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https://mathoverflow.net/questions/407086
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1
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Let $[\omega]^{<\omega}$ denote the collection of finite subsets of the integers, and let us call $E\subseteq [\omega]^{<\omega}$ *non-nested* if $a\not\subseteq b$ whenever $a\neq b\in E$.
Is there a non-nested set $E \subseteq [\omega]^{<\omega}$ with the following properties?
1. Every $n\in \omega$ is contained in infinitely many members of $E$, and
2. For every bijection $\varphi:\omega\to\omega$ and for every $a\neq b\in E$ we have $\sum\_{n\in a} \varphi(n) \neq \sum\_{n\in b} \varphi(b)$.
|
https://mathoverflow.net/users/8628
|
Non-summable subsets of $[\omega]^{<\omega}$
|
Theorem: There is no such $E$.
Claim: for each $a\in E$ there exists $b\in E$ with $|b\setminus a|\ge 2$.
Proof of Claim: Let $a$ be a counterexample. Then all $b\ne a$ contain exactly one element each that's not in $a$. Let $n\not\in a$. Since $n$ is in infinitely many $b$'s, some $b$'s must be equal, contradiction.
Proof of Theorem:
Fix any $a\in E$, and a corresponding $b$ as in the Claim.
To show that some $\varphi$ exists with $\sum\_{n\in a}\varphi(n)=\sum\_{n\in b}\varphi(n)$, we may assume $a\cap b=\emptyset$ since $a\cap b$ contributes the same to both sides.
Let $a=\{\alpha\_1,\dots,\alpha\_n\}$ and $b=\{\beta\_1,\dots,\beta\_m\}$.
Let $\varphi(\alpha\_i)$ for $i<n$ be distinct and arbitrary.
Let $\varphi(\beta\_i)$ for $i\le m$ be distinct, with each $\varphi(\beta\_i)> \sum\_{j<n}\varphi(\alpha\_j)$.
Then for any $k$ and $t\ne k$ we have
$$\sum\varphi(\beta\_i)\ge \varphi(\beta\_t)+\varphi(\beta\_k) >\varphi(\beta\_t) + \sum\_{j<n}\varphi(\alpha\_j)$$
so $\sum\varphi(\beta\_i)-\sum\varphi(\alpha\_i)> \varphi(\beta\_t)$.
Therefore we are free to define
$$\varphi(\alpha\_n)=\sum\varphi(\beta\_i)-\sum\_{j<n}\varphi(\alpha\_i),$$
and we have defined an injective function from $a\cup b$ to $\omega$. This can now be extended to a bijection of $\omega$ arbitrarily.
This completes the construction of the desired $\varphi$ showing that no such $E$ exists. $\Box$
|
2
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https://mathoverflow.net/users/4600
|
407091
| 166,813 |
https://mathoverflow.net/questions/407110
|
4
|
Let $G$ be a planar graph, which we may assume to be a triangulation, with vertex set $V$ and edge set $E$. Suppose the minimum vertex degree is at least 3, and suppose any two distinct edges share at most one vertex.
**Definition.** Given an edge $e \in E$ with vertices $v\_1$ and $v\_2$, define $D(e) = \mathrm{deg}(v\_1). \mathrm{deg}(v\_2)$, the product of the degrees of the vertices of $e$.
I have been calling this quantity the *density* of the edge $e$, but:
**Question 1:** Has this quantity $D(e)$ been previously studied? Does it already have a name?
This quantity is likely to have properties akin to those of the *weight* $w(e) = \mathrm{deg}(v\_1) + \mathrm{deg}(v\_2)$. It is a result of Borodin [1] that every planar graph under consideration has a $(5,6)$ edge, a $(4,7)$ edge, a $(3,10)$ edge, or an edge with degrees less than one of these. As a direct consequence of this:
**Theorem.** Every planar graph $G$ has an edge $e$ with $D(e) \leq 30$.
**Question 2:** Is there a reference for this result?
I am interested in the values of $D$ that can and must appear in a graph with $n$ vertices.
**Definition.** Let $f(n)$ be the least natural number $N$ such that every planar graph with $n$ vertices has an edge $e$ with $D(e) \leq N$.
The theorem above implies that for all $n$, $f(n) \leq 30$. For example, $f(4) = 9$ because the only possible graph is the tetrahedron with all vertices degree 3, and so each edge has $D(e) = 9$. I think $f(5)=12$, $f(6)=f(7)=f(8)=16$ and $f(12)=25$. I suspect that $f(n)=30$ for all $n \geq 32$.
**Question 3:** Has this quantity $f(n)$ been studied for any or all $n$?
Thanks for any help.
[1] Borodin, O. V., Joint generalization of the theorems of Lebesgue and Kotzig on the combinatorics of planar maps. (Russian) Diskret. Mat. 3 (1991), no. 4, 24–27.
|
https://mathoverflow.net/users/6040
|
Product of vertex degrees of an edge in a planar graph
|
Regarding **Question 3**, here is a proof that $f(n)=30$ for all $n \geq 40$. Let $H$ be a $2$-connected planar graph with minimum degree $5$. Let $G$ be obtained from $H$ by adding a new vertex inside each face of $H$, and making it adjacent to all vertices of the face. Let $V(G)=X \cup Y$, where $X=V(H)$, and $Y$ are the newly added vertices. Since $H$ has minimum degree $5$, $\deg\_G(x) \geq 10$ for all $x \in X$. Moreover, $\deg\_G(y) \geq 3$ for all $y \in Y$ and no two vertices of $Y$ are adjacent. Thus, $\deg(u)\deg(w) \geq 30$ for all $uw \in E(G)$. Note that $G$ is a planar graph with $|V(H)|+|F(H)|$ vertices, where $F(H)$ is the number of faces of $H$. By Euler's formula, $|V(G)|=|E(H)|+2$. Since there is a $2$-connected planar graph with minimum degree $5$ and $m$ edges for all $m \geq 38$, we are done.
|
4
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https://mathoverflow.net/users/2233
|
407113
| 166,819 |
https://mathoverflow.net/questions/407133
|
9
|
Let $X$ be a complex variety. By Poincare's lemma, its singular cohomology can be computed as hypercohomology of the holomorphic de Rham complex (viewing $X$ as a complex manifold)
$$\text{H}^\cdot(X,\mathbf{C})\ \simeq\ \text{H}^\cdot(X,\Omega^\bullet).$$
By GAGA the right side is the same as the hypercohomology of the complex of *algebraic* de Rham complex.
In any case, you can compute the hypercohomology $ \text{H}^\cdot(X,\Omega^\bullet)$ using Cech cohomology, in terms of an open cover $\{U\_\alpha\}$ of $X$. For instance, if $X$ has dimension $n$, a representative of $\text{H}^n(X,\Omega^\bullet)$ is a collection of $n$ forms
$$\omega\_{n,\alpha}\ \in\ \text{H}^0(U\_\alpha,\Omega^n)$$
which do not quite glue to a global $n$-form, but there are $n-1$ forms
$$\omega\_{n-1,\alpha,\beta}\ \in\ \text{H}^0(U\_\alpha\cap U\_\beta,\Omega^{n-1})$$
such that $d\omega\_{n-1,\alpha,\beta}=\omega\_\alpha-\omega\_\beta$ on the intersection. Similarly, there are $n-2$ forms $\omega\_{n-2,\alpha,\beta,\gamma}$ such that $d\omega\_{n-2,\alpha,\beta,\gamma}=\omega\_{n-1,\alpha,\beta}-\omega\_{n-1,\alpha,\gamma}+\omega\_{n-1,\beta,\gamma}$, and so on.
For example, on $X=\mathbf{P}^1$ with its standard cover by two affine lines, $\text{H}^2(\mathbf{P}^1)$ is generated by the meromorphic differential form
$$dx/x\ \in\ \text{H}^0(\mathbf{A}^1\_0\cap\mathbf{A}^1\_\infty, \Omega^1). $$
---
**Question:** Assume that $X$ is smooth and proper. We then know that an element of $\text{H}^\cdot(X,\mathbf{C}) \simeq \text{H}^\cdot(X,\Omega^\bullet)$ can be integrated. What does integration look like in terms of these hypercohomology classes?
---
Looking at the $\mathbf{P}^1$ example, one might guess that the answer has something interesting to do with residues.
|
https://mathoverflow.net/users/119012
|
Integrating hypercohomology classes
|
One way to see this is algebraically. Let $\omega\_X = \Omega^n\_X$. The map you seek to describe is
$$
\int\_X \colon \mathrm{H}^n(X,\omega\_X) \to \mathbb{C}
$$
Abstractly, it arises as the counit of the Serre duality adjunction
$$
\mathrm{H}^n(X,-) \dashv -\otimes \omega\_X
$$
By the compatibility between local and global duality, there is a commutative diagram
$\require{AMScd}$
\begin{CD}
\otimes\_P \mathrm{H}^n\_P(\omega\_X)@>can>> \mathrm{H}^n(X,\omega\_X)\\
@V \otimes\_P \mathrm{res}\_P V V @VV\int\_XV\\
\mathbb{C} @= \mathbb{C}
\end{CD}
and the maps $\mathrm{res}\_P$ that arise by local duality, are computed through higher dimensional residues. This is explained with great detail in J. Lipman's blue book:
*Dualizing sheaves, differentials and residues on algebraic varieties*. Astérisque, No. **117** (1984).
For the comparison between the algebraic and the analytic case and the $\frac{1}{(2 \pi i)^n}$ factor (up to a sign), see
Sastry & Tong, The Grothendieck Trace and the de Rham Integral, *Canad. Math. Bull.* Vol. **46** (3), 2003 pp. 429–440.
To make things more explicit, you may assume that your covering $\{U\_i\}\_{i \in I}$ is finite and each $U\_i$ is the complement of a certain section $f\_i$ of a line bundle. In this case a Czech cocycle is defined by a residual symbol
$$
\left[ \frac{ w } { f\_0 \dots f\_n } \right]
$$
with $w$ a meromorphic $n$-form. This is described on page 195 of
Hartshorne, *Residues and duality*. Lecture Notes in Mathematics, **20**, Springer-Verlag, 1966.
So the objective is to compute
$$
\int\_X \left[ \frac{ w } { f\_0 \dots f\_n } \right]
$$
The Czech-cocycle/symbol induces a certain element
$$
\left[ \frac{ w } { f\_0 \dots \hat{f\_i} \dots f\_n } \right]\_P
$$
in $\mathrm{H}^n\_P(\omega\_X)$ where $p\in U\_i$ so $(f\_i)\_P$ is invertible in $\mathcal{O}\_{X,P}$. This element is describable using a Koszul complex to compute $\mathrm{H}^n\_P(\omega\_X)$. Finally the computation of the integral is reduced to compute
$$
\mathrm{res}\_P \left[ \frac{ w } { f\_0 \dots \hat{f\_i} \dots f\_n } \right]\_P
$$.
This can be done explicitly, over the points $P$ that are *poles* of the differential form $w$. Locally at $P$
$$
w = \phi f\_0 \wedge \hat{f\_i} \wedge f\_n
$$
with $\phi$ in the fraction field of $\mathcal{O}\_{X,P}$. Then,
$$
\mathrm{res}\_P \left[ \frac{ w } { f\_0 \dots \hat{f\_i} \dots f\_n } \right]\_P = a\_{(-1, \dots, -1)}
$$
where $a\_{(-1, \dots, -1)} \in \mathbb{C}$ denotes the corresponding coefficient in the Laurent expansion of $\phi$. This is described using the identification of $\widehat{\mathcal{O}}\_{X,P}$ with the power series ring $\mathbb{C}[[f\_0 \dots \hat{f\_i} \dots f\_n]]$.
|
8
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https://mathoverflow.net/users/6348
|
407136
| 166,824 |
https://mathoverflow.net/questions/407139
|
0
|
Consider an arbitrary simplex $\mathcal{S} \subseteq \mathbb{R}^n$ ($\mathcal{S}$ is a polytope in $\mathbb{R}^n$ with $n+1$ vertices and non-empty interior). Let ${\bf P} \in \mathbb{R}^{m \times n}, m<n$ be an orthogonal matrix (in the sense that ${\bf P} {\bf P}^T = {\bf I}$, where ${\bf I} \in \mathbb{R}^{m \times m}$ is the identity matrix) defining a projection over a subspace of dimension $m$ of $\mathbb{R}^n$. Let $\mathcal{P} = {\bf P} \mathcal{S}$ be the projection of the simplex over such subspace.
My question is: what is the maximum amounts of vertices of $\mathcal{P}$?.
I am specially interested in low-dimensional cases ($m = 2$, $3$ or $4$ for example). I am also interested in special cases like the maximum amount for regular simplices.
Thanks in advance.
|
https://mathoverflow.net/users/106178
|
Maximum vertex amount of low-dimensional simplex projection
|
For every simplex $\Delta\subset\Bbb R^n$ with $n+1$ vertices and every number $m\le n$ there is an $m$-dimensional subspace $W\subseteq\Bbb R^n$ so that the orthogonal projection of $\Delta$ onto $W$ has $n+1$ vertices.
*Proof*.
Fix a polytope $P\subset\Bbb R^m$ with $n+1$ vertices.
First, we show that $P$ is the orthogonal projection of *some* simplex $\Delta\_P \subset\Bbb R^n$: if $P$ has vertices $v\_0,...,v\_n\in\Bbb R^m$, then for $i\in\{0,...,n\}$ the
$$w\_i:=\underset{\!\!\!\!\!\!\!\!\begin{array} .\uparrow\\ \!\!\!\!\!\!\!\!\text{at index $m$}\end{array}}{\begin{cases}
(v\_i,0,...,0) & \text{for $i\le m$} \\
(v\_i,0,...,1,...,0) & \text{for $i > m$}
\end{cases}}\quad\in\Bbb R^n$$
are $n+1$ points in $\Bbb R^n$ in convex position.
They necessarily form the vertices of some simplex $\Delta\_P$.
Also, the projection onto the first $m$ coordinates (which is orthogonal) yields the initial polytope $P$.
Alternatively, we can say that the intersection of the subspace $\Bbb R^m\subset\Bbb R^n$ with the cylinder $C:=P+\Bbb R e\_{m+1}+...+\Bbb R e\_n$ is $P$.
Now, any two simplices in $\Bbb R^n$ with $n+1$ vertices are affinely equivalent, and so we can choose an affine transformation $X$ with $X\Delta\_P= \Delta$. Clearly, the intersection of the transformed cylinder $X C$ with the transformed subspace $W':=X\Bbb R^m$ is still a polytope with $n+1$ vertices. The same is true if you replace $W'$ by a subspace $W\subseteq\Bbb R^n$ orthogonal to $Xe\_i,i\in\{m+1,...,n\}$, as all cross-sections of the cylinder are affinely equivalent. But this intersection can be interpreted as the *orthogonal* projection of $\Delta=X\Delta\_P$ onto $W$, which then has $n+1$ vertices. $\quad$
$\square$
As a side note, the intersection $XC\cap W$ is an affine transformation of $P$. Thus, given any simplex $\Delta\subset\Bbb R^n$ and any polytope $P\subset\Bbb R^m$ with $n+1$ vertices, $P$ is affinely equivalent to an orthogonal projection of $\Delta$.
|
1
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https://mathoverflow.net/users/108884
|
407161
| 166,827 |
https://mathoverflow.net/questions/162593
|
13
|
Normally the theory of (elliptic) differential operators between vector bundles (or $\mathbb{R}^n$) is presented in the language of Sobolev spaces. I'm searching for a book (or something similar) which exposes the theory completely in the setting of smooth sections, that is using Fréchet space techniques.
In particular, I'm interested in how the Fredholm theory translates into Fréchet spaces. The basic results should transfer without big modifications, however some results are not so straightforward (like "Fredholm operators are open in the space of all continuous linear maps" since the theory of dual spaces is more complicated).
|
https://mathoverflow.net/users/17047
|
Reference Request: Elliptic differential operators in the Fréchet setting
|
A few bits and pieces of Fredholm theory in the locally convex (and, in particular, Fréchet) setting are discussed in the literature. The most comprehensive treatment I could find were two old articles by Schaefer:
* [Über singuläre Integralgleichungen und eine Klasse von Homomorphismen in lokalkonvexen Räumen, Math. Z., Springer Nature, 1956, 66, 147-163](https://link.springer.com/article/10.1007%2FBF01186604)
* [On the Fredholm alternative in locally convex linear spaces, Studia Math., 1959, 18, 229-245](https://www.impan.pl/pl/wydawnictwa/czasopisma-i-serie-wydawnicze/studia-mathematica/all/18/3/94447/on-the-fredholm-alternative-in-locally-convex-linear-spaces)
There he discusses how the basic elements of Fredholm theory like invertibility modulo compact operators, existence of a parametrix and the Fredholm alternative generalize from Banach to locally convex spaces.
However, most applications of Fredholm theory use families of operators and things like the invariance of the index under small deformations etc. These results do not directly generalize to locally spaces since the set of invertible operators fails to be open, leading to many pathological (?) counterexamples. In order to discuss families of Fredholm operators and extend the important results about them to the locally convex setting, in [my Phd Thesis](https://arxiv.org/abs/1909.00744), I introduced the notation of a "uniformly regular" family of operators. Roughly speaking, these are families of operators that posses a family of (generalized) inverses. This allows to extend most classical results to this setting. For example, if you have a uniformly regular family of operators and one of them happens to be Fredholm, then all of them are Fredholm with the same index. Finally, a classical result about elliptic operators (Theorem II.3.3.3 in [Hamilton, R. S., The inverse function theorem of Nash and Moser](https://www.ams.org/journals/bull/1982-07-01/S0273-0979-1982-15004-2/)) can be used to show that families of elliptic operators are uniformly regular so that these results apply in particular to elliptic operators.
A streamlined (but perhaps less pedagogical) exposition of uniformly regularity and Fredholm theory can be found in the article
* [Diez, T. & Rudolph, G., Normal form of equivariant maps in infinite dimensions, Ann. Glob. Anal. Geom., 2021](https://link.springer.com/article/10.1007%2Fs10455-021-09777-2)
|
1
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https://mathoverflow.net/users/17047
|
407162
| 166,828 |
https://mathoverflow.net/questions/407153
|
2
|
Let $U\to X$ be an open immersion of schemes and denote by $D$ the (say reduced) complement. Then by applying the de Rham functor, we get morphisms
$$U\_{dR}\to X\_{dR}\leftarrow D\_{dR}$$
of the associated de Rham spaces. Then in [this](https://people.math.harvard.edu/%7Egaitsgde/GL/Crystalstext.pdf) (in 2.5.2) Gaitsgory and Rozenblyum say that $U\_{dR}\to X\_{dR}$ is an open embedding.
It looks like they are proving this using the functorial description of open embeddings via $QCoh$, but instead for $IndCoh$. Its not clear to me that we also have a description of openness via $IndCoh$, do you know of a good reference for it? Once we have that, I think using $IndCoh(X\_{dR})=\text{Crys}(X)$ and Kashiwara shows that the map is an open embedding.
My question is what do we know $D\_{dR}\to X\_{dR}$? Is it also a closed embedding? Naively you would think so, since we can see via the functor of points that $D\_{dR}$ is the complement of $X\_{dR}$, and the complement of a open substack ought to be a closed substack. Is this correct?
|
https://mathoverflow.net/users/152554
|
Open/closed embeddings and the de Rham space
|
I believe the following is a counter example to the claim in the last paragraph. Let $X$ be a smooth affine curve for simplicity, and let $pt \to X$ be a closed point. Consider the canonical map $X \to X\_{dR}$. Then $D := X \times\_{X\_{dR}} pt$ is the formal completion of the point, and the map
$$D \longrightarrow X $$
is not a closed immersion.
|
4
|
https://mathoverflow.net/users/101861
|
407163
| 166,829 |
https://mathoverflow.net/questions/407115
|
4
|
I was reading the book "Differentiable Periodic Maps" by P.E. Conner (1979). I am stuck at the following problem given at the end of section 21:
Let $\xi\to V^n$ be a $k$-plane bundle over a closed n-manifold then $[\mathbb{R}P(\xi)]\_2=0$ if either $k=2$ or $n=1$.
I was able to solve the $k=2$ case just by using the expression of total Stiefel Whitney class for the projectivization of the bundle given at the starting of the section.
But I could not complete the second case using a similar approach.
Can anyone suggest an alternate approach?
|
https://mathoverflow.net/users/126899
|
Cobordism class of projectivization of a bundle
|
First, we can assume wlog that $\xi$ has an inner product.
Next, given a $2$-dimensional real vector space $U$ with inner product, let $QU$ be the space of self-adjoint endomorphisms of trace zero. Given a unit vector $u$, define $\phi(u)\in QU$ by $\phi(u)(v)=\sqrt{2}(\langle v,u\rangle u - \langle u,u\rangle v/2)$. This satisfies $\phi(-u)=\phi(u)$ and $\|\phi(u)\|=\|u\|^2$. It therefore induces a map from the projective space $PU$ to the unit circle $SQU$, which is a diffeomorphism.
Now suppose that $\xi$ is a $2$-dimensional real bundle over a base $V$. We can apply the above fibrewise to identify $P\xi$ with the circle bundle associated to $Q\xi$, which is the boundary of the corresponding disc bundle. Thus, $P\xi$ is nullbordant.
Now consider instead the case of a vector bundle $\xi$ (with inner product) of dimension $k$ over a closed $1$-manifold $V$. By dividing $V$ into path components, we reduce to the case where $V=S^1$. Here standard arguments identify the total space $E\xi$ with $(\mathbb{R}\times\mathbb{R}^k)/\sim$, where $(t,v)\sim(t+1,Av)$ for some fixed matrix $A\in O(k)$. The isomorphism type of $\xi$ depends only on the path-component of $A$ in $O(n)$. Because $-I$ acts as the identity on $\mathbb{R}P^{k-1}$, we also see that the diffeomorphism type of $P\xi$ depends only on the path component of $\pm A$ in $O(k)/\{\pm I\}$. If $k$ is odd then $O(k)/\{\pm I\}$ is connected so $P\xi\simeq S^1\times \mathbb{R}P^{k-1}$, which is the boundary of $B^2\times \mathbb{R}P^{k-1}$. The same applies if $k$ is even but $A$ lies in the base component.
The case where $A$ is not in the base component is less obvious. When $k=2$ we get the Klein bottle, which is diffeomorphic to the connected sum of two copies of $\mathbb{R}P^2$. This is nullbordant by the first part of this answer, or by general properties of connected sums. I think that there should be a direct geometric construction for $k>2$, but I don't see one just now.
If I remember rightly, there are various constructions like this in Stong's book on cobordism, but I don't have a copy.
|
4
|
https://mathoverflow.net/users/10366
|
407169
| 166,832 |
https://mathoverflow.net/questions/407067
|
4
|
I'm trying to understand some things about quotients of braid groups, and particularly I'd like to solve the word problem for some elements of these quotients. I'm using MAGMA to try to access this, but I can't seem to find whether MAGMA's `GrpBrd` has a quotient construction. The more general finitely-presented group category `GrpFP` does not have an implementation for the word problem.
To be more specific about the groups considered, I'm looking at the quotients of the braid groups $B\_n(d)=B\_n/\langle \sigma\_i^d \rangle$ where each Artin generator has some fixed finite order. Do these things fit into any category of groups in MAGMA that has a word problem implementation? I don't care which category these land in, just that I can compute with them somehow.
|
https://mathoverflow.net/users/151664
|
Good algorithmic properties for quotients of braid groups
|
Patrick Dehornoy's handle reduction [1] technique that solves the word problem in the braid group can be modified so that it seems to solve the word problem in the groups of the form $B\_{n}(d)$. In practice, for all the braid words $w$ that I have tested that are in the normal subgroup generated by $\sigma\_{i}^{d}$, the algorithm is able to quickly show that $w$ is equivalent in $B\_{n}(d)$ to the identity $e$ by applying the braid group identities together with the identity $\sigma\_{i}^{d}=e$. If the algorithm gets stuck and is unable to reduce $w$ to $e$ after enough time, then we can be confident that $w$ is not equivalent to the identity $e$.
We say that a braid word is a $\sigma\_{i}$-handle if it is of the form
$\sigma\_{i}^{c}u\sigma\_{i}^{-c}$ where $u$ is a braid word that does not contain any instance of $\sigma\_{j}$ or $\sigma\_{j}^{-1}$ whenever $j\geq i$. A good $\sigma\_{i}$-handle is a $\sigma\_{i}$-handle that does not contain a subword that is a $\sigma\_{i-1}$-handle; equivalently, a good $\sigma\_{i}$-handle is a $\sigma\_{i}$-handle that either has no instance of $\sigma\_{i-1}$ or has no instance of $\sigma\_{i-1}^{-1}$.
Consider the following moves that relate braid words to other braid words.
Move 1: (handle reduction) Suppose that $z=u\sigma\_{i}^{r}v\sigma\_{i}^{-r}w$, $r\in\{-1,1\}$, and
$\sigma\_{i}^{r}v\sigma\_{i}^{-r}$ is a good $\sigma\_{i}$-handle where $r\in\{-1,1\}$. Then perform the replacement $z\mapsto uv'w$ where $v'$ is the word obtained from $v$ by replacing each instance of $\sigma\_{i-1}^{s}$ where $s\in\{-1,1\}$ with
$\sigma\_{i-1}^{-r}\sigma\_{i}^{s}\sigma\_{i-1}^{r}$. Observe that the resulting braid word obtained after applying this move represents the same braid.
Move 2: Replace each $\sigma\_{i}^{r}$ where $r\in\{1,-1\}$ with $\sigma\_{n-i-1}^{r}$. This move conjugates the braid with the half-twist $\Delta\_{n}$. This move together with Move 1 allows us to reduce right handles as well as left handles.
Move 3a: $u\sigma\_{i}^{r}v\mapsto u\sigma\_{i}^{d-r}v$ whenever $|d-r|\leq|r|$
Move 3b: $u\sigma\_{i}^{r}v\mapsto u\sigma\_{i}^{d+r}v$ whenever $|d+r|\leq|r|$
Observe that the resulting braid word obtained after applying Move 3 represents the same element in $B\_{n}(d)$. We observe that if we obtain the empty word from $w$ by applying Moves 1-3, then $w$ represents the identity in $B\_{n}(d)$.
Patrick Dehornoy has proven that the process of simply applying Move 1 (handle reduction) always terminates, and a braid word $w$ represents the trivial braid in $B\_{n}$ if and only if this process terminates with the identity braid. In practice, handle reduction solves the word problem for braid groups very quickly. Moves 1-2 are both applied in order to heuristically minimize the length of a braid word representing a braid, and this braid word minimization technique has been used to attack braid based cryptosystems [2].
In all the words $w$ that I have shown to be equivalent in $B\_{n}(d)$ to the identity, I was able to reduce $w$ to the identity simply by haphazardly applying Moves 1-3. Unfortunately, I do not know how to formally prove that one can solve the word problem in $B\_{n}(d)$ by demonstrating that Moves 1-3 always simplifies a word $w$ to the empty word in a reasonable amount of time whenever $w$ represents the identity in $B\_{n}(d)$.
[1] <https://www.lmno.cnrs.fr/archives/dehornoy/Surveys/Dhn.pdf>
[2] A practical attack on a braid group based cryptographic protocol. Alexei Myasnikov, Vladimir Shpilrain, and Alexander Ushakov. 2005
<https://iacr.org/archive/crypto2005/36210085/36210085.pdf>
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1
|
https://mathoverflow.net/users/22277
|
407181
| 166,837 |
https://mathoverflow.net/questions/407159
|
2
|
Consider the specific element of the corresponding Lie algebra $\mathbb{1}\_3 \times \sigma^3$, where $\mathbb{1}\_3$ is the unit matrix in 3 dimensions, $\sigma^3$ is the 3rd Pauli matrix and $\times$ stands for Kronecker product. I want to find the space
$$
\left\{ U\in U(6) \; \text{ such that} \; U\left( \mathbb{1}\_3 \times \sigma^3\right)U^\dagger = \mathbb{1}\_3 \times \sigma^3 \right\}.
$$
I am particularly interested in whether this set is connected or disjoint.
|
https://mathoverflow.net/users/159498
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Set of $U(6)$ elements which leave a Lie-algebra element invariant under conjugation
|
It is a theorem (of Hopf, I believe) that the centralizer of any member of the Lie algebra (not just $\mathbb{1}\_3 \otimes \sigma^3$) is connected. See Bourbaki, Lie groups, Chap. 9, §2, nº2, [Corollary 5](https://books.google.com/books?id=oCO0xOzNLhAC&pg=PA290).
In your case $\mathbb{1}\_3 \otimes \sigma^3$ is conjugate to $\smash[b]{\begin{pmatrix}\mathbb{1}\_3&0\\0&-\mathbb{1}\_3\end{pmatrix}}$ and so the centralizer is conjugate to the diagonal $U(3)\times U(3)$.
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4
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https://mathoverflow.net/users/19276
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407199
| 166,842 |
https://mathoverflow.net/questions/407209
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8
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In [Between $T\_1$ and $T\_2$](https://doi.org/10.2307/2316017), Albert Wilansky mentioned in **6.** that it was not known whether or not every locally compact US space is $T\_2$.
Is this matter still an open problem?
|
https://mathoverflow.net/users/146942
|
Is this question about US spaces still an open problem?
|
No, it is not. S. Franklin gave an example in his [review of Wilansky's paper](https://mathscinet.ams.org/mathscinet-getitem?mr=208557): take a compact space with a point $x$ that is not the limit of a non-trivial sequence, for example the ordinal $\omega\_1+1$ with $x=\omega\_1$, and double that point. The set of our space is $\omega\_1\cup\{\omega\_1^a, \omega\_1^b\}$; the points other than $\omega\_1^a$ and $\omega\_1^b$ have their normal neighbourhoods and the (basic) neigbourhoods of $\omega\_1^i$ ($i=a,b$) are $(\alpha,\omega\_1)\cup\{\omega\_1^i\}$ with $\alpha<\omega\_1$.
The result is US and (locally compact), but not Hausdorff.
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19
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https://mathoverflow.net/users/5903
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407210
| 166,845 |
https://mathoverflow.net/questions/402225
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21
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For all primes up to $p=89$ there exists a product $Q=\prod\_{j=1}^d(x-a\_j)$ involving $d\geq (p-1)/4$ distinct linear factors $x-a\_j$ in $\mathbb F\_p[x]$ such that $Q'$ has all its roots in $\mathbb F\_p$. Since $Q$ has only simple roots,
the product $QQ'$ has also only simple roots.
(It is obvious that a factorization into distinct linear factors of $QQ'$ implies that $Q$ has degree at most $(p+1)/2$. My computations suggest however that it seems to be hard to go substantially beyond $p/4$.)
An example for $p=53$ is given by
$$Q=x(x^2-5^2)(x^2-8^2)(x^2-10^2)(x^2-12^2)(x^2-14^2)(x^2-16^2)$$
with derivative $$Q'=13(x^2-4^2)(x^2-13^2)(x^2-19^2)(x^2-20^2)(x^2-21^2)(x^2-22^2).$$
Similar examples of the form $x\prod\_{j=1}^7(x^2-a\_j^2)$ for $59$ and $61$ are given by $\{a\_1,\ldots,a\_7\}=\{1,3,12,15,17,18,22\}$ for $59$ and by $\{1,3,6,8,11,21,22\}$ for $61$.
For $p=67$ we have $x\prod\_{j=1}^8(x^2-a\_j^2)$ with $\{a\_1,\ldots,a\_8\}=\{1,2,5,9,15,23,31,32\}$.
For $p=71$ and $73$, we have $\prod\_{j=1}^9(x^2-a\_j^2)$ with $\{a\_1,\ldots,a\_9\}$ given by $\{1,2,3,5,18,20,29,30,33\}$ for $p=71$ and by $\{1,2,5,17,19,22,27,32,33\}$ for $p=73$.
For $p=79$, we have $\prod\_{j=1}^{10}(x^2-a\_j^2)$ with $\{a\_1,\ldots,a\_{10}\}=\{1,2,6,10,15,17,18,20,33,37\}$.
For $p=83$, we have $x\prod\_{j=1}^{10}(x^2-a\_j^2)$ with $\{a\_1,\ldots,a\_{10}\}=\{1,2,3,5,12,16,21,33,36,41\}$.
For $p=89$, we have $\prod\_{j=1}^{11}(x^2-a\_j^2)$ with $\{a\_1,\ldots,a\_{11}\}=\{1,10,14,15,18,33,34,36,37,43,44\}$.
(I did a search over all polynomials up to $p=23$ and restricted then my attention to polynomials which are either even or odd for all primes up to $89$.)
Adopting a very naive viewpoint, the existence of such a polynomial of degree $\geq \lambda p$ for some $\lambda>0$ and for almost all primes would be somewhat surprising: There are ${p\choose \lambda p}$ such polynomials of degree $\lambda p$
given as a product of $\lambda p$ distinct linear factors
in $\mathbb F\_p[x]$. Adopting the very
naive viewpoint that derivatives of these polynomials behave randomly with respect to decomposition into irreducible factors,
the probability for such a polynomial $Q\in\mathbb F\_p[x]$ (with $QQ'$ factorizing completely into distinct linear factors) to exist should be roughly given by $$\left(\lambda^{2\lambda}(1-\lambda)^{2(1-\lambda)}p^\lambda\right)^{-p}$$ (up to polynomial contributions) which decays exponentially fast.
**Question:** Can this naive argument be made rigourous or
do such polynomials of degree at least $(p-1)/4$ (or perhaps of degre at least $\lambda p$ for some $\lambda >0$) exist for all primes?
**Remark (added after adding additional examples up to $83$):** The outlook for the existence of such polynomials (of degre at least $(p-1)/4$) for all primes starts to look better: Indeed, for $p=83$, the naive
argument given above (slightly improved) yields a fairly small
probability:
$${41\choose 10}{31\choose 10}83^{-10}\sim 0.0032\ .$$
This suggests that we should have ended up empty-handed at this point.
(End of added remark.)
**Motivation:** The analogous question is open for degree $7$ over $\mathbb Z[x]$ (existence of a polynomial $Q\in\mathbb Z[x]$ of degree $7$ such that $QQ'$ has $13$ distinct roots in $\mathbb Z$), see proposition 27 of [Proposals for polymath projects](https://mathoverflow.net/questions/219638/proposals-for-polymath-projects).
A stronger question (over finite fields) is to ask for
polynomials of maximal degree in $\mathbb F\_p[x]$ such that $Q$ and all its derivatives $Q^{(1)},Q^{(2)},\ldots$ factorize into distinct linear factors.
One can either admit or forbid common roots among different derivatives. Forbidding them gives of course the trivial upper bound
$O(\sqrt{p})$ on the maximal degree. I guess that admitting them does not allow substantially higher degrees.
Given an integer $d$, I guess that such polynomials of degree $\geq d$ exist in $\mathbb F\_p[x]$ except for a finite number of small primes. (A bug in my code made me think that there are no such polynomials of degree $\geq 6$ which prompted me to post a now deleted question on MO.)
It would of course be interesting to have information on the maximal degree $d\_p$ for (the two versions of) this stronger question. (Interestingly, the stronger question is computationally slightly easier since the set of all such polynomials (not necessarily of maximal degree) is closed under derivation.)
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https://mathoverflow.net/users/4556
|
Existence of a polynomial $Q$ of degree $\geq (p-1)/4$ in $\mathbb F_p[x]$ such that $QQ'$ factorizes into distinct linear factors
|
Such polynomials always exist, examples are [Dickson polynomials](https://en.wikipedia.org/wiki/Dickson_polynomial) of the first kind (with parameter $1$). For a positive integer $n$ these degree $n$ polynomials $D\_n$ are most conveniently defined implicitly by $D\_n(z+1/z)=z^n+1/z^n$.
For $p=4m\pm1$ the polynomial $D\_m(x)$ has the required property, which is to say that $D\_m(x)\cdot D\_m'(x)$ divides $x^p-x$.
In fact, using $D\_n'(z+1/z)=n\frac{z^n-1/z^n}{z-1/z}$ we verify
the identity (which holds in any characteristic)
\begin{equation}
D\_m(x)\cdot D\_m'(x)\cdot D\_{2m\pm1}'(x)\cdot(x^2-4)=m\cdot(2m\pm1)\cdot(D\_{4m\pm1}(x)-x),
\end{equation}
and the claim follows since $D\_p(x)=x^p$ in characteristic $p$.
*Remark:* I have no idea about the strong version of the other question for the higher derivatives. The polynomials which achieve maximal degrees for given $p$ do not seem to be of a particular form (even in those cases when they are unique up to trivial transformations).
|
15
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https://mathoverflow.net/users/18739
|
407212
| 166,846 |
https://mathoverflow.net/questions/398269
|
1
|
let $X,Y:M\to TM$ be vector fields on $M$. $\nabla\_XY$ is the change in $Y$ along the flow curves of $X$. so for a point $p \in M$ let $\phi^X(t):\mathbb{R}\to M$ be a flow curve of $X$ passing through $p$ :
$$(\phi^X)'=X \; \; ; \; \; \phi^X(0)=p$$
since I can't measure the change in $Y$ at two different points directly. I'll define a parallel transport map $\Pi\_{tX}:T\_pM\to T\_{\phi^X(t)}M$ taking a vector at $p$ to its equivalent at $\phi^X(t)$. Then the derivative at $p$ is:
$$\nabla\_XY|\_p=\lim\_{t\to 0}\frac{\delta\_XY|\_{\phi^X(t)}}{t}=\lim\_{t\to 0}\frac{Y\_{\phi^X(t)}-\Pi\_{tX}(Y\_p)}{t}$$
I tried to do the usual approximation trick by assuming that for a "small" enough $t$ :
$$\nabla\_XY|\_p \approx \frac{\delta\_XY|\_{\phi^X(t)}}{t}$$
but there is a problem that LHS and RHS are not at the same point.
here is what I did "correct me if i'm wrong":
since the points $p$ and $\phi^X(t)$ are so close:
$$\Pi\_{tX}(\nabla\_XY|\_p) \approx \frac{\delta\_XY|\_{\phi^X(t)}}{t}$$
in other words:
$$\Pi\_{tX}(\nabla\_XY|\_p)=\frac{\delta\_XY|\_{\phi^X(t)}}{t}+\vec\epsilon\_1$$
$\vec\epsilon$ being a really small difference vector. Another thing that follows is that the values of the vector field $\nabla\_XY$ at the two points are close so:
$$\Pi\_{tX}(\nabla\_XY|\_p)=\nabla\_XY|\_{\phi^X(t)}+\vec\epsilon\_2$$
then
$$\delta\_XY|\_{\phi^X(t)}=t\nabla\_XY|\_{\phi^X(t)}+t\vec\epsilon\_2-t\vec\epsilon\_1$$
since $\vec\epsilon$ errors are already small. $t\vec\epsilon$ would be even much smaller "second order error" and could be neglected:
$$\delta\_XY|\_{\phi^X(t)} \approx t\nabla\_XY|\_{\phi^X(t)}$$
so:
$$\Pi\_{tX}(Y\_p) \approx (-t\nabla\_XY+Y)|\_{\phi^X(t)}$$
so is there anything wrong with this attempt?
|
https://mathoverflow.net/users/311976
|
Approximating the parallel transport map on a curve with the covariant derivative
|
Defining
\begin{equation}
\nabla\_XY|\_p := \lim\_{t\to 0} \frac{\Pi\_{tX}^{-1}(Y\_{\phi^X(t)})-Y\_p}{t}
\end{equation}
the approximation
$\nabla\_XY|\_p \approx \frac{\Pi\_{tX}^{-1}(Y\_{\phi^X(t)})-Y\_p}{t}$, for $t$ small, holds and now both sides are tangent vectors in $T\_pM$. It also follows that $\Pi\_{tX}^{-1}(Y\_{\phi^X(t)})\approx t\,\nabla\_XY|\_p+Y\_p$, namely $Y\_{\phi^X(t)}\approx \Pi\_{tX}(t\,\nabla\_XY|\_p+Y\_p)$ and therefore
\begin{equation}
\Pi\_{tX}(Y\_p)\approx Y\_{\phi^X(t)}-t\,\Pi\_{tX}(\nabla\_XY|\_p).
\end{equation}
|
1
|
https://mathoverflow.net/users/332652
|
407224
| 166,847 |
https://mathoverflow.net/questions/396890
|
38
|
Alain Connes and Caterina Consani seem to be currently working on "absolute algebraic geometry", which is a kind of "algebraic geometry over the sphere spectrum" (<https://arxiv.org/abs/1909.09796>, <https://arxiv.org/abs/1502.05585>).
They seem to be mainly motivated by the idea that this helps with the Riemann hypothesis (see chapter 5 of Connes' essay on the Riemann hypothesis: <https://arxiv.org/abs/1509.05576>).
There is an approach to the Riemann hypothesis via $\mathbb{F}\_1$-geometry ([Riemann hypothesis via absolute geometry](https://mathoverflow.net/questions/69389/riemann-hypothesis-via-absolute-geometry)),
and Connes and Consani identify $\mathbb{F}\_1$ with the sphere spectrum.
Completely independently of that Jacob Lurie is developing his "[Spectral algebraic geometry](https://www.math.ias.edu/%7Elurie/papers/SAG-rootfile.pdf)", which is also a kind of "algebraic geometry over the sphere spectrum".
But here there is absolutely no mention of $\mathbb{F}\_1$-geometry, Arakelov theory or the Riemann hypothesis.
On the surface level the two theories sound quite similar, because both are "geometry over the sphere spectrum", and both are about studying spaces (Connes–Consani: topological spaces; Lurie: $\infty$-topoi) equipped with sheaves of ring spectra (Connes–Consani: $\Gamma$-sets with $S$-algebra structure; Lurie: $E\_{\infty}$-ring spectra).
My question is: How are these two theories related? Can Lurie's spectral algebraic geometry be useful for the Riemann hypothesis?
It seems like each spectral scheme of Connes–Consani can be made into a spectral scheme of Lurie.
If I have an affine spectral scheme à la Connes–Consani, then it is $\operatorname{Spec}$ of a $\Gamma$-set with $S$-algebra structure. This then gives rise to a connective spectrum with $E\_{\infty}$-structure, and taking $\operatorname{Spec}$ of that gives an affine spectral scheme à la Lurie.
For non-affine schemes à la Connes–Consani, like $\overline{\operatorname{Spec}(\mathbb{Z})}$ we can take an affine cover, then make each of those affine
opens into a spectral scheme à la Lurie, and then glue them back together in Lurie's formalism.
So it looks like one can construct the Arakelov-theoretic objects like $\overline{\operatorname{Spec}(\mathbb{Z})}$ that Connes–Consani cares about also in Lurie's formalism.
Since Lurie's formalism seems much further developed already it is maybe easier to prove the Riemann-Roch theorem that Connes–Consani strategy requires in the setting of Lurie.
Can the two theories be usefully connected, or are there obvious problems with this?
|
https://mathoverflow.net/users/143968
|
Connes–Consani's absolute geometry and Lurie's spectral algebraic geometry
|
I know very little about the absolute/algebraic geometry side, but I think I understand the gist of the category theory going on here. I guess this answer might require one to know a bit of both the stable homotopy story and the $\mathbb{F}\_1$-geometry story.
tl;dr is that, no, algebra over $\mathbb{S}$ requires that the underlying objects be spectra, which are a derived version of abelian groups, in the sense that "discrete $\mathbb{S}$-modules" are exactly abelian groups; but algebra over $\mathfrak{s}$ is (i) entirely discrete or $1$-categorical and not derived at all (i.e. there's no notion of *weak equivalence* at all, only of isomorphism) and (ii) even if we took $\mathfrak{s}$-modules to be the discrete version of some framework for derived algebra, it wouldn't be $\mathbb{S}$-modules because $\mathfrak{s}$-modules contain things like hypergroups, tropical rings, and commutative monoids, so their "derived theory" will need to be based in $(\infty,n)$-categories rather than $(\infty,1)$-categories, as $\mathbb{S}$-modules is.
I'll try to follow Connes and Consani and say $\mathfrak{s}$-modules instead of $\Gamma$-sets, but I'll almost certainly just start using them interchangeably at some point. So anyway, we've got $\mathfrak{s}$-modules and we've got $\mathbb{S}$-modules. First let's clarify the difference between $\mathfrak{s}$ and $\mathbb{S}$, as, in a sense, they are both the sphere spectrum, but I think this point of view is misleading. What we're calling $\mathfrak{s}$ here is the inclusion $\Gamma\hookrightarrow Set\_\ast$, whereas, for the time being at least, $\mathbb{S}$ is the unit of the $\infty$-category of spectra, if you like, or the suspension spectrum of $S^0$.
Recall that there is a model structure on $\Gamma$-spaces, called the stable model structure, whose fibrant objects are the very special $\Gamma$-spaces. Namely, those that satisfy the Segal condition, roughly $X[n\_+]\simeq X[1\_+]^n$ where the right hand side has $n$ factors, and such that $X[1\_+]$ is an abelian group (here I'm using $n\_+$ to denote the set with $n+1$ elements, one of which is the basepoint). These model "commutative topological groups up to homotopy," and therefore grouplike $\mathbb{E}\_\infty$-spaces, and therefore connective spectra. Now, notice that the inclusion $\Gamma\to FinSet\_\ast\hookrightarrow Set\_\ast\hookrightarrow sSet\_\ast$ does *not* define a special $\Gamma$-space. To see this, consider $1\_+\times 1\_+=\{\ast,1\}\times\{\ast,1\}=\{(\ast,\ast),(\ast,1),(1,\ast),(1,1)\}\cong3\_+\neq 2\_+$. So $\mathfrak{s}$ itself is not fibrant in $\Gamma$-spaces with the stable model structure and therefore doesn't correspond to any connective spectrum. Of course the way that we get around this in model categories is that we fibrantly replace. It's a theorem that when we fibrantly replace $\mathfrak{s}$ (thought of as a discrete $\Gamma$-space) we get $\mathbb{S}$.
There's a concrete way to describe fibrant replacement in this category, which is to formally invert the so-called Segal maps, i.e. localize at them, and group complete $X[1\_+]$ (thanks to Chris Schommer-Pries for explaining this part to me). In a very real way this deserves to be called "group completion." The first part is, basically, "commutative monoid completion" since it forces the multiplication to be defined everywhere and to be singly defined (an arbitrary $\Gamma$-space can be thought of as a sort of $\mathbb{E}\_\infty$-monoid object in which the multiplication is either multi-valued or only partially defined). The second part is formally adding inverses. The second part is what we usually think of as group completion of course, but if you're starting with something that's not even a commutative monoid, you've got to do a bit more.
So, if we believe that "fibrant replacement in the stable model structure" can reasonably be called group completion, then that's exactly what $\mathbb{S}$ is, it's the group completion of $\mathfrak{s}$ (or, if you accept Connes and Consani's contention that $\mathfrak{s}=\mathbb{F}\_1$, then $\mathbb{S}$ is the group completion of $\mathbb{F}\_1$, which is consistent with interpreting the Barratt-Priddy-Quillen as saying $K(\mathbb{F}\_1)\simeq \mathbb{S}$; notice that although Barratt-Priddy-Quillen says that $K(FinSet)\simeq \mathbb{S}$, it's still true that $K(\mathfrak{s}-Mod)\simeq \mathbb{S}$).
The category of $\Gamma$-spaces, or $\Gamma$-sets, has a symmetric monoidal structure coming from Day convolution. In this monoidal structure, $\mathfrak{s}$ is the monoidal unit. So everything in $\Gamma$-set deserves to be called an $\mathfrak{s}$-module, or an $\mathbb{F}\_1$-module. But now, given an $\mathfrak{s}$-module $M$, one can further say that $M$ itself is a monoid with respect to the Day convolution monoidal structure, and this is what Connes and Consani call an $\mathfrak{s}$-algebra. The key insight here for them is that most of the existing models of "$\mathbb{F}\_1$-algebra" are subsumed by this construction. In particular, commutative monoids give $\mathfrak{s}$-modules and semirings give $\mathfrak{s}$-algebras; a large class of abelian hypergroups and hyperrings embed into $\mathfrak{s}$-modules and $\mathfrak{s}$-algebras (for the two preceding examples we're using this fact that $\Gamma$-sets can be thought of as having either a partially defined or multivalued abelian group structure); and perhaps most excitingly, Durov's algebraic monads embed into $\Gamma$-sets, so we get things like tropical rings, and these generalized rings that Durov uses to do things like Arakelov geometry (Connes and Consani have also written about a connection between their $\mathfrak{s}$-modules and Borger's $\Lambda$-ring model for $\mathbb{F}\_1$-geometry, but I haven't tried to understand that; it's also not at all clear to me, or anyone else as far as I know, that Lorscheid's blueprint stuff, and all the things related to it, can be described in terms of Connes and Consani's work).
I should mention that one can take Durov's model of $\mathbb{F}\_1$ and map it (via the so-called assembly map of Lydakis that Connes and Consani use) to $\Gamma$-sets, and one does, as an object, get $\mathfrak{s}$ (I think). But, if I'm not mistaken, the modules over Durov's $\mathbb{F}\_1$, in Durov's setup, are less general than $\mathfrak{s}$-modules.
One can of course talk about "simplicial $\mathfrak{s}$-modules," but if we think about what a simplicial $\mathfrak{s}$-module is going to be, well it's the same as a $\Gamma$-space (where by space I mean simplicial set) so if one wishes to do homotopy theory in the usual way, I think one is still going to end up back at something like $\mathbb{E}\_\infty$-spaces, which loses all the examples that Connes and Consani are interested in (for instance, they seem pretty interested in this adèle class space construction, which gives a hyperring, so gets destroyed if you force the addition to be single-valued).
Moreover, recall that if we have an $\mathbb{E}\_\infty$-space $X$ then we get the associated $\Omega$-spectrum, i.e. connective $\mathbb{S}$-module, by taking the infinite delooping $HX=\{X,BX,B^2X,B^3X,\ldots\}$. This doesn't even make sense if $X$ isn't a very special $\Gamma$-space. So to do "homotopy theory over $\mathbb{F}\_1$," we have to be a bit more clever. I.e., prima facie, "derived $\mathbb{F}\_1$-algebra" or "derived $\mathbb{F}\_1$-geometry" is going to be very different from spectral algebra(ic geometry).
I think the way to get to derived $\mathbb{F}\_1$-algebra, and have it be more general than spectral algebra (i.e. avoiding group completion), is that we're going to have to work with something like complicial $\Gamma$-sets, rather than simplicial. The idea here is that simplicial sets are pretty good at modeling $(\infty,1)$-categories, but pretty bad at modeling $(\infty,n)$-categories, and if you want to take the infinite delooping of, say, a commutative monoid $M$, you're going to need $B^nM$ to be an $n$-category of some sort, because you're going to have one object with an $M$'s worth of non-invertible $n$-cells, and complicial sets seem like a decent way to model categories like this (although there are other options too of course, and I haven't worked out any of the details at all). But whatever your category of $\mathbb{F}\_1$-spectra is, there should be a group completion functor landing in $\mathbb{S}$-modules.
So if we're following Connes and Consani's proposed path to a proof of the Riemann Hypothesis, we can't work over $\mathbb{S}$, because all of the machinery that Connes and Consani want to use disappears (because we're group completing). In particular, with respect to your plan of taking an affine scheme over $\mathbb{F}\_1$ and producing the associated spectral scheme, what you're basically doing is group completing an $\mathbb{F}\_1$-algebra to some commutative connective ring spectrum $A$, then taking spec of that. But, again, this is destroying most of the tools Connes and Consani are using or are interested in.
If you wanted to start from scratch and say "Let's figure out how to rewrite Deligne's proof of the function fields Riemann Hypothesis by thinking of $H\mathbb{Z}$ as a curve over $\mathbb{S}$," well, I think you're going to struggle. I certainly wouldn't go on the record as saying it's impossible, but the sorts of machinery you'd want to use either don't exist over $\mathbb{S}$ or, at the time being at least, don't make sense (cf. Tyler Lawson's answer here: [Dedekind spectra](https://mathoverflow.net/questions/82485/dedekind-spectra/82624)). Although you should take this last assertion of mine with a large grain of salt. Might make sense to ask Lars Hesselholt or Bjørn Dundas or somebody about this.
|
18
|
https://mathoverflow.net/users/11546
|
407227
| 166,849 |
https://mathoverflow.net/questions/407213
|
14
|
For a model category $C$, I'm denoting $h\_\infty(C)$ the associated $\infty$-category (for example its Dwyer-Kan localization at weak equivalences, or if $C$ is simplicial the simplicial nerve of the category of bifibrant objects, or any other equivalent construction...).
If $C$ is a combinatorial model structure and $I$ is any small category, then the category $C^I$ carries projectives and injectives model structures ( that are clearly equivalent).
It then happens that (in this case) we have an equivalence $h\_\infty(C^I) \simeq h\_\infty(C)^I$ using one of these model structures on $C^I$.
So informally this model structure on $C^I$ does model all homotopy coherent I-diagram in $C$. I feel this is used in many places, but I never got a good understanding of that result. It seems like a very non-trivial "rectification" theorem (where you turn homotopy coherent diagram $ I \to C$ into actual functors). For example if I replace $C$ by a general relative category this is obviously not the case.
I know proofs of this fact, but they are all fairly indirect, with some technical steps, and I feel they don't really explain why such a result is true, or at least I don't get the explanation.
So, my question : can someone give some sort of intuition of why this is true or maybe a more "high level" proof that this is true ? Maybe someone just knows a simpler/more direct proof than the ones I may have seen ?
Basically, I'd be interested in hearing any good answer to the question why is it the case that $h\_\infty(C^I) \simeq h\_\infty(C)^I$ ?
Maybe a more precise way to ask this : What I'd be especially interested in is an explanation that would give some intuition of when more general statement of this kind are true. For e.g, I have little intuition on the following questions :
* Is it true if $C$ is a model category, that is not combinatorial, but for which I do have a satisfying model structure on $C^I$ ? Is there some condition I can add to make it true ?
* what if $C$ is only a fibration category or a cofibration category and $I$ is nice enough ? ( Okay here the answer is clearly No, so, I'm refining it in the next point)
* What if $C$ is a cofibration or fibration category with some additional conditions on infinite limits / colimits that among other things guarantee that $h\_\infty(C)$ has infinite limits/colimits ?
* What kind of conditions on a general relative category $C$ might make this true ?
|
https://mathoverflow.net/users/22131
|
On diagrams in model categories and rectification
|
If we look at the proof that I know of in the case this is true - see below - we need in fact that, for any small $1$-category $I$, $h\_\infty(C^I)$ has small (co)limits and that they can be computed termwise, and we need to prove the particular case where $I$ is discrete as a first step, which usually comes from a variant of calculus of fractions ($C$ could be an $\infty$-category to begin with). The proof can be formulated in a model free way once we have a good theory of pointwise Kan extensions.
Anyway, this is true for any model category (in Quillen's original sense) $C$ which has either small colimits or small limits. For a category with fibrations and weak equivalences $C$ (such as a category of fibrant objects à la Brown), and if small products of (trivial) fibrations between fibrant objects exist and are (trivial) fibrations (this already implies that $h\_\infty(C)$ has small limits), this is true if one of the following further conditions is satisfied:
* limits of countable towers of (trivial) fibrations between fibrant objects are (trivial) fibrations;
* the factorization of any map with fibrant codomain into a weak equivalence followed by a fibration can be made functorially.
In fact the last two conditions are only there to ensure that $C^I$ is a category with weak equivalences and fibrations defined level-wise (the tricky part being the existence of factorizations if ever these were not functorial in $C$ to begin with, but this is documented in the work of [Radulescu-Banu](https://arxiv.org/abs/math/0610009)). The principle of all the proofs I know of is that one proves that both $\infty$-categories $h\_\infty(C^I)$ and $h\_\infty(C)^I$ have small limits (it sufficices to prove this is true for $h\_\infty(C)$ for a generic $C$), and to prove that the comparison functor $h\_\infty(C^I)\to h\_\infty(C)^I$ commutes with small limits (=with small products and with pullbacks). One can then check fully faithfulness on objects which are homotopy right Kan extensions of objects of $C$ along an object of $I$ (seen as a functor from the terminal category), and prove through cofinality arguments that any $I$-indexed functor is an homotopy limit of such thing to finish the proof. This kind of approach survives very well if we allow $C$ to be an $\infty$-category to begin with (but $I$ must be a $1$-category), so that any reasonnable notion of model $\infty$-category (e.g. [Mazel-Gee's](https://arxiv.org/abs/1412.8411)) is eligible for such rectification theorem. The case where $I$ is a finite direct category holds with a much greater generality (no need to have small homotopy (co)limits, finite ones are good enough), but the idea of the proof is similar (with simpler technical arguments, though). References can be found in my [book](http://www.mathematik.uni-regensburg.de/cisinski/CatLR.pdf) on higher categories: Theorem 7.9.8 (with Example 7.9.6 and Remark 7.9.7. to get easy sufficient conditions in practice) if $I$ is small and Theorem 7.6.17 if $I$ is a direct finite category. It is striking to me that this kind of rectification is the best explanation of why homotopy limits in $C$ coincide with limits in $h\_\infty(C)$; see Remark 7.9.10. In the same direction, there is [Balzin's work](https://arxiv.org/abs/1803.00681) which is more general because it deals with sections of general fibred model structures but also less general because it restricts to the case where the indexing small category is Reedy. The case of a cofibration categories is also documented in [this paper](https://msp.org/agt/2018/18-6/p12.xhtml) of Tobias Lenz with a different proof from the one refered to above (he does this in the language of derivators, but it is easy to extract an $\infty$-categorical statement from his proof).
|
11
|
https://mathoverflow.net/users/1017
|
407229
| 166,850 |
https://mathoverflow.net/questions/407231
|
5
|
For a simple complex Lie algebra $\frak{g}$, let $V$ be an irreducible $\frak{g}$-module. Is it true that the weights of the non-zero weight vectors in $V$ are less than the highest weight vector and greater than the lowest weight vector with respect to the partial order on weights? If not, what is a simple counterexample?
|
https://mathoverflow.net/users/371382
|
Do weight vectors live between the highest and lowest weights?
|
One nice way to see this is using the [PBW theorem](https://en.wikipedia.org/wiki/Poincar%C3%A9%E2%80%93Birkhoff%E2%80%93Witt_theorem). Write $\mathfrak{g} = \mathfrak{n}\_- \oplus \mathfrak{h} \oplus \mathfrak{n}\_+$ in the usual way. Take an ordered basis of $\mathfrak{g}$ consisting of, first, a basis for $\mathfrak{n}\_-$, then a basis for $\mathfrak{h}$ and then a basis for $\mathfrak{n}\_+$. Every PBW monomial thus factors as a product $x\_- x\_0 x\_+$ where $x\_- \in U(\mathfrak{n}\_-)$, $x\_0 \in U(\mathfrak{h})$ and $x\_+ \in U(\mathfrak{n}\_+)$.
Now, let $v\_0$ be a highest weight vector in $V$. Then $U(\mathfrak{g}) v\_0$ is a subrep of $V$ which, since $V$ is simple, must equal $V$. So $V$ is spanned by $x\_- x\_0 x\_+ v$ for $x\_-$, $x\_0$ and $x\_+$ as above.
Now, if $x\_+$ is a positive degree monomial in $\mathfrak{n}\_+$, then $x\_+ v\_0=0$ since $v\_0$ is highest weight, so we can consider just the span of $x\_- x\_0 v$. And $v$ is an eigenvector for every $x\_0 \in U(\mathfrak{h})$, so we can consider just the span of $x\_- v\_0$ for $x\_- \in U(\mathfrak{n}\_-)$. In short, we have proven that $V = U(\mathfrak{n}\_-) v\_0$. But it is clear that acting on $v\_0$ by anything in $\mathfrak{n}\_-$ lowers the weight.
Incidentally, we have only used the easy part of the PBW theorem here, which is that the PBW monomials span $U(\mathfrak{g})$; we didn't need the hard part, which is that they are linearly independent.
|
6
|
https://mathoverflow.net/users/297
|
407236
| 166,852 |
https://mathoverflow.net/questions/407207
|
-1
|
Can someone explain to me the proof on page 7/20 of the original paper about potential games ([https://www.cs.tau.ac.il/~mansour/sem-game-02-03/monderer-potential-96.pdf](https://www.cs.tau.ac.il/%7Emansour/sem-game-02-03/monderer-potential-96.pdf))?
It is about why two potentials of a game G can only differ by a constant.
|
https://mathoverflow.net/users/434074
|
Why do two potentials of a game only differ by a constant?
|
This is what it looks like to me: to get $H(y)$, we change strategy profile $z$ into $y$ one player at a time, summing the changes in utility. For any exact potential $P$ (which is actually equation 2.2 with weights $1$), each change in utility is equal to $P(a\_{i-1}) - P(a\_{i})$. This is a telescoping sum, so $H(y) = P(y) - P(z)$. This didn't depend on which exact potential $P$ we chose, so $P\_1(y) - P\_1(z) = P\_2(y) - P\_2(z)$ for all $y$. In other words $P\_1(y) = P\_2(y) + c$ where $c = P\_1(z) - P\_2(z)$.
|
1
|
https://mathoverflow.net/users/29697
|
407241
| 166,854 |
https://mathoverflow.net/questions/407187
|
1
|
Consider the symplectic group $\text{Sp}\_{2g}(\mathbb{Z})$ over the integers. It has a classical root system $C\_g$ and associated root subgroups $U\_\varphi$ for $\varphi\in C\_g$. These subgroups are defined by the [Bourbaki tables](https://books.google.nl/books?id=oCO0xOzNLhAC&pg=PA205&redir_esc=y#v=onepage&q&f=false).
Denote by $2U\_\varphi$ the set of matrices $A^2$ where $A\in U\_\varphi$. The symplectic group also has a maximal torus $T$, consisting of the diagonal matrices in $\text{Sp}\_{2g}(\mathbb{Z})$.
The kernel $\Gamma(2)$ of the canoncical reduction map $\text{Sp}\_{2g}(\mathbb{Z})\to\text{Sp}\_{2g}(\mathbb{Z}/2\mathbb{Z})$ is the principal congruence subgroup of level $2$.
**The question**
I want to show that $\Gamma(2)$ is the group generated by $T$ and the $2U\_\varphi$. The inclusion $\langle T,2U\_\varphi:\varphi\in C\_g\rangle\subset \Gamma(2)$ is trivial, by definition of the root subgroups and because the only nonzero entries of any matrix in $T$ are $1$ and $-1$. How do I prove the other inclusion?
|
https://mathoverflow.net/users/nan
|
The principal congruence subgroup of the symplectic group over the integers
|
If we assume that $g \geq 2$, then it is known by a Theorem of Tits
( Tits, Jacques :
Systèmes générateurs de groupes de congruence. (French. English summary)
C. R. Acad. Sci. Paris Sér. A-B 283 (1976), no. 9, Ai, A693–A695)
that the group $\Delta$ generated by $T$ and $2U\_{\phi}$ (your notation) has finite index in $Sp\_{2g}(\mathbb Z)$. You can now use the fact that $Sp\_{2g}(\mathbb Z)$ has the congruence subgroup property, to conclude that $\Delta$ is a congruence subgroup; it is easily seen that its closure in the congruence completion $Sp\_{2g}(\widehat{\mathbb Z})$ of $Sp\_{2g}(\mathbb Z)$ contains $\prod Sp\_{2g}(\mathbb Z\_p)$, where the product is over all odd primes. At the prime 2, the group $\Delta$ and $\Gamma (2)$ both have the same closure; we then conclude that $\Gamma (2)=\Delta$ (if two congruence subgroups have the same congruence closure, then they are the same).
When $g=1$, this is a well known result (and is actually proved in Ahlfors' book on complex analysis; the group $\Gamma (2)$ modulo centre is interpreted there as the fundamental group of ${\mathbb P}^1\setminus \{0,1,\infty \}$).
|
3
|
https://mathoverflow.net/users/23291
|
407244
| 166,855 |
https://mathoverflow.net/questions/405335
|
4
|
While browsing through Davenport's lecture notes "Analytic methods for diophantine equations and Diophantine inequalities", near the end of chap 8 I came across the statement that for the equation
$c\_1 x\_1^k + ... + c\_s x\_s^k = 0$,
if the $c\_i$ are non-zero and, if $k$ is even, the sign of $c\_i$ are not all the same, then this equation has infinitely many integer solutions if $s\ge k^2+1$. Davenport then noted that the bound $k^2+1$ is optimal if $k+1$ is prime, and that "for most values of $k$ smaller value than $k^2+1$ will suffice.
Question: What's the current record for $s$ for small $k = 3, 5, 7, 8, ...$ " ? What about conjectural/expected "optimal" value for $s$ for small $k$?
THANKS!
|
https://mathoverflow.net/users/66397
|
bounds for circle method
|
In general, what the circle method gives you is an asymptotic formula for the number of solutions to the equation, which will be of the form $C P^{s-k}$ for some non-negative $C$. We are currently (with the help of recent advances on Vinogradov's mean value theorem) able to establish such asymptotic formulae when $s>k^2 + O(k)$ (see Wooley 2012 IMRN).
If one doesn't require an asymptotic formula and is instead happy with infinitely many solutions, it is possible to restrict the base set from the integers to smooth numbers (i.e. no large prime divisors). One can then establish a similar asymptotic formula for the number of solutions in as few as $s>k \log k +O(k \log \log k)$ variables.
Generally, the expectation is that such an asymptotic formula should hold whenever $s>2k$, and possibly (with a diverging factor $C$) when $s>k$. However, as things stand this is very much pie in the sky territory!
The elephant in the room, however, is the constant $C$. In Davenport's book you can see that $C$ has a representation as the product over the densities of solutions to your equation over all local fields $\mathbb R$ and $\mathbb Q\_p$, and this is where the real bottleneck sits. The general rule of thumb is that you can expect a value that is smaller than $k^2+1$ whenever you can prove that there are no $p$-adic obstructions.
Understanding the details and particulars of this last question is in general still open, but Mike Knapp and Hemar Godinho have published papers on this, which you could check out.
|
4
|
https://mathoverflow.net/users/435788
|
407257
| 166,859 |
https://mathoverflow.net/questions/407253
|
2
|
A polynomial in the complex variable $z$, whose coefficients are themselves complex polynomials in another complex variable $a$, looks like
$$
f\in\Bbb C[A][Z],\;\;f(z,a)=c\_0(a)+c\_1(a)z+\cdots+c\_n(a)z^n
$$
with $c\_j\in\Bbb C[A]$ of degree less or equal than $1$. Assume not all $c\_j$ are constant and the leading coefficient $c\_n\equiv1$.
I need to prove that the roots of $z\mapsto f(z,a)$ are simple for all $a\in\Bbb C$ except for a finite number of $a$.
A root of $f(\cdot,a)$ is simple if and only if it is not a root of $f'(\cdot,a)$ and this happens if and only if the [resultant](https://en.wikipedia.org/wiki/Resultant) $P(a)$ of $f(\cdot,a),f'(\cdot,a)$ doesn't vanish.
Now $P\in\Bbb C[A]$, hence if it is not constantly $0$, then the claim is true (if $P$ is non constant then it vanishes for a finite number of values of $a$, while if it is constant and non zero, $f(\cdot,a)$ and $f'(\cdot,a)$ don't share a zero for any $a$).
Is it always possible to exclude the case $P\equiv0$?
|
https://mathoverflow.net/users/70148
|
Simple zeroes of complex polynomial $f(\cdot,a)$: condition on $P(a)=\operatorname{Res}_z(f,f')$
|
No, it's not true, as is shown by the polynomial $f(z,a):=(z-a)(z-1)^2$.
|
3
|
https://mathoverflow.net/users/24309
|
407271
| 166,862 |
https://mathoverflow.net/questions/407266
|
4
|
Let $A$ be a differentially graded augmented algebra. Then $\mathbf{B}A$ can be equipped with the structure of a coalgebra. This is proved in, for example, Loday and Vallette's book on *Algebraic Operads.*
The $n$-lab page <https://ncatlab.org/nlab/show/bar+and+cobar+construction> points out that the coalgebra structure can be understood as showing all the possible ways to decompose a path of length $n$ into pairs of shorter paths of lengths $(p,q)$ with $p+q=n$. This helps but I still don't really have intuition.
Many books *define* $\mathbf{B}A$ as follows. First, we can construct the free (conilpotent) coalgebra on $\mathbf{B}A$. Now, define the differential using the universal property of the free coalgebra, and check that it happens to satisfy the law $d^2=0$; so we have a DG coalgebra.
This makes sense, of course. However even if you had no idea that $\mathbf{B}A$ was supposed to be a coalgebra, there's still a very natural differential to put on that chain complex, which comes from applying the Moore normalization to the simplicial object constructed using this process. <https://ncatlab.org/nlab/show/two-sided+bar+construction>
So my question is, if I look at $\mathbf{B}A$ as the Moore normalization of the two-sided bar construction $B(1,A,1)$ (where $A$ acts on $1$ as a module on both the left and the right) why should we expect this to have a coalgebra structure. Or, alternatively, if you prefer to view $\mathbf{B}A$ as a coalgebra a priori on top of which we can put a unique compatible differential, why does this differential agree with the one given by $B(1,A,1)$?
I think I am either looking for geometric intuition for why this should be true, or ideally kind of nice categorical argument by which we can construct the comultiplication and counit on $B(1,A,1)$.
|
https://mathoverflow.net/users/112756
|
Why is the bar construction of a DG algebra a coalgebra?
|
This is the type of question with multiple correct answers, because, as you say, it depends very much on what you think the bar construction "is" initially.
You say you want to think of $\mathbf{B}A$ as $B(1,A,1)$. Well, I don't know what you think $B(X,A,Y)$ "is", but one thing that it "is" is a specific nice resolution of the derived tensor product $X \otimes\_A^{\mathbb{L}} Y$.
From this perspective, your question is a version of: is there a good reason for $1 \otimes\_A 1$ to be a coalgebra? In other words, can we produce a map $$1\otimes\_A 1 \to (1 \otimes\_A 1) \otimes (1 \otimes\_A 1)?$$
Yes. The left hand side is equivalent to $1 \otimes\_A A \otimes\_A 1$, whereas the right hand side is equivalent to $1 \otimes\_A 1 \otimes\_A 1$. The (augmentation) ring homomorphism $\epsilon: A \to 1$ that you start with is in particular an $A$-bimodule homomorphism, so you do indeed have the map
$$ 1 \otimes\_A \epsilon \otimes\_A 1 : 1 \otimes\_A A \otimes\_A 1 \to 1 \otimes\_A 1 \otimes\_A 1.$$
Then coassociativity is just that you apply $\epsilon$ in "different spots". Oh, and it requires associativity of $\otimes\_A$, which I have already suppressed from the notation.
Actually, because $B(-,A,-)$ is a derived tensor product, the associativity of $\otimes\_A$ could a priori require lots of coherence data — it could have been merely $\mathsf{A}\_\infty$ rather than associative. The result is that you could have concluded that $B(1,A,1)$ was merely a co$\mathsf{A}\_\infty$-coalgebra, rather than a coassociative coalgebra. That it is indeed coassociative is a niceness property of the specific resolution that you chose.
|
6
|
https://mathoverflow.net/users/78
|
407272
| 166,863 |
https://mathoverflow.net/questions/407261
|
3
|
Let $M$ be a Riemannian manifold and $S \subset M$ a compact submanifold of strictly lower dimension. Does every smooth function on $S$ extend to a harmonic function on a neighborhood of $S$?
|
https://mathoverflow.net/users/409915
|
Dirichlet-type condition on Riemannian manifold
|
This is not possible in general, for example because of constraints related to the analyticity of the extension.
For a concrete example, let $M = \mathbf{R}^2$, and $S \subset \mathbf{R}^2$ be a smooth, simple closed curve in the plane so that in the unit disc
\begin{equation}
S \cap D\_1 = \{ x\_2 = 0 \} \cap D\_1 = (-1,1) \times \{ 0 \}.
\end{equation}
On $S$ define a function $u\_0$ so that on the portion lying inside $D\_1$,
\begin{equation}
u\_0: x\_1 \in D\_1 \cap S \mapsto \mathrm{e}^{-1/x\_1^2}.
\end{equation}
This is not analytic, whereas a harmonic extension to a neighbourhood of $S$, say $u$ would be.
The existence of such a harmonic extension $u$ is therefore absurd. In conclusion: no matter how small $\delta > 0$ is chosen, there is no harmonic function $u: D\_\delta \to \mathbf{R}$ extending $u\_0$, let alone a function defined on a neighbourhood of $S$.
|
5
|
https://mathoverflow.net/users/103792
|
407278
| 166,864 |
https://mathoverflow.net/questions/407218
|
6
|
According to the prescription of [functorial quantum field theory](https://ncatlab.org/nlab/show/functorial+field+theory), a quantum field theory can be viewed as a monoidal functor from some monoidal category of $n$-cobordisms to some monoidal category of vector spaces (typically a category of Hilbert spaces).
Now, what does the functor category of QFTs look like? More precisely, I am having trouble even conceptualizing what "natural transformation between QFTs" would be. I am also interested in the specific case of the category of CFTs (a CFT here would be a monoidal functor from a conformal cobordism category).
|
https://mathoverflow.net/users/432464
|
Functor category of quantum field theories?
|
The question of what "natural transformation of QFTs" should be is a somewhat subtle one. The issue is most apparent if you work with TQFTs, but it doesn't completely go away if you work with dynamical theories.
Suppose, for example, that you have two $n$D TQFTs $\cal{V},\cal{W}$ and an $(n-1)$D closed manifold $\Sigma$, and let $\bar{\Sigma}$ be a copy of $\Sigma$ in which the direction of time has been reversed. Then you can evaluate your TQFTs to produce vector spaces $\cal{V}(\Sigma), \cal{W}(\Sigma)$. Recall that these are finite-dimensional vector spaces because of the existence of "elbow" cobordisms $\eta\_\Sigma : \emptyset \to \Sigma \sqcup \bar{\Sigma}$ and $\epsilon\_\Sigma : \bar{\Sigma} \sqcup \Sigma \to \emptyset$. If you apply $\cal V$ to these, you build an isomorphism $\cal V(\bar\Sigma) \cong V(\Sigma)^\*$, the linear dual.
Now consider a natural transformation of monoidal functors $f : \cal{V} \to \cal{W}$. Evaluated on $\Sigma$, you would see a linear map $f(\Sigma) : \cal{V}(\Sigma) \to \cal{W}(\Sigma)$, and evaluated on $\bar\Sigma$ would would see a linear map $f(\bar\Sigma) : \cal{V}(\bar \Sigma) \to \cal{W}(\bar \Sigma)$. Symmetric monoidality of $f$ tells you furthermore that
$$ f(\Sigma \sqcup \bar\Sigma) = f(\Sigma) \otimes f(\bar\Sigma) : \cal V(\Sigma)\otimes \cal V(\bar\Sigma) \to \cal W(\Sigma)\otimes \cal W(\bar\Sigma).$$
Now, naturality of $f$ tells you that any time you have a cobordism, you should get an equation. So, for example, $f(\epsilon\_\Sigma)$ is the equation
$$ (\*)\quad \cal W(\epsilon\_\Sigma) \circ (f(\Sigma) \otimes f(\bar\Sigma)) = \cal V(\epsilon\_\Sigma). $$
Finally, note that $f(\bar\Sigma) : \cal V(\Sigma)^\* \cong \cal V(\bar\Sigma) \to \cal W(\bar\Sigma) \cong \cal W(\Sigma)^\*$ has an "adjoint" map $f(\bar\Sigma)^\* : \cal W(\Sigma) \to \cal{V}(\Sigma)$.
After unpacking, the above translations then tell you that:
$$ f(\bar\Sigma)^\* \circ f(\Sigma) : \cal{V}(\Sigma) \to \cal{V}(\Sigma)$$
is the identity, and similarly on the other side.
As a result, the natural transformation $f$ assigns isomorphisms to all $(n-1)$-manifolds. Thus it is a natural isomorphism $\mathcal{V} \cong \mathcal{W}$.
Note that actually we used nothing about "vector spaces". The problem would any time you have some amount of duality data in your source category. In particular, the problem does not go away if you allow your target to be an even-higher category. If the target is a higher category, then "naturality" would impose not an equality in (\*) but an isomorphism, but you would still end up producing an isomorphism $f(\bar\Sigma)^\* \circ f(\Sigma) \cong \mathrm{id}\_{\cal V(\Sigma)}$.
Nontopologically you might not have elbow cobordisms of "zero length", but you will have some of positive length, and using them still lets you get molified versions of the equation $f(\bar\Sigma)^\* \circ f(\Sigma) = \mathrm{id}\_{\cal V(\Sigma)}$ where "$f(\bar\Sigma)^\*$" gets replaced by a length-depended object, and $\mathrm{id}\_{\cal V(\Sigma)}$ gets replaced by the evolution operator for that amount of time.
All together, you see that in order to have a really good (i.e. nontrivial) notion of "natural transformation of QFTs", you really need a version of natural transformations in which "naturality" is imposed only by a morphism, and not an isomorphism.
The bicategory theorists invented such a notion decades ago: there are "lax natural transformations" and "oplax natural transformations". In a lax natural transformation $f : \cal V \to \cal W$ of functors, if you have a 1-morphism $\alpha : X \to Y$, then you get a 2-morphism $f(\alpha) : {\cal W}(\alpha) \circ f(X) \Rightarrow f(Y) \circ {\cal V}(\alpha)$. In an oplax natural transformation, the 2-morphism goes the other direction. Note that "lax" and "oplax" are the two resolutions of a slight conflict: should $f(\alpha)$ go "in the $f$-direction" (i.e. with a $\cal V$ in the domain and a $\cal W$ in the codomain) or "in the $\alpha$ direction" (i.e. with an $X$ in the domain and a $Y$ in the codomain)?
This same conflict propagates to higher categories, and the combinatorics was pretty fun to work out. Scheimbauer and I did so in our paper [(Op)lax natural transformations, twisted quantum field theories, and "even higher" Morita categories](https://arxiv.org/abs/1502.06526).
|
8
|
https://mathoverflow.net/users/78
|
407280
| 166,865 |
https://mathoverflow.net/questions/407219
|
4
|
Consider the following wave-type equation,
$$u\_{tt}-\frac{2}{t}u\_t-\Delta u=g(t,x)$$
where $(t,x)\in [\epsilon, 1]\times \mathbb{R}^3$ for some $\epsilon>0.$ Furthermore assume that $(u,u\_{t})=(0,0)$ at $t=\epsilon.$ Then define the energy of the solution $u$ as follows,
$$E(t) = E(u(t))=t^2 \int (u\_t)^2+|\nabla u|^2 dx.$$
My **goal** is to derive a time decay estimates on the energy $E$.
As usual, we compute the time derivative of $E$. Thus,
$$\frac{d}{dt}E(t) = \frac{2E}{t} + 2t^2 \int u\_t u\_{tt} + \nabla u\cdot \nabla u\_t.$$
Using integration by parts we get,
$$\frac{d}{dt}E(t) = \frac{2E}{t} + 2t^2 \int u\_t \left(g+\Delta u + \frac{2}{t}u\_t\right) - u\_t\Delta u \\
=\frac{2E}{t} + 2t^2 \int u\_t g + 4t \int (u\_t)^2.$$
For the third term, we can simply control it as follows,
$$4t \int (u\_t)^2 \leq 4t\int (u\_t)^2 + |\nabla u|^2\leq \frac{4E(t)}{t}.$$
For the second term, using the above estimate, we get
$$ 2t^2 \int u\_t g\leq 2t^2 \|u\_t\|\_{L^2}\|g\|\_{L^2}\leq 2 t \sqrt{E(t)}\|g\|\_{L^2}$$
which implies,
$$\frac{dE}{dt}\leq \frac{6E}{t} + 2t \sqrt{E}\|g\|\_{L^2}.$$
I am not sure how to proceed further. I would like to apply Gronwall's inequality but the right-hand side does not seem very nice. I started with $g\in L^2$, but perhaps I need more regularity for the function $g.$ Any ideas here will be much appreciated.
|
https://mathoverflow.net/users/68232
|
Energy estimates for nonlinear wave type equation
|
Define using $H(t) = \int (u\_t)^2 + |\nabla u|^2 ~dx $ the standard energy.
Taking the time derivative you find
$$ \frac{d}{dt}H(t) = 2 \int u\_t( g + \frac{2}{t} u\_t)$$
Writing $\|\cdot \|$ for the $L^2$ integral, you have then
$$ \frac{d}{dt} H(t) \geq - 2 \|u\_t\|\cdot \|g\| + \frac{4}{t} \|u\_t\|^2 \tag{A}$$
by Cauchy-Schwarz, and then by AM-GM you get
$$ \frac{d}{dt} H(t) \geq - \frac{4}{t} \|u\_t\|^2 - \frac{t}{4} \|g\|^2 + \frac{4}{t} \|u\_t\|^2 = - \frac{t}{4} \|g\|^2 $$
So integrating you find
$$ H(1) + \int\_\epsilon^1 \frac{t}{4} \|g\|^2 ~dt \geq H(\epsilon) $$
So if you assume that $$\tag{\*} \int\_0^1 \int\_{\mathbb{R}^3} |g|^2 t ~ dx~dt $$is finite, the above shows that $H(\epsilon)$ is uniformly bounded as $\epsilon \searrow 0$ and hence what you defined as the weighted energy $E(\epsilon)$ converges to zero at rate $\epsilon^2$.
---
If you only assume (\*), then it is impossible to prove that the standard energy decays with any rate like $\epsilon^\beta$. This is because of the ODE examples
$$ y(t) = t^{1+\alpha}$$
with $\alpha\in (0,1)$. You find that
$$ \ddot{y} - \frac{2}{t} \dot{y} = (1+\alpha)(\alpha - 2) t^{\alpha - 1} =: g(t)$$
and this $g(t)$ satisfies $\int\_0^1 t |g(t)|^2 ~dt < \infty$. So the energy ($(\dot{y})^2$) cannot be proven to go to zero as $t^\beta$ for any $\beta > 0$. By spatial truncation this ODE example can be upgraded to a bona fide solution of the wave equation you wrote down.
---
On the other hand, if instead of (\*) you know something stronger about $g$, then some decay may be concluded.
For example: suppose you know that $\int\_0^1 \|g\|^2 ~dt$ is bounded. Then starting from (A) you can instead find
$$ \frac{d}{dt} H(t) \geq \left( \frac4t-1\right) \|u\_t\|^2 - \|g\|^2 $$
This would give you
$$ H(1) + \int\_0^1 \|g\|^2 ~dt \geq H(\epsilon) + \int\_{\epsilon}^1 \frac{3}t\|u\_t\|^2 ~dt $$
This gives you an integrated decay of time component of the standard energy.
But this would depend on what you want to know and what you are willing to provide for (in terms of properties of $g$).
|
2
|
https://mathoverflow.net/users/3948
|
407284
| 166,866 |
https://mathoverflow.net/questions/281771
|
10
|
I couldn't add to the well-written $n$Lab page about [relative adjoints](https://ncatlab.org/nlab/show/relative+adjoint+functor), so let me start taking that definition for granted.
I have a few questions about how the classical results on adjoints remain true, even in this fairly asymmetric setting.
I would like to understand better the meaning of this structure (e.g. when, how and why does it arise "in practice"). I collect here a few questions I have no clue how to attack.
1. Let's assume that there are relative adjunctions $f\dashv\_j g$ and $f' \,{}\_{j'}\!\!\dashv g'$; we can then arrange a pair of commutative triangles
$$
\begin{array}{ccc}
\bullet & \overset{f}\to & \bullet\\
\hskip-4mm g\uparrow & \nearrow \\
\bullet &&
\end{array}
\qquad \epsilon : fg \Rightarrow j
$$
$$
\begin{array}{ccc}
&&\bullet \\
&\nearrow& \hskip5mm \uparrow g'\\
\bullet &\underset{f'}\to& \bullet
\end{array}
\qquad \eta : j' \Rightarrow g'f'
$$
assume, now, that there exists a natural transformation $j\Rightarrow j'$; under which assumptions the resulting pasting square is filled by an invertible 2-cell?
2. Assume that
$$
g\_1 \dashv\_{j\_1} f \;{}\_{j\_2}\!\!\dashv g\_2
$$
are relative adjoints; in this particular case it is possible to build a 2-cell $j\_2 g\_1\Rightarrow g\_2 j\_1$ pasting the unit of $f \;{}\_{j\_2}\!\!\dashv g\_2$ with the counit of $g\_1 \dashv\_{j\_1} f$. Under which conditions is this 2-cell an iso? In the same situation, the composition $j\_1 j\_2$ is parallel to $f$; is there a "comparison 2-cell" to/from $f$, and if there is, under which conditions it is invertible?
3. The $n$Lab says that $f \; {}\_j\!\!\dashv g$ if $f\cong \text{LIFT}\_gj$ (absolute lift); is this an if**f**? This relation means that $g$ uniquely determines $f$; the converse is not true in general: does it mean that there can be two non-isomorphic $g,g'$ such that $f \; {}\_j\!\!\dashv g$ *and* $f \; {}\_j\!\!\dashv g'$? Is there an instructive example of this? What is the structure, if any, of the class $\{g\mid f \; {}\_j\!\!\dashv g\}$?
4. If $j$ is an isomorphism and $f \;{}\_{j}\!\!\dashv g$, then this means that $f\dashv j^{-1}g$; is this condition any different from the particular case $f\dashv g = f \;{}\_1\!\!\dashv g$? Is there a nontrivial, instructive example of this situation? I am confused by the loss of uniqueness outlined in the previous point...
5. One of the reasons why relative adjunctions are useful is that they model weighted co/limits: writing the weight in a suitable form, one has that $j\otimes f \dashv\_j \hom(f,1)$, which is, well, quite obvious if you write
$$
{\bf hom}(j\otimes f,1) \cong {\bf hom}(j, \hom(f,1))
$$
how far can this analogy be pushed? (write $j\otimes(j'\otimes f) \cong \dots$: what does it mean for the associated relative adjunction?)
6. What is the meaning of relations like $1\_A \dashv\_j g$ for suitable functors? such a situation gives an isomorphism of profunctors $\hom \cong \hom(j,g)$, so in some sense finding "all $j$-right adjoints of the identity" means finding all factorizations of the identity profunctor $\hom\_A$. How far can this equivalence be pushed? What is the formal meaning of this analogy in terms of profunctor theory?
7. What does the invertibilty of the relative unit $\eta : j\Rightarrow gf$ imply for $g,f$ ($f$ is "relatively fully faithful"?!). Dual question for the counit.
|
https://mathoverflow.net/users/7952
|
What is known about relative adjunctions?
|
1. Using the answer to (7), if the natural transformation $J \Rightarrow J'$ is invertible, and $G$ and $F'$ are fully faithful, then the pasting square will commute up to natural isomorphism (assuming $J$ is dense and fully faithful).
2. This condition was called being a "symmetric lift" in Lewicki's [Categories with New Foundations](https://www.repository.cam.ac.uk/handle/1810/311541). Again using the answer to (7), if $F$ and $G\_1$ are fully faithful, then the composite natural transformation will be invertible (assuming $J$ is dense and fully faithful). As far as I can tell, the second part of the question is ill-formed.
3. Yes, $F$ is a $J$-relative left-adjoint to $G$ if and only if $f \cong \mathrm{lift}\_G J$ and this left lifting is absolute (this is an easy exercise). However, this is not true for enriched relative adjunctions. An example of non-unique relative right adjoints is given in Ulmer's *Properties of Dense and Relative Adjoint Functors*: let $J : \mathrm{AbGrp}\_f \to \mathrm{AbGrp}$ be the inclusion of finite Abelian groups in all Abelian groups and let $L : \mathrm{AbGrp}\_f \to \mathrm{Vect}\_{\mathbb Q}$ be the zero functor. Then $L$ is $J$-relative left adjoint both to $0 : \mathrm{Vect}\_{\mathbb Q} \to \mathrm{AbGrp}$ and to the forgetful functor $V : \mathrm{Vect}\_{\mathbb Q} \to \mathrm{AbGrp}$. I'm not aware of any nice structure on the set of $J$-relative right adjoints to a given functor.
4. Adjunctions relative to isomorphisms are the same as adjunctions relative to the identity. There is no lack of uniqueness here, since isomorphisms are dense, and $J$-relative right adjoints are unique when $J$ is dense.
5. Weighted colimits as relative adjunctions are treated in §4 of Street–Walter's *Yoneda structures on 2-categories*, where they are called "weak indexed colimits". See there for various useful lemmas making use of this characterisation.
6. If the identity on $\mathbf A$ is $J$-relative left coadjoint to $G : \mathbf B \to \mathbf A$, then there is a natural isomorphism $\mathbf A(a, Jb) \cong \mathbf A(a, Gb)$ which, by Yoneda, just says that $J \cong G$.
7. Assume $J$ is dense and fully faithful. If $L$ is $J$-relative left adjoint to $R$, then the unit $\eta : J \Rightarrow RL$ is an isomorphism if and only if $L$ is fully faithful (easy exercise). Dually, assume $J$ is codense and fully faithful. If $L$ is $J$-relative left coadjoint to $R$, then the counit $\epsilon : LR \Rightarrow J$ is an isomorphism if and only if $R$ is fully faithful.
|
3
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https://mathoverflow.net/users/152679
|
407287
| 166,868 |
https://mathoverflow.net/questions/407250
|
3
|
A ternary $C^{\ast}$-ring is a complex Banach space $X$, equipped with a ternary product $[\cdot,\cdot,\cdot]:X^3 \to X$ which is linear in outer variables and conjugate linear in middle variable. Also $X$ is associative i.e. $$[[a,b,c],d,e]=[a,[d,c,b],e]=[a,b, [c,d,e]].$$ Moreover, $\lVert[a,a,a]\rVert= \lVert a\rVert^3$ and $\lVert[a,b,c]\rVert \leq \lVert a \rVert \lVert b\rVert\lVert c\rVert$.
>
> Does there exist a ternary $C^{\ast}$-ring which is not an operator space?
>
>
>
One obvious class of examples of ternary $C^{\ast}$-rings is the class of ternary rings of operators but they all are operator spaces.
|
https://mathoverflow.net/users/129638
|
Example of a ternary $C^{\ast}$-ring which is not an operator space
|
According to Zettl [1], a *ternary ring of operators* (TRO) is a
ternary $C^\*$-ring which is isomorphic to a closed
subspace $X\subseteq B(H)$, such that $XX^\*X\subseteq X$, equipped with the ternary multiplication
$$
[x,y,z] := xy^\*z.
$$
On the other hand, an *anti-TRO* is a ternary $C^\*$-ring defined as above, except that the multiplication operation is
$$
[x,y,z] := -xy^\*z.
$$
It is a fundamental result of Zettl [1] that every ternary $C^\*$-ring $X$ decomposes uniquely as
$$
X=X\_+\oplus X\_-,
$$
where $X\_+$ is a TRO, and $X\_-$ is an anti-TRO .
It seems to me that the reading of the question posed by the OP that makes the most sense is by taking the expression
"operator space" to mean a TRO. In this case the answer is yes, there does exist a ternary $C^\*$-ring which is not a
TRO: just take any non-zero anti-TRO. For an even more concrete example, take $X=M\_{n\times m}({\bf C})$, with ternary
multiplication $[x,y,z] := -xy^\*z$.
On the other hand, if one takes the expression "operator space" for its face value, Zettl's result implies that every
ternary $C^\*$-ring is an operator space in an even more canonical form than suggested by user @YemonChoi: write
$X=X\_+\oplus X\_-$, embedd $X\_+$ in $B(H\_+)$, and $X\_-$ in $B(H\_-)$, whence
$$
X\subseteq B(H\_-\oplus H\_+).
$$
This embeding preserves the operator space structure (norms on matrix algebras) that a TRO canonical possesses.
It is interesting to remark that if you change the (binary) multiplication operation on a $C^\*$-algebra by
$$
x\circ y := -xy,
$$
then the resulting object is strictly speaking a new C\*-algebra, but it is isomorphic to the old one. The isomorphism
is simply $a\mapsto -a$.
However, if you change the (ternary) multiplication on a ternary $C^\*$-ring by inserting a minus sign as above, then the
map $a\mapsto -a$ is no longer an isomorphism, essentially because 2 is even and 3 is odd!
Indeed, Zettl's uniqueness result tells you that the new ternary $C^\*$-ring
might not be isomorphic to the old one at all!
---
EDIT: Here are some details of Zettl's proof which might shed some light into the reason an anti-TRO not isomorphic to a TRO.
Given a ternary $C^\*$-ring $X$, let $A$ be the closed linear span within $B(X,X)$ (bounded operators on $X$) of the set
of operators of the form
$$
T\_{y, z}:x\in X\mapsto [x,y,z]\in X,
$$
as $y$ and $z$ range in $X$. It is easy to see that $A$ is a Banach algebra, and Zettl proves that $A$ is indeed a
$C^\*$-algebra for a unique involution operation "$^\*$" satisfying
$$
T\_{y, z}^\* = T\_{z, y}.
$$
Given this, it is clear that an operator of the form $T\_{y,y}$ is self-adjoint but the key question is whether or not
this is moreover positive.
If $X$ is a TRO, then $T\_{y, y}\geq 0$, while in the anti-TRO case, one has that $T\_{y, y}\leq 0$.
In other words, the positivity of $T\_{y, y}$ is a signature of TRO's not shared by their anti-TRO cousins.
[1] *Zettl, Heinrich*, [**A characterization of ternary rings of
operators**](http://dx.doi.org/10.1016/0001-8708(83)90083-X), Adv. Math. 48, 117-143
(1983). [ZBL0517.46049](https://zbmath.org/?q=an:0517.46049).
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4
|
https://mathoverflow.net/users/97532
|
407306
| 166,874 |
https://mathoverflow.net/questions/407302
|
1
|
Here is the definition of Hajós construction.
>
> Let $G$ and $H$ be two undirected graphs, $vw$ be an edge of $G$, and $xy$ be an edge of $H$.
> Then the Hajós construction forms a new graph that combines the two graphs by identifying vertices $v$ and $x$ into a single vertex, removing the two edges $vw$ and $xy$, and adding a new edge $wy$.
>
>
>
In many papers about Hajós construction, they consider the following statement as a 'fact':
>
> If $G$ and $H$ are $k$-critical, then applying Hajós construction to $G$ and $H$ obtains a $k$-critical graph.
>
>
>
But I have no idea with how to prove it.
Would you help me?
|
https://mathoverflow.net/users/384338
|
If $G$ and $H$ are $k$-critical, then applying Hajós construction to $G$ and $H$ makes $k$-critical graph
|
Let $G$ and $H$ be $k$-critical graphs and $G +\_h H$ be the Hajós construction applied to $G$ and $H$ with respect to $vw \in E(G)$ and $xy \in E(H)$. Since $G$ and $H$ are $k$-critical, $G-vw$ and $H-xy$ both have $(k-1)$-colourings. Permute the colours so that the colour of $v$ is the same as the colour of $x$. Then recolour $w$ with the new colour $k$. This yields a $k$-colouring of $G +\_h H$. Towards a contradiction, suppose $c$ is a $(k-1)$-colouring of $G +\_h H$. Since $\chi(G)=k$, we must have $c(v)=c(w)$. Similarly, since $\chi(H)=k$, $c(x)=c(y)$. This implies $c(w)=c(y)$, which contradicts that $c$ is a proper colouring. Thus, $\chi(G +\_h H)=k$.
It remains to show that for every edge $e$ of $G +\_h H$, removing $e$ decreases the chromatic number of $G +\_h H$. If $e=wy$, then the previous argument (without the recolouring step) shows that $\chi( (G +\_h H) \setminus e) \leq k-1$. Therefore, by symmetry, we may assume that $e \in E(G)$. Since $G$ is $k$-critical, $G \setminus e$ has a $(k-1)$-colouring $c$. Since $H$ is $k$-critical, $H \setminus xy$ has a $(k-1)$-colouring $c'$. Since $\chi(H)=k$, $c'(x)=c'(y)$. Since $vw \in E(G \setminus e)$, $c(v) \neq c(w)$. Therefore, permuting the colours so that $c(v)=c'(x)$ gives a $(k-1)$-colouring of $(G +\_h H) \setminus e$.
|
1
|
https://mathoverflow.net/users/2233
|
407307
| 166,875 |
https://mathoverflow.net/questions/407289
|
15
|
Is it true that in the category of connected smooth manifolds equipped with a compatible field structure (all six operations are smooth) there are only two objects (up to isomorphism) - $\mathbb{R}$ and $\mathbb{C}$?
|
https://mathoverflow.net/users/148161
|
Are there only two smooth manifolds with field structure: real numbers and complex numbers?
|
Here is a series of standard arguments.
Let $(\mathbb{F},+,\star)$ be such a field. Then $(\mathbb{F},+)$ is a finite-dimensional (path-)connected abelian Lie group, hence $(\mathbb{F},+) \cong \mathbb{R}^n \times (\mathbb{S}^1)^m$ as Lie groups. Since $\mathbb{F}$ is path-connected, there is in particular a path $\gamma: [0,1] \to \mathbb{F}$ with $\gamma(0) = 0\_{\mathbb{F}}$ and $\gamma(1) = 1\_{\mathbb{F}}$. Now consider the homotopy $H: \mathbb{F} \times [0,1] \to \mathbb{F}$, $(x,t) \mapsto \gamma(t) \star x$. This gives a contraction of $\mathbb{F}$ and so we can exclude all the circle factors.
Now, fix $y\_0 \in \mathbb{F}$ and consider the map $\widehat{y\_0}: \mathbb{R}^n \to \mathbb{R}^n$, $x \mapsto x \star y\_0$. Then $\widehat{y\_0}$ is an **additive** map (but at the moment not necessarily linear with respect to the natural vector space structure on $\mathbb{R}^n$). It is not too difficult to see that by additivity we have $\forall q\in \mathbb{Q}: \widehat{y\_0}(qx) = q \widehat{y\_0}(x)$. Since $\widehat{y\_0}$ is continuous (as being smooth), it *now* follows that it's actually $\mathbb{R}$-**linear**.
Thus $\mathbb{F}$ is an $\mathbb{R}$-algebra. From this point on one can finish either by the Frobenius theorem on the classification of finite-dimensional *associative* $\mathbb{R}$-algebras or invoke a theorem of Bott and Milnor from algebraic topology that $\mathbb{R}^n$ can be equipped with a bilinear form $\beta$ turning $(\mathbb{R}^n,\beta)$ into a division $\mathbb{R}$-algebra (not necessarily associative) only in the cases $n=1,2,4,8$.
**EDIT:** Another finishing topological argument is a theorem of Hopf saying that $\mathbb{R}$ and $\mathbb{C}$ are the only finite-dimensional commutative division $\mathbb{R}$-algebras. This is less of an overkill compared to invoking Frobenius or Bott–Milnor as the proof is a rather short and cute application of homology, see p.173, Thm. 2B.5 in Hatcher's "Algebraic Topology".
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22
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https://mathoverflow.net/users/1849
|
407309
| 166,876 |
https://mathoverflow.net/questions/407200
|
2
|
In Neukirch’s book “Algebraic Number Theory”, Proposition II.5.7, the following is insisted: for a mixed characteristic local field $K$ with a residue field $\mathbb{F}\_q$, $q = p^f$, then one has an isomorphism $K^\times \cong \mathbb{Z} \times \mathbb{Z}/(q-1)\mathbb{Z} \times \mathbb{Z}/p^a\mathbb{Z} \times \mathbb{Z}\_p^{[K:\mathbb{Q}\_p]}$ for some $a$. [enter image description here](https://i.stack.imgur.com/Q0qEq.jpg)
I agree to the following statements: algebraically and topologically, $K^\times \cong \mathbb{Z} \times \mathbb{Z}/(q-1)\mathbb{Z} \times U^{(1)}$; algebraically $U^{(1)} \cong \mathbb{Z}/p^a\mathbb{Z} \times \mathbb{Z}\_p^{[K:\mathbb{Q}\_p]}$ for some $a$. However I don’t have any idea to prove that $U^{(1)}$ is topologically isomorphic to the right hand side. Does anyone have good idea?
|
https://mathoverflow.net/users/433725
|
Topology of multiplication groups of local fields
|
Let $\mathfrak m$ be the maximal ideal of the ring of integers (valuation ring) of $K$, so your $U^{(1)}$ is $1 + \mathfrak m$. You want to prove $1+\mathfrak m$ looks like $\mathbf Z/p^a\mathbf Z \times \mathbf Z\_p^d$ topologically for some $a \geq 0$ and $d = [K:\mathbf Q\_p]$. You said you agree there is such an isomorphism "algebraically". What do you mean by that: an isomorphism as what kind of algebraic structures? It's not just as groups.
The key point is to be thinking about the isomorphism $1+\mathfrak m \to \mathbf Z/p^a\mathbf Z \times \mathbf Z\_p^d$ by viewing the multiplicative group $1 + \mathfrak m$, which is a $\mathbf Z$-module by exponentiation, as a $\mathbf Z\_p$-module by exponentiation: for each $u \in 1 + \mathfrak m$, the powers $u^n$ for $n \in \mathbf Z$ are $p$-adically uniformly continuous in $n$, so we can extend ordinary integral powers of $u$ to $p$-adic integer powers $u^x$ with $x \in \mathbf Z\_p$ by $p$-adic continuity, and the effect of $\mathbf Z\_p$-exponents on elements of $1+\mathfrak m$ is how $1+\mathfrak m$ turns into a $\mathbf Z\_p$-module.
A torsion element $u$ of $1+\mathfrak m$ as a $\mathbf Z\_p$-module is a $p$-power root of unity: if $u^x = 1$ where $x \not= 0$ then writing
$x = p^nv$ with $n \geq 0$ and $v \in \mathbf Z\_p^\times$, we get
$1 = u^{p^nv} = (u^{p^n})^v$, so (raising both sides to the $1/v$-power) we get $1 = u^{p^n}$. Conversely, a $p$-power root of unity in $K$ must lie in $1 + \mathfrak m$ (basically because the only $p$-power roots of unity in a finite field is 1), so the torsion submodule of $1+\mathfrak m$ (as a $\mathbf Z\_p$-module) is the $p$-power roots of unity in $K$. That is a finite group because a finite extension of $\mathbf Q\_p$ has only finitely many $p$-power roots of unity in it: a root of unity of order $p^r$ has degree $p^{r-1}(p-1)$ over $\mathbf Q\_p$ and that exceeds $[K:\mathbf Q\_p]$ for large enough $r$. Finite subgroups of $K^\times$ are cyclic, so the torsion submodule $T$ of $1+\mathfrak m$ is a finite cyclic $p$-group.
Set $|T| = p^a$, so $T \cong \mathbf Z/p^a\mathbf Z$ as $\mathbf Z\_p$-modules
($\mathbf Z\_p$ acting by exponents on the left and by multiplication after reducing $p$-adic integers modulo $p^a$ on the right).
Provided $1 + \mathfrak m$ is a finitely generated $\mathbf Z\_p$-module, the classification of finitely generated modules over a PID (such as the PID $\mathbf Z\_p$) implies $1 + \mathfrak m \cong T \times \mathbf Z\_p^d \cong \mathbf Z/p^a\mathbf Z \times \mathbf Z\_p^d$ as $\mathbf Z\_p$-modules for some $d\geq 0$. Of course $d \geq 1$ since $1+\mathfrak m$ is not finite. You said in your post that you already agree $1 + \mathfrak m \cong \mathbf Z/p^a\mathbf Z \times \mathbf Z\_p^d$ algebraically, where $d = [K:\mathbf Q\_p]$. I don't know if you meant that isomorphism exists as abelian groups or as $\mathbf Z\_p$-modules. If you grant there is such an isomorphism as $\mathbf Z\_p$-modules, then here is a root of unity $\zeta$ of order $p^a$ in $1+\mathfrak m$ and $d$ units
$u\_1, \ldots, u\_d$ in $1 + \mathfrak m$ that are $\mathbf Z\_p$-independent (if $u\_1^{x\_1} \cdots u\_d^{x\_d} = 1$ then each $p$-adic integer $x\_i$ is $0$) such that $1+\mathfrak m = \langle \zeta\rangle \times u\_1^{\mathbf Z\_p} \times \cdots \times u\_d^{\mathbf Z\_p}$. Then the mapping
$$
f \colon
\mathbf Z/p^a\mathbf Z \times \mathbf Z\_p^d \to 1 + \mathfrak m
$$
given by $f(k,x\_1,\ldots,x\_d) = \zeta^ku\_1^{x\_1} \cdots u\_d^{x\_d}$ is a $\mathbf Z\_p$-module isomorphism. This function $f$ is *continuous* using the product topology on its domain and the subspace topology on its codomain ($1+\mathfrak m$ as a subset of $K$) because (i) $u^x$ is $p$-adically continuous in $x$ for $u \in 1+\mathfrak m$ and $x \in \mathbf Z\_p$ and (ii) multiplication in $1+\mathfrak m$ is continuous. The domain and codomain of $f$ are *compact Hausdorff* spaces, and a continuous bijection of compact Hausdorff spaces is a homeomorphism. Therefore if $f$ is an isomorphism of $\mathbf Z\_p$-modules ("algebraically"), it is also a homeomorphism, which is what you wanted to understand.
|
4
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https://mathoverflow.net/users/3272
|
407320
| 166,878 |
https://mathoverflow.net/questions/407188
|
3
|
Let $\Delta$ be the Laplacian on a smooth domain $\Omega\subset \mathbb{R}^2$ with Dirichlet boundary conditions. I am interested in whether the implication
\begin{align}
\Omega \text{ is asymmetric } \Rightarrow \Delta \text{ only has single eigenvalues}
\end{align}
holds, where by asymmetric I mean that the only planar rotation $\varphi$ such that $\varphi(\Omega) = \Omega$ is the identity.
It is well known that the set of Dirichlet eigenvalues $\{\lambda\_i\}\_{i\in \mathbb{N}}$ forms a sequence
\begin{align}
0<\lambda\_1< \lambda\_2 \leq \lambda\_3 \leq \dots \nearrow \infty
\end{align}
with the first eigenvalue being single.
Examples of symmetric domains with multiple eigenvalues (see [1]) include a square, an equilateral triangle or a circular disk.
An example of a symmetric domain with only simple eigenvalues would be a rectangle with side lengths $l\_1$, $l\_2$ such that $\left(\frac{l\_1}{l\_2}\right)^2$ is irrational [1, 3.1].
I have carried out some numerical experiments with asymmetric shapes:
<https://i.stack.imgur.com/uwroB.png>
<https://i.stack.imgur.com/6VJk6.png>
<https://i.stack.imgur.com/v1suq.png>
but up to the 200th eigenvalue there are only single ones.
Does anyone know whether the above claim is true or false, or has ideas for counterexample shapes to check (i.e., asymmetric shapes with multiple eigenvalues)?
[1] Grebenkov, Denis S., and B-T. Nguyen. "Geometrical structure of Laplacian eigenfunctions." siam REVIEW 55.4 (2013): 601-667.
|
https://mathoverflow.net/users/430002
|
Multiplicity of Dirichlet Laplacian eigenvalues of asymmetric domains
|
A counterexample is given by $\Omega = (0,2\pi)^2\setminus [0,\pi]^2$. Then $u\_{mn}(x,y)=\sin mx\sin ny$ is an eigenfunction with eigenvalue $m^2+n^2$, and so is $u\_{nm}$.
You can build many other examples in this way by looking at the zero set of eigenfunctions of the square (in particular, the reflection symmetry of the region can be removed also) and perhaps also starting from other regions with multiple eigenvalues.
|
1
|
https://mathoverflow.net/users/48839
|
407325
| 166,881 |
https://mathoverflow.net/questions/407322
|
3
|
Suppose that we are given an AF-algebra $A$ and a sequence of finite-dimensional subalgebras $\mathbb{C}=A\_0\subset A\_1\subset A\_2\subset\ldots$ such that $A=\overline{\bigcup\limits\_{n\geq 0}A\_n}$. Let me denote this dense subalgebra of $A$ by $A^{LS}$, i.e. $A^{LS}= \bigcup\limits\_{n\geq 0}A\_n$.
Next, we define the positive elements as
$$A^+=\left\{h^2\ \middle|\ h\in A,\ h^\*=h \right\}$$
and
$$(A^{LS})^+=\left\{h^2\ \middle|\ h\in A^{LS},\ h^\*=h \right\}.$$
Then for any $y\in A^+$ we can find a sequence $\{y\_n\}\_{n\geq 1}\subset (A^{LS})^+$ such that $\lim\limits\_{n\to\infty}y\_n=y$.
**Question:**
Can we pick this $\{y\_n\}$ in such a way that $y\_n\leq y$ for any $n$ in the sense of the partial order defined by the cone $A^+$?
**Note**: existence of the desired sequence is equivalent to existence of the sequence $\{y\_n\}$ such that $y\_n\to y$ and for any $n$ there exists $M$ such that for any $m\geq M$ we have $y\_n\leq y\_{m}$.
|
https://mathoverflow.net/users/170524
|
Monotone approximation of elements in AF-algebras
|
No, this is not possible in general. $A^{LS}$ might have trivial intersection with a non-zero hereditary $C^\ast$-subalgebra of $A$, and thus any non-zero positive element in such a hereditary $C^\ast$-subalgebra cannot be approximated from below by elements in $A^{LS}$.
For a concrete example, I will give a non-unital example. You can simply unitise this example to get a unital counter-example (but I leave this to the reader).
Consider $\mathbb C^n \subseteq \ell^2(\mathbb N)$ in the obvious way, and let $A = \mathcal K(\ell^2(\mathbb N))$ with finite dimensional $C^\ast$-subalgebras $A\_n = \mathcal K(\mathbb C^n)$. Fix any unit vector $\xi \in \ell^2(\mathbb N) \setminus \bigcup\_n \mathbb C^n$ and let $p$ be the projection onto the span of $\xi$. Then $pAp \cap \bigcup\_n A\_n = \{0\}$, and thus $0$ is the only positive element in $\bigcup A\_n$ which is below $p$.
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8
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https://mathoverflow.net/users/126109
|
407329
| 166,883 |
https://mathoverflow.net/questions/407283
|
5
|
Let $t\_1$ and $t\_2$ be lower semicontinuous semifinite densely defined traces on a $C^\*$-algebra $A$. Let us denote by $\mathcal{R}\_1$ and $\mathcal{R}\_2$ their ideals of definition, i.e. $\mathcal{R}\_i=\left\{x\in A\ \middle|\ t\_i(x^\*x)<+\infty \right\}$. Next, let $B$ be an involutive dense subalgebra of $A$ such that $B\subset \mathcal{R}\_1,\mathcal{R}\_2$ and $t\_1(b^\*b)=t\_2(b^\*b)$ for any $b\in B$. Is it true that then $t\_1=t\_2$ on the whole algebra $A$?
I have found a similar statement in “C\*-algebras” by J.Dixmier 1977: Lemma 6.5.3 on page 139. But he formulates this claim in terms of bitraces, which look artificial to me. Can anyone provide a straightforward and concise argument without bitraces? Is there a better reference for this?
|
https://mathoverflow.net/users/170524
|
Two densely defined traces on a $C^*$-algebra coinciding on a dense subalgebra are equal
|
Yes, this is true.
Let $a\in A\_+$. By lower semicontinuity it suffices to show that $t\_1((a-\delta)\_+) = t\_2((a-\delta)\_+)$ for all $\delta>0$ (where $(a-\delta)\_+$ is the positive part of $a-\delta 1$ in the unitisation of $A$ (which is an element of $A$ and not actually in the unitisation)). Fix $\delta>0$.
The key trick for getting the approximations to work, is to use that $t\_i$ extends canonically to a positive linear functional on $\mathrm{span} R\_i^\ast R\_i$. Hence if $x\in R\_i$ and $b,c\in A\_+$ then
\begin{equation}
- \| b-c\| t\_i(x^\ast x) \leq t\_i(x^\ast(b-c)x) \leq \| b-c\| t\_i(x^\ast x)
\end{equation}
and thus
\begin{equation}
|t\_i(x^\ast b x) - t\_i(x^\ast c x) | \leq \| b-c\| t\_i(x^\ast x).
\end{equation}
This will be used a couple of times (in particular, using that $(a-\delta)\_+^{1/2}\in R\_i$, since $(a-\delta)\_+$ is in the Pedersen ideal; the smallest dense two-sided ideal of $A$).
Let $\epsilon>0$. Pick $e\in A\_+$ be such that $(a-\delta)\_+ e = (a-\delta)\_+$, and let $x\in B$ be close enough to $e^{1/2}$ that $\| e - x^\ast x\| \max\{ t\_1((a-\delta)\_+) , t\_2((a-\delta)\_+)\} < \epsilon$. By the trick mentioned above we have
\begin{equation}
t\_i((a-\delta)\_+) = t\_i((a-\delta)\_+^{1/2} e (a-\delta)\_+^{1/2}) \approx\_\epsilon t\_i((a-\delta)\_+^{1/2} x^\ast x (a-\delta)\_+^{1/2}).
\end{equation}
Now pick $y\in B$ close enough to $(a-\delta)\_+^{1/2}$ so that
\begin{equation}
\| y y^\ast - (a-\delta)\_+ \| \max\{ t\_1(x^\ast x) , t\_2(x^\ast x)\} < \epsilon.
\end{equation}
Then by the above trick
\begin{equation}
t\_i((a-\delta)\_+^{1/2} x^\ast x (a-\delta)\_+^{1/2}) = t\_i(x(a-\delta)\_+ x^\ast) \approx\_\epsilon t\_i (xyy^\ast x^\ast).
\end{equation}
As $xy \in B$ we get $t\_1(xyy^\ast x^\ast) = t\_2(xyy^\ast x^\ast)$ by assumption, and therefore
\begin{equation}
| t\_1((a-\delta)\_+) - t\_2((a-\delta)\_+) | < 4 \epsilon.
\end{equation}
As $\epsilon>0$ was arbitrary we get $t\_1((a-\delta)\_+) = t\_2((a-\delta)\_+)$, and thus $t\_1(a) = t\_2(a)$ by lower semicontinuity.
|
5
|
https://mathoverflow.net/users/126109
|
407333
| 166,884 |
https://mathoverflow.net/questions/407332
|
0
|
Let $E$ be a (right) Hilbert module over the $C^\*$-algebra $B$. Let $\phi$ be a state on the $C^\*$-algebra $B$. Then consider
$$N\_\phi:= \{x \in E: \phi(\langle x,x\rangle)=0\}.$$
I want to show that $N\_\phi$ is a submodule of $E$, but for this I need to show that
$$b \in B, x \in N\_\phi \implies \phi(b^\*\langle x,x\rangle b)=\phi(\langle xb,xb\rangle)= 0.$$
Why is this true?
|
https://mathoverflow.net/users/216007
|
Is $N_\phi = \{x \in E: \phi(\langle x,x\rangle)=0\}$ a Hilbert submodule of $E$?
|
It is not true. Take $B= M\_2(\mathbb C)$ (with standard matrix units $e\_{i,j}$), $E= B$ as a Hilbert $B$-module in the usual way, and let $\phi \in B^\ast$ be compression to the $(1,1)$-corner. Then $x = e\_{2,2} \in E$ and $b = e\_{2,1} \in B$ satisfy $x\in N\_\phi$ and $xb \notin N\_\phi$.
|
5
|
https://mathoverflow.net/users/126109
|
407335
| 166,886 |
https://mathoverflow.net/questions/407347
|
1
|
Let $f(x)\in \mathbb{Z}[X]$ be a polynomial of degree at least $2$. We denote the set of primes $p$ for which $f(x)$ is injective modulo $p$ as $\mathcal{T}$. Then, can we say something about the proportion of polynomials $f(x)$ for which cardinality of the set
$$\#\mathcal{T}(y)\ll \frac{y}{(\log{y})^2}.$$
|
https://mathoverflow.net/users/160943
|
Estimating the size of set of primes $p$ for which the polynomial is bijective in $\mathbb{F}_p[X]$
|
I'll write $T\_f$ for the set of primes $p$ such that $f(x)$ is a bijection $\mathbb{F}\_p \to \mathbb{F}\_p$. I claim that $T\_f$ is always either finite or else $\# \{ p \in T\_f : p \leq y \} \sim c \tfrac{y}{\log y}$ for some $c>0$.
The following result is known as Schur's conjecture; a flawed proof was given by [Fried](https://mathscinet.ams.org/mathscinet-getitem?mr=257033), with corrected versions by [Turnwald](https://mathscinet.ams.org/mathscinet-getitem?mr=1329867) and [Müller](https://mathscinet.ams.org/mathscinet-getitem?mr=1429041):
**Theorem** Let $f(x) \in \mathbb{Z}[x]$. If $T\_f$ is infinite, then $f(x)$ is a composition of linear polynomials and [Dickson polynomials](https://en.wikipedia.org/wiki/Dickson_polynomial).
One should note that the Dickson polynomial $D\_n(x,0)$ is just $x^n$, so this includes the possibility of including monomials in our composition.
A composition of functions $\mathbb{F}\_p \to \mathbb{F}\_p$ will be bijective if and only if all the functions composed are bijective. Linear functions are always bijective; the monomial $D\_n(x,0)=x^n$ is bijective iff $GCD(n,p-1)=1$ and, for $a \neq 0$, the Dickson polynomial $D\_n(x,a)$ is bijective iff $GCD(n,p^2-1)=1$ or, equivalently, $GCD(n,p-1) = GCD(n,p+1)=1$. (This last statement is copied from Lemma 1.4 in Turnwald; I didn't check it.) In short, imposing that our composition is bijective imposes finitely many conditions on the residue class of $p$ modulo various integers.
If these modular conditions can be satisfied by infinitely many primes, then they are satisfied by $\sim c \tfrac{y}{\log y}$ primes $\leq y$, by the PNT in arithmetic progressions.
---
Polynomials where $T\_f$ is infinite are [exceptional polynomials](https://en.wikipedia.org/wiki/Permutation_polynomial#Exceptional_polynomials). (The definition of "exceptional" is that there are infinitely many prime powers $q$ such that $f$ is bijective on $\mathbb{F}\_q$, so it also allows examples like $x^2$ which is bijective on $\mathbb{F}\_{2^k}$ but not on $\mathbb{F}\_p$ for any odd $p$; imposing bijectivity for infinitely many primes rather than
prime powers is obviously even more restrictive.)
As the name suggests, exceptional polynomials are very rare for degree $\geq 5$. For example, they induced maps $\mathbb{C} \to \mathbb{C}$ all have solvable monodromy, where as almost all degree $n$ polynomials have monodromy group $S\_n$. (Indeed, as Will Sawin points out, this is even a Zariski open condition.)
In degrees $2$, $3$ and $4$, your condition that $T\_f$ be infinite is still very rare (although polynomials which are bijective on infinitely many $\mathbb{F}\_{2^k}$ or infinitely many $\mathbb{F}\_{3^k}$ are not so rare). Indeed, a composition of Dickson polynomials of degree $\leq 4$ must be a composition of $D\_2(x,\alpha)$, $D\_3(x,\alpha)$ and $D\_4(x, \alpha)$ for various values of $\alpha$. The condition that these are bijective imposes either that $GCD(p-1,2)=1$ (impossible for odd $p$), that $GCD(p^2-1,3)=1$ (impossible for $p \neq 3$) or that $GCD(p-1,3)=1$ (this case can happen). So the only case which occurs is the pre-and-post-composition of $D\_3(x,0)=x^3$ with linear polynomial. But then the monodromy of $f : \mathbb{C} \to \mathbb{C}$ is $A\_3$ (in other words, $f'(x)$ has a double root), and this does not happen on a Zariski open set.
|
5
|
https://mathoverflow.net/users/297
|
407353
| 166,891 |
https://mathoverflow.net/questions/407158
|
4
|
$\DeclareMathOperator\gr{gr}$Let $A = \cup\_{i=0}^\infty F\_i A$ be a filtered commutative ring, $I \subseteq A$ an ideal. Then we have a canonical surjection
$$ \gr(A)/\gr(I) \to \gr(A/I).$$
Under what conditions is this surjection an isomorphism?
I most wish to know about the following special case: $A = \mathbb C[z\_1,\ldots, z\_n]$ equipped with the total degree filtration, $I$ an ideal of finite codimension. The ring $\gr(A/I)$ in this case should be functions on the scheme "$\lim\_{t \to 0} t\cdot V(I)$", and I'd like to know when the defining ideal is $gr(I)$.
|
https://mathoverflow.net/users/125523
|
Quotients and associated graded
|
I follow user @Z. M's comment.
If $M = \cup\_{i=0}^\infty F\_i M$ is a filtered module and
$$0 \to M' \to M \to M'' \to 0$$
is a short exact sequence such that $M', M''$ have the *induced filtrations*, then we get a short exact sequence of Rees modules
$$0 \to R\_hM' \to R\_hM \to R\_hM'' \to 0,$$
which yields the short exact sequence $0 \to \mathrm{gr}(M') \to \mathrm{gr}(M) \to \mathrm{gr}(M'') \to 0$ after taking the Snake Lemma (since our filtration is by submodules).
Now in our situation, $A/I$ and $I$ have the induced filtrations, so we get a short exact sequence
$ 0 \to \mathrm{gr}(I) \to \mathrm{gr}(A) \to \mathrm{gr}(A/I) \to 0$.
|
0
|
https://mathoverflow.net/users/125523
|
407362
| 166,896 |
https://mathoverflow.net/questions/407311
|
3
|
I read Neukirch’s book “Algebraic Number Theory”, and its remark following to proposition VI.2.3, there is an assertion that natural map $\mathbb{A}\_K \otimes\_K L \to \mathbb{A}\_L$ is isomorphism. How can it be proved ?
|
https://mathoverflow.net/users/433725
|
base change of adele rings
|
This is Theorem 1 in Chapter IV-1 of Weil: Basic Number Theory. See also Corollaries 1-2 after the proof of this theorem.
|
4
|
https://mathoverflow.net/users/11919
|
407364
| 166,897 |
https://mathoverflow.net/questions/407155
|
8
|
Let $D$ be a central division (or maybe just simple) algebra over $\mathbb{Q}$. Let $\mathcal{O} \subset \mathcal{O}\_m$ be an order inside a fixed maximal order and denote by $\mathcal{O}^1$ its group of units with norm $1$. If $[\mathcal{O}\_m : \mathcal{O}] = N$, then can we say in general that $[\mathcal{O}^1\_m : \mathcal{O}^1]$ is also of size $N$? More precisely, can we show that $N^{1-\epsilon} \leq [\mathcal{O}^1\_m : \mathcal{O}^1] \leq N^{1+\epsilon}$ for (arbitrarily) small $\epsilon$?
This seems to be the case for $D = M\_2(\mathbb{Q})$ (e.g. for Hecke congruence subgroups) and also more generally for quaternion algebras by the volume formulas (see e.g. Voight, relating the volume of the corresponding locally symetric space to the discriminant of the order). Are there more general volume formulas for all degrees? Is there a softer way to compute the "order of magnitude" of the index?
Edit: Using approximation theorems, it should be enough to show this locally. So we could assume to be working over the $p$-adic numbers. Are there additional classification results we could use now?
|
https://mathoverflow.net/users/168129
|
Covolumes of unit groups of division algebras
|
First, you need to avoid definite quaternion algebras over $\mathbb{Q}$: in this case, the unit groups are finite, so the index cannot grow with $N$.
With that out of the way, your algebra $D$ satisfies strong approximation (chapter 28 of my book, I'm sorry I only discuss the quaternion case, but see Remark 28.6.11; I also recommend the book "Units in skew fields" by Ernst Kleinert)! This means that the global index $[\mathcal{O}\_m^1 : \mathcal{O}^1]$ is equal to the product of the local indices $[(\mathcal{O}\_m)\_p^1 : \mathcal{O}\_p^1]$ over (finitely many) primes $p$, so it is enough to understand the local situation.
In the local case, an argument like the one in Lemma 26.6.7 applies. It starts out the same, and we are left to determine the ratio
$[\mathcal{O}\_m^1 : 1+p\mathcal{O}\_m]/[\mathcal{O}^1 : 1+p\mathcal{O}]=\#(\mathcal{O}\_m/p\mathcal{O}\_m)^1/\#(\mathcal{O}/p\mathcal{O})^1$. Each term (numerator and denominator) is the calculation of the units in an $\mathbb{F}\_p$-algebra, and already in the case of quaternions you can see that there are cases needed if we want an exact formula.
Actually, I think there's a more uniform thing to reduce to the semisimple case, using the Jacobson radical $J=\mathrm{rad} \mathcal{O}$ (20.4): we have a filtration $\mathcal{O} \supseteq J \supseteq J^2 \supseteq \dots \supseteq p\mathcal{O} \supseteq J^r$ for some $r \geq 1$ which gives $\mathcal{O}^\times \geq 1+J \geq 1+J^2 \geq \dots \geq 1+p\mathcal{O} \geq 1+J^r$ and $(1+J^i)/(1+J^{i+1}) \simeq J^i/J^{i+1}$ for $i \geq 1$. So you only need to count $(\mathcal{O}/J)^1$. Now $\mathcal{O}/J$ is semisimple (that's the job of the Jacobson radical), and we're over $\mathbb{F}\_p$, so $\mathcal{O}/J$ is the product of matrix rings over finite fields, so the exact count can be given in terms of the number of factors, their degree, and the cardinality of the residue field.
Combining these should give you an explicit formula (with cases) or an estimate, like what you asked for above. (I didn't work out the details, hoping that you would be motivated to do so!) This kind of calculation shows up in the computation of zeta functions of orders in division algebras, but I couldn't quickly find a reference...
|
11
|
https://mathoverflow.net/users/4433
|
407367
| 166,899 |
https://mathoverflow.net/questions/407032
|
6
|
Let $\mathcal{U}$ be an ultrafilter over $\omega$, and let $\mathcal{X} \subseteq [\omega]^\omega$. In two separate texts, there are two possible interpretations of a $\mathcal{U}$-Ramsey set, as described below (Definition 7.37 and Definition 3). My question is:
>
> Do these two definitions coincide? What if we restrict $\mathcal{U}$ to be a Ramsey ultrafilter?
>
>
>
---
**Interpretation 1.** The first interpretation is found in Stevo Todorcevic's book *Introduction to Ramsey Spaces*.
**Definition 7.29.** For any ultrafilter $\mathcal{U}$ over $\omega$ (not necessarily Ramsey), we define a **$\mathcal{U}$-tree** to be a subtree $T$ of ${}^{<\omega}\omega$ (finite subsets of $\omega$, not sequences) with the property that for every finite subset $t \subseteq \omega$ such that $\operatorname{stem}(T) \subseteq t$, we have:
$$
\{n \in \omega : t \cup \{n\} \in T\} \in \mathcal{U}
$$
**Definition 7.30.** For two $\mathcal{U}$-trees $T'$ and $T$, we say that $T'$ is a **pure refinement** of $T$ if $\operatorname{stem}(T') = \operatorname{stem}(T)$ and $T' \subseteq T$.
**Definition 7.37.** We then say that $\mathcal{X}$ is **$\mathcal{U}$-Ramsey** if for every $\mathcal{U}$-tree $T$, there is a pure refinement $T'$ of $T$ such that $[T'] \subseteq \mathcal{X}$ or $[T'] \subseteq \mathcal{X}^c$.
---
**Interpretation 2.** I first saw this interpretation in the paper [*Happy and mad families in $L(\mathbb{R})$*](http://pi.math.cornell.edu/%7Ezbnorwood/files/hmlr.pdf), but I believe this is quite a standard definition.
**Definition 1.** If $y\_0 \supseteq y\_1 \supseteq y\_2 \supseteq \cdots$ is a decreasing sequence of subsets of $\omega$, then we call a set $y\_\infty \in [\omega]^\omega$ a **diagonalisation** of the sequence $\langle{y\_n : n < \omega}\rangle$ iff $f(n+1) \in y\_{f(n)}$ for every $n < \omega$, where $f : \omega \to \omega$ is the increasing enumeration of $y\_\infty$.
**Definition 2.** A $\mathcal{H} \subseteq [\omega]^\omega$ is called a **coideal** if is satisfies the conditions:
1. (Upward-closure) If $x \in H$ and $y \supseteq x$, then $y \in H$.
2. (Pigeonhole) If $x\_0 \cup \cdots \cup x\_n \in H$, then $x\_k \in H$ for some $k$.
Furthermore, $\mathcal{H}$ is said to be **selective** if it satisfies the condition:
3. (Selectivity) Every decreasing sequence $y\_0 \supseteq y\_1 \supseteq \cdots$ of members of $\mathcal{H}$ has a diagonalisation in $H$.
A selective coideal is also called a **happy family**.
**Definition 3.** If $H$ is a coideal (typically a happy family) and $\mathcal{X} \subseteq [\omega]^\omega$ is a set of reals, then we say $\mathcal{X}$ is **$\mathcal{H}$-Ramsey** if there is $M \in \mathcal{H}$ such that $[M]^\omega \subseteq X$ or $[M]^\omega \subseteq X^c$.
Note that if $\mathcal{H}$ is an ultrafilter, then it is a happy family iff it is a Ramsey ultrafilter.
|
https://mathoverflow.net/users/146831
|
Are these two definitions of $\mathcal{U}$-Ramsey set equivalent?
|
If I understand your question correctly, you ask if **Defintion 7.37.** and **Definition 3.** are equivalent for a Ramsey ultrafilter $\mathcal{U}(=\mathcal{H})$. The short answer is no, but let me elaborate.
Since for a $\mathcal{U}$-tree $T$, the set $[T]$ is technically a subset of $\omega^\omega$, we will only consider $\mathcal{U}$-trees $T$ such that $\text{stem}(T)$ is increasing. For such a tree $T$ there exists a pure refinement $T' \subseteq T$ such that every $x \in [T']$ is increasing, hence $\text{ran}(x) \in [\omega]^\omega$ can uniquely be identified with $x \in T'$.
**Fact:** If $\mathcal{U}$ is a Ramsey ultrafilter, then $\mathbb{M}\_\mathcal{U}$ (Mathias forcing with an ultrafilter $\mathcal{U}$) is dense in $\mathbb{L}\_\mathcal{U}$ (Laver forcing with an ultrafilter $\mathcal{U}$ such that for every $T \in \mathbb{L}\_\mathcal{U}$ we have that $\text{stem}(T)$ is increasing). In particular, for every $T \in \mathbb{L}\_\mathcal{U}$ there exists $A \in \mathcal{U}$ such that $\{\text{ran}(\text{stem}(T)) \cup x \,\, \colon \,\ x \in [A]^\omega \} \subseteq [T]$.
Now **Defintion 7.37.** obviously implies **Definition 3.** : Let $\mathcal{X} \subseteq [\omega]^\omega$ be arbitrary such that $\mathcal{X}$ is $\mathcal{U}$-Ramsey. Let $T$ be a pure refinement of $\omega^\omega$ such that either $[T] \subseteq \mathcal{X}$ or $[T] \cap \mathcal{X} = \emptyset$. By the above Fact we can find $A \in \mathcal{U}$ such that $\{\emptyset \cup x \,\, \colon \,\ x \in [A]^\omega \} \subseteq [T]$. Hence $\mathcal{X}$ is also $\mathcal{H}$-Ramsey.
On the other hand **Definition 3.** does not imply **Defintion 7.37.** : By an induction of length $\mathfrak{c}$ we can construct a $\mathcal{X} \subseteq [\omega]^\omega$ which is not $\mathcal{H}$-Ramsey. W.l.o.g. we can assume $0 \notin x$ for every $x \in \mathcal{X}$. Now define $\mathcal{X}':=\{\{0\} \cup x \,\, \colon \,\, x \in \mathcal{X}\}$, which is obviously $\mathcal{H}$-Ramsey. Now if $T$ is a $\mathcal{U}$-tree such that $\text{stem}(T)=\langle 0 \rangle$, there cannot exist a pure refinement $T' \subseteq T$ such that $[T'] \subseteq \mathcal{X}'$ or $[T'] \cap \mathcal{X}' = \emptyset$, because there does not exist an $A \in \mathcal{U}$ such that $\{\{0\} \cup x \,\, \colon \,\ x \in [A]^\omega \} \subseteq \mathcal{X}'$ or $\{\{0\} \cup x \,\, \colon \,\ x \in [A]^\omega \} \cap \mathcal{X}' = \emptyset$.
|
3
|
https://mathoverflow.net/users/134910
|
407372
| 166,901 |
https://mathoverflow.net/questions/407350
|
2
|
* [Li-Yau 1983\_Article](https://link.springer.com/content/pdf/10.1007/BF01213210.pdf)
* The second part of above paper used the discrete eigenvalues of $\frac{-\Delta}{q}$ where $q>0$ to proof the the number of non-positive eigenvalues of
Schrödinger operator $-\Delta+V$ can be bounded by the $L\_{\frac{n}{2}}$-norm of $V^-$.
* My question is: under what condition of $q$ can we proof the spectrum of $\frac{-\Delta}{q}$ is discrete.
|
https://mathoverflow.net/users/166368
|
On the Schrödinger equation and the eigenvalue problem
|
Assuming $q>0$ the Schroedinger operator $-\Delta/q$ is associated to the form $a(u,v)=\int\_{\mathbb R^n} \nabla u \cdot \nabla v$ in $L^2(\mathbb R^n, q\, dx)$. The form domain consists of all $u\in L^2(\mathbb R^n, q\, dx)$ such that $u \in \dot H^1:=\{u \in L^{2^\*}(\mathbb R^n), \nabla u \in L^2(\mathbb R^n)\} $ and the discreteness of the spectrum is equivalent to the compactness of the embedding of the form domain into $L^2(\mathbb R^n, q\, dx)$.
This follows when the map $$T:\dot H^1 \to L^2(\mathbb R^n), \quad Tu=q^{1/2}u$$ is compact, which is true whenever $q \in L^{n/2}$.
|
2
|
https://mathoverflow.net/users/150653
|
407383
| 166,905 |
https://mathoverflow.net/questions/407259
|
3
|
Let $F$ be a $p$-adic field with residue field $k$ and let $G$ be a connected reductive group over $F$. Let us assume that $G$ is simply connected as an algebraic group over an algebraic closure of $F$. If I am not mistaken, it will insure that maximal compact subgroups and maximal parahoric subgroups of $G(F)$ are the same thing.
Given a maximal compact subgroup $K$ of $G(F)$, one can ask whether it is special or hyperspecial. This is defined in terms of the affine building of $G$ ; and there is a characterization of hyperspecial subgroups in terms of integral models of $G$.
On the other hand, given a parahoric subgroup $K$ in $G(F)$, one can form its pro-p radical $K^+$ (I also saw it called pro-unipotent radical, but I assume it is the same thing). The quotient $\mathcal K := K/K^+$ can be seen as the group of $k$-rational points of a reductive group over the finite field $k$. Therefore we call it the finite reductive quotient of $K$. Usually, when $G$ is a classical group then $\mathcal K$ is a product of two classical groups of the same type as $G$.
Both the adjectives "special", "hyperspecial" and the finite reductive quotient only depend on the conjugacy class of $K$.
>
> When $K$ is a maximal parahoric subgroup, is there a way to detect
> when $K$ is special or hyperspecial solely in terms of conditions on
> the finite reductive quotient $\mathcal K$ ?
>
>
>
It's a naive question and I don't know what to expect. It just came to my mind while looking at the case of unitary groups. That is, if $E/F$ is an unramified quadratic extension, $V$ is a nondegenerate hermitian space over $E/F$ of dimension $n$ and $G = \mathrm U(V)$. Then, finite reductive quotients of maximal parahoric subgroups look like $\mathrm U\_{t}(k) \times \mathrm U\_{n-t}(k)$ where $1\leq t \leq n$ is some integer. Unless I am mistaken, we have
* When $n$ is odd, then $t$ must be odd and the corresponding parahoric is special for $t = 1$ and $t = n$. It is hyperspecial when $t = n$.
* When $n$ is even and $G$ is not unramified, then $t$ must be odd and the corresponding parahoric is special for $t = 1$ and $t = n-1$.
* When $n$ is even and $G$ is unramified, then $t$ must be even and the corresponding parahoric is special for $t = 0$ and $t = n$. In fact, they are hyperspecial in both cases.
Thus, it looks like the parahorics are special when one summand of the finite quotient is "as small as possible". Furthermore they are hyperspecial when this summand actually vanishes.
As a sidenote, maybe not related to the problem above, let me denote by $K\_t$ a parahoric subgroup having finite quotient $\mathrm U\_t(k) \times \mathrm U\_{n-t}(k)$ where $t$ is odd or even depending on the cases we described. They form a system of representatives for each conjugacy class of maximal parahoric subgroups of $G$ (the number of such classes being precisely the Witt index of $V$). In particular, when $n$ is even the parahorics $K\_t$ and $K\_{n-t}$ are not conjugate in $G$ even though their finite reductive quotients are isomorphic.
But I noticed that it is "fixed" when we consider unitary similitude groups $G = \mathrm{GU}(V)$ instead. In this case, we still have maximal parahoric subgroups $K\_t$ with finite quotient isomorphic to $\mathrm{G}(\mathrm{GU}\_t(k)\times \mathrm{GU}\_{n-t}(k))$, however when $n$ is even then the parahorics $K\_t$ and $K\_{n-t}$ are conjugate in $G$.
|
https://mathoverflow.net/users/125617
|
Is it possible to detect when a maximal parahoric subgroup is (hyper)special from its finite reductive quotient?
|
(I write this answer in quite a haste, so there will probably be some inaccuracies. Sorry for that and, please, let me know, when you find them.)
First, I think, that if the group is not almost simple, then the question makes few sense. For example, each of the split groups $G\_1 = Sp\_4(F)$ and $G\_2 = {\rm SL}\_2(F) \times {\rm SL}\_2(F)$ have a parahoric subgroup of the same type, namely ${\rm SL}\_2(\mathcal{O}\_F) \times {\rm SL}\_2(\mathcal{O}\_F)$ (where $\mathcal{O}\_F$ are the integers of $F$), obviously with the same finite reductive quotient. For $G\_1$, the corresponding parahoric is non-special, whereas for $G\_2$ it is hyperspecial.
For simplicity, I will even assume that $G$ is absolutely almost simple over $F$, which means that the absolute Dynkin diagram is connected. We have the affine Dynkin diagram $\widetilde \Delta$ of $G(F)$, whose vertices are in 1-1 bijection with the vertices in a fixed base alcove of the Bruhat--Tits building of $G(F)$. Alternatively, the affine Dynkin diagram can be obtained from the usual one by adding one vertex. Conjugacy classes of maximal parahorics $K = K\_x$ of $G(F)$ are in 1-1 bijection with vertices $x$ of $\widetilde \Delta$ (use that $G$ is simply connected). The Dynkin diagram of the finite reduction quotient $\mathcal{K}\_x = K\_x/K\_x^+$ is then just $\widetilde \Delta \backslash \{x\}$ (with all edges beginning/ending in $v$ removed). Thus the number of simple factors of $\mathcal{K}\_x$ is equal to the number of connected components of $\widetilde \Delta \backslash \{x\}$. For example, if you remove the middle point in (affine) type $E\_6$ (but also in some other cases) it is possible for $\widetilde \Delta \backslash \{x\}$ to have $3$ connected components, so that there are three factors.
Now, $x$ is special, iff $\widetilde \Delta \backslash \{x\}$ equals $\Delta$, the (usual, not affine) Dynkin diagram of $G$. To answer the question, observe (by looking at the finite and affine Dynkin diagrams, which can be found for example in wikipedia) that $\widetilde \Delta \backslash \{x\}$ will tend to be of type $\Delta$ when $x$ tends to be an extremal vertex of $\widetilde \Delta$. For example, if $G = Sp\_4$, i.e. with $\Delta$ of type $C\_2$ and $\widetilde \Delta$ being $\cdot = \cdot = \cdot$, then $\widetilde \Delta$ without the left or the right vertex will be of type $C\_2$ (so these are hyperspecial), whereas $\widetilde \Delta$ without the middle vertex is of type $A\_1 \times A\_1$ (and it is not hyperspecial).
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2
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https://mathoverflow.net/users/148992
|
407385
| 166,906 |
https://mathoverflow.net/questions/407388
|
6
|
**For reference, my motivation:** It's of interest to classify free actions of groups on spheres of positive even dimension. Establishing such a classification up to homotopy is not too difficult: Every free group action on a sphere of even dimension is homotopic to either the trivial action by the trivial group or the antipodal action by $\mathbb{Z}/2$. The question in the title is, by the reduction that follows, equivalent to whether this classification is conserved if the "homotopic" in the above sentence is strengthened to "homeomorphic".
Suppose that every space doubly covered by $S^{2n}$ has the homeomorphism type of $\mathbb{P}^{2n}\_{\mathbb{R}}$ and let $\tau\ \colon S^{2n}\to S^{2n}$ be some continuous involution lacking fixed points. Then:
1. By the compactness of $S^{2n}$, $$x\mapsto\text{dist}\_{\text{standard subspace Euclidean metric on }S^{2n}}\left(x,\tau\left(x\right)\right)\colon S^{2n}\to\mathbb{R}\_{\geq 0}$$ attains a nonzero mminimum on its domain.
2. By (1), the projection map $$\gamma\ \colon S^{2n}\to\text{coeq}\left(S^{2n}\substack{\overset{\text{id}}{\longrightarrow}\\ \underset{\tau}{\longrightarrow}}S^{2n}\right)$$ is a covering map.
3. By (2), there exists an isomorphism $$\psi\ \colon \text{coeq}\left(S^{2n}\substack{\overset{\text{id}}{\longrightarrow}\\ \underset{\tau}{\longrightarrow}}S^{2n}\right)\to\mathbb{P}\_{\mathbb{R}}^{2n}.$$
4. By (2) and the lifting theorem for covering spaces, the $\psi$ of (3) lifts to an isomorphism $$\tilde{\psi}\ \colon S^{2n}\to S^{2n}$$ such that $$\text{anti}\circ\tau = \tau\circ \tilde{\psi}$$ (with $\text{anti}\ \colon S^{2n}\to S^{2n}$ the antipodal involution), precisely the desideratum.
|
https://mathoverflow.net/users/131309
|
Do all spaces doubly covered by $S^{2n}$ have the homeomorphism type of $\mathbb{P}^{2n}_{\mathbb{R}}$?
|
One approach to a homeomorphism classification of closed manifolds simply homotopy equivalent to a closed manifold $X$ of dimension $>4$ is to compute the topological structure set $\mathcal S^s\_\text{TOP}(X)$ and the group of homotopy classes of simple homotopy equivalences $\text{Aut}\_s(X)$. Then $\text{Aut}\_s(X)$ acts on $\mathcal S^s\_\text{TOP}(X)$ by composition and the quotient is the desired set of homeomorphism classes of closed manifolds simply homotopy equivalent to $X$.
For $X=\mathbb RP^n$ the topological structure set is [known](http://www.map.mpim-bonn.mpg.de/Fake_real_projective_spaces#Homeomorphism_classification_of_topological_actions), and the structure set always has at least 4 elements, and it is finite unless $n-3$ is divisible by $4$.
The group $\text{Aut}\_s(\mathbb RP^n)$ is trivial if $n$ is even and has order $2$ if $n$ is odd. See [Corollary 6](http://www.numdam.org/item/CM_1973__26_2_119_0.pdf) in "Coverings of fibrations" by Becker and Gottlieb.
Thus there are lots of fake even-dimensional real projective spaces in dimensions $>4$.
By Freedman's work this can be extended to dimension $4$, see [Invariant knots of free involutions on $S^4$](https://core.ac.uk/download/pdf/82800539.pdf) by Ruberman for examples of fake $\mathbb RP^4$.
|
16
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https://mathoverflow.net/users/1573
|
407390
| 166,907 |
https://mathoverflow.net/questions/407407
|
2
|
Let $\frak{g}$ be a complex simple Lie algebra and let $\frak{k}$ be a non-zero semisimple Lie subalgebra of $\frak{g}$. Is it possible to realize every simple $\frak{k}$-module $W$ as a $\frak{k}$-submodule of a some $\frak{g}$-module $V$. (Of course we are thinking about $V$ as a $\frak{k}$-module by restriction). I guess semsimplicity is necessary here, since if I start to look at solvable subalgebras I can imagine problems arising . . .
Edit: I would like to add that I am really interested in the finite dimensional case: If $V$ is f.d. will it extend to a f.d. representation of $\frak{g}$?
|
https://mathoverflow.net/users/371382
|
Extending representations of Lie subalgebras to the whole Lie algebra
|
Suppose you have an inclusion of algebraic objects $A \subset B$. In this case, $A = \mathfrak{k}$ and $B = \mathfrak{g}$ are Lie algebras, but it doesn't make a big difference — you can read "algebraic object" as "group" or "Lie group" or "algebraic group" or "Hopf algebra" or many other things. What's important is that algebraic objects have a natural notion of "module".
Now, whatever "modules" are, surely there will be a way to restrict modules: you will have a restriction functor $\mathrm{Res} : \mathrm{Mod}(B) \to \mathrm{Mod}(A)$.
This functor might or might not have an adjoint, and it depends on what "algebraic object" and "module" mean in your case. The simplest case is where a $B$-module is a vector space $V$ and a homomorphism $B \to \mathrm{End}(V)$ without any "size" constraints. In this case, abstract nonsense will promise you that $\mathrm{Res}$ has both left and right adjoints (which might or might not agree). If there are size constraints built into the meaning of "module" (for example, if you are talking about integrable modules of algebraic groups), then you might have only one adjoint. Anyway, in the case you care about of Lie algebras, there are no real issues, and $\mathrm{Res}$ has a left adjoint
$$\mathrm{Ind} : \mathrm{Mod}(A) \to \mathrm{Mod}(B),$$
called "induction". (The right adjoint to $\mathrm{Res}$, if it exists, is called "coinduction".)
Just because you have an adjoint pair, you in particular find, for every $V \in \mathrm{Mod}(A)$, a canonical natural-in-$V$ homomorphism
$$ V \to \mathrm{Res}(\mathrm{Ind}(V)).$$
Most of the time, this homomorphism will be an injection. This is in particular true when $A = \mathfrak{k}$ and $B = \mathfrak{g}$ are semisimple Lie algebras, as in your question.
|
4
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https://mathoverflow.net/users/78
|
407409
| 166,912 |
https://mathoverflow.net/questions/407369
|
0
|
Let $\mathcal{X}\_n=\{ X\_{n,\lambda}, \lambda \in \Lambda\}$ be a collection of random variables (defined on the same probability space) indexed by a deterministic index $\lambda$ over an index space $\Lambda$. Assume that for any $\lambda \in \Lambda$ it is known that $X\_{n,\lambda}\to 0$ almost surely, i.e. for any $\epsilon>0$
$$
P\{|X\_{n,\lambda}|>\epsilon \text{ i.o. }\}=0.
$$
Under which assumptions can we conclude that $P\{\sup\_{\lambda \in\Lambda}|X\_{n,\lambda}|>\epsilon \text{ i.o. }\}=0$ for all $\epsilon>0$?
I know that this problem is well studied in the case where $Y\_1, Y\_2, \ldots$ are independent and identically distributed random vectors, with common distribution $Q$, and each $X\_{n,\lambda}$ is of the form
$$
\frac{1}{n}\sum\_{i=1}^nf\_\lambda(Y\_i)-\int f\_\lambda(y)\text{d}Q(y),
$$
where $\mathcal{F}=\{f\_\lambda, \lambda \in \Lambda\}$ constitutes a class of measurable functions. In this context, appropriate metric entropy conditions yield a functional generalisation of the strong law of large numbers. Are you aware of results which hold true beyond the case of random variables in the form of centered means?
|
https://mathoverflow.net/users/148849
|
Almost sure convergence of the supremum over a class of random variables
|
The fundamental issue here is bounding the distribution of the supremum of a collection of random variables. The book "Upper and Lower Bounds for Stochastic Processes" by Michel Talagrand is largely devoted to this issue, <https://link.springer.com/book/10.1007/978-3-642-54075-2>
as is his earlier related book "the generic chaining".
|
1
|
https://mathoverflow.net/users/7691
|
407411
| 166,913 |
https://mathoverflow.net/questions/407230
|
3
|
The other day I was playing a game called Trans Europa (or Trans America) which is quite graph theoretic in flavour. The game takes place on a triangular lattice graph with certain distinguished nodes (called cities). Each player is given five cities at random. The players then take it in turns to 'claim' an edge by laying a 'rail' on it (ie. by assigning an edge as being a distinguished edge). However, a player can only set down a rail by moving their player piece from one node to another, with the starting node being assigned at random.
If there is an unbroken path of rails connecting two of a player's cities, then the two cities are said to be connected, and the first person to create a railed path which connects all five of their cities wins.
The information is not perfect, since the other player does not know which cities the other player is trying to connect. Does this game already have a known name in the mathematics literature, or is there something similar? What would the optimal strategy be? The game is essentially who can be first to move along a path which connects five random nodes on a triangular lattice, with at least one other player attempting to do the same and being able to use the other player's path. An optional variation allows a player to lay a few rails which cannot be used by the other player.
Edit: I did a literature search and it looks as if might be possible to understand the game in the framework of this [article](https://www.sciencedirect.com/science/article/pii/0097316573900058) of Erdös and Selfridge.
|
https://mathoverflow.net/users/119114
|
Game theory approach to Trans Europa
|
This is a non-answer that's growing a little large for a comment.
First, a mathematical statement of a slightly generalised version of the game. The board is a graph. Each player has a starting vertex and a set of goal vertices. At any stage there is a subgraph of marked edges. A legal move for a player is to mark any edge incident on the connected component containing their starting vertex. Players take it in turns to move, and a player wins as soon at their goal vertices are in one connected component of the marked subgraph. (This does mean that draws are possible.)
Trans Europa/America are played on a graph which is a subgraph of the triangular lattice. Goal vertices are assigned secretly and randomly from a smaller subset split into 5 colours, with each player receiving one of each colour (with no repeats). The starting vertex is chosen by each player in turn order. Players make two moves each round, but some edges are double-weighted so that they take up a whole turn, which sets up some situations in which you have to be careful about parity.
I don't think Erdős-Selfridge type positional games are the way to think about this problem, as that framework typically supposes that claimed objects belong to the claiming player. Most of the results I know about building structures in graphs are in that framework or the closely-related world of maker-breaker games. I'm pleasantly surprised by how much people take to Trans Europa and have wondered idly myself about the maths behind it to no particular end.
|
1
|
https://mathoverflow.net/users/25485
|
407414
| 166,916 |
https://mathoverflow.net/questions/407368
|
4
|
For a compact manifold $M$ the space of smooth functions $C^{\infty}(M)$ is a Fréchet space where the seminorms are the suprema of the norms of all partial derivatives. Is there some way to characterise those Fréchet subalgebras coming from a differential structure? As a naive guess, how about the following:
Conjecture: We have a bijective correspondence between smooth structures on a manifold $M$ and Fréchet algebras that are $\|-\|\_{\infty}$ norm dense in the continuous functions $C(M)$ and are maximal with respect to these properties.
Sorry if this guess if obviously wrong, but its purpose to demonstrate the type of result I am looking for.
\**Edit: I would also be interested in analogous results for topological spaces with more structure: For example, topological groups with the requirement that our Fréchet algebra contained the representable functions.*
|
https://mathoverflow.net/users/438034
|
A Fréchet space characterization of smooth structures on topological spaces?
|
The book *Smooth Manifolds and Observables* by the pseudonymous Jet Nestruev may be of interest, as it defines and studies smooth manifolds using only the algebra of smooth functions. However, their definition of smooth structure is somewhat disappointing, as it requires the algebra to be locally isomorphic to $C^\infty(\mathbb{R}^n)$. It would be nice to have an algebraic characterization of smooth structures that doesn't require us to already have the smooth structure on $\mathbb{R}^n$ in-hand. I don't know of a solution, but I can list some more properties one might impose on the algebra.
1. It should contain the constant function $1$.
2. It should separate points, or more strongly, satisfy an Urysohn lemma.
3. It should be local. Given a continuous function $f$, if every point has a neighborhood such that $f$ coincides with a function in the algebra on that neighborhood, then $f$ itself is in the algebra. I think this condition is redundant for compact manifolds. For non-compact manifolds, it rules out algebras like $C^\infty\_b(M)$, the bounded smooth functions. You might consider this also unnecessary, since $C^\infty\_b(M)$ has enough information to recover the smooth structure.
These first three properties give lower bounds, in the sense that "small" subalgebras of $C(M)$ tend to violate them, but they do hold for the entire $C(M)$. So you could try to look for algebras which are *minimal* with respect to these properties. On the other hand, there is an upper-bounding property:
4. Dimensionality. You can define the tangent space of a point as the space of derivations of the algebra, composed with evaluation at that point. Then you can require that the tangent space of every point have full dimension $n$. (By derivation I mean a linear map from the algebra to itself which satisfies the Leibniz rule. In this sense $C(M)$ has no nontrivial derivations.)
|
4
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https://mathoverflow.net/users/226696
|
407418
| 166,917 |
https://mathoverflow.net/questions/407404
|
19
|
$f \in C^2([0,1])$ with $f''$ convex and $f(0) = f'(0) = f''(0) = 0$.
Is it true that : $f''(1)+6f(1)\geq 4f'(1)$ ?
---
Source: [AoPS](https://artofproblemsolving.com/community/c7h1570936)
|
https://mathoverflow.net/users/110301
|
Strange result about convexity
|
The answer is yes, and it follows immediately from a simple description of the extreme rays of the convex cone of the functions with a convex second derivative.
Indeed, for $x\in(0,1]$, let $f'''(x)$ denote the left derivative of the convex function $f''$ at $x$, so that $f'''$ is non-decreasing on $(0,1]$.
Take any $x\in(0,1]$. Then, taking into account the conditions $f(0)=f'(0)=0$, one has the Taylor-like formula (cf. e.g. [formula (4.12)](https://projecteuclid.org/journals/banach-journal-of-mathematical-analysis/volume-10/issue-4/Convex-cones-of-generalized-multiply-monotone-functions-and-the-dual/10.1215/17358787-3649788.full) or [formula (3.2)](https://projecteuclid.org/journals/annals-of-statistics/volume-22/issue-1/Extremal-Probabilistic-Problems-and-Hotellings-T2-Test-Under-a-Symmetry/10.1214/aos/1176325373.full)):
\begin{equation\*}
f(x)=\frac{f''(0)}2\,x^2+\frac{f'''(0+)}6\,x^3+\frac16\int\_{(0,1)}df'''(t)(x-t)\_+^3, \tag{1}
\end{equation\*}
where $z\_+:=\max(0,z)$ and the integral is a Lebesgue--Stieltjes one
(see a detailed derivation of this formula at the end of this answer.)
So, $f$ is a "mixture" of the functions $x\mapsto g\_2(x):=x^2$, $x\mapsto g\_3(x):=x^3$, and $x\mapsto f\_t(x):=(x-t)\_+^3$ for $t\in(0,1)$, and the "coefficients" of the functions $f\_t$ in this mixture are nonnegative.
Therefore, it remains to note that (i) $f''(1)+6f(1)=4f'(1)$ if $f=g\_2$ or $f=g\_3$ and (ii) $f''(1)+6f(1)\geq 4f'(1)$ if $f=f\_t$ for some $t\in(0,1)$ (in the latter case, the inequality $f''(1)+6f(1)\ge 4f'(1)$ is the obvious inequality $6z+6z^3\ge12z^2$ with $z:=1-t$). $\qquad\Box$
The condition $f''(0)=0$ was not needed or used here.
---
**Details on the Taylor-like formula (1):**
Take any $x\in(0,1]$. Then
\begin{equation\*}
\begin{aligned}
f''(x)&=f''(0)+\int\_0^x du\,f'''(u) \\
&=f''(0)+\int\_0^x du\,\Big(f'''(0+)+\int\_{(0,u)}df'''(t)\Big) \\
&=f''(0)+f'''(0+)x+\int\_0^x du\,\int\_{(0,u)}df'''(t) \\
&=f''(0)+f'''(0+)x+\int\_{(0,x)}df'''(t)\int\_t^x du\, \\
&=f''(0)+f'''(0+)x+\int\_{(0,x)}df'''(t)(x-t) \\
&=f''(0)+f'''(0+)x+\int\_{(0,1)}df'''(t)(x-t)\_+;
\end{aligned}
\end{equation\*}
the fourth equality in the above display is an instance of the Fubini--Tonelli theorem.
Therefore and by a Taylor formula and the conditions $f(0)=f'(0)=0$,
\begin{equation\*}
\begin{aligned}
f(x)&=\int\_0^x du\,f''(u)(x-u) \\
&=\int\_0^x du\,\Big(f''(0)+f'''(0+)x+\int\_{(0,1)}df'''(t)(x-t)\_+\Big)(x-u) \\
&= \frac{f''(0)}2\,x^2+\frac{f'''(0+)}6\,x^3+\frac16\int\_{(0,1)}df'''(t)(x-t)\_+^3.
\end{aligned}
\end{equation\*}
One can also check this Taylor-like formula by taking the integral $\int\_{(0,1)}df'''(t)(x-t)\_+^3$ "by parts" three times -- that is, by applying the Fubini--Tonelli theorem to this integral three times.
|
15
|
https://mathoverflow.net/users/36721
|
407431
| 166,922 |
https://mathoverflow.net/questions/407377
|
4
|
If $f$ is any monic polynomial/$\mathbb{Z}$ with non-zero constant coefficient. I wish to study the quantities
$$t\_n=\sum\_{i}\theta\_i^n\in\mathbb{Z}$$
where $(\theta\_i)\_{i=1}^{d}$ are the roots of $f$ counted with multiplicity.
The main question I am interested is finding all the primes $p$ such that $p\mid t\_n$ for all large enough $n$. I have proved that this can only happen in the case that either $p\mid c\_n$ $\forall$ $n$ where $c\_n$ are the non-leading coefficients of $f$, or $p\mid t\_n$ $\forall$ $n\geq0$.
The condition that $p\mid c\_n$ is easy to check, so my interest has turned to some algebraic interpretation of the condition that $p\mid t\_n$ for all $n$. My current idea is to let $K=\mathbb{Q}(\theta\_i)\_{i=1}^{d}$ be the field attached to $f$, and analyse the ideal $I$ of $\mathcal{O}\_K$ generated by $(\theta\_i)\_{i=1}^{d}$. My idea is that it seems as if $p\mid t\_n$ $\forall$ $n$ implies that any element of $I \cap \mathbb{Z}$ will be a multiple of $p$. The converse is obvious.
Is the above observation correct? If so, is there some known property of such primes (i.e. $I\cap \mathbb{Z} \subset p\mathbb{Z}$ iff $p$ ramifies in $\mathcal{O}\_K$)?
|
https://mathoverflow.net/users/159298
|
A condition such that $p\mid\sum_{f(\theta)=0}\theta^n$ for all $n$?
|
You have $t\_k=0$ for all $k$ if and only if $f(x) \bmod p$ is a $p$-th power.
Let $g(x)$ be the image of $f(x)$ in $\mathbb{F}\_p[x]$; let $\alpha\_1$, $\alpha\_2$, ..., $\alpha\_n$ be the roots of $g$ (with multiplicity) in $\overline{\mathbb{F}\_p}$, let $e\_k$ be the $k$-th elementary symmetric function in the roots, and let $p\_k = \sum \alpha\_i^k$. You want a criterion for when all the $p\_k$ are $0$.
If each $\alpha\_i$ occurs with multiplicity divisible by $p$ then, clearly, $p\_k=0$.
Conversely, suppose all the $p\_k$ are $0$. Then, by Newton's identities, $k e\_k = 0$ for all $k$. So, whenever $p$ does not divide $k$, we have $e\_k=0$. But this means that the coefficient of $x^{(\deg g(x))-k}$ in $g(x)$ vanishes whenever $p$ does not divide $x$, so $g(x)$ is of the form $h(x^p) x^m$, where $m$ is equivalent mod $p$ to the degree of $g$. Since you imposed that $p\_0=0$ as well, the degree of $g(x)$ is $0 \bmod p$.
|
7
|
https://mathoverflow.net/users/297
|
407434
| 166,923 |
https://mathoverflow.net/questions/407391
|
2
|
Let $f$ vanishes on an open set containing 0. So there exists $l>0$ such that $f$ vanishes on $B(0,2l).$ So we can choose $g\in C\_c^\infty (\mathbb{R}^n)$ (supported on $B(0,l)$) such that $f\*g$ vanishes on an open set ( vanishes on $B(0,l)$).
My question: Is it remains true if we replace open set by set of positive Lebesgue measure i.e. if $f$ vanishes on an positive Lebesgue measure set around 0, then can we find a nonzero $g\in C\_c^\infty (\mathbb{R}^n)$ such that $f\*g$ vanishes on a set of positive Lebesgue measure?
|
https://mathoverflow.net/users/184109
|
Problem regarding vanishing set of convolution
|
As noted in Pietro Majer's comment, you can trivially take $g=0$ to get the "yes" answer.
To avoid this and make the question less trivial, one may additionally require that the function $g$ be nonnegative and nonzero. Then the answer becomes "no".
Indeed, e.g. let
$$f:=1\_{D},$$
where $D:=\mathbb R\setminus C$ and $C$ is a [fat Cantor subset](https://en.wikipedia.org/wiki/Smith%E2%80%93Volterra%E2%80%93Cantor_set) of the interval $[0,1]$, so that $C$ is a closed nowhere dense set of positive (Lebesgue) measure and hence $D$ is an everywhere dense open set.
So, $f=0$ on the set $C$ of measure $|C|>0$.
Now, take any nonnegative nonzero continuous function $g\colon\mathbb R\to\mathbb R$, so that for some real $c>0$ we have $g\ge c$ on a nonempty open interval $I$. Then, for any real $x$,
$$(f\*g)(x)=\int\_{\mathbb R}f(x-y)g(y)\,dy \\
\ge
c\int\_I f(x-y)\,dy=c|D\cap(x-I)|>0,$$
because $D$ is an everywhere dense open set and hence $D\cap(x-I)$ is a nonempty open interval.
Thus, $f>0$ on $\mathbb R$.
|
0
|
https://mathoverflow.net/users/36721
|
407459
| 166,931 |
https://mathoverflow.net/questions/407455
|
1
|
Let $A$ be a $C^\*$-algebra and $(a\_{ij}) \in M\_n(A)$ be a positive matrix. Does there exist a constant $C \ge 0$ (not depending on the $a\_{ij}$) such that
$$\lVert(a\_{ij})\rVert \le C \Bigl\lVert\sum\_i a\_{ii}\Bigr\rVert?$$
**Attempt**: Choose a faithful representation $A \subseteq B(H)$. Then $M\_n(A) \subseteq M\_n(B(H)) = B(H^{n})$ so we have
$$\lVert(a\_{ij})\rVert = \sup\_{(x\_1, \dotsc, x\_n) \in (H^n)\_1} \sum\_j \Bigl\lVert\sum\_i a\_{ji} x\_i\Bigr\rVert.$$
How to proceed?
|
https://mathoverflow.net/users/216007
|
$C\lVert\sum_i a_{ii}\rVert \ge \lVert(a_{ij})\rVert$ for matrices with entries in a $C^*$-algebra
|
This answer arose by a discussion with @JamieGabe in the comments.
One can prove that
the map
$$\Phi: M\_n(A) \to M\_n(A): A \mapsto n \operatorname{Diag}(A)-A$$
is completely positive [Paulsen, "Completely bounded maps and operator spaces", exercise 3.6].
In particular, it is positive. Hence, writing $A= (a\_{i,j})$ as in the OP, we obtain
$$A \le n \operatorname{Diag}(A)$$
and taking norms leads to
$$\|A\| \le n \lVert\operatorname{Diag}(A)\rVert=n \max\_{i=1}^n \|a\_{i,i}\| \le n \left\|\sum\_{i=1}^n a\_{i,i}\right\|.$$
|
3
|
https://mathoverflow.net/users/216007
|
407461
| 166,932 |
https://mathoverflow.net/questions/407464
|
1
|
Motivated by the answer to this [question](https://mathoverflow.net/questions/407407/extending-representations-of-lie-subalgebras-to-the-whole-lie-algebra), I will ask the following question: Let $\mathcal{A}$ and $\mathcal{B}$ be small semisimple abelian categories. Let $U:\mathcal{A} \to \mathcal{B}$ be a functor that preserves and reflects exact sequences.
Is this enough in general to give an adjoint functor such that the unit of the adjunction is injective? If not, then what are examples of assumptions we can add to make this true?
Edit: In response to the comments below, the definition of semisimple is the one taken from nLab, so yes I am assuming finite direct sums.
>
> An abelian category is called semisimple if every object is a
> semisimple object, hence a direct sum of finitely many simple objects.
>
>
>
Edit: I am happy to assume that my categories are linear, that is enriced over vector spaces.
|
https://mathoverflow.net/users/371382
|
Adjoints of exact functors between semisimple abelian categories
|
This was too big to fit as a comment. Here is a cute, completely trivial, but incredibly useful fact.
Suppose $(f^\*, f\_\*)$ is an adjoint pair of functors between additive categories (not necessary abelian or anything of that nature).
If $X$ is an object such that $f^\*X \neq 0$, then the unit map $\eta\colon X \to f\_\*f^\*X$ is not zero. This is because this map occurs in the defining relation for unit/counit of an adjunction: the composition
$$f^\*X \to f^\*f\_\*f^\*X \to f^\*X$$
is the identity.
There is a dual result using $f\_\*$ too.
The relevance of this is that if, additionally, $X$ is a "simple object" (I leave it to you to pin the meaning of this down), then $\eta$ will be automatically injective. If in your category "every object is finite length", then induction will give injectivity on everything.
There are all sorts of fun games you can play with this completely trivial fact and get non-trivial results (for eg., you can deduce auto-equivalences at the level of derived categories using the complex $id \to f\_\*f^\*$ from just observing what the functors are doing at the $K\_0$ level - I am assuming a lot of things about the category in question, abelian, finite length, etc - so don't take this too literally).
I digress though. Coming back to the original question. The interesting bit about the question, in my opinion, is the existence of the adjoint in the first place. As I hope is clear from the above, it won't take much more additional hypothesis to make the unit of adjunction to be injective (or even an isomorphism - assume $f^\*$ preserves "simples").
In this vein, the "obstruction" to the adjoint existing has a lot to do with if the category "has enough objects". That's very vague, so let me amplify with the example I alluded to in the comments.
Consider the category $\mathcal{C}$ of finite dimensional representations for a semisimple Lie algebra over $\mathbb{C}$ ($\mathfrak{sl}\_2$ will do just fine). Every representation is completely reducible, the endomorphism ring of simples is $\mathbb{C}$, etc. - everything is about as nice as you can get.
Now consider the forgetful functor from $\mathcal{C}$ to vector spaces. This is about as nice as can be: exact, reflects exact sequences, etc. I don't think this has a left adjoint. Morally, such an adjoint would be the free module for the universal enveloping algebra (this is not finite dimensional at all) over a vector space (this is the adjoint in the big category of all modules).
This isn't a proof that such an adjoint can't exist, but should indicate the difficulty in producing one if there aren't "enough objects".
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4
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https://mathoverflow.net/users/392998
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407467
| 166,935 |
https://mathoverflow.net/questions/407465
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5
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For infinite-dimensional Gaussian measures, we often see the definition of Gaussian random variables like this:
>
> Let $H(\Omega;\mathbb{R})$ be a separable Hilbert space. A random
> variable $u \in H$ is said to be a Gaussian random variable if
> $\langle u,v \rangle$ is Gaussian (i.e., $\langle u,v \rangle$ follows
> the normal distribution on $\mathbb{R}$) for all $v \in H$.
>
>
>
Even more generally, I found in this [lecture note (Definition 2.1)](https://arxiv.org/pdf/1607.03591.pdf) that Gaussian r.v. in topological vector space is defined as follows
>
> Let $W$ be a topological vector space, and $\mu$ a Borel probability
> measure on $W$. $\mu$ is Gaussian iff, for each continuous linear
> functional $f\in W\_\*$, the pushforward $\mu \circ f^{−1}$ is a Gaussian
> measure on $\mathbb{R}$.
>
>
>
My question might be a bit vague: I would like to know the intuition and motivation of such definition. It seems to me that the infinite-dimensional Gaussian measure is defined via its finite-dimensional analogy. I can think of the analogy from characteristic function $\mathbb{E}[e^{i \langle u,v \rangle}]$ or that linear combinations of Gaussians $\Leftrightarrow$ Gaussian. What benefits does this kind of definition bring? Are there other definitions of Gaussian r.v.s in infinite-dimensional spaces?
---
Found a similar [question](https://mathoverflow.net/questions/123493/what-is-a-gaussian-measure) from searching.
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https://mathoverflow.net/users/170508
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Definition of infinite-dimensional Gaussian random variable
|
Even in finite dimensions this definition is more convenient since it is independent of coordinates. If you are interested in geometric applications this is what you need.
This definition has the advantage that clarifies the nature of the the various invariants. Here are some more details.
A Gaussian measure on a a topological vector space is a Borel measure $\mu$ such that any continuous linear functional $\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\bE}{\mathbb{E}}$$\newcommand{\bC}{\mathbb{C}}$ $\xi: V\to\bR$, viewed as a random variable has a Gaussian distribution. The Gaussian measures on $\bR$ are characterized by their mean and variance. Denote by $m(\xi)$ and respectively $V(\xi)$ the mean and respectively variance of $\xi$. Observe that the map
$$
m:V^\*\to\bR,\;\; \xi \mapsto m(\xi)\in \bR
$$
is linear so it naturally lives in $V^{\*\*}$, the bidual of $V$. With a bit of luck $m$ lands in $V\subset V^{\*\*}$. This explains why in many applications some form of reflexivity is assumed about $V$.
As for the variance $V(-)$, note that it is defined by the covariance form
$$
C: V^\*\times V^\*\to \bR,\;\;C(\xi,\eta)=\bE\big[\big(\,\xi-m(\xi)\,\big)\big(\,\eta-m(\eta)\,\big)\big].
$$
The covariance form is a symmetric nonnegative definite form again on the dual $V^\*$ and $V(\xi)=C(\xi,\xi)$.
The Fourier transform (a.k.a. the characteristic function) is then the function $\newcommand{\ii}{\boldsymbol{i}}$
$$
\widehat{\mu}:V^\*\to\bC,\;\;\widehat{\mu}(\xi)= \bE\_\mu\big[ e^{\ii \xi}\,\big]=\int\_V e^{\ii \xi(v)} \mu[dv].
$$
The characteristic function $f\_\xi$ of $\xi$ is
$$
f\_\xi(t)= \widehat{\mu}(t\xi),\;\;t\in\bR.
$$
The book *Gaussian measures* by V. Bogachev adopts this point and it is worth consulting it. One source that I like very much is the fourth volume of the treatise on generalized functions by Gelfand
>
> I. M. Gelfand, N.Ya. Vilenkin: *Generalized Functions. Volume 4. Applications of harmonic Analysis*.
>
>
>
If $V$ is finite dimensional then for any $m\in V^{\*\*}$ and $C$ symmetric nonnegative definite form on $V^\*$ there exists a unique Gaussian measure with mean $m$ and covariance form $C$. This is no longer universally true in infinite dimensions. This a rather subtle issue. Things work out nicely if $V$ is the dual of nuclear space, for example if $V=C^{-\infty}(\bR^n)$ the dual of $C^\infty\_0(
\bR^n)$.
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https://mathoverflow.net/users/20302
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407468
| 166,936 |
https://mathoverflow.net/questions/407439
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16
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Let $\mathcal P$ be the set of finite subsets of $\mathbb Z\_{\geq 0}$ , each of them contains $0$. We say that $A \in \mathcal P$ is *indecomposable* if it is not $B+C$ (the sum set of $B,C$) with $B,C\in \mathcal P$ and $B,C\neq \{0\}$.
It is easy to see which small cardinality sets are indecomposable. If $|A|=2$, then it is indecomposable. If $A=\{0, x, y\}$ with $0<x<y$, A is indecomposable iff $y\neq 2x$. If $A=\{0<x<y<z\}$, A is indecomposable iff $z\neq x+y$.
My questions are:
>
> Questions: have these sets been studied before? Are there nice characterizations? Do they have interesting counts, for instance, the number of indecomposable sets whose largest element is $n$?
>
>
>
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https://mathoverflow.net/users/2083
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Sets that are not sum of subsets
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There are three questions in the OP, and I'll try to address each of them in as comprehensive a way as I can. I'll be glad to add in further details (and references) if requested.
◇ **Preliminaries on free monoids.**
We write $\mathscr F(X)$ for the [free monoid](https://en.wikipedia.org/wiki/Free_monoid) on an alphabet $X$. We refer to the elements of $\mathscr F(X)$ as *$X$-words* (or simply *words*), and to the identity of $\mathscr F(X)$, denoted by $\varepsilon\_X$, as the *empty word*. We use the symbol $\ast$ for the operation of $\mathscr F(X)$, and assume that $\mathscr F(X)$ has been realized in such a way that $X \subseteq \mathscr F(X)$. In so doing, a non-empty $X$-word $\mathfrak z$ can be uniquely factored as $z\_1 \ast \cdots \ast z\_n$ for some $z\_1, \ldots, z\_n \in X$, where $n$ is a positive integer called the *length* of $\mathfrak z$ and denoted by $\|\mathfrak z\|$. Besides, we set $\|\varepsilon\_X\| := 0$.
◇ **Pills of factorization theory.**
Let $H$ be a multiplicatively written monoid; in particular, one may want to think about the case where $H$ is the multiplicative monoid of a (unital) ring, or the multiplicative monoid of the non-zero elements of a domain (note that $H$ need not be commutative).
An *atom* of $H$ is a non-unit $a \in H$ with the property that $a \ne xy$ for all non-units $x, y \in H$ (see Note (0)). Atoms are basically a monoid-theoretic abstraction of what (most) people in commutative algebra keep calling *irreducible elements*; the origins of the term are commonly traced back to the early work of P.M. Cohn on unique factorization in non-commutative domains (see Note (1)).
Let $\mathscr A(H)$ be the set of atoms of $H$. An *atomic factorization* of an element $x \in H$ is a word $\mathfrak a \in \mathscr F(\mathscr A(H))$ such that $\pi\_H(\mathfrak a) = x$, where $\pi\_H$ is the unique monoid homomorphism $\mathscr F(H) \to H$ such that $\pi\_H(z) = z$ for every $z \in H$ (see Note (2)); we write $\mathsf L\_H(x)$ for the *set of* (*atomic*) *lengths* of $x$, that is, the set of all integers $k \ge 0$ for which there is an atomic factorization $\mathfrak a$ of $x$ with $\|\mathfrak a\| = k$. We refer to
$$
\mathscr L(H) := \{\mathsf L\_H(x): x \in H\} \setminus \{\emptyset\}
$$
as the *system of sets of* (*atomic*) *lengths* of $H$; this is probably the most fundamental invariant commonly used in factorization theory to describe the departure of $H$ from a condition of "factoriality".
◇ **Power monoids.**
Given a monoid $H$, the set of all subsets of $H$ is itself a monoid, herein denoted by $\mathcal P(H)$ and called the *full power monoid* of $H$, when endowed with the binary operation of setwise multiplication
$$
(X, Y) \mapsto \{xy: x \in X,\, y \in Y\}.
$$
Apart from trivial cases, the structure of $\mathcal P(H)$ is rather complicated, and a first step towards understanding it better is to consider submonoids of this object that are in some sense tamer, including the *reduced power monoid*
$$
\mathcal P\_{{\rm fin}, 1}(H) := \{X \in \mathcal P(H): 1\_H \in X \text{ and } |X| < \infty\}.
$$
In a way, "power monoids" have been around for a while; after all, they are the "natural framework" beyond some of the main questions of interest in arithmetic combinatorics (see [Seva's answer](https://mathoverflow.net/a/407447/16537) for some references). However, it's only very recently that, to the best of my knowledge, power monoids have been *explicitly* considered as (algebraic) structures in their own right; in particular, this is definitely true if we restrict attention to the study of their properties from the point of view of factorization theory.
The OP is about the special case where $H$ is the additive monoid $(\mathbf N, +)$ of the non-negative integers; in fact, what the OP is calling an *indecomposable subset* of $\mathbf N$ is but an atom of the restricted power monoid of $(\mathbf N, +)$, which I'll denote herein by $\mathcal P\_{{\rm fin},0}(\mathbf N)$ and write additively (in contrast with the previous section). One open problem in this area is the conjecture formulated in Sect. 5 of
* Fan and T., *Power monoids: A bridge between factorization theory and arithmetic combinatorics*, J. Algebra 512 (Oct 2018), 252-294.
The conjecture states that every non-empty finite subset $L$ of $\mathbf N\_{\ge 2}$ belongs to the system of sets of lengths of $\mathcal P\_{{\rm fin},0}(\mathbf N)$; that is, there exists a finite subset $X$ of $\mathbf N$ with $0 \in X$ such that $k \in L$ if and only if $X = A\_1 + \cdots + A\_k$ for some atoms $A\_1, \ldots, A\_k$ of $\mathcal P\_{{\rm fin},0}(\mathbf N)$. Not much is known about this problem, apart from the fact that the conjecture is true when $L$ is either the (discrete) interval $[\![2, n ]\!]$, the singleton $\{n\}$, or the two-element set $\{2, n+1\}$, where $n$ is an arbitrary integer $\ge 2$. Related questions are discussed in
* Antoniou and T., *On the Arithmetic of Power Monoids and Sumsets in Cyclic Groups*, Pacific J. Math. 312 (2021), No. 2, 279-308,
where, in particular, one can read a lot more about *minimal* atomic factorizations in the restricted power monoid of a *finite* group (see Note (3)).
◇ **Questions (freely quoted from the OP).**
Q1: *Have indecomposable subsets of the non-negative integers been studied before?*
All sorts of questions about atomic factorizations require, in the first place, an adequate understanding of the atoms of the monoid under consideration. So I'd say that the answer to this question is rather yes than no.
Q2: *Are there nice characterizations of the indecomposable subsets of the non-negative integers?*
I'm afraid this question is hopeless, at least if taken literally. In a sense (see the next question), there are way too many atoms in $\mathcal P\_{{\rm fin}, 0}(\mathbf N)$ for any *sensible* characterization to be feasible. However, it's possible to construct (infinite) families of "well-structured atoms", which can often be used to prove a number of cute little things; the two papers cited in the section "Power monoids" discuss some of these constructions.
Q3: *Is there anything in the know about the number of indecomposable subsets of the non-negative integers whose maximum is $\leq n$?*
Given $n \in \mathbf N$, let $\alpha(n)$ be the number of atoms of $\mathcal P\_{{\rm fin}, 0}(\mathbf N)$ in the interval $[\![0, n ]\!]$. [Qinghai Zhong](https://imsc.uni-graz.at/zhong/) and I have a manuscript (resting in peace in our drawers for a couple of years or so) where we prove a lower bound on $\alpha(n)$ that, in particular, implies that $\alpha(n)/2^n \to 1$ as $n \to \infty$. This is a "density result", suggesting that most elements of $\mathcal P\_{{\rm fin}, 0}(\mathbf N)$ are in fact atoms (note that $2^n$ is the total number of subsets of $[\![0, n ]\!]$ containing $0$).
◇ **Notes.**
(0) A *unit* of $H$ is an element $u \in H$ with $uv = vu = 1\_H$ for some $v \in H$, where $1\_H$ is the identity of $H$; and a *non-unit* is, unsurprisingly, an element that is not a unit.
(1) Out of the "comfort zone" of integral domains (and, more generally, of commutative "nearly cancellative" monoids), it is useful and, to some extent, even necessary to distinguish atoms from the related notion of "irreducible", where we take an *irreducible* of $H$ to be a non-unit element $a \in H$ such that, if $a = xy$ for some $x, y \in H$, then either $a$ and $x$ generate the same principal $s$-ideal (i.e., $HaH = HxH$), or so do $a$ and $y$. To my knowledge, the significance of this distinction was first realized by D.D. Anderson and S. Valdes-Leon in
* *Factorization in commutative rings with zero divisors*, Rocky Mountain J. Math. 26 (1996), No. 2, 439-480.
However, Anderson and Valdes-Leon's paper (as most of the papers in this business) is entirely about the case where $H$ is a *commutative* monoid (more precisely, the multiplicative monoid of a commutative ring). For the non-commutative part of the story and beyond, I can only address the interested reader to a recent preprint of mine on monoids and preorders ([arXiv link](https://arxiv.org/abs/2102.01598)).
(2) In this context, $\pi\_H$ is usually referred to as the *factorization homomorphism* of $H$.
(3) Power monoids are "highly non-cancellative" monoids, so much so that the classical theory of factorization breaks down dramatically when applied to them (if, for instance, $H$ is a finite monoid, then $XH = HX = H$ for every non-empty $X \subseteq H$); this "conceptual breakdown" doesn't even spare the notion of atomic factorization defined in the section "Pills of factorization theory", which has eventually led to the introduction of the refined notion of *minimal* atomic factorization in the PJM paper I'm referring to. The good news is that $\mathcal P\_{{\rm fin}, 0}(\mathbf N)$ is commutative and "nearly cancellative", which implies that, in this particular case, atomic factorizations and minimal atomic factorizations are one and the same thing.
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https://mathoverflow.net/users/16537
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407471
| 166,937 |
https://mathoverflow.net/questions/407476
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1
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Definition of 'replication of $v$' is
>
> Suppose $v \in V(G)$. Replication of $v$ is constructing $G'$ by adding a new vertex $v'$ such that $N\_{G'}(v')=N\_G(v) \cup \{v\}$.
>
>
>
And the following statement is well-known fact(you can find a proof by googling).
>
> Replication of a single vertex conserves perfectivity, i.e., $G'$ obtained from a perfect graph $G$ by replication of $v \in V(G)$ is still perfect.
>
>
>
But I am curious whether the following statement is also true:
>
> $G''$ obtained from $G$ by adding a new vertex $v''$ such that $N\_{G''}(v'')=N\_G(v)$ is perfect.
>
>
>
Adding $v''$ looks similar to the replication, but it does not contain $v$ in neighborhood of $v''$.
I could not find any counterexample to disprove it.
And proving it seems to be different from the proof of the first statement, so I cannot follow its steps.
Would you help me?
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https://mathoverflow.net/users/384338
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Pseudo-replication of a vertex in a perfect graph
|
If we pass to the complement, this is just a replication. But passing to a complement preserves the class of perfect graphs (a theorem of Lovasz, previously conjecture of Berge).
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2
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https://mathoverflow.net/users/4312
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407477
| 166,938 |
https://mathoverflow.net/questions/407480
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11
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In "[Toward a Galoisian interpretation of homotopy theory](https://arxiv.org/abs/math/0007157)" (2000), B. Toën wrote:
>
> Pour expliquer notre point de vue sur la notion de champs rappelons une construction (non conventionnelle) du topos de l’espace $X$ (i.e. d’une catégorie qui est naturellement équivalente à la catégorie des faisceaux sur X) qui n’utilise pas directement la notion de faisceaux. Pour cela, soit $Pr(X)$ la catégorie des préfaisceaux d’ensembles sur l’espace topologique $X$. Dans cette catégorie on considère l’ensemble $W$ des morphismes qui induisent des isomorphismes fibre à fibre, et on forme la catégorie $W^{−1}Pr(X)$, obtenue à partir de $Pr(X)$ en inversant formellement les
> morphismes de $W$. On peut alors vérifier que $W^{−1}Pr(X)$ est naturellement ́equivalente à la catégorie des faisceaux sur $X$. Il faut remarquer que les objets de $W^{−1}Pr(X)$ sont les préfaisceaux sur $X$, mais ses ensembles de morphismes sont en réalité isomorphes aux ensembles de morphismes entre faisceaux associés. Aussi surprenant que cela puisse paraître, cette construction montre qu’il n’est pas nécessaire de connaître la notion de faisceaux pour pouvoir parler de la catégorie des faisceaux sur $X$.
>
>
>
I would like to know if this is the first apparition of this non-conventional definition of topos.
and
Can $X$ be a general site?
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https://mathoverflow.net/users/429204
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A non-conventional definition of topoi
|
This idea originates in homotopy theory and is due to Jardine, "Simplicial presheaves", JPAA **47** Issue 1 (1987) pp 35–87, <https://doi.org/10.1016/0022-4049(87)90100-9> ([pdf](https://core.ac.uk/download/pdf/82485559.pdf)).
One can not take for $X$ any site, this definition only works in a topos with "enough points".
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13
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https://mathoverflow.net/users/1310
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407482
| 166,940 |
https://mathoverflow.net/questions/407427
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3
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The definition of a line graph is as follows:
>
> Given a graph $G$, its line graph $L(G)$ is a graph such that
>
>
> 1. each vertex of $L(G)$ represents an edge of $G$.
> 2. two vertices of $L(G)$ are adjacent if and only if their corresponding edges share a common endpoint ("are incident") in $G$.
>
>
>
I am curious about what kind of graph satisfies $\lvert E(L(G))\rvert < \lvert E(G) \rvert$.
It is a well-known fact that $$\lvert E(L(G)) \rvert=\sum\_{i=1}^n {d\_i \choose 2}$$ where the degree sequence of $G$ is $d\_1,\dotsc,d\_n$.
Now I have $$\lvert E(L(G))\rvert - \lvert E(G) \rvert<0 \Leftrightarrow \sum\_{i=1}^n d\_i(d\_i-2) < 0.$$
Obviously, isolated vertices and cycles can be ignored since $d\_i(d\_i-2)=0$.
So remaining cases are $d\_i=1$ or $d\_i \geq 3$.
I observed that if $d\_j \geq 3$ then $$d\_j < \sqrt{n}+1.$$
Otherwise $d\_j(d\_j-2) \geq n-1$ so that $$\sum\_{i=1}^n d\_i(d\_i-2) \geq d\_j(d\_j-2)+(-1)\cdot(n-1) \geq 0.$$
So I considered the union of stars with the degree of ‘center’ less than $\sqrt{n}+1$, but it didn't work well.
I have no idea how to proceed from here to find the properties of $G$.
Would you give me advice?
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https://mathoverflow.net/users/384338
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What kind of graph has more edges than its line graph?
|
For the case when $G$ is connected, we can argue as follows:
Since $\lvert E(G)\rvert=\lvert V(L(G))\rvert$, the inequality $\lvert E(L(G))\rvert<\lvert E(G)\rvert$ can be read as "$L(G)$ has more vertices than edges". Since $L(G)$ is connected as well, it must therefore be a tree. In particular, $L(G)$ cannot contain cycles. Therefore, $G$ cannot contain a vertex of degree 3 or higher and $G$ must be a path.
If $G$ is not connected, then every component which is a path (of length at least 1) loses one edge when transitioning to the line graph. If you have enough such path components, you can make up for arbitrarily many other components whose numbers of edges rise.
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5
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https://mathoverflow.net/users/108884
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407483
| 166,941 |
https://mathoverflow.net/questions/407365
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6
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For a function $x:\omega\to\mathbb R$ let $|x|$ denote the function $|x|:\omega\to[0,\infty)$, $|x|:n\mapsto|x(n)|$.
It is well-know that a series $\sum\_{n\in\omega}r\_n$ of real numbers converges unconditionally if and only if $\sum\_{n\in\omega}|r\_n|<\infty$.
On the other hand, there exists an unconditionally convergent series $\sum\_{n\in\omega}x\_n$ in $\ell\_2$ such that the series $\sum\_{n\in\omega}|x\_n|$ is divergent.
I am interested in finding conditions on an unconditionally convergent series $\sum\_{n\in\omega}x\_n$ in $\ell\_2$ guaranteeing that the series $\sum\_{n\in\omega}|x\_n|$ converges.
>
> **Question.** Let $I:\ell\_1\to\ell\_2$ be the identity operator and $(x\_n)\_{n\in\omega}$ be a sequence of elements of $\ell\_1$ such that $\sum\_{n\in\omega}\|x\_n\|^2<\infty$ and the series $\sum\_{n\in\omega}I(x\_n)$ unconditionally converges in $\ell\_2$. Is the series $\sum\_{n\in\omega}|I(x\_n)|$ convergent in $\ell\_2$?
>
>
>
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https://mathoverflow.net/users/61536
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Unconditionally convergent series in $\ell_2$ consisting of $\ell_1$-small vectors
|
If I understand correctly what you are asking, the answer is "certainly not".
Consider $k$ orthogonal vectors $v\_j$ with $k$ coordinates $\pm 1$ (the Hadamard matrix). Now multiply them by $t$ and take $n$ such identical bunches. Then the sum of $\ell^1$ norms squared is $nt^2k^3$, the squared $\ell^2$ norm of the sum of absolute value vectors is $n^2t^2k^3$ and the maximum of the squared $\ell^2$ norms of subset sums is $n^2t^2k^2$. Then playing this game on disjoint bunches of coordinates, we see that the question is whether the conditions $\sum\_m n\_mt\_m^2k\_m^3<+\infty$ and $\sum\_m n\_m^2t\_m^2k\_m^2<+\infty$ imply $\sum\_m n\_m^2t\_m^2k\_m^3<+\infty$. Now just take $n\_m=k\_m=2^m$ and $t\_m=2^{-2m}/m$, say.
But perhaps you meant something else? :-)
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6
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https://mathoverflow.net/users/1131
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407485
| 166,942 |
https://mathoverflow.net/questions/407491
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5
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Let $M^n$ be an $n$-manifold with nonempty boundary and let $\partial\_0 M$ be a specific connected component of $\partial M$. I am interested in the set of continuous maps $f : [0,1] \to M$ such that $f^{-1}(\partial M) = \{ 0,1\}$ and such that $f(0),f(1) \in \partial\_0 M$. Actually, I am interested in this set considered up to homotopies satisfying the same conditions. For the sake of having a name for it, I'll call it $\text{Arc}(M, \partial\_0 M)$.
Now, pick a base point $\ast$ in $\partial\_0 M$. There is a map $w : \text{Arc}(M, \partial\_0 M) \to \pi\_1(\partial\_0 M) \backslash \pi\_1(M) / \pi\_1(\partial\_0 M)$ given by taking whiskers from $\ast$ to the two feet $f(0)$ and $f(1)$ and pre/post-composing $f$ with these whiskers.
Is this map a bijection? Any references are welcome since I'd imagine this is well known if true.
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https://mathoverflow.net/users/99414
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Set of proper homotopy classes of arcs in a manifold
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Yes, this map is a bijection.
The condition $f^{-1}(\partial M)=\{0,1\}$ is superfluous. Using a collar neighborhood on $\partial M$ you can show that the space of continuous maps that you defined is a deformation retract of the space of maps $f\colon [0, 1]\to M$ that satisfy just the condition $f(0), f(1)\in \partial\_0 M$.
Let us denote the space of such maps by $(M, \partial\_0 M)^{(I, \partial I)}$. It is the homotopy pullback of the diagram
$$ \partial\_0 M \rightarrow M \leftarrow \partial\_0 M.$$
You are asking about $$\pi\_0 \left((M, \partial\_0 M)^{(I, \partial I)}\right).$$ The homotopy pullback square gives rise to a long exact sequence in homotopy. The relevant section of the LES is the following (implicitly choosing a basepoint in $\partial\_0 M$)
$$ \pi\_1(\partial\_0 M)\times \pi\_1(\partial\_0 M) \to \pi\_1(M) \to \pi\_0\left((M, \partial\_0 M)^{(I, \partial I)}\right)\to \pi\_0(\partial\_0 M)\times \pi\_0(\partial\_0 M)=\{1\}.$$
This is not an exact sequence of groups. Rather, there is an action of $ \pi\_1(\partial\_0 M)\times \pi\_1(\partial\_0 M)$ on $\pi\_1(M)$, where the two copies of $ \pi\_1(\partial\_0 M)$ act on the left and on the right. $\pi\_0\left((M, \partial\_0 M)^{(I, \partial I)}\right)$ is the quotient set of $\pi\_1(M)$ by this action.
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7
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https://mathoverflow.net/users/6668
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407497
| 166,945 |
https://mathoverflow.net/questions/407475
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8
|
Let $\Omega$ be a bounded domain in $\mathbb R^n$ with a smooth boundary and let $g$ be a smooth Riemannian metric on $\Omega$. Let $f\_1,f\_2,\ldots,f\_n$ be non-constant smooth functions on $\partial \Omega$ that are linearly independent from each other. Given any $j=1,2,\ldots,n$, we denote by $u\_j$, the unique harmonic function in $\Omega$ (i.e $\Delta\_gu=0$ on $\Omega$) with Dirichlet data $f\_j$.
Let us consider the set of points $p$ where the set $u\_1,\ldots,u\_n$ fails to give a coordinate system near $p$ (that is to say, $\det J=0$ where $J\_{jk}=\partial\_j u\_k(p)$). Can we say that there is only a finite number of such points in $\Omega$? If not, can we say that the measure of such points is zero?
|
https://mathoverflow.net/users/50438
|
Points where harmonic functions fail to give a coordinates system
|
In dimension two, the [Rado-Kneser-Choquet theorem](https://en.wikipedia.org/wiki/Rad%C3%B3%E2%80%93Kneser%E2%80%93Choquet_theorem) explains how to choose boundary data to obtain a non vanishing Jacobian in the interior. [Lewy's Theorem](https://projecteuclid.org/journals/bulletin-of-the-american-mathematical-society/volume-42/issue-10/On-the-non-vanishing-of-the-Jacobian-in-certain-one/bams/1183499319.full) shows that harmonic one-to-one mappings have non vanishing jacobians.
[J. C. Wood](https://www.mscand.dk/article/view/12375/10391) one page paper called "Lewy's Theorem Fails in Higher Dimensions" gives a counter-example in dimensions larger or equal to three (with an hyperplane where the jacobian cancels). Choosing the boundary data to be the trace of the map on the boundary of the domain of your choice gives you a counter example.
|
4
|
https://mathoverflow.net/users/40120
|
407503
| 166,948 |
https://mathoverflow.net/questions/407235
|
2
|
$\DeclareMathOperator\SU{SU}$Consider the Lie group $\SU(6)$, its Lie algebra $\mathfrak{su}(6)$ and the $\mathfrak{su}(2)$ subalgebra spanned by $\mathbb{1}\_3 \otimes \sigma^i$, where $\sigma^i$ are the 3 Pauli matrices, $\mathbb{1}\_3$ is the unit matrix in 3 dimensions and $\otimes$ stands for Kronecker product. I want to find the space
$$
\left\{ U\in \SU(6) \; \text{ such that} \; U\left( \mathbb{1}\_3 o\times \sigma^3\right)U^\dagger \in \mathfrak{su}(2) \right\}.
$$
i.e. the set of $U$ such that $U\left( \mathbb{1}\_3 \otimes \sigma^3\right)U^\dagger = \mathbb{1}\_3 \otimes \sigma^i \alpha^i$, for $\alpha^i \in \mathbb{R}$.
I am mostly interested in whether this set is connected or disjoint.
|
https://mathoverflow.net/users/159498
|
Set of $\mathrm{SU}(6)$ matrices which conjugate $\mathbb{1}_3 \otimes \sigma^3$ subalgebra element into $\mathfrak{su}(2)$
|
The set in question is isomorphic to the fiber product $F\times^SH$ where $F=SU(2)$, $H=S(U(3)\times U(3))$ and $S=F\cap H=U(1)$. In particular, it is a connected manifold.
This can be seen as follows: Let $\xi=1\_3\otimes\sigma^3$ und $U\in\mathfrak{su}(6)$ with $U\xi U^{-1}\in\mathfrak{su}(2)$. Since $\xi$ and $U\xi U^{−1}$ are elements of $\mathfrak{su}(2)$ with the same norm there is an $A\in SU(2)$ such that $U\xi U^{-1}=A\xi A^{-1}$ (because $\mathrm{Ad}SU(2)=SO(3)$). Thus $B:=A^{-1}U$ is in the centralizer $H$ of $\xi$ in $\mathfrak{su}(6)$. So $U=AB\in FH$. Now $A$ and $B$ are unique up to an element of $F\cap H=S$ which proves the claim.
|
5
|
https://mathoverflow.net/users/89948
|
407507
| 166,949 |
https://mathoverflow.net/questions/407505
|
-1
|
If $n\in\mathbb{N}$ is a non-negative integer, we consider it as a cardinal, so $n = \{0, \ldots, n-1\}$. If $X$ is a set, and $\kappa$ is a cardinal, we let $[X]^\kappa$ be the collection of subsets of $X$ having cardinality $\kappa$.
If $H=(V,E)$ is a [hypergraph](https://en.wikipedia.org/wiki/Hypergraph) and $\kappa\neq \emptyset$ is a cardinal, a map $c:V\to \kappa$ is said to be a *coloring* if the restriction $c\restriction\_e$ is non-constant whenever $e\in E$ has more than $1$ element. The smallest cardinal $\kappa$ such that there is a coloring map $c:V\to \kappa$ is said to be the *chromatic number* of $H=(V,E)$, and we denote it by $\chi(H)$.
**Question.** Given integers $n\geq k \geq 3$, what is the value of $\chi(n,[n]^k)$ in terms of $n$ and $k$?
|
https://mathoverflow.net/users/8628
|
Chromatic number of $(n, [n]^k)$
|
It is $\lceil \frac{n}{k-1}\rceil$, simply since every color class must contain at most $k-1$ elements.
|
2
|
https://mathoverflow.net/users/4312
|
407510
| 166,950 |
https://mathoverflow.net/questions/407521
|
1
|
Let $\{E\_j\}$ be measurable subsets in $B\_1\subset\mathbb{R}^n$ and $\exists$ $A>0$, such that
$\int\_{B\_1}\chi\_{E\_j}(x)dx\geq A$ for any $j=1,2,3,...$. Can we select a subsequence of functions $\{\chi\_{E\_j}(x)\}$ such that $\chi\_{E\_j}(x)\rightarrow\chi\_E(x)$ a.e. $x\in B\_1$, for some measurable set $E$ in $B\_1$.
Note that $\chi\_{E\_j}\in L^2(B\_1)$, then $\exists$ subsequence (we still use $E\_j$) $\chi\_{E\_j}$ converge weakly to $f\in L^2(B\_1)$. From the weakly convergence, it not hard to see $0\leq f(x)\leq 1$ for $a.e. x\in B\_1$ and $\int\_{B\_1}f(x)dx\geq A$.
The main question is that whether $f$ is a characteristic function for some measurable set in $B\_1$. Maybe there exists a counterexample. I don't know how to solve this problem.
Thanks!
|
https://mathoverflow.net/users/99411
|
a.e. convergence of characteristic functions
|
The answer is no. E.g., let $n=1$ and let $B\_1$ be the set of all irrational numbers in the interval $[0,1]$. For each $x\in B\_1$, let $b\_j(x)$ be the $j$th binary digit of $x$. For each natural $j$, let $E\_j:=\{x\in B\_1\colon b\_j(x)=1\}$, so that $\chi\_{E\_j}=b\_j$ on $B\_1$.
Then no subsequence of the sequence $(\chi\_{E\_j})$ [converges in measure](https://en.wikipedia.org/wiki/Convergence_in_measure). Indeed, otherwise there would exist an increasing sequence $(j\_k)$ of natural numbers such that the sequence $(b\_{j\_k})$ converges in measure and hence $b\_{j\_{k+1}}-b\_{j\_k}\to0$ in measure (as $k\to\infty$), so that
$$\tfrac12=\lambda(\{x\in B\_1\colon b\_{j\_{k+1}}(x)\ne b\_{j\_k}(x)\}) \\
=\lambda(\{x\in B\_1\colon|b\_{j\_{k+1}}(x)-b\_{j\_k}(x)|\ge1\})\to0,$$ where $\lambda$ is the Lebesgue measure -- a contradiction.
Thus, no subsequence of the sequence $(\chi\_{E\_j})$ converges almost everywhere.
|
1
|
https://mathoverflow.net/users/36721
|
407525
| 166,953 |
https://mathoverflow.net/questions/407488
|
3
|
Let $K$ be a number field and consider a finite Galois extension $L|K$. Moreover let $X$ be a projective, regular, integral variety over $K$. After a base change we obtain a morphism of varieties $f:X\_L\to X$. Assume that $\mathcal L$ is a $\operatorname{Gal}(L|K)$ invariant line bundle on $X\_L$; why is it true that there exists $n\in \mathbb N$ and a line bundle $M$ on $X$ such that $\mathcal L^n\cong f^\ast M$? Clearly the Galois invariance here plays a crucial role, but I cannot figure out how to use it.
Many Thanks in advance
|
https://mathoverflow.net/users/65980
|
Galois invariant line bundle and base change
|
I am posting my comment as an answer. This result is discussed in many sources. I do not have Serre's "Galois cohomology" with me at this moment, but I am certain that it is discussed there. It should be discussed also in "Dix exposes sur le Groupe de Brauer".
In fact the reference where I first learned this is Igor Dolgachev's textbook on invariant theory, Section 2 of Chapter 7. He writes the long exact sequence of the spectral sequence in a slightly different setting of a group $G$ acting on $X\_L$ through morphisms of $L$-schemes, rather than the "twisted" action of the Galois group of $L/K$ acting through morphisms of $K$-schemes. However, the long exact sequence of the short exact sequence is the same in both cases.
$$
0\to \text{Pic}(X)\to \text{Pic}(X\_L)^{\text{Aut}(L/K)} \to \text{Br}(K)
$$
Since the Brauer group of $K$ is a torsion group, every $\text{Aut}(L/K)$-invariant element of $\text{Pic}(X\_L)$ is in the image of $\text{Pic}(X)$ after replacing the invertible sheaf by all sufficiently divisible tensor powers.
|
4
|
https://mathoverflow.net/users/13265
|
407531
| 166,955 |
https://mathoverflow.net/questions/407516
|
3
|
I try to calculate the following series
\begin{align\*}
S\_{n,m}(z)=\sum\_{k=0}^{\infty} {\frac{(-1)^{k} \Gamma(\frac{2k+n+m}{2})\,\Gamma(\frac{2k+m-1}{2})}{k!\Gamma(k+\frac{m}{2})}} \, z^{2k},
\end{align\*}
where $\Gamma(z)$ is the Gamma function and where $n,m\in \mathbb Z^+$ (positive integers).
More precisely I would like to show that the series $S\_{n,m}(z)$ is an **elementary function**.
I used the Legendre's duplication formula:
\begin{align\*}
\Gamma (z)\;\Gamma \left(z+{\frac {1}{2}}\right)=2^{1-2z}\;{\sqrt {\pi }}\;\Gamma (2z),
\end{align\*}
to simplify the expression of the series $S\_{n,m}(z)$, but without success.
Otherwise, I thought of the hypergeometric functions:
$$\displaystyle {}\_{2}F\_{1}(a,b;c;z)=\frac{\Gamma(c)}{\Gamma(a)\Gamma(b)}\sum\_{n=0}^{\infty} \frac{\Gamma(a+n)\Gamma(b+n)}{\Gamma(c+n)} \frac{z^{n}}{n!}.$$
Then, returning back to $S\_{n,m}(z)$, keeping in mind the expression of the hypergeometric functions $\displaystyle {}\_{2}F\_{1}(a,b;c;z)$, we get
\begin{align\*}
S\_{n,m}(z)&=\sum\_{k=0}^{\infty} {\frac{(-1)^{k} \Gamma(\frac{2k+n+m}{2})\,\Gamma(\frac{2k+m-1}{2})}{k!\Gamma(k+\frac{m}{2})}} \, z^{2k}\\
&=\sum\_{k=0}^{\infty} {(-1)^{k} \frac{\Gamma(k+\frac{n+m}{2})\,\Gamma(k+\frac{m-1}{2})}{\Gamma(k+\frac{m}{2})}} \, \frac{(z^{2})^k}{k!}\\
&=\frac{\Gamma(\frac{n+m}{2})\Gamma(\frac{m-1}{2})}{\Gamma(\frac{m}{2})}\, {}\_{2}F\_{1}\left(\frac{n+m}{2},\frac{m-1}{2};\frac{m}{2};-z^2\right)\\
&=\frac{\Gamma(\frac{n+m}{2})\Gamma(\frac{m-1}{2})}{\Gamma(\frac{m}{2})}\, {}\_{2}F\_{1}\left(\frac{m}{2}+\frac{n}{2},\frac{m}{2}-\frac{1}{2};\frac{m}{2};-z^2\right)?
\end{align\*}
If that's right, how can I show that $S\_{n,m}(z)$ is an elementary function?
|
https://mathoverflow.net/users/84558
|
$\sum_{k=0}^{\infty} {\frac{(-1)^{k} \Gamma(\frac{2k+n+m}{2})\,\Gamma(\frac{2k+m-1}{2})}{k!\Gamma(k+\frac{m}{2})}} z^{2k}$ is an elementary function
|
As Carlo noted, for $n$ an even integer, $S\_{n,m}(z)$ is an elementary function of $z$.
What about $n$ odd?
When $n,m$ are both odd, I get something in terms of arcsinh, also elementary.
---
But for $n$ odd and $m$ even, Maple gets complete elliptic integrals, which are not elementary... Examples
$$
S\_{1,2}(z) = \frac{1}{\sqrt{z^2+1}}\;E\left(\frac{z}{\sqrt{z^2+1}\;}\right)
$$
and
$$
S\_{7,6}(z) =
3{\frac {24\,{z}^{10}+148\,{z}^{8}+398\,{z}^{6}+669\,{z}^{4}-280
\,{z}^{2}-35}{ 16\left( {z}^{2}+1 \right) ^{11/2}{z}^{4}}{E}
\left( {\frac {z}{\sqrt {{z}^{2}+1}}} \right) }-{\frac {72\,{z}^{8}+
381\,{z}^{6}+864\,{z}^{4}-1575\,{z}^{2}-210}{32\, \left( {z}^{2}+1
\right) ^{11/2}{z}^{4}}{K} \left( {\frac {z}{\sqrt {{z}^{
2}+1}}} \right) }
$$
|
6
|
https://mathoverflow.net/users/454
|
407536
| 166,957 |
https://mathoverflow.net/questions/407535
|
1
|
This question can be seen as a variant of the post [Bounded density for diffusions with diffusion coefficients bounded away from $0$](https://mathoverflow.net/questions/405624/bounded-density-for-diffusions-with-diffusion-coefficients-bounded-away-from-0) by Iosif Pinelis. Namely, consider the diffusion
$$X\_t=\int\_0^t a(s,X\_s){\bf 1}\_{\{|X\_s|<1\}}dW\_s,\quad \forall t\ge 0,$$
where $(W\_t)\_{t\ge 0}$ is a standard Brownian motion and $a$ is smooth s.t. $\inf\_{(t,x)}a(t,x)\ge c>0$. Denote by $\mu\_t$ the distribution of $X\_t$. Can we write (under suitable conditions)
$$\mu\_t(dx) = q^+\_t\delta\_{1}(dx)+q^-\_t\delta\_{-1}(dx)+ p\_t(x)dx?$$
Here $q^{\pm}\_t=\mathbb P[X\_t=\pm 1]$ and $p\_t$ (up to a normalization) is the conditional density function of $X\_t$ knowing $\{\tau>t\}$, where $\tau:=\inf\{t\ge 0: |X\_t|\ge 1\}$. So an alternative formulation is whether $X\_t$ admits a density on the event $\{\tau>t\}$.
Any solution, references or comments are appreciated.
|
https://mathoverflow.net/users/261243
|
On the marginal distributions of an absorbed diffusion
|
The answer is yes. Indeed, let
$$Y\_t=\int\_0^t a(s,Y\_s)\,dW\_s\quad \forall t\ge 0.$$
Then $X\_t=Y\_t$ on the event $\{\tau>t\}$. So, for any Borel set $A\subseteq(-1,1)$, we have
$$P(X\_t\in A)=P(Y\_t\in A,\tau>t)\le P(Y\_t\in A).$$
So, the distribution of $X\_t$ is absolutely continuous with respect to the distribution of $Y\_t$. By the [previous answer](https://mathoverflow.net/a/405686/36721), for $t>0$, the distribution of $Y\_t$ has a density (with respect to the Lebesgue measure). Thus, for $t>0$, the distribution $\mu\_t$ of $X\_t$ has a density $p\_t$ on the interval $(-1,1)$ as well, so that indeed
$$\mu\_t(dx) = q^+\_t\delta\_{1}(dx)+q^-\_t\delta\_{-1}(dx)+ p\_t(x)dx$$
for some nonnegative $q^\pm\_t$.
|
1
|
https://mathoverflow.net/users/36721
|
407538
| 166,958 |
https://mathoverflow.net/questions/407540
|
4
|
Let $t$ be a natural number. For a unitary matrix $U$ let $U\_{1,1}$ be the top left matrix element of $U$. I am trying to figure out the value of $\int |U\_{1,1}|^{2t} dH(U)$ where $H$ is the Haar measure over the unitary group.
I know that this integral could be expressed as $$\sum\_{\sigma,\tau\in S\_t} \mbox{Wg}\_d(\sigma \tau^{-1}) = t!\sum\_{\sigma\in S\_t} \mbox{Wg}\_d(\sigma)$$ where $d$ is the implicit dimension the unitary group acts on $\mbox{Wg}\_d$ is the $d$-dimensional [Weingarten function](https://en.wikipedia.org/wiki/Weingarten_function), but I have no idea how to calculate this sum.
What I can say is that (applying Weingarten's theorem in reverse) $$\sum\_{\sigma\in S\_t} \mbox{Wg}\_d(\sigma) = \int U\_{1,2} U^\dagger\_{2,1}\ldots U\_{1,t+1} U^\dagger\_{t+1,1} dH(U),$$ and for $d \gg t$ I think this should become well approximated by $$\int U\_{1,2} U^\dagger\_{2,1} dH(U) \ldots \int U\_{1,t+1} U^\dagger\_{t+1,1}dH(U)$$ (intuitively, having a large dimension makes the correlation between a small subset of matrix elements weaker). The value of $\int U\_{1,2} U^\dagger\_{2,1} dH(U)$ is exactly the expected absolute value of some element of a random complex unit vector, which again I am not sure how to calculate.
|
https://mathoverflow.net/users/42993
|
Expected even power of absolute value of an element of a random unitary matrix
|
We have
$$\mathbb{E}\_{U(N)} |U\_{1,1}|^{2t} = \binom{t+N-1}{t}^{-1}$$
for all integers $t \ge 1$. A reference is Corollary 1.2 of J. Novak, ``Truncations of random unitary matrices and Young tableaux'', Electron. J. Combin. 14 (2007), no. 1, Research Paper 21. He indeed uses a Weingarten-type approach, as part of a much more general result than this specific corollary. He studied moments of $\mathrm{Tr}(U')$ for $U'$ a submatrix of $U$ (so $U'$ could be $U$, or it could be a single entry, corresponding to your question).
This moment result in particular implies that $\sqrt{N}U\_{1,1}$ tends to a standard complex Gaussian, see his Corollary 1.3.
However, although I couldn't find a precise reference, there is a shorter and easier proof, avoiding Weingarten calculus. I'll explain this in the case of the orthogonal group $O(N)$, which goes way back and for which I have references, and all the ideas adapt to $U(N)$.
In $O(N)$, the first column of a random Haar matrix is distributed like a random point on the sphere $\mathbb{S}^{N-1}$, and Borel studied the limiting behavior of the coordinates of such points already in 1906. A standard computation of the moments of these coordinates is given in Proposition 2.5 of Elizabeth Meckes' book ``The Random Matrix Theory of the Classical Compact Groups". It yields
$$ \mathbb{E}\_{O(N)} O\_{1,1}^{2t} = \frac{\Gamma(t+\frac{1}{2})\Gamma(\frac{N}{2})}{\Gamma(t+\frac{N}{2})\Gamma(\frac{1}{2})}.$$
|
5
|
https://mathoverflow.net/users/31469
|
407545
| 166,960 |
https://mathoverflow.net/questions/407517
|
5
|
For two functions $x,y:\omega\to\mathbb R$ let $xy:\omega\to\mathbb R$, $xy:n\mapsto x(n)y(n)$, be their pointwise product.
It is clear that for any elements $x,y\in\ell\_2$ their pointwise product $xy$ is an element of $\ell\_2$.
>
> **Problem.** Let $\sum\_i x\_i$ be an unconditionally convergent series in $\ell\_2$. Is it true that for any sequence $\{y\_i\}\_{i\in\omega}\subseteq\{y\in\ell\_2:\|y\|\le 1\}$, the series $\sum\_i x\_iy\_i$ converges in $\ell\_2$?
>
>
>
|
https://mathoverflow.net/users/61536
|
A perturbation of an unconditionally convergent series in $\ell_2$
|
Volodymyr Kadets kindly informed me that the answer to this problem is affirmative. His argument easily generalizes to prove the following
>
> **Theorem.** For any $p\in[1,\infty)$, any unconditionally convergent series $\sum\_{n\in\omega}x\_n$ in the Banach space $\ell\_p$ and any sequence $\{y\_n\}\_{n\in\omega}\subseteq\{y\in\ell\_2:\|y\|\le 1\}$, the series $\sum\_{n\in\omega}x\_ny\_n$ converges unconditionally in $\ell\_p$.
>
>
>
*Proof.* Let $q\in(1,\infty]$ be such that $\frac1q+\frac1p=1$. By Bessaga-Pelczynski Theorem (6.4.3 in [this book](https://link.springer.com/book/10.1007/978-3-0348-9196-7)), to show that the series $\sum\_{n\in\omega}x\_ny\_n$ is unconditionally convergent in $\ell\_p$, it suffices to check that it is weakly absolutely convergent, i.e., for every $v\in\ell\_q=\ell\_p^\*$ we have $\sum\_{n\in\omega}|\langle v,x\_ny\_n\rangle|<\infty$.
Holder's inequality ensures that for every $x\in\ell\_p$ the product $vx$ belongs to $\ell\_1$ and the operator $T\_v:\ell\_p\to\ell\_1$, $T\_v:x\mapsto vx$, is continuous. Then the series $\sum\_{n\in\omega}vx\_n$ is unconditionally convergent in $\ell\_1$.
The absolute summability of the identity inclusion $\ell\_1\to\ell\_2$ (which follows from Grothendieck's Theorem 4.3.2 in [this book](https://link.springer.com/book/10.1007/978-3-0348-9196-7)) implies that $\sum\_{n\in\omega}\|vx\_n\|\_{\ell\_2}<\infty$.
Now observe that Cauchy-Bunyakovsky-Schwartz inequality ensures that
$$\sum\_{n\in\omega}|\langle v,x\_ny\_n\rangle|=\sum\_{n\in\omega}|\langle vx\_n,y\_n\rangle|\le\sum\_{n\in\omega}\|vx\_n\|\_{\ell\_2}\cdot\|y\_n\|\_{\ell\_2}\le\sum\_{n\in\omega}\|vx\_n\|\_{\ell\_2}<\infty.$$
|
3
|
https://mathoverflow.net/users/61536
|
407551
| 166,962 |
https://mathoverflow.net/questions/407553
|
26
|
Differential equations are at the heart of applied mathematics - they are used to great success in fields from physics to economics. Certainly, they are very useful in modelling a wide range of phenomena.
Integral equations, on the other hand, do not receive such attention. While I have seen some integral equations crop up in physics (Boltzmann equation or the tautochrone problem) or biology (population dynamics), their importance pales in comparison to differential equations.
Why is it that differential equations are so much more popular than integral ones? Or am I just ignorant of the matter and there actually are many examples of integral equations in applied mathematics?
It also seems that when an integral equation appears, one immediately wants to reduce it to a differential equation. So examples, where this is not possible or not done for different reasons would be welcome.
|
https://mathoverflow.net/users/114143
|
Importance of integral equations
|
One important point is that differential equations encode *local* behaviour of a system, while integral equations typically endcode *global* behaviour. Local behaviour is often easier to model and to grasp intuitively. In many cases, it can also be described by much simpler formulae.
More specifically:
* Let us consider the simple example where $p(t)$ describes the population of a species which reproduces without any resource limit. It is very intuitive to make the assumption that the growth of the population will be proportional to the size of the population, i.e., one has
$$
(\*) \quad
\begin{cases}
\dot p(t) & = c p(t), \\
p(0) & = p\_0
\end{cases}
$$
where $c$ is a constant, and $p\_0$ is the initial size of the population.
The reason why this behaviour is easy to model is that we have an intuitive understanding of *growth*, which is a local (with respect to time) quantity (and modelled by a derivative).
The integral equation
$$
(\*\*) \qquad p(t) = p\_0 + c \int\_0^{t} p(s) \, ds
$$
is mathematically equivalent to $(\*)$, but its intuitive meaning is more difficult to understand, since it involves the behaviour of the population over time intervals rather than only at single instances in time.
* The local character of differential equation is reflected by the fact that initial and boundary conditions can be taken into account separately. In the initial value problem $(\*)$, the initial condition $p(0) = p\_0$ is separated from the differential equation, and has a clear intuitive meaning. The equivalent integral equations $(\*\*)$ on the other hand, encodes both the dynamical behaviour of $p(t)$ and the initial condition in the same equation, which makes it more difficult to distinguish between the two effects.
* These phenomena get even more pronounced when one consides partial differential equations. For instance, the heat equation is very easy to heuristically derive locally. The behaviour at the boundary (fixed temperature = Dirichlet boundary conditions, thermal isolation = Neumann boundary conditions) can then be taken into account separately.
Reformulating the equation as an integral equations (which, for homogeneous boundary conditions, essentially comes down to computing the resolvent of the Laplace operator with the given boundary conditions) means that ones has to include the boundary conditions in the integral equation. By corollary, such an integral formulation would also need to take the geometry of the domain into account, which can be arbitrarily complicated.
On a related note, this also explains why it is impossible to explicitly compute the integral kernel of the resolvent (= Green function) of the Laplace operator on any but the most simple domains.
|
24
|
https://mathoverflow.net/users/102946
|
407557
| 166,965 |
https://mathoverflow.net/questions/406831
|
18
|
This question was motivated by [a recent MO post.](https://mathoverflow.net/q/406753/11260) You know $n$ elements of the $N\times N$ matrix $M$ and you do *not* know $n$ elements of the inverse $M^{-1}$ (but you know the other $N^2-n$ elements of $M^{-1}$). Equating $(M^{-1})^{-1}=M$ gives $n$ nonlinear equations in $n$ unknowns, which in general will have multiple solutions. Under which additional condition can one reconstruct the matrix $M$ uniquely? Does it matter where in the matrix are the $n$ elements located?
**Conjecture:** *A positive definite $N\times N$ matrix is uniquely determined by its diagonal elements and by the off-diagonal elements of its inverse.*
For $N=2$ it is true,$^\ast$ and some experimentation$^{\ast\ast}$ for larger $N$ suggests it is true for all $N$.
---
$^\ast$ For $N=2$ one has $M = \begin{pmatrix}a & b \\ b & c
\end{pmatrix}$, $M^{-1} = \frac{1}{a c - b^2} \begin{pmatrix}c & -b \\ -b & a
\end{pmatrix}$, we know $a,c$ and we know $\beta=b/(ac−b^2)$. There are two solutions for the unknown $b$, $b\_\pm=(\pm\sqrt{4ac\beta^2+1}−1)/2\beta$, only $b\_+$ gives a positive definite $M$.
$^{\ast\ast}$ [Mathematica test](https://ilorentz.org/beenakker/MO/pdtest.nb) for $N=3,4,5$, when there are, respectively, up to $5,14,22$ solutions for the unknown matrix elements, but only one of these gives a positive definite $M$.
|
https://mathoverflow.net/users/11260
|
Uniquely reconstruct a matrix $M$ from its inverse $M^{-1}$ if $n$ elements of $M^{-1}$ are unknown and $n$ elements of $M$ are given
|
The conjecture is true. More precisely, here is what I will prove:
**Theorem** Partition $\{ (i,j) : 1 \leq i \leq j \leq n \}$ into two disjoint sets, $A \sqcup B$. Let $X$ and $Y$ be positive definite $n \times n$ matrices and suppose that $X\_{ij} = Y\_{ij}$ for $(i,j) \in A$ and $(X^{-1})\_{ij} = (Y^{-1})\_{ij}$ for $(i,j) \in B$. Then $X = Y$.
**Lemma 1** Consider the function $f(M) = \log \det(M)$ on the space of $n \times n$ matrices $M$ with positive determinant. Then we can think of the gradient, $\nabla(f)$, as an $n \times n$ matrix as well. In this sense, we have $\nabla(f)|\_M = (M^{-1})^T$.
**Proof:** Let $E\_{ij}$ be the matrix with a $1$ in position $(i,j)$ and $0$'s everywhere else. Then $\nabla(f)\_{ij} = \tfrac{d}{dt} \log \det(M+t E\_{ij})$. This is $\tfrac{1}{\det(M)} \ \tfrac{d\ \det(M+t E\_{ij})}{dt} = \tfrac{1}{\det(M)} \cdot (-1)^{i+j} \det \widehat{M\_{ij}}$. Here $\widehat{M}\_{ij}$ is the $(n-1) \times (n-1)$ matrix where we delete row $i$ and column $j$. Using the formula for the inverse matrix in terms of the adjugate matrix, we see that $\nabla(f)\_{ij} = (M^{-1})\_{ji}$. $\square$
**Lemma 2** The function $f$ is strictly concave on the cone of positive definite matrices.
**Proof** Let $P$ be a positive definite matrix and let $Q$ be a symmetric matrix. We will show that $f(P+tQ)$ is concave for small $t$.
Since $P$ is positive definite, we can factor $P = ZZ^T$ for some invertible $Z$. Then $\log \det(P+Q t) = \log P + \log \det (\mathrm{Id} + t Z^{-1} Q (Z^{-1})^T)$. Since $Z^{-1} Q (Z^{-1})^T$ is symmetric, it is diagonalizable, say with eigenvalues $\lambda\_1$, $\lambda\_2$, ..., $\lambda\_n$. So $\log \det (\mathrm{Id} + t Z^{-1} Q (Z^{-1})^T) = \sum \log (1+t \lambda\_i)$. As long as $|t| < \min(|\lambda\_i|^{-1})$, this is a sum of concave functions. $\square$
Now, let $A$, $B$, $X$ and $Y$ be as in the statement of the theorem. Let $V = \mathrm{Span}\_{(i,j) \in B} (E\_{ij}+E\_{ji})$.
The condition that $X\_{ij} = Y\_{ij}$ for $(i,j) \in A$ means that $Y$ is in
the affine space $X+V$.
From Lemma 1, the condition that $(X^{-1})\_{ij} = (Y^{-1})\_{ij}$ for $(i,j) \in B$ means that $f$, when restricted to the affine space $X+V$, has the same gradient at $X$ and at $Y$.
Now, the space of positive definite matrices is convex, so its intersection with the affine space $X+V$ is convex, so we can draw a line segment from $X$ to $Y$ and the line segment will stay in the
space of positive definite matrices. By Lemma 2, our function $f$ restricted to this line segment will be strictly concave. But by our observation in the previous paragraph, this function has the same derivative at $X$ and at $Y$. This is only possible if the line segment has length $0$, so $X=Y$. $\square$
---
This was a uniqueness result. I will now prove an existence result. Let $A$ and $B$ be as above, and let $P$ and $Q$ be positive definite matrices. I will now show that there is positive definite matrix $X$ with $X\_{ij} = P\_{ij}$ for $(i,j) \in A$ and $(X^{-1})\_{ij} = Q\_{ij}$ for $(i,j) \in B$.
As above, let $V=\mathrm{Span}\_{(i,j) \in B} (E\_{ij}+E\_{ji})$. Let $K$ be the intersection of the affine space $P+V$ with the cone of positive definite matrices. Let $g(X) = \log \det X - \mathrm{Tr}(QX)$. We will show that there is a point $X$ in $K$ where $\nabla(g)=0$. Since $X \in K$, the matrix $X$ is positive definite with $X\_{ij} = P\_{ij}$ for $(i,j) \in A$. By the computations in the first part, the equation $\nabla(g)=0$ shows that $(X^{-1})\_{ij} = Q\_{ij}$.
So we are reduced to the goal of showing that $g$ has a local maximum in $K$. Fortunately, $P \in K$, so $K$ is nonempty. Unfortunately, $K$ is not compact. So we need to replace it by a compact version. Let $\overline{K}$ be the set of positive semi-definite matrices in $P+V$. So $\overline{K}$ is closed, but not yet compact.
**Lemma** There is a constant $c>0$ such that, for all positive definite matrices $X$, we have $|X| \leq c \mathrm{Tr}(QX)$, where $|X|$ is the Frobenius norm.
**Proof** The statement of the Lemma is invariant under conjugating $Q$ and $X$ by an orthogonal matrix, so we may assume that $Q$ is diagonal, with diagonal entries $0 < q\_1 < q\_2 < \cdots < q\_n$. So
$$\mathrm{Tr}(QX)= \sum\_i q\_i X\_{ii} \geq q\_1 \sum\_i X\_{ii}.$$
We also have
$$|X| = \sqrt{ \sum\_{i,j} X\_{ij}^2} \leq \sqrt{\sum\_{i,j} X\_{ii} X\_{jj}} = \sum\_i X\_{ii}.$$
(We have used that $X$ is positive definite to deduce the Cauchy-Schwartz inequality $X\_{ij}^2 \leq X\_{ii} X\_{jj}$.) So we can take $c=q\_1$. $\square$.
Now, if the eigenvalues of $X$ are $\lambda\_1$, $\lambda\_2$, \dots, $\lambda\_n$, then $\det(X) = \prod \lambda\_i$ and $|X| = \sqrt{\sum \lambda\_i^2}$ so we have $\det(X) \leq |X|^n$ and thus $\log \det(X) \leq n \log |X| \leq n (\log \mathrm{Tr}(QX)+\log c)$. So $g(X)$ is bounded above by $n \log \mathrm{Tr}(QX) - \mathrm{Tr}(QX) + n \log c$.
We have $\lim\_{z \to \infty} n \log z - z=- \infty$ so, if $\mathrm{Tr}(QX) \to \infty$ then $g(X) \to - \infty$. Thus, we can choose some large $M$ such that the value of $g$ at $P$ is larger then its value on the half space $\{ X : \mathrm{Tr}(QX) \geq M \}$.
Now, consider the function $X$ on $\overline{K} \cap \{ X : \mathrm{Tr}(QX) \leq M \}$. This, finally, is a compact set, so $g$ achieves a maximum somewhere on it. The maximum is not at the positive semidefinite points, since $g$ is $-\infty$ there, and it is not on the hyperplane $\{\mathrm{Tr}(QX) = M \}$ by the choice of $M$. So the maximum is in the interior and we win. $\square$
|
16
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https://mathoverflow.net/users/297
|
407563
| 166,967 |
https://mathoverflow.net/questions/405552
|
2
|
Let $X(t)$ be a $C^1$ (continuously differentiable) path in the Lie algebra (actually I just need finite-dimensional matrices). It is well-known (from [Wikipedia page of Derivative of the exponential map](https://en.wikipedia.org/wiki/Derivative_of_the_exponential_map), also in many Lie algebras/groups textbooks) that
$$\mathrm{Ad}\_{e^{X}} = e^{\mathrm{ad}\_{X}}$$
and that
$$
\frac{d}{dt}e^{X(t)} = e^{X(t)}\frac{1 - e^{-\mathrm{ad}\_{X}}}{\mathrm{ad}\_{X}}\frac{dX(t)}{dt}.
$$
I am wondering, **is there a formula of the adjoint action on exponential map**
$$
\frac{d}{dt} \mathrm{Ad}\_{e^{X(t)}} Y= {?}
$$
where $Y$ is in Lie algebra (or just a matrix).
Please refer to [Wikipedia page of Derivative of the exponential map](https://en.wikipedia.org/wiki/Derivative_of_the_exponential_map) for the notations for exponential map $e^X$ and adjoint action $\mathrm{Ad}\_{e^X}$:
* $e^X = \sum\_{k=0}^\infty \frac{1}{k!} X^k$
* $\mathrm{Ad}\_{e^X} Y= e^X Y e^{-X}$
* $\mathrm{ad}\_{X} Y= X Y - Y X$.
---
I found in a previous question [On the derivative of the exponential of adjoint action on a Lie algebra](https://mathoverflow.net/questions/270460/on-the-derivative-of-the-exponential-of-adjoint-action-on-a-lie-algebra) in which an [answer](https://mathoverflow.net/a/298882) stated without derivation that (rephrased in notations):
$$
\frac{d}{dt} e^{\mathrm{ad}\_{X(t)}}Y = e^{\mathrm{ad}\_{X(t)}} \left( \mathrm{ad}\_{\frac{d}{dt}X(t)} Y \right)
$$
If such formula is correct, then by the equation (proved as a Lemma in [Derivative of the exponential map](https://en.wikipedia.org/wiki/Derivative_of_the_exponential_map))
$$\mathrm{Ad}\_{e^{X}} = e^{\mathrm{ad}\_{X}},$$
the answer to my question would simply be:
$$
\frac{d}{dt} \mathrm{Ad}\_{e^{X(t)}} Y
=
\mathrm{Ad}\_{e^{X(t)}} \left( \mathrm{ad}\_{\frac{d}{dt}X(t)} Y \right).
$$
However, I am wondering, is such simple formula too good to be true? **Is there any reference asserts this formula?**
---
I am trying to derive this formula, since this formula was stated without derivation. I start with the original formula for the derivative of the exponential map:
$$
\frac{d}{dt}e^{X(t)} = e^{X(t)}\frac{1 - e^{-\mathrm{ad}\_{X}}}{\mathrm{ad}\_{X}}\frac{dX(t)}{dt}
$$
Let $\tilde{X}(t) = \mathrm{ad}\_{X(t)}$ which is a linear operatior on Lie algebra. Then, with direct substitution:
$$
\begin{aligned}
\frac{d}{dt}e^{\tilde{X}(t)}
&=
e^{\tilde{X}(t)}\frac{1 - e^{-\mathrm{ad}\_{\tilde{X}}}}{\mathrm{ad}\_{\tilde{X}}}\frac{d\tilde{X}(t)}{dt}
\\
&=
e^{\mathrm{ad}\_{X(t)}}\frac{1 - e^{-\mathrm{ad}\_{\tilde{X}}}}{\mathrm{ad}\_{\tilde{X}}} \mathrm{ad}\_{\frac{d}{dt}X(t)}
\end{aligned}
$$
The middle term is explicitly
$$
\frac{1 - e^{-\mathrm{ad}\_{\tilde{X}}}}{\mathrm{ad}\_{\tilde{X}}}
\mathrm{ad}\_{\frac{d}{dt}X(t)}
=
\sum\_{k = 0}^\infty \frac{(-1)^k}{(k + 1)!}(\mathrm{ad}\_\tilde{X})^k
\mathrm{ad}\_{\frac{d}{dt}X(t)}
$$
If this middle term is indeed identity, we would have the previous simple formula $\frac{d}{dt} e^{\mathrm{ad}\_{X(t)}} = e^{\mathrm{ad}\_{X(t)}} \mathrm{ad}\_{\frac{d}{dt}X(t)}$. In other word, the composition $\mathrm{ad}\_\tilde{X} (\mathrm{ad}\_{\frac{d}{dt}X(t)})$ is zero. To see when it is zero, I expand this composition:
$$
\begin{aligned}
\left(\mathrm{ad}\_\tilde{X} (\mathrm{ad}\_{\frac{d}{dt}X(t)})
\right)Y
&=
\left(
\mathrm{ad}\_{X(t)} \circ \mathrm{ad}\_{\frac{d}{dt}X(t)} - \mathrm{ad}\_{\frac{d}{dt}X(t)} \circ \mathrm{ad}\_{X(t)}
\right)Y
\\
&=
[X,[\frac{d}{dt}X, Y]]
-
[\frac{d}{dt}X,[X, Y]]
\\
&=
[X,[\frac{d}{dt}X, Y]]
+
[\frac{d}{dt}X,[Y, X]]
\\
&=
-
[Y,[X, \frac{d}{dt}X]] \text{ by Jacobi identity}.
\end{aligned}
$$
which requires $[Y,[X, \frac{d}{dt}X]]$ is zero. I guess it is generally not true, unless, for example, $X(t) = t X$, or we can restrict the $X$ and $Y$ satisfy this equation.
At this point, I know that if $[Y,[X, \frac{d}{dt}X]] = 0$, then we have that simple formula, otherwise, I am not sure $\frac{1 - e^{-\mathrm{ad}\_{\tilde{X}}}}{\mathrm{ad}\_{\tilde{X}}}\frac{d\tilde{X}(t)}{dt}$ could be simplified. Did I go into a bad direction in deriving the formula?
|
https://mathoverflow.net/users/13838
|
Derivative of adjoint action of exponential map
|
Long story short, here is the answer:
$$
\begin{aligned}
\frac{d}{dt}\exp{X(t)}
&=
\exp{X}\frac{1 - \exp(-\mathrm{ad}(X))}{\mathrm{ad}(X)}\frac{dX}{dt}
\\
\frac{d}{dt} \mathrm{Ad}(\exp{X(t)})
&=
\mathrm{Ad}(\exp{X}) \mathrm{ad} \left(\frac{1 - \exp(-\mathrm{ad}(X))}{\mathrm{ad}(X)}\frac{dX}{dt}\right)
\end{aligned}
$$
It is indeed a beautiful formula. I will come back to add more details on the derivation.
|
0
|
https://mathoverflow.net/users/13838
|
407573
| 166,969 |
https://mathoverflow.net/questions/407562
|
3
|
What is an example of a real solvable simply-connected Lie group $G$ whose nilradical does not have a complement (that is, $G$ is not a semidirect product of the nilradical and another subgroup)? Is it possible to produce such a group if $G$ is supersolvable (that is, $G$ can be embedded in a group of upper triangular matrices)? Or the nilradical of a triangular real Lie group always split?
I know that such examples cannot be isomorphic to an algebraic group, because by the Jordan-Chevalley decomposition the unipotent radical (and therefore the nilradical) has a complement.
|
https://mathoverflow.net/users/442553
|
Example of a supersolvable Lie group/algebra whose nilradical does not have a complement
|
$\newcommand{\mk}{\mathfrak}$Let $\mk{h}\_{2n+1}$ be the $(2n+1)$-dimensional Lie algebra (basis $x\_1,\dotsc,x\_n,y\_1,\dotsc,y\_n,z$, nonzero brackets $[x\_i,y\_i]=z$).
The $n$-dimensional abelian Lie algebra $\mk{a}\_n$ (basis $a\_1,\dotsc,a\_n$) acts on it by $a\_i\cdot x\_i=x\_i$, $a\_i\cdot y\_i=-y\_i$, rest of the action being zero.
Through the quotient map $\mk{h}\_{2n+1}\to\mk{a}\_{2n}$ one thus gets an action of $\mk{h}\_{2n+1}$ on $\mk{h}\_{4n+1}$.
Then define the semidirect product $\mk{h}\_{2n+1}\ltimes\mk{h}\_{4n+1}$, and identify the central basis elements of the two.
This thus defines a $(6n+1)$-dimensional Lie algebra with basis $x\_i,y\_i,t\_i,u\_i,v\_i,w\_i,z$ ($1\le i\le n$) and nonzero brackets
\begin{gather\*}
[x\_i,t\_i]=t\_i,\quad [x\_i,v\_i]=-v\_i,\quad [y\_i,u\_i]=u\_i, \quad[y\_i,w\_i]=-w\_i, \\
[x\_i,y\_i]=[t\_i,v\_i]=[u\_i,w\_i]=z.
\end{gather\*}
This is supersolvable. The nilpotent radical has basis $t\_i,u\_i,v\_i,w\_i,z$ ($1\le i\le n$), so the quotient is abelian with basis $x\_i,y\_i$ ($1\le i\le n$). Observe that the isomorphic image of $\mk{h}\_{2n+1}$, i.e., with basis $x\_i,y\_i,z$ ($1\le i\le n$) is a Cartan subalgebra (= nilpotent, equal to its normalizer).
This does not split (for $n\ge 1$). Indeed, suppose by contradiction it does: let $\mk{a}$ be a splitting subalgebra. Then the subspace generated by $\mk{a}$ and $z$ is an abelian $(2n+1)$-dimensional subalgebra and is a Cartan subalgebra as well, unlike the previous one which is abelian. But in a solvable Lie algebra all Cartan subalgebras are conjugate (cf. Bourbaki) under automorphisms of the Lie algebra. This is a contradiction (for $n=1$ it can be directly checked: indeed $[x\_1+a,y\_1+b]$ is never zero for any $a,b$ in the nilpotent radical).
---
Edit 1. Wondering if the above 7-dimensional example is of minimal dimension: ~~I check that the minimal dimension of an example is 6.~~ Here's ~~the~~ a 6-dimensional example (it depends in a nonzero parameter $\lambda$), without details (which are analogous to the previous one): basis $(x,y,t,u,v,w)$, nonzero brackets
$$[x,t]=\lambda t,\quad [x,u]=u,\quad [x,v]=-v,\quad [u,v]=[x,y]=z.$$
[By the next edit, this is not the smallest example. I think this becomes the smallest example if we require in addition that $[\mk{g},\operatorname{nil}(\mk{g})]=\operatorname{nil}(\mk{g})$, where $\operatorname{nil}(\mk{g})$ means the nilpotent radical.]
---
Edit 2. Looking for a reference I found Bourbaki, Lie I, §5, exercise 6:
Consider the 5-dimensional Lie algebra with basis $(u,v,w,x,y)$ and nonzero brackets
$$[u,v]=w,\quad [u,x]=x,\quad[v,y]=y.$$
Check that the subspace with basis $(w,x,y)$ is the nilpotent radical, and that $\mathfrak{g}$ is not the semidirect product of an abelian subalgebra with a nilpotent ideal (hence the nilpotent radical doesn't split, since the quotient is abelian).
|
1
|
https://mathoverflow.net/users/14094
|
407576
| 166,971 |
https://mathoverflow.net/questions/407492
|
8
|
By constructive mathematics in this matter we mean [intuitionistic ZF](https://en.wikipedia.org/wiki/Constructive_set_theory#Intuitionistic_Zermelo%E2%80%93Fraenkel) (\*).
In the language of [locales](https://ncatlab.org/nlab/show/locale), the Jordan curve can be defined as $f\colon S^1 \to X$ such that "if $U \cap V = \varnothing$, then $f(U) \cap f(V) = \varnothing$" (coincides with the classical definition for Hausdorff spaces).
So, is [Jordan's theorem](https://en.wikipedia.org/wiki/Jordan_curve_theorem#Definitions_and_the_statement_of_the_Jordan_theorem) true for locales in constructive mathematics?
(\*) I like Martin-Löf's intuitionistic theory of types more and it seems that in a sense "this is the same question", but I have not studied the type theory systematically yet, so I am formulating the question in a more familiar language.
|
https://mathoverflow.net/users/148161
|
What is the status of Jordan's theorem in constructive mathematics in the language of locales?
|
Let me first clarify some confusion in the comments to the original question. To be clear : I'm not at all saying the persons making them were confused, as far as I can tell all the comments were correct, my point is rather than the mixture of the "topological" and "localic" point of view created some confusion (which I might very well be responsible for), and I would like to clarify the distinction before we go any further.
The big point that need clarification is whether we talk about the "Topological space $\mathbb{R}$" which is constructed by starting from the set of (for exemple) Dedekind (continuous) real numbers, putting a topology on them and looking at the associated locales, or about the "formal locale $\mathbb{R}$" which is constructed as the classifying topos (or rather classifying locale) for Dedekind real number.
Many problems in non-localic constructive analysis (of the kind mentioned by Andrej Bauer in his comments) are closely related to the fact that topological space $\mathbb{R}$ is not (always) locally compact, so for example, not every map on it to $\mathbb{R}^n$ has to be bounded on $[0,1]$ or uniformly continuous on $[0,1]$.
On the other hand, the formal locale $\mathbb{R}$ is always locally compact. So if $I = [0,1]$ denote the closed sublocale $[0,1]$, any map $I \to \mathbb{R}^n$ (where I'm using the formal locale on both side) is automatically bounded and uniformly continuous.
In fact, the two definitions coincide if and only if the topological space $\mathbb{R}$ is locally compact and a map between the topological $\mathbb{R}^n$ (or open subsets of them) extend into a map between the localic version if and only if it is uniformly continuous on closed bunded subsets. So this really caputres the difference.
As the question explicitely refers to "the language of locales" I will be (from now on) only refering to the localic version when talking about $\mathbb{R}^n$ and its subspace.
Once we know that all the locale under consideration are (locally) compact and Haussdorf (By Hausdorff I mean with a closed diagonal), one can decude that all the important maps are proper from the fact that any map from a compact locale to a Hausdorff locale is proper.
In particular (and that is another difference between the topological and localic point of view), the map $I \to S^1$ that sends $t$ to $(\cos(2\pi t),sin(2 \pi t) )$ is a proper morphism, it is also a surjection (its image is a closed sublocale that contains a dense set of point), hence it is an effective descent morphisms and hence the equalizer of its Kernel pair. Its kernel pair is exactly $\Delta \coprod \{(0,1) , (1,0) \}$ (where $\Delta$ denote the diagonal), and so $S^1$ can be obtained by gluing the end point of $I$. For any locale $X$, a map $S^1 \to X$ is the same as a map $f: I \to X$ such that $f(0)=f(1)$. Note that - as observed in the comments - this is in contrast with what happen with topological spaces, where the map $I \to S^1$ fail to even be surjective on points constructively.
Finally, any map $S^1 \to \mathbb{R}^2$ is proper and uniformly continuous, in particular, any monomorphism $S^1 \to \mathbb{R}^2$ is automatically a closed embeddings (proper injections are closed embeddings). Now I'm completely convinced by the paper Andrej Bauer linked that the following is constructively\* valid:
**Theorem :** Let $j:S^1 \to \mathbb{R}^2$ be any monomorphisms of locales (between the formale locales). Then $j$ is a closed embeddings and it's open complement $U$ can be written uniquely as a union of two opens $U = U\_b \cup U\_u$ with $U\_b \cap U\_u = \emptyset$, $U\_b$ and $U\_u$ inhabited and connected (and path connected) with $U\_b$ bounded and $U\_u$ unbounded. Moreover the image of $j$ is included in the closure of both $U\_u$ and $U\_b$.
I guess a complete proof of this would require writing a (short?) paper, and I couldn't find a simple enough Barr covering argument to get something like this (at least a cheap version of it) for free (at least not without doing substancial work first to, for example, to define the index of a curve at a point, but at this point, we can as well do the full proof). But I claim that the same methods as in the paper linked by Andrej Bauer [The constructive Jordan Curves theorem](https://projecteuclid.org/download/pdf_1/euclid.rmjm/1250130636) can be applied to prove the theorem :
The fact that we look at map of locale gives us all the boundeness and uniformity assumption they need, and because we look at the "open complement" of the closed curve, we get all the condition of "being at bounded distance from the curve". Finally, while all the proof in the paper explicitely refers to points, I believe they can all be interpreted as reffering to generalized elements : when they talk about taking a point in $U$, interpret this as working internally in the topos $Sh(U)$ and using the generic point of $U$ that you have there. Of course, in order to this, some additional work is requiered to show that many construction they do can be transfered from one topos to another (they are all "geometric") but I looked at it and so far I see no problem for doing this.
\*Note : regarding the type of foundation/constructivism, I'll go to my safe place and say this is valid in any elementary topos with a natural number object. In particular it is a theorem of intuitionistic ZF. Though I'm sure if some care is taken when talking about locales, this can be made into a completely predicative statement as well, so weaker systems like constructive set theory or some form of type theory should be ok too.
|
4
|
https://mathoverflow.net/users/22131
|
407589
| 166,977 |
https://mathoverflow.net/questions/407552
|
5
|
Let $f(n)$ denote the proposition "There exists some $k>1$ such that
$$
\sum\_{m=k}^{k+n-1}\tau(m) < \sum\_{m=1}^n\tau(m)
$$
where $\tau(m)$ is the number of the divisors of $m$." (This is like the so-called second Hardy-Littlewood conjecture on $\pi(x)+\pi(y)$ vs. $\pi(x+y)$ in testing the initial interval, which has guaranteed good behavior, vs. arbitrary intervals, which can perhaps be chosen cleverly.)
$f$ seems like a natural object, so I imagine this is not a novel question. What is known about $f(n)$ in terms of truth, falsity, or conjecture? The analogy with 2HL makes it look like $f(n)$ could be true for large $n$, but I can't find any.
I've convinced myself that $f(n)$ is false for $n<1000,$ but I don't have a proof. It's easy for small $n$ but it gets increasingly inconvenient for larger $n$.
|
https://mathoverflow.net/users/6043
|
Analogue of the second Hardy-Littlewood conjecture for numbers of divisors?
|
This is false, as you suspect.
By definition, $$ \sum\_{m=k}^{k+n-1} \tau(m)$$ is the sum of the number of divisors of integers $m$ in $[k, k+n)$. Equivalently, it is the sum over $d$ of the number of $m\in [k, k+n)$ such that $d$ divides $m$.
For each $d$, the number of $m\in [k, k+n)$ such that $d$ divides $m$ is at least $\lfloor \frac{n}{d} \rfloor $ since a half-open interval of length $\frac{n}{d}$ contains at least $\lfloor \frac{n}{d} \rfloor $ integers.
For the interval $[1,n)$, this inequality is an equality - there are exactly $\lfloor \frac{n}{d} \rfloor$ integers from $1$ to $n$ that are multiples of $d$.
Thus
$$ \sum\_{m=k}^{k+n-1} \tau(m) \geq \sum\_{d=1}^n \left\lfloor \frac{n}{d} \right\rfloor = \sum\_{m=1}^n \tau(m).$$
|
7
|
https://mathoverflow.net/users/18060
|
407590
| 166,978 |
https://mathoverflow.net/questions/407592
|
2
|
If $Y\_n=\sum\_{i=1}^n X\_i$ is a martingale, where $X\_i$ is a martingale difference sequence, $\mathbb{E}[X\_n\mid \mathcal{F}\_{n-1}]=0$ for all $n$, we know that
$$ \mathbb{E}\big[Y\_n^2-Y\_{n-1}^2\big]=\mathbb{E}X\_n^2.$$
A similar property, but now as an inequality, holds if we replace the square with the absolute value,
$$ \mathbb{E}\big[|Y\_n|-|Y\_{n-1}|\big]\le\mathbb{E}|X\_n|.$$
Does something analogous hold for other powers? Namely, something along the lines of
$$ \mathbb{E}\big[|Y\_n|^r-|Y\_{n-1}|^r\big]\le C\mathbb{E}|X\_n|^r,$$
for $1<r<2$ and some $C>0$?
I guess if the distribution of $X\_n$ given $Y\_{n-1}$ were symmetric about zero, then this would be a direct consequence of von Bahr-Essen bounds (*Inequalities for the rth Absolute Moment of a Sum of Random Variables*, $1 \le r\le 2$, The Annals of Mathematical Statistics, 36), with $C=1$. Does it also hold under weaker assumptions? The von Bahr-Essen paper states the first equation (equality in case of $r=2$) as a special case.
|
https://mathoverflow.net/users/443327
|
Inequality for increments of $r$th absolute moments of martingales, $1<r<2$
|
Yes, this inequality, with the best possible $C$ ($\le 2$), was proved in [this paper](https://projecteuclid.org/journals/annals-of-functional-analysis/volume-6/issue-4/Best-possible-bounds-of-the-von-Bahr--Esseen-type/10.15352/afa/06-4-1.full); see e.g. inequality (1.11) there.
Indeed, that inequality implies that
$$E\Big|\sum\_{i=1}^m U\_i\Big|^r\le E|U\_1|^r+C\_r\sum\_{i=2}^m E|U\_i|^r,$$
where the $U\_i$'s are martingale differences and $C\_r\in[1,2]$. Taking here $m=2$, $U\_1:=Y\_{n-1}$, and $U\_2:=X\_n$, we get
$$E|Y\_n|^r\le E|Y\_{n-1}|^r+C\_rE|X\_n|^r,$$
as desired.
|
3
|
https://mathoverflow.net/users/36721
|
407595
| 166,979 |
https://mathoverflow.net/questions/407584
|
2
|
We consider any non-negative integer $n$ as a cardinal, so $0 = \emptyset$, and $n=\{0,\ldots,n-1\}$ for positive $n$. Given $n,k\in \mathbb{N}$, let $[n]^k$ denote the collection of $k$-element subsets of $n$. We say ${\cal L}\subseteq [n]^k$ is *linear* if $|a\cap b|\leq 1$ whenever $a\neq b\in {\cal L}$, and we call ${\cal L}$ *maximal linear* if ${\cal M}\subseteq [n]^k$ is no longer linear whenever ${\cal M}\supseteq {\cal L}$ and ${\cal M}\neq {\cal L}$.
Given integers $n\geq k\geq 1$, what is the minimum cardinality of any maximal linear subset of $[n]^k$, and what is the maximum cardinality, in terms of $n,k$?
|
https://mathoverflow.net/users/8628
|
Cardinalities of maximal linear $k$-subsets of $n = \{0,\ldots,n-1\}$
|
For the minimum question, this is answered in Theorem 4 of
*Erdős, Paul; Füredi, Zoltán; Tuza, Zsolt*, [**Saturated (r)-uniform hypergraphs**](http://dx.doi.org/10.1016/0012-365X(91)90035-Z), Discrete Math. 98, No. 2, 95-104 (1991). [ZBL0766.05060](https://zbmath.org/?q=an:0766.05060).
The answer is $\frac{n^2}{k(k-1)^2}+\Theta(n)$ (which is roughly a $\frac{1}{k-1}$ proportion of the maximum possible $\frac{n(n-1)}{k(k-1)}$), and you can get something a bit more precise from their proof if you need.
|
4
|
https://mathoverflow.net/users/17798
|
407608
| 166,983 |
https://mathoverflow.net/questions/407603
|
7
|
There is a lot of important recent work on the construction of Calabi-Yau metrics on non-compact complex manifolds, such as $\mathbb{C}^n$. For example:
[1] Li, Y. *A new complete Calabi-Yau metric on $\mathbb{C}^3$.* Invent. Math. 217 (2019), no. 1, 1–34.
[2] Székelyhidi, G. *Degenerations of $\mathbb{C}^n$ and Calabi-Yau metrics.* Duke Math. J. 168 (2019), no. 14, 2651–2700.
Being a complete outsider, I have a very hard time understanding what exactly they mean by a Calabi-Yau metric on a non-compact manifold such as $\mathbb{C}^n$ (they don't seem to define it in their papers). For compact manifolds, one usually defines a Calabi-Yau metric to be a metric with holonomy in $\mathrm{SU}(n)$ (see, e.g., Joyce's book *Riemannian holonomy groups and calibrated geometries* on page 54). For simply connected manifolds, this is equivalent to a Ricci-flat Kähler metric. So is a Calabi-Yau metric on $\mathbb{C}^n$ the same as a Ricci-flat Kähler metric?
What is the definition of a Calabi-Yau metric on $\mathbb{C}^n$? Are there several equivalent definitions?
|
https://mathoverflow.net/users/392184
|
What is the definition of a Calabi-Yau metric on a non-compact manifold?
|
There are two slightly different definitions. The first is that it is a Kähler metric that is Ricci-flat, and the second is that it is a Kähler metric on a (usually connected) complex $n$-manifold with holonomy in $\mathrm{SU}(n)$. They are equivalent in the simply-connected case, but not always in the non-simply connected case.
Some people reserve the term Calabi-Yau to mean that the holonomy is exactly equal to $\mathrm{SU}(n)$, so that, for example, the flat metric on $\mathbb{C}^n$ would not be Calabi-Yau in their terminology. However, this is not universal.
|
6
|
https://mathoverflow.net/users/13972
|
407610
| 166,985 |
https://mathoverflow.net/questions/407612
|
1
|
Let $k=(k\_1,k\_2) \in \mathbb{Z}^2$. Let $\lambda=(\lambda\_1,\lambda\_2)\in [0,2\pi]^2$ and $F(\lambda)$ be a bounded real function of $\lambda\in [0,2\pi]^2$.
I am interested in the following equation:
$$
\frac{1}{(2\pi)}\int\_{[0,2\pi]}\sum\_{l=-\infty}^\infty \lvert l\rvert \bigg\lvert\frac{1}{2\pi}\int\_{[0,2\pi]} e^{il\lambda\_1 }F(\lambda)d\lambda\_1 \bigg\rvert^2{d\lambda\_2}\\
+\frac{1}{(2\pi)}\int\_{[0,2\pi]}\sum\_{l=-\infty}^\infty \lvert l\rvert \bigg\lvert\frac{1}{2\pi}\int\_{[0,2\pi]} e^{il\lambda\_2 }F(\lambda)d\lambda\_2 \bigg\rvert^2{d\lambda\_1}
\\=\sum\_{k\in \mathbb{Z}^2}|k\_1|\bigg|\frac{1}{(2\pi)^2}\int\_{[0,2\pi]^2}e^{i(k\_1\lambda\_1+k\_2\lambda\_2)}F(\lambda)d\lambda\bigg|^2\\
\\+\sum\_{k\in \mathbb{Z}^2}|k\_2|\bigg|\frac{1}{(2\pi)^2}\int\_{[0,2\pi]^2}e^{i(k\_1\lambda\_1+k\_2\lambda\_2)}F(\lambda)d\lambda\bigg|^2
$$
Can anyone give me a brief explanation as to why the statement is wrong or give me a hint for the solution?
|
https://mathoverflow.net/users/151758
|
2-dimensional Fourier transform
|
$\newcommand\Z{\mathbb Z}$Suppose that $F$ is a bounded measurable complex-valued function on $[0,2\pi]^2$, so that $F\in L^2([0,2\pi]^2)$. Then
$$F(s,t)=\sum\_{(m,n)\in\Z^2}c\_{m,n}e^{i(ms+nt)} \tag{1}$$
for some complex $c\_{m,n}$'s such that $\sum\_{(m,n)\in\Z^2}|c\_{m,n}|^2<\infty$ (the above double series converges in $L^2([0,2\pi]^2)$).
Now direct calculations (see details below) show that the left-hand side (lhs) of your identity is $\sum\_{(l,n)\in\Z^2}|l|(|c\_{l,n}|^2+|c\_{n,l}|^2)$, whereas the right-hand side (rhs) of your identity is $\sum\_{(k\_1,k\_2)\in\Z^2}(|k\_1|+|k\_2|)||c\_{k\_1,k\_2}|^2$, which is obviously the same as the left-hand side.
Thus, your identity does hold.
---
**Details:** By (1),
$$I\_l:=\frac1{2\pi}\int\_0^{2\pi}e^{ils}F(s,t)ds
=\sum\_{(m,n)\in\Z^2}c\_{m,n}e^{int}1(m=l)
=\sum\_{n\in\Z}c\_{l,n}e^{int},$$
and hence
$$|I\_l|^2=I\_l\overline{I\_l}
=\sum\_{(m,n)\in\Z^2}c\_{l,m}\overline{c\_{l,n}}e^{i(m-n)t}$$
and
$$\frac1{2\pi}\int\_0^{2\pi}|I\_l|^2 dt
=\sum\_{(m,n)\in\Z^2}c\_{l,m}\overline{c\_{l,n}}1(m=n)
=\sum\_{n\in\Z}|c\_{l,n}|^2.$$
So, the first summand on the lhs of your identity is $\sum\_{(l,n)\in\Z^2}|l||c\_{l,n}|^2$ and, similarly, the second summand there is $\sum\_{(l,n)\in\Z^2}|l||c\_{n,l}|^2$, so that the lhs of your identity is $\sum\_{(l,n)\in\Z^2}|l|(|c\_{l,n}|^2+|c\_{n,l}|^2)$.
Next, again by (1),
$$\frac1{(2\pi)^2}\int\_{[0,2\pi]^2}e^{i(k\_1s+k\_2t)}F(s,t)ds\,dt \\
=\frac1{(2\pi)^2}\int\_{[0,2\pi]^2}e^{i(k\_1s+k\_2t)}\sum\_{(m,n)\in\Z^2}c\_{m,n}e^{i(ms+nt)}ds\,dt \\
=\sum\_{(m,n)\in\Z^2}c\_{m,n}1(m=k\_1,n=k\_2)=c\_{k\_1,k\_2}.$$
So, the rhs of your identity is $\sum\_{(k\_1,k\_2)\in\Z^2}(|k\_1|+|k\_2|)|c\_{k\_1,k\_2}|^2$.
|
2
|
https://mathoverflow.net/users/36721
|
407618
| 166,988 |
https://mathoverflow.net/questions/407646
|
9
|
What is an example of a ring $R$ for which the abelian category of left $R$-modules is not isomorphic to the category of right $R$-modules?
|
https://mathoverflow.net/users/371382
|
A ring for which the category of left and right modules are distinct
|
Let $Q$ be a finite acyclic quiver and $K$ a field. Let $Q^{op}$ be the opposite quiver (where all the arrows are reversed). Let $KQ$ be the path algebra. Then the category of left $KQ$-modules is equivalent to the category of right $KQ^{op}$-modules. It is well known that if $Q,Q'$ are acyclic quivers, then the categories of right $KQ$- and $KQ'$-modules are equivalent if and only if $Q$ and $Q'$ are isomorphic as directed graphs. You can recover the quiver from the Ext between simple modules. So if $Q$ is quiver not isomorphic to its opposite then its categories of left and right modules are not equivalent. One can find a suitable orientation of the Dynkin quiver $D\_4$ with this property.
**Added Nov 4, 2021.** In light of @BugsBunny's answer and the comments there, maybe it is worth making things more explicit. Let $K$ be your favorite field and consider the subalgebra of $4\times 4$-upper triangular matrices over $K$:
$$A=\begin{bmatrix} K & K & K & K\\ 0 & K & K & K\\ 0 & 0& K&0\\ 0& 0&0&K\end{bmatrix}.$$ This algebra is isomorphic to the path algebra of the Dynkin quiver $D\_4$ with a suitable orientation. This algebra has $4$ simple right modules up to isomorphism (since the radical is the ideal of strictly upper triangular matrices in $A$ and the semisimple quotient is isomorphic to the subalgebra of diagonal matrices). Exactly two isomorphism classes of simple modules are projective. The simple right projectives correspond to $$\begin{bmatrix} 0 & 0& K& 0\end{bmatrix}$$ and $$\begin{bmatrix} 0 & 0 & 0 &K\end{bmatrix}$$ acted on by right multiplication. In general the right projective indecomposables are given by the rows "in" $A$. But there is only one simple projective left module $$\begin{bmatrix} K\\ 0 \\ 0\\ 0\end{bmatrix}$$ acted on by left multiplication. In general the left projective indecomposables are given by the columns "in" $A$. So the categories of left and right modules are not equivalent.
**Added Nov. 5, 2021.** As @JeremyRickard pointed out in the comments, an easier example is the algebra
$$B=\begin{bmatrix} K & 0 & K\\ 0 & K & K\\ 0 & 0& K \end{bmatrix}.$$ Arguing as above, this has one right simple projective (coming from the action on vectors of the form of the last row) and two left simple projectives (acting on vectors of the form of the first column and on vectors of the form of the second column).
|
19
|
https://mathoverflow.net/users/15934
|
407648
| 166,994 |
https://mathoverflow.net/questions/407649
|
-1
|
Let $n$ be a positive integer. Clearly $\mathbb{R}^{n-1}$ and the interior of the $n$-simplex $
\delta\_n := \{x \in [0,1]^n:\,\Sigma\_k x\_n =1, (\forall i)\,x\_i>0\}
$ are homeomorphic. What I'm looking for is much weaker, however. Namely, does anyone know of an explicit continuous surjection:
$$
f:\mathbb{R}^{n-1}\rightarrow \delta\_n,
$$
which has a *continuous right-inverse*?
**Update:**
Consider the maps
$$
\begin{aligned}
f\_1:\mathbb{R}^{n-1} & \rightarrow (0,1)^{n-1}\\
x & \mapsto \frac1{2}\cdot \left(1+ \frac{x\_k}{1+|x\_k|}\right)\_{k=1}^{n-1}\\
f\_2:(0,1)^{n-1} & \rightarrow \delta\_n\\
x & \to \frac{\|x\|\_{\infty}}{\|x\|\_1} x I\_{x\neq 0};
\end{aligned}
$$
then, isn't $f\_2\circ f\_1$ a homomorphism; so in particular setting $f=f\_2\circ f\_1$ satisfies the conditions we are looking for.
|
https://mathoverflow.net/users/36886
|
Continuous surjection of $\mathbb{R}^{n-1}$ onto the interior of the $n$-simplex with continuous right inverse
|
Let $\Delta\subset \mathbb R^{n-1}$ be an isometric copy of $\delta\_n$ with barycenter in the origin. Let $j:\mathbb R^{n-1}\to[0,+\infty)$ be the associated Minkowski functional, defined by $x\in j(x) \partial\Delta $ for all $x\in \mathbb R^{n-1}$. Then $x\mapsto \frac {j(x)}{j(x)+1}x$ is a homeo $\mathbb R^{n-1}\to \Delta$.
|
2
|
https://mathoverflow.net/users/6101
|
407650
| 166,995 |
https://mathoverflow.net/questions/377843
|
4
|
Let $s>0$, $1<p<\infty$ and let $\Omega\subset\mathbb R^n$ be a bounded Lipschitz domain. Set $H^{s,p}(\Omega)=\{u\in L^p(\Omega):\exists\tilde u\in L^p(\mathbb R^n),\tilde u|\_{\Omega}=u,(I-\Delta)^\frac s2\tilde u\in L^p(\mathbb R^n)\}$ and $\|u\|\_{W^{s,p}(\Omega)}=\inf\limits\_{\tilde u}\|(I-\Delta)^\frac s2\tilde u\|\_{L^p(\mathbb R^n)}$ be the Bessel potential spaces.
**Question:** If $s>1$, how can we show from the definition that $\|u\|\_{W^{s,p}(\Omega)}\approx\|u\|\_{W^{s-1,p}(\Omega)}+\sum\_{i=1}^n\|\partial\_iu\|\_{W^{s-1,p}(\Omega)}$?
And how does the existence of extension operator guarantee this fact?
Indeed for each $\partial\_iu$ by definition we can only find a $W^{s-1,p}$-extension $\tilde v\_i$, but $\tilde v\_i$ may not be the gradient of some function.
If $s$ is integer, I know this is true because of the classical characterization of Sobolev norm is just the L^p-norm of the derivatives inside $\Omega$, which does not require the extension.
|
https://mathoverflow.net/users/118469
|
Equivalent norms of fractional Sobolev spaces on bounded Lipschitz domain
|
The result is mentioned in Triebel's *Function Spaces and Wavelets on Domains* Proposition 4.21. Also see Theorem 1.4 for my paper <https://arxiv.org/abs/2110.14477>
|
0
|
https://mathoverflow.net/users/118469
|
407659
| 166,997 |
https://mathoverflow.net/questions/407641
|
0
|
It is well known that the Riemann Hypothesis implies the following:
$|\theta(x) - x| = O(x^{1/2 + \epsilon})$ for all $\epsilon > 0$.
where $\theta$ is the first Chebyshev function; that is, $\theta(x)$ is the sum of the logs of all the primes up to $x$. Is it known whether the reverse implication holds? If so, could you please supply the reference? Thank you.
|
https://mathoverflow.net/users/130113
|
Question about $\theta$ and the Riemann Hypthesis - reference request
|
More is true. Let $c\in [1/2,1)$ be a constant. It is well-known that the Riemann zeta function $\zeta(s)$ has no zeros in the half plane $\Re(s)> c$ if and only if $|\theta(x)-x|=O(x^{c+\epsilon})$ for all $\epsilon>0$.
You can find this proved in most texts on the Riemann zeta function, such as Titchmarsh's book.
|
5
|
https://mathoverflow.net/users/3199
|
407664
| 167,000 |
https://mathoverflow.net/questions/394037
|
8
|
Fix even $n$ and consider the boolean function $f : \{0, 1\}^n \rightarrow \{0, 1\}$, $f : (x\_0, \ldots , x\_{n - 1}) \mapsto (x\_0 \vee x\_1) \wedge (x\_2 \vee x\_3) \wedge \cdots \wedge (x\_{n - 2} \vee x\_{n - 1})$. Fix a field $K$ and any affine hyperplane $A \subset K^n$.
**Conjecture.** If $A \cap \{0, 1\}^n \subset f^{-1}(1)$, then $\left| A \cap \{0, 1\}^n \right| \leq 2^{\frac{n}{2}}$.
**Notes.**
* Because $\left| f^{-1}(1) \right| = 3^{\frac{n}{2}}$, the claimed bound on $\left| A \cap \{0, 1\}^n \right|$ is exponentially smaller than the *a priori* maximum.
* The claimed upper bound can easily be attained: indeed, set $K := \mathbb{F}\_p$ for $p > n$ (e.g.) and set $A$ as the hyperplane $x\_1 + x\_3 + \cdots + x\_{n - 1} = \frac{n}{2}$. It's easy to check that $A \cap \{0, 1\}^n \subset f^{-1}(1)$ and $\left| A \cap \{0, 1\}^n \right| = 2^{\frac{n}{2}}$. Thus the claim is that this is the best you can do.
* Viewed as a poset with the natural ordering inherited from $\{0, 1\}^n$, $f^{-1}(1)$ is exactly the $\frac{n}{2}$-dimensional "cubical lattice" of e.g. [Metropolis and Rota, 1978](https://www.jstor.org/stable/2100984) (i.e., the facets of the $\frac{n}{2}$-cube, ordered by inclusion). $2^{\frac{n}{2}}$ is exactly the number of *vertices* (minimal elements) of this lattice.
* The problem can also be phrased "dually" in terms of subset sums. It says that if an array of field elements $a\_0, \ldots , a\_{n - 1}, a$'s subset sums "lacks adjacent elements", in the sense that every $\{i\_0, \ldots , i\_{k - 1}\} \subset \{0, \ldots , n -1 \}$ such that $\sum\_{j = 0}^{k - 1} a\_{i\_j} = a$ also satisfies $\{2 j, 2 j + 1 \} \cap \{i\_0, \ldots , i\_{k - 1}\} \neq \emptyset$ for each $j \in \{0, \ldots , \frac{n}{2} - 1\}$, then there can be at most $2^{\frac{n}{2}}$ subsets $\{i\_0, \ldots , i\_{k - 1}\} \subset \{0, \ldots , n - 1\}$ such that $\sum\_{j = 0}^{k - 1} a\_{i\_j} = a$.
* The problem is related to certain Littlewood–Offord-type problems. Indeed, a [1993 paper of Griggs](https://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.104.5579&rep=rep1&type=pdf) shows that any $A$ with nonzero coefficients satisfies $A \cap \{0, 1\}^n \leq {n \choose \frac{n}{2}}$. Thus the goal is to instead assume that $A \cap \{0, 1\}^n \subset f^{-1}(1)$, and then again try to upper-bound the size of $A \cap \{0, 1\}^n$.
* The problem seems possibly related to the "Eventown" problem (see [Babai and Frankl](https://people.cs.uchicago.edu/%7Elaci/CLASS/HANDOUTS-COMB/BaFrNew.pdf), Ex. 2.3.11). That theorem states that if a family $\mathcal{F} = \{F\_0, \ldots , F\_{m - 1}\}$ of subsets of the powerset $\mathcal{P}(\{0, \ldots , n - 1\})$ satisfies $\left| F\_i \cap F\_j \right| \equiv 0 \pmod{2}$ for each $i, j \in \{0, \ldots , m - 1\}$, then $m \leq 2^{\frac{n}{2}}$. I have been unable to reduce the problem at hand to the Eventown problem.
**EDIT:** Thanks @Antoine Labelle for the nice answer in characteristic 2. I think the general case is much harder, so I will ideally wait for that before accepting.
I care most about the case $\mathbb{F}\_p$, for $p$ a "large" prime (say, in $p \in \{2^{n-1}, \ldots , 2^n - 1\}$), and that's what I have in mind for the bounty. **I believe** this should be true for any field!
|
https://mathoverflow.net/users/92003
|
a Littlewood–Offord-type problem concerning the "cubical lattice"
|
An asymptotic reformulation of this conjecture has now been solved by Diamond and Yehudayoff in the paper *Explicit Exponential Lower Bounds for Exact Hyperplane Covers*. Preprint is available [here](https://eccc.weizmann.ac.il/report/2021/148/). The sharp form of the conjecture is still open.
|
0
|
https://mathoverflow.net/users/92003
|
407665
| 167,001 |
https://mathoverflow.net/questions/407667
|
1
|
Hellow.
I don't understand why the following formula is valid.
Can you please tell me the proof?
Let $f(x,y)$ be a function on star-shaped domain of $\mathbb{R}^2$, and let $c(x,y):=\int^1\_0tf(tx,ty)dt$.
Then
$f(x,y)=\partial\_x(xc)+\partial\_y(yc)$ holds.
|
https://mathoverflow.net/users/152099
|
An integral representation of a two-variable function
|
$$f(x,y) = \int\_0^1 \frac{d}{dt}[t^2f(tx,ty)]\, dt$$
$$=^{\ast} 2\int\_0^1 tf(tx,ty)\, dt+x\frac{\partial}{\partial x}\int\_0^1 tf(tx,ty)\, dt+y\frac{\partial}{\partial y}\int\_0^1 tf(tx,ty)\, dt$$
$$=\partial\_x(xc)+\partial\_y(yc),\;\;\text{for}\;\;c(x,y)=\int^1\_0tf(tx,ty)\,dt.$$
In the $^\ast$ equation I used that
$df(tx,ty)/dt=(x/t)\partial f(tx,ty)/\partial x+(y/t)\partial f(tx,ty)/\partial y.$
|
2
|
https://mathoverflow.net/users/11260
|
407670
| 167,002 |
https://mathoverflow.net/questions/407653
|
3
|
Let $T$ be a measure preserving bijection of a probability space $(X,\nu)$. Consider the Koopman representation of $\mathbb{Z}$ on $L^2(X,\nu)$ given by $[z.f](x) = f(T^{-z}(x))$. The question is: can I tell from the representation whether $(X,\nu,T)$ has zero entropy?
|
https://mathoverflow.net/users/23661
|
Zero entropy and the Koopman representation
|
There is an example (due to [Newton - Parry](https://mathscinet.ams.org/mathscinet-getitem?mr=206209) and also attributed by [Rokhlin](http://www.mathnet.ru/links/8b6eec06952e8e0c9d023e3caa73e40b/rm5788.pdf) to Girsanov) of a zero entropy measure preserving transformation with a countable Lebesgue spectrum (and mixing of all orders). Therefore, the associated Koopman operator is unitarily equivalent to that of any Bernoulli shift.
|
3
|
https://mathoverflow.net/users/8588
|
407683
| 167,007 |
https://mathoverflow.net/questions/407677
|
7
|
Do there exists rational numbers $x$ and $y$ such that
$$
y^3 = x^4 + x + 2 ?
$$
Context: There are a lot of publications about computing rational points on elliptic and hyperelliptic curves, and these problems has been solved in a number of special cases. The "next simplest" case are Picard curves, which can be described by equations in the form $y^3=P(x)$, where $P(x)$ is a polynomial of degree 4. An important parameter measuring the complexity of this problem is the rank of the Jacobian of the curve. The rank can be computed using RankBounds function in Magma, and this particular curve has rank 0. There is a magma function (Chabauty0) for computing rational points on rank 0 hyperelliptic curves, but Picard curves are not hyperelliptic. On the other hand, Hashimoto and Morrison <https://arxiv.org/abs/2002.03291> recently showed how to compute rational points on Picard curves of rank 1. Based on this, it looks like the case of Picard curves of rank 0 should be tractable, but I cannot find any reference, and also cannot solve this myself, hence the question. The particular equation above is chosen because it is the simplest example of such curve for which there is a rational point everywhere locally but there are no obvious rational points.
|
https://mathoverflow.net/users/89064
|
$y^3 = x^4 + x + 2$, and existence of rational points on rank 0 Picard curves
|
The below Magma code determines the size of $J\_C(\mathbb{F}\_p)$ for various primes $p$, and finally compute the GCD of their orders, which gives you a bound on the size of $J\_C(\mathbb{Q})$. For a discussion of this, see Section 4.1 of [this paper](https://arxiv.org/pdf/2007.13929.pdf). To determine $J\_C(\mathbb{F}\_p)$, you can just evaluate the numerator of the zeta function of the reduction of $C$ mod $p$ (see e.g., Hindry--Silverman *Diophantine Geometry* Exercise C.4).
The below code does this for primes up to 30 and tells you that the GCD of $\#J\_C(\mathbb{F}\_p)$ for these primes is 1, and hence $J\_C(\mathbb{Q})$ is just the trivial group. The code is general and can work for any equation of this form of rank 0. If you do have some non-trivial torsion in $J\_C(\mathbb{Q})$, then you will need to work harder to determine the rational points.
I will also note that you can run this on the online magma calculator.
```
P2<x,y,z> := ProjectiveSpace(Rationals(),2);
C := Curve(P2,z*y^3 - (x^4 + x*z^3 + 2*z^4));
bad := {2,3};
bool := false; p := 2;
torsOrders := {@@};
while p le 30 do
p := NextPrime(p);
p := p in bad select NextPrime(p) else p;
Cp := Curve(Reduction(C,p));
torsOrders := torsOrders join {@ Evaluate(Numerator(ZetaFunction(Cp)),1) @};
end while;
GCD(torsOrders); //1
```
|
4
|
https://mathoverflow.net/users/56667
|
407686
| 167,009 |
https://mathoverflow.net/questions/407655
|
8
|
This question is related to [Monotone version of one-dimensional Whitney extension theorem](https://mathoverflow.net/q/215153/121665).
Let $m$ be a positive integer or $m=\infty$.
>
> Suppose that $E\subset\mathbb{R}$ is a closed set and $f:E\to\mathbb{R}$ is a non-decreasing (strictly increasing) function such that there is a function $F\in C^m(\mathbb{R})$ that coincides with $f$ on $E$. Does it follow that we can choose $F$ to be non-decreasing (strictly increasing)?
>
>
>
This looks like a reasonable conjecture, but I could not find it anywhere in the literature and I would like to know if this is a known fact.
**Edit.** This problem was not well thought. I have to think about it again and try to formulate it in a more reasonable way.
|
https://mathoverflow.net/users/121665
|
Whitney extension theorem preserving monotonicity
|
No also in the strictly increasing case.
Let $E = [0,1]$ and $f: x\mapsto x^2$ is strictly increasing, and is the restriction of a $C^\infty$ function. Any $C^2$ extension of this function must have $F'(0) = 0$ and $F''(0) = 2$, and so for some $\epsilon > 0$ must have $F'(-\epsilon) < 0$.
---
On the other hand, I believe the following refined conjecture is true:
>
> If $f = F|\_E$ where $F\in C^m$, $m\geq 1$. Suppose $f$ is strictly increasing and $F'|\_E > 0$. Then $f$ admits a monotone extension.
>
>
>
|
9
|
https://mathoverflow.net/users/3948
|
407688
| 167,010 |
https://mathoverflow.net/questions/407593
|
4
|
After a long story of dancing around the effective topos $ \mathcal{Eff}$, I finally resolved to get to the bottom of it. To this effect, working as it were backward, I ended up revisiting Kleene's original definition of realizability for Heyting Arithmetic (1949), the "*Mother of all realizabilities*" .
To my surprise, I found something which makes me a bit uneasy. The definition for connectives and quantifiers are what one would expect, keeping in mind the so-called $\mathbf{BHK} $interpretation.
However, here is the definition of ground zero realizability, ie how a number realizes a basic sentence:
>
> A number $n$ realizes an atomic formula $s=t$ if and only if $s=t$ is true.
> Thus every number realizes a true equation, and no number realizes a false equation.
>
>
>
You might ask: what is wrong with that? The answer is, obviously nothing. A definition is a definition.
However, keeping in mind the overarching goal, which is to create a constructive semantics, it makes me balk a little. Say that you have two (closed ) terms in the language of $\mathbf{HA}$, and you intend to show that they denote the same number. What do you do? *You calculate*. In other words, you reduce the first to the second, by either appealing to direct calculation, or some short cuts. Thus, this calculation, suitably encoded, is a realizer for the equality. Appealing to "truth " in $N$ goes against the spirit of realizability. Even if you are not a skeptic of crystal clear $N$, in this era of $\mathbf{HoTT}$, infinity groupoids, and generally *replacing deprecated equality for equivalences*, one would hope for an amendment of the above.
If you like, an "**homotopical realizability**", where the above gets morphed into something like
>
> $e$ realizes $t=s$ iff $e$ is the code of a reduction path from t to $s$.
>
>
>
Questions:
1. **Does there exist already some kind of modified realizability in the lines mentioned above?**
2. **Assuming the first answer is yes, how does this reflects within the toposophical standpoint?**
Comment: it is clear that if we work with homotopical realizability, the truth value of $t=s$ is not the entire set of natural numbers, but rather the set of coded paths between the two terms (their path space). Thus, even not jumping directly into realizability topoi, and assuming some more simple minded approach like the category of assemblies, adjustments have to be made. Very curious to know which ones.
**ADDENDUM**
Reflecting a bit further, and also after the little exchange with Andreas, I have realized a few things:
1. if one swaps KLeene's with my proposed realizability, *the topos remains exactly the same*. Why? because the realizability topos depends exclusively on the partial combinatory algebra, which in this case is still what it was before. The two notions of realizability change only the way sentences are evaluated, the truth values, nothing else.
2. Thus, the only major change is this: in the Effective topos we have of course the internal language, and when it is applied to the Natural Number Object, it maps precisely Kleene's. It is a theorem, which for instance is proved in [Vermeeren, 2009](https://stijnvermeeren.be/download/mathematics/essay.pdf).
3. What next? Well, the obvious question is, is there another object in the topos such that its internal logic corresponds to the realizability I have sketched? Open question, though I do have a hunch...
|
https://mathoverflow.net/users/15293
|
Homotopical realizability
|
The reference to "truth in $\mathbb{N}$" is a mirage. As Andreas Blass points out, there is a computable procedure that determines whether $s = t$ holds for closed term $s$ and $t$ of HA. One needs only run a certain algorithm to find out whether $s = t$, there is no "faith in the model".
And because equality is decidable, nothing is gained by providing a realizer of $s = t$. (We can just throw it away and reconstruct it at will whenever needed.)
In my opinion you have gone too far back in history. Kleene's work was about *first-order arithmetic*, and since equality of natural numbers is decidable in HA, nothing interesting can be discovered by considering its realizers.
The right historical moment to pay attention to is the invention of the effective topos, because that is the first realizability model with effective quotients. There equality *does* in general have computational content and so is informative. (For example, a realizer of equality of truth values $p, q \in \Omega$ is a pair $\langle a, b \rangle$ such that $a$ realizers $p \Rightarrow q$ and $b$ realizers $q \Rightarrow p$.)
We can conjure up an object of the effective topos in which a realizer of $s = t$ does in fact encode a "path" or a "reduction sequence", but for such an example to be non-trivial, the reduction sequence must not be computable from $s$ and $t$ alone (otherwise it will end up being $\lnot\lnot$-stable).
P.S. There is homotopical realizability, but it may not be what you asked for. See Uemura's [cubical assemblies](https://arxiv.org/abs/1803.06649).
#### Addendum
We may indeed define an object in the effective topos such that equality is realized by (encodings of) reduction sequences. Note that any such object, in order not to be isomorphic to one whose equality has trivial realizers, must *not* be an assembly. This rules out any notion of reduction for which the predicate "$p$ is (the code of) a reduction step from $s$ to $t$" is decidable. Indeed, if such a predicate is decidable, then we may reconstruct a reduction sequence from $s$ to $t$ whenever it exists, by simply enumerating all possibilities and looking for one that works. Consequently, the resulting equality relation will be $\lnot\lnot$-stable, and the object isomorphic to one with trivial realizers of equality.
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4
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https://mathoverflow.net/users/1176
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407690
| 167,011 |
https://mathoverflow.net/questions/407691
|
22
|
First of all sorry for this non-research post.
I was watching [*Jeffrey Blitz* Lucky](https://rads.stackoverflow.com/amzn/click/com/B004JM5S9I) documentary movie and it was interesting to me that a winner of Lottery was a math Ph.D. from Berkeley.
In the movie he said:
>
> **Robert:** ... when I was in Berkeley, I had good fortune to take a course in differential geometry from professor S. S. Chern because loving the class but more to the point I was getting inspired about differential geometry because **the way Chern taught**.
>
>
>
There are lots of articles on the web about Chern's research, works, life and I remember someone called him a legendary mathematician.
But I want to know what is so special about Chern's way of teaching?
Or the above compliment is because of the intrinsic enjoyment of differential geometry that he is relating to Chern? I am curious to know some examples of Chern's teaching that you are aware of.
|
https://mathoverflow.net/users/90655
|
What is so special about Chern's way of teaching?
|
[Louis Auslander](https://en.wikipedia.org/wiki/Louis_Auslander) has described his experiences on [S.S. Chern as a teacher:](https://academic.hep.com.cn/skld/EN/chapter/9781571462411/chapter14)
>
> Somehow Chern conveyed the philosophy that making mistakes was normal
> and that passing from mistake to mistake to truth was the doing of
> mathematics. And somehow he also conveyed the understanding that once
> one began doing mathematics it would naturally flow on and on. Doing
> mathematics would become like a stream pushing one on and on. If one
> was a mathematician, one lived mathematics.... and so it has turned out [for me].
>
>
>
[Robert Uomini](https://www.berkeley.edu/news/media/releases/96legacy/releases.96/14351.html) set aside money from the $22 million lottery jackpot he won in 1995 to endow a visiting professorship in the Department of Mathematics in honor of mathematics professor Shiing-Shen Chern, who he credits for getting him into graduate school.
This is how he remembers Chern:
"I loved his lectures," Uomini said. He found Chern's undergraduate course in differential geometry so exciting and stimulating that "by the end of the class I felt I wanted to become a differential geometer."
As he completed his coursework for an A.B. in mathematics in 1969, Uomini applied to graduate school at Berkeley, but despite a letter from Chern he was turned down, primarily because of his poor grades. Chern urged him to reapply, though, and this time, with Chern's strong support, he got in. Uomini completed his PhD work in 1976.
"I got in only because of Chern," Uomini said. "Since my graduation, in order to recognize my indebtedness to Chern, I wanted to create a chair in his name."
|
37
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https://mathoverflow.net/users/11260
|
407692
| 167,012 |
https://mathoverflow.net/questions/407700
|
15
|
In teaching my linear algebra students about Gram-Schmidt orthogonalization, I found a curious sequence of polynomials. They are closely related to Legendre polynomials, but they also appear to be related to Catalan numbers. (Several of the statements below are conjectural and I am not an expert on orthogonal polynomials, so please bear with me.)
Recall that if we apply the Gram-Schmidt process to the sequence $\{1,t,t^2,t^3,...\}$, where the inner product is given by $\left<f,g\right>=\int\_{-1}^1 f(t)g(t)dt$, then one obtains a the Legendre polynomials.
Doing Gram-Schmidt for the first time is always a pain, and I wanted to make the problem easier to do by hand by choosing the initial basis in such a way to avoid a lot of uncomfortable fractions. I gave my students the set $\{1,2t,6t^2,20t^3\}$ and told them to use the inner product $\left<f,g\right>=\int\_{0}^1 f(t)g(t)dt$. (This yields a sort of "shifted" version of the Legendre polynomials.) Notice that the coefficients of these monomials are the "central" binomial coefficients $\frac{(2n)!}{n!^2}$. If we apply Gram-Schmidt to these, then we get the polynomials $\{1,2t-1,6t^2-6t+1,20t^3-30t^2+12t-1\}$. (In spite of my efforts, one of my students declared that, upon finding these, he could no longer feel joy.)
The new polynomials that I want to know about are obtained by writing these "shifted" Legendre polynomials as linear combinations of the initial polynomials. Thus, in this instance, we have
$$\left[\begin{array}{c}
1 \\
t-1 \\
6t^2-6t+1 \\
20t^3-30t^2+12t-1 \\
\end{array}\right]
=\left[\begin{array}{cccc}
1 & 0 & 0 & 0 \\
-1 & 1 & 0 & 0 \\
1 & -3 & 1 & 0 \\
-1 & 6 & -5 & 1 \\
\end{array}\right]
\left[\begin{array}{c}
1 \\
2t \\
6t^2 \\
20t^3 \\
\end{array}\right].$$
We use the coefficient matrix here to define a new sequence by
$$\left[\begin{array}{c}
f\_0 \\
f\_1 \\
f\_2 \\
f\_3 \\
\end{array}\right]
=\left[\begin{array}{cccc}
1 & 0 & 0 & 0 \\
-1 & 1 & 0 & 0 \\
1 & -3 & 1 & 0 \\
-1 & 6 & -5 & 1 \\
\end{array}\right]
\left[\begin{array}{c}
1 \\
t \\
t^2 \\
t^3 \\
\end{array}\right]
=\left[\begin{array}{c}
1 \\
t-1 \\
t^2-3t+1 \\
t^3-5t^2+6t-1 \\
\end{array}\right].$$
I hope that the reader now understands how to define $f\_n$ for all $n$. Use the coefficients that are required to write the orthogonal polynomials with respect to the monomials $\frac{(2n)!}{n!^2}t^n$.
My first thought about these was that they should also be a sequence of orthogonal polynomials. They seem to have the root-interlacing property, although the roots appear to be unbounded. This makes me think that they are related to Laguerre polynomials. (The roots of the Legendre polynomials are necessarily between -1 and 1.) Also, these polynomials seem to obey the (very nice) 3-term recurrence $tf\_n=f\_{n-1}+2f\_n+f\_{n+1}$ for $n=1,2,3,4,....$
This is where the Catalan numbers seem to appear. Recall that the $n$th Catalan number is $C\_n=\frac{(2n)!}{n!(n+1)!}.$ If we had an inner product on the space of polynomials such that $\left<t^i,t^j\right>=C\_{i+j}$, then applying the Gram-Schmidt process to the sequence $\{1,t,t^2,t^3,t^4,...\}$ appears to yield the sequence $f\_n$. Is there a function $g$ such that $\int\_0^\infty t^n g(t)dt=C\_n$? I am guessing that such $g$ should be defined on $[0,\infty)$ because of the behavior of the roots of $f\_n$.
What are these polynomials called?
|
https://mathoverflow.net/users/11264
|
Do you recognize this sequence of polynomials?
|
Let $b\_n(t)$ be the Morgan-Voyce polynomial defined by $$\begin{eqnarray}b\_0(t) &=& 1 \\
b\_1(t) &=& t + 1 \\
b\_n(t) &=& (t+2) b\_{n-1}(t) - b\_{n-2}(t)
\end{eqnarray}$$
Then $f\_n(t) = (-1)^n b\_n(-t)$ fits your recurrence.
See [Rising diagonal polynomials associated with Morgan-Voyce polynomials](https://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.376.3755&rep=rep1&type=pdf), Swamy, M. N., Fibonacci Quarterly 38.1 (2000) pp61-69 for an overview of these polynomials which includes the following observations:
* $b\_n(t) = \sum\_{k=0}^n \binom{n+k}{n-k} t^k$ (2.10)
* The zeros of $b\_n(t)$ are $-4 \sin^2 \left(\frac{2r-1}{2n+1} \frac{\pi}2 \right)$ for $r = 1, 2, \ldots, n$ (2.14)
* $b\_n(t)$ is orthogonal over $(-4,0)$ with respect to the weight function $\sqrt{-(t+4)/t}$ (2.26)
---
In light of that the following is largely superfluous, but...
From $b\_n(t) = \sum\_{k=0}^n \binom{n+k}{n-k} t^k$ we can test that we get the original "shifted" Legendre polynomials, and since each transformation in the process which gives $f\_n$ is invertible that would validate the conjectured identity of the $f\_n$. Denote the polynomials which come from the Gram-Schmidt process as $u\_n$, so $$u\_n(t) = \binom{2n}{n}t^n - \sum\_{j=0}^{n-1} \frac{\left<u\_j,\binom{2n}{n}t^n\right>}{\left<u\_j,u\_j\right>} u\_j(t)$$
Then the conjecture is that $$\begin{eqnarray\*}u\_n(t) \stackrel?= \tilde{f\_n}(t) &:=& \sum\_{k=0}^n \binom{2k}{k} t^k [t^k] f\_n(t) \\
%&=& (-1)^n \sum\_{k=0}^n \binom{2k}{k} t^k [t^k] b\_n(-t) \\
&=& \sum\_{k=0}^n (-1)^{n+k} \binom{2k}{k} \binom{n+k}{n-k} t^k \\
\end{eqnarray\*}$$
#### Product with original basis function
$$\begin{eqnarray\*}
\left<\tilde{f\_j}, \binom{2n}{n}t^n\right> &=& \int\_0^1 \tilde{f\_j} \binom{2n}{n}t^n dt \\
&=& \binom{2n}{n} \int\_0^1 \sum\_{k=0}^j (-1)^{j+k} \binom{2k}{k} \binom{j+k}{j-k} t^{k+n} dt \\
&=& \binom{2n}{n} \sum\_{k=0}^j \frac{(-1)^{j+k}}{k+n+1} \binom{2k}{k} \binom{j+k}{j-k} \\
&=& \frac{1}{n+j+1} \binom{2n}{n-j}
\end{eqnarray\*}$$
with [some help from Sage](https://sagecell.sagemath.org/?z=eJxdizkOgDAMwHZekY2kTQVl5ytIsDWhAZXj_ZQRRlu2sLHCCPdcsJUXWmr2kuzE48qIIdKE4pXAwZJsy2lecXDG9jPK36Y-LKG6DtC8-kgMytAzCNEDAbMhMw==&lang=sage&interacts=eJyLjgUAARUAuQ==).
#### Self-product
$$\begin{eqnarray\*}
\left<\tilde{f\_j}, \tilde{f\_j}\right> &=& \int\_0^1 \tilde{f\_j}^2 dt \\
&=& \int\_0^1 \tilde{f\_j} \sum\_{k=0}^j (-1)^{j+k} \binom{2k}{k} \binom{j+k}{j-k} t^k dt \\
&=& \sum\_{k=0}^j (-1)^{j+k} \binom{j+k}{j-k} \int\_0^1 \tilde{f\_j} \binom{2k}{k} t^k dt \\
&=& \sum\_{k=0}^j (-1)^{j+k} \binom{j+k}{j-k} \frac{1}{k+j+1} \binom{2k}{k-j}
\end{eqnarray\*}$$
and the term in the sum has support only when $j=k$, yielding $$\left<\tilde{f\_j}, \tilde{f\_j}\right> = \frac{1}{2j+1}$$
#### Gram-Schmidt induction
If we assume that $u\_j = \tilde{f\_j}$ for all $j < n$ then
$$\begin{eqnarray\*}
u\_n(t) &=& \binom{2n}{n}t^n - \sum\_{j=0}^{n-1} \frac{\left<\tilde{f\_j},\binom{2n}{n}t^n\right>}{\left<\tilde{f\_j},\tilde{f\_j}\right>} \tilde{f\_j}(t) \\
&=& \binom{2n}{n}t^n - \sum\_{j=0}^{n-1} \frac{2j+1}{n+j+1} \binom{2n}{n-j} \tilde{f\_j}(t) \\
&=& \binom{2n}{n}t^n - \sum\_{j=0}^{n-1} \frac{2j+1}{n+j+1} \binom{2n}{n-j} \sum\_{k=0}^j (-1)^{j+k} \binom{2k}{k} \binom{j+k}{j-k} t^k \\
&=& \binom{2n}{n}t^n + \sum\_{k=0}^{n-1} (-1)^k \binom{2k}{k} t^k \sum\_{j=k}^{n-1} (-1)^{j+1} \frac{2j+1}{n+j+1} \binom{2n}{n-j} \binom{j+k}{j-k} \\
% subst j: = n-j
&=& \binom{2n}{n}t^n + \sum\_{k=0}^{n-1} (-1)^{n+k} \binom{2k}{k} t^k \sum\_{j=1}^{n-k} (-1)^{j+1} \frac{2n-2j+1}{2n-j+1} \binom{2n}{j} \binom{n-j+k}{n-j-k} \\
&=& \binom{2n}{n}t^n + \sum\_{k=0}^{n-1} (-1)^{n+k} \binom{2k}{k} t^k
\binom{n+k}{n-k} \\
&=& \sum\_{k=0}^n (-1)^{n+k} \binom{2k}{k} \binom{n+k}{n-k} t^k
\end{eqnarray\*}$$
with [some more help from Sage](https://sagecell.sagemath.org/?z=eJxVizsSgCAMBXtPkY6AyTjQexVnsCNIdPBzfkUru7c7-4QyKYxwxYpGGhjbdVtNeuB-FkT2dkLpvQUHc9K1pLhgcEryM8rSP2cWzs23goN7f8NHbRMIgSfQp7I3VNogng==&lang=sage&interacts=eJyLjgUAARUAuQ==) for one of the sums.
|
11
|
https://mathoverflow.net/users/46140
|
407706
| 167,015 |
https://mathoverflow.net/questions/407676
|
0
|
I've been casually reading about optimal transport, and I was intrigued by the Wasserstein metric, in which we define the distance between two measures $\mu$ and $\nu$ on a metric space $X$ by
$$
W\_p(\mu, \nu) = \left(\inf\_\gamma \int\_{X\times X} \mathrm{dist}(x,y)^p \mathrm{d}\gamma(x,y)\right)^{1/p},
$$
where the infimum is taken over all measures $\gamma$ on $X\times X$ whose projections onto the respective factors give $\mu$ and $\nu$.
It occurred to me that this appears to provide a framework to generalize the notion of functions altogether. In particular, for a measure space $X,\mu$, we could define a generalized function from $X$ to $Y$ as a measure $\gamma$ on $X\times Y$ such that $\pi\_\* \gamma = \mu$, where $\pi:X\times Y\rightarrow X$ is the projection to the first factor.
We can then try to define all the usual operations on functions in this new framework. For instance, for generalized funcitons $\gamma\_1$ from $X$ to $Y$ and $\gamma\_2$ from $Y$ to $Z$, we could try to define $\gamma\_2\circ \gamma\_1$ by
$$
\gamma\_2\circ \gamma\_1 (A\times C) = \sum\_{y\in Y}\gamma\_1(A\times \{y\})\gamma\_2(\{y\}\times C).$$
Of course, this only works if $Y$ is finite (or countable). I believe there should be a generalization of this to arbitrary $Y$ using integrals, but I am having trouble coming up with a good formulation. [Apparently you can use Markov chains](https://mathoverflow.net/questions/351214/composition-of-couplings-as-a-pullback-construction), but I am still trying to understand how to write this down cleanly.
In any case, I am interested in whether this notion provides a useful way to generalize functions. How much of the usual theory carries over? Could this be used e.g. to study weak solutions to PDEs, to study the monodromy of solutions to algebraic equations, or to study backwards iteration in otherwise non-invertible dynamical systems?
|
https://mathoverflow.net/users/445256
|
Couplings as generalized functions
|
Since you refer to transportation metrics, it seems to be fair to assume that your probability spaces are Lebesgue (=standard). Then it is the same to talk either about "lifting" your measure $\mu$ to $\gamma$ or about the **family of the conditional measures** of the projection $\gamma\to\mu$ (or, which is also the same, about the corresponding **Markov kernel**). Ordinary functions from $X$ to $Y$ correspond then to the situation when all transition measures are delta-measures.
This construction is known as **mutivalued maps with invariant measure** or **polymorphisms** (the term introduced by [Vershik](https://mathscinet.ams.org/mathscinet-getitem?mr=2166671), also see [Schmidt - Vershik](https://mathscinet.ams.org/mathscinet-getitem?mr=2408396) or [Neretin](https://mathscinet.ams.org/mathscinet-getitem?mr=1944085)).
|
1
|
https://mathoverflow.net/users/8588
|
407714
| 167,019 |
https://mathoverflow.net/questions/407727
|
0
|
This is a related question to an [older one](https://mathoverflow.net/questions/362115/are-complete-regular-linear-hypergraphs-on-omega-isomorphic).
If $H\_i = (V\_i, E\_i)$ are [hypergraphs](https://en.wikipedia.org/wiki/Hypergraph) for $i=1,2$ then we say they are *isomorphic* if there is a bijection $f: V\_1 \to V\_2$ such that for $A \subseteq V\_1$ we have $$A\in E\_1 \text{ if and only if } f(A) \in E\_2.$$
We say that $H=(\omega, E)$ is a *strongly complete regular linear* hypergraph on $\omega$ if
1. $|e| = \aleph\_0$ for all $e\in E$,
2. $e\_1\neq e\_2\in E \implies |e\_1\cap e\_2| = 1$, and
3. for all $m,n\in\omega$ there is $e\in E$ such that $\{m,n\}\subseteq e$.
**Question.** If $H\_i = (\omega, E\_i)$ are strongly complete regular linear hypergraphs for $i = 1,2$, are $H\_1$ and $H\_2$ necessarily isomorphic?
|
https://mathoverflow.net/users/8628
|
Are strongly complete regular linear hypergraphs on $\omega$ isomorphic?
|
If $H\_1= P^2(\mathbb{Q})$ (the projective plane defined on $\mathbb{Q}$ with the points as vertices and lines as hyper-edges), and $H\_2$ the [Moulton plane](https://en.wikipedia.org/wiki/Moulton_plane) defined on $\mathbb{Q}$, then $H\_1$ and $H\_2$ are both strongly complete regular linear hypergraphs but they are not isomorphic.
The reason is that, if $H\_1$ and $H\_2$ are isomorphic, Desargues' theorem would hold on both of them or neither of them, but Desargues' theorem holds on $H\_1$ and not $H\_2$.
|
3
|
https://mathoverflow.net/users/125498
|
407744
| 167,030 |
https://mathoverflow.net/questions/407740
|
1
|
I am looking for a reference/a hint to the following problem:
We are given $f\_1(x),f\_2(x)$ convex functions (say, on $\mathbb R^d$) such that $f\_1(x) \to\infty$ for $\|x\|\to\infty$. Also there is an $\alpha > 0$ such that $f\_1(x) - \alpha f\_2(x) \geq 0$.
A first consequence is that for any $\epsilon > 0$,
$$f\_1(x) - (\alpha - \epsilon)f\_2(x) \to \infty, \quad \|x\|\to \infty$$
because $f\_1(x) - (\alpha - \epsilon)f\_2(x) = \frac{\epsilon}{\alpha }f\_1(x) + \frac{a-\epsilon}{\alpha }(f\_1 - \alpha f\_2(x)) \geq \frac{\epsilon}{\alpha}f\_1(x) \to \infty$ for $\|x\|\to \infty$.
Now **my question** is whether (and maybe under which circumstances) it is possible to find a local approximation $g\_2(x)$ such that $|g\_2(x)-f\_2(x)| \to 0$ for $\|x\|\to \infty$ and
$$f\_1(x) - \beta g\_2(x)$$ is **convex** for some $\beta \in (0,\alpha]$.
---
The concrete example I have in mind is the following: $X = \mathbb R^d$ and $f\_1(x) = \sum\_{k=1}^d \frac{x\_k^2}{\sigma\_k^2}$ (a weighted squared $l^2$-norm) and $f\_2(x) = \|x\|\_1^2$ (the squared $l^1$-norm).
It can be shown that $f\_1(x) - \alpha f\_2(x) \geq 0$ for $\alpha \leq (\sum\_{k=1}^d \sigma\_k^2)^{-1}$.
Unfortunately, $f\_1(x) - \beta\cdot f\_2(x)$ is not convex for any $\beta\in (0,\alpha]$. My question in this specific example corresponds to finding a function $g\_2(x)$ such that $|g\_2(x)-f\_2(x)|$ is "asymptotically small", e.g. $g\_2(x) \geq f\_2(x)$ and $|g\_2(x)-f\_2(x)|\to 0$ for $\|x\|\_1\to\infty$. In addition for some $\beta \in (0,\alpha]$, I want
$$f\_1(x) -\beta f\_2(x)$$
to be **convex**. Note again that $f\_1(x)$ is convex, and I only need convexity of $f\_1 - \beta f\_2$ for an arbitrarily small $\beta \in (0,\alpha)$.
---
PS: I have thought a bit more about this problem and it looks like the following might work (at least in this specific setting):
$g\_2(x) := \left(\sum\_{k=1}^d (\sqrt{\sigma\_k^2 + |x\_k|^2} - \sigma\_k)\right)^2$
|
https://mathoverflow.net/users/88505
|
Convexification of difference of convex functions
|
$\newcommand\R{\mathbb R}$In general, the answer is no.
E.g., let $d=1$,
\begin{equation}
f\_1(x):=\sum\_{j=1}^\infty(|x|-2j)\_+,\quad f\_2(x):=\sum\_{j=1}^\infty(|x|-2j-1)\_+,
\end{equation}
where $u\_+:=\max(0,u)$ --
so that $f\_1$ and $f\_2$ are convex functions, $f\_1(x)\to\infty$ as $|x|\to\infty$, and $f\_1\ge f\_2$.
However, for any $b\in(0,1)$, for the function
$$h:=f\_1-bf\_2,$$
and for any natural $n$, we have
\begin{equation}
h(2n)-2h(2n+1)+h(2n+2)=-b<0.
\end{equation}
So, if a function $g\_2$ is such that $g\_2(x)-f\_2(x)\to0$ as $|x|\to\infty$, then for $g:=f\_1-bg\_2$ we have $g(x)-h(x)\to0$ as $|x|\to\infty$, whence
\begin{equation}
g(2n)-2g(2n+1)+g(2n+2)<0
\end{equation}
for all large enough natural $n$.
So, $g$ is not convex for any $b\in(0,1)$.
---
In your concrete example, the answer is no as well. Indeed, let $\sigma\_k=1$ for all $k$, and let $x:=(-1,u,0,\dots,0)\in\R^d$, $y:=(1,u,0,\dots,0)\in\R^d$, and $z:=(x+y)/2=(0,u,0,\dots,0)$, with $u\to\infty$. Then
\begin{equation}
f\_1(x)-2f\_1(z)+f\_1(y)=2(1+u^2)-2u^2=2,
\end{equation}
whereas
\begin{equation}
f\_2(x)-2f\_2(z)+f\_2(y)=2(1+u)^2-2u^2\to\infty.
\end{equation}
So, if $g\_2(v)-f\_2(v)\to0$ as $\|v\|\to\infty$, then for any $b\in(0,1)$ and for the function
$g:=f\_1-bg\_2$, we have $g(x)-2g(z)+g(y)\to-\infty$, so that $g$ is not convex for any $b\in(0,1)$.
|
2
|
https://mathoverflow.net/users/36721
|
407746
| 167,031 |
https://mathoverflow.net/questions/407704
|
1
|
For $m\in \mathbb{N}$ and $a=(a\_0,a\_1,\ldots,a\_{m}) \in \mathbb{R}^{m+1}$, consider the polynomial $P\_{a}$ defined by
$$
P\_{a} (x):= a\_0 + a\_1 x^2 + \ldots + a\_{m}x^{2m}\text{, for $x \in \mathbb{R}$.}
$$
Then there is $C\_m >0$ such that for every $a \in \mathbb{R}^{m+1}$, we have
$$
\int\_{[0,1]} P\_{a}(x)^2 dx \geq C\_m \|a\|^2,
$$
where $\|\cdot\|$ is the standard Euclidean norm. The existence of $C\_m$ is guaranteed by the following compactness argument. Define the (continuous) function $\phi : \mathbb{R}^{m+1} \rightarrow \mathbb{R}$ by
$$
\phi(a) := \int\_{[0,1]} P\_{a}(x)^2 dx.
$$
Then for $r \in \mathbb{R}$, $\phi(ra)= r^2 \phi(a)$ and hence $\phi(a)= \|a\|^2 \phi(a/\|a\|)$. So $C\_m$ can be taken to be $C\_m = \min\_{v \in \mathbb{S}^m} \phi(v)$.
My questions are the following:
1. Can we get a lower bound for $C\_m$ (in terms of $m$)?
2. Do we know which value of $a \in \mathbb{S}^{m}$ minimizes the function $\phi$?
My guess is that the minimizer $a\_{\text{min}}$ must have coordinates whose signs alternate so that it minimizes the value of $a\_0 + a\_1 x^2 + \ldots + a\_{m}x^{2m}$, but $a\_j$'s should not have the same absolute value. Rather, they should increase in some way.
Thanks!
|
https://mathoverflow.net/users/118316
|
Comparing different norms of a polynomial
|
(Note: there is no need to consider only polynomials of even degree $n=2m$, since the same works for any degree $n\in\mathbb N$).
The map $Q\_{n+1}:\mathbb R^{n+1}\ni a\mapsto \int\_0^1 \big(\sum\_{k=0}^na\_kx^k\big)^2dx= \sum\_{h,k}\frac1{k+h+1}a\_ka\_h=(Ha\cdot a)$ is a positive definite quadratic form on $\mathbb R^{n+1}$, corresponding to the [Hilbert matrix](https://en.wikipedia.org/wiki/Hilbert_matrix) $H$ of order $n+1$.
So you want the minimum eigenvalue $C=\min\_{a\neq 0}\frac {(Ha\cdot a)}{(a\cdot a)}$ of the Hilbert matrix $H$ of order $n+1$. These are computed and tabulated (I haven't a suitable link). Note that the maximum eigenvalue gives the best bound for the opposite problem, $\max\_{a\ne0} \frac{(Ha\cdot a)}{(a\cdot a)}$.
|
1
|
https://mathoverflow.net/users/6101
|
407754
| 167,034 |
https://mathoverflow.net/questions/407767
|
-1
|
We define an embedding of the set of prim numbers into the Cantor set as follows:
First we recall that the cantor set $\mathcal{C}$ is homeomorphic to $(\mathbb{Z}/10\mathbb{Z})^\omega $ since the latter is a compact metrizable space without any isolated point. So according to [topological characterization of the Cantor set](https://en.wikipedia.org/wiki/Cantor_space) the classical Cantor set is homeomorphic to $(\mathbb{Z}/10\mathbb{Z})^\omega $.
The space of prime numbers is denoted by $\mathcal{Prime}$.
We define the embedding $\mathcal{E}:\mathcal{P}\to (\mathbb{Z}/10\mathbb{Z})^\omega $ as follows:
$$\mathcal{E}(p)=(a\_1,a\_2,\ldots,a\_n,\ldots)$$
where the decimal expansion of $\sqrt{p}=b\_nb\_{n-1}\ldots b\_{1}b\_0/a\_1a\_2\ldots a\_n\ldots$
So in this way we may consider the space of prime numbers $\mathcal{Prime}$ as a subspace of the Cantor set $\mathcal{C}$.
>
> Is $\mathcal{Prime}$ a compact set? Is it an open subset of $\mathcal{C}$?
>
>
>
What would be a number theoretical interpretations for these topological questions?
|
https://mathoverflow.net/users/36688
|
The set of prime numbers as a subspace of the Cantor set
|
Your set is a countable dense subset of $(\mathbb{Z}/10\mathbb{Z})^\omega$; cf. Lucia's response [here](https://mathoverflow.net/questions/292044/distribution-of-square-roots-mod-1). Hence it is neither open, nor compact.
|
2
|
https://mathoverflow.net/users/11919
|
407770
| 167,038 |
https://mathoverflow.net/questions/407760
|
8
|
This seems such a simple question that I fear I must have missed some elementary maths.
I am looking for a way to solve $x+x^a = y$ by reference to an already defined function, $a,x,y > 0$ are real.
Failing that a reasonable approximation for $a$ in $(0,1)$.
Many thanks!
|
https://mathoverflow.net/users/446964
|
Is there a specific named function that is the inverse of $x+x^a$ for $x$ real?
|
The answer is yes indeed. It is a special case of Fox-H function, a variation of the confluent Fox-Wright $\_{1}\Psi\_{1}$ function (a generalization of the confluent hypergeometric function $\_{1}F\_{1}$) providing the inverse function. See a previous answer [here](https://mathoverflow.net/questions/406250/transcendental-formulas-for-roots-of-polynomials/406340#406340) for details and references. For this particular case solution is
(Setting $\alpha = a$), for **$\alpha>1$**
$$x = y\cdot\,\_{1}\Psi\_{1}([1,\alpha];[2,\alpha-1];-y^{\alpha-1})$$ which can be set as $$x=y+\sum\_{n=1}^\infty\binom{n\alpha}{n-1}\frac{(-1)^ny^{n(\alpha-1)+1}}{n}$$ whose convergence region is $|y^{\alpha-1}|<|(\alpha-1)^{\alpha-1}\alpha^{-\alpha}|$. For non integer $\alpha>1$ binomials must be set in terms of $\Gamma$ function.
Since Fox-Wright's generalized function can be expressed in terms of [Fox-H function](https://gitlab.com/RZ-FZJ/hypergeom/-/blob/master/Hypergeom.pdf) we have $$\,\_{1}\Psi\_{1}([1,\alpha];[2,\alpha-1];-y^{\alpha-1})=H\_{1,2}^{1,1}([(0,\alpha)];[(0,1),(-1,\alpha-1)];y^{\alpha-1})$$ $$\,\_{1}\Psi\_{1}([1,\alpha];[2,\alpha-1];-y^{\alpha-1})=H\_{2,1}^{1,1}([(1,1),(2,\alpha-1)];[(1,\alpha)];y^{1-\alpha})$$ for this particular case, [Wolfram's Mathematica 12.3](https://reference.wolfram.com/language/ref/FoxH.html) provides an explicit inverse as
**$x=y\cdot$FoxH[{{{0,$\alpha$}},{{}}},{{{0,1}},{{-1,$\alpha$-1}}},$y^{\alpha-1}$]**
For **$0<\alpha<1$** the solution is $$x = y\cdot(1-\,\_{1}\Psi\_{1}([1,\alpha^{-1}];[2,\alpha^{-1}-1];-y^{\alpha^{-1}-1}))$$
and the above relationships are turned in replacing $\alpha$ by $\alpha^{-1}$ and Fox-Wright function $\,\_{1}\Psi\_{1}$ by $1-\,\_{1}\Psi\_{1}$. In this case Mathematica's expression is
**$x=y\cdot(1-$FoxH[{{{0,$\alpha^{-1}$}},{{}}},{{{0,1}},{{-1,$\alpha^{-1}$-1}}},$y^{\alpha^{-1}-1}$])**
Finally, just to complement this answer, general trinomial equation solutions are developed in section 4 of the following
**Reference**
Miller A.R., Moskowitz I.S. Reduction of a Class of Fox-Wright Psi Functions for Certain Rational Parameters. Computers Math. Applic. Vol. 30, No. 11, pp. 73-82, (1995). Pergamon
A preprint can be found [here](https://apps.dtic.mil/sti/pdfs/ADA465435.pdf). (Document has mis-embedded fonts, isolated commas must be replaced by $\Gamma$ symbol)
|
16
|
https://mathoverflow.net/users/141375
|
407777
| 167,039 |
https://mathoverflow.net/questions/407784
|
1
|
Let $\Omega \subset \mathbb{R}^n$ be convex. We write points of $\mathbb{R}^n$ as $(x\_1, x\_2, \dots, x\_n)$. Set $p(x) = m(\Omega \cap \{x\_1 = x\})$, where $m$ is the $n-1$ dimensional Lebesgue measure and $\{x\_1 = x\}$ is the hyperplane $\{(x\_1, x\_2, \dots, x\_n) \in \mathbb{R}^n : x\_1 = x\}$.
Geometrically, it is not hard to see that this function is concave (i.e. $p(\lambda x + (1-\lambda)y) \ge \lambda p(x) + (1-\lambda) p(y)$, but I can't figure out how to make a formal proof of that fact.
This is stated as a fact in page 197 of this paper: <https://www.jstor.org/stable/1193994>, but I could not figure out how to prove it or find a reference for it.
|
https://mathoverflow.net/users/447371
|
Measure of intersection of convex set with hyperplane is concave function
|
It is not true in general that the function $p$ is concave. For instance, let $\Omega$ be the conical hull of the ball of radius $1$ centered at the point $(2,0,\dots,0)\in\mathbb R^n$. Then $p(x)=c\_nx^{n-1}$ for some real $c\_n>0$ depending only on $n$ and for all real $x\ge0$. So, $p$ is not concave even on the interval $[0,\infty)$ if $n\ge3$.
However, for $n\ge2$, it is true that the function $p^{1/(n-1)}$ is concave on the interval where it is positive. This follows immediately from the [Brunn--Minkowski inequality](https://en.wikipedia.org/wiki/Brunn%E2%80%93Minkowski_theorem#Remarks).
|
0
|
https://mathoverflow.net/users/36721
|
407787
| 167,042 |
https://mathoverflow.net/questions/407732
|
6
|
Suppose $S$ a noetherian base scheme, $X \to S$ is projective and $F, G$ are coherent $\mathcal O\_X$-modules. Then by EGA (7.7.8) and (7.7.9) there exists a scheme $H = \underline{\operatorname{Hom}}\_X(F, G)$, affine over $S$, which represents the functor
$$(f: T \to S) \mapsto \operatorname{Hom}\_{X\_T}(f^\*F, f^\*G).$$
On $X\_H$ there is a universal homomorphism $\varphi: F\_H \to G\_H$. Considering the exact sequence
$$0 \to \ker \varphi \to F\_H \xrightarrow{\varphi} G\_H \to \operatorname{coker} \varphi \to 0,$$
one can check that the open set $$\underline{\operatorname{Isom}}\_X(F, G):= H \setminus (\operatorname{supp}(\ker \varphi) \cup \operatorname{supp}(\operatorname{coker} \varphi)) \subset H$$ represents the functor
$$(f: T \to S) \mapsto \operatorname{Isom}\_{X\_T}(f^\*F, f^\*G).$$
In [1, Section 2.1], Jason Starr and Johan de Jong write that $\underline{\operatorname{Isom}}\_X(F, G)$ is affine over the base $S$. Why is that true? In general, opens of affine schemes are not affine.
[1] Jason Starr, Johan de Jong; *Almost proper GIT stacks and Discriminant Avoidance*
|
https://mathoverflow.net/users/111897
|
Why is the scheme of isomorphisms of sheaves affine over the base?
|
So I realized Jason Starr and Johan de Jong only claim that $H = \underline{\operatorname{Hom}}\_S(F, G)$ is affine if $F$ and $G$ are locally free. In that case, if $U = \operatorname{Spec}(A) \subset H$ is such that $F$ and $G$ are free of rank $n$ on $U$, we get
$$H\_U = Gl\_n(A) = \operatorname{Spec}A[X\_{ij}|i,j = 1, \dotsc, n][\frac{1}{\det}],$$
so $H$ is affine over $S$.
|
5
|
https://mathoverflow.net/users/111897
|
407802
| 167,044 |
https://mathoverflow.net/questions/407779
|
4
|
I am learning the idea of "gradient" of a functional in Otto's calculus. It is defined as follows.
Suppose the space we are thinking about is $(\mathcal{P}\_{2,AC}(\mathbb{R}^d),W\_2)$, the space of probability measures with finite second moment that is absolutely continuous w.r.t. Lebesgue measure, and equipped with 2-Wasserstein distance. So later on we use the density $\rho$ instead of measure in this space. The "tangent space" at $\rho$ is defined to be $T\_{\rho}\mathcal{P}\_{2,AC}(\mathbb{R}^d)=\{f:\int fdx=0\}$. A vector $v=\nabla\phi$ is said to be "coupled" with some $f\in T\_{\rho}\mathcal{P}\_{2,AC}(\mathbb{R}^d)$ if they satisfy the equation $$-\nabla\cdot(\rho\nabla\phi)=f.$$
This definition of "couple" is consistent with the idea of "continuity equation" in optimal transport. The Otto's metric tensor is then defined as $$<f,f'>\_{\rho}:=\int \rho\nabla\phi\cdot\nabla\phi'dx,$$
with $\phi,\phi'$ coupled with $f,f'\in T\_{\rho}\mathcal{P}\_{2,AC}(\mathbb{R}^d)$ resp.
Having these in hand we can definethe gradient of a functional $F(\rho)$: $\mathop{grad} F(\rho)\in T\_{\rho}\mathcal{P}\_{2,AC}(\mathbb{R}^d)$ is the function such that $$<\mathop{grad}F(\rho),f>\_{\rho}=d\_{\rho}F(f)=\frac{d}{dt}\mid\_{t=0}F(\rho\_t),$$
for every $f\in T\_{\rho}\mathcal{P}\_{2,AC}(\mathbb{R}^d)$, and $\rho\_t$ is any curve such that $\rho\_0=\rho$ and $\rho'(0)=f$.
By this definition we can easily calculate out that eg. when $F(\rho)=\int\rho\log\rho dx$ being the entropy functional, $\mathop{grad}F(\rho)=-\Delta\rho$.
My question is now what is the gradient of the functional $F(\rho)=W\_2^2(\rho dx,\eta dx)$ for any given $\eta dx\in \mathcal{P}\_{2,AC}(\mathbb{R}^d)$. My guess is that, if $\phi\_\*$ is the **unique** Kantorovich potential(the solution to the dual Kantorovich problem between $\rho dx$ and $\eta dx$ with quadratic cost), then we may have $$\mathop{grad}W\_2^2(\rho,\eta)=-\nabla\cdot(\rho\nabla\phi\_\*),$$
since if we represent the Waserstein distance in terms of dual problem, then $$W\_2^2(\rho,\eta)=\max\_{\phi(x)+\phi^c(y)\leq |x-y|^2}\int\phi\rho dx+\int\phi^c\eta dx.$$
If there is no maximum in the formula(which means $\phi$ is fixed), then the above formula surely holds. Now I am thinking that whether we can use the stability of optimal transport map or something to prove that the guess is correct, but I don't know how to make that work.
|
https://mathoverflow.net/users/174600
|
Gradient of Wasserstein distance in the sense of Otto's calculus
|
Yes this is true, formally this follows by the envelope theorem. In an abstract and very smooth setting, the envelope theorem says that for an objective functional depending on a parameter $t$
$$
F(t)=\max\limits\_z f(t,z),
$$
then the derivative of the optimal value can be computed as
$$
\frac{dF}{dt}(t)=\partial\_t f(t,z\_t)
\qquad
\mbox{for any smooth selection of a maximizer }z\_t \mbox{ of }F(t).
$$
This can be seen easily: for any such choice of a maximizer, just apply a chain rule and use the optimality condition of $z\_t$ in the maximization problem for fixed $t$:
$$
F'(t)=\frac d{dt}f(t,z\_t)=\partial\_t f(t,z\_t)+\underbrace{\partial\_zf(t,z\_t)}\_{=0}\frac {dz\_t}{dt}.
$$
This means, roughly speaking, that one can simply forget that the minimizer varies, only the variation of the functional matter.
---
In your specific context, you are trying to differentiate (w.r.t $\rho$) the optimal value of the optimization problem given by the Kantorovich dual formulation
$$
W^2(\rho,\eta)
=F\_\eta(\rho)
=\max\limits\_\phi f\_\eta(\rho,\phi)
=\max\limits\_\phi \left\{\int \rho\phi+\int\eta\phi^c\right\}
$$
(here $\eta$ is fixed once and for all, I'm mimicking my $F,f$ notations above to give some perspective and I hope the notation is sufficiently self-explanatory).
Although the Kantorovich potential $\phi$ from $\rho$ to $\eta$ (the optimizer) varies when $\rho$ varies, the envelope theorem strongly suggests that you can actually argue as it did not vary at leading order (same for its $c$-transform $\phi^c$), and one can simply differentiate the functional w.r.t. the varying "parameter" $\rho$. Since the Kantorovich functional is linear in $\rho$, the conclusion is indeed that the first variation is given by $ \frac{\partial f\_\eta}{\partial\_\rho}(\rho,\phi)=\phi$.
Of course various subtle problems may arise owing essentially to the infinite-dimensional setting and functional-analytic details, but this is the rough idea.
For a completely rigorous statement and proof I can recommend Filippo Santambrogio's book [1], in particular chapter 7 and Proposition 7.17
[1] Santambrogio, Filippo. "Optimal transport for applied mathematicians." Birkäuser, NY 55.58-63 (2015): 94.
|
5
|
https://mathoverflow.net/users/33741
|
407806
| 167,046 |
https://mathoverflow.net/questions/407795
|
4
|
Suppose $\pi:U(n)\rightarrow GL(V)$ is a positive-dimensional irreducible representation of the unitary group. Given $\varepsilon>0$, how could one rigorously show that the probability that $|\det(I-\pi(g))|<\varepsilon$ as $g$ is chosen in $U(n)$ (with respect to the Haar probability measure) is small (in a quantifiable way, hopefully)?
Edit: In light of David's answer, what if $n$ is large? Under what conditions do irreducible representations lead to the above phenomenon?
|
https://mathoverflow.net/users/447643
|
Irreducible representations of U(n) and probability of being close to having fixed points
|
This isn't true. Let $V$ be the standard two dimensional representation of $U(2)$ and let $W = (\det V)^{-1} \otimes \mathrm{Sym}^2(V)$. If the eigenvalues of $g$ acting on $V$ are $z\_1$ and $z\_2$, then the eigenvalues of $g$ acting on $W$ are $z\_1 z\_2^{-1}$, $1$ and $z\_1^{-1} z\_2$. So $\det (\mathrm{Id} - \pi\_W(g))=0$ for all $g$.
---
Okay, general nonsense about representation theory of $U(n)$ first.
Let $T$ be the torus of diagonal matrices inside $U(n)$, and write $(z\_1, \ldots, z\_n)$ for coordinates on $T$. Irreducible representations of $U(n)$ are called $V\_{\lambda}$ where $\lambda$ is a weakly decreasing $n$-tuples of integers: $\lambda\_1 \geq \lambda\_2 \geq \cdots \geq \lambda\_n$. When we decompose $V\_{\lambda}$ into characters of $T$, the character $\prod z\_j^{\mu\_j}$ occurs if and only if $(\mu\_1, \ldots, \mu\_n)$ is in the convex hull of the $n!$ permutations of $(\lambda\_1, \ldots, \lambda\_n)$.
The example I gave above is $\lambda = (1,-1)$, so $(0,0)$ is in the convex hull of $(1,-1)$ and $(-1, 1)$.
**Lemma** Let $(a\_1, \ldots, a\_n) \in \mathbb{R}^n$. Then $(0,0,\ldots,0)$ is in the convex hull of the $n!$ permutations of $(a\_1, \ldots, a\_n)$ if and only if $\sum a\_i=0$.
**Proof** If $\sum a\_i=0$, then the average of the permutation is the $0$-vector. If $\sum a\_i = h \neq 0$, then all the permutations are in the affine hyperplane where the coordinates sum to $h$. $\square$
So you want to study representations where $\sum \lambda\_i = h \neq 0$. The integer $h$ can be thought be described by saying that the central matrix $z \mathrm{Id}\_n$ acts by $z^h$, so you want the center of $U(n)$ to act nontrivially.
---
Given this, what can we say? The multiplicity of the character $(z\_1, \ldots, z\_n) \mapsto \prod z\_j^{\mu\_j}$ in the representation $V\_{\lambda}$ is called the Kotska number $K\_{\lambda \mu}$. So
$$\det (1-\pi(g)) = \prod\_{\mu} (1-\prod\_j z\_j^{\mu\_j})^{K\_{\lambda \mu}}.$$
Every conjugacy class of $U(n)$ meets the torus $T$, and conjugacy classes correspond to unordered $n$-tuples $\{ z\_1, z\_2, \ldots, z\_n \}$. The volume of the conjugacy class through $(z\_1, \ldots, z\_n)$ is proportional to $\prod\_{i \neq j} |z\_i - z\_j| = \prod\_{i<j} |z\_i-z\_j|^2$ (remember that the $z\_i$ are complex numbers of norm $1$).
So you want some way to control the integral of $\prod\_{i<j} |z\_i-z\_j|^2$ over the region where $\prod\_{\mu} (1-\prod\_j z\_j^{\mu\_j})^{K\_{\lambda \mu}}$ is small.
If you instead were looking the eigenvalue of $\pi(g)$ closest to $1$, so you wanted $\min\_{\mu} |1-\prod\_j z\_j^{\mu\_j}|$ to be large, I think you might have a chance. The assumption that $\sum \mu\_j = h \neq 0$ would mean that multiplying $(z\_1, \ldots, z\_n)$ by $w$ would multiply $\prod\_j z\_j^{\mu\_j}$ by $w^h$ while leaving the volume of the conjugacy class unchanged. The probability for an individual $\mu$ would just be the probability that a random element of $S^1$ wasn't too small, and then you could use a union bound.
But the different terms $(1-\prod\_j z\_j^{\mu\_j})$ will be correlated in complicated ways, so I don't know how you'll control their product.
|
10
|
https://mathoverflow.net/users/297
|
407816
| 167,049 |
https://mathoverflow.net/questions/407788
|
0
|
I have a question that occurred to me and has been bothering me, because maybe graphically it seems obvious but I don't know how to get there. It has to do with the distribution function and monotone rearrangment.
Given a bounded function $f\colon [a,b] \rightarrow \mathbb{R}$, the (right-continuous) distribution of $f$ is the function $D\_{f}$ definded by
\begin{align}
D\_{f}(y):= \mu(\{t\in I\colon f(t)\leq y \}),\; \forall y\in [\inf f, \sup f].
\end{align}
$\mu$ denotes de Lebesgue-measure. The (left-continuous) rearrangment of $f$, denoted by $f^{\ast}$, is the function:
\begin{align}
f^{\ast}(t):= \inf\{y\in [\inf f, \sup f] \colon D\_{f}(y)\geq t \},\; \forall t\in [0,b-a].
\end{align}
It turns out that $f^{\ast}$ is the left inverse function of $D\_{f}$, moreover it is the generalized inverse. They are both monotone increasing functions. Here is my question:
Let $t'\in [0,b-a]$ be a point such that $f^{\ast}$ is discontinuous in $t'$, this means $D\_{f}$ has a constant segment over $[y\_{j},y\_{j+1})$ of value $t'$, so
\begin{align}
t' = D\_{f}(y\_{j}) = D\_{f}(y), \; \forall y\in [y\_{j},y\_{j+1}).
\end{align}
>
> **How can I proof that:
> \begin{align}
> \lim\_{t\to t'^{+}} f^{\ast}(t) = y\_{j+1}
> \end{align}**
>
>
>
Clearly:
\begin{align}
\lim\_{t\to t'^{-}} f^{\ast}(t) = f^{\ast}(t') = f^{\ast}(D\_{f}(y\_{j}))=y\_{j}.
\end{align}
I would appreciate your help very much! If I am not being clear on something or you need more details please let me know.
|
https://mathoverflow.net/users/447433
|
Evaluating a limit at a discontinuity of a monotone rearrangment (distribution function)
|
$\newcommand{\ep}{\varepsilon}$Let
\begin{equation\*}
c:=\inf f,\quad d:=\sup f,\quad I:=[a,b],\quad F:=D\_f,
\end{equation\*}
so that
\begin{equation\*}
F(y)=\mu(\{t\in I\colon f(t)\le y \})\quad \forall y\in[c,d].
\end{equation\*}
Also introduce the set
\begin{equation\*}
E\_t:=\{y\in[c,d]\colon F(y)\ge t\}\quad \forall t\in[0,b-a].
\end{equation\*}
Since $F$ is nondecreasing and right-continuous, for all $t\in[0,b-a]$
\begin{equation\*}
f^\*(t)=\inf E\_t=\min E\_t,\quad E\_t=[f^\*(t),d],
\end{equation\*}
and, for all $y\in[c,d]$,
\begin{equation\*}
F(y)\ge t\iff y\ge f^\*(t). \tag{1}
\end{equation\*}
Now take any $t'\in[0,b-a]$ such that $f^\*$ is discontinuous at $t'$. Since $f^\*$ is nondecreasing and left-continuous, we have $f^\*(t'+)>f^\*(t')$. So, $t'<b-a$, and for some real $\ep>0$ and all $t\in(t',b-a]$ we have $f^\*(t)>\ep+f^\*(t')$; so, by (1), $F(\ep+f^\*(t'))<t$ for all $t\in(t',b-a]$ and hence $F(\ep+f^\*(t'))\le t'$. On the other hand, again by (1), $F(f^\*(t'))\ge t'$. Since $F$ is nondecreasing, it follows that $F(y)=t'$ for all $y\in[f^\*(t'),\ep+f^\*(t')]$. So, introducing
\begin{equation\*}
Y:=\{y\in[c,d]\colon F(y)=t'\},
\end{equation\*}
\begin{equation\*}
y\_1:=\inf Y,\quad y\_2:=\sup Y,
\end{equation\*}
we do have $y\_1<y\_2$ and $F(y)=t'$ for all $y\in[y\_1,y\_2)$.
Moreover, for any $y\in[y\_1,y\_2)$ and any $t\in(t',b-a]$, we have $F(y)=t'<t$ and hence, once again by (1), $y<f^\*(t)$. Letting here $y\uparrow y\_2$ and $t\downarrow t'$, we get $f^\*(t'+)\ge y\_2$.
Suppose now that $f^\*(t'+)\ne y\_2$. Then for some real $\ep>0$ we have $f^\*(t'+)>\ep+y\_2$. So, for all $t\in(t',b-a]$ we have $f^\*(t)>\ep+y\_2$, whence, once again by(1), $t>F(\ep+y\_2)$. So, $t'\ge F(\ep+y\_2)\ge F(y\_1)=t'$ and hence $F(\ep+y\_2)=t'$, which contradicts the definition $y\_2:=\sup Y$.
Thus, indeed $f^\*(t'+)=y\_2$.
|
0
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https://mathoverflow.net/users/36721
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407824
| 167,052 |
https://mathoverflow.net/questions/353340
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4
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I'm confused by the definition of a "cusp" as found in
>
> V.S. Guba, *Conditions for the embeddability of semigroups into groups*, Math. Notes **56** (1994), Nos. 1-2, 763-769 ([link](https://link.springer.com/article/10.1007/BF02110736)).
>
>
>
In the words of Mark Sapir (from an answer that has meanwhile been removed from this thread), "cusp" is a «weird translation» into English of the Russian term "кусок" (piece) used by Guba in the [original version](http://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=mzm&paperid=2237&option_lang=eng) of his paper. However, this is not the (main) point here: The point is rather that I have seen the same definition repeated elsewhere (with the terms "$s$-piece" or "$S$-piece" used in place of Guba's "кусок"), but continue to find it "strange". Let me try and explain my problem with Guba's definition, before asking a question.
Suppose that $S = \langle X \mid R \rangle$ is a semigroup presentation, and denote by $\mathscr F(X)$ the free monoid over $X$ and by $\ast$ the operation of word concatenation in $\mathscr F(X)$. Following Guba, it seems that
* a *defining word* (relative to $S$) is, as usual, any word in the set $\bigcup\_{(\mathfrak a, \mathfrak b) \in R} \{\mathfrak a, \mathfrak b\}$;
* a *cusp* (*кусок*) is a word $\mathfrak u \in \mathscr F(X)$ such that there exist
$\mathfrak p, \mathfrak p^\prime, \mathfrak q, \mathfrak q^\prime \in \mathscr F(X)$, with $\mathfrak p \ne \mathfrak q$ or
$\mathfrak p^\prime \ne \mathfrak q^\prime$, such that $
\mathfrak p \ast \mathfrak u \ast \mathfrak q$ and $\mathfrak p^\prime \ast \mathfrak u \ast \mathfrak q^\prime$ are defining words.
But this doesn't look very natural. For one thing, what would prevent us from taking $\mathfrak p = \mathfrak p^\prime$ and $\mathfrak q = \mathfrak q^\prime$? Maybe I'm misunderstanding something and what Guba really means is that
* a cusp (кусок) is a word $\mathfrak u \in \mathscr F(X)$ such that there exist
$\mathfrak p, \mathfrak p^\prime, \mathfrak q, \mathfrak q^\prime \in \mathscr F(X)$, with $\mathfrak p \ne \mathfrak q$ or
$\mathfrak p^\prime \ne \mathfrak q^\prime$, such $
(\mathfrak p \ast \mathfrak u \ast \mathfrak q, \mathfrak p^\prime \ast \mathfrak u \ast \mathfrak q^\prime) \in R$;
or maybe there is a typo in the paper. In fact, I checked the original version of Guba's article, and at least for what concerns the definition of a "cusp" ("кусок"), there hasn't been any typo introduced in the translation process. I've also tried to follow the proof of Theorem 1 in Guba's paper, but there are some details I'm still trying to demystify. On the other hand, it appears that Guba's notion of "cusp" ("кусок") is borrowed from E.V. Kashintsev's paper
>
> *Small cancellation conditions and embeddability of semigroups in groups*, Int. J. Alg. Comp. **2** (1992), No. 4, 433-441;
>
>
>
except that Kashintsev uses the term "$s$-piece" instead of "cusp" ("кусок") and lets an $s$-piece be a word $\mathfrak u \in \mathscr F(X)$ for which there exist
$\mathfrak p, \mathfrak p^\prime, \mathfrak q, \mathfrak q^\prime \in \mathscr F(X)$, with $\mathfrak p \ne \mathfrak p^\prime$ or
$\mathfrak q \ne \mathfrak q^\prime$, such that $
\mathfrak p \ast \mathfrak u \ast \mathfrak q$ and $\mathfrak p^\prime \ast \mathfrak u \ast \mathfrak q^\prime$ are defining words (as remarked by Kashintsev, these two words need not be distinct). This is one reason making me think that there is a typo in Guba's definition. The following excerpt if from the English translation of Guba's paper:
>
> The classes $K\_p^q$ were studied in [[1](https://link.springer.com/article/10.1007/BF02110736)]. By definition, a semigroup $S$ belongs to the class $K\_p^q$ if it can be specified by a corepresentation (1) that satisfies the small cancellation conditions $C\_s(p)$ and $D(q)$.
>
>
>
Here, item [[1](https://link.springer.com/article/10.1007/BF02110736)] in the bibliography of Guba's paper is Kashintsev's paper (the same one mentioned in the above) and a semigroup is of class $K\_p^q$ if it is isomorphic to a semigroup presentation $S$ with finitely many generators such that (i) no defining word can be expressed as the concatenation of less than $p$ "cusps" and (ii) the left and the right graphs of $S$ (that is, some undirected multigraphs naturally associated with the presentation) have both girth $\ge q$. Then, Guba's embedding theorem (that is, Theorem 1 in Guba's paper) is the statement that every semigroup of class $K\_3^2$ embeds into a group. So I would expect Guba's notion of "cusp" ("кусок") to coincide with Kashintsev's notion of "$s$-piece", and yet the two definitions are different (and probably non-equivalent). With all this in mind, I'd like to ask the following:
>
> **Question.** Can anyone familiar with these things clarify the situation or suggest a reference where Guba's embedding theorem or generalizations of it are discussed as part of a more systematic treatment of the subject (e.g., a chapter in a book concerned with the embedding of a semigroup into a group)?
>
>
>
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https://mathoverflow.net/users/16537
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What is a "cusp" ("кусок") in relation to Guba's embedding theorem?
|
**Update:** I had an email exchange with Victor Guba. He has kindly confirmed that there is indeed a typo in (the Russian and English versions of) his paper: a "кусок" (as per his paper) and an "$s$-piece" (as per Kashintsev's paper) are meant to be one and the same thing.
The part below was written before hearing from Guba.
---
(A recent comment of user TT\_ has reminded me of this question, which has also entered a private exchange with [Laura Cossu](https://sites.google.com/view/laura-cossu) these days. So I've thought to post an answer in the hope that it can also be useful to someone else or those more familiar than me with the subject can shed further light on the story.)
As far as I can tell, there are two possible *interpretations* of the facts described in the OP:
* It is *implicitly* understood in Guba's paper (*as suggested by* TT\_ *in their comment under the OP*) that either the defining words $w = \mathfrak p \ast \mathfrak u \ast \mathfrak q$ and $w' = \mathfrak p' \ast \mathfrak u \ast \mathfrak q'$ are distinct, or $\mathfrak u$ occurs twice as a subword of $w$.
* There is a typo in Guba's paper (and the typo has unfortunately spread through the literature).
In the first case, a cusp (as per Guba's paper) is always an $s$-piece (as per Kashintsev's paper); and in both cases, an $s$-piece need not be a cusp. E.g., consider the monoid presentation $H := \langle X \mid R \rangle$, where $X$ is the $3$-element set $\{a, b, c\}$ and the only defining relation in $R$ is the pair $(a \ast b \ast a, c \ast b \ast c)$: The only $s$-pieces of $H$ are the empty $X$-word $\varepsilon\_X$ and the generators $a$, $b$, and $c$; while, *in any case*, the only cusps are $\varepsilon\_X$, $a$, and $c$ (in particular, $b$ is not a cusp because it only appears in defining words of the form $\mathfrak z \ast b \ast \mathfrak z$).
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1
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https://mathoverflow.net/users/16537
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407827
| 167,053 |
https://mathoverflow.net/questions/407803
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3
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Consider a twice differentiable 1-strongly convex function $f:\mathbb{R}^n \to \mathbb{R}$.
Is it true that there exists $\alpha>0$ **independent of $n$** such that, for all $x \in \mathbb{R}^n$:
\begin{equation}
\label{prop}
\tag{P}\qquad \alpha \lVert x-x^\*\rVert\_{\infty} \leq \lVert\nabla f(x)\rVert\_{\infty},
\end{equation}
where $x^\*$ is the unique global minimizer of $f$.
If the answer is no, what would be a sufficient condition to verify property \eqref{prop}?
The reason why I am asking the question is that, using the equivalence of norms, it is simple to show that (P) holds with $\alpha = \frac{1}{\sqrt{n}}$. But I think it is possible to do better, but cannot prove it. For example, it holds with $\alpha = 1$ for $n=1$, and for $f=x\mapsto \frac{1}{2}\lVert x\rVert\_{2}^2$ for every $n$.
Edit. ***Reminder***:
* As $f$ is twice differentiable, it is 1-strongly convex iff $\nabla^2 f \succcurlyeq I\_{n \times n}$.
* $\lVert x\rVert\_{\infty} \triangleq \max\_{1 \leq i \leq n}\lvert x\_i\rvert$.
|
https://mathoverflow.net/users/447802
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Strong convexity inequality w.r.t. infinity norm $\lVert\cdot\rVert_{\infty}$
|
I think it is not possible to do much better than $\frac{1}{\sqrt n}$. More precisely, I believe the best $\alpha$ is $\frac{1}{\sqrt n}$ whenever $n$ is a power of $2$, and therefore (since $\alpha$ is non-decreasing in $n$) at most $\frac{\sqrt{2}}{\sqrt n}$ in general.
Indeed, consider $f(x)=\frac 1 2 \|A^{-1} x\|\_2^2$ for a linear invertible map $A$ on $\mathbf{R}^n$. If I understand correctly the definition of $1$-strongly convex function, $f$ is $1$-strongly convex iff the $2\to 2$ norm of $A$ is $\leq 1$. For this specific class of examples, your question reduces to "is there a constant independent of $n$ such that $\alpha \|A\|\_{\infty \to \infty} \leq \|A\|\_{2\to 2}$ for every linear map $A$ on $\mathbf{R}^n$". This is clearly false, for example for a [Hadamard matrix](https://en.wikipedia.org/wiki/Hadamard_matrix): its $2 \to 2$ norm is $\sqrt{n}$ and $\infty\to \infty$ norm is $n$. The claim in the first paragraph of the answer therefore follows from the existence of Hadamard matrices of size $n$ whenever $n$ is a power of $2$.
|
3
|
https://mathoverflow.net/users/10265
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407830
| 167,055 |
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