriksson in [1]. See also the survey on the mathematics of lecture hall partitions by Carla Savage in [2]. The fact that there are the same number of lecture hall partitions of $n$ into at most $s$ parts as partitions of $n$ into odd parts less than $2s$ was the main result of that first paper of Bousquet-Mélou and Eriksson, and they gave a couple of different proofs: one based on affine Coxeter groups, the other more direct and combinatorial. At the end of their paper they gave a recursive bijection between $\mathcal{B}\_{n,s}$ and $\mathcal{O}\_{n,s}$ based on their combinatorial argument. A simple bijection was later obtained by Niklas Eriksen; see the FPSAC 2002 conference article [3].
[1] *Bousquet-Mélou, Mireille; Eriksson, Kimmo*, [**Lecture hall partitions**](http://dx.doi.org/10.1023/A:1009771306380), Ramanujan J. 1, No. 1, 101-111 (1997). [ZBL0909.05008](https://zbmath.org/?q=an:0909.05008).
[2] *Savage, Carla D.*, [**The mathematics of lecture hall partitions**](http://dx.doi.org/10.1016/j.jcta.2016.06.006), J. Comb. Theory, Ser. A 144, 443-475 (2016). [ZBL1343.05032](https://zbmath.org/?q=an:1343.05032).
[3] *Eirksen, Niklas*, **[A simple bijection between lecture hall partitions and partitions into odd integers](http://igm.univ-mlv.fr/~fpsac/FPSAC02/ARTICLES/Eriksen2.pdf)**, [FPSAC 2002](https://igm.univ-mlv.fr/~fpsac/FPSAC02/articles.html).
| 3 | https://mathoverflow.net/users/25028 | 427048 | 173,263 |
https://mathoverflow.net/questions/427053 | 0 | Suppose $X$ is a random variable with a density $f(x)$ such that $f(x)$ is a convolution of some density $g$ with some other density $q$:
$$
f = g\ast q.
$$
~~Under what conditions does $X=h(Y)$, where $Y\sim g(y)$ and $h$ is some function?~~
Under what conditions does there exist a function $h$ such $h(Y)$ has the same density as $X$, where $Y\sim g(y)$?
This problem is equivalent to the following: Given independent random variables $(Y,U)$, when does there exist an $h$ such that $h(Y)\sim U+Y$? (In this re-formulation, $U\sim q(u)$.)
In my use case, $Y$ is simple (e.g. a Gaussian), $f$ and $q$ are very general (but smoothness/regularity conditions are fine if necessary), and $h$ is allowed to be general as well. So this precludes counterexamples like $X=Y$, etc.
Edit: Clarified that I did not mean $h(Y)=U+Y$, which of course implies $U$ is a.s. constant.
| https://mathoverflow.net/users/486206 | When can a convolution be written as a change of variables? | Suppose that $Y,U$ are independent random variables (r.v.'s) such that $h(Y)=U+Y$ for some (Borel-measurable) function $h$. Then $U=Z:=h(Y)-Y$ and $U,Z$ are independent. So, $U$ is independent of itself. So, $U$ is constant almost surely (a.s.): $P(U=u)=1$ for some real $u$. (Indeed, if $U$ is not constant a.s., then the support of the distribution of $U$ contains at least two real numbers $a,b$ such that $a<b$. Take any $c\in(a,b)$. Then the independence of $U$ of itself implies $0=P(U<c,U>c)=P(U<c)P(U>c)>0$, a contradiction.)
Vice versa, if $P(U=u)=1$ for some real $u$, then $h(Y)=U+Y$ a.s. with $h(Y):=u+Y$.
Thus, if $Y,U$ are independent, then there exists a (Borel-measurable) function $h$ such that $h(Y)=U+Y$ if and only if $U$ is constant a.s.
---
Unfortunately, the OP has changed the question, thus invalidating the above answer. The changed question admits a trivial answer as well, though, and after the change the "almost never" answer becomes "always".
Indeed, the r.v. $Y$ is assumed to have a density $g$. So, the distribution of $Y$ is atomless. So,
$$T:=F\_Y(Y)$$
has the uniform distribution on the interval $(0,1)$, where $F\_V$ is the cdf of a r.v. $V$. So, $Z:=U+Y$ equals $F\_Z^{-1}(T)$ in distribution. That is, $U+Y$ equals $h(Y):=F\_Z^{-1}(F\_Y(Y))$ in distribution, as desired.
(As usual, $F^{-1}(t):=\inf\{x\in\mathbb R\colon F(x)\ge t\}.$)
| 2 | https://mathoverflow.net/users/36721 | 427055 | 173,264 |
https://mathoverflow.net/questions/427060 | 6 | I am interested in knowing which natural categories of its representations the étale fundamental group of a scheme can be recovered from.
Suppose $X$ is a scheme. Let $\pi\_1^\text{ét}(X)$ be its étale fundamental group. Thinking along the lines of Tannakian reconstruction of a pro-algebraic group from the category of its representations over some field $k$ as the automorphism group of the forgetful functor to the category of $k$-vector spaces, I am wondering if there is a natural category of objects associated to $X$ (e.g., sheaves) on which $\pi\_1^\text{ét}(X)$ acts and from which it can be reconstructed in a manner analogous to Tannakian reconstruction.
Also, what role does the choice of the field $k$ play?
| https://mathoverflow.net/users/486214 | Tannakian-type reconstruction of étale fundamental group | Yes, this is possible. Since finite groups are algebraic groups, pro-finite groups are pro-algebraic groups. So one can recover $\pi\_1$ in exactly the way you say from the category of algebraic representations of $\pi\_1$, i.e. finite-dimensional representations that are continuous for the discrete topology of $k$. (These all factor through a finite group).
In geometric terms, these are locally-constant finite-rank sheaves of $k$-vector spaces in the étale topology, if $X$ is normal - if $X$ is not normal there may be locally constant sheaves where the $\pi\_1$-action is not continuous.
For these purposes, the choice of field $k$ doesn't matter very much.
But it's usually more interesting to study a different construction, where you consider finite-dimensional representations of $\pi\_1$ that are continuous for the $\ell$-adic topology on $k$ for some $\ell$-adic field $k$. These form a Tannakian category, and the resulting pro-algebraic group is usually not pro-finite, and thus much larger than $\pi\_1$. But its component group recovers $\pi\_1$, basically because the representations of the component group are the category we discussed above.
Geometrically, this is (at least for $X$ normal) the category of lisse $\ell$-adic sheaves defined the usual (slightly complicated) way.
For this construction, the field matters a lot - it must be $\ell$-adic, and we get very different groups for different primes $\ell$ (though there are known to be some relationships between their representations, at least for schemes over finite fields).
| 13 | https://mathoverflow.net/users/18060 | 427061 | 173,266 |
https://mathoverflow.net/questions/426917 | 5 | I am looking for a local integral domain $(D, m)$ with $Spec(D)=\{0,m\}\cup\{ P\_i\}\_i$ such that $P\_i's$ are incomparable (that is, $P\_i\not\subseteq P\_j$ and $P\_j\not\subseteq P\_i$ for $i\not= j$) and $\cap\_i P\_i=0$ and $\cup\_i P\_i\not=m$, where $Spec(D)$ is the set of all prime ideals of $D$ ($D$ is not Noetherian in general).
| https://mathoverflow.net/users/338309 | An example of a local integral domain with special spectrum | Let $k$ be a field. Put $R=k[[x,y]]$ and let $D\subset R$ be the subring $k+xR$. It consists of power series without any term $ay^n$ ($a\in k^\times$, $n>0$), or (equivalently) series $f(x,y)$ such that $f(0,y)\in k$. It is easy to see that $D$ is local with maximal ideal $m=\{f\in D\mid f(0,0)=0\}$.
I claim that $D$ is as required.
First, $D[x^{-1}]\to R[x^{-1}]$ is an isomorphism: it is clearly injective by localization, and surjective because for $g\in R$ and $m\in\mathbb{N}$, we have $x^{-m}g=x^{-m-1}(xg)$ and $xg\in D$. Thus, the nonzero primes $P\_i$ of $D$ not containing $x$ correspond bijectively to the primes of height one in $R$ distinct from $xR$, and there are infinitely many of these. On the other hand, we have $m^2\subset xD$ (check!), so the only prime of $D$ containing $x$ is $m$. Since $x$ is not in any of the $P\_i$'s, their union is not $m$.
Here is a geometric explanation: we can view $D$ as the fibre product ring $R\times\_{R/xR}k$. It follows that the natural diagram
$$\begin{array}{rcl}
\mathrm{Spec}(R/xR) & \longrightarrow & \mathrm{Spec}(R) \\
\downarrow\;&&\;\downarrow\\
\mathrm{Spec}(k) & \longrightarrow & \mathrm{Spec}(D)
\end{array}$$
is a pushout of ringed spaces; this is Theorem 5.1 in [this paper](http://www.numdam.org/item/10.24033/bsmf.2455.pdf) by Ferrand. In other words, $\mathrm{Spec}(D)$ is obtained from $\mathrm{Spec}(R)$ by ``crushing'' the closed subset $\mathrm{Spec}(R/xR)\cong \mathrm{Spec}(k[[y]])$ to the point $\mathrm{Spec}(k)$.
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