parent_url
stringlengths 37
41
| parent_score
stringlengths 1
3
| parent_body
stringlengths 19
30.2k
| parent_user
stringlengths 32
37
| parent_title
stringlengths 15
248
| body
stringlengths 8
29.9k
| score
stringlengths 1
3
| user
stringlengths 32
37
| answer_id
stringlengths 2
6
| __index_level_0__
int64 1
182k
|
|---|---|---|---|---|---|---|---|---|---|
https://mathoverflow.net/questions/408437
|
9
|
Consider the following statement (in $\mathsf{ZF}+\text{AC}\_\omega (\mathbb{R})$):
>
> There exists $(\varphi\_\alpha)\_{\alpha\in\omega\_1}$ with $\varphi\_\alpha : \alpha \rightarrow \mathbb{N}^\mathbb{N}$ injective and $\text{ran}(\varphi\_\alpha)$ is closed and $\text{rank}\_{CB}(\text{ran}(\varphi\_\alpha)) = 1$, i.e. $\text{ran}(\varphi\_\alpha)$ is made of isolated points, for all $\alpha\in \omega\_1$.
>
>
>
where $\mathbb{N}^\mathbb{N}$ is the Baire space (the space of infinite sequences of natural numbers with its usual topology) and $\text{rank}\_{CB}$ is the Cantor-Bendixon rank. The statement trivially holds if we assume $\text{AC}\_{\omega\_1}$.
My questions are:
1. How much choice do we need (at least) to prove the above statement? Or more generally what known "weak" axiom (on top of $\mathsf{ZF}+\text{AC}\_\omega (\mathbb{R})$) can be assumed to prove the statement?
2. Is it equivalent to the statement "There exists an $\omega\_1$ subset of the reals"? In case it is not, is it consistent $\mathsf{ZF}+\text{AC}\_\omega (\mathbb{R})+$ The above statement $+$ "It does not exists an $\omega\_1$ subset of the reals"?
Thanks!
|
https://mathoverflow.net/users/141146
|
How much choice is necessary to prove this statement?
|
Your statement is equivalent to the assertion that there is a function choosing an enumeration of every countable ordinal. From an enumeration of $\alpha,$ you can easily inject it into a countable set of isolated points in Baire space. For the other direction, if $\varphi\_{\alpha}$ injects $\alpha$ into Baire space, with the range a set of isolated points, there is a surjective partial map from basic open sets onto $\alpha,$ by sending $U \mapsto \beta$ if $\beta$ is unique such that $\varphi\_{\alpha}(\beta) \in U.$ From this we can easily enumerate $\alpha.$ Note that by the Cantor-Bendixson analysis, any closed countable set of reals can be canonically enumerated, by sending basic open sets to the unique point in $U$ of maximal rank if there is such a point.
This principle implies there is an $\omega\_1$-sequence of reals, since an enumeration of an ordinal can be canonically coded by a real. It is strictly stronger than the existence of an $\omega\_1$-sequence of reals. For example, it clearly implies $\omega\_1$ is regular, while in Figura's model ($\mathcal{M}36$ in Consequences of the Axiom of Choice), there is an $\omega\_1$-sequence of reals yet $\omega\_1$ is singular.
Note that everything above is purely in ZF. The possibility remains that countable choice for reals plus an $\omega\_1$-sequence of reals implies your principle, though that seems unlikely. I would guess adding $\omega\_1$ Cohen reals to the Solovay model would result in a model with an $\omega\_1$-sequence of reals and DC but no choice of enumerations for every countable ordinal. I have not verified this however.
|
11
|
https://mathoverflow.net/users/109573
|
408442
| 167,257 |
https://mathoverflow.net/questions/406858
|
0
|
I need to solve the following equation for the matrix $P \in\mathbb{R}^{r\times d}:$
$$
((PAP^\top)^{-1} P S P^\top (PAP^\top)^{-1})^2 = I\_r,
$$
where $S$ is a symmetric $d\times d$ matrix, $A$ is a PSD $d\times d$ matrix, and $I\_r$ is the identity matrix of dimension $r$.
Is there any easy way to solve this equation?
|
https://mathoverflow.net/users/156139
|
Product of matrices equal identity
|
$\DeclareMathOperator\diag{diag}\DeclareMathOperator\rank{rank}$Here is how we can construct solutions in $P$.
Necessarily, $r\leq d$, $\rank(P)=r$, $\rank(S)=r\_1\geq r$.
There is $Q$ invertible s.t. $QAQ^T=I\_d$, $QSQ^T=\diag((\lambda\_i)\_{i\leq p\_1},(-\mu\_j)\_{j\leq q\_1},0\_{s})$, where $\lambda\_i,\mu\_j>0$ and $p\_1+q\_1=r\_1$, $p\_1+q\_1+s=d$.
Let $p$, $q$ be s.t. $p\leq p\_1$, $q\leq q\_1$, $p+q=r$. Even if it means changing the ordering of the diagonal of $QSQ^T$, we may assume that the first $p$ elements of this diagonal are $>0$ and the following $q$ are $<0$.
Let $D\in M\_{r,d}$ be the "diagonal" matrix $\diag((d\_i)\_{i\leq r})$.
Then $U=DQAQ^TD^T=\diag((d\_i)^2)$, $V=DQSQ^TD^T=\diag((\lambda\_i d\_i^2),(-\mu\_j d\_j^2))$.
Finally $U^{-1}VU^{-1}=\diag\left(\left(\dfrac{\lambda\_i}{d\_i^2}\right)\_{i\leq p},\left(\dfrac{-\mu\_j}{d\_j^2}\right)\_{j\leq q}\right)$ and we choose $d\_i=\sqrt{\lambda\_i},d\_j=\sqrt{\vphantom\lambda\mu\_j}$.
For $P=DQ$, the considered expression is $\diag(I\_p,-I\_q)$ and we are done.
|
0
|
https://mathoverflow.net/users/9091
|
408446
| 167,258 |
https://mathoverflow.net/questions/408458
|
9
|
Recall that if we have a metric space $X$ then we can consider the set of its nonempty compact subsets and equip this with a metric called the Hausdorff distance. Denote the resulting metric space $\mathcal{H}(X)$. We of course get an inclusion $X \hookrightarrow \mathcal{H}(X)$. Are any of the homotopy-theoretic properties of this map known? For instance, if $X$ is connected, is $\mathcal{H}(X)$ connected? (I think the answer is yes). Note if $X$ consists of $n$ discrete points, then $\mathcal{H}(X)$ consists of $2^n - 1$ discrete points, so certainly the map is not a weak homotopy equivalence (perhaps it would be fruitful to consider only connected or path-connected compact subsets). Furthermore, I wonder if there are any insights to be had about more categorical structure of this map - $\mathcal{H}$ is clearly a functor; is it a monad? If so, is there something meaningful to be said about it?
|
https://mathoverflow.net/users/461492
|
Homotopy type of the Hausdorff metric
|
In J. Andres, M. Väth, *Calculation of Lefschetz and Nielsen Numbers in Hyperspaces for Fractals and Dynamical Systems*, Proc. Amer. Math. Soc. **135** (2007), 479-487, it was shown (esssentially, the result was already implicitly shown in D.W. Curtis, *Hyperspaces of noncompact metric spaces*, Compositio Math **40** (1980) (2), 139-152, without explicitly noting it):
1. ${\mathcal H}(X)$ is a union of disjoint open ARs (being its connected components) if and only if $X$ is locally continuum-connected.
2. ${\mathcal H}(X)$ is an AR if and only if $X$ is locally continuum-connected and connected.
(Locally continuum-connected is a property between locally path-connected and locally connected.)
In particular, ${\mathcal H}(X)$ is contractible in all cases of 2 while $X$, in general, is not.
|
8
|
https://mathoverflow.net/users/165275
|
408464
| 167,263 |
https://mathoverflow.net/questions/408406
|
5
|
Consider the following well-known statement:
>
> Let $C$ be a small category, $E$ a cocomplete category, and $F\colon C\to E$ a functor. Then there is a (up to isomorphism) unique cocontinuous functor $F'\colon \mathbf{Set}^{C^\mathrm{op}}\to E$ such that $F'\circ y \cong F$, where $y\colon C\to \mathbf{Set}^{C^\mathrm{op}}$ is the Yoneda embedding.
>
>
>
*Question:* Is there a way to phrase this as an adjunction (between bicategories, of course)?
*Idea:* Consider the bicategory $\mathbf{SmallCat}$ of small categories and functors and the bicategory $\mathbf{CocompCat}$ of cocomplete categories and cocontinuous functors. Then the above statement almost states that $C\mapsto \mathbf{Set}^{C^\mathrm{op}}$ is a left adjoint to the forgetful functor $\mathbf{CocompCat}\to \mathbf{SmallCat}$, *except* that this forgetful functor isn't well-defined: the underlying category of a cocomplete category isn't small.
How can one fix this size issue?
|
https://mathoverflow.net/users/442261
|
Size issue in exhibiting the free cocompletion as a left adjoint
|
As Denis-Charles says in the comments, the best way to handle this is to replace the presheaf category $\mathbf{Set}^{C^{\mathrm{op}}}$ by the full subcategory $\hat{C}$ of small presheaves. By definition, a presheaf is **small** if it satisfies any of these equivalent conditions:
* it is a small colimit of representables;
* it is the left Kan extension of its restriction to some small full subcategory of $C$;
* it is the left Kan extension of some presheaf on some small category along some functor into $C$.
Every presheaf on a small category is small. But, for instance, a presheaf on a large discrete category is small iff its support is small; hence the terminal presheaf is not small.
The functor $C \mapsto \hat{C}$ is left adjoint to the forgetful functor
$$
(\text{cocomplete locally small categories}) \to (\text{locally small categories}),
$$
in a suitable 2-categorical sense.
A standard reference for this is:
>
> Brian J. Day and Stephen Lack. [Limits of small functors.](https://arxiv.org/abs/math/0610439) *Journal of Pure and Applied Algebra* 210 (2007), 651–663.
>
>
>
But it goes back further than 2007. The introduction to Day and Lack's paper recounts some of the history.
|
14
|
https://mathoverflow.net/users/586
|
408476
| 167,265 |
https://mathoverflow.net/questions/408473
|
7
|
Let $f: A \to B\ $ be an abelian group homomorphism. Are there abelian groups $G,\ H,\ K$ such that $K \subseteq H \subseteq G$ and the map
$$\pi \circ i: H \to G/K$$
which is the composition of projection and inclusion is isomorphic to $f$? By isomorphic to $f\ $I mean there exist isomorphisms $\tau: H \to A\ $ and $\sigma: G/K \to B$ such that $f = \sigma \circ \pi \circ i \circ \tau^{-1}$.
I think it is a quite natural question, since the case where $f$ is epimorphism is a consequence of the fundamental homomorphism theorem. However, I can't prove the existence nor the uniqueness of the group extension. I'm not familiar at all with the general theory of group extension, but maybe the case where $A,\ B\ $are abelian could be easier.
The question has been at MSE ([link](https://math.stackexchange.com/questions/4304157/existence-of-abelian-group-extension-relative-to-group-homomorphism)) for two days, but there was no answer.
|
https://mathoverflow.net/users/166613
|
Existence of abelian group extension relative to group homomorphism
|
The factorization always exists, but in general is not unique.
Let me identify $A$ with $H$. Clearly, $K$ can be identified with $\ker(f)$. Let $f(A)$ be the image of $A$ in $B$. Then $A$ is an abelian extension of $f(A)$ by $K$. If I understand correctly, you are asking whether given an inclusion $f(A)\hookrightarrow B$, it is possible to find an extension of $B$ by $K$, extending the given extension of $f(A)$. Equivalently, you are asking if it is possible to construct a diagram of the following form, where the rows are short exact, and all the vertical homomorphisms are monomorphisms:
$$\require{AMScd}
\begin{CD}
K @>>> A @>f>> f(A)\\
@V=VV @VVV @VVV \\
K @>>> G @>>> B
\end{CD}
$$
The set of isomorphism classes of abelian extension of $B$ by $K$ is in bijective correspondence with $\operatorname{Ext}(B, K)$, where Ext is taken in the category of abelian groups. The inclusion of groups $f(A)\hookrightarrow B$ induces a **surjective** homomorphism $$\operatorname{Ext}(B, K)\twoheadrightarrow \operatorname{Ext}(f(A), K).$$ Because of the surjectivity, a group extension of $f(A)$ can always be extended to a group extension of $B$.
The reason that the homomorphism is surjective is that the cokernel would be a subgroup of $\operatorname{Ext}^2(B/f(A), K)$, but the category of abelian groups has projective dimension one, so Ext$^2$ is always zero.
On the other hand, the homomorphism of ext groups is not always injective. The kernel is a quotient of $\operatorname{Ext}(B/f(A), K)$. So the lift is not unique.
You can construct an extension of $B$ explicitly as follows. It is a standard result of homological algebra that you can construct a map of free resolutions of the inclusion $f(A)\hookrightarrow B$.
$$\require{AMScd}
\begin{CD}
0@>>> R\_A @>>> F\_A @>>> f(A)\\
@. @VVV @VVV @VVV \\
0 @>>> R\_B @>>> F\_B @>>> B
\end{CD}
$$
Such that the homomorphisms $R\_A\to R\_B$ and $F\_A\to F\_B$ are split injections. An extension of $f(A)$ by $K$ induces a homomorphism $R\_A\to K$. Use the splitting to get a homomorphism $R\_B\to K$. Now you have a surjective homomorphism $F\_B \oplus\_{R\_B} K \twoheadrightarrow B$, whose kernel is $K$. This is the desired extension. In your notation, $G=F\_B \oplus\_{R\_B} K$.
|
9
|
https://mathoverflow.net/users/6668
|
408487
| 167,271 |
https://mathoverflow.net/questions/408495
|
3
|
Let $\mathcal{F}$ be a coherent sheaf on a projective manifold $X$. It is well known that one can construct a resolution of $\mathcal{F}$ by holomorphic vector bundles (locally free sheaves).
Are two such resolutions homotopic? Any reference would be much appreciated.
|
https://mathoverflow.net/users/102114
|
When are two resolutions of a coherent sheaf homotopic
|
If this were true, then any short exact sequence of vector bundles would split. Indeed, if $0 \to \mathscr E\_1 \to \mathscr E\_2 \to \mathscr E\_3 \to 0$ is a short exact sequence of vector bundles, then both
\begin{align\*}
K^\bullet = \cdots \to 0 \to \mathscr E\_1 \to \mathscr E\_2 \to 0 \to \cdots
\end{align\*}
and $L^\bullet = \mathscr E\_3[0]$ are resolutions of $\mathscr E\_3$. If $g \colon L^\bullet \to K^\bullet$ is a homotopy equivalence (or even a quasi-isomorphism!), then the map $g^0 \colon \mathscr E\_3 \to \mathscr E\_2$ induces an isomorphism
$$\phi \colon \mathscr E\_3 \stackrel\sim\to H^0(K^\bullet) = \mathscr E\_3.$$
Then $g^0 \circ \phi^{-1} \colon \mathscr E\_3 \to \mathscr E\_2$ is a splitting of $0 \to \mathscr E\_1 \to \mathscr E\_2 \to \mathscr E\_3 \to 0$. $\square$
(On the other hand, there does always exist a quasi-isomorphism $K^\bullet \to L^\bullet$ in this case, just not in the other direction.)
An example of a short exact sequence of vector bundles that doesn't split is the Koszul sequence
$$0 \to \mathcal O\_{\mathbf P^1}(-2) \stackrel{\left(\begin{smallmatrix}-y \\ x\end{smallmatrix}\right)}\longrightarrow \mathcal O\_{\mathbf P^1}(-1) \oplus \mathcal O\_{\mathbf P^1}(-1) \stackrel{\left(x\ \ y\right)}\longrightarrow \mathcal O\_{\mathbf P^1} \to 0$$
on $X = \mathbf P^1$. Indeed, $\operatorname{Hom}(\mathcal O\_{\mathbf P^1}, \mathcal O\_{\mathbf P^1}(-1) \oplus \mathcal O\_{\mathbf P^1}(-1)) = 0$.
|
7
|
https://mathoverflow.net/users/82179
|
408501
| 167,276 |
https://mathoverflow.net/questions/408441
|
1
|
$\DeclareMathOperator\Pr{P}\newcommand\cPr[2]{\Pr(#1 \mid #2)}$I have a $J \times J$ matrix:
$$
M:= \begin{bmatrix}
\cPr{X=1}{Y=1} & \cPr{X=2}{Y = 1} & \cdots & \cPr{X=J}{Y = 1} \\
\cPr{X=1}{Y=2} & \ddots & & \vdots \\
\vdots & & \ddots & \vdots \\
\cPr{X=1}{Y=J} &\cdots & \cdots & \cPr{X=J}{Y=J}\end{bmatrix},
$$
where $X$ and $Y$ are two discrete variables taking values in $\{1, \dotsc, J\}$, and $\cPr{X=j}{Y=k}$ are the conditional probabilities that $X=j$ knowing $Y=k$.
I want to find a "simple" condition (if it exists) on these conditional probabilities $\cPr x y$ under which $\det(M) \neq 0$.
**When $J=2$**, it is in fact very simple, we have:
$$\det(M) \neq 0 \iff \frac{\cPr{X=2}{Y=2}}{\cPr{X=1}{Y= 2}} \lessgtr \frac{\cPr{X=2}{Y=1}}{\cPr{X=1}{Y= 1}}.$$
Since the conditional probabilities sum to one for any $y$, we get that
$$\det(M) \neq 0 \iff \frac{\cPr{X=2}{Y=2}}{1-\cPr{X=2}{Y= 2}} \lessgtr \frac{\cPr{X=2}{Y=1}}{1-\cPr{X=2}{Y= 1}}.$$
And since the function $f(x) = 1/(1-x)$ is strictly increasing on $[0,1]$ where our probabilities lie, we simply get that
$$\det(M) \neq 0 \iff \cPr{X=2}{Y=2} \lessgtr \cPr{X=2}{Y=1}.$$
I'm trying to find a similar condition (but obviously more complex) under which it is true for the general case $J \times J$. I don't know if such a condition exists but I think it should exist, yet I've not been able to find it.
|
https://mathoverflow.net/users/91969
|
Condition on the probabilities for the $J\times J$ matrix $[ \Pr(X=j \mid Y=k) ]$ to be invertible
|
Indeed, the condition $\det M\ne0$ can be expressed as a certain non-independence condition, as follows.
For $i$ and $j$ in $[J]:=\{1,\dots,J\}$, let
\begin{equation\*}
p\_{i|j}:=P(X=i|Y=j),
\end{equation\*}
so that $M=[p\_{i|j}]\_{i,j\in[J]}$.
Suppose that $\det M=0$. Then for some real $c\_1,\dots,c\_J$ not all of which are $0$ and for all $i\in[J]$ we have
\begin{equation\*}
\sum\_{j\in[J]}c\_j p\_{i|j}=0. \tag{1}
\end{equation\*}
Summing both sides of (1) in $i\in[J]$ and noting that $\sum\_{i\in[J]}p\_{i|j}=1$ for each $j\in[J]$, we get $\sum\_{j\in[J]}c\_j=0$ and hence
\begin{equation\*}
\sum\_{j\in A}c\_j=-\sum\_{j\in A^c}c\_j=\frac12\sum\_{j\in[J]}|c\_j|>0, \tag{2}
\end{equation\*}
where
\begin{equation\*}
A:=\{j\in[J]\colon c\_j>0\},\quad A^c:=[J]\setminus A.
\end{equation\*}
For $j\in[J]$, let
\begin{equation\*}
p\_j:=
\frac{|c\_j|}{\sum\_{k\in A}|c\_k|}.
\end{equation\*}
Let then $Y$ be a random variable (r.v.) with values in $[J]$ such that for all $j\in[J]$
\begin{equation\*}
P(Y=j)=p\_j;
\end{equation\*}
clearly, such a r.v. exists.
Then, in view of (2), we have $P(Y\in A)=P(Y\in A^c)=1/2$ and
(1) can be rewritten as $\sum\_{j\in A}p\_j p\_{i|j}=\sum\_{j\in A^c}p\_j p\_{i|j}$ and then as
\begin{equation\*}
P(X=i,Y\in A)=P(X=i,Y\in A^c)[=\tfrac12\,P(X=i)],
\end{equation\*}
for each $i\in[J]$, which means that the r.v.'s $X$ and $1(Y\in A)$ are independent.
This reasoning is invertible, so that we get
>
> **Proposition 1:** Let $M=[p\_{i|j}]\_{i,j\in[J]}$ be the transpose of any stochastic matrix. Then
> $\det M=0$ if and only if there exist a subset $A$ of $[J]$ and a r.v. $Y$ with values in $[J]$ such that $P(Y\in A)=1/2$ and the r.v.'s $X$ and $1(Y\in A)$ are independent, where $X$ is any r.v. with values in $[J]$ such that $P(X=i|Y=j)=p\_{i|j}$ for all $j\in[J]$ with $P(Y=j)\ne0$.
>
>
>
In the particular case when $J=2$, Proposition 1 reduces to this: $\det M=0\iff p\_{1|1}=p\_{1|2}\iff p\_{2|1}=p\_{2|2}$, as desired.
|
2
|
https://mathoverflow.net/users/36721
|
408503
| 167,277 |
https://mathoverflow.net/questions/408471
|
2
|
Let $G$ be a subcubic graph.
Suppose that $G$ has an edge coloring $\varphi$ using colors from $\{1,2,3,4,5\}$ such that
* each edge is colored with a set of two elements from $\{1,2,3,4,5\}$ (e.g., $\varphi(e)=\{1,2\}$ for some edge $e$),
* if $e\_1$ and $e\_2$ are adjacent, then $\lvert\varphi(e\_1)\cap \varphi(e\_2)\rvert=1$,
* if the distance between two edges $e\_1$ and $e\_2$ is 2 (i.e, $e\_1$ and $e\_2$ are not adjacent, and there is an edge $e\_3$ adjacent to both $e\_1$ and $e\_2$), then $\varphi(e\_1)\neq \varphi(e\_2)$.
Then, does $G$ have a 2-distance vertex 4-coloring (i.e., a proper vertex 4-coloring of $G$ such that every two vertices at distance 2 receive distinct colors)?
My feeling is that the answer is YES as no conterexample had been constructed yet.
Note that if $G$ admits a 2-distance vertex 4-coloring, then one can easily construct the above mentioned edge coloring.
|
https://mathoverflow.net/users/375270
|
The equivalence of a kind of 2-fold edge coloring and the 2-distance vertex coloring for subcubic graphs
|
The answer is **False**. Let $G$ be the [Möbius ladder](https://en.wikipedia.org/wiki/M%C3%B6bius_ladder) on 12 vertices with every edge subdivided, or in SageMath code,
```
>G=Graph('K?AEF@oM?w@o') #Mobius ladder on 12 vertices
>G.subdivide_edges(G.edges(),1)
```
$G$ is a subcubic graph.
```
>max(G.degree())
3
```
If the line graph of $G$ admits a 2-distance vertex 4-coloring $\psi$ with colors in $\{1,2,3,4\}$, then the graph $G$ admits the edge coloring you have mentioned, by coloring each edge $e$ by the pair $\{\psi(e), 5\}$.
SageMath shows that there's such a coloring.
```
>l=G.line_graph()
>l=l.distance_graph([1,2])
>l.chromatic_number()
4
```
But the graph $G$ has no 2-distance vertex 4-coloring.
```
>d=G.distance_graph([1,2])
>d.chromatic_number()
5
```
|
2
|
https://mathoverflow.net/users/125498
|
408507
| 167,278 |
https://mathoverflow.net/questions/407964
|
25
|
Every undergraduate in mathematics learns about proofs by *mathematical induction*. Moreover, every undergraduate taking a course in theoretical computer science or logic learns about *inductive definitions* (of sets, i.e., types if you will), such as the inductively defined set of first-order formulas for a given signature, and that one can do [*structural induction*](https://en.wikipedia.org/wiki/Structural_induction) on these inductively defined sets. (Functional programmers call these inductively defined sets *algebraic data types*.) Similarly, one can define functions from these inductively defined sets by *structural recursion*.
I recently picked up that all these inductive and recursion principles can be dualized: there are [coinductive data types](https://wiki.portal.chalmers.se/agda/ReferenceManual/Codatatypes) and there's [coinduction](https://ncatlab.org/nlab/show/coinduction) and [corecursion](https://ncatlab.org/nlab/show/corecursion).
It seems to me these concepts are only known to niche computer scientists and type theorists and almost no "normal" mathematician uses them.
*Question:* Is this because coinduction, while formally being dual to induction, is less fundamental and less important?
Are there any opportunities we miss because of hiding coinduction? For instance, are there topics in the undergraduate or graduate curriculum that could be simplified by telling students about coinduction?
I found a [blog post](https://golem.ph.utexas.edu/category/2011/07/coinductive_definitions.html) arguing that coinduction might clarify some concepts in the study of $\infty$-categories, but my question is a bit broader: is coinduction fundamental enough that *every* pure mathematician should know about it?
---
*En passant:* One example the above blog post gives is the definition of $\omega$-categories as follows:
>
> An $\omega$-category is a category enriched over $\omega$-categories.
>
>
>
It is then explained that this can be considered as the definition of $\omega$-categories as the terminal coalgebra for an endofunctor of some category. But I'd complain that *even if the foundation one is using admits coinductive data types*, in this case one needs to construct this terminal coalgebra by hand, because this endofunctor isn't an endofunctor on "types", but on "categories one can enrich over", for which the "principle of definitions of coinductive data types" doesn't apply. So I'm not sure if coinduction is really useful here (and everybody can talk about terminal coalgebras without knowing about coinduction).
|
https://mathoverflow.net/users/442261
|
Coinduction for all?
|
This is a question that I've puzzled about myself, and I don't pretend to have The Answer. But here's one thought that I've found illuminating. Let's start by comparing the behavior of induction and coinduction in a programming language that has them both.
Inductive datatypes are static. They store information in a particular configuration, with constructors that put that information together and destructors that inspect that information recursively. The constructors of an inductive datatype compute immediately to a value, unless they get stuck on an axiom or a variable; this is the content of "canonicity" results for type theory.
By contrast, coinductive codatatypes are dynamic. According to the modern "copattern matching" perspective on computation with codata, the constructor of a codatatype doesn't compute until a destructor is applied to it, in which case it computes only one step to expose the result and another instance of the constructor. Thus, an inhabitant of a codatatype is like an "object" in the OOP sense, which maintains an internal state (the domain of an application of the corecursor) and can be sent messages (the destructors) that return results and alter the internal state. More precisely, the return values can include versions of the object with altered state, so this is "immutable OOP".
I was first clued into this viewpoint by a remark in Conor McBride’s "[Let’s see how things unfold](https://personal.cis.strath.ac.uk/conor.mcbride/pub/JUnfold/junfold.pdf)":
>
> As a total programmer, I am often asked ‘how do I implement a server as a program in your terminating language?’, and I reply that I do not: a server is a coprogram in a language guaranteeing liveness.
>
>
>
I found this remark rather cryptic at first, but recently I’ve come to appreciate it much more. In particular, I prefer this perspective on codata to its common description as "infinite objects". The infinite object underlying an element of a codatatype is essentially the trace of all possible execution paths of all possible method calls that might be applied to it in all possible sequences. This can be helpful to have in mind (particularly when constructing categorical semantics), but it has little to do with how we actually use codata. On the one hand, the "infinite object" description is static and fails to correctly describe the dynamic behavior of codata; while on the other hand, if what we *actually* care about is an infinite object, there are much more flexible and direct ways to represent them.
The simplest kind of codata is of course a stream, which corresponds to an "infinite list". But another way to represent an infinite list is with a function $\mathbb{N} \to A$. From the "static" perspective of "representing an infinite list" the two are equivalent, and the latter can even be preferable because it is more flexible in what we can do with it, and that style can represent more different kinds of infinite objects. But from a dynamic perspective, if we view a stream as a *program* that computes one result every time it is queried and moving into a new state suitable for computing the next result, then the representation as $\mathbb{N} \to A$ is terribly inefficient: every time we ask for the $n$th result, the program has to start from 0 and compute all the previous results on its way to the $n$th one, even if all those results were already computed. Can you imagine if the only way a web server could respond to a single request is by being passed, every time, the entire sequence of all the requests it had received up until that point, and replaying from square one all the calculations and updates it had to do in response to them?
Why, then, does coinduction get short shrift? From this perspective I would suggest that:
1. In programming, coinduction is implicitly present everywhere, particularly in object-oriented programming. But the prevalence of imperative programming languages and mutable state, which achieve essentially the same effect in an "easier" way (but without a conceptual framework and in a way that's hard to reason formally about), means that it is rarely explicitly recognized.
2. In mathematics, coinduction makes occasional appearances, but it is just not as useful, because mathematics (at least as it's generally done nowadays) is basically static. Apart from some constructivists and type theorists trying to change the world, mathematicians don't really care about computational interpretations of what they write; so it's natural to prefer the more flexible static representations of infinite objects over the coinductive versions that are more tailored to their computational behavior.
|
19
|
https://mathoverflow.net/users/49
|
408509
| 167,279 |
https://mathoverflow.net/questions/405658
|
2
|
I am trying to prove (or disprove) the following assertion:
Consider a probability triple $(X,\mathcal{B},\mu)$, $X$ separable Banach space (complete), $\mathcal{B}$ the Borel $\sigma-$algebra and $\mu$ a countably additive probability measure there on. Let $Y$ be a different separable Banach space.
Given a Baire 1 function $f:X\rightarrow Y$, there exists a $\mu-$almost everywhere continuous function $g:X\rightarrow Y$, such that $f(x)=g(x)$, $\mu-$almost everywhere.
---
I couldn't find a counterexample so far, even in $\mathbb{R}$, since all the major examples of Baire 1 functions that I know, seems to have this property: the characteristic function of the integers, the characteristic function of the Cantor set, etc.
It is interesting that it is true also for some Baire 2 functions. For example the Dirichlet function, which is equal to 0 $\lambda-$almost everywhere (where $\lambda$ is the Lebesgue measure).
|
https://mathoverflow.net/users/401135
|
Baire 1 function equivalence in measure
|
Let $F$ be a closed subset of $[0,1]$ (say) with empty interior and $\mu(F)>0$ (where $\mu$ is Lebesgue measure), for example given by a fat Cantor set. Clearly, $1\_F$ is Baire class $1$. I claim that $1\_F$ cannot be almost everywhere equal to an almost everywhere continuous function.
Indeed, assume that $f = 1\_F$ almost everywhere, say $f(x) = 0$ if $x \not\in F \cup N$ and $f(x) = 1$ if $x \in F \setminus N'$ with $N,N'$ having measure zero. I need to prove that $f$ is discontinuous on a set of positive measure. Note that $\mu(F \setminus N') > 0$ so I am done if I show that $f$ is discontinuous on $F \setminus N'$.
Assume toward a contradiction that $x \in F \setminus N'$ and that $f$ is continuous at $x$. Since $f(x) = 1$, there is an open neighborhood $V$ of $x$ such that $f(x) \neq 0$ on $V$, and in particular, $V \subseteq F \cup N$.
But since $F$ has empty interior, $x$ is in the closure of the complement of $F$, so $V$ meets this (open) complement, so there is a nonempty open interval $I \subseteq V$ which is disjoint from $F$. So we have $I \subseteq N$, which is a contradiction as $N$ has zero measure and $I$ has positive measure.
|
3
|
https://mathoverflow.net/users/17064
|
408515
| 167,281 |
https://mathoverflow.net/questions/408512
|
7
|
Let $a(n)$ be [A214973](https://oeis.org/A214973), number of terms in greedy representation of $n$ using Fibonacci and Lucas numbers.
Let $b(n)$ be [A329320](https://oeis.org/A329320), sequence which arises from attempts to simplify computing of [A329319](https://oeis.org/A329319). Here
$$b(n)=b\left(\left\lfloor\frac{n}{2}\right\rfloor\right)+1-f(n+1), b(0)=0$$
where $f(n)$ is [A035263](https://oeis.org/A035263), trajectory of $1$ under the morphism $0$ -> $11$, $1$ -> $10$; parity of $2$-adic valuation of $2n$.
Let $c(n)$ be [A048679](https://oeis.org/A048679), compressed fibbinary numbers ([A003714](https://oeis.org/A003714)), with rewrite $0$->$0$, $01$->$1$ applied to their binary expansion.
Then [sequencedb.net conjecture](http://sequencedb.net/s/A214973) that
$$a(n)=b(c(n))$$
Is there a way to prove it?
|
https://mathoverflow.net/users/231922
|
One conjecture by sequencedb.net
|
Sure. Start with the Zeckendorf representation of $n$, $$ n = \sum\_{ i} F\_{j\_i} $$ where $F\_j$ are the Fibonacci numbers and $j\_i \geq j\_{i-1} + 2$.
Then the fibbinary number associated to $n$ is $\sum\_i 2^{j\_i}$ and $c(n)$ is obtained from this by removing a zero in front of each $1$, i.e. $$c(n) = \sum\_i 2^{ j - (i-1)}$$
Now the $b(m)$ function is written in a confusing way. It's defined using the function 1-AO35263, where A035263(n) counts the number of trailing zeroes of $2n$, modulo $2$, so 1- A036263(n) simply counts the number of trailing zeroes of the binary expansion of $n$, modulo $2$. So $b(m)$ is the sum over $k$ of the number of trailing zeroes of $\lfloor \frac{m}{2^k} \rfloor+1$, mod 2. In other words, $b(m)$ is the sum over $k$ of the number of trailing ones of the binary expansion $ \lfloor \frac{m}{2^k} \rfloor$, mod $2$. If we break the binary expansion of $m$ into blocks of consecutive ones separated by zeroes, we see that each block of length $\ell$ produces one quotient with $\ell$ trailing ones, one with $\ell-1$ trailing ones, etc. down to one with $1$ trailing ones, for a total contribution of $\lceil \frac{\ell}{2} \rceil$.
So $b(m)$ is the sum over blocks of consecutive ones in the binary expansion of $m$ of half the number of ones in the block, rounded up.
Now we make the following observation. For $n = \sum\_{i=1}^k F\_{j\_i}$ in Zeckendorf representation, if $ i\_{k-1} = i\_k-2$ then the largest Fibonacci-Lucas number $\leq n$ is $F\_{i\_k} + F\_{i\_k-2} = L\_{i\_k-1}$ and otherwise the largest Fibonacci-Lucas number $\leq n$ is $F\_{i\_k}$.
This can be checked quickly by properties of Fibonacci and Lucas numbers. We have $F\_n \leq L\_{n-1} \leq F\_{n+1}$ so it suffices to prove in the first case that $F\_{i\_k+1} $ is too large and in the second case that $L\_{i\_k-1}$ is too large. In the first case, this claim follows from [the Lemma used in the proof of the Zeckendorf decomposition](https://en.wikipedia.org/wiki/Zeckendorf%27s_theorem) and in the second case $L\_{i\_k-1} = F\_{i\_k} + F\_{i\_k-2}$ is too large by the same lemma after subtracting $F\_{i\_k}$ from both sides.
Using this and the definition of greedy algorithm, we can see that if $i\_{k-1} = i\_{k}-2$ then $a(n) = a(n- F\_{i\_k} - F\_{i\_k-2}) + 1$ and if $i\_{k-1}< i\_k-2$ then $a(n) = a(n-F\_{i\_k})-1$.
We can now prove $a(n) = b(c(n))$ by induction. In the first case, passing from $n$ to $n- F\_{i\_k} - F\_{i\_{k}-2}$ removes a leading $11$ from $c(n)$, and thus subtracts $1$ from $b(c(n))$ since it reduces the length of a block of ones by two, while in the second case, passing from $n$ to $n-F\_{i\_k}$ removes a leading $1$ followed by one or more $0$s from $c(n)$, and thus subtracts $1$ from $b(c(n))$ since it removes a block of ones of length one.
Since we passed to a lower number and shifted both $a(n)$ and $b(c(n))$ by the same amount, the two sequences are equal as long as they are equal in the base case $n=0$, which is automatic: $a(0) =0 = b(0 ) =b(c(0))$
|
20
|
https://mathoverflow.net/users/18060
|
408516
| 167,282 |
https://mathoverflow.net/questions/408520
|
2
|
Suppose that $\Omega=[0,1]^2$. I will say that a real valued function $u$ on $\Omega$ satisfies periodic boundary values if
$$u(x,0)=u(x,1), \;u(0,y)=u(1,y),\;\;\;\text{ for all }x,y\in[0,1].$$
Now let $f\in L^2(\Omega)$, and suppose that I want to look for a solution of the following problem
$$ \begin{cases}-\Delta u+u=f, \;\;\text{ on }\Omega=(0,1)^2,\\
u,\; u\_x,\text{ and }u\_y\text{ satisfy periodic boundary values.}\end{cases}$$
The question is how to show existence (if possible) of $u$.
---
**My input to the problem:** The weak formulation of the problem with test space taken to be $$X=\{v\in H^1(\Omega):\text{ $v$ satisfies periodic boundary values}\},$$
is the same as the $H^1\_0$ formulation, ie
$$a(u,v):=\int\_\Omega\nabla u\cdot \nabla v+\int\_\Omega uv =\int\_\Omega fv=: F(v),\;\;\;\text{ for all }v\in X.$$
This is because the boundary integral vanishes due to $u\_x$, $u\_y$ and $v$ being in $X$. On the other hand, $X$ is a closed subspace of $H^1(\Omega)$ and therefore it is a Hilbert space with the $\|\cdot\|\_{H^1}$ norm. I think we can use the Lax-Milgram theorem on $X$, because $a$ seems coercive on $X$, $a$ and $F$ are continuous on $X$.
However, the Lax-Milgram theorem gives uniqueness, which is weird because unlike $H^1\_0(\Omega)$, the space $X$ does not exactly prescribe the boundary values (it just says that they are periodic). In any case, even if I do get the solution $u\in X$ and then by regularity $u\in X\cap H^2(\Omega)$, I should still show that $u\_x$ and $u\_y$ satisfy periodic boundary values, and this I don't know how to.
|
https://mathoverflow.net/users/345667
|
Elliptic equation on square with periodic boundary values for the solution and it's partial derivatives
|
You are using the wrong space, that's all. The correct space is
$$
X=\{u \in H^1\_{\textrm{loc}}(\mathbb R^2): u(\cdot+n)=u(\cdot) \quad \forall n=(n\_1,n\_2)\in \mathbb Z^2\}.
$$
Then you apply Lax-Milgram and you are good to go. Look at
Asymptotic Analysis for Periodic Structures, by Bensoussan, Lions, Papanicolaou (1979) chapter 1, for example.
---
Answering your comment. $f$ is likewise extended periodically, namely $f$ is extended by zero into $f\_0$, and then identified with $\sum\_{n\in \mathbb Z^2} f\_0(\cdot + n)$ which is periodic on $\mathbb R^2$ and $L^2\_{\textrm{loc}} (\mathbb R^2)$.
Now, on say $(-2,2)^2$, $-\Delta u + u=f$ as a weak solution, and therefore interior regularity shows $u\in H^2((-\frac32,\frac32)^2)$. All in all, $u\in H^2\_{\textrm{loc}} (\mathbb R^2)$. Consequently, $\partial\_x u \in X$, and $\partial\_y u \in X$.
Incidentally, because all coefficients are constant, you can write write $u$ explicitely as a (double) Fourier series, its coefficients being that of $f$, divided by $n\_1^2+n\_2^2+1$. It saves you the trouble of using Lax-Milgram.
|
1
|
https://mathoverflow.net/users/40120
|
408522
| 167,285 |
https://mathoverflow.net/questions/408524
|
1
|
I am considering the minimizing movement scheme related to the gradient of entropy functional in 2-Wasserstein space. The problem is to minimize the following functional for each fixed $\eta$ which is a probability density w.r.t. $(\mathbb{R}^d,Leb)$ with finite second moments: $$\int\rho\log\rho
dx+W\_2^2(\rho,\eta),$$ among all probability densities $\rho$(so $\rho dx\ll Leb$) with finite second moments. Now I need to show the existence of a minimizer to this problem.
So first we choose a minimizing sequence $\rho\_n$, which gives that $W\_2^2(\rho\_n,\eta)$ are uniformly bounded. Since the second moments can be bounded by the 2-Wasserstein distance, we know the second moments of $\rho\_n$ are uniformly bounded, so they are tight(and also uniformly integrable). This gives a subsequence $\rho\_{n\_k}$ converging weakly to some probability measure $\mu$. Now we need to show $\mu\ll Leb$ and has finite second moment.
For the second part I used Skorokhod's theorem to find $X\_n\sim\rho\_n$ and $X\sim\mu$ with $X\_n\overset{a.s.}{\rightarrow}X$. Then Fatou's lemma gives $\mathbb{E}X^2\leq\liminf\_{n\rightarrow\infty}\mathbb{E}X\_n^2<\infty$.
But I have no idea how to show $\mu\ll Leb$: we can find counterexamples if we only have $X\_n$ converges a.s. and in $L^1$. We might need other observations; or it is possible that the limit of the minimizing sequence of this problem is not absolutely continuous w.r.t. Lebesgue measure?
|
https://mathoverflow.net/users/174600
|
How to prove the limit of minimizing sequence of measures is again absolutely continuous(w.r.t. Lebesgue) in the minimizing movement scheme?
|
$\newcommand{\ep}{\varepsilon}\newcommand\R{\mathbb R}$Yes, the minimizer $\mu$ is absolutely continuous (w.r. to the Lebesgue measure $|\cdot|$).
Indeed, you showed that
\begin{equation\*}
F(\rho\_n)\to m:=\inf\_\rho F(\rho) \tag{-1}
\end{equation\*}
and
\begin{equation\*}
\mu\_{\rho\_n}\to\mu \tag{-0.5}
\end{equation\*}
weakly for some sequence $(\rho\_n)$ of probability densities and some probability measure $\mu$, where
\begin{equation\*}
F(\rho):=\int\rho\ln\rho\,dx+W\_2^2(\rho,\eta) \tag{0}
\end{equation\*}
and $\mu\_\rho(dx):=\rho(x)\,dx$.
Take any set $E\subseteq\R$ with $|E|=0$. We have to show that then $\mu(E)=0$.
Take any real $\ep>0$. By the regularity of the Lebesgue measure, there is an open set $G\_\ep\subset\R$ such that
\begin{equation\*}
\text{$E\subseteq G\_\ep$ and $|G\_\ep|<\ep$.} \tag{0.5}
\end{equation\*}
By (-0.5) and the [Portmanteau theorem](https://en.wikipedia.org/wiki/Convergence_of_measures#Weak_convergence_of_measures),
\begin{equation\*}
\mu(G\_\ep)\le\liminf\_n\mu\_{\rho\_n}(G\_\ep). \tag{1}
\end{equation\*}
Next, for each real $a>1$,
\begin{equation\*}
\mu\_{\rho\_n}(G\_\ep)=K\_n+L\_n, \tag{2}
\end{equation\*}
where
\begin{equation\*}
K\_n:=\int\_{G\_\ep\cap[\rho\_n\le a]}\rho\_n\,dx,\quad L\_n:=\int\_{G\_\ep\cap[\rho\_n>a]}\rho\_n\,dx,
\end{equation\*}
$[\rho\_n\le a]:=\rho\_n^{-1}((-\infty,a])$, $[\rho\_n>a]:=\rho\_n^{-1}((a,\infty))$.
Further,
\begin{equation\*}
K\_n\le a|G\_\ep|<a\ep \tag{3}
\end{equation\*}
by (0.5), and
\begin{equation\*}
L\_n\le \int\_{[\rho\_n>a]}\rho\_n\,dx
\le\frac1{\ln a}\int \rho\_n\ln\rho\_n\,dx\le \frac{m+1}{\ln a} \tag{4}
\end{equation\*}
for all large enough $n$, by (-1) and (0).
By (0.5), (1), (2), (3), (4),
\begin{equation\*}
\mu(E)\le\mu(G\_\ep)\le a\ep+\frac{m+1}{\ln a},
\end{equation\*}
for all real $\ep>0$ and all real $a>1$. Letting now $\ep\downarrow0$ and then $a\to\infty$, we get $\mu(E)=0$, as desired.
|
2
|
https://mathoverflow.net/users/36721
|
408533
| 167,289 |
https://mathoverflow.net/questions/408508
|
12
|
In algebraic topology, relative (co)homology is very useful. For example, we have a long exact sequence which is often helpful for lots of calculations.
In algebraic geometry, we have local cohomology, which is basically the same thing and has the same long exact sequence. However, while the commutative algebra community seems to use this a lot, it seems to be rarely used in algebraic geometry. Is indeed local cohomology more useful in commutative algebra than it is in algebraic geometry? If so, why?
(I'm primarily talking about the Zariski topology, but we also have local cohomology in any context where we have six functors; in étale cohomology, in de Rham cohomology, etc. I would also like to know something about local cohomology in these contexts.)
|
https://mathoverflow.net/users/131975
|
Why doesn't local cohomology seem to be used as much in algebraic geometry?
|
I don't agree with the premise of this question. Local cohomology (per se and not in the wider context of the six functor formalisms) and its consequences are still very much used in algebraic geometry. If you just glance at SGA2, you will find that the following results are proved using local cohomology:
$\bullet$ Lefschetz Theorems for the Picard groups and étale fundamental groups,
$\bullet$ Samuel conjecture on factoriality for complete intersections with small singular loci,
$\bullet$ Comparison Theorems between formal and algebraic geometry, which lead to Grothendieck's proof of Zariski Main Theorem.
$\bullet$ Grothendieck and Fulton-Hansen connectedness results, which have been at the origin of the whole industry studying (higher) secant varieties of projective varieties.
And more recently:
$\bullet$ Bounds for the étale cohomological dimensions of toroidal and determinantal varieties, which imply new bounds on the arithmetical rank of such varieties,
$\bullet$ Improved bounds for Castelnuevo-Mumford regularity of specific projective verieties.
Hence, I would be rather inclined to reformulate your question as "**Are there significant areas in Algebraic Geometry where people do not use local cohomology in any way?**"
|
20
|
https://mathoverflow.net/users/37214
|
408534
| 167,290 |
https://mathoverflow.net/questions/408555
|
4
|
$\DeclareMathOperator\GL{GL}$Let $(\pi,V)$ be an infinite dimensional irreducible admissible representation of $\GL\_2(\mathbb{Q}\_p)$. Let us fix an element $v\_0\in V$ and define a vector space
$$V\_{v\_0} := \left\{\pi\begin{pmatrix}a& \\ &1\end{pmatrix}v\_0,\;\text{s.t.}\;a\in \GL\_1(\mathbb{Q}\_p)\right\}.$$
We define the representation
\begin{align}\GL\_1(\mathbb{Q}\_p)&{}\to \operatorname{Aut}(V\_{v\_0}),\\ a&{}\mapsto \pi\begin{pmatrix}a& \\ &1\end{pmatrix}.\end{align}
It is irreducible and admissible since
$$\begin{pmatrix}\mathbb{Z}\_p^{\times}& \\ &1\end{pmatrix}\subseteq GL\_2(\mathbb{Z}\_p).$$
On the one hand applying Schur's lemma for admissible representations (Lemma 4.2.4 of "[Automorphic forms and representations](https://doi.org/10.1017/CBO9780511609572)" of Bump) then the representation of $\GL\_1(\mathbb{Q}\_p)$ should satisfy that
$$\pi\begin{pmatrix}a& \\ &1\end{pmatrix}v\_0 = \chi(a)v\_0,$$
where $\chi$ is a character of $\GL\_1(\mathbb{Q}\_p)$.
On the other hand, let us suppose that $\pi\simeq \operatorname{Ind}\_{P\_{\GL\_2}}^{\GL\_2}\xi$, where $\xi$ is a character defined on the diagonal elements of $\GL\_2(\mathbb{Q}\_p)$ such that $\pi$ is irreducible. We denote their Satake parameters by $\alpha\_1,\;\alpha\_2$ and the Whittaker functional by $W(\cdot)$. Theorem 4.6.5 of "[Automorphic forms and representations](https://doi.org/10.1017/CBO9780511609572)" of Bump states that
$$W\left(\pi\begin{pmatrix}p^k& \\ &1\end{pmatrix}v\_0\right) = W(v\_0)\frac{\alpha\_1^{k+1}-\alpha\_2^{k+1}}{\alpha\_1-\alpha\_2}.$$
The function
$$p^k\to \frac{\alpha\_1^{k+1}-\alpha\_2^{k+1}}{\alpha\_1-\alpha\_2},$$
is not a character for $(p)$. This contradicts the previous Schur's lemma argument. Where is the mistake in those computations?
|
https://mathoverflow.net/users/173538
|
Schur lemma and Whittaker functions
|
$\DeclareMathOperator\GL{GL}$Let me try to clarify. The formula for the Whittaker functional in Theorem 4.6.5 of "[Automorphic forms and representations](https://doi.org/10.1017/CBO9780511609572)" of Bump states that
$$W\left(\pi\begin{pmatrix}p^k& \\ &1\end{pmatrix}v\_0\right) = W(v\_0)\frac{\alpha\_1^{k+1}-\alpha\_2^{k+1}}{\alpha\_1-\alpha\_2},$$
where $v\_0$ is a $\GL\_2(\mathbb{Z}\_p)$-stable vector in the induced representation.
Your claim is then that the cyclic representation generated by $\GL\_1(\mathbb{Q}\_p)$-translates of $v\_0$ is always irreducible as a $\GL\_1(\mathbb{Q}\_p)$ representation. As Aurel points out in the [comments](https://mathoverflow.net/questions/408555/schur-lemma-and-whittaker-functions#comment1047991_408555), this is certainly false in general. In the case related to the above formula, the fact that $v\_0$ is $\GL\_2(\mathbb{Z}\_p)$-stable does imply that $\mathbb{Z}\_p^\times$ acts by a character since
$$\begin{pmatrix}\mathbb{Z}\_p^{\times}& \\ &1\end{pmatrix}\subseteq GL\_2(\mathbb{Z}\_p),$$
but this says nothing about the action of
$$
\begin{pmatrix}p^{k}& \\ &1\end{pmatrix}
$$
for $k>0$. Those values are determined by the formula.
|
9
|
https://mathoverflow.net/users/62154
|
408564
| 167,300 |
https://mathoverflow.net/questions/408545
|
2
|
Let $P(z)$ be a complex polynomial of degree $n.$ I am working on the class of polynomials assiociated to $P(z)$ such that their moduli are identical with that of $P(z)$ on the imaginary axis.
For example if $Q(z)$ is a polynomial obtained by the replacement of coefficients of $P(z)$ by their complex conjugates and $z$ is replaced by $-z,$ then $|P(iy)|=|Q(iy)| $ where $y$ is any real.
The above is one such associated polynomial. May I request you to share your thoughts on this class of polynomials with respect to $P(z).$ Do there exist any other such polynomials which behave similarly on the imaginary axis?
|
https://mathoverflow.net/users/128472
|
A polynomial having same modulus that of an associated polynomial on the imaginary axis
|
The set of all polynomials associated with a given one is described
as follows:
Let the given polynomial be
$$P(z)=c(z-z\_1)\ldots(z-z\_n).$$
Then any associated polynomial is of the form
$$Q(z)=\lambda c(z-\sigma\_1(z\_1))\ldots(z-\sigma\_n(z\_n)),$$
where each $\sigma\_j(z)=z$ or $-\overline{z}$, and $|\lambda|=1$.
So, besides the continuous parameter $\lambda$ you have at most $2^n$
possible sets of zeros in a group of associated polynomials ($\sigma$ does not change imaginary roots).
The proof is simple. Let $P,Q$ be associated, and assume for simplicity that they are monic (the coefficients at top degree are equal to $1$). Then the polynomials $P(z)\overline{P(-\overline{z})}$
and $Q(z)\overline{Q(-\overline{z})}$ have the same zeros
and are monic, and coincide on the imaginary axis (because they are both non-negative and have equal absolute values on the imaginary axis). Therefore they are equal. So each zero of $Q$ is either a zero of $P$ or symmetric to
a zero of $Q$ with respect to the imaginary axis.
|
7
|
https://mathoverflow.net/users/25510
|
408570
| 167,303 |
https://mathoverflow.net/questions/408542
|
3
|
Let $\mathcal P$ be the set of probability densities on $[0,1]$ with mean $1/2$, i.e. $p\in \mathcal P$ iff
$$\int\_0^1 p(x)dx=1,\quad \int\_0^1 xp(x)dx=\frac{1}{2}\quad \mbox{and}\quad p(x)\ge 0, ~~\forall x\in [0,1].$$
How to solve the minimization problem below ?
$$\min\_{p\in\mathcal P}~ \left\{V(p) ~:=~ \int\_0^1 \log\big(p(x)\big)p(x)dx + \int\_0^1 \big(x\log(x)+(1-x)\log(1-x)\big)p(x)dx\right\}.$$
|
https://mathoverflow.net/users/nan
|
Minimization of an entropy type functional
|
As in the comment by leo monsaingeon, let
$$p\_\*(x):=e^{h(x)}/c,$$
where $h(x):=-x\ln x-(1-x)\ln(1-x)$ and $c:=\int\_0^1 e^{h(x)}\,dx$, so that $p$ is a pdf on $(0,1)$ with mean $1/2$, and
$$V(p)=\int\_0^1(p(x)\ln p(x)-h(x)p(x))\,dx.$$
For any pdf $q$ on $(0,1)$ with $V(q)<\infty$, the directional derivative of $V$ at $p\_\*$ in the direction of $q-p\_\*$ is
$$\begin{aligned}
&\frac d{dt}\,V(p\_\*+t(q-p\_\*))\Big|\_{t=0} \\
&=\int\_0^1 (1+\ln p\_\*(x)-h(x))(q(x)-p\_\*(x))\,dx \\
&=\int\_0^1 (1+h(x)-\ln c-h(x))(q(x)-p\_\*(x))\,dx=0,
\end{aligned}$$
since $q$ and $p\_\*$ are pdf's on $(0,1)$.
The crucial point is that the function $V$ is convex (since $u\ln u$ is convex in $u\ge0$, with $0\ln0:=0$). So, $p\_\*$ is indeed a minimizer of $V$.
|
1
|
https://mathoverflow.net/users/36721
|
408571
| 167,304 |
https://mathoverflow.net/questions/408553
|
1
|
I happen to have the heat kernel on the two-dimensional hyperbolic space and I need to take partial derivatives in order to check that it satisfies the heat equation as expected. The problem is I can not apply the Leibniz formula because I get zero in the denominator. The function is
$$P\_2(x,t)=\frac{\sqrt{2}e^{-t/4}}{(4\pi t)^{3/2}}\int\_x^\infty\frac{se^{-s^2/4t}ds}{\sqrt{\cosh(s)-\cosh(x)}}.$$
I would be very grateful if you could help me take the partial derivative $\frac{\partial P\_2}{\partial x}$. I need it symbolically, not numerically, because I want to use it in the heat equation.
|
https://mathoverflow.net/users/411616
|
Partial derivative of the heat kernel
|
A partial integration can remove the singularity:
$$P\_2(x,t)=\frac{\sqrt{2}e^{-t/4}}{(4\pi t)^{3/2}}\int\_x^\infty\frac{se^{-s^2/4t}ds}{\sqrt{\cosh s -\cosh x }}=$$
$$\qquad =\frac{\sqrt{2}e^{-t/4}}{(4\pi t)^{3/2}}\int\_x^\infty\frac{2\sqrt{s-x}\,s e^{-s^2/4t}}{\sqrt{\cosh s -\cosh x}}\left(\frac{d}{ds}\sqrt{s-x}\right)\,ds$$
$$\qquad=-\frac{\sqrt{2}e^{-t/4}}{(4\pi t)^{3/2}}\int\_x^\infty\sqrt{s-x}\left(\frac{d}{ds}\frac{2\sqrt{s-x}\,s e^{-s^2/4t}}{\sqrt{\cosh s -\cosh x}}\right)\,ds$$
$$\qquad=\frac{\sqrt{2}e^{-t/4}}{(4\pi t)^{3/2}}\int\_x^\infty e^{-s^2/4 t}\frac{ \left(s^3-s^2 x-3 s t+2 t x\right) (\cosh s-\cosh x)+s t (s-x) \sinh s}{t (\cosh s-\cosh x)^{3/2}}\,ds.$$
In the final expression the integrand vanishes$^\ast$ as $(s-x)^{1/2}$ when $s\rightarrow x$, so there are no contributions from the integration bounds when we differentiate the integral with respect to $x$.
---
$^\ast$ The numerator expands around $s=x$ as
$$\left(s^3-s^2 x-3 s t+2 t x\right) (\cosh s-\cosh x)+s t (s-x) \sinh s$$
$$\qquad=-xt(s-x)\sinh s+xt(s-x)\sinh s+{\cal O}(s-x)^2={\cal O}(s-x)^2.$$
The denominator is of order $(s-x)^{3/2}$, so the ratio is of order $(s-x)^{1/2}$.
|
3
|
https://mathoverflow.net/users/11260
|
408574
| 167,305 |
https://mathoverflow.net/questions/408595
|
3
|
Given two non-negative Borel measures $\mu$, $\nu$ on $\mathbb{R}^n$, that are finite on compact sets, such that $\nu\ll\mu$, it is well known that
$$\frac{d\nu}{d\mu}(x)= \lim\_{\epsilon\to 0} \frac{\nu(B\_\epsilon(x))}{\mu(B\_\epsilon(x))}$$
holds $\mu$ a.e., where $B\_\epsilon(x)$ is the open ball of radius $\epsilon$ centered at $x\in\mathbb{R}$. Can this be generalized to measures on arbitrary metric spaces, or even Banach spaces?
I am particularly interested in the case of probability measures (for which the quantities in the above fraction are finite). Additionally, if necessary, these could also be Radon.
|
https://mathoverflow.net/users/131718
|
Is the ball ratio theorem for Radon–Nikodým derivative known for general metric spaces?
|
Arbitrary metric spaces, no.
A classic text discussing such "derivation" is:
*Hayes, C. A.; Pauc, C. Y.*, Derivation and martingales, Ergebnisse der Mathematik und ihrer Grenzgebiete. 49. Berlin-Heidelberg-New York: Springer-Verlag. VII, 203 p. (1970). [ZBL0192.40604](https://zbmath.org/?q=an:0192.40604).
I think you get trouble even in $\mathbb R^2$ if you define your metric so that the $r$-ball centered at $(x,y)$ is something like
$$
[x-r,x+r] \times [y-e^{-1/r},y+e^{-1/r}]
$$
---
*plug*
This is covered in Chapter 7 of my text with Sucheston:
*Edgar, G. A.; Sucheston, Louis*, [**Stopping times and directed processes.**](http://dx.doi.org/10.1017/CBO9780511574740), Encyclopedia of Mathematics and Its Applications 47. Cambridge: Cambridge University Press (ISBN 978-0-521-13508-5/pbk). xii, 428 p. (2010). [ZBL1189.60074](https://zbmath.org/?q=an:1189.60074).
|
3
|
https://mathoverflow.net/users/454
|
408606
| 167,313 |
https://mathoverflow.net/questions/401232
|
4
|
Boutot's theorem says that if $X$ is a variety over a field of characteristic 0 with rational singularities, and if $G$ is a reductive group acting on $X$, then the quotient $X/G$ has rational singularities as well.
Is it known whether an analog of this result is true for log terminal singularities?
Namely suppose $X$ is a variety over a field of characteristic 0 with log terminal singularities and an action of a reductive group $G$. Then must $X/G$ also have log terminal singularities?
|
https://mathoverflow.net/users/334560
|
Does the quotient of a variety with log terminal singularities also have log terminal singularities?
|
Now I can give you a definite answer. In general, the quotient of a klt singularity by a reductive group is not klt, because for instance, the canonical divisor of the quotient may not be $\mathbb{Q}$-Cartier.
However, one can define a broader notion: klt type. A singularity $(X;x)$ is said to be of klt type if there exists a boundary $B$ through $x$ for which $(X,B;x)$ is klt.
Then, one gets the following theorem:
**Theorem:** Let $X$ be an affine variety with klt type singularities over an algebraically closed field of characteristic zero and $G$ be a reductive group acting on $X$. Then $X/\!/G$ is of klt type again.
The previous is Theorem 1 in <https://arxiv.org/abs/2111.02812>.
I should also mention that the klt type property is an etale property, i.e., if you can check it in an etale cover of $X$ then it holds for $X$. This is Proposition 4.1. in <https://arxiv.org/abs/2111.02812>.
Furthermore, the klt type condition
|
3
|
https://mathoverflow.net/users/37338
|
408611
| 167,316 |
https://mathoverflow.net/questions/408387
|
3
|
$\DeclareMathOperator\Ext{Ext}\DeclareMathOperator\Tr{Tr}\DeclareMathOperator\coker{coker}\DeclareMathOperator\Hom{Hom}\DeclareMathOperator\Tor{Tor}$I am investigating the interplay between freeness criteria and Ext vanishing. A nice example is a vast literature around the Auslander-Reiten conjecture (ARC) (in the local case):
(ARC) For a Noetherian local ring $R$ and a finitely generated $R$-module $M$ such that $\Ext^i\_R(M,M\oplus R)=0$ for all $i>0$, then $M$ is free.
Let $R$ be a Noetherian local ring and $M$ be finitely generated $R$-modules. The Auslander transpose of $M$ is defined as $$\Tr M:=\coker(F^\*\rightarrow G^\*)$$ where $G\rightarrow F\rightarrow M\rightarrow0$ is a presentation of $M$ and $\\_^\*=\Hom\_R(\\_,R)$. An interplay between freeness and Tor vanishing is the Yoshino freeness criterion: if $\Tor\_1^R(Tr M,M)=0$, then $M$ is free.
I wonder if there is also an interplay between freeness criteria and vanishing of $\Ext\_R^i(\Tr M,N)$ for some $R$-module $N$. Or even any sufficient condition for having $\Ext^i\_R(\Tr M,N)=0$.
|
https://mathoverflow.net/users/115603
|
Vanishing of $\operatorname{Ext}_R(\operatorname{Tr} M,N)$ and freeness criteria
|
I am not sure if this is the kind of thing you are interested in, but let me at least state the easiest to prove criteria that I know for free-ness of $M$ in terms of vanishing of certain $\text{Ext}\_R^i(\text{Tr}M, -)$ .
Proposition: If $M$ is a finitely generated module over a Noetherian local ring $R$ such that $M^\*\ne 0$, and $\text{Ext}\_R^{1,2}(\text{Tr} M, M^\*)=0$, then $M$ is free. (See <https://arxiv.org/abs/1805.04568> Lemma 4.12)
Proof: By definition, one have exact sequences $0\to M^\*\to F\to K \to 0$ and $0\to K \to G\to \text{ Tr} M\to 0$ for some modules $K,F,G$, where $F$ and $G$ are free. Now $\text{Ext}\_R^{1}(K, M^\*)\cong \text{Ext}\_R^{2}(\text{Tr} M, M^\*)=0$, hence $0\to M^\*\to F\to K \to 0$ splits, so $M^\*$ is a free module. Since $M^\*$ is non-zero, and now we know it is free, so $\text{Ext}\_R^{1,2}(\text{Tr} M, M^\*)=0$ yields $\text{Ext}\_R^{1,2}(\text{Tr} M, R)=0$, hence $M$ is reflexive, so $M\cong M^{\*\*}$ is free.
If you have other specific kind of situations in mind, I probably will be able to help you better ...
|
2
|
https://mathoverflow.net/users/135253
|
408615
| 167,318 |
https://mathoverflow.net/questions/408601
|
40
|
As all analytic number theorists know, iterated logarithms ($\log x$, $\log \log x$, $\log \log \log x$, etc.) are prevalent in analytic number theory. One can give countless examples of this phenomenon. My question is, can someone give an intuitive account for why this is so? Specifics regarding any of the famous theorems involving iterated logarithms are welcome. Many thanks!
EDIT: Thank you so much for the answers so far! I'm still trying to get a better intuition on how a $\log \log \log$ or $\log \log \log \log$ arises, especially in Littlewood's 1914 proof that $\pi(x)-\operatorname{li}(x) = \Omega\_{\pm} \left(\frac{\sqrt{x}\log \log \log x}{\log x}\right) \ (x \to \infty)$ or Montgomery's conjecture that $\limsup\_{x \to \infty}\dfrac{\lvert\pi(x)-\operatorname{li}(x)\rvert}{\;\frac{\sqrt{x}\, (\log \log \log x)^2}{\log x}\;}$ is finite and postive. I admit to knowing nothing (yet) about sieve theory, so I will have to dive into the proof of the prime gap theorem by Tao, Maynard, et. al. Can someone give a more precise account of how the $\log \log \log$ or $\log \log \log\log$ arises in the proof? I'm very familiar with why occurrences of $\log \log$ happen, but once you get to $\log \log \log$, I'm still a bit mystified. Also, is there a good introduction to sieve theory where I could start, or should I just dive right in to the papers on large prime gaps?
FURTHER EDIT: Can someone also explain intuitively the reason for the $\log \log \log$ in Littlewood's theorem? Historically, was this the first occurrence of a triple log in number theory?
|
https://mathoverflow.net/users/17218
|
Iterated logarithms in analytic number theory
|
There are two main sources of repeated logs. (These sources can be further refined into natural subcategories, but I'll only mention a couple of those subcategories.) Those two main sources are:
**Type 1**: Repeated logs occur because that is just the truth of the matter.
One of my favorite examples is a 2008 theorem of Kevin Ford, solving the multiplication table problem. The theorem states that
$$
|\{a\cdot b\, : a,b\in \{1,2,\ldots,N\}\}|\asymp \frac{N^2}{\log(N)^c(\log\log(N))^{3/2}},
$$
where $c=1-\frac{1+\log\log(2)}{\log(2)}$.
Lest you believe that the $(\log\log(N))^{3/2}$ factor is a consequence of this being a 2-dimensional problem, it also shows up in the other dimensions. See [this other question](https://mathoverflow.net/questions/108912/number-of-elements-in-the-set-1-cdots-n-cdot-1-cdots-n?lq=1) for more information.
In some cases it is much easier to see where these extra log's come from. For instance, when turning sums over integers into sums over primes, this often leads to an extra log coming into force, just from the nature of the problem at hand and asymptotics with primes.
For instance, we have
$$
\sum\_{n=1}^{N}\frac{1}{n}=\log(N)+\gamma+o(1)\ \text{ while }\ \sum\_{p\leq N,\ p\text{ prime}}\frac{1}{p}=\log\log(N)+B+o(1),
$$
where $\gamma$ and $B$ are well-known constants. These two asymptotics can be thought of as discrete version of the integral equalities
$$
\int\frac{1}{x}\, dx = \log(x) \ \text{ while }\ \int\frac{1}{x\log(x)}\, dx=\log\log(x).
$$
Since primes occur all over number theory, and they also come weighted with an extra log factor, this often contributes extra double-log factors.
**Type 2**: Repeated logs occur as an artifact of our current best machinery.
For example, Rankin showed in 1938 that the largest prime gap below $N$, for $N\gg 0$, is at least
$$
\frac{1}{3}\frac{\log(N)\log\log(N)\log\log\log\log(N)}{(\log\log\log(N))^2}.
$$
These extra logs happen when optimizing inequalities, and when using the known machinery of the day. But they do not represent a fundamental truth about the problem. The constant $\frac{1}{3}$ has been slowly improved. Recently, in 2014, Ford, Green, Konyagin, and Tao improved this bound, and Maynard also did so independently, by replacing the fraction $\frac{1}{3}$ by an arbitrary number. See [this preprint](https://arxiv.org/pdf/1408.4505.pdf) and [this other preprint](https://arxiv.org/pdf/1408.5110v1.pdf) for more details. Later, these five mathematicians together removed the square from the denominator.
If you read the proofs, the logs are coming from the current state of the art sieve methods, together with bounding techniques. When you solve for the best fit functions to undo some of the exponentiation that occurs in calculations, the logs just fall out.
In these types of problems, it is not inconceivable (and actually occurs quite regularly) that one new idea is applied to the problem, and the asymptotic changes (sometimes involving *more* multi-log factors, to account for the small additional room for improvement that was gained). What is surprising about Rankin's bound is that even though it is far from the predicted asymptotic, each extra idea only changed the constant out front---at least until recently.
---
**Edited to add:** Working through a well-written proof will, of course, give a deeper understanding of where iterated logs arise in the problem at hand. That is certainly true for the three examples above.
However, if you are not yet familiar with sieve theory, or the circle method, I wouldn't recommend working through the big proofs of those theorems mentioned above and in your question (at least, not initially). Rather, I would recommend starting with an introductory text on sieves, such as Cojocaru and Murty's book "An Introduction to Sieve Methods and their Applications". Double-logs occur almost at the very beginning. Triple-logs show up in the exercises in Chapter 5 (and perhaps earlier). Indeed, problem 25 is a typical example of how a triple log is introduced to improve an asymptotic.
|
36
|
https://mathoverflow.net/users/3199
|
408620
| 167,320 |
https://mathoverflow.net/questions/408614
|
7
|
Let $(R, \mathfrak m)$ be a complete local normal domain of dimension $2$ with residue field $R/\mathfrak m$ algebraically closed and characteristic $0$. Assume Spec$(R)$ has rational singularity, let $\pi: X \to \text{Spec}(R)$ be minimal resolution of singularities with exceptional divisor $E=\pi^{-1}(\mathfrak m)$. If $E$ has exactly one irreducible component, then must it be true that $R$ is a [cyclic quotient singularity](https://www.jstage.jst.go.jp/article/kyotoms1969/10/1/10_1_293/_article/-char/en)?
|
https://mathoverflow.net/users/386496
|
$2$-dimensional complete local normal domain with rational singularity that has exactly one exceptional curve
|
The answer is *yes*, at least over $\mathbb{C}$, since $2$-dimensional (cyclic) quotient singularities are *taut* (*starr*, in German), namely, they are uniquely characterized, up to biholomorphisms, by their resolution graph.
In other words, every $2$-dimensional normal singularity, having the same resolution graph of a (cyclic) quotient singularity, is itself a (cyclic) quotient singularity.
See Korollar 2.12 in
E. Brieskorn: [Rationale Singularitäten komplexer Flächen](http://dx.doi.org/10.1007/BF01425318), *Invent. Math.* **4**, 336-358 (1968). [ZBL0219.14003](https://zbmath.org/?q=an:0219.14003).
|
3
|
https://mathoverflow.net/users/7460
|
408634
| 167,326 |
https://mathoverflow.net/questions/408514
|
2
|
I'm searching for some software or open source project which is able to prove propositions of predicate logic of first order in the way of natural deduction introduced for example in the book of Lemmon (beginning logic) and can put the results in LaTeX code.
(You can find a short introduction [here](https://plato.stanford.edu/entries/natural-deduction/).)
|
https://mathoverflow.net/users/462601
|
Software to prove statements in the way of natural deduction (tabular form introduced by Lemmon)
|
Here are two tools for teaching natural deduction proofs that were developed in London in the 1990s.
**Jape**
Developed by
[Bernard Suffrin](http://www.cs.ox.ac.uk/bernard.sufrin/personal/), at Worcester College Oxford, and
[Richard Bornat](http://www.eis.mdx.ac.uk/staffpages/r_bornat/), formerly at Queen Mary and now at Middlesex.
[Downloads and source code at GitHub](https://github.com/RBornat/jape)
**Pandora**
[Main page](https://www.doc.ic.ac.uk/pandora/) orginally developed by students at Imperial College in connection with a course taught by [Krysia Broda](http://wp.doc.ic.ac.uk/kb/teaching/) and others.
**Proof boxes in LaTeX**
I forgot to mention my own work! This is not a piece of software of the above kind, but a way of typesetting box-style natural deduction proofs in LaTeX.
[My LaTeX macros](http://paultaylor.eu/proofs/)
|
2
|
https://mathoverflow.net/users/2733
|
408638
| 167,328 |
https://mathoverflow.net/questions/408623
|
3
|
Let $\{\phi(n)\}\_{n\in\mathbb Z}$ be a sequence of complex numbers with the following properties:
1. $\phi(0)=0$ and $|\phi(n)|\leq \frac{C\_1}{|n|}$ for all $n\neq 0$ and $C\_1>0$ is independent of $n.$
2. $|\phi(n+1)-\phi(n)|\leq \frac{C\_2}{n^2}$ for all $n\neq 0$ and $C\_2>0$ is independent of $n.$
3. $\sum\_{-N}^N\phi(n)$ converges as $N\to\infty.$
Consider
$$
K(x)=\sum\_{n\in\mathbb Z}\phi(n)\chi\_{\left[n-\frac{1}{2},n+\frac{1}{2}\right)}(x),
$$ which exists as a function in $L\_p(\mathbb R).$ Can anyone prove that
$$
\lim\_{\epsilon\to \infty}\int\limits\_{\frac{1}{\epsilon}<|x|<\epsilon}K(x)\,\mathrm{d}x
$$ exists? Also is the following true?
$|K(x)-K(x-y)|\leq C\_3\frac{|y|}{|x|^2}$ for $|x|>2|y|.$ We also have by <http://matwbn.icm.edu.pl/ksiazki/cm/cm66/cm66211.pdf> that $|K(x)|\leq C\_4|x|^{-1}$ for $x\neq 0.$ That is $K$ is standard Calderon-Zygmund kernel.
|
https://mathoverflow.net/users/136860
|
Discrete singular integrals
|
$\newcommand{\Z}{\mathbb{Z}}\newcommand{\ep}{\epsilon}$Let $a\_n:=\phi(n)$. Then
\begin{equation}
K(x)=\sum\_{n\in\Z}a\_n 1(n-1/2\le x<n+1/2).
\end{equation}
So, $K(x)=a\_0=0$ if $1/2\le x<1/2$. So, for $\ep\in(0,1/2)$,
\begin{equation}
I\_\ep:=\int\_{1/\ep<|x|<\ep}K(x)\,dx=\int\_{|x|<\ep}K(x)\,dx
=\sum\_{n\in\Z}a\_n J\_n,
\end{equation}
where
\begin{equation}
J\_n:=\int dx\,1(-\ep\le x<\ep,n-1/2\le x<n+1/2).
\end{equation}
Let now $N:=\lfloor\ep+1/2\rfloor$, so that $N-1/2\le\ep<N+1/2$. Then $J\_n=1$ if $|n|\le N-1$ and $J\_n=0$ if $|n|\ge N+1$. Also, $0\le J\_n\le1$ for all $n\in\Z$. So,
\begin{equation}
I\_\ep
=\sum\_{|n|\le N-1}a\_n +O(|a\_N|+|a\_{-N}|).
\end{equation}
So, $I\_\ep$ converges, since $N\to\infty$, $\sum\_{|n|\le N-1}a\_n$ converges, and $|a\_N|+|a\_{-N}|=O(1/N)\to0$.
This provides the positive answer to your question.
The answer to your second question is no, in general. Indeed, take $x=3/2$ and let $y\downarrow0$. Then $K(x)=a\_2$ and, eventually (for all small enough $y>0$), $K(x-y)=a\_1$ and $|x|>2|y|$. However, for any real $C\_3$, the inequality
$|K(x)-K(x-y)|\le C\_3\frac{|y|}{|x|^2}$ will eventually fail to hold if $a\_2\ne a\_1$. However, it is not hard to see that the answer to your second question will be yes under an additional restriction such as $|y|\ge1$. (I will add details to this later -- now have to do something else.)
|
1
|
https://mathoverflow.net/users/36721
|
408650
| 167,332 |
https://mathoverflow.net/questions/408646
|
3
|
**Summary:** I have (I think) a generalisation of Hölder's inequality for positive reals that I can neither prove nor find references to. Pointers would be appreciated or, indeed, a counterexample. Thank you.
The original [enquiry at math stackexchange](https://math.stackexchange.com/questions/4305820) met with no response, hence cross-posting here.
The rest of this post describes Hölder's inequality for positive reals, the conjectured "array power-means" generalisation, and some experimental evidence for the conjecture.
---
**Hölder's inequality** for real vectors $a=(a\_i)\_{i=1,\ldots,n}$ up to, say, $z=(z\_i)\_{i=1,\ldots,n}$ and weights $\Lambda=(\lambda\_{a,\ldots,z})$ s.t. $\lambda\_a+\ldots+\lambda\_z=1$ states that
$$
(a\_1+\ldots+a\_n)^{\lambda\_a}\ldots(z\_1+\ldots+z\_n)^{\lambda\_z}
\ge a\_1^{\lambda\_a}\ldots z\_1^{\lambda\_z}+\ldots+a\_n^{\lambda\_a}\ldots z\_n^{\lambda\_z}
$$
Write those vectors as rows in a matrix, $M$. The LHS of Hölder is obtained by replacing each row with its arithmetic mean ($\text{rowAM}$), taking the $\Lambda$-weighted geometric mean ($\text{colGM}\_\Lambda$) of the resulting column vector, then multiplying by $n$. The RHS is obtained by applying $\text{colGM}\_\Lambda$ to each column, taking the $\text{rowAM}$ of the resulting row vector, then multiplying by $n$. Concisely, then, Hölder is equivalent to:
$$
\text{colGM}\_\Lambda(\ \text{rowAM}(M)\ )\ \ge\ \text{rowAM}(\ \text{colGM}\_\Lambda(M)\ )
$$
Motivated by the AM-GM inequality, we might say very informally that "big mean followed by little mean beats little mean followed by big mean", where the big mean is AM along the rows and the little mean is weighted GM down the columns.
---
**The generalisation** is that the "big mean" and "small mean" need not be AM and weighted-GM; they can be any pair of weighted power means with different exponents. The weights don't seem to matter. The only evidence I have, however, is from experimentation.
Formally, define the $\Lambda$-weighted $k$-power mean $\mathcal{P}\_\Lambda^k$ by its action on a vector of positive reals $x=(x\_\star)$ thus:
$$
\mathcal{P}\_\Lambda^k(x) = \left( \sum\_i \lambda\_i {x\_i}^k \right)^{1/k},k\ne 0,
\quad\text{and}\quad
\mathcal{P}\_\Lambda^0(x) = \prod\_i {x\_i}^{\lambda\_i}
$$
The claim is then that for arbitrary weight-vectors $\Lambda\_1$ and $\Lambda\_2$ and exponents $k\_2\ge k\_1$
$$
\text{col}\mathcal{P}\_{\Lambda\_1}^{k\_1}(\ \text{row}\mathcal{P}\_{\Lambda\_2}^{k\_2}(M)\ )
\ \ge\
\text{row}\mathcal{P}\_{\Lambda\_2}^{k\_2}(\ \text{col}\mathcal{P}\_{\Lambda\_1}^{k\_1}(M)\ )
$$
A trivial point in favour of the claim is that we have equality when $k\_1=k\_2$, irrespective of the potentially different weights $\Lambda\_1$ and $\Lambda\_2$.
---
**Python-3 code** to demonstrate the conjectured inequality is attached. Alter the final line to change the exponents in the two power means and to alter the shape of the matrix. It's set up here to fail with what I assume is merely loss of precision (the exponents are deliberately close and the matrix small). I haven't yet seen it produce an error that genuinely invalidates the inequality.
```
from random import random
def PM(x,p,w):
'''Weighted power mean. p==0 (GM) not implemented.'''
assert len(x) == len(w)
return sum(ww*xx**p for ww,xx in zip(w,x)) ** (1/p)
def rand_wts(n):
'''Random weights. They don't need to sum to 1.'''
return [random() for _ in range(n)]
def doit(H_p,V_p,nrows,ncols):
'''Run the experiment on prescribed powers and matrix shape.'''
assert H_p > V_p, 'programmer must supply H_p > V_p'
# horizontal and vertical power-mean operations
def big_H(wts,M):
return [ [PM(m,H_p,wts)] for m in M ]
def small_V(wts,M):
J = range(len(M[0]))
return [[ PM([m[j] for m in M],V_p,wts) for j in J ]]
# loop "forever"
for trial in range(1,10**10):
# random weights, random matrix
H_w = rand_wts(ncols)
V_w = rand_wts(nrows)
M = [ [random() for _ in range(ncols)] for _ in range(nrows) ]
# power-mean reduction, both orderings
hv = big_H(H_w, small_V(V_w, M ) )[0][0]
vh = small_V(V_w, big_H(H_w, M ) )[0][0]
# big-then-small (vh) should win; if not, report the error
if hv > vh:
print(f'\nFailed on trial {trial} with M =')
for m in M:
print(f' {m}')
print(f'and H_w = {H_w}')
print(f'and V_w = {V_w}')
print(f'\nH(V(M)) = {hv} > {vh} = V(H(M))')
print(f'H(V(M))/V(H(M)) = {hv/vh}\n')
exit(1)
# reassure the user that the experiment is actually running
if trial>1000 and trial&(trial-1) == 0:
print(f'OK after {trial} trials ...')
# should fail after a few seconds, only(?) due to precision issues
doit( 4.00000001,4, 3,2 )
```
|
https://mathoverflow.net/users/461993
|
Array power-means / generalisation of Hölder inequality
|
This inequality is known. Indeed, [inequality (1.1)](https://projecteuclid.org/journals/annals-of-probability/volume-19/issue-1/Hypercontraction-Methods-in-Moment-Inequalities-for-Series-of-Independent-Random/10.1214/aop/1176990550.full) by Kwapień and Szulga states that
$$\Big(\int\_S\Big(\int\_T|f(s,t)|^q\mu(dt)\Big)^{p/q}\nu(ds)\Big)^{1/p} \\
\le
\Big(\int\_T\Big(\int\_S|f(s,t)|^p\nu(ds)\Big)^{q/p})\mu(dt)\Big)^{1/q}$$
if $0<q<p$, where $\mu$ and $\nu$ are measures.
The case $p<q<0$ is obtained from this by replacing $f$ (which is without loss of generality $>0$) by $1/f$, at the same time replacing $p,q$ by $-p,-q$.
The case $q=0$ is now obtained by the limit transition as $q\to0$, which then yields the case $q<0<p$.
|
3
|
https://mathoverflow.net/users/36721
|
408657
| 167,337 |
https://mathoverflow.net/questions/408644
|
5
|
$\require{AMScd}$I am currently thinking about [(strict) henselisations](https://stacks.math.columbia.edu/tag/0BSK) but I don't know too much literature about the topic. So I am wondering if there is a natural way to restrict maps between strict henselisations to henselisations:
Let $A$, $B$ be local rings with an injective homomorphism $h:A\to B$. If I have a homomorphism $f:A^\text{sh} \to B^\text{sh}$, is there always a homomorphism $g:A^\text h \to B^\text h$ such that the following diagram commutes?
\begin{CD}
A^\text{sh} @>f>> B^\text{sh}\\
@AAA @AAA\\
A^\text h @>>g> B^\text h
\end{CD}
Equivalently, I am asking if you start with $x \in A^\text h \subset A^\text{sh}$, is $f(x) \in B^\text h$? It feels like this should be related to Galois theory, where Galois extensions fix the base field.
(Note: There might be requirements on $A$ and $B$ like being normal, or $B$ being a field. I'm also interested in answers with more assumptions than stated above.)
**Clarifications:**
The underlying injective homomorphism $h: A \to B$ is *not necessarily a local map* but $f$ commutes with $h$.
\begin{CD}
A^\text{sh} @>f>> B^\text{sh}\\
@AAA @AAA\\
A @>>h> B
\end{CD}
The example I have in mind is the following: Pick a ring $R$ and a maximal ideal $\mathfrak{m} \in R$ and a map $\operatorname{Frac}(R) \to \operatorname{Frac}(R)^\text{sep}$ (i.e. a geometric point over the generic point of my curve $\operatorname{Spec}(R)$). Then $A := R\_{\mathfrak{m}}$, $B:=\operatorname{Frac}(R)$, the map $h: A \to B$ is not local and $B \to B^\text{sh} = \operatorname{Frac}(R) ^\text{sep}$ is given by above chosen geometric point.
|
https://mathoverflow.net/users/103737
|
Restricting maps between strict henselisations
|
No. Let $A= k[x]\_{(x)}$, the localization of the ring of polynomials in one variable at the maximal ideal $(x)$, and $B = k(x)$.
Assume (for simplicity) that the characteristic of $k$ is not $2$.
Then there exists $y \in A^h$ satisfying $y^2 = 1+x$, as that polynomial splits into distinct linear factors modulo $x$.
But there exists no $y \in B^h = B$ satisfying that equation, as writing $y = f/g$ we would have $f^2 = g^2(1+x)$ so $$2 \deg f = \deg f^2= \deg (g^2 (1+x)) = 2\deg g+1$$
So there can be no homomorphism $A^h \to B^h$ sending $x$ to $x$ (as it must to form a commutative diagram with the induced map on strict henselizations).
|
4
|
https://mathoverflow.net/users/18060
|
408661
| 167,339 |
https://mathoverflow.net/questions/408666
|
2
|
Suppose I am working over a field $\mathbb{F}$ and have $n$ points in the point-value representation $(x\_0,x\_1,\cdots,x\_{n-1})$. What is the fastest way to do polynomial interpolation and convert this to the coefficient form, that is, obtain the coefficients of polynomial $f()$ such that $f(i)=x\_i, i \in \{0,\cdots,n-1\}$ ? Based on my understanding, it would take $O(n^2)$ computations using the Lagrange interpolation method. If the point-value representations were for the roots of unity then we could use FFT to get the coefficients in $O(nlogn)$ but in the general case which is the most efficient method? More specifically, can we do better than $O(n^2)$?
|
https://mathoverflow.net/users/437976
|
Fastest Implementation of polynomial interpolation?
|
Polynomial interpolation can be done via multiplying a Vandermonde matrix (or its inverse) by your coefficient/evaluation vector --- it is a change of basis on the vector space of polynomials of bounded degree. If this matrix has special structure (such as when the evaluation points are roots of unity --- here the matrix is Toeplitz) this matrix-vector multiplication can be done faster (in $O(n\log n)$ time) this is precisely the FFT. But any choice of evaluation points such that you have fast(er) matrix-vector multiplication algorithms suffices. This might give you a slightly more general question to investigate, which might be useful.
|
3
|
https://mathoverflow.net/users/101207
|
408667
| 167,342 |
https://mathoverflow.net/questions/408680
|
2
|
I have $A$ an associative algebra and $B$ at least an alternative algebra. Is there a sufficient condition on $A$ or $B$ to have $A \otimes B$ an alternative algebra?
|
https://mathoverflow.net/users/83165
|
Alternativity on $A \otimes B$
|
The monograph “Alternative Loop Rings” by Goodaire, Jespers and Polcino Milies (North Holland Mathematics Studies **184**, 1996) contains, in chapter I (“Alternative Rings”), §5 (“Tensor Products”), the following proposition (5.13; I'm changing the notation to match yours):
>
> Let $B$ be an alternative algebra over a field $F$ and suppose $A$ is a commutative associative algebra over $F$. Then the tensor product $A\otimes\_F B$ is alternative.
>
>
>
So $A$ being commutative is a sufficient condition.
|
4
|
https://mathoverflow.net/users/17064
|
408689
| 167,352 |
https://mathoverflow.net/questions/408702
|
4
|
Let $\mathfrak{S}\_\mathbb{N}$ be the symmetric group of all positive integers. Let $\ell^\infty(\mathbb{N})^\*$ be the dual space of $\ell^\infty(\mathbb{N})$ equipped with weak\*-topology. There is a natural group action of $\mathfrak{S}\_\mathbb{N}$ on $\ell^\infty(\mathbb{N})^\*$, that is to define
$$\sigma(u)\big(x\_i\big):=u\big((x\_{\sigma(i)})\big)$$
for any $\sigma\in\mathfrak{S}\_\mathbb{N}$, $u\in\ell^\infty(\mathbb{N})^\*$ and $(x\_i)\in \ell^\infty(\mathbb{N})$.
Now let's equip $\mathfrak{S}\_\mathbb{N}$ with the permutation topology, i.e. the topology generated by the basis of neighborhood at identity element in form of $V(F)=\left\{\sigma\in\mathfrak{S}\_\mathbb{N}|\sigma(n)=n\text{ for all }n\in F\right\}$, where $F$ is a finite subset of $\mathbb{N}$.
**My question**: is there a chance that the group action defined as above will be continuous under the setting of permutation topology?
My *bad idea* is to start with $\ell^1(\mathbb{N})\subset \ell^\infty(\mathbb{N})^\*$ and the permutation $\sigma$ acts on $\ell^1(\mathbb{N})$ as usual. If a permutation $\sigma$ sends a closer position to a very far one (as modelling $\sigma$ not in a neighborhood of identity), then $\sigma(f)-f$ differs a lot in a sufficiently far position, and this will yield a great difference after applying an element in $\ell^\infty(\mathbb{N})=\ell^1(\mathbb{N})^\*$. But this does not provide a rigorous proof, besides one knows that $\ell^\infty(\mathbb{N})^\*=\ell^1(\mathbb{N})\oplus c^\perp\_0(\mathbb{N})$.
So I wonder if anyone can give a proof or disproof for it?
|
https://mathoverflow.net/users/140578
|
The symmetric group of positive integers acting on $\ell^\infty(\mathbb{N})^*$
|
No, it's not continuous.
Indeed, fix a non-principal ultrafilter $U$ supported by the set of even numbers, and define $m\in\ell^\infty(\mathbf{N})^\*$ by $m(f)=\lim\_{n\to U}f(n)$.
Now define $\tau\_n$ as the transposition $(2n,2n+1)$ and $s\_n=\prod\_{k\ge n}\tau\_k$. Then $s\_n$ tends to the identity map for the permutation topology.
I claim that $s\_n\cdot m$ does not tend to $m$. Indeed, let $f$ be the indicator function of the set of even numbers. It is enough to see that $(s\_n\cdot m)(f)$ does not tend to $m(f)$. Indeed, $m(f)=1$, while $ (s\_n\cdot m)(f)=m(s\_n^{-1}\cdot f)=0$ since $s\_n\cdot f$ is eventually supported by odd numbers.
|
3
|
https://mathoverflow.net/users/14094
|
408706
| 167,357 |
https://mathoverflow.net/questions/408705
|
3
|
*Note: This is a concrete case of the following question: [Are almost all measure-preserving flows on compact manifolds ergodic?](https://mathoverflow.net/questions/408676/are-almost-all-measure-preserving-flows-on-compact-manifolds-ergodic.)*
Let $M$ be a Riemannian manifold with its natural Riemannian measure, and $V$ a $C^1$ vector field on $M$ whose associated flow is measure preserving.
Does there exist, for every $\varepsilon > 0$, a $C^1$ vector field $W$ that is $\varepsilon$-close to $V$ in $C^0$ norm such that the flow generated by $W$ is measure preserving and ergodic?
|
https://mathoverflow.net/users/173490
|
Are $C^1$ vector fields generating an ergodic flow $C^0$ dense?
|
You can't always approximate by ergodic flows, because ergodic flows might not even exist.
For example, on $S^2$ the Poincare-Bendixson theorem rules out ergodic flows, but there are many measure-preserving flows.
|
3
|
https://mathoverflow.net/users/1227
|
408707
| 167,358 |
https://mathoverflow.net/questions/408704
|
2
|
Let $\Gamma: \mathcal{Y} \twoheadrightarrow \mathcal{X}$ be an upper hemicontinuous correspondence with non-empty values and a closed graph. A continuous selection function in general does not exist, but is it always possible to construct a finite number of continuous functions, $f\_i: \mathcal{Y} \to \mathcal{X}, i=\{1,...,K\}$, such that $(\cup\_{i=1}^K f\_i(y)) \ \cap \ \Gamma(y) \neq \emptyset $?
|
https://mathoverflow.net/users/466713
|
Collection of continuous selections for upper hemicontinuous map
|
$\newcommand{\Ga}{\Gamma}\newcommand{\N}{\mathbb N}$Not in general.
E.g., let $Y:=\mathcal Y:=[0,1]$, $X:=\mathcal X:=[-1,1]$, $\Ga(0):=[-1,1]$, and $\Ga(y):=\{g(y)\}$ for $y\in(0,1]$, where $g(y):=\sin\frac1y$.
Then $\Ga\colon Y\twoheadrightarrow X$ is an [upper hemicontinuous](https://en.wikipedia.org/wiki/Hemicontinuity) correspondence with non-empty values and a closed graph.
Suppose now that, for some natural $K$, there is a family $(f\_i)\_{i\in[K]}$ of continuous functions from $Y$ to $X$ such that $\bigcup\_{i\in[K]}\{f\_i(y)\}\cap\Ga(y)\ne\emptyset$ for all $y\in Y$. Here, as usual, $[n]:=\{1,\dots,n\}$. Then for each $y\in(0,1]$ there is some $I(y)\in[K]$ such that
\begin{equation\*}
f\_{I(y)}(y)=g(y). \tag{1}
\end{equation\*}
Take any natural $L>K$ and let $a\_1,\dots,a\_L$ be any pairwise distinct points in $X=[-1,1]$. Then for each $j\in[L]$ there is a sequence $(y\_{j,1},y\_{j,2},\dots)$ in $(0,1]$ such that $y\_{j,k}\to0$ and
$$g(y\_{j,k})\to a\_j$$
as $k\to\infty$.
For each $j\in[L]$, $\bigcup\_{i\in[K]}R\_{i,j}=\N$, where $R\_{i,j}:=\{k\in\N\colon I(y\_{j,k})=i\}$, where $I(\cdot)$ is as in (1). So,
for each $j\in[L]$ there is some $i\_j\in[K]$ such that the set
\begin{equation\*}
Q\_j:=R\_{i\_j,j}=\{k\in\N\colon I(y\_{j,k})=i\_j\}
\end{equation\*}
is infinite. Since $L>K$, the map $[L]\ni j\mapsto i\_j\in[K]$ cannot be injective (the pigeonhole principle). So, $i\_{j\_1}=i\_{j\_2}=:i\_\*\in[K]$ for some distinct $j\_1,j\_2$ in $[L]$. So, for each $r\in\{1,2\}$ and all $k\in Q\_{j\_r}$
\begin{equation\*}
f\_{i\_\*}(y\_{j\_r,k})=f\_{i\_{j\_r}}(y\_{j\_r,k})=f\_{I(y\_{j\_r,k})}(y\_{j\_r,k})=g(y\_{j\_r,k})\to a\_{j\_r}
\end{equation\*}
as $Q\_{j\_r}\ni k\to\infty$, and also $y\_{j\_r,k}\to0$ as $k\to\infty$.
Therefore and because $a\_{j\_1}\ne a\_{j\_2}$, the function $f\_{i\_\*}$ is not continuous at $0\in Y$, a contradiction. $\quad\Box$
*Remark:* The proof would be much simplified if, in accordance with the mentions of "continuous selection" in the title and body of the posted question, we assumed in addition that the graphs of the functions $f\_i$ are contained in the graph of $\Ga$.
|
1
|
https://mathoverflow.net/users/36721
|
408713
| 167,360 |
https://mathoverflow.net/questions/408597
|
0
|
In a [2014 article](https://link.springer.com/article/10.1007%2FJHEP04%282014%29186) by Chapman, Hoyos and Oz, the authors study non-equilibrium fluid dynamics and describe a method for deriving [Kubo formulas](https://en.wikipedia.org/wiki/Kubo_formula) for thermal transport coefficients of superfluids (the method relies on the equilibrium partition function).
Near the end of the article, a few possible generalisations of their results are mentioned and I was wondering what work had been done on these in the meantime. These include Kubo formulas for first order non-dissipative transport coefficients of anomalous fluids, Kubo formulas for relativistic Rindler hydrodynamics at second order, and Kubo formulas for superfluids with multiple broken charges.
Another generalisation would be for superfluids with multiple unbroken non-abelian charges rather than abelian charges.
|
https://mathoverflow.net/users/119114
|
Generalising results on superfluid Kubo formulas
|
Not quite sure what you are asking, but Shukla and Kovtun have provided all Kubo formulas for non-dissipative transport coefficients [here](https://link.springer.com/article/10.1007%2FJHEP10%282018%29007)
|
3
|
https://mathoverflow.net/users/466792
|
408715
| 167,362 |
https://mathoverflow.net/questions/408663
|
5
|
Let $\mathbb{G}$ be a compact quantum group with function algebra $(C(\mathbb{G}), \Delta)$ (in the sense of Woronowicz). Let $X \in M(B\_0(H) \otimes C(\mathbb{G}))$ be a (possibly infinite-dimensional) representation of the quantum group $\mathbb{G}$, and let $K$ be an $X$-invariant subspace of $H$, i.e. if $p\in B(H)$ is the projection on the closed subspace $K$, then $(p\otimes 1)X (p\otimes 1) = X(p\otimes 1).$
Where one encounters an invariant subspace $K$, one hopes to define a subrepresentation $X\_K \in M(B\_0(K)\otimes C(\mathbb{G}))$. It looks like there is an obvious way to do this:
Consider the surjective strict completely positive map
$$\Psi: B\_0(H) \to B\_0(K): x \mapsto pxp^\*$$
The strict completely positive map $$\Psi \otimes \iota: B\_0(H) \otimes C(\mathbb{G}) \to B\_0(K) \otimes C(\mathbb{G})$$ extends uniquely to a bounded linear map
$$\Psi \otimes \iota: M(B\_0(H) \otimes C(\mathbb{G}) \to M(B\_0(K)\otimes C(\mathbb{G}))$$
which is strictly continuous on bounded subsets and we define
$$X\_K:= (\Psi \otimes \iota)(X)$$
as our candidate for a subrepresentation. Everything works out nicely, for example, it is easily verified that
$$(\iota \otimes \Delta)(X\_K) = (X\_K)\_{12}(X\_K)\_{13}.$$
However, what is not clear to me is why $X\_K$ must be invertible (I consider representations to be invertible elements in the multiplier algebra by definition, and representations do not need to be unitary). Does invertibility of $X \in M(B\_0(H)\otimes C(\mathbb{G}))$ imply invertibility of $X\_K \in M(B\_0(K) \otimes C(\mathbb{G}))?$ I have a feeling that the answer might be negative because compressing with a projection can make things non-invertible.
|
https://mathoverflow.net/users/216007
|
Subrepresentations of C*-algebraic compact quantum groups
|
The following was my original answer, dealing with the case where $X$ is unitary.
It is a nontrivial fact that the orthogonal complement of an invariant subspace is again an invariant subspace. Thus, the projection $p$ in the question will automatically satisfy the stronger property $(p \otimes 1)X = X(p \otimes 1)$. This was part of Woronowicz' original approach to the representation theory of compact quantum groups. You can also find this result as [Proposition 6.2 in the expository notes of Maes and Van Daele](https://arxiv.org/abs/math/9803122).
When $X$ is merely an invertible representation and the orthogonal projection $p$ of $H$ onto $K$ satisfies $(p \otimes 1) X (p \otimes 1) = X(p \otimes 1)$, it is still true that the restriction of $X$ to the invariant subspace $K$ is an invertible representation of $\mathbb{G}$. The only tricky point is that the equality $X(p \otimes 1) = (p \otimes 1)X$ need not hold. The point is that the complementary invariant subspace $L \subset H$ is not the orthogonal complement of $K$, but another complement of $K$.
For every continuous functional $\omega$ on $C(\mathbb{G})$, denote $X(\omega) = (\text{id} \otimes \omega)(X)$. The assumptions say that $X(\omega) K \subset K$ for every $\omega$. The unitarizability means that we can find an invertible $u \in B(H)$ (not necessarily unitary though) such that $Y := (u \otimes 1) X (u^{-1} \otimes 1)$ is a unitary representation. Writing $K' = u(K)$, we have that $Y(\omega)K' \subset K'$ for all $\omega$. Denoting by $q$ the orthogonal projection onto $K'$, the first paragraph says that $Y (q \otimes 1) = (q \otimes 1) Y$.
We then define $e$ as the, potentially nonorthogonal, projection $e = u^{-1}qu$. Then, $X(e \otimes 1) = (e \otimes 1) X$ is invertible. The projection $e$ is still a projection onto $K$. So, $X(e \otimes 1)$ is the restriction of $X$ to $K$, which thus is invertible. The complementary invariant subspace is $L = (1-e)(H)$.
|
4
|
https://mathoverflow.net/users/159170
|
408728
| 167,365 |
https://mathoverflow.net/questions/408716
|
4
|
Let $\mathrm{Bundle}$ be the category whose objects are [smooth vector fiber bundles](https://en.wikipedia.org/wiki/Fiber_bundle) over $\mathbb{R}$, and morphisms are fiberwise smooth linear map (that is, the base is not assumed to be fixed).
Let $Base \colon \mathrm{Bundle} \to \mathrm{Diff}$ be the functor returning the base of a given bundle (and the morphism between bases of a given bundle morphism, respectively). Let's define $ \mathrm{FunctorialBundle}$ as the full subcategory of $ \mathrm{Func}(\mathrm{Diff}, \mathrm {Bundle})$ on those functors that are section to $Base$ (that is, $ F \in \mathrm{Ob}~\mathrm{FunctorialBundle} \ \Leftarrow:\Rightarrow\ F \circ Base = id $).
**Question 1**: Classify objects of $ \mathrm{FunctorialBundle} $ up to isomorphism.
Of course, my question is rather "is such a classification possible?". Or is the class of such functors immense and the classification problem does not make sense (just as the problem of classifying all finite magmas, semigroups, groups does not make sense)? My next question is an example of an answer:
Starting with tangent bundle $TM$ and one-dimensional trivial bundle $\mathbb{R} \times M$ and applying operations direct sum, tensor product, dual we obtain all tensor bundle functors (whose sections are tensor fields). If there are some standard operations that extend the functor class, then add them to this list.
**Question 2**: are there functorial vector bundles (that is, objects of $ \mathrm{FunctorialBundle} $ ) that are not in this class (up to isomorphism, of course)?
|
https://mathoverflow.net/users/148161
|
Classification of functorial smooth vector fiber bundles
|
This is called [natural bundle](https://ncatlab.org/nlab/show/natural+bundle). Apparently, all known information is in [Kolár, Slovák, Michor: Natural operations in differential geometry](https://www.mat.univie.ac.at/%7Emichor/kmsbookh.pdf) (recommended by Stefan Waldmann).
From the description, it is also the best categorical textbook on differential geometry and topology.
>
> *Third* in the beginning of this book we try to give an introduction to the fundamentals of differential geometry (manifolds, flows, Lie groups, differential forms, bundles and connections) which stresses naturality and functoriality from the beginning and is as coordinate free as possible. Here we present the Frölicher–Nijenhuis bracket (a natural extension of the Lie bracket from vector fields to vector valued differential forms) as one of the basic structures of differential geometry, and we base nearly all treatment of curvature and Bianchi identities on it. This allows us to present the concept of a connection first on general fiber bundles (without structure group), with curvature, parallel transport and Bianchi identity, and only then add G-equivariance as a further property for principal fiber bundles. We think, that in this way the underlying geometric ideas are more easily understood by the novice than in the traditional approach, where too much structure at the same time is rather confusing.
>
>
>
|
2
|
https://mathoverflow.net/users/148161
|
408731
| 167,367 |
https://mathoverflow.net/questions/408741
|
6
|
By Kuratowski's theorem, every nonplanar graph contains a (topological) minor of $K\_5$ or $K\_{3,3}$.
But I observed that every time I construct a $4$-connected nonplanar graph, it always contains not only a $K\_{3,3}$-minor but also a $K\_5$-minor.
Moreover, although I tried many times, I cannot construct a $4$-connected nonplanar graph containing only $K\_{3,3}$-minors!
So I want to know whether the following statement is true:
>
> Every $4$-connected nonplanar graph contains a $K\_5$-minor.
>
>
>
Unfortunately, I could not find any references about this topic.
If the statement is true, can you give me a proof?
Or if it's not, can you show me a counterexample?
Thanks a lot.
|
https://mathoverflow.net/users/384338
|
Does every $4$-connected nonplanar graph contain a $K_5$-minor?
|
Yes, this is true and follows from [Wagner's theorem](https://en.wikipedia.org/wiki/Wagner%27s_theorem). Wagner's theorem asserts that every graph with no $K\_5$ minor can be built from $0$-, $1$-, $2$-, and $3$-sums from planar graphs and a fixed $8$ vertex non-planar graph called the [Wagner graph](https://en.wikipedia.org/wiki/Wagner_graph). Since the Wagner graph is not $4$-connected, this implies that every $4$-connected graph with no $K\_5$ minor is planar.
Alternatively, there is a short proof provided you are happy to assume Kuratowski's theorem. The proof idea is to start with a model of $K\_{3,3}$ and then use $4$-connectivity to 'augment' the model of $K\_{3,3}$ to a model of $K\_5$. See the paper [A Quick Proof of Wagner's Equivalence Theorem](https://londmathsoc.onlinelibrary.wiley.com/doi/10.1112/jlms/s2-3.4.661) by Young.
|
10
|
https://mathoverflow.net/users/2233
|
408744
| 167,371 |
https://mathoverflow.net/questions/408747
|
1
|
Let $f(z)$ be a holomorphic function in the angle $A=\{0<\arg z<\frac{\pi}2\}$, continuous in $\bar A$, satisfying $|f(z)|\le M$ on $\partial A$ and satysfying the following growth condition:
$$
|f(x+i y)|\le Ce^{y^2}\quad\hbox{in $A$}. \label{1}\tag{$\ast$}
$$
**Question**. Does it follow that $f$ is bounded in $A$?
If in \eqref{1} we consider $y^\rho$ with $\rho<2$, then we have $|f(z)|\le C e^{|z|^\rho}$ and by the [Phragmén–Lindelöf principle](https://en.wikipedia.org/wiki/Phragm%C3%A9n%E2%80%93Lindel%C3%B6f_principle#Phragm%C3%A9n%E2%80%93Lindel%C3%B6f_principle_for_a_sector_in_the_complex_plane) $f$ would be bounded in $A$. For $\rho=2$ it's not the case as the example $f(z)=e^{-i z^2}$, where $|f(x+i y)|=e^{2xy}$ shows. But for this function the condition \eqref{1} doesn't hold.
|
https://mathoverflow.net/users/14551
|
Phragmén–Lindelöf principle for the critical exponent
|
For a small $\alpha > 0$, write $$\beta = \frac{\sin^2 \alpha}{\sin(2 \alpha)} = \frac{\tan \alpha}{2} ,$$ and define $$g(z) = f(z) \exp(i \beta z^2).$$ Then $$|g(z)| \leqslant |f(z)| \leqslant C \exp(|z|^2)$$ for $z \in A$, $$|g(i r)| = |f(i r)| \leqslant M$$ for $r > 0$, and $$\begin{aligned}|g(r e^{i \alpha})| & = |f(r e^{i \alpha}))| \exp(-\beta r^2 \sin(2 \alpha)) \\ & \leqslant C \exp(r^2 \sin^2 \alpha - \beta r^2 \sin(2 \alpha)) = C \end{aligned}$$ for $r > 0$. By the Phragmén–Lindelöf principle, $g$ is bounded by $C + M$ in $A\_\alpha = \{z \in \mathbb C : \alpha < z < \tfrac\pi2\}$. Passing to a limit as $\alpha \to 0^+$, we find that $f$ is bounded by $C + M$ in $A$, and by another application of Phragmén–Lindelöf principle, $f$ is in fact bounded by $M$.
|
1
|
https://mathoverflow.net/users/108637
|
408767
| 167,383 |
https://mathoverflow.net/questions/408669
|
6
|
Let $\kappa$ and $\lambda$ be cardinals. A thin $(\kappa,\lambda)$-list is a function $L:[\lambda]^{<\kappa}\longrightarrow [\lambda]^{<\kappa}$ such that for all $x\in[\lambda]^{<\kappa}$, $L(x)\subseteq x$ and $\{L(y)\;|\;y\subseteq x\}$ has cardinality $<\kappa$.
Say that $(\kappa,\lambda)$-STP holds iff whenever $L$ is a thin $(\kappa,\lambda)$-list, there is some $b\subseteq\lambda$ such that $\{x\;|\;x\cap b=L(x)\}$ is stationary and that $\kappa$ satisfies the super tree property iff $(\kappa,\lambda)$-STP holds for all $\lambda\geq\kappa$.
Say that $(\kappa,\lambda)$-SSTP holds iff for every sequence $(L\_{\alpha})\_{\alpha\in\mu}$ (where $\mu<\kappa$), if every $L\_{\alpha}$ is a thin $(\kappa,\lambda)$-list, there is a sequence $(b\_{\alpha})\_{\alpha\in\mu}$ such that $\{x\;|\;\forall\alpha\in\mu(b\_{\alpha}\cap x=L\_{\alpha}(x))\}$ is stationary. Say that $\kappa$ satisfies the simultaneous super tree property iff $(\kappa,\lambda)$-SSTP holds for all $\lambda\geq\kappa$.
Let $\kappa$ be supercompact, $\lambda\geq\kappa$ and $U$ a normal $\kappa$-complete ultrafilter on $[\lambda]^{<\kappa}$. For a thin $(\kappa,\lambda)$-List $L$ one can show that there is some $b$ such that $\{x\;|\;x\cap b=L(x)\}\in U$, therefore, using $\kappa$-completeness, we have that supercompact cardinals satisfy the simultaneous super tree property, so the simultaneous super tree property is consistent, at least modulo the existence of a supercompact cardinal (and I suspect $\omega\_2$ has the simultaneous super tree property if PFA holds). This begs the following questions:
1. Are the principles $(\kappa,\lambda)$-STP and $(\kappa,\lambda)$-SSTP equivalent for all $\lambda$ (or some fixed $\lambda$ depending on $\kappa$)?
2. Is the super tree property equivalent to the simultaneous super tree property?
|
https://mathoverflow.net/users/138274
|
Are the following two "tree properties" equivalent?
|
The $(\kappa,\lambda)$-STP and $(\kappa,\lambda)$-SSTP are equivalent for any uncountable cardinals $\kappa\leq\lambda$: Let $\mu<\kappa$ and $(L\_\gamma)\_{\gamma<\mu}$ a sequence of thin $(\kappa,\lambda)$-lists. We can then "amalgamate" these lists into one list $L$ as follows:
Let $h:\lambda\times\mu\rightarrow\lambda$ be a bijection. Let
$$C=\{x\in[\lambda]^{<\kappa}\mid \mu\subseteq x\wedge\ x \text{ is closed under both } h \text{ and }h^{-1}\}$$
Then $C$ is club in $[\lambda]^{<\kappa}$ so for all our intents and purposes it is enough to define $L$ on $C$. $L$ puts the information of $L\_\gamma(x)$ onto the "$\gamma$th slice" of $x$, i.e. for all $\beta, \gamma$:
$$h(\beta,\gamma)\in L(x)\Leftrightarrow \beta\in L\_\gamma(x)$$
If $x\in C$ then indeed $L(x)\subseteq x$ and every $L\_\gamma(x)$, $\gamma<\kappa$ is coded into $L(x)$ in a uniform way. Using $\mu<\kappa$ it should be easy to see that $L$ is a thin $(\kappa,\lambda)$-list (we need here that $\kappa$ is inaccessible but this follows from $(\kappa, \lambda)$-STP, see the remark at the end).
If $(\kappa,\lambda)$-STP holds true, there is $b\subseteq \lambda$ and $S\subseteq C$ stationary with $L(x)=x\cap b$ for all $x\in S$. Now for $\gamma<\mu$ define $b\_\gamma$ as the "$\gamma$th slice" of $b$, i.e.:
$$b\_\gamma=\{\beta<\lambda\mid h(\beta,\gamma)\in b\}$$
We get that for $x\in S$ and $\beta\in x$, $\gamma<\mu$:
$$\beta\in x\cap b\_\gamma\Leftrightarrow h(\beta,\gamma)\in x\cap b\Leftrightarrow h(\beta, \gamma)\in L(x)\Leftrightarrow \beta\in L\_\gamma(x)$$
So $L\_\gamma(x)=x\cap b\_\gamma$.
It is also worth noting that $(\kappa, \kappa)$-STP (and $(\kappa, \kappa)$-SSTP for that matter) are both equivalent to $\kappa$ being ineffable.
|
4
|
https://mathoverflow.net/users/125703
|
408769
| 167,385 |
https://mathoverflow.net/questions/408687
|
2
|
Let $A$ and $B$ be $C^\*$-algebras. Given $f \in B^\*$, we can form the right slice map
$$\iota \otimes f: A \otimes B \to A: a \otimes b \mapsto af(b)$$
which extends uniquely to a bounded linear map
$$\iota \otimes f: M(A \otimes B) \to M(A)$$
that is strictly continuous on the unit ball.
Assume $X \in M(A \otimes B)$ satisfies $(\iota \otimes f)(X)=0$ for all $f \in A^\*$. Can we conclude that $X=0?$
**Attempt**: When $B$ is unital, we can proceed as follows: if $a\in A$
$$0 = (\iota\otimes f)(X)a = (\iota \otimes f)(X(a \otimes 1\_B))$$ for all $f \in A^\*$, so since $X(a \otimes 1\_B) \in A \otimes B$ we conclude that $X(a \otimes 1\_B)=0$. Hence, $X(A \otimes B) = 0$ which implies $X=0$.
How to deal with the case that $B$ is non-unital?
|
https://mathoverflow.net/users/216007
|
$(\iota \otimes f)(X) = 0$ for all $f \in B^*$ implies $X=0$
|
From comments, it seems that the OP is using the "abstract" definition of multipliers (compare below). A good reference is indeed the appendix of [arXiv:funct-an/9707009](https://arxiv.org/abs/funct-an/9707009). Let's use some remarks from there (bottom of page 38) to show that $\iota\otimes f:A\otimes B\rightarrow A$ is indeed strict. By Cohen--Hewitt factorisation, we can find $g\in B^\*, c\in B$ with $f = cg$, and so
$$ (\iota\otimes f)(a\otimes b) = f(b) a = g(bc) a = (\iota\otimes g)(a\otimes bc) $$
Thus if $(u\_i)$ is a bounded net in $A\otimes B$ converging strictly to $0$, for $a\in A$ we have that
$$ (\iota\otimes f)(u\_i) a = (\iota\otimes g)(u\_i(a\otimes c)) \rightarrow 0 $$
because $u\_i(a\otimes c)\rightarrow 0$ in norm.
So we form the strict extension to $M(A\otimes B)$. If $(u\_i)$ is a bounded net in $A\otimes B$ converging strictly to $X\in M(A\otimes B)$ then by definition of the strict extension (or by strict continuity),
$$ (\iota\otimes f)(X) a = \lim\_i (\iota\otimes f)(u\_i) a \qquad (a\in A). $$
However, this is equal to
$$ \lim\_i (\iota\otimes g)(u\_i(a\otimes c)) = (\iota\otimes g)(X(a\otimes c)). $$
So if $(\iota\otimes f)(X)=0$ for all $f$, then $(\iota\otimes g)(X(a\otimes c)) = 0$ for all $g,c$ and $a$, and so $X(a\otimes c)=0$ for all $a,c$ so $X=0$.
(I think of this as the "factorisation trick". Aside from CP maps, most examples of strict linear maps seem to feature some notion of "factorisation".)
---
Taka's comment was to use a representation of $A\otimes B$ on a Hilbert space. This is the "centraliser" picture of multipliers: if $A\subseteq\mathcal B(H), B\subseteq\mathcal B(K)$ acting non-degenerately, then $A\otimes B\subseteq\mathcal B(H\otimes K)$ non-degenerately and
$$ M(A\otimes B) \cong \{ T\in\mathcal B(H\otimes K) : Tu, uT\in A\otimes B \ (u\in A\otimes B) \}. $$
This is independent of the representations chosen, so let's suppose that $f\in B^\*$ is the restriction of $\omega\_{\xi,\eta}$ to $B$. Then we have a natural notion of what $(\iota\otimes f)$ is acting on $M(A\otimes B)$: just the restriction of $\iota\otimes\omega\_{\xi,\eta}:\mathcal B(H\otimes K) \rightarrow \mathcal B(H)$. Of course, you'd need to check that this gave the same definition as before. The required result is now obvious.
|
5
|
https://mathoverflow.net/users/406
|
408776
| 167,387 |
https://mathoverflow.net/questions/408740
|
3
|
Let $n\geq2$. Are there sets $A, B \subseteq \mathbb{N}$ such that $|A|=|B|=n$ and all numbers in $A+B$ are primes?
A well-known conjecture is that there are infinite set $A$ and finite set $|B|=n$ such that all numbers in $A+B$ are primes.
Are there infinite sets $A, B \subseteq \mathbb{N}$ such that all numbers in $A+B$ are primes?
I conjecture that the sets do not exist. Someone told me this question seems to be open.
|
https://mathoverflow.net/users/345221
|
Are there infinite sets $A$ and $B$ such that all numbers in $A+B$ are primes?
|
Yes, conditional on the Hardy-Littlewood prime tuples conjecture.
Let $A$ and $B$ be two finite sets such that $A +B$ consists of primes, and such that for all primes $p$, there are residue classes $x\_p$ and $y\_p$ mod $p$ with $x\_p+y\_p \neq 0 \mod p$ such that $A$ does not contain any numbers congruent to $x\_p$ mod $p$ and $B$ does not contain any numbers congruent to $y\_p$ mod $p$.
Conditionally on the Hardy-Littlewood conjecture, it is possible to add one new element to $A$ or $B$, whichever you prefer, while preserving all these properties.
Given this, the existence of infinite $A$, $B$ follows by starting with the empty set, adding an element to $A$, then an element to $B$, then an element to $A$, then an element to $B$, etc. So it suffices to prove this step.
Without loss of generality, we may assume that we are trying to increase $B$.
First, enlarge $A$ to a set $\overline{A}$ such that $\overline{A}$ still does not contain any numbers congruent to $x\_p$ mod $p$, and, in addition, for all $p< |B|+3$, $\overline{A}$ contains all residue classes modulo $p$ except $x\_p$. This is easy to do with the Chinese remainder theorem (and may require adjusting $x\_p$ for $p$ large).
Then $\overline{A}$ is an admissible tuple so, by the Hardy-Littlewood prime tuples conjecture, there is some large $z$ such that $a +z$ is prime for all $a\in \overline{A}$.
Setting $B' = B \cup \{z\}$, we can see that $A = B'$ consists of primes. For each prime $p$, if $p \geq |B|+3$ then there are at least $2$ choices of residue class $y\_p$ such that $B'$ does not contain any numbers congruent to $y\_p$ mod $p$, so at least $1$ such class satisfying $x\_p +y\_p \neq 0$ mod $p$. If $p < |B|+3$ then, taking $z$ to be sufficiently large, for every $a\in \overline{A}$, $a + z$ is not $p$ and thus, because it is prime, is not a multiple of $p$. Thus, by assumption on $\overline{A}$, $z$ is congruent to $-x\_p$ mod $p$. So for $p <|B|+3$, the same $y\_p$ works for $B'$ as worked for $B$.
So indeed $A,B'$ satisfy all the conditions, completing the induction step.
---
This implication was earlier established by Andrew Granville in [A Note on Sums of Primes](https://doi.org/10.4153/CMB-1990-073-7).
|
11
|
https://mathoverflow.net/users/18060
|
408782
| 167,389 |
https://mathoverflow.net/questions/408743
|
5
|
Let $Sh^\infty(\mathsf{Man})$ denote the $\infty$-category of sheaves of $\infty$-groupoids over the site $\mathsf{Man}$ of smooth manifolds (if you prefer, that's the model category of simplicial sheaves on $\mathsf{Man}$),
and let $\mathcal S$ denote the $\infty$-category of $\infty$-groupoids (the usual model category of simplicial sets).
The inclusion $\mathcal S\to Sh^\infty(\mathsf{Man})$
admits a left adjoint
$$
\|\cdot\|:Sh^\infty(\mathsf{Man})\to\mathcal S
$$
called *geometric realisation*.
>
> Given two morphisms $f,g:X\to Y$ in $Sh^\infty(\mathsf{Man})$
>
>
>
> >
> > let us write $f\sim g$ if there exists a (necessarily invertible) 2-morphism $f\Rightarrow g$ in $Sh^\infty(\mathsf{Man})$, and
> >
> >
> >
>
>
>
> >
> > let us write $f\approx g$ if there exists a map $h:X\times\mathbb R\to Y$ such that $h|\_{X\times\{0\}}\sim f$ and $h|\_{X\times\{1\}}\sim g$.
> >
> >
> >
>
>
>
Is it true that for all $M\in\mathsf{Man}$, and all $X\in Sh^\infty(\mathsf{Man})$, the obvious map
$$
\qquad\quad
Hom\_{Sh^\infty(\mathsf{Man})}(M,X)/\approx\quad \to \quad
Hom\_{\mathcal S}(\|M\|,\|X\|)/\sim\qquad\quad(\*)
$$
is bijective?
[In the RHS of (\*), the symbol ∼ just means "homotopic" (and there's only one notion of two morphisms in $\mathcal S$ being homotopic)]
What can be said about the class of objects $M\in Sh^\infty(\mathsf{Man})$ with the property that $\forall X\in Sh^\infty(\mathsf{Man})$ the map $(\*)$ is bijective?
|
https://mathoverflow.net/users/5690
|
Geometric realisation of smooth $\infty$-stacks
|
The case when $M$ is a smooth manifold follows from the [smooth Oka principle](https://ncatlab.org/nlab/show/shape+via+cohesive+path+%E2%88%9E-groupoid#ConsequenceSmoothOkaPrinciple).
See there for an expository account of the argument and references to additional sources.
Indeed, the left side of (\*) is $$\def\Hom{\mathop{\rm Hom}} π\_0(ʃ\Hom(M,X)),$$
whereas the right side of (\*) is $$π\_0(\Hom(ʃM,ʃX)),$$
where $ʃ$ denotes the [shape functor](https://ncatlab.org/nlab/show/shape+modality),
which is called “geometric realization” in the main post and is denoted by $‖{-}‖$ there.
(From my point of view, a geometric realization functor converts a categorical object like a simplicial set to a geometric object like a topological space, whereas a shape functor converts a geometric object like a sheaf of simplicial sets on manifolds to a categorical object like a simplicial set.)
Concerning the case of a general $M$,
not much can be expected if $M$ has homotopy groups in degree 1 or higher
(meaning $M$ is not weakly equivalent to a sheaf of sets on manifolds).
For example, taking $M=N/\!/G$ to be the stacky quotient of a manifold by an action of a Lie group, and $X$ to be the stack given by the homotopy group completion of the sheaf of symmetric monoidal groupoids of vector bundles,
the left side of (\*) computes the equivariant K-theory of $M$ (basically, it boils down to Segal's model),
whereas the right side computes the Borel equivariant K-theory of $M=N/\!/G$.
These two are different in general.
|
6
|
https://mathoverflow.net/users/402
|
408790
| 167,392 |
https://mathoverflow.net/questions/408725
|
4
|
I was working on finding a series expression for a function $f: \mathbb{C} \rightarrow \mathbb{C}$ such that $f(x^y) = f(x)^{f(y)}$ along the way for construction of such a function I came across a limit that I don't seem to have any tools to evaluate:
If we use the usual notation for tetration as $^{n}x$ then I am interested in evaluating
$$ \lim\_{\epsilon \rightarrow 1} \frac{^{\epsilon}2-2}{\epsilon - 1}$$
How can such a limit be "made sense of" and then evaluated? (I realize a closed form is probably out of question but hopefully a series representation might still be possible).
I did some digging around and it seems like [Hellmuth Kneser](https://en.wikipedia.org/wiki/Hellmuth_Kneser)'s tetration is the commonly accepted answer for tetration with fractional real values. At least as per wikipedia.
It also seems there are some formulas for tetration such as the one given [here](https://mathoverflow.net/questions/259278/an-explicit-series-representation-for-the-analytic-tetration-with-complex-height) by what appears to be James D Nixon.
I could try to evaluate the limit in terms of one of those formulae and see if anything comes out but before I went through that I wanted to see if some hyperoperator experts might be able to suggest something better.
|
https://mathoverflow.net/users/46536
|
Finding closed forms/related constants to a limit involving tetration
|
There is an implementation around (Pari/GP; in the [tetration-forum](https://math.eretrandre.org/tetrationforum/showthread.php?tid=1017)) which claims to have a Kneser-implementation. It is a bit difficult to handle, so I'll show here a simpler version (essentially polynomial) of a tetration-function which seems to approximate that Kneser function when the polynomial's order is increased.
With a polynomial of order 32 I got for your problem a value of about $a=1.2329216$
***Added***: *I got the mentioned Kneser-implementation work; it gives a taylorseries in $h$ (where the spurious imaginary parts of the coefficients are deleted) as*
```
f(h)= 2.0
+ 1.2329216 21568706952691849*h
+ 0.3920521 821375754199047770*h^2
+ 0.2175339 103974922699983933*h^3
+ 0.087772 79346256013703882103*h^4
+ 0.0409164 3821897543029382908*h^5 + O(h^6)
```
*where the coefficient at $h^1$ is the first derivative with respect to $h$ which I've approximated by my method.
To compare, the method described below gives (when the same parameters are inserted)*
```
q&d(h) = 2
+ 1.2329216 07*h
+ 0.3920521 614*h^2
+ 0.2175339 098*h^3
+ 0.087772 80375*h^4
+ 0.0409164 4709*h^5 + O(h^6)
```
*[end addition]*
---
This is how I got my approximate solution:
* get `pc` as the vector of first 32 coefficients of the power series for $f(x)=2^x$
* From pc generate the (truncated!) Carlemanmatrix `F` for the function $f(x)$.
The vector `pc` and thus `F` should have infinite size, but of course we must cut it to something we can handle and which gives not too much crap, at least in a certain range of data.
* By the principle of (infinite) Carlemanmatrixes, powers of `F` give iterates of $f(x)$. Fractional powers give fractional iterates- of course, practically this is always only approximated.
* Fractional powers of `F` can be achieved, when we diagonalize `F = M * D * W`(where I write `W`for `M^-1` for simplicity) and compute $F^h = M\*D^h\*W$
The approximation runs now
```
z_0=1;
m_z0 = vector(32,j, sum(k=0,31, z_0^k * M[1+k,j] ) )
h=1.1
z_h=sum(k=0,31, mz_0[1+k] * D[1+k]^h * W[1+k,2])
```
$z\_h$ is now (approximately) the $h$'th iterate from $z\_0$ to $z\_h$.
Letting $h$ go downwards to $h=1$ it seems that we approximate the value I've given above. And because I've tested earlier this method against the (claimed) Kneser's method and have empirically found improving convergence to that results when increasing the size of the matrix (which means increased order of the underlying polynomial) I think, that we are near the true Kneser's solution here.
*Remark: the matrices `M` and `W` look ugly in the view of giving coefficients for a polynomials (ideally we should have power series with infinitely many coefficients) and should surely be evaluated only for $z\_0$ and $h$ near zero. But tests with $z\_0=1$ and various heights $h$ gave meaningful results such that $z\_{h\_1+h\_2} = (z\_{h1})\_{h\_2}$ .*
***{update***
A much better way to compute the derivative of *d/dh F^h* is to use the derivative of the diagonalmatrix $D^h$ with repect to *h*. The code
```
z_0=1;
m_z0=vector(32,j, sum(k=0,31, z_0^k * M[1+k,j] ) )
h=1
z_h=sum(k=0,31, m_z0[1+k] * log(D[1+k]) * D[1+k]^h * W[1+k,2])
```
$\hspace{120 px}$ represents the derivative of `F` with respect to $h$ and gives immediately the first derivative with respect to $h$ at $z\_0=1$ and $h=1$ ***}***
This Q&D-method can easily be reconstructed using Pari/GP. Note that the diagonalization needs large value for the internal decimal precision, I just used *prec=800* (dec digits) for the matrixsize *32x32*.
Short protocol ("a" means your sought value $a={z\_h-2 \over h-1}$):
```
h z_h a
1.0001, 2.0001233, 1.2329608
1.000001, 2.0000012, 1.2329220
1.000000001, 2.0000000, 1.2329216
1.0000000000001, 2.0000000, 1.2329216
```
---
I did not find some closed form for this via google, W|A and inverse symbolic calculator, so possibly the same coefficient, just for the base $e$ instead, might be better known:
The computation for the base $b=e$ instead of $b=2$ gives the following approximation:
```
1.1000000, 3.0407711, 3.2248931
1.0100000, 2.7481969, 2.9915037
1.0010000, 2.7212519, 2.9700850
1.0001000, 2.7185786, 2.9679609
1.0000100, 2.7183115, 2.9677487
1.0000010, 2.7182848, 2.9677274
1.0000001, 2.7182821, 2.9677253
1.0000000, 2.7182819, 2.9677251
1.0000000, 2.7182818, 2.9677251
1.0000000, 2.7182818, 2.9677251
```
*But no success for finding a closed form representation so far.*
---
***Added:*** With that idea of a derivative with respect to the height-parameter the idea of constructing a powerseries from the sequence of higher derivatives comes to mind. With the adapting of my derivative-formula (see par. "update") I construct a powerseries in $h$ which gives the $h$'th iterates starting from $z\_0=0$. This are the first *32* coefficients:
$$ g(h)=
\small \begin{array} {rl}
1 & \\
+0.8893649182531790 & \*h \\
+0.008676688896772925 & \*h^{2} \\
+0.09523868310780171 & \*h^{3} \\
-0.005752348511600163 & \*h^{4} \\
+0.01296662374986303 & \*h^{5} \\
-0.002196108907293045 & \*h^{6} \\
+0.001996803031014021 & \*h^{7} \\
-0.0005633994435825211 & \*h^{8} \\
+0.0003482750168520301 & \*h^{9} \\
-0.0001285546189885988 & \*h^{10} \\
+0.00006709638266032520 & \*h^{11} \\
-0.00002830779288402342 & \*h^{12} \\
+0.00001380563705669089 & \*h^{13} \\
-0.000006205175034368587 & \*h^{14} \\
+0.000002957461339961192 & \*h^{15} \\
-0.000001369764406300663 & \*h^{16} \\
+0.0000006496675894767698 & \*h^{17} \\
-0.0000003055056195438239 & \*h^{18} \\
+0.0000001451337771274319 & \*h^{19} \\
-0.00000006884715898369447 & \*h^{20} \\
+0.00000003282131706893498 & \*h^{21} \\
-0.00000001565967274219413 & \*h^{22} \\
+0.000000007493152321891950 & \*h^{23} \\
-0.000000003590500482888162 & \*h^{24} \\
+0.000000001723905030908289 & \*h^{25} \\
-0.0000000008288804738313791 & \*h^{26} \\
+0.0000000003991564395933366 & \*h^{27} \\
-1.924678487785590E-10 & \*h^{28} \\
+9.292387572960278E-11 & \*h^{29} \\
-4.491503965074198E-11 & \*h^{30} \\
+2.173332915755159E-11 & \*h^{31}
\end{array} $$
This is then good at least for $h= 0 \ldots 1$ and should reproduce the Kneser-solution by $z\_h=\exp\_2^{\circ h}(0)=g(h-1)$ .
*(This part is just current experimenting and might have errors)*
|
3
|
https://mathoverflow.net/users/7710
|
408796
| 167,395 |
https://mathoverflow.net/questions/408797
|
2
|
When can you reconstruct the power series of a function by taking the limits of the coefficients of the polynomials that interpolate its values at $0,1,2,\dots,j$?
More precisely:
Let $f\colon\mathbb{R}\to\mathbb{R}$. For all nonnegative integers $j$, let $p\_j$ be the unique polynomial of degree $j$ with real coefficients such that $p\_j(n)=f(n)$ for $n=0,\dots,j$. Let $a\_{j,k}$ be the coefficient of $x^k$ in $p\_j$, where $a\_{j,k}=0$ if $k>j$. Under what circumstances do we have that $\lim\_{j\to\infty} a\_{j,k}$ exists for all $j$? In this case, let $c\_k=\lim\_{j\to\infty} a\_{j,k}$. When $f$ is analytic, under what circumstances do we have that $f(x)=c\_0+c\_1 x+c\_2 x^2 + c\_3 x^3 +\cdots$?
|
https://mathoverflow.net/users/132459
|
Power series whose coefficients are limits of coefficients of polynomial interpolations
|
In the paper, A note on convergence of Newton interpolating polynomials,
by D. Dimitrov and J. Philipps, Journal of Computational and Applied Mathematics
Volume 51, Issue 1, 30 May 1994, Pages 127-130, the following simple criterion is mentioned: $f$ is entire of exponential type less than $\log 2$,
then the sequence of interpolation polynomials interpolating at non-negative integers converges to $f$.
|
3
|
https://mathoverflow.net/users/25510
|
408802
| 167,397 |
https://mathoverflow.net/questions/408818
|
2
|
Conway constructed a field of characteristic 2 whose elements are the finite Nim values, indexed by the natural numbers. What is known about nontrivial automorphisms of this field? Do any of them correspond to arithmetically tractable bijections from the set of natural numbers to itself?
|
https://mathoverflow.net/users/3621
|
Does Conway’s field of finite nim values have arithmetically tractable isomorphisms?
|
The field of natural numbers under nim operations is precisely the quadratic closure $\mathbb{F}\_{2^{2^\infty}}$ of $\mathbb{F}\_2$, viꝫ. the inductive limit (“union”) of the subfields $\mathbb{F}\_{2^{2^d}}$ given by the nim multiplication on the integers $0,\ldots,2^{2^d}-1$.
The Galois group $\operatorname{Gal}(\mathbb{F}\_{2^{2^\infty}}/\mathbb{F}\_2)$ of automorphisms of $\mathbb{F}\_{2^{2^\infty}}$ is the projective limit of the Galois groups $\operatorname{Gal}(\mathbb{F}\_{2^{2^d}}/\mathbb{F}\_2)$ which are none other than the cyclic groups $\mathbb{Z}/2^d\mathbb{Z}$ generated by the Frobenius or “squaring” map $\sigma\colon x \mapsto x^2$ (which is of order precisely $d$ on $\mathbb{F}\_{2^{2^d}}$); the map $\mathbb{Z}/2^d\mathbb{Z} \to \mathbb{Z}/2^{d'}\mathbb{Z}$ for $d'\leq d$ is, as one might expect, reduction mod $2^{d'}$. So this projective limit is $\mathbb{Z}\_2$ (the $2$-adic integers), i.e., we have $\operatorname{Gal}(\mathbb{F}\_{2^{2^\infty}}/\mathbb{F}\_2) = \mathbb{Z}\_2$, progenerated by $\sigma$, meaning that every automorphism of $\mathbb{F}\_{2^{2^\infty}}$ can be described as $\sigma^z$ for some uniquely defined $2$-adic integer $z$ (we might write this as raising to the power $2^z$ but it's an abuse of notation).
Certainly the $\sigma^k$ for $k\in\mathbb{Z}$ are tractable, since they are just $x \mapsto x^{2^k}$. More generally, if $z$ is tractable as a $2$-adic integer, meaning we have some way to compute its residue $\bar z$ in $\mathbb{Z}/2^d\mathbb{Z}$, then we can compute $\sigma^z$, by applying $\sigma^{\bar z}$ whenever the given $x$ is in $\mathbb{F}\_{2^{2^d}}$ (concretely on the nimbers: is $<2^{2^d}$ as a natural number).
As a very concrete example, $z = 1/3$ is certainly a tractable $2$-adic integer, so this gives an automorphism $\sigma^{1/3}$ which, when applied three times, is the squaring automorphism $\sigma$.
|
4
|
https://mathoverflow.net/users/17064
|
408826
| 167,404 |
https://mathoverflow.net/questions/408463
|
7
|
Can we upper bound the convergence rate of
$$\max\_{\textbf{v}: \left\Vert \textbf{v}\right\Vert\_2=1} \left\{ \left\Vert \textbf{T}^n \textbf{v}\right\Vert^2\_2 - \left\Vert \textbf{T}^{n+1} \textbf{v}\right\Vert^2\_2 \right\}~,$$
where $\textbf{T}\in \mathbb{R}^{d \times d}$ is a contraction operator
($\left\Vert\textbf{T}\right\Vert\_2\le1$)
of rank $r<d$?
---
For example, $\textbf{T}$ can be *nilpotent* with index $q$ (e.g., $\left[\begin{array}{cc}
0 & 1\\
0 & 0
\end{array}\right]
$) and then the maximal decrease *can* be fixed as long as $n<q$, and afterwards it is $0$.
---
I found a Toeplitz operator whose rate is $d/en$:
$$\boldsymbol{T}=\left[\begin{array}{ccccc}
& 1\\
& & \ddots\\
& & & 1\\
1-\epsilon
\end{array}\right]$$ and I have an intuition this should be the upper bound, but I was not able to prove that this is indeed the worst case.
|
https://mathoverflow.net/users/100796
|
Bounding the decrease after applying a contraction operator $n$ vs $n+1$ times
|
You can easily get a slightly cruder bound $d/n$ (or, if you want, $r/n$) as follows.
Let $A\_n=(T^\*)^nT^n$ and $B\_n=A\_n-A\_{n+1}$. Then $(B\_nv,v)=\|T^nv\|^2-\|T^{n+1}v\|^2\ge 0$, so $B\_n$ is positive definite. Also $B\_{n+1}=T^\*B\_nT$, so, since $T$ is a contraction, $Tr B\_{n+1}\le Tr B\_n$ (this is obvious if $T$ is diagonal but in general $T=R\_1DR\_2$ where $R\_j$ are orthogonal and $D$ is a diagonal contraction and conjugation by an orthogonal matrix doesn't change either trace, or positive definiteness).
So, $n\, Tr B\_n\le \sum\_{k=1}^n Tr B\_k=Tr A\_1-Tr A\_{n+1}\le Tr A\_1\le r$ and we are done because the trace dominates the norm for positive definite matrices.
|
5
|
https://mathoverflow.net/users/1131
|
408834
| 167,405 |
https://mathoverflow.net/questions/408839
|
6
|
Let $T$ be a compact Hausdorff extremally disconnected set (so $T$ is a compact Hausdorff space, such that the closure of each open subset is again open). Let $S \subseteq T$ be a closed subset.
**Question:** Is $S$ extremally disconnected?
For me, this looks like a very natural question about extremally disconnected sets. However, on the spot, I could neither prove this, nor find a counterexample. Also, I was not able to find anything on this in the literature.
|
https://mathoverflow.net/users/148992
|
Is a closed subset of an extremally disconnected set again extremally disconnected?
|
No, the Stone-Cech compactification $\beta\mathbf{N}$ of $\mathbf{N}$ is extremally disconnected, but not the Stone-Cech boundary $\beta\mathbf{N}\smallsetminus\mathbf{N}$.
To see this, it is enough to find an increasing sequence $(F\_n)$ of clopen subsets with no supremum (=least upper bound) in the Boolean algebra of clopen subsets, or equivalently of the Boolean algebra of subsets modulo symmetric difference finite subsets.
For this, it is more convenient to work with $\mathbf{N}^2$: then $F\_n$ is just the boundary of $\mathbf{N}\times\{0,\dots,n\}$. An upper bound for this sequence is just a subset $Y$ intersecting each horizontal line in a cofinite subset. By given any such $Y$ one can remove one point in each horizontal layer and get another upper bound $Y'\subset Y$ with $Y\smallsetminus Y'$ infinite. So there is no least upper bound.
|
10
|
https://mathoverflow.net/users/14094
|
408844
| 167,408 |
https://mathoverflow.net/questions/408810
|
5
|
Consider the equation $\bar{\partial} f=g$ on the complex plane. We may assume $g$ is compactly supported, but we need the case that $g$ is only assumed to be continuous. Is there a solution to this equation? (I mean classical solution.) If yes, is it the solution given by Cauchy integral formula?
1. If $g$ is $C^1$, then we can find the answer from almost all textbooks.
2. I tried to use an approximation method, but failed to find the correct estimate.
It seems to me that the answer is yes and is widely known to experts.
|
https://mathoverflow.net/users/27205
|
Inhomogeneous Cauchy–Riemann equation on complex plane with continous right hand side
|
The integral operator
$$Ph(z)=-\frac{1}{\pi}\int\int h(\zeta)\left(\frac{1}{\zeta-z}-\frac{1}{z}\right)dxdy$$
acts on $L^p$, $p>2$, and the result satiafies $(Ph)\_{\overline{z}}=h$
in the sense of distributions. For continuous $h$, this equation may not
have a classical $C^1$ solution.
Edit. The following example was suggested by user @Fedja.
It is known that $(Ph)\_z=Th,$ where $T$ is the following operator:
$$Th(z)=\lim\_{\epsilon\to 0}\left(-\frac{1}{\pi}\int\int\_{|\zeta-z|>\epsilon}\frac{h(\zeta)}{(z-\zeta)^2}dxdy\right).$$
So taking
$f=Ph$, with
$$h(z)=\frac{z^2}{|z|^2\log|z|},$$
which is continuous,
we obtain $f\_z(0)=\infty$, so $f$ is not in $C^1$.
Reference for these two operators $P$ and $T$ is L. Ahlfors, Lectures on quasiconformal mappings, Chap. V, A.
|
5
|
https://mathoverflow.net/users/25510
|
408847
| 167,409 |
https://mathoverflow.net/questions/408846
|
3
|
Let $\mu$ and $\nu$ be probability measures on $\mathbb{R}^n$ with first moment and suppose that both $\mu$ and $\nu$ have a densities with respect to the $n$-dimensional Lebesgue measure. Fix some positive integer $k$.
Are the "simple and broad conditions" guaranteeing that exist a class $C^k$ Monge map $T:\mathbb{R}^n\rightarrow \mathbb{R}^n$ i.e.:
$$
T\_{\#}\mu=\nu
\mbox{, }
\int \|T(x)-x\|\mu(dx) = \mathcal{W}\_1(\mu,\nu)
\mbox{, and } T\in C^k(\mathbb{R}^n,\mathbb{R}^n)?
$$
|
https://mathoverflow.net/users/36886
|
Regularity of transport map
|
This is an open question. The difficulty is that Monge's cost function is very degenerate, so doesn't satisfy the assumptions of the standard regularity theory of optimal transport.
The first difficulty is that the solutions to this problem need not be unique. However, the transport will occur along disjoint line segments (called transport rays) and in order for the transport to be continuous along each line segment, it is generally necessary to assume that the transport is monotonic along transport rays. With this additional assumption, you do obtain a unique solution to the Monge problem (for reasonable measures). In order to obtain any sort of regularity for this transport, you then need to assume that the support of the target measure is convex (which was originally shown by Caffarelli when the cost is the squared distance).
However, even among smooth measures with convex supports, it is possible to find examples where the ray monotone solution fails to be Lipschitz continuous [1]. On the other hand, in two dimensions, there is a result which shows that the monotone optimal mapping is continuous in the interior of the transfer set (i.e., the union of all transfer rays), under the assumptions that the densities of $\mu$ and $\nu$ are positive, continuous, and have compact, convex and disjoint supports [2]. It seems that the optimal estimate, even for smooth measures, might be a $C^{1/2}$ estimate [3], but this is still an open problem.
[1] *Li, Qi-Rui; Santambrogio, Filippo; Wang, Xu-Jia*, [**Regularity in Monge’s mass transfer problem**](http://dx.doi.org/10.1016/j.matpur.2014.03.001), J. Math. Pures Appl. (9) 102, No. 6, 1015-1040 (2014). [ZBL1304.49094](https://zbmath.org/?q=an:1304.49094).
[2] *Fragalà, Ilaria; Gelli, Maria Stella; Pratelli, Aldo*, [**Continuity of an optimal transport in Monge problem**](http://dx.doi.org/10.1016/j.matpur.2005.02.002), J. Math. Pures Appl., IX. Sér. 84, No. 9, 1261-1294 (2005). [ZBL1075.49018](https://zbmath.org/?q=an:1075.49018).
[3] *Colombo, Maria; Indrei, Emanuel*, [**Obstructions to regularity in the classical Monge problem**](http://dx.doi.org/10.4310/MRL.2014.v21.n4.a6), Math. Res. Lett. 21, No. 4, 697-712 (2014). [ZBL1305.49068](https://zbmath.org/?q=an:1305.49068).
|
4
|
https://mathoverflow.net/users/125275
|
408855
| 167,411 |
https://mathoverflow.net/questions/408861
|
16
|
Let $f$ be a meromorphic function on $\mathbb{C}$ which is algebraic over the field of rational functions $\mathbb{C}(z)$ (i.e. satisfies a nontrivial equation $\sum a\_i(z)f(z)^{i}=0$, with $a\_i(z)\in \mathbb{C}(z)$). Is $f$ actually rational?
|
https://mathoverflow.net/users/40297
|
Meromorphic function on $\mathbb{C}$ algebraic over $\mathbb{C}(z)$
|
The following argument is based on Christian Remling's proof (given in a [comment](https://mathoverflow.net/questions/408861/meromorphic-function-on-mathbbc-algebraic-over-mathbbcz#comment1048802_408861)), but is more elementary. Let us examine the behavior of $f(1/z)$ as $z\to 0$. The function $f(1/z)$ is algebraic over $\mathbb{C}(z)$, hence there are complex polynomials $p\_n(z)$ such that
$$\sum\_{n=0}^N p\_n(z)f(1/z)^n=0.$$
Here $N$ is a positive integer. Without loss of generality, $p\_N(z)$ is not identically zero, and the
the constant terms $p\_n(0)$ are also not all zero. Rewriting the
equation as
$$p\_N(z)=-\sum\_{n=0}^{N-1} p\_n(z)f(1/z)^{n-N},\qquad f(1/z)\neq 0,$$
we see that the set of poles of $f(1/z)$ is contained in the set of zeros of $p\_N(z)$. In particular $f(1/z)$ is holomorphic in some punctured disk $\dot D(0,r)$ around the origin. Let $(z\_k)\subset\dot D(0,r)$ be any sequence tending to zero such that $f(1/z\_k)$ tends to a finite limit $w\in\mathbb{C}$. Then we have
$$\sum\_{n=0}^N p\_n(0)w^n=0.$$
That is, there are at most $N$ possible values for the finite limit $w\in\mathbb{C}$. By the [Casorati-Weierstrass theorem](https://en.wikipedia.org/wiki/Casorati%E2%80%93Weierstrass_theorem), we conclude that $f(1/z)$ does not have an essential singularity at $z=0$. That is, both $f(z)$ and $f(1/z)$ are meromorphic on $\mathbb{C}$, which implies that $f(z)$ is a rational function.
**Remark.** The last sentence is also elementary and can be explained as follows. The poles of $f(z)$ are contained in the disk $D(0,1/r)$, so there are finitely many poles, and we can subtract from $f(z)$ the principal parts of its Laurent series at the various poles. The resulting function $g(z)$ is entire, and $g(1/z)$ does not have an essential singularity at $z=0$. Expanding $g(z)$ into a power series around $z=0$, and then replacing $z$ by $1/z$, we see that $g(z)$ is a polynomial. Returning to $f(z)$, we conclude that $f(z)$ is a rational function.
|
15
|
https://mathoverflow.net/users/11919
|
408869
| 167,415 |
https://mathoverflow.net/questions/408865
|
1
|
Consider the vector field $V:\mathbb{R}^4\rightarrow\mathbb{R}^4$, defined by
\begin{equation}
V(x,v,M\_0,M\_1)=(v,\kappa^{-1}(\beta M\_0-v-kx),-M\_0+v M\_1,-M\_1+1-vM\_0),
\end{equation}
such that $\beta,\kappa,k$ are constants. The only equilibrium point occurs at $P^\*=(0,0,0,1)$ and the Jacobian matrix of $V$ at $P^\*$ is
\begin{equation}
JV\_{P^\*}=
\begin{pmatrix}
0 & 1 & 0 & 0 \\
-\frac{k}{\kappa} & -\frac{1}{\kappa} & \frac{\beta}{\kappa} & 0 \\
0 & 1 & -1 & 0 \\
0 & 0 & 0 & -1
\end{pmatrix}.
\end{equation}
So, we have $\lambda\_0=-1$ and using Cardano's formula
\begin{equation\*}
\begin{split}
\Delta\_0&=\left(\frac{\kappa+1}{\kappa}\right)^2-\frac{3(k+1-\beta)}{\kappa},\\
\Delta\_1&=2\left(\frac{\kappa+1}{\kappa}\right)^3-\frac{9(k+1-\beta)(\kappa+1)}{\kappa^2}+\frac{27k}{\kappa},\\
\mathcal{C}&=\sqrt[3]{\frac{\Delta\_1\pm \sqrt{\Delta\_1^2-4\Delta\_0^3}}{2}},\\
\lambda\_i&=-\frac{1}{3}\left(\frac{\kappa+1}{\kappa}+\xi^{i-1} \mathcal{C}+\frac{\Delta\_0}{\xi^{i-1}\mathcal{C}}\right).
\end{split}
\end{equation\*}
The motivation is to find a solution to the linearised system $$\vec{\dot{x}}=JV\_{P^\*}\text{ }\vec{x},$$
whereby $\vec{x}=C\_1 e^{\lambda\_1 t}\vec{v}\_1+...+C\_4 e^{\lambda\_4 t}\vec{v}\_4$ iff $JV\_{P^\*}$ is diagonalisable. Due to the Hartman–Grobman Theorem, we rely on the property that $\Re(\lambda\_j)\neq0$, $\forall j$. Is it possible then to isolate the real part of $\lambda\_{1,2,3}$?
|
https://mathoverflow.net/users/167822
|
Finding the eigenvalues and eigenvectors of Jacobian at equilibrium point of nonlinear ODEs
|
The characteristic polynomial is
$$P(\lambda) = (\lambda+1)\left(\lambda^3 + \frac{\kappa+1}{\kappa} \lambda^2 + \frac{k-\beta+1}{\kappa} \lambda + \frac{k}{\kappa}\right) $$
Since $\kappa \lambda^3 + (\kappa+1) \lambda^2 + (k-\beta+1) \lambda + k$ is irreducible over the rationals, there's no further factorization possible: if you want explicit expressions for the roots, you will indeed have to use the formulas for roots of a
cubic. Of course, for numerical values of the constants you can use numerical methods to get approximate eigenvalues, or you might try series expansions.
|
1
|
https://mathoverflow.net/users/13650
|
408873
| 167,417 |
https://mathoverflow.net/questions/408830
|
2
|
Let us take a relative category $(\mathcal{C},\mathcal{W})$, and consider its hammock localization $L\_H \mathcal{C}$. It seems to me that for every two objects $X,Y \in \mathcal{C}$ the mapping simplicial set $L\_H \mathcal{C}(X,Y)$ has the (strict) right lifting property against all inclusions $\partial \Delta^n \hookrightarrow \Delta^n$ for $n \geq 3$. The reason is the following: let us consider the case $n=3$ for simplicity, the general case is analogous. A diagram of shape $\partial \Delta^n$ means a collection of $n+1$ hammocks of width $n-1$ such that some selected $(n-2)$-faces of these correspond. Therefore, up to unreducing these hammocks, they can all be assumed to be of the same length, so let us consider only one column at a time. For each $i=0,1,2,3$ we have four columns of the form
$$\require{AMScd}
\begin{CD}
\bullet @>f\_1^0>> \bullet\\
@Vv\_{12}^0VV @VVw\_{12}^0V\\
\bullet @>f\_2^0>> \bullet\\
@Vv\_{23}^0VV @VVw\_{23}^0V\\
\bullet @>f\_3^0>> \bullet
\end{CD}
\hspace{30pt}
\begin{CD}
\bullet @>f\_0^1>> \bullet\\
@Vv\_{02}^1VV @VVw\_{02}^1V\\
\bullet @>f\_2^1>> \bullet\\
@Vv\_{23}^1VV @VVw\_{23}^1V\\
\bullet @>f\_3^1>> \bullet
\end{CD}
\hspace{30pt}
\begin{CD}
\bullet @>f\_0^2>> \bullet\\
@Vv\_{01}^2VV @VVw\_{01}^2V\\
\bullet @>f\_1^2>> \bullet\\
@Vv\_{13}^2VV @VVw\_{13}^2V\\
\bullet @>f\_3^2>> \bullet
\end{CD}
\hspace{30pt}
\begin{CD}
\bullet @>f\_0^3>> \bullet\\
@Vv\_{01}^3VV @VVw\_{01}^3V\\
\bullet @>f\_1^3>> \bullet\\
@Vv\_{12}^3VV @VVw\_{12}^3V\\
\bullet @>f\_2^3>> \bullet
\end{CD}
$$
I'm omitting names of objects, a superscript index $i$ just refers to the $i$-th labeled 2-face, a subscript index $i$ refers to the $i$-th row, and a subscript index $ij$ refers to the arrow going from the $i$-th row to the $j$-row. Now let us look at what it means for these wannabe $2$-faces to be assembled into a diagram of shape $\partial \Delta^3$.
For instance, the 0-th face of the 0-th diagram and the 0-th face of the 1st diagram should coincide, and similarly for the other compatibility condtions. This allows us to remove the superscript indices from all $f$'s and from all $v\_{ij}$'s and $w\_{ij}$'s whenever $i$ and $j$ are consecutive.
Moreover, the 1st face of the 0-th diagram and the 0-th face of the 2nd diagram should coincide, and similarly for others. This means that $v\_{13}^2 = v\_{23} \circ v\_{12}$ and analogously for $w\_{13}^2$. This means that we can also remove the subscript indices $ij$ when they are not consecutive and write the corresponding $v$'s and $w$'s in terms of those with consecutive indices. In other words, the above columns may be rewritten as
$$\require{AMScd}
\begin{CD}
\bullet @>f\_1>> \bullet\\
@Vv\_{12}VV @VVw\_{12}V\\
\bullet @>f\_2>> \bullet\\
@Vv\_{23}VV @VVw\_{23}V\\
\bullet @>f\_3>> \bullet
\end{CD}
\hspace{30pt}
\begin{CD}
\bullet @>f\_0>> \bullet\\
@Vv\_{12}\circ v\_{01}VV @VVw\_{12} \circ w\_{01}V\\
\bullet @>f\_2>> \bullet\\
@Vv\_{23}VV @VVw\_{23}V\\
\bullet @>f\_3>> \bullet
\end{CD}
\hspace{30pt}
\begin{CD}
\bullet @>f\_0>> \bullet\\
@Vv\_{01}VV @VVw\_{01}V\\
\bullet @>f\_1>> \bullet\\
@Vv\_{23} \circ v\_{12}VV @VVw\_{23} \circ w\_{12}V\\
\bullet @>f\_3>> \bullet
\end{CD}
\hspace{30pt}
\begin{CD}
\bullet @>f\_0>> \bullet\\
@Vv\_{01}VV @VVw\_{01}V\\
\bullet @>f\_1>> \bullet\\
@Vv\_{12}VV @VVw\_{12}V\\
\bullet @>f\_2>> \bullet
\end{CD}
$$
Now consider the column
$$\require{AMScd}
\begin{CD}
\bullet @>f\_0>> \bullet\\
@Vv\_{01}VV @VVw\_{01}V\\
\bullet @>f\_1>> \bullet\\
@Vv\_{12}VV @VVw\_{12}V\\
\bullet @>f\_2>> \bullet\\
@Vv\_{23}VV @VVw\_{23}V\\
\bullet @>f\_3>> \bullet
\end{CD}
$$
whose faces are clearly the four columns above. Reasoning like this for each column, we have constructed a diagram of shape $\Delta^3$ whose restriction to $\partial \Delta^3$ is what we started with.
There should be a mistake in the above argument, in that we know that there are relative categories with arbitrary mapping simplicial sets, but I fail to find it. Could someone please point it out?
|
https://mathoverflow.net/users/134438
|
Are hammock localizations locally truncated?
|
Your calculation is correct. For every two objects $X, Y \in \mathcal{C}$, the hom space $L^H\mathcal{C}(X,Y)$ has the right lifting property against $\partial \Delta^n \to \Delta^n$ for $n \geq 3$.
First note that the nerve of any category has the strict right lifting property against $\partial \Delta^n \to \Delta^n$ for $n \geq 3$. The space $L^H\mathcal{C}(X,Y)$ is not the nerve of a category, but it is a quotient of the nerve a category (See [here](https://ncatlab.org/nlab/show/simplicial+localization#DwyerKanCalculating)). Call this nerve $N\mathcal{D}(X,Y)$. Your argument can be summarized by saying that any map $\partial \Delta^n \to L^H\mathcal{C}(X,Y)$ can be factored through $N\mathcal{D}(X,Y)$, where we can solve the lifting problem and then project back into $L^H\mathcal{C}(X,Y)$. (This is the "unreducing" part of your argument).
Note that what I just said does not yet prove that the lift to $L^H\mathcal{C}(X,Y)$ is unique. This would require a further argument about what happens when you have two different lifts $\partial \Delta^n \to N\mathcal{D}(X,Y)$. This might be possible - I am not 100% sure though.
In any case, the simplicial set $L^H\mathcal{C}(X,Y)$ is not a Kan complex (usually). So, as pointed out in the comments by Zhen Lin, having the right lifting property against $\partial \Delta^n \to \Delta^n$ for $n \geq 3$ does not tell us about the higher homotopy groups of $L^H\mathcal{C}(X,Y)$. Indeed, as you correctly observe, there are relative categories which allow you to realize $L^H\mathcal{C}(X,Y)$ as any homotopy type you like.
|
2
|
https://mathoverflow.net/users/184
|
408887
| 167,420 |
https://mathoverflow.net/questions/408872
|
2
|
I have a question that is clearly not research level, but it's confusing me so I will ask anyway.
There must be some little logic flaw I am missing. Take $\Omega$ a bounded smooth domain in $\mathbb R^N$ and assume $ \lambda\_k$ is the $k$ eigenvalue of $ -\Delta$ in $H^1\_0(\Omega)$.
Let $v$ denote a smooth solution of
$$-\Delta v - \lambda^2 v = \lambda^2 \mbox{ in } \Omega$$ with $ v=0$ on $ \partial \Omega$ and we assume $ \lambda^2 \neq \lambda\_k$ for any $k$ but with $ \lambda^2> \lambda\_1$. Then we know that $v$ must be negative somewhere.
Now consider $u$ given by $v= e^{\lambda u}-1$ and note that $ v \ge 0$ in $\Omega$ exactly when $ u \ge 0$ in $ \Omega$. So we expect that $u$ must be negative somewhere. Also note that $u$ must be smooth since $v$ is smooth. Also note that $u$ satisfies
$$-\Delta u = \lambda ( \lvert \nabla u\rvert^2+1) \mbox{ in } \Omega$$ with $u=0$ on $ \partial \Omega$ and hence we can apply the maximum principle to see that $u \ge 0$ in $ \Omega$.
Clearly I am missing something.
|
https://mathoverflow.net/users/66623
|
Simple elliptic pde problem
|
If you apply the maximum principle, at a point $p$ where the function $v$ reaches its minimum, you get $-\lambda^2 v(p) \geq \lambda^2$ so $v(p) \leq -1$. In particular, the function $u$ is not globally defined as it has to go to $-\infty$ at least at $p$.
|
5
|
https://mathoverflow.net/users/24271
|
408890
| 167,422 |
https://mathoverflow.net/questions/408819
|
1
|
In one of Soundararajan's papers, he claims without proof that it is a standard exercise to show that the number $N(X)$ of positive square-free integers $d \equiv 1 \; \bmod \; 8$ less than $X$, with at least two primes factors all of which is congruent to $\pm1 \; \bmod \; 8$ is asymptotically bounded by $\frac{X}{\sqrt{\log(X)}}$ up to a constant. I am new to techniques in analytical number theory, and I find it hard to see this as a standard exercise. Based on my limited knowledge of such techniques I believe this may follow from some kind of Tauberian theorem. But, I couldn't come up with that sort of proof to the above fact. So I would like to see a standard proof of this fact.
Instead, I did some hand-wavy computations that lead me to a much stronger lower bound $\frac{X}{\sqrt{\log(\log X)}}$ (up to a constant). For simplicity, I present my ideas to just counting those $d$ with the stated properties but with prime factors congruent to $1 \; \bmod \; 8$ only, the general case is a simple extension of the same ideas. There are $M = M(x,r) = \frac{X^{1/r}}{8\log(X^{1/r})} = \frac{rX^{1/r}}{8\log(X)}$ number of primes less than $X^{1/r}$ and congruent to $1 \; \bmod \; 8$. So there are $\binom{M}{r}$ number of such $d<X$. Now summing them from $r = 2$ onwards we have $$N(X) = \sum\_{r \geq 2} \binom{M(x,r)}{r} \gg \sum\_{r \geq 2} {\left(\frac{rX^{1/r}}{8\log(X)}\right)}^r \frac{1}{r!} \gg X \sum\_{r \geq 2} \frac{1}{{(8\log(X))}^r} \frac{r^r}{r!}.$$ And now using Stirling's approximation, $\frac{r^r}{r!} \approx \frac{e^r}{\sqrt{2 \pi r}}$. So $N(X) \gg X \sum\_{r \geq 2} {\left(\frac{e}{8\log(X)}\right)}^r r^{-1/2}$, approximating the sum as an integral $$N(X) \gg \int\_2^{\infty}a^x x^{-1/2} dx \approx \frac{\Gamma(1/2, -2 \log(a))}{\sqrt{-\log(a)}}\gg> \frac{1}{\sqrt{\log(\log X)}},$$ where $a = \frac{e}{8\log(X)}$. Hence we have the result. (I use "$\gg$" above to hide constants in the inequalities)
Since my method gives a somewhat stronger bound, I am skeptical if there is some flaw in my ideas. Please feel free to comment on what I have done.
|
https://mathoverflow.net/users/167999
|
Asymptotic lower bound for the number of square free with at least two prime factors
|
Let us introduce a multiplicative function supported on squarefrees by $\alpha(p)=\mathbf{1}\_{p \equiv \pm 1 \bmod 8}$.
We are interested in
$$\sum\_{\substack{n \le x\\ n \equiv 1 \bmod 8\\ n \text{ has }\ge 2 \text{ prime factors}}} \alpha(n) \gg \frac{x}{\sqrt{\log x}}.$$
The primes have density $1/\log x = o(1/\sqrt{\log x})$, so we may omit the condition of having at least two primes factors.
Let us, for the moment, forget about the condition $n \equiv 1 \bmod 8$. The sum $\sum\_{n \le x} \alpha(n)$ can be estimated by Wirsing's Theorem (the main theorem of 'Das asymptotische Verhalten von Summen über multiplikative Funktionen. II' (Acta Math. Acad. Sci. Hungar. 18 (1967), 411–467), giving
$$ \sum\_{n \le x} \alpha(n) \sim C\frac{x}{\log x} \prod\_{p\le x} \left( 1+ \frac{\alpha(p)}{p}\right) = C\frac{x}{\log x} \prod\_{\substack{p\equiv \pm 1 \bmod 8\\ p \le x}} \left( 1+ \frac{1}{p}\right) \sim D \frac{x}{\sqrt{\log x}}$$
for some positive constants $C,D$ depending on $\alpha$. An arithmetic input to these results is $\sum\_{p \equiv a \bmod q, \, p \le x} 1/p = \log \log x/\phi(q) + O(1)$ with $q=8$.
A few remarks:
* One alternative approach is the L-S-D (Landau-Selberg-Delange) method, as suggested by Lucia. Its advantage is that it gives an asymptotic expansion and not only a main term. Tenenbaum's book has good exposition on that. A guided exercise about Landau's original proof, mentioned by Lucia as well, appears in page 187 of Montgomery and Vaughan's book. Granville and Koukoulopoulos have a paper that gives results similar to the L-S-D method, but avoiding complex analysis ('Beyond the LSD method for the partial sums of multiplicative functions
').
* A second alternative approach, which is elementary but very sophisticated, is Iwaniec's half-dimensional sieve.
* There might be a simple (and non sieve-theoretic) elementary way to study $\sum\_{n \le x} \alpha(n)$; see pages 182-184 of Selberg's Collected Papers, Vol II. There he derives quickly and elementarily a variant of Landau's result on sums of two squares.
* If you are only interested in upper bounds, you could use a special case of Shiu's Theorem (`A Brun-Titschmarsh theorem for multiplicative functions'), where again the input will be $\sum\_{p \le x, \, p \equiv a \bmod q} 1/p$.
Now, let us introduce back the condition $n \equiv 1 \bmod 8$. Let us define a multiplicative Möbius-like function:$$ \widetilde{\mu}(p) = \begin{cases} -1 & p \equiv -1 \bmod 8, \\ 1 & p \equiv 1 \bmod 8, \\ 0 & \text{otherwise.}\end{cases}$$
If $n$ is in the support of $\alpha$, $n \equiv (-1)^{\#\{ p \mid n : p \equiv -1 \bmod 8\}} \bmod 8$, so the condition we need to add is equivalent to $\widetilde{\mu}(n) = 1$. In other words, you are interested in evaluating $$\sum\_{n \le x} \alpha(n) \frac{1+\widetilde{\mu}(n)}{2}$$
asymptotically. Since our Möbius-like function is oscillating, we would expect cancellation in $$\sum\_{n \le x} \widetilde{\mu}(n)\alpha(n)$$
over the original sum. Such cancellation can be exhibited again by the L-S-D method. However, in the same paper of Wirsing mentioned above, there is a theorem (Satz 1.2.2) guarantying $$\sum\_{n \le x} \widetilde{\mu}(n)\alpha(n) = o\left( \sum\_{n \le x} \alpha(n)\right).$$
More generally, $\sum\_{n \le x} \beta(n)/\sum\_{n \le x} \alpha(n)$ can be asymptotically estimated if $\beta$ is dominated by $\alpha$ and $\alpha$ is `nice' enough.
|
4
|
https://mathoverflow.net/users/31469
|
408918
| 167,430 |
https://mathoverflow.net/questions/408917
|
2
|
Let $m \in \mathbb{R}$.
Let $f(n)$ be [A007814](https://oeis.org/A007814), exponent of highest power of $2$ dividing $n$, a.k.a. the binary carry sequence, the ruler sequence, or the $2$-adic valuation of $n$.
Let $g(n)$ be [A129760](https://oeis.org/A129760), bitwise $\operatorname{AND}$ of binary representation of $n-1$ and $n$. Here
$$g(n) = n - 2^f(n)$$
Then we have a sequence
$$a(n) = \sum\limits\_{j=0}^{n}(\binom{n}{j}\operatorname{mod} 2)m^j$$
I conjecture that for $n>0$
$$a(n)=(m^{2^{f(n)}}+1)a(g(n))$$
Is there a way to prove it?
|
https://mathoverflow.net/users/231922
|
Modulo $2$ binomial transform of $m^n$
|
If $n\in \mathbb N$ has binary expansion $n=\sum\_{k\in S}2^k$ for a finite subset $S\subset \mathbb N$, then
$$ \sum\_{j=0}^n\Big[\big( {n \atop j}\big)\text{mod}\, 2 \Big]x^jy^{n-j}=\prod\_{k\in S} (x^{2^k}+y^{2^k})$$
More generally, but not needed here (see below): for sums of $r$ indeterminates $x\_i$, the binary coefficient homogeneous polynomial obtained as expansion of $\Big( \sum\_{i=1}^r x\_i\Big)^n$ with coefficients reduced $\text{mod}\,2$ has the factorisation (in $\mathbb Z[x\_1,\dots,x\_r]$)
$$ \sum\_{\alpha\in\mathbb N^r,\, |\alpha|=n}\Big[\big( {n \atop \alpha }\big)\text{mod}\,2 \Big]x^\alpha=\prod\_{k\in S} \sum\_{i=1}^r x\_i^{2^k}. $$
In particular (for $x=1$ and $y=m$) by definition of $f$ and $g$, this implies immediately the relation $a(n)=(1+m^{2^f})a(g(n))$, just because $f\in S$ and $\{2^k\}\_{ k\in S\setminus\{f\}}$ are the powers of the binary expansion of $n-2^f$.
**Details** on the above factorization: immediately by induction on $k$ we have $$\Big( \sum\_{i=1}^r x\_i\Big)^{2^k}= \sum\_{i=1}^r x\_i^{2^k}\text{mod}\,2.$$ Then,
from $n=\sum\_{k\in S}2^k$ one has
$$\Big( \sum\_{i=1}^r x\_i\Big)^n=\Big( \sum\_{i=1}^r x\_i\Big)^{\sum\_{k\in S}2^k}= \prod\_{k\in S} \Big( \sum\_{i=1}^r x\_i\Big)^{2^k} = \prod\_{k\in S} \Big( \sum\_{i=1}^r x\_i^{2^k}\Big) \,\text{mod}\,2.$$
On the other hand, again by induction (on $|S|$) if we do the expansion $$\prod\_{k\in S}\sum\_{i\in [r]}c(k,i)=\sum\_{\iota\in [r]^S}\prod\_{k\in S}c(k,\iota(k))$$ for $\prod\_{k\in S} \Big( \sum\_{i=1}^r x\_i^{2^k}\Big)$, we see that all the $r^{|S|}$ terms of the expansion are different from each other, so that it is a $\{0,1\}$-coefficient polynomial, namely $\displaystyle \sum\_{\alpha\in\mathbb N^r,\, |\alpha|=n}\Big[\big( {n \atop \alpha }\big)\text{mod}\,2 \Big]x^\alpha$.
**Notation:** As usual, for $x=(x\_1,\dots,x\_r)\in \mathbb R^r$ and $\alpha=(\alpha\_1,\dots,\alpha\_r)\in \mathbb N^r$ we denote $x^\alpha:=x\_1^{\alpha\_1}\cdots x\_r^{\alpha\_r}$, $|\alpha|:=\alpha\_1+\dots+\alpha\_r$, and $\big({|\alpha| \atop \alpha }\big):=\frac{|\alpha|!}{\alpha\_1!\cdots\alpha\_r!}$.
|
3
|
https://mathoverflow.net/users/6101
|
408920
| 167,431 |
https://mathoverflow.net/questions/408900
|
4
|
Let $ G $ be a Lie group, $ H $ a closed subgroup, and $ G/H $ compact. Under what conditions do we have that
$$
G/H \cong K/(K\cap H)
$$
where $ K $ is a maximal compact subgroup of $ G $? Obviously the result is trivial if $ G $ is compact.
If $ G=\mathbb{R} $ and $ H=\mathbb{Z} $ this is not true. I guess that is because $ \mathbb{Z} $ is not the real points of a linear algebraic group. As long as $ G $ and $ H $ are real linear algebraic groups and $ G/H $ is compact do we have $ G/H \cong K/(K\cap H) $ ?
EDIT:
I poked around more on the internet and it looks like my example of $ \mathbb{Z} $ in $ \mathbb{R} $ generalizes to a whole class of cocompact but not Zariski closed lattices in nilpotent lie groups and these things called (compact) nilmanifolds which are all iterated torus bundles that can be realized as a nilpotent Lie group mod a cocompact lattice (e.g. Heisenberg group mod integer Heisenberg group). Nilpotent Lie groups are not compact so the maximal compact has strictly smaller dimension and thus cannot act transitively on a compact nilmanifold (since it is just $ G $ mod a lattice and thus has same dimension as $ G $).
I think a more complicated but vaguely similar story holds when generalizing to all solvable lie groups and (compact) solvmanifolds. So lots of solvable Lie groups have non Zariski closed cocompact subgroups and the resulting homogeneous spaces are not acted on transitively by the maximal compact subgroup.
Anyway that's a bunch of no go results...now let's talk sufficient conditions.
Thanks to Ycor for the argument that if $ H $ is Zariski closed cocompact then $ K $ still acts transitively on $ G/H $.
There was claim in this question
<https://math.stackexchange.com/questions/2043479/homogeneous-space-of-sl3-r/2043623#2043623>
that if a compact simply connected manifold is homogeneous for a Lie group $ G $ then it is also homogeneous for the maximal compact subgroup of $ G $. I'm a bit surprised by this fact. Can anyone provide an argument or reference for this fact?
|
https://mathoverflow.net/users/387190
|
Does the maximal compact subgroup always act transitively on a compact homogeneous space?
|
(Comment converted to answer per request:)
The “surprising result” about simply connected homogeneous spaces in your (currently) last paragraph is Montgomery's Theorem ([1950](//ams.org/mathscinet-getitem?mr=0037311)): generally (in your notation) if $G/H$ is compact and $H$ closed ***connected***, then $K$ is transitive on $G/H$. The theorem is also discussed in Samelson ([1952](//mathscinet.ams.org/mathscinet-getitem?mr=45129), p. 17).
|
4
|
https://mathoverflow.net/users/19276
|
408937
| 167,436 |
https://mathoverflow.net/questions/408901
|
1
|
Let $\{\phi(n)\}\_{n\in\mathbb Z}$ be a sequence of complex numbers with the following properties:
1. $\phi(0)=0$ and $|\phi(n)|\leq \frac{C\_1}{|n|}$ for all $n\neq 0$ and $C\_1>0$ is independent of $n.$
2. $|\phi(n+1)-\phi(n)|\leq \frac{C\_2}{n^2}$ for all $n\neq 0$ and $C\_2>0$ is independent of $n.$
3. $\sum\_{-N}^N\phi(n)$ converges as $N\to\infty.$
Denote
$$
R(x):=
\begin{cases}
1-|x| & \text{for }|x|<1\\
0 &\text{otherwise}
\end{cases}.
$$
and consider
$$
K(x)=\sum\_{n\in\mathbb Z}\phi(n)R(x-n),
$$ which exists as a function on $\mathbb R.$
* Can anyone prove that
$$
\lim\_{\epsilon\to \infty}\int\limits\_{\frac{1}{\epsilon}<|x|<\epsilon}K(x)\,\mathrm{d}x
$$ exists?
* Also are the following true?
$$
\begin{align}
|K(x)-K(x-y)| &\leq C\_3\frac{|y|}{|x|^2} &\text{for }|x|>2|y|>0\\
|K(x)| & \leq C\_4|x|^{-1}&\text{for }x\neq 0.
\end{align}
$$ Also it is okay to have the above conditions true almost everywhere.
|
https://mathoverflow.net/users/136860
|
Extending a discrete singular kernel
|
$\newcommand\R{\mathbb R}\newcommand{\Z}{\mathbb{Z}}\newcommand{\ep}{\epsilon}\newcommand{\fl}[1]{\lfloor#1\rfloor}$The answer is yes to each of your two questions.
Let $a\_n:=\phi(n)$. Then
\begin{equation\*}
K(x)=\sum\_{n\in\Z}a\_n R(x-n).
\end{equation\*}
Note that for all $j\in\Z$ we have $K(j)=a\_j$ and $K$ linear (or, more exactly, affine) on the interval $[j,j+1]$. Also, $K$ is continuous. So, $K$ is obtained by the linear interpolation of the function $\Z\ni j\mapsto a\_j$. In particular, $K$ is bounded.
So, for real $\ep>0$,
\begin{equation\*}
\int\_{1/\ep<|x|<\ep}K(x)\,dx=I\_\ep+O(1/\ep),
\end{equation\*}
where
\begin{equation\*}
I\_\ep:=\int\_{|x|<\ep}K(x)\,dx
=\sum\_{n\in\Z}a\_n J\_n,
\end{equation\*}
\begin{equation\*}
J\_n:=\int\_{-\ep}^\ep dx\,R(x-n).
\end{equation\*}
Let now $N:=\lfloor\ep\rfloor$, so that $N\le\ep<N+1$. Then $J\_n=\int\_\R dx\,R(x-n)=1$ if $|n|\le N-1$ and $J\_n=0$ if $|n|\ge N+2$. Also, $0\le J\_n\le1$ for all $n\in\Z$. So,
\begin{equation\*}
I\_\ep
=\sum\_{|n|\le N-1}a\_n +O(|a\_N|+|a\_{-N}|+|a\_{N+1}|+|a\_{-N-1}|).
\end{equation\*}
So, $I\_\ep$ converges, since $N\to\infty$ (as $\ep\to\infty$), $\sum\_{|n|\le N-1}a\_n$ converges, and $|a\_N|+|a\_{-N}|+|a\_{N+1}|+|a\_{-N-1}|=O(1/N)\to0$. So, $\int\_{1/\ep<|x|<\ep}K(x)\,dx$ converges.
This provides the positive answer to your first question.
The answer to your second question is also positive, that is, for some real $C\_3$ and $C\_4$,
\begin{equation\*}
|K(x)|\le C\_3|x|^{-1} \text{ if }x\ne 0 \tag{1}
\end{equation\*}
and
\begin{equation\*}
|K(x)-K(x-y)| \le C\_4\frac{|y|}{|x|^2} \text{ if }|x|>2|y|>0. \tag{2}
\end{equation\*}
Indeed, since $K$ is obtained by the linear interpolation of the function $\Z\ni j\mapsto a\_j$ and the $a\_j$'s are bounded, we see that the function $K$ is bounded and Lipschitz, so that without loss of generality $|x|>8$ in (1) and (2).
Now take indeed any real $x$ with $|x|>8$ and any real $y$ as in (2). Let
\begin{equation}
j:=\fl{|x|},\quad m:=\fl{|x-y|},
\end{equation}
so that $|j|\ge|m|\ge1+|x|/4$ and also $|j+1|\ge|x|/4$.
So, by the linear interpolation observation and the condition $|a\_n|\le C\_1/|n|$ for $n\ne0$,
\begin{equation\*}
|K(x)|\le|a\_j|+|a\_{j+1}|\le C\_1(|j|^{-1}+|j+1|^{-1})\le8C\_1|x|^{-1},
\end{equation\*}
which verifies (1).
Next, by the linear interpolation observation and the condition $|a\_{n+1}-a\_n|\le C\_2/n^2$ for $n\ne0$, the function $K$ is Lipschitz on $[m,\infty)$ and on $(-\infty,-m]$ with Lipschitz constant $C\_2/(m-1)^2\le4C\_2/x^2$.
Also, $|x|\ge j\ge m$, $|x-y|\ge m$, and, by the condition $|x|>2|y|>0$ in (2), $x$ and $x-y$ are of the same sign. So, either both $x$ and $x-y$ are in $[m,\infty)$ or they are both in $(-\infty,-m]$.
Therefore and because the function $K$ is Lipschitz on $[m,\infty)$ and on $(-\infty,-m]$ with Lipschitz constant $4C\_2/x^2$, (2) follows.
|
1
|
https://mathoverflow.net/users/36721
|
408941
| 167,437 |
https://mathoverflow.net/questions/408700
|
9
|
There are the complex p-adic numbers.
But what is the p-adic analogue of the Cayley–Dickson construction?
Or more important: What is the p-adic analogue of the octonions?
It would be nice if the (unit)-multiplication table of such a p-adic analogue corresponds to the projective plane over the finite field with p elements.
Is there any interesting mathematical structure with such a multiplication table for a general p?
Is there any research work about this?
|
https://mathoverflow.net/users/466686
|
p-adic analogue of octonions
|
Defining and classifying the octonion algebras (composition algebras of dimension $8$) over fields $k$, or, in more sophisticated terms, computing the Galois cohomology set $H^1(k, G\_2)$, is the topic of the book *Octonions, Jordan Algebras and Exceptional Groups* by T. A. Springer & F. D. Veldkamp (2000, appropriately published by Springer). Among other things, after properly defining what this means, they explain (§1.10(vi), page 22) that over the $p$-adic fields, just like over the finite fields or totally imaginary number fields, every octonion algebra is split, meaning it is isomorphic to the obvious one.
[These notes by P. Gille](http://math.univ-lyon1.fr/homes-www/gille/prenotes/octonions_beamer.pdf) discussing the more general problem of the classification of octonion algebras over rings, might also be worth looking at.
|
12
|
https://mathoverflow.net/users/17064
|
408946
| 167,439 |
https://mathoverflow.net/questions/408966
|
0
|
Consider a Brownian motion $B$ and let $f(r)=\sqrt{2r \ln(\ln(r))}.$
Is it true that $\lim\_{r \to +\infty}\frac{1}{f(r)}(B\_r-B\_{\left \lfloor{f(r)}\right \rfloor})= 0$ a.s. ?
If so, how to prove it? Otherwise what counter-example do you suggest ?
|
https://mathoverflow.net/users/172528
|
$\lim_{r \to +\infty}\frac{1}{\sqrt{2r \ln(\ln(r))}}(B_r-B_{\left \lfloor{\sqrt{2r \ln(\ln(r))}}\right \rfloor})= 0$ a.s.?
|
no, the second term is basically $B\_t/t$ which (proabably) does go to 0, but the first is governed by the law of the iterated logarithm and does not. In fact, you know that limsup of that term is 1.
|
1
|
https://mathoverflow.net/users/143907
|
408971
| 167,444 |
https://mathoverflow.net/questions/408763
|
6
|
Let $E$ be a locally convex topological vector space over $\mathbb{R}$. The projectivization $PE$ is the quotient of $E\backslash\{0\_{E}\}$ with respect to the equivalence relation $e\sim f$ if $e=\lambda f$.
>
> Is $PE$ a Tychonoff (i.e. completely regular Hausdorff) space?
>
>
>
As far as I can tell, the theorems about the quotient uniform spaces do not apply. On the other hand, it is plausible to expect that this is a known fact.
I can show that $PE$ is completely Hausdorff, i.e. any two points can be separated by a real-valued continuous function. Indeed, if $e\not\sim f$, take $\mu,\nu\in E^{\*}$ such that $\left<\mu,e\right>=1, \left<\nu,e\right>=0, \left<\mu,f\right>=0, \left<\nu,f\right>=1$, and consider the map $\mu\oplus \nu:E\to \mathbb{R}^2$. By the definition of the quotient, this map induces a map $\varphi: PE\to P\mathbb{R}^2=S^1$. Since the latter is Tychonoff, we can separate the images of the classes of $e$ and $f$ by a continuous function.
|
https://mathoverflow.net/users/53155
|
Is the projectivization of a topological vector space Tychonoff?
|
The projective space $PE$ of a topological vector space $E$ is Hausdorff but in general is not Tychonoff, not functionally Hausdorff and even not Urysohn (let us recall that a topological space is *Urysohn* if any distinct points have disjoint closed neighborhoods).
As a suitable counterexample, consider the countable product of lines $E=\mathbb R^\omega$. The projective space $P\mathbb R^\omega$ is *superconnected* in the sense that for any non-empty open sets $U\_1,\dots,U\_n$ in $P\mathbb R^\omega$ the intersection of their closures $\overline U\_1\cap\dots\cap\overline U\_n$ is not empty. This pathological property of the projective space $P\mathbb R^\omega$ was first noticed by [Gelfand and Fuks](http://www.mathnet.ru/php/getFT.phtml?jrnid=faa&paperid=2842&what=fullt&option_lang=rus) in 1967.
A countable counterpart of the projective space $P\mathbb R^\omega$ is the projective space $\mathbb QP^\infty$, whose topology has been characterized in [this paper](https://doi.org/10.1016/j.topol.2021.107909).
|
5
|
https://mathoverflow.net/users/61536
|
408973
| 167,445 |
https://mathoverflow.net/questions/408979
|
8
|
I am interested in smooth nonedegenerate surfaces $X\subset\mathbb{P}^n$, $n\geq 5$, whose secant variety $\sigma(X)$ has dimension $4$. Clearly, the second Veronese embedding of $\mathbb{P}^2$ is such an example. I would be happy about an answer to any of the following
>
> **Questions:** Which other examples exist? Is there a characterization of all such
> surfaces?
>
>
>
Given a projective surface $X$ and a very ample divisor $D$ on $X$, are there convenient criteria on $D$ that guarantee that the secant variety of the embedding of $X$ via the complete linear system $|D|$ has dimension $4$? Maybe in terms of cohomology and/or some intersection product?
|
https://mathoverflow.net/users/36563
|
Smooth surfaces with defective secant variety
|
The Veronese surface in $\mathbb{P}^5$ is indeed the only secant defective surface that is not a cone. This is a classical (and non-trivial) result by F. Severi, see p. 6 and Theorem 10.1 in
C. Ciliberto, F. Russo: [Varieties with minimal secant degree and linear systems of maximal dimension on surfaces](http://dx.doi.org/10.1016/j.aim.2004.10.008), *Adv. Math.* **200**, No. 1, 1-50 (2006). [ZBL1086.14043](https://zbmath.org/?q=an:1086.14043)
ad some of the references cited in the Introduction of the same paper, notably [14, 54, 57].
|
10
|
https://mathoverflow.net/users/7460
|
408980
| 167,447 |
https://mathoverflow.net/questions/408982
|
3
|
**Could you please recommend a reference to or supply a proof of the following identity \eqref{combin-ID-Maclaurin}, or \eqref{first-equiv-form}, or \eqref{combin-ID-Mac-Equiv}, or \eqref{combin-ID-Mac-Reform} for $m\ge2$?**
For non-negative integers $k,n\ge0$, I guess that
\begin{equation}\label{combin-ID-Maclaurin}\tag{Q1}
\sum\_{\ell=0}^{n}\binom{2n+1}{2\ell+1} \binom{n-\ell}{n-k}
=2^{2k}\frac{2n+1}{2k+1}\binom{n+k}{2k}.
\end{equation}
>
> The identity \eqref{combin-ID-Maclaurin} is equivalent to
> \begin{equation}\label{first-equiv-form}\tag{Q2}
> \sum\_{\ell=0}^{k} \binom{2n+1}{2k-2\ell+1} \binom{\ell+n-k}{\ell}
> =2^{2k}\frac{2n+1}{2k+1}\binom{n+k}{2k}, \quad k,n\ge0
> \end{equation}
> and is equivalent to
> \begin{equation}\label{combin-ID-Mac-Equiv}\tag{Q3}
> \sum\_{\ell=0}^{n}\binom{2n+1}{2\ell+1} \binom{n-\ell}{m}
> =2^{2(n-m)}\frac{2n+1}{2(n-m)+1}\binom{2n-m}{m}, \quad n\ge0,\quad n\ge m\in\mathbb{Z}.
> \end{equation}
> The identity \eqref{combin-ID-Mac-Equiv} can be further reformulated as
> \begin{equation}\label{combin-ID-Mac-Reform}\tag{Q4}
> \sum\_{\ell=0}^{n}\binom{2n+1}{2\ell} \binom{\ell}{m}
> =2^{2(n-m)}\frac{2n+1}{2(n-m)+1}\binom{2n-m}{m},\quad n\ge0,\quad n\ge m\in\mathbb{Z}.
> \end{equation}
>
>
>
>
>
> >
> > When $m<0$, it is trivial that both sides of \eqref{combin-ID-Mac-Reform} are equal to $0$.
> >
> >
> >
>
>
>
>
>
> >
> > When $m=0$, the identity \eqref{combin-ID-Mac-Equiv} or \eqref{combin-ID-Mac-Reform} becomes
> > \begin{equation}\label{m=0-(1.93)Sprug}\tag{Q5}
> > \sum\_{\ell=0}^{n}\binom{2n+1}{2\ell}
> > =\sum\_{\ell=0}^{n}\binom{2n+1}{2\ell+1}
> > =2^{2n}, \quad n\ge0.
> > \end{equation}
> > This is just the identity (1.93) on page 38 in the monograph [1] below.
> >
> >
> >
>
>
>
>
>
> >
> > When $m=1$, the identity \eqref{combin-ID-Mac-Equiv} or \eqref{combin-ID-Mac-Reform} reduces to
> > \begin{equation}\tag{Q6}
> > \sum\_{\ell=0}^{n}(n-\ell)\binom{2n+1}{2\ell+1}
> > =\sum\_{\ell=0}^{n}\binom{2n+1}{2\ell}\ell
> > =2^{2(n-1)}(2n+1), \quad n\ge1.
> > \end{equation}
> > This follows from combining \eqref{m=0-(1.93)Sprug} with the identity
> > \begin{equation}\tag{Q7}
> > \sum\_{\ell=1}^{n}\binom{2n+1}{2\ell+1}\ell=(2n-1)4^{n-1}, \quad n\ge1,
> > \end{equation}
> > which can be found in (1.100) on page 40 of the monograph [1] below.
> >
> >
> >
>
>
>
**Could you please recommend a reference to or provide a proof of the above identity \eqref{combin-ID-Maclaurin}, or \eqref{first-equiv-form}, or \eqref{combin-ID-Mac-Equiv}, or \eqref{combin-ID-Mac-Reform} for $m\ge2$?**
References
1. R. Sprugnoli, *Riordan Array Proofs of Identities in Gould’s Book*, University of Florence, Italy, 2006.
2. F. Qi, *Diagonal recurrence relations for the Stirling numbers of the first kind*, Contrib. Discrete Math. **11** (2016), no. 1, 22--30; available online at <https://doi.org/10.11575/cdm.v11i1.62389>.
3. F. Qi, *Diagonal recurrence relations, inequalities, and monotonicity related to the Stirling numbers of the second kind*, Math. Inequal. Appl. **19** (2016), no. 1, 313--323; available online at <https://doi.org/10.7153/mia-19-23>.
4. F. Qi and B.-N. Guo, *A diagonal recurrence relation for the Stirling numbers of the first kind*, Appl. Anal. Discrete Math. **12** (2018), no. 1, 153--165; available online at <https://doi.org/10.2298/AADM170405004Q>.
|
https://mathoverflow.net/users/147732
|
Ask for a reference or a proof of a combinatorial identity $\sum_{k=0}^n\binom{2n+1}{2k}\binom {k}{m} =2^{2(n-m)}\frac{2n+1}{2(n-m)+1}\binom{2n-m}{m}$
|
We will prove the identity **(A8)** that qifeng618 alluded to. The method is called the Wilf-Zeilberger methodology. To this end, define the two functions (suppressing $x$)
$$F(n,k):=\frac{\binom{2x+1}{2k+1}\binom{x-k}{n-k}(2n+1)}{(2x+1)\binom{x+n}{2n}4^n}
\qquad \text{and} \qquad
G(n,k):=-\frac{F(n,k)\,2k\,(2k+1)}{2(x+n+1)(n+1-k)}.$$
Next, verify (preferably using a math software) that
$$F(n+1,k)-F(n,k)=G(n,k+1)-G(n,k).$$
Now, sum both sides of this equation over all integers $k$. The right-hand side then vanishes, showing that $f(n+1)-f(n)=0$ where $f(n):=\sum\_kF(n,k)$. Since $f(0)=1$, it follows that $f(n)=1$ for all $n$. That is,
$$\sum\_{k=0}^n\frac{\binom{2x+1}{2k+1}\binom{x-k}{n-k}(2n+1)}{(2x+1)\binom{x+n}{2n}4^n}=1 \qquad \text{or} \qquad
\sum\_{k=0}^n\binom{2x+1}{2k+1}\binom{x-k}{n-k}=\frac{2x+1}{2n+1}\binom{x+n}{2n}4^n,$$
as desired. $\,\,\,\,\square$
|
2
|
https://mathoverflow.net/users/66131
|
408996
| 167,453 |
https://mathoverflow.net/questions/408773
|
10
|
This question has to do with some experimental phenomenon in groups generated by involutions that I cannot explain.
Let $G$ be a finite, undirected graph, and let $W$ be the corresponding *right-angled Coxeter group*, i.e., the generators of $W$ are the vertices $v \in G$, and we have relations $v^2=1$ for all $v$ and $wv=vw$ if $v$ and $w$ are **non-adjacent** in $G$.
Recall that a *Coxeter element* $c$ in $W$ is a product of the generators $v$ in some order. It is well-known that Coxeter elements in such a right-angled Coxeter group correspond to acyclic orientations of $G$: for an acyclic orientation $\mathcal{O}$ of $G$, the corresponding Coxeter element $c$ is $c\_{\mathcal{O}} = v\_1v\_2\cdots v\_n$ where if $(v\_i,v\_j)$ is an arc of $\mathcal{O}$ then $i < j$. (This is enough to specify the Coxeter element because we can commute non-adjacent vertices.)
Furthermore, two Coxeter elements $c\_{\mathcal{O}}$ and $c\_{\mathcal{O}'}$ are conjugate in $W$ iff $\mathcal{O}'$ can be obtained from $\mathcal{O}$ by a series of *sink-source reversals*: choose a sink in $\mathcal{O}$ and reverse the direction of all arcs incident to that sink so that it becomes a source. In fact, there is even an explicit numerical criterion for deciding if $\mathcal{O}$ and $\mathcal{O}'$ are related by a series of sink-source reversals, having to do with choosing a basis of the space of cycles of $G$ (see Pretzel, "On reorienting graphs by pushing down maximal vertices", <https://doi.org/10.1007/BF00390104>).
My question is about a different kind of reversal, corresponding to something different than conjugacy in $W$.
Let $\mathcal{O}$ be an acyclic orientation of $G$; we say that a subset $A$ of vertices of $G$ is *autonomous* with respect to $\mathcal{O}$ if all vertices in $A$ have the same relationship to any vertex outside of $A$, i.e., for $a, a' \in A$ and $x \notin A$: we have an arc $(a,x)$ in $\mathcal{O}$ iff we have an arc $(a',x)$ in $\mathcal{O}$; we have an arc $(x,a)$ in $\mathcal{O}$ iff we have an arc $(x,a')$ in $\mathcal{O}$; and $a$ and $x$ are non-adjacent in $G$ iff $a'$ and $x$ are non-adjacent in $G$. For such an autonomous subset, let $\mathcal{O}\_{-A}$ denote the orientation obtained from $\mathcal{O}$ by reversing all arcs inside of $A$ (i.e., between two vertices in $A$).
**Question**: Suppose that $W$ acts on a finite set $X$. Let $\mathcal{O}$ be an acyclic orientation of $G$ and let $A$ be an autonomous subset with respect to $\mathcal{O}$. Set $c := c\_{\mathcal{O}}$ and $c' := c\_{\mathcal{O}\_{-A}}$. Is it true that the orbit structures of $c$ and $c'$ acting on $X$ are the same? In other words, is it true that, viewing the action as a homomorphism $\pi\colon W \to \mathfrak{S}\_X$, we have that $\pi(c)$ and $\pi(c')$ are conjugate in $\mathfrak{S}\_X$?
Probably it is inessential that $X$ be finite here, but for my purposes that would be enough.
Note that $c$ and $c'$ need not be conjugate in $W$: for example, with $A=G$, then $c'=c^{-1}$, and in general a Coxeter element will not be conjugate to its inverse. But certainly $c$ and $c^{-1}$ have the same orbit structure for any action.
**EDIT**: To make this a little more concrete, let me give one example of one example of an action of $W$ on a finite set.
Let $X$ be the set of independent sets of $G$. Recall that the generators of $W$ are the vertices $v \in G$. For $I\in X$ we define
$$ v\cdot I = \begin{cases} I \cup \{v\} &\textrm{if $v\notin I$ and $I\cup \{v\}$ is an independent set}; \\ I\setminus \{v\} &\textrm{if $v\in I$}; \\ I &\textrm{otherwise}.\end{cases}$$
It's easy to check that this gives an action of $W$ on $X$, and in this case I can give an ad hoc argument that $c$ and $c'$ (related by a reversal of an autonomous subset) have the same orbit structure. But I suspect this is a more general phenomenon.
|
https://mathoverflow.net/users/25028
|
Reversals of autonomous subsets in right-angled Coxeter groups
|
Okay, following the ideas from the comments, I do think there is a counterexample.
Let $G=K\_4$, so $W$ is generated by involutions $v\_1,\ldots,v\_4$ subject to no other relations. Consider the Coxeter element $c=v\_1v\_2v\_3v\_4$. We can reverse the autonomous subset $\{v\_1,v\_2\}$ here to get $c'=v\_2v\_1v\_3v\_4$.
Now consider the homomorphism $\phi\colon W\to S\_3$ defined by $v\_1\mapsto (12)$, $v\_2 \mapsto (13)$, $v\_3\mapsto (13)$, $v\_4\mapsto (23)$. Then $\phi(c)=(12)(13)(13)(23) = (231)$ while $\phi(c') = (13)(12)(13)(23) = e$, so obviously not conjugate.
Clearly what I was thinking could be the case was the wrong thing, but I still believe there should be a broader explanation for this phenomenon of autonomous subset reversal.
|
2
|
https://mathoverflow.net/users/25028
|
409007
| 167,458 |
https://mathoverflow.net/questions/409009
|
4
|
Let $\phi$ be an $N$-function, (i.e. $\phi : \mathbb{R}\_{\geq 0} \to \mathbb{R}\_{\geq 0}$ is convex and satisfies $\lim\_{t \to 0} \frac{\phi(t)}{t} = 0, \lim\_{t\to \infty} \frac{\phi(t)}{t} = \infty$).
We can define the associated Luxemburg norm on the appropriate subspace of $\mathbb{R}$-valued random variables (Orlicz space $L\_\phi$), by $\|Z\|\_{\phi} := \inf \{ \lambda : \mathbb{E} \phi(|Z|/\lambda) \leq 1\}$. The space $L\_\phi$ is just a space of all random variables for which this infimum is finite.
For example, when $\phi(t) = t^p$ for $p > 1$, this is just the $L\_p$ norm, and when $\phi(t) = e^{t^2} -1$ the corresponding Orlicz space is the space of sub-gaussian random variables.
**Question**
Is the following statement true:
Consider a sequence $Z\_1, Z\_2, \ldots Z\_n$ of independent random variables with $\mathbb{E} Z\_i=0$ and $\|Z\_i\|\_\phi \leq 1$ for all $i$. Then
$$ \|\frac{1}{n} \sum\_{i \leq n} Z\_i\|\_\phi \leq \lambda\_\phi(n)$$
for some $\lambda\_\phi$ s.t. $\lambda\_\phi(n) \to 0$ as $n\to \infty$.
**A bit more context**
I am mostly interested in the scenario in which $\phi$ grows only slightly faster than linear, say $\phi(t) = t \ln(1+t)$.
Together with Markov inequality, this would imply that for any $\varepsilon, \delta$, there is some $n\_\phi(\varepsilon, \delta)$ , s.t. $\mathbf{Pr}(|\sum\_{i\leq n} \frac{1}{n} X\_i| > \varepsilon) \leq \delta$. Law of large numbers asserts that as long as the first moment is bounded, the sample average $\sum \frac{1}{s} X\_i$ converges to $0$ - and I hope to be able to quantify the speed of convergence under just slightly stronger assumption.
Note that this is true in $L\_p$ spaces: for any $1< p \leq 2$, we have $\|\frac{1}{n}\sum\_i Z\_i\|\_p \lesssim n^{1/p - 1}$, so the interesting case is $\phi$ growing slower than $t^{1+\gamma}$ for any $\gamma$.
|
https://mathoverflow.net/users/468679
|
Weak concentration bounds for averages of independent random variables in Orlicz spaces
|
In general, the answer is no. Moreover, the answer is no even if
\begin{equation}
\phi(t)=t\ln(1+t). \tag{1}
\end{equation}
Indeed, suppose that $P(Z\_i=0)=1-2p$ and $P(Z\_i=b)=p=P(Z\_i=-b)$ for all $i$, where
\begin{equation\*}
p:=\frac1{2\phi(b)},
\end{equation\*}
$\phi$ is as given by (1),
and $b$ is a large enough positive real number so that $p\in(0,1/2)$.
Then for all $i$ we have $EZ\_i=0$ and $E\phi(|Z\_i|)=1$, so that $\|Z\_i\|\_\phi\le1$. On the other hand, for all real $c>0$ and all natural $n\ge2$
\begin{equation\*}
\begin{aligned}
&E\phi\Big(\Big|\frac1n\sum\_{i=1}^n Z\_i\Big|/c\Big) \\
&\ge\sum\_{i=1}^n \phi\Big(\frac b{cn}\Big)P(|Z\_i|=b,\ Z\_j=0\ \forall j\ne i) \\
&=n \phi\Big(\frac b{cn}\Big)2p(1-2p)^{n-1} \\
&=\frac{2pb}c\,\ln\Big(1+\frac b{cn}\Big)(1-2p)^{n-1}\to\frac1{2c}>1
\end{aligned}
\end{equation\*}
as $n\to\infty$, if $b=n^2$ and $c\in(0,1/2)$. So, for all large enough $n$ we have $E\phi\big(\big|\frac1n\sum\_{i=1}^n Z\_i\big|/c\big)>1$ and hence $\|\frac1n\sum\_{i=1}^n Z\_i\|\_\phi\ge c$ and hence
\begin{equation\*}
\Big\|\frac1n\sum\_{i=1}^n Z\_i\Big\|\_\phi\not\to0
\end{equation\*}
as $n\to\infty$.
---
More generally, the answer will remain no if $\phi(t)=t \ell(t)$, where $\ell$ is any function such that $\ell(t)$ is [slowly varying](https://en.wikipedia.org/wiki/Slowly_varying_function) as $t\to\infty$. Yet more generally, the answer will remain no if $\phi(t)=t L(t)$, where $L$ is any function such that $\sup\limits\_{K\in(0,\infty)}\limsup\limits\_{t\to\infty}\dfrac{L(Kt)}{L(t)}<\infty$.
|
3
|
https://mathoverflow.net/users/36721
|
409031
| 167,468 |
https://mathoverflow.net/questions/408781
|
4
|
Are there examples of
1. a non-zero C\*-algebra which is
2. universally generated by
3. finitely many projections (not all commuting) together with a unit and plus
4. necessarily satisfying some additional relations such that
5. there remain no traces?
In other words, is there a non-zero quotient
$$C^\*(Z/2\*\ldots\*Z/2)\to B\to 0$$
which carries no traces?
So basically the question is about the representation theory of $C^\*(Z/2\*\ldots\*Z/2)$:
For instance, can there arise purely infinite projections?
A particular instance I'm curious about is the following:
I remember vaguely that (under certain circumstances) one can surprisingly build partial isometries out of projections solely, and for example, as Yemon Choi mentioned, even obtain the Cuntz algebra $\mathcal{O}\_2$. So a particular question I have in mind is:
How can this happen resp. does someone remember a reference for this?
Basically I'm curious about how interesting such quotients can generally look like.
And also more generally, what kind of classes of groups are there with some traceless quotients and what kind of interesting phenomena can happen there?
(Unfortunately, a more narrowed down question might be stepping on someone's toes, and I really wouldn't want to cause some troubles for my colleagues here. So my big apologies that I can't be more specific here. I hope that is understandable.)
|
https://mathoverflow.net/users/45494
|
Example: traceless C*-algebra universally generated by projections
|
I'm not sure if the following is exactly what the OP was looking for, but it definitely solves the question. The following lemma implies that $\mathcal O\_2$ is a quotient of $C^\*(\underbrace{(\mathbb Z/2\mathbb Z) \* \dots \* (\mathbb Z/2\mathbb Z)}\_{4\textrm{ times (edited)}} )$.
>
> **Lemma (edited):** Let $A$ be a unital $C^\ast$-algebra which is generated as a unital $C^\ast$-algebra by $n$ arbitrary elements. Then $M\_2(A)$ is generated by $n+2$ self-adjoint unitaries.
>
>
>
>
> **Proof (edited):** Let $x\_1,\dots,x\_n$ be arbitrary generators. We may assume $\|x\_i\| \leq 1$ for all $i$. Then the dilation $u\_i = \begin{pmatrix} \sqrt{1-x\_i^\*x\_i} & x\_i^\*\\ x\_i & -\sqrt{1-x\_ix\_i^\*} \end{pmatrix}$ is a self-adjoint unitary. Let $u\_{n+1} = \begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix}$ and $u\_{n+2} = \begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}$ which are self-adjoint unitaries generating $M\_2(\mathbb C) \subseteq M\_2(A)$. Let $B$ be the unital $C^\ast$-subalgebra of $M\_2(A)$ generated by $u\_1,\dots, u\_{n+2}$. Then $B$ contains $M\_2(\mathbb C)$ and thus also
> \begin{equation}
> \begin{pmatrix} 0 & 0 \\ x\_i & 0 \end{pmatrix} =
> \begin{pmatrix} 0 & 0 \\ 0 & 1 \end{pmatrix} u\_i
> \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}
> \end{equation}
> for $i=1,\dots, n$. As $A= C^\ast(x\_1, \dots, x\_n, 1)$ it follows that $B$ contains both $A\oplus 0 \subseteq M\_2(A)$ and $M\_2(\mathbb C) \subseteq M\_2(A)$, and hence $B= M\_2(A)$. QED.
>
>
>
If $s\_1,s\_2\in \mathcal O\_2$ are the canonical generators of $\mathcal O\_2$. There is an isomorphism $M\_2(\mathcal O\_2) \to \mathcal O\_2$ given by $(z\_{i,j})\_{i,j=1}^2 \mapsto \sum\_{i,j=1}^2 s\_i z\_{i,j} s\_j^\ast$. Thus by the above, $M\_2(\mathcal O\_2) \cong \mathcal O\_2$ is generated by $4$ (edited) self-adjoint unitaries, so $\mathcal O\_2$ is a quotient of $C^\*(\underbrace{(\mathbb Z/2\mathbb Z) \* \dots \* (\mathbb Z/2\mathbb Z)}\_{4\textrm{ times (edited)}} )$.
Finally, I don't know the smallest number of self-adjoint unitaries generating $\mathcal O\_2$. Note also that $\mathcal O\_2$ is actually singly generated and so even generated by $2$ self-adjoint elements instead of 2 isometries (see my answer for [Endomorphisms of the Cuntz algebra](https://mathoverflow.net/questions/375242/endomorphisms-of-the-cuntz-algebra/375270#375270)). So while the above got the number down to 4 (edited), I strongly believe it can be reduced further.
|
6
|
https://mathoverflow.net/users/126109
|
409043
| 167,469 |
https://mathoverflow.net/questions/406758
|
6
|
Let $\{e\_1,\ldots, e\_n\}$ be the standard basis of $\mathbb{C}^n$. Consider the $m$-multilinear form $$v=\sum\_{i=1}^n e\_i^{\otimes m}\in (\mathbb{C}^n)^{\otimes m}$$
and consider the action of $\text{GL}\_n(\mathbb{C})$ on
$(\mathbb{C}^n)^{\otimes m}$.
Question: What is the stabilizer of $v$ in $\text{GL}\_n(\mathbb{C})$ where $m>2$?
If $m=2$ this is just the orthogonal group. If $m>2$ then this must contain the subgroup $S\_n\wr C\_m$, where the $i$-th copy of $C\_m$ acts nontrivialy only on $e\_i$. Does the stabilizer equal to this wreath product, or can it be bigger?
|
https://mathoverflow.net/users/41644
|
Stabilizers of multilinear forms
|
It is equal.
Since the ground field is of characteristic zero and $v$ is symmetric one may as well compute the stabilizer of the polynomial $f:=\sum\_{i=1}^nx\_i^m$.
Assume $g\in\mathrm{GL}\_n(\mathbb C)$ stabilizes $f$. Then it also stabilizes the Hessian $$\det\nolimits\_{ij}(\partial\_i\partial\_jf)\in\mathbb C^\*(x\_1\cdots x\_n)^{m-2}$$ up to a factor. Hence it permutes the irreducibel factors $x\_1,\ldots,x\_n$ up to scalars, i.e., $g\in\mathbb C^\*\wr S\_n$. Plugging this back into $f$ we conclude $g\in C\_m\wr S\_n$.
|
5
|
https://mathoverflow.net/users/89948
|
409061
| 167,474 |
https://mathoverflow.net/questions/409058
|
11
|
Define the *divisibility graph* of a set of positive integers as the graph whose vertices are the integers, two of which are joined by an edge if one divides the other.
For all *N*, is it true that integers less than or equal to *N* whose proper divisors have divisibility graph which is planar are more numerous than those that don't?
Using SAGE, Freddy Barrera determined those not greater than 1000 which are not planar:
32, 36, 48, 60, 64, 72, 80, 84, 90, 96, 100, 108, 112, 120, 126, 128, 132, 140, 144, 150, 156, 160, 162, 168, 176, 180, 192, 196, 198, 200, 204, 208, 210, 216, 220, 224, 225, 228, 234, 240, 243, 252, 256, 260, 264, 270, 272, 276, 280, 288, 294, 300, 304, 306, 308, 312, 315, 320, 324, 330, 336, 340, 342, 348, 350, 352, 360, 364, 368, 372, 378, 380, 384, 390, 392, 396, 400, 405, 408, 414, 416, 420, 432, 440, 441, 444, 448, 450, 456, 460, 462, 464, 468, 476, 480, 484, 486, 490, 492, 495, 496, 500, 504, 510, 512, 516, 520, 522, 525, 528, 532, 540, 544, 546, 550, 552, 558, 560, 564, 567, 570, 572, 576, 580, 585, 588, 592, 594, 600, 608, 612, 616, 620, 624, 630, 636, 640, 644, 648, 650, 656, 660, 666, 672, 675, 676, 680, 684, 688, 690, 693, 696, 700, 702, 704, 708, 714, 720, 726, 728, 729, 732, 735, 736, 738, 740, 744, 748, 750, 752, 756, 760, 765, 768, 770, 774, 780, 784, 792, 798, 800, 804, 810, 812, 816, 819, 820, 825, 828, 832, 836, 840, 846, 848, 850, 852, 855, 858, 860, 864, 868, 870, 876, 880, 882, 884, 888, 891, 896, 900, 910, 912, 918, 920, 924, 928, 930, 936, 940, 944, 945, 948, 950, 952, 954, 960, 966, 968, 972, 975, 976, 980, 984, 988, 990, 992, 996, 1000
<https://puzzling.stackexchange.com/questions/112686/the-divisibility-graph-again/112731#112731>
|
https://mathoverflow.net/users/60732
|
Is the divisibility graph of the proper divisors of n more often planar than not?
|
No, because almost all numbers have at least $4$ distinct prime factors, making the divisibility graph contain a hypercube and thus be nonplanar.
|
27
|
https://mathoverflow.net/users/18060
|
409064
| 167,476 |
https://mathoverflow.net/questions/409080
|
4
|
Suppose we are working in the language of a binary operation symbol $\*$. Let $S$ be a set of equations which generate precisely the same equational theory generated by the set containing the commutative and associative equations: $\{ x\*y=y\*x, (x\*y)\*z=x\*(y\*z) \}$. Suppose further that $S$ has at least three elements. Must $S$ have at least one redundant equation? I asked this on MSE, but got no response, so I am asking it here.
|
https://mathoverflow.net/users/43439
|
A question regarding equational bases of the theory of the commutative and associative properties
|
Yes it's true.
Write $S=(s\_i=t\_i)\_{i\in I}$ with $s\_i,t\_i$ in some free magma on $k\_i$ generators, where each of the $k\_i$ variables appears in either $s\_i$ or $t\_i$.
First, since $\mathbf{N}$ with addition satisfies this theory, we see that the length of $s\_i$ and $t\_i$ (number of letters) are the same. Call it $n\_i$. We suppose no relation is redundant.
If $n\_i\ge 3$ for all $i$, then we have a non-commutative model of the theory ($\{x,y,u,v,0\}$ with $xy=u$, $yx=v$, other products equal $0$), contradiction.
So there exists $j$ with $n\_j\le 2$. If $k\_j=4$, we have the relation $x\_1x\_2=x\_3x\_4$ which obviously doesn't hold in every commutative semigroup. Same if $k\_j=3$ (relation $x\_1x\_2=x\_1x\_3$ or $x\_1x\_2=x\_3x\_1$). Also $k\_j=1$ would mean either the redundant relation $x\_1=x\_1$, or the relation $x\_1=x\_2$ not true in every commutative semigroup. So $k\_j=2$ and we have the commutativity relation (since $x\_1x\_2=x\_1x\_2$ would be redundant). For the same reason, $n\_i\ge 3$ for all $i\neq j$ since otherwise we would obtain a redundant relation.
If $n\_i\ge 4$ for all $i\neq j$ there is a non-associative model $\{x,y,z,u,0\}$ with $xy=yx=z$, $xz,zx=u$ all other products equal 0.
Hence $n\_i=3$ for some $i$. So $k\_i\le 6$. We claim that if $k\_i\ge 4$, the relation $s\_i=t\_i$ does not hold in every commutative semigroup. Indeed, among commutative semigroup, such a relation reduces to one of the following:
$$x\_1x\_2x\_3=x\_4x\_5x\_6,\;x\_1x\_2x\_3=x\_1x\_4x\_5,\; x\_1x\_2x\_3=x\_4x\_5x\_5,$$ $$x\_1x\_2x\_2=x\_3x\_4x\_4,\;x\_1x\_2x\_3=x\_1x\_4x\_4,\;x\_1x\_2x\_3=x\_4x\_4x\_4,\;x\_1x\_2x\_3=x\_1x\_1x\_4,$$
none of which holds in every commutative semigroup.
If $k\_i=3$, for commutative semigroup the relation $s\_i=t\_i$ yields either $x\_1x\_2x\_3=x\_1x\_2x\_3$, or a nontrivial equality which is not true in every commutative semigroup. So $s\_i=t\_i$ is either the associativity relation (modulo commutative changes), or a trivial (hence redundant) equality. So we have the associativity relation, and all other ones are redundant.
It remains to deal with $k\_i\le 2$. If $k\_i=1$, since $xxx$ has only one possible meaning in a commutative magma, we have no possibility. Suppose $k\_i=2$. So we have a relation yielding, in the associative case, $xxy=xxy$, i.e., the relation $x(xy)=(xx)y$. By the same argument, this is the only relation of length $3$ we can get. But then the equational theory is satisfied by the non-associative commutative model $\{x,y,z,X,Y,Z,u,v,0\}$, with (commutativity being implicit) $xy=Z$, $yz=X$, $xz=Y$, $xX=u$, $yY=v$, all other products being zero.
Note: one can't always remove from an infinite equational theory, a non-redundant one. However one can (a) choose a single representative for all equation (modulo changing variables, reversing equalities. For instance if we have $x\_1x\_2=x\_2x\_1$ and $x\_1x\_3=x\_3x\_1$ we merge into a single one. After this, we have only finitely many equalities of the same total length (say $s=t$ has length the sum of lengths of $s$ and $t$), and then we can remove redundant ones of total length $\le 6$. Then the above argument works, since it only assumes that there is no redundant equality of length $\le 6$.
|
5
|
https://mathoverflow.net/users/14094
|
409081
| 167,480 |
https://mathoverflow.net/questions/400628
|
0
|
This question was inspired by Joel David Hamkins's excellent [question](https://mathoverflow.net/questions/205985/is-there-a-leibnizian-model-with-no-definable-elements-in-a-finite-language) on Leibnizian structures with no definable elements. Let $n$ be a positive integer. Is there an infinite structure in a finite language which is Leibnizian, and has exactly $n$ definable elements? The structure has to be infinite, because otherwise we can just take the linear order with $n$ elements.
|
https://mathoverflow.net/users/43439
|
An infinite Leibnizian structure in a finite language with precisely $n$ definable elements
|
*Turning my comment into an answer to move this off the unanswered queue:*
Take a Leibnizian structure with no definable elements and just add $n$ constants to it (and make the structure "otherwise boring" on those elements - e.g. no relation involving one of those elements should hold, and any function involving one of those elements as an input should do something boring like output the first coordinate always). The result will be a Leibnizian structure with exactly $n$ definable elements.
One natural follow-up question is whether there is a relational Leibnizian structure $\mathcal{A}$ with exactly one definable element $a$ such that $\mathcal{A}\setminus\{a\}$ is **not** Leibnizian. The answer to this is **yes**: following [Joel's construction](https://mathoverflow.net/a/206058/8133), let $A\subseteq\mathbb{Z}$ be such that $(\mathbb{Z};<,A)$ is Leibnizian and has no definable elements. We now replace the unary predicate $A$ with a binary predicate $R$ and a fresh element $\*$: let $\mathcal{A}$ be the $\{R,<\}$-structure with domain $\mathbb{Z}\sqcup\{\*\}$, where $<$ is interpreted as usual and $$R^\mathcal{A}=\{(x,y): x=\*\wedge y\in A\}.$$ Clearly $\mathcal{A}$ is Leibnizian with unique definable element $\*$ and removing $\*$ from $\mathcal{A}$ results in *(something equivalent to)* the very non-Leibnizian structure $(\mathbb{Z};<)$. A similar trick can produce for any $k<n$ a Leibnizian structure with exactly $n$ definable elements which becomes non-Leibnizian if we remove at least $k$ of those definable elements.
Of course in a sense all of these are cheats. My next guess at the "right" question is whether there is a Leibnizian structure which is not mutually interpretable with any Leibnizian structure without definable elements. I suspect the answer to that question is negative, but I'm not sure.
|
2
|
https://mathoverflow.net/users/8133
|
409082
| 167,481 |
https://mathoverflow.net/questions/265570
|
6
|
This is essentially what Exercise 5.4 in
Boucheron, Lugosi, Massart *Concentration Inequalities* boils down to:
For real $a,b$ and $0<p<1$,
\begin{align\*}
&pa^2\log( \frac{a^2}{b^2+pa^2-pb^2}) +
(1-p)b^2\log(\frac{b^2}{b^2+pa^2-pb^2})
\\
&\le
\frac{p(1-p)(a-b)^2}{1-2p}\log\frac{1-p}p
.
\end{align\*}
This is supposed to be provable by elementary means and in finite time. Any ideas?
|
https://mathoverflow.net/users/12518
|
Exercise related to log-Sobolev inequalities
|
By scaling if necessary, we may assume without loss of generality that $a^2p + b^2\bar p = 1$ Substituting $u = a^2p$, we can rewrite the 1-dimensional inequality as
\begin{align\*}
f(u) := u\log \frac{u}{p} + \bar u \log \frac{\bar u}{\bar p} - c(p) (\sqrt{\bar u p} - \sqrt{u \bar p})^2 \le 0.
\end{align\*}
We calculate the first two derivatives of $f(u)$. They are
\begin{align\*}
f'(u) &= \log \frac{u}{\bar u} - \log \frac{p}{\bar p} - c(p)(1-2p) + c(p)(1-2u)\sqrt{\frac{p\bar p}{u\bar u}}
\end{align\*}
and
\begin{align\*}
f''(u) &= \frac{1}{u\bar u} - c(p)\frac{\sqrt{p\bar p}}{2(u\bar u)^{3/2}}.
\end{align\*}
We first claim that $f''$ goes to $+\infty$ at 0 and 1, and precisely 2 zeros between 0 and 1 (they may be repeated). The first part is easy to check. To check the number of zeros, we need $2\sqrt{u\bar u} = c(p) \sqrt{p\bar p}$. As $u$ ranges from $[0,1]$, the left hand side ranges from $[0, 1]$ as well. It is not difficult to check that the right hand side is $\le 1$ (for example, by differentiating it and checking that the derivative is 0 exactly at $p=1/2$, which is a maxima where the value is 1).
Having established this, we verify that $f'$ has the following zeros: $p, 1/2, \bar p$, and moreover, $f'$ is $+\infty$ at $0$ and $1$. Since $f''$ has exactly two zeros, $f'$ cannot have any more zeros. Recall that the points where $f'$ is 0 is precisely the list of maxima or minima of $f$. Moreover, the shape of $f''$ ensures that the local maxima are at $p$ and $\bar p$ whereas the local minimum is at $1/2$. Evaluating $f$ at the local maxima, we see that
\begin{align\*}
f(p) = 0,
\end{align\*}
and
\begin{align\*}
f(\bar p) = \bar p \log \frac{\bar p}{p} + p \log \frac{p}{\bar p} - (\bar p - p) \log \frac{\bar p}{p} = 0.
\end{align\*}
Thus, the maximum of $f$ is 0, and the result is proved.
|
1
|
https://mathoverflow.net/users/20062
|
409091
| 167,485 |
https://mathoverflow.net/questions/408955
|
6
|
I’d like to know if a sharp version of Craig’s interpolation theorem for $L\_{\omega\_1 \omega}$ is already known or exists in the literature. By a “sharp” version of this theorem, I mean something like the following statement: if $\phi$ implies $\psi$ then the interpolant $\theta$ is of the same syntactic complexity as $\phi$. For example, if $\phi$ is $\Pi\_\alpha$ then $\theta$ is also $\Pi\_{\alpha}$.
I am aware of similar “sharp” versions of the Craig interpolation theorem, such as that appearing in [this paper](https://arxiv.org/pdf/1808.01588.pdf), but it is not exactly the kind I need. I am pretty certain the “sharp” version above is true, and would be tedius (but not difficult) to prove using a similar proof to the original interpolation theorem. But it also feels to me like something that is known and likely to have been proven before, and that is what I’m curious about.
|
https://mathoverflow.net/users/42949
|
Sharp Craig interpolation theorem for $L_{\omega_1 \omega}$
|
It actually isn't true that the complexity of $\theta$ could be meaningfully bounded in the term of complexities of $\varphi$ and $\psi$.
Let us fix an arbitrary recursive ordinal $\alpha$. Below I sketch a construction of infinitary $\Pi\_n$ formulas $\varphi,\psi$, for some finite $n$ such that there are no $\Pi\_\alpha$ interpolant for them.
Consider the standard model $\mathbb{N}$ of $\mathsf{PA}$. We fix some $\Delta^1\_1$-property $F(X)$ of sets $X\subseteq \mathbb{N}$ that it is not $\boldsymbol\Pi\_\alpha$. For example, $F(X)$ could be the property of $X$ to encode an isomorphic copy of $\omega^{\alpha+1}$. Next we fix first-order arithmetical formulas $\varphi'(X,Y)$ and $\psi'(X,Y)$ depending on free unary predicates such that $$F(X)\iff \mathbb{N}\models\_2 \exists Y\; \varphi'(X,Y)\iff \mathbb{N}\models\_2 \forall Z\; \psi'(X,Z).$$
We put $$\varphi(X,Y) \mathrel{:{=}} \bigwedge \mathsf{Q} \land \forall x \bigvee\limits\_{n<\omega} x=S^n(0)\to \varphi'(X,Y)\text{ and}$$
$$\psi(X,Z) \mathrel{:{=}} \bigwedge \mathsf{Q} \land \forall x \bigvee\limits\_{n<\omega} x=S^n(0)\to \psi'(X,Z),$$
where $\bigwedge \mathbf{Q}$ is the conjuction of the axioms of Robinson's arithmetic $\mathsf{Q}$.
Observe that any interpolant $\theta(X)$ for this pair of $\varphi(X,Y)$ and $\psi(X,Z)$ should express the property $F(X)$ in $\mathbb{N}$. Thus $\theta(X)$ couldn't be a $\Pi\_\alpha$ infinitary formula.
|
5
|
https://mathoverflow.net/users/36385
|
409109
| 167,488 |
https://mathoverflow.net/questions/409118
|
2
|
Let $Pr\_{(X,Y)}$ be a probability distribution of a random vector $(X,Y)$. Let $F$ be the cumulative distribution function of $(X,Y)$. Define
$$
\mathcal{A}\equiv \{(x,y): x\leq 2 \text{ and }x-y\leq 3\}
$$
Is there a way to express $Pr\_{(X,Y)}(\mathcal{A})$ as
$$
(\*)\quad \sum\_{k=1}^K F(a\_k, b\_k)\times c\_k
$$
for some finite $K$ and $\{a\_k, b\_k, c\_k\}\_{k=1}^K$?
---
Note: if $(X,Y)$ had a uniform distribution, then we could write
$$
Pr\_{(X,Y)}(\mathcal{A})=Pr\_{(X,Y)}((-\infty,2]\times [-1,\infty))+\frac{1}{2}Pr\_{(X,Y)}((-\infty,2]\times (-\infty,-1])
$$
which can be rewritten as $(\*)$.
However, I wonder whether we could do something similar for the general case without uniformity.
|
https://mathoverflow.net/users/42412
|
Probability measure of trapezoidal area
|
No, this cannot be done in general. Indeed, let $A:=\mathcal A$. You want to express
$$P((X,Y)\in A)=Ef(X,Y)$$
as
$$\sum\_{k=1}^K c\_k F(a\_k,b\_k)=Eg(X,Y),$$
where
$$f(x,y):=1(x\le2,x-y\le3)$$
and
$$g(x,y):=\sum\_{k=1}^K c\_k 1(x\le a\_k,y\le b\_k).$$
However, for any choice of the numbers $a\_k,b\_k,c\_k$, there will be some $(x\_\*,y\_\*)\in\mathbb R^2$ such that $f(x\_\*,y\_\*)\ne g(x\_\*,y\_\*)$. Letting the random pair $(X,Y)$ take value $(x\_\*,y\_\*)$ with probability $1$, we get
$$P((X,Y)\in A)=Ef(X,Y)=f(x\_\*,y\_\*)\ne g(x\_\*,y\_\*)=Eg(X,Y)=\sum\_{k=1}^K c\_k F(a\_k,b\_k),$$
so that $P(A)\ne\sum\_{k=1}^K c\_k F(a\_k,b\_k)$.
|
2
|
https://mathoverflow.net/users/36721
|
409123
| 167,491 |
https://mathoverflow.net/questions/409122
|
5
|
Let $m, n\in \mathbb{N}$ and $|x| < 1$. I look for hints to derive an analytic formula for
$$f\_{m,n}(x) = \sum\_{k \in \mathbb{N}} {n + k \choose k} {m + k \choose k} x^{k}. $$
|
https://mathoverflow.net/users/3441
|
Generating function of product of binomial coefficients
|
This is Gauss' hypergeometric function
$F(n+1,m+1,1;x)$. You can then apply the huge theory of hypergeometric functions to derive further expressions. For instance, Euler's transformation formula gives the alternative expression
$$\frac 1{(1-x)^{m+n+1}}\,F(-m,-n,1;x)=\frac 1{(1-x)^{m+n+1}}\sum\_{k=0}^{\min(m,n)}\binom mk\binom nk x^k$$
for the same series as a finite sum.
|
8
|
https://mathoverflow.net/users/10846
|
409124
| 167,492 |
https://mathoverflow.net/questions/409115
|
4
|
Let $T$ be an **injective** operator system and $U$ be an arbitrary operator system. Let $\varphi: T \to U$ be a unital completely positive map and $\psi: U \to T$ be a unital completely positive map with $\psi \varphi = \iota\_T$. Is it true that $\varphi$ is a complete isometry?
**Attempt**: I think yes. Here is an argument:
It is clear that $\varphi: T \to \varphi(T)\subseteq U$ is a unital complete order isomorphism, as its inverse $\psi: \varphi(T) \to T$ is also a completely positive map. However, it then follows that $\varphi(T)$ is an injective operator system. By a result of Choi and Effros, we can equip $T$ and $\varphi(T)$ with multiplications such that these operator systems become $C^\*$-algebras such that the identity maps become unital completely isometric maps. But then the map $\varphi: T \to \varphi(T)$ becomes a $\*$-isomorphism (because a unital complete order isomorphism between $C^\*$-algebras automatically preserves the $C^\*$-algebra structures, this is a result by Choi). In particular, $\varphi$ is completely isometric.
Is the above argument correct? And can we prove it using more "elementary" results?
---
**Context of the question**: I try to understand the fact that an injective rigid extension is automatically also an essential extension (see for instance Paulsen's book, theorem 15.8). I tracked down the original proof in the paper "Injective envelopes of operator systems" by Hamana, lemma 3.7, where they end the proof by asserting that a certain composition is the identity map, which motivates my question above. In other papers by Hamana, one can encounter similar statements.
|
https://mathoverflow.net/users/216007
|
If a completely positive unital map admits a completely positive unital left inverse, it is a complete isometry
|
Injectivity of $T$ plays no role. The point is that unital completely positive maps are contractive, so if a strict inequality $\|\phi(a)\|< \|a\|$ holds for some $a\in T$, then $\|a\|=\|\psi(\phi(a))\|\leqslant \|\phi(a)\| < \|a\|$, which is a contradiction. This shows that $\phi$ is an isometry and the same argument applied to matrix amplifications proves that it is in fact a complete isometry.
|
9
|
https://mathoverflow.net/users/24953
|
409128
| 167,493 |
https://mathoverflow.net/questions/409119
|
8
|
In McLarty's *The Rising Sea: Grothendieck on simplicity and generality* I found the following quote:
>
> The same, Grothendieck knew, would work for cases yet unimagined. He notes that Tohoku [Grothendieck 1957] already gave foundations for the cohomology of any topos [Grothendieck 1985–1987, p. P41n.]. That context was hardly foreseen as he wrote Tohoku in 1955. This is one more proof that it was the right idea of cohomology.
>
>
>
In which sense gave Tohoku a foundation for the cohomology of any topos? In particular, which theorem in Tohoku proves or constructs the cohomology of toposes?
[Grothendieck 1985–1987, p. P41n.] is *Récoltes et Semailles*. (I can spot the passage in which Grothendieck refers to Tohoku, but this doesn't answer my question.)
I could swear I heard the claim before that Tohoku is the *only* place in the literature which shows that toposes have cohomology (of course without mentioning the word "topos"), although I can't recall at the moment where I heard that.
|
https://mathoverflow.net/users/469007
|
Tohoku and cohomology of toposes
|
As requested:
By Theorem 1.10.1 in Tohoku, an Grothendieck abelian category has enough injectives. Sheaves of abelian groups on a Grothendieck topos form a Grothendieck abelian category. By Theorem 2.2.2 in Tohoku, one may then take derived functors of global sections.
As mentioned in the comments, though, sheaf cohomology on toposes is developed quite a bit elsewhere (e.g. in SGA 4).
|
9
|
https://mathoverflow.net/users/6936
|
409139
| 167,496 |
https://mathoverflow.net/questions/409073
|
8
|
There are two languages endow the theory of coherent sheaves with a six functor formalism (that I "know" of), one being formulated in $\text{ProCoh}(X)$ by Deligne and the other being $D(\mathcal{O}\_{X,\blacksquare})$ of Clausen-Scholze.
I'm curious as to the relation between the two. Every coherent sheaf gives rise to a solid module, and so we have a natural functor
$$\text{Coh}(X)\to D(\mathcal{O}\_{X,\blacksquare}).$$
A naive extension of this to a functor $\text{ProCoh}(X)\to D(\mathcal{O}\_{X,\blacksquare})$ is given by sending the "limit" of coherent sheaves to the actual limit of their solid modules. Is this the "right" extension to consider?
From the "right" extension you would expect that for any morphism $f:X\to Y$ we have a commutative diagram
$\require{AMScd}$
\begin{CD}
\text{ProCoh}(X) @>>> D(\mathcal{O}\_{X,\blacksquare})\\
@V Rf\_! V V= @VV f\_! V\\
\text{ProCoh}(Y) @>>> D(\mathcal{O}\_{Y,\blacksquare}).
\end{CD}
However I've gotten confused by a toy-computation: if $f:\mathbb{A}^1\_k\to \text{Spec } k$, $j:\mathbb{A}^1\_k\to \mathbb{P}^1\_k$, then since $\mathcal{O}\_{\mathbb{P}^1\_k}$ is an extension of $\mathcal{O}\_{\mathbb{A}^1\_k}$, we have $j\_!\mathcal{O}\_{\mathbb{A}^1\_k}=''\lim"(\mathcal{O}\_{\mathbb{P}^1\_k}(-n\infty))\_{n}$. In particular
$$R^1f\_!\mathcal{O}\_{\mathbb{A}^1\_k}=\lim\_nH^1(\mathbb{P}^1\_k,\mathcal{O}\_{\mathbb{P}^1\_k}(-n\infty))\_{n}=\lim\_n \left(\frac{1}{x\_0x\_1}k[\frac{1}{x\_0},\frac{1}{x\_1}]\right)\_{-n}.$$
However, $f\_!$ is supposed to respect compact objects, and this does look non-compact. Is my computation wrong or is this indeed compact?
|
https://mathoverflow.net/users/152554
|
Relation between ProCoh and solid modules
|
This is a long comment addressing the functor in question but not the lower shriek. I would like to say that it is close to a fully faithful embedding. For example, we claim that the pro-objects in the category of abelian groups of finite presentation (or equivalent, of finite type) is a full subcategory of solid abelian groups.
To see this, let $R=\mathbb Z$. Consider the diagram $\require{AMScd}\newcommand\Ind{\operatorname{Ind}}\newcommand\Pro{\operatorname{Pro}}\newcommand\Perf{\operatorname{Perf}}\newcommand\op{\operatorname{op}}\newcommand\colim{\operatorname{colim}}\newcommand\RHom{\operatorname{RHom}}\newcommand\iRHom{\underline{\operatorname{RHom}}}$
\begin{CD}
\Ind(\Perf\_R)^{\op}@>\colim^{\op}>\simeq>D(R)^{\op}\\
@V\simeq VV@VV\iRHom\_{R\_\blacksquare}(-,R)V\\
\Pro(\Perf\_R)@>\lim>>D(R\_\blacksquare)
\end{CD}
where the left vertical arrow is induced by the self anti-equivalence $\RHom\_R(\cdot,R)\colon\Perf\_R^{\op}\to\Perf\_R$. The top horizontal arrow is an equivalence since $\Perf\_R\subseteq D(R)$ is a set of compact generators. It follows that the bottom functor $\lim$ is fully faithful precisely where the right functor $\iRHom\_{R\_\blacksquare}(\cdot,R)$ is fully faithful.
We note that the right vertical arrow is fully faithful when restricting to the full subcategory $D^b(R)^{\op}\subseteq D(R)^{\op}$ of bounded objects (which was asked before: [Duality between $D^b(\mathbb{Z})$ and $D(\mathrm{Solid})^\omega$](https://mathoverflow.net/q/384600/)). In particular, given a cofiltered diagram of uniformly bounded perfect complexes on the bottom left, since $R=\mathbb Z$ has finite global dimension, the corresponding object on the top left is also uniformly bounded, the result follows.
If I am not mistaken, we only need to assume that $R$ has finite global dimension to make the preceding argument works, and could be slightly strengthened. For example, the restriction of the functor $\iRHom\_{R\_\blacksquare}(\cdot,R)$ to the full subcategory of bounded *above* complexes is known to be fully faithful, inducing an anti-equivalence between bounded above complexes and *pseudocoherent* objects in $D(R\_\blacksquare)$ when $R$ has finite global dimension.
|
4
|
https://mathoverflow.net/users/176381
|
409142
| 167,497 |
https://mathoverflow.net/questions/409093
|
8
|
The classical Calderon-Zygmund decomposition says that if $f\geq 0$ is $L^1$ on a cubes $B$, with average value $\alpha$, then there is a sequence of disjoint cubes $B\_j$, such that the average of $f$ on each $B\_j$ is in between $\alpha$ and $2^n \alpha$, and $f\leq \alpha$ a.e. away from $\bigcup\_j B\_j$.
I am wondering if there is a similar result on (compact) manifolds, for example with a volume doubling condition. If one uses a Vitali type argument, it seems there is no control on the radius (they could be too small), so no upper bound in the average value of $f$ on (small) balls obtained by Vitali - in contrast to the $2^n\alpha$ upper bound of Calderon-Zygmund.
|
https://mathoverflow.net/users/130379
|
Calderon-Zygmund decomposition on manifolds?
|
Calderon-Zygmund theory generalises without much difficulty to doubling metric measure spaces (or more generally to "spaces of homogeneous type"). See for instance Chapter 1 of
*Stein, Elias M.*, Harmonic analysis: Real-variable methods, orthogonality, and oscillatory integrals. With the assistance of Timothy S. Murphy, Princeton Mathematical Series. 43. Princeton, NJ: Princeton University Press. xiii, 695 pp. (1993). [ZBL0821.42001](https://zbmath.org/?q=an:0821.42001).
The generalisation of the Calderon-Zygmund decomposition in this setting is given in Section 1.4.
In more recent years, most of Calderon-Zygmund theory has also been extended to the non-doubling case, though one has to make some natural modifications to the statements in order to avoid trivial counterexamples. One reference is Chapter 2 of
*Tolsa, Xavier*, [**Analytic capacity, the Cauchy transform, and non-homogeneous Calderón-Zygmund theory**](http://dx.doi.org/10.1007/978-3-319-00596-6), Progress in Mathematics 307. Cham: Birkhäuser/Springer (ISBN 978-3-319-00595-9/hbk; 978-3-319-00596-6/ebook). xiii, 396 p. (2014). [ZBL1290.42002](https://zbmath.org/?q=an:1290.42002).
The Calderon-Zygmund decomposition is given as Lemma 2.14 of that book, though the statement may look slightly different from the classical one.
One common trick in this subject is to replace balls by roughly equivalent "dyadic cubes" that have good geometric properties, such as nesting. See for instance
*Hytönen, Tuomas; Kairema, Anna*, [**What is a cube?**](http://dx.doi.org/10.5186/aasfm.2013.3838), Ann. Acad. Sci. Fenn., Math. 38, No. 2, 405-412 (2013). [ZBL1288.30066](https://zbmath.org/?q=an:1288.30066).
for a brief introduction to this topic.
|
10
|
https://mathoverflow.net/users/766
|
409145
| 167,499 |
https://mathoverflow.net/questions/409079
|
7
|
If $R$ is a ring and $M$ an $R$-module, $M$ is *uniserial* if its lattice of submodules is a chain. Over an Artinian $R$, the chain will be finite. From what I understand, deciding when two uniserial modules are isomorphic is an open problem. D'Este, Kaynarca, and Tutuncu point out in the introduction to their paper *Isomorphism problem for uniserial modules over an arbitrary ring,* arXiv:1910.06173v1[math.RT], that B. Huisgen-Zimmerman (*The geometry of uniserial representations of finite dimensional algebras,* J. Pure Appl. Alg. 127 (1998), 39-72) solved the problem for finite dimensional algebras over algebraically closed fields. D'E-K-T also provide an example of an Artin algebra having two non-isomorphic uniserial modules of length two with the same socle and top.
However, what I am more interested in is knowing whether an Artinian ring has *finite uniserial type,* that is, only finitely many isomorphism classes of uniserial modules, especially in the case that the Artinian ring is also a principal ideal ring (two-sided Artinian and two-sided principal). Whether the answer is positive, or negative, it will be of use to me as I put the final touches on an article. If someone has a reference for the answer, that would be great.
|
https://mathoverflow.net/users/8027
|
Rings of finite uniserial type
|
It is an open problem which Artin algebras have only finitely many uniserial indecomposable modules.
This is stated for example as problem 2 in the open problems section in the book "Representation theory of Artin algebras" by Auslander-Reiten-Smalo.
No good characterisation exists as far as I know or an algorithm to check that property for a given algebra.
|
1
|
https://mathoverflow.net/users/61949
|
409148
| 167,501 |
https://mathoverflow.net/questions/408976
|
3
|
Can someone please give me an example of a Noetherian normal local domain of dimension two such that there exists a prime ideal $P$ of height one having the property $P^{(n)}$ is not a principal ideal for any $n \geq 1$. Here $P^{(n)}$ is the [symbolic $n$-power](https://en.wikipedia.org/wiki/Symbolic_power_of_an_ideal).
|
https://mathoverflow.net/users/97962
|
Symbolic powers of a prime ideal of height one
|
Take $E$ an elliptic curve $zy^2 - x(x-z)(x-tz)$ say over $\mathbb{C}$ and choose a point $Q$ of infinite order (or so for instance the divisor $Q - O$ has infinite order in the divisor class group, here $O$ is the point at infinity).
It follows that in the graded ring of dimension $2$,
$$\mathbb{C}[x,y,z]/(zy^2 - x(x-z)(x-tz))$$
that the homogeneous ideal corresponding to $Q$, call it $P$, has the property that $P^{(n)}$ is never principal. Indeed, if $P^{(n)} = (f)$, then $(f)$ is homogeneous and the corresponding divisor $\mathrm{Div}\_E(f/z^{\deg f})$ is linearly equivalent to 0. This contradicts the infinite order of $P$.
One should point out that for any rational double point, the divisor class group is finite by a result of Lipman (if and only if under some hypotheses), so one has to leave the setting of rational singularities. The cone over an elliptic curve is the probably the simplest singularity that is not rational.
|
4
|
https://mathoverflow.net/users/3521
|
409155
| 167,505 |
https://mathoverflow.net/questions/408816
|
4
|
Let $P$ be a convex simplicial polytope in $\mathbb{R}^n$. Can we find a convex simplicial polytope $P\_0$ in $\mathbb{R}^n$ combinatorially equivalent to $P$, satisfying the following condition: The vertices of $P\_0$ are lattice points and for every facet $F$ of $P\_0$ its vertices $v\_1,\dots,v\_n$ span $\mathbb{Z}^n$?
In other words, is a simplicial polytope combinatorially equivalent to a simplicial polytope such that the vertices of the facets and the origin form a unimodular simplex?
|
https://mathoverflow.net/users/38983
|
Simplicial polytope with regular cones
|
The conditions you pose on $P\_0$ imply that it is a *reflexive* polytope. (That is, a lattice polytope with the origin in its interior and such that its polar dual is also a lattice polytope).
There are finitely many reflexive polytopes in each dimension (modulo $GL(\mathbb Z,n)$), which implies that the answer to your question is negative.
For example, in dimension two you can easily construct $P\_0$ for a triangle, quadrilateral, pentagon, and hexagon, but there exists no reflexive heptagon.
|
2
|
https://mathoverflow.net/users/22608
|
409163
| 167,511 |
https://mathoverflow.net/questions/409107
|
10
|
Let $X,Y$ be Hilbert spaces and $P$ a topological space$^1$ and $p\_0\in P$.
Let $f:X\times P\to Y$ be a *continuous* map such that
for any parameter $p\in P$, $f\_p:= f|\_{X\times \{p\}}:X\to Y$ is *smooth*.
Suppose also that $f\_p\to f\_{p\_0}$ in $C^3\_{loc}(X,Y)$ for $p\to p\_0$ (i.e. once we fix an arbitrary compact $K\subset X$, we have uniform convergence in $C^3$-norm over it)
Suppose that $Df\_{p\_0}(x\_0):X\to Y$ is an isomorphism, we would like to prove that under these assumptions there exists a continuous implicit function, i.e.
>
> Show that exists $U\_{x\_0}\times U\_{p\_0}\subset X\times P$ product neighbourhood of $(x\_0,p\_0)$ and *continuous* function $x:U\_{p\_0}\to U\_{x\_0}$ such that the zero locus of $f|\_{U\_{x\_0}\times U\_{p\_0}}$ is the graph of $x$, $\{(x(p),p)\ | \ p\in U\_{p\_0}\}$.
> If false please provide an example
>
>
>
Notice that the existence of the neighbourhood is already a difficult part. Indeed $C^1\_{loc}$ convergence implies that there is $U\_{p\_0}$ such that for any $p\in U\_{p\_0}$, $Df\_{p}(x\_0)$ is isomorphism so we can apply the inverse function theorem but the size of the "inversion"-neighbourhood can shrink to zero as $p\to p\_0$.
In order to prevent this, I thought to use the $C^3\_{loc}$ convergence, because we can estimate the radius of the neighbourhood using the second derivative in a neighbourhood of $x\_0$, however the convergence is only on compact sets therefore I do not know how to use it to produce a bound for $D^2f\_p(x)$ for $\|x-x\_0\|<r$.
For the continuity I hope that somebody knows a version of the inverse function theorem or contraction principle that works in this non-smooth setting, unfortunately $C^r\_{loc}(X,Y)$ is not even a metric space for $X,Y$ infinite dimensional.
$^1$ I am interested in the case where $P$ is a finite dimensional manifold so feel free to assume so if you can say something about this situation.
|
https://mathoverflow.net/users/99042
|
Implicit function theorem with continuous dependence on parameter
|
Assuming $P$ first countable, the standard contraction principle and elementary bounds are sufficient to conclude. You do not need higher regularity:
*Let $X$, $Y$ be Banach spaces, $P$ a topological space, $f:X\times P\to Y$. Assume that*
*i. $f:X\times P\to Y$ is continuous;*
*ii. $f(\cdot,p):X\to Y$ is differentiable, for any $p\in\ P$.*
*iii. $f(x\_0,p\_0)=0$ and $D\_1f(x\_0,p\_0)$ is invertible.*
*iv. $D\_1 f :X\times P \to L(X,Y)$ is continuous *in the pair* at $(x\_0,p\_0)$.*
**Proof:** Let $L:=[D\_1f(x\_0,p\_0)]^{-1}\in L(Y,X)$, and define $g:X\times P\to X$ by $g(x,p):=x-Lf(x,p)$, so that $f(x,p)=0$ iff $x=g(x,p)$. Since $D\_1g(x\_0,p\_0)=0$, and (by iv) $D\_1g$ is continuous at $(x\_0,p\_0)$, there are $r>0$ and a nbd $U$ of $p\_0$ in $P$ such that $\|D\_1g(x,p)\|\le 1/2$ for all $x\in \overline B(x\_0,r)$ and $p\in U$, which implies that for all $p\in U$ the maps $g(\cdot,p)$ are $1/2$ Lipschitz. Since $g(x\_0,p\_0)=x\_0$ and $g$ is continuous, choosing a smaller $U$ we may also assume $\|g(x\_0,p) -x\_0\|\le r/2$. Then for all $x\in \overline B(x\_0,r)$ and $p\in U$ we have
$$\|g(x,p)-x\_0\|\le \|g(x,p) -g(x\_0,p)\|+\|g(x\_0,p) -x\_0\|\le \frac12\|x-x\_0\|+\frac r2\le r.$$
So for all $p\in U$, the closed ball $\overline B(x\_0,r)$ is invariant for the contraction $g(\cdot,p)$, which has therefore a unique fixed point $x=\xi(p)$ in $\overline B(x\_0,r)$. In other words, for any $(x,p)\in \overline B(x\_0,r)\times U$ one has $f(x,p)=0$ iff $x=\xi(p)$. To show that $\xi:U\to \overline B(x\_0,r)$ is continuous at any $p\in U$, write for $q\to p$ in $U$
$$\|\xi(p)-\xi(q)\|=\|g(\xi(p),p)-g(\xi(q),q)\|\le \|g(\xi(p),p)-g(\xi(p),q)\|+\|g(\xi(p),q)-g(\xi(q),q)\|\le$$
$$\le \|g(\xi(p),p)-g(\xi(p),q)\|+\frac12\|\xi(p) -\xi(q) \|,$$
so $\|\xi(p)-\xi(q)\|\le 2\|g(\xi(p),p)-g(\xi(p),q)\|=o(1)$ as $q\to p.$ $\qquad\square$
$$ \* $$
Note that the assumption *iv* on the map $F:=D\_1g$ follows from your hypotheses, provided $P$ is assumed first countable:
*Let $X,P$ be first countable spaces, $(E,\|\cdot\|)$ a normed space, $F:X\times P\to E$. Assume that $F(\cdot,p\_0):X\to E$ is continuous at $x\_0$ and $P\ni p\mapsto F(\cdot,p)$ is continuous at $p\_0$ w.r.to the uniform convergence on compact sets (that is, $\|F(\cdot,p)-F(\cdot,p\_0)\|\_{\infty,K}=o(1)$ as $p\to p\_0$). Then $F$ is continuous at $(x\_0,p\_0).$*
Indeed, for any sequence $(x\_k,p\_k)\to (x\_0,p\_0)$ in $X\times P$ we have
$$\|F(x\_k,p\_k)-F(x\_0,p\_0)\|\le \|F(x\_k,p\_k)-F(x\_k,p\_0)\|+\|F(x\_k,p\_0)-F(x\_0,p\_0)\|\le$$
$$\le \sup\_{j\in\mathbb{N}}\|F(x\_j,p\_k)-F(x\_j,p\_0)\|+\|F(x\_k,p\_0)-F(x\_0,p\_0)\|=o(1).\qquad \square$$
|
5
|
https://mathoverflow.net/users/6101
|
409166
| 167,512 |
https://mathoverflow.net/questions/392731
|
8
|
Suppose $\lambda\vdash n$ is a partition and $S^\lambda$ is the associated irreducible representation of $S\_n$. As we know from the branching rule, we have an isomorphism of $S\_{n-1}$ modules
$$S^\lambda\cong\bigoplus\_{\mu}S^{\mu}$$
where $\mu$ are the partitions obtained by deleting an outer corner from $\lambda$.
In particular, I am concerned with the case when $S^\lambda$ is spanned by basis elements $\{B\_Q\mid Q\in\text{SYT}(\lambda)\}$ of the Kazhdan-Lusztig basis (with the relevant left-action of the generators). Can the isomorphism from the branching rule be realised in such a way that preserves the KL-basis in some form? That is, if $\phi$ is a realisation of the isomorphism, then
$$\phi\cdot B\_Q=\sum\_{\mu}\sum\_{P\in\text{SYT}(\mu)}z\_pB\_P$$
for some constants $z\_P$. What can we say about these constants $z\_P$?
For example, when $\lambda$ is a rectangular partition (and so $S^\lambda\cong S^\mu$), then the isomorphism can be realised as $B\_Q\mapsto B\_{d(Q)}$, where $d(Q)$ is the tableau obtained by deleting the $n$-box from $Q$. I am unsure as to the theory for arbitrary partitions.
Any help would be appreciated!
|
https://mathoverflow.net/users/172429
|
Branching rule for Specht modules over Kazhdan-Lusztig basis
|
It turns out that this is indeed true. The relevant proofs are given in Chapter 3 of [this extended abstract](https://arxiv.org/abs/2111.09510).
|
4
|
https://mathoverflow.net/users/172429
|
409168
| 167,513 |
https://mathoverflow.net/questions/409165
|
-4
|
Is the underlying set of every renormalization group countable and finite？
Suppose A is a renormalization group, and the elements of it compose of the set B. Is B the set countable and finite?
|
https://mathoverflow.net/users/14024
|
Is the underlying set of every renormalization group countable and finite？
|
No, the renormalization group of a continuum field theory contains continuously parameterized scale-changing transformations—hence an uncountable number of them.
|
4
|
https://mathoverflow.net/users/170778
|
409169
| 167,514 |
https://mathoverflow.net/questions/408575
|
9
|
My problem seems elementary. However a post in [SE](https://math.stackexchange.com/questions/4304278/least-common-multiple-of-three-integers) has not got an answer.
Let $a\_1,a\_2,a\_3\geq1$ be integers and let $A=\mathrm{lcm}(a\_1,a\_2,a\_3)$ be their least common multiple. I want to show the following.
>
> If $m\_1,m\_2,m\_3\geq0$ are natural numbers satisfying $m\_1a\_1+m\_2a\_2+m\_3a\_3=2A$, then there exist integers $m\_1',m\_2',m\_3'$ satifying $0\leq m\_i'\leq m\_i$ for each $i$ and $m\_1'a\_1+m\_2'a\_2+m\_3'a\_3=A$.
>
>
>
The motivation for this problem is the following problem I conjecture but can not prove either.
>
> Let $\mathbb Q[x,y,z]$ be the polynomial ring of three variables. Let $a,b,c\geq1$ be the degrees of $x,y,z$ respectively. Let $A=\mathrm{lcm}(a,b,c)$. Then $\mathbb Q[x,y,z]\_{(A)}:=\bigoplus\_{d\geq0}\mathbb Q[x,y,z]\_{Ad}$ is generated by $\mathbb Q[x,y,z]\_A$. The subscript in $\mathbb Q[x,y,z]\_A$ denotes the degree $A$ subspace, i.e. $\mathbb Q[x,y,z]\_A=\mathrm{span}\big\{x^uy^vz^w:u,v,w\geq0,\;ua+vb+wc=A\big\}$.
>
>
>
The first statement will prove the second. Consider $x^uy^vz^w\in\mathbb Q[x,y,z]\_{Ad}$ with $ua+vb+wc=Ad$.
* It suffices to show that there exist $\begin{cases}0\leq u'\leq u\\0\leq v'\leq v\\0\leq w'\leq w\end{cases}$ and $u'a+v'b+w'c=A$. This is because we then have $x^uy^vz^w=\big(x^{u'}y^{v'}z^{w'}\big)\big(x^{u-u'}y^{v-v'}z^{w-w'}\big)\in\mathbb Q[x,y,z]\_A\cdot\mathbb Q[x,y,z]\_{A(d-1)}$, and we reduce $d$ to $d-1$.
* Also if $ua\geq A$ (or $vb\geq A$ or $wc\geq A$), we just take $\begin{cases}u'=A/a\\v'=0\\w'=0\end{cases}$ as a solution.
* So the difficult situation is when $\begin{cases}0\leq u<A/a\\0\leq v<A/b\\0\leq w<A/c\end{cases}$. In this case $dA=ua+vb+wc<3A$, so $d<3$. This reduces to the $d=2$ case, which is the first statement.
Thanks for any comments.
|
https://mathoverflow.net/users/105537
|
Certain property of the least common multiple of three integers
|
1. Without loss of generality $d:=\gcd(a\_1,a\_2,a\_2)=1$, else divide $a\_1,a\_2,a\_3$ and $A$ by $d$.
2. Denote $d\_{ij}=\gcd(a\_i,a\_j)$. Now $d\_{12}, d\_{13},d\_{23}$ are mutually coprime, we may write $a\_1=d\_{12}d\_{13}b\_1$ etc, $A=d\_{12}d\_{13}d\_{23}b\_1b\_2b\_3$, $m\_1$ must be divisible by $d\_{23}$ etc, say $m\_1=n\_1d\_{23}$ and we get $n\_1b\_1+n\_2b\_2+n\_3b\_3=2b\_1b\_2b\_3$ for mutually coprime $b\_1$, $b\_2$, $b\_3$.
3. Assume the contrary. Denote $n\_1=k\_1b\_2+r\_1$ where $0\leqslant r\_1<b\_2$, $n\_2=k\_2b\_1+r\_2$ where $0\leqslant r\_2<b\_1$. Note that all numbers divisible by $b\_1b\_2$ and not exceeding $(k\_1+k\_2)b\_1b\_2$ are represented in the necessary form $t\_1b\_1+t\_2b\_2+t\_3b\_3$, $0\leqslant t\_i\leqslant n\_i$. Thus, assuming the contrary, we get $k\_1+k\_2\leqslant b\_3-1$, and $$n\_1b\_1+n\_2b\_2\leqslant (k\_1+k\_2)b\_1b\_2+(b\_2-1)b\_1+(b\_1-1)b\_2\\ \leqslant (b\_3-1)b\_1b\_2+(b\_2-1)b\_1+(b\_1-1)b\_2.$$
Summing up three such bounds we get $$4b\_1b\_2b\_3=2(n\_1b\_1+n\_2b\_2+n\_3b\_3)\leqslant 3b\_1b\_2b\_3+b\_1b\_2+b\_1b\_3+b\_2b\_3-2b\_1-2b\_2-2b\_3,$$
and $$(b\_1-1)(b\_2-1)(b\_3-1)<0,$$
a contradiction.
|
3
|
https://mathoverflow.net/users/4312
|
409179
| 167,516 |
https://mathoverflow.net/questions/409135
|
5
|
Let $C$ be a reduced, connected, projective and purely one-dimensional scheme of finite type over a field $k$.
Suppose that $C$ is rational, i.e. that the normalisation of $C$ is a disjoint union of copies of $\mathbb{P}^1\_k$.
Let $T\_C = \mathcal{H}om(\Omega^1\_C,\mathcal{O}\_C)$ be the tangent sheaf of $C$.
**Question:** Is it true that $H^1(C,T\_C)$ vanishes?
You may additionally assume that $C$ has only planar singularities, but I'm not sure this is needed.
The question is true if $C$ is smooth.
For context, a positive answer would imply that $C$ has no locally trivial deformations.
|
https://mathoverflow.net/users/110362
|
First cohomology of tangent sheaf of rational curve
|
Let $C$ be the union of 5 lines in general position in $\mathbb{P}^2$ (hence with 10 pairwise intersection points $P\_{ij}$, $1 \le i < j \le 5$) and let $F$ be the equation of $C$. We have the standard exact sequence
$$
0 \to \mathcal{O}\_C(-5) \stackrel{dF}\to \Omega\_{\mathbb{P}^2}\vert\_C \to \Omega\_C \to 0.
$$
Taking its dual we obtain an exact sequence
$$
0 \to T\_C \to T\_{\mathbb{P}^2}\vert\_C \stackrel{dF}\to \mathcal{O}\_C(5) \to \bigoplus\_{1 \le i < j \le 5} \mathcal{O}\_{P\_{ij}} \to 0.
$$
Now an easy computation of Euler characteristics gives
$$
\chi(T\_{\mathbb{P}^2}\vert\_C) = 5,
\quad
\chi(\mathcal{O}\_C(5)) = 20,
\quad
\chi(\mathcal{O}\_{P\_{ij}}) = 1,
$$
hence $\chi(T\_C) = 5 - 20 + 10 = -5$, and since $C$ is 1-dimensional, it follows that $H^1(C,T\_C)$ is non-zero.
|
10
|
https://mathoverflow.net/users/4428
|
409185
| 167,518 |
https://mathoverflow.net/questions/409013
|
7
|
In classical homotopy theory, there are a number of spaces which are important because they represent an interesting functor on $\operatorname{Ho(Top)}$; for example, $K(G,n)$ represents singular cohomology and $BG$ represents principal $G$-bundles. However, modern homotopy theorists know that the homotopy category is merely the truncation of a much richer structure, namely the $\infty$-category $\operatorname{Spaces}$, and both of these examples then yield representable functors from this $\infty$-category to itself. The conventional interpretation would appear to suggest that these spaces should be thought of as an "enriched" or "coherent" version of cohomology and vector bundles. Is this the case? What do these spaces look like, and are there any interesting results known about them?
|
https://mathoverflow.net/users/158123
|
The contravariant mapping space represented by a homotopical classifying space (e.g. BG)
|
Let $G$ be a topological group and $X$ be a paracompact Hausdorff topological space. For simplicity let us assume that $G$ has the homotopy type of a CW complex, although a lot of this answer does not need it. Then we can define the simplicial category of principal $G$-bundles over $X$ in the following way:
* Its objects are principal $G$-bundles $p:P\to X$
* Given two objects $p:P\to X$ and $p':P'\to X$, the simplicial set $\operatorname{Map}(p,p')$ has as $n$ simplices the equivariant continuous maps $f:P\times |\Delta^n|\to P'\times|\Delta^n|$ over $X\times|\Delta^n|$, where $|\Delta^n|=\{(t\_0,\dots,t\_n)\mid \sum\_i t\_i=1\}$ is the topological $n$-simplex.
Note that all the mapping simplicial sets are Kan complexes. Therefore we can take its simplicial nerve and we get an $\infty$-category. In fact it is easily seen to be an $\infty$-groupoid, a.k.a. a space, $\operatorname{Bun}\_G(X)$, which I will call the **space of principal $G$-bundles over $X$**. Note that it is canonically pointed by the trivial bundle.
With some effort this can be made controvariantly functorial in $X$, with the functoriality given by the pullback of bundles (as usual, this is accomplished by constructing a suitable fibration over the category of topological spaces -- the details are left as an exercise).
I am going to claim that $\operatorname{Bun}\_G(X)$ is equivalent to $\operatorname{Map}(X,BG)$ where $BG$ is the classifying space of $G$.
The quickest proof I know of this fact is by proving that $\operatorname{Bun}\_G(-)$ is a sheaf on paracompact Hausdorff spaces for the open covering topology (this requires some work -- in particular you have to show that the restriction map $\operatorname{Map}(p,p')\to \operatorname{Map}(p|\_U,p'|\_U)$ to an open subset is a Kan fibration).
Then it is easy to see that its sheaf $\pi\_0$ is trivial and that $\Omega\operatorname{Bun}\_G(-)$ is the constant sheaf at $G$ (this follows from the fact that if $L$ is a space homotopy equivalent to a CW-complex, the constant sheaf at the weak homotopy type of $L$ is given by the simplicial mapping space $U\mapsto \operatorname{Map}(U,L)$, by combining Corollary 7.1.4.4 and Proposition 7.1.5.1 in Higher Topos Theory). Therefore the recognition theorem for loopspaces in an $\infty$-topos shows that $\operatorname{Bun}\_G(-)$ is the constant sheaf at $BG$. Finally, arguing as in [*Higher Algebra* Remark A.1.4] we see that the constant sheaf at $BG$ is exactly $\operatorname{Map}(-,BG)$.
---
A similar argument shows that the space of fiber bundles with fiber $F$ is equivalent to the space $\operatorname{Map}(X,B\operatorname{Homeo}(F))$, where $\operatorname{Homeo}(F)$ is the simplicial group of homeomorphisms of $F$.
|
5
|
https://mathoverflow.net/users/43054
|
409187
| 167,519 |
https://mathoverflow.net/questions/409202
|
2
|
Let $(M^2,g)$ be a 2-dimensional Riemannian manifold. For any point $p\in M^2$ can we always find coordinates $(u,v)$ in a neighborhood $U$ of $p$ such that the Gaussian curvature is only a function of $u\pm v$, i.e, $K=K(u\pm v)$ ?
|
https://mathoverflow.net/users/171439
|
Can we always find coordinates on a surface such that $K=K(u-v)$?
|
Curvature is a smooth function on the surface, and locally, any smooth negative function can serve as a curvature of some surface (M. S. Berger, Riemannian structure of prescribed Gaussian curvature for compact 2-manifolds, J. Differential Geom. 5 (1971), 325-332.)
So the question is whether for an arbitrary smooth function there exists
a coordinate system such that the function depends on only one coordinate. The answer is certainly negative, since the function can have critical points.
|
2
|
https://mathoverflow.net/users/25510
|
409203
| 167,520 |
https://mathoverflow.net/questions/409200
|
8
|
This question asks whether there exists an analogue of the Jordan decomposition for an arbitrary ring $R$. This analogue is not necessarily the Jordan-Chevalley decomposition, which is unnecessarily strong. This follows from [this question](https://mathoverflow.net/questions/408349/matrix-decompositions-as-monoid-isomorphisms-ever-considered-before), but you don't need to read it.
Given a ring $R$, let $J(R)$ be the monoid whose elements are all square matrices over $R$, and where the monoid operation is $\oplus$ denoting direct sum of matrices. Let $A { \sim\_\text{S}} B$ mean that there exists an invertible matrix $P$ such that $PAP^{-1} = B$. Is the monoid $J(R)/{ \sim\_\text{S}}$ *always a free abelian monoid*?
For perfect fields, the answer is yes by the Jordan-Chevalley decomposition. What about for every ring? Is there a counterexample?
|
https://mathoverflow.net/users/75761
|
For every ring R, is there a block-diagonal canonical form for a square matrix over R?
|
Your question is equivalent to whether the category $\mathcal{E}$ of pairs $(V,f)$ consisting of a finitely generated free (right) $R$-module and an endomorphism $f$ of $V$ is a *Krull-Schmidt category*, i.e., an additive category where every object decomposes as a direct sum of finitely many indecomposable objects and the decomposition is unique up to isomorphism and reordering. (The category $\cal E$ is also equivalent to the category of right $R[t]$-modules which are f.g. free $R$-modules, and it is rarely Krull-Schmidt, but I won't use this point of view.)
Here is one possible counterexample, phrased using the category $\mathcal{E}$ of pairs $(V,f)$ above: Take $R$ to be a Dedekind domain admitting a non-free rank-$1$ projective module $L$ such that $L\oplus L\cong R\oplus R$ (equivalently, $L$ represents an element of order $2$ in the Picard group of $R$). Let $f\_L : L^2\to L^2$ be defined by $f\_L(x,y)=(0,x)$, and define $f\_R:R^2\to R^2$ similarly. The isomorphism $L\oplus L\cong R\oplus R$ gives rise to an isomorphism
$$(L^2,f\_L)\oplus (L^2,f\_L) \cong (R^2,f\_R)\oplus (R^2,f\_R)$$
in $\cal E$.
One readily checks that ${\rm End}\_{\cal E}(L^2,f\_L)\cong R[\epsilon|\epsilon^2=0]$ and ${\rm End}\_{\cal E}(R^2,f\_R)\cong R[\epsilon|\epsilon^2=0]$, so the endomorphism rings of $(L^2,f\_L)$ and $(R^2,f\_R)$ contain no nontrivial idempotents. This means that these objects are indecomposable in $\cal E$. On the other hand, $(L^2,f\_L)\ncong (R^2,f\_R)$ because $\ker f\_L\cong L\ncong R\cong \ker f\_R$. Consequently, the monoid $(\cal E/\cong, \oplus)$ (which is isomorphic to $(J(R)/\sim, \oplus)$ in your question) is not a free abelian monoid (because $2x=2y$ implies $x=y$ in a free abelian monoid).
One the other hand, a sufficient (but not necessary) condition for the category $\cal E$ to be Krull-Schmidt is that the endomorphism ring of every object $(V,f)$, i.e., the centralizer of $f$ in ${\rm End}\_R(V)\cong {\rm M}\_n(R)$, is a *semiperfect ring*.
For example, if $R$ is commutative and aritinian as in Benjamin Steinberg's comment, then ${\rm End}\_{\cal E}(V,f)$ will be an $R$-subalgebra of ${\rm End}\_R(V)$, hence artinian, and in particular semiperfect.
The semiperfectness ${\rm End}\_{\cal E}(V,f)$ for all f.g. free $V$ is actually true even if $R$ is non-commutative one-sided artinian, and even if $R$ is just *semiprimary*. This appears implicitly in a [paper](https://arxiv.org/pdf/1212.2124.pdf) of mine (page 20 & Thm. 8.3(iii) & Remark 2.9).
When $R$ is commutative noetherian and local, one can use a [theorem of Azumaya](https://projecteuclid.org/journals/nagoya-mathematical-journal/volume-2/issue-none/On-maximally-central-algebras/nmj/1118764746.full) (Theorem 22) and a little work to show that this property is equivalent to $R$ being henselian.
|
12
|
https://mathoverflow.net/users/86006
|
409211
| 167,522 |
https://mathoverflow.net/questions/409193
|
1
|
I am working with the heat kernel on the hyperbolic space explicitly (as you may guess by my previous questions) and I got the desired results when the curvature is $-\kappa=-1$. Now I am trying to do the same for a fixed but arbitrary curvature $-\kappa<0$, so I need to generalize the explicit formulas for the heat kernel that I got in $\mathbb{H}^n(1)$ (see [''The Heat Kernel on Hyperbolic Space''](https://www.math.uni-bielefeld.de/%7Egrigor/nog.pdf) or [''Heat kernel bounds on hyperbolic space and Kleinian space''](https://londmathsoc.onlinelibrary.wiley.com/doi/epdf/10.1112/plms/s3-57.1.182)) to a more general $\mathbb{H}^n(\kappa)$.
I achieve to get a general formula when $n$ is odd, that is $$
p\_{n}(\rho, t)=\frac{(-1)^{m}}{2^{m} \pi^{m}} \frac{1}{(4 \pi t)^{\frac{1}{2}}}\left(\frac{\kappa}{\sinh (\kappa\rho)} \frac{\partial}{\partial \rho}\right)^{m} e^{-\kappa^2m^{2} t-\frac{\rho^{2}}{4 t}}
$$ with $n=2m +1$, but I did it by just adding $\kappa$ in the formula and then cheking that it satisfies the equation, that is now $$\frac{\partial^2}{\partial \rho^2}p\_{n}(\rho, t)+(n-1)\kappa
\coth(\kappa \rho)\frac{\partial}{\partial \rho}p\_{n}(\rho, t)-\frac{\partial}{\partial t}p\_{n}(\rho, t).$$
When I try to do the same to even $n$, it fails because I can not compute the derivatives so easily.
Could someone help me find a general formula when $n=2m+2$? I have some intuition but I can not check if they are, in fact, the fundamental solution that I want.
|
https://mathoverflow.net/users/411616
|
Heat kernel on hyperbolic space of variable curvature
|
The sectional curvature scales like the inverse of the metric.
So fixing a coordinate system on $\mathbb{H}^n(1)$, with metric $g$, the scaled metric $\kappa^{-1} g$ has sectional curvature $-\kappa$.
If $u(t,x)$ solves the heat equation you have
$$ u\_t = \Delta\_g u \iff \kappa u\_t = \kappa \Delta\_g u \iff \kappa u\_t = \Delta\_{\kappa^{-1} g} u $$
and so $u(\kappa t,x)$ solves the heat equation for the $-\kappa$ curvature.
The function $\rho$ being the geodesic distance scales like $\kappa^{1/2}$: that is $\rho\_g = \kappa^{1/2} \rho\_{\kappa^{-1} g}$.
And the volume form scales like $\mathrm{dvol}\_g = \kappa^{n/2} \mathrm{dvol}\_{\kappa^{-1} g}$.
This tells you that if you write $\tilde{p}$ for the heat kernel when the section curvature equals $-\kappa$, and $p$ for the heat kernel when the sectional curvature equals $-1$, you should have
$$ \tilde{p}\_n(\rho,t) = \kappa^{n/2} p\_n(\kappa^{1/2}\rho, \kappa t)$$
So first: your formula for the odd dimensional case is **wrong**. Where you have $\kappa$ you should have $\kappa^{1/2}$ instead (unless you are actually looking at the case where sectional curvature equals $-\kappa^2$). The correct formula should be
$$ \tilde{p}\_{n}(\rho, t)=\frac{(-1)^{m}}{2^{m} \pi^{m}} \frac{1}{(4 \pi t)^{\frac{1}{2}}}\left(\frac{\kappa^{1/2}}{\sinh (\kappa^{1/2}\rho)} \frac{\partial}{\partial \rho}\right)^{m} e^{-\kappa m^{2} t-\frac{\rho^{2}}{4 t}} $$
In the even case (n = 2m+2) you should have
$$ \tilde{p}\_n(\rho, t) = \frac{(-1)^m}{2^{m+5/2} \pi^{m+3/2}} \kappa^{-1/2} t^{-3/2} e^{-\frac{(2m+1)^2}{4} \kappa t} \left( \frac{\kappa^{1/2}}{\sinh \kappa^{1/2}\rho} \partial\_\rho \right)^m \int\_{\kappa^{1/2}\rho}^\infty \frac{s e^{- s^2/(4\kappa t)}}{(\cosh s - \cosh \kappa^{1/2}\rho)^{1/2}} ds
$$
|
2
|
https://mathoverflow.net/users/3948
|
409214
| 167,524 |
https://mathoverflow.net/questions/368883
|
9
|
I'm searching for a copy of an old paper made by Edmund Landau:
>
> *Zur relativen Wertbemessung der Turnierresultate*, Deutsches Wochenschach, 11. Jahrgang (1895), 366–369.
>
>
>
However, I can't find it anywhere. I looked at some old books written by Landau and did not succeed in my quest.
The Princeton University has the volume 50-1895 of the magazine *"Deutsche Schachzeitung"* where this paper should appear: [HathiTrust record](https://catalog.hathitrust.org/Record/009008387).
Landau published this early work in a Chess magazine. However, it seems to be that paper has several implications in the theory of Ranking and positive matrices and worked as a motivation for the next works of Landau.
|
https://mathoverflow.net/users/109828
|
Math history research: a copy of "Zur relativen Wertbemessung der Turnierresultate" , eigenvector centrality by Edmund Landau
|
The paper which actually was Landau's first scientific paper written at the tender age of 18, was published in his Collected Works, vol. 1. In it, he proposes to rank chess players having played a round robin tournament according to an eigenvector of the results matrix . A much more comprehensive analysis of this method with the help of the (then new) Perron-Frobenius theorem is given in his 1915 paper "Über Preisverteilung bei Spielturnieren", to be found in his Collected Works, vol. 6.
The latter paper can be downloaded here: <https://iris.univ-lille.fr/handle/1908/2031>
|
6
|
https://mathoverflow.net/users/96921
|
409217
| 167,525 |
https://mathoverflow.net/questions/409212
|
6
|
This question may be related to [this one](https://math.stackexchange.com/q/4310801/498394).
Now I try to make some statistical estimator using Laplace transform, but I face the following serious problem.
Let $f$ be some one-sided probability distribution defined on $[0,\infty)$, and $\hat{f}$ be its Laplace transform.
Now assume that we only have information of $\hat{f}(s)$ near $s=0$ (for example, $\hat{f}(s) = \hat{g}(s) / (2-\hat{g}(s))$ with known probability distribution $g(x)$), and using this, we want to find the convergence speed of the tail:
\begin{align\*}
\int\_x^{\infty} f(x)\, dx = O(?).
\end{align\*}
According to Tauberian remainder theory in [J. Korevaar's book](https://books.google.co.kr/books?hl=en&lr=&id=aWfsCAAAQBAJ&oi=fnd&pg=PA1&dq=Tauberian+theory&ots=Z5Nl1wlXq5&sig=JyZpb6E6H8gCrPRIIIekyRYwFu0&redir_esc=y#v=onepage&q=Tauberian%20theory&f=false) (Examples 2.3., page 348), he said
* If $|\hat{f}(s) - \hat{f}(0)| \le Cs^{\alpha}$ with some $\alpha > 0$, then the tail rate is $O(1/\log x)$;
* If $|\hat{f}(s) - \hat{f}(0)| \le Ce^{-\alpha/s}$ with some $\alpha > 0$, then the tail rate is $O(1/\sqrt{x})$.
In fact, this is a disaster for statisticians because any higher-order Taylor approximation of $\hat{f}(s)$ near $s=0$ cannot guess whether the tail is light or super super heavy ($1/\log x$).
So the question is the following.
>
> **Question.** What other condition on $\hat{f}$ near $0$ is required to guarantee the tail has at least a power-tail $O(x^{-\beta})$? If the power tail cannot be guaranteed by any of information about $\hat{f}$ near $s=0$, then what condition is required for $g$?
>
>
>
Any help would be appreciated. (suggesting books, papers, or anything!)
Thanks for the reading,
|
https://mathoverflow.net/users/159685
|
Convergence speed of the tail of distribution using Tauberian remainder theorem
|
Let $f$ be a pdf on $[0,\infty)$. Let $\hat f$ be the Laplace transform of $f$, so that
\begin{equation}
\hat f(s)=\int\_0^\infty f(x)e^{-sx}\,dx
\end{equation}
for real $s\ge0$.
Suppose that
\begin{equation}
|\hat f(s)-\hat f(0)|\le Cs^a
\end{equation}
for some real $a,C>0$ and all real $s\ge0$.
Then for all real $s>0$
\begin{equation}
Cs^a\ge|\hat f(s)-\hat f(0)|=\int\_0^\infty f(u)(1-e^{-su})\,du
\ge(1-e^{-1})\int\_{1/s}^\infty f(u)\,du,
\end{equation}
whence for all real $x>0$
\begin{equation}
\int\_x^\infty f(u)\,du\le C\_1/x^a,
\end{equation}
where $C\_1:=C/(1-e^{-1})\in(0,\infty)$, as desired.
(Your difficulty was due to the fact that you tried to use general Tauberian theorems, valid without the nonnegativity condition. On the other hand, all pdf's are of course nonnegative. Once this nonnegativity condition is taken into account, everything becomes much simpler and better.)
|
3
|
https://mathoverflow.net/users/36721
|
409219
| 167,527 |
https://mathoverflow.net/questions/409233
|
11
|
To quote [Kerodon](https://kerodon.net/tag/0004):
>
> In fact, it is possible to develop the theory of algebraic topology in entirely combinatorial terms, using simplicial sets as surrogates for topological spaces.
>
>
>
A similar quote can be found in the mathscinet review for Kan's *On c. s. s. complexes*:
>
> In recent years it has become evident that for most purposes in homotopy theory it is more convenient to use semi-simplicial complexes instead of topological spaces.
>
>
>
For instance, I [know](https://kerodon.net/tag/00V2) that to special simplicial sets called Kan complexes one can assign higher homotopy groups and prove and analogue of Whitehead's theorem. This certainly demonstrates that one can do *some* homotopy theory with simplicial sets / Kan complexes.
*If I open an introductory book on algebraic topology or homotopy theory (such as Hatcher's), do all the main theorems admit analogues in the world of simplicial sets or Kan complexes (replacing topological spaces)?*
I'd be totally happy if you could give me, say, four theorems in algebraic topology / homotopy theory that can be phrased for simplicial sets, together with the original reference. I'd also be interested in whether these theorems are more algebraic topology or more homotopy theory (I don't really know the difference).
|
https://mathoverflow.net/users/469290
|
Algebraic topology and homotopy theory with simplicial sets instead of topological spaces
|
It depends on what you mean by "all results". Of course results regarding manifolds or vector bundles do not admit statements completely internal to the world of simplicial sets (although most of them are just an application of $\operatorname{Sing}$ away from the world of simplicial sets).
But if one concentrates oneself to the "purely homotopical" statements (like, say, the Freudenthal suspension theorem, the Whitehead theorem, the Brown representability theorem and the Blakers-Massey theorem) they can all be stated in terms of simplicial sets (or, better, Kan complexes).
Indeed there is a [textbook](https://link.springer.com/book/10.1007/978-3-0346-0189-4) by Goerss and Jardine that does most elementary homotopy theory in terms of simplicial sets.
|
16
|
https://mathoverflow.net/users/43054
|
409236
| 167,533 |
https://mathoverflow.net/questions/409140
|
2
|
$\DeclareMathOperator\Hom{Hom}$Assume $F$ and $M$ are respectively right and left modules over a ring $R$ and let $I^\bullet$ be a left-bounded exact complex of $R$-$R$-bimodules. We know there is a natural map of complexes $\varphi: F\otimes\_R \Hom(I^\bullet, M)\longrightarrow \Hom\_R(\Hom\_R(F, I^\bullet), M)$ which is defined in degree $i$ as $(\varphi\_i(x\otimes g))(h)=g(h(x))$ for every $x\in F, g\in \Hom\_R(I^i, M)$, and $h\in \Hom\_R(F, I^i)$. In certain cases, I know the exactness of the complex $\Hom\_R(\Hom\_R(F, I^\bullet), M)$. I want to know if this implies the exactness of $F\otimes\_R \Hom(I^\bullet, M)$. I can also add the hypothesis that $F$ is $R$-flat and $I^\bullet$ is a complex of injective $R$-$R$-bimodules, except for the first nonzero entry ( because that would make $I^\bullet$ split).
I tried to show that this map $\varphi$ is a quasi-isomorphism under the extra assumptions mentioned; however I'm suspicious about this. Might mapping cone arguments be of any help? I appreciate any comment.
|
https://mathoverflow.net/users/466540
|
Is a certain map a quasi-isomorphism?
|
This isn’t true when $R=\mathbb{Z}$, $F=\mathbb{Q}$, $M=\mathbb{Z}$ and $I^\bullet$ is the complex
$$\cdots\to0\to\mathbb{Z}\to\mathbb{Q}\to\mathbb{Q}/\mathbb{Z}\to0\to\cdots$$
with $\mathbb{Z}$ in degree zero.
Then $\varphi$ is a map from $\mathbb{Q}$, as a complex concentrated in degree zero, to the zero complex.
|
2
|
https://mathoverflow.net/users/22989
|
409241
| 167,536 |
https://mathoverflow.net/questions/409237
|
5
|
Suppose I have a genus 1 curve $C$ over a field $k$. If $C$ has a point, then we can embed it into the projective plane by a Weierstrass equation. Now let us suppose that $C$ does not have a point (so that it is a non trivial torsor for it's Picard group).
Can I still embed $C$ into the projective plane? I guess not but there is apparently a theorem of Lang-Tate that we can always find an effective divisor of some degree over $k$ (what is a reference?) so we can embed it into some high dimensional projective space.
Can we always embed C into a Severi Brauer variety of dimension $2$?
|
https://mathoverflow.net/users/58001
|
Embedding torsors of elliptic curves into projective space
|
Suppose that $C \subset X$ is a smooth projective curve of genus $1$ embedded in a Brauer-Severi surface over a field $k$. We have $C^2 = 9$ since this holds after passing to the algebraic closure, where it is embedded as a curve of degree $3$ in the projective plane. So we deduce that $C$ admits a divisor of degree $9$.
It thus just suffices to write down a curve of genus $1$ without a divisor of degree $9$. The example of Piotr Achinger works here. The given curve has a divisor of degree $4$ and cannot have a divisor of degree $9$, since otherwise it would have a divisor of degree $1$ as $\gcd(4,9)=1$.
|
11
|
https://mathoverflow.net/users/5101
|
409248
| 167,539 |
https://mathoverflow.net/questions/409252
|
3
|
In the plane, two figures are called congruent exactly if one can be transformed into the other by translation, rotation, and reflection. What if reflection is excluded, that is, preservation of orientation is required? Is there a term for the resulting equivalence relation?
|
https://mathoverflow.net/users/25527
|
What is the term for two figures being congruent and of same orientation?
|
While perhaps not widespread, the term “direct congruence” is used for this equivalence relation.
|
2
|
https://mathoverflow.net/users/25527
|
409254
| 167,540 |
https://mathoverflow.net/questions/409208
|
4
|
A tower $\Delta\_g:=\{(x,n)\in X \times \{0,1,2,\cdots\}: n < R(x)\}$
where $R:X \to \{1,2,3,\cdots\}$ is a $L^1$ function on a probability space $(X,\mu)$, $g: X \to X$ is mixing and $\gcd \{R\}=1$,
A map $f: \Delta \to \Delta$ is defined as: $f(x,n)=(x,n+1)$ if $n < R(x)-1$ and $f(x,n)=(g(x),0)$ if $n=R(x)-1$.
An invariant probability on $\mu\_{\Delta}$ is $\mu\_{\Delta}=(\int R d\mu)^{-1}\sum\_{i\ge 0}f^i\_{\*}(\mu|\_{R>i})$.
Is $(\Delta, f, \mu\_{\Delta})$ mixing? or have a counter-example?
|
https://mathoverflow.net/users/124254
|
a mixing property on a tower
|
There are counterexamples. The easiest way to “cheat” is to let the height function be cohomologous to a constant.
As an example, let $T$ be an ergodic transformation of a space $X$. Let $A$ be a subset of $X$ such that $A$ and $T^{-1}A$ are disjoint. Now define $g(x)=1$ if $x\in A$, $g(x)=3$ if $x\in T^{-1}A$ and $2$ otherwise.
Then $T\_g$ has an eigenfunction: $h(x,0)=1$ if $x\in A$, $h(x,0)=h(x,2)=-1$ and $h(x,1)=1$ if $x\in T^{-1}A$, and $h(x,0)=1$ and $h(x,1)=-1$ for other $x$’s. Then $h\circ T\_g=-h$.
The way that this works is that $g$ is cohomologous to the constant function 2: the difference $g-2$ can be expressed as $j\circ T-j$ where $j(x)=1$ if $x\in A$ and 0 otherwise.
|
3
|
https://mathoverflow.net/users/11054
|
409259
| 167,542 |
https://mathoverflow.net/questions/409269
|
3
|
I am looking into a question of how many points can be put on a plane that pairwise distances between them are as close to a constant as possible. As the first step, it was rather easy to figure out that only 3 points can be put on a plane so that all pairwise distances between them are equal. As the next step, I think about 4 points.
Formally, if there are four distinct points $A\_1(x\_1, y\_1)$, $A\_2(x\_2, y\_2)$, $A\_3(x\_3, y\_3)$, $A\_4(x\_4, y\_4)$, and we define pairwise distances $d\_{ij} = d(A\_i, A\_j) = \sqrt{(x\_i - x\_j)^2 + (y\_i - y\_j)^2}$ and $d\_\text{min} = \min\_{i \neq j} d\_{ij}$, $d\_\text{max} = \max\_{i \neq j} d\_{ij}$, what is the minimum ratio between the largest and smallest distances:
$$
\min\_{\text{distinct } A\_1, \dotsc, A\_4} \frac{d\_\text{max}}{d\_\text{min}} = ?
$$
At the moment, I cannot even figure out if it is in fact a minimum or infimum (i.e., this ratio can get as close to 1 as you want).
Geometry is not really my area, so perhaps it is even some known result...
|
https://mathoverflow.net/users/101533
|
4 points on a plane with (almost) equal pairwise distances
|
It is $\sqrt{2}$. For 4 vertices of a square, you get this value. For proving that it is always not less than $\sqrt{2}$, note that one of angles $\angle A\_iA\_jA\_k$ is not less than $\pi/2$ (if $A\_1A\_2A\_3A\_4$ is a convex quadrilateral, the sum of angles equals $2\pi$, thus one of them is at least $\pi/2$; if $A\_4$ lies in a a triangle $A\_1A\_2A\_3$, the sum of angles $\angle A\_iA\_4A\_j$, $1\leqslant i<j\leqslant 3$, equals $2\pi$, thus one of them is at least $2\pi/3>\pi/2$). Now $A\_iA\_k^2\geqslant A\_iA\_j^2+A\_jA\_k^2$, hence $\max (d\_{ik}/d\_{ij},d\_{ik}/d\_{jk})\geqslant \sqrt{2}$.
|
6
|
https://mathoverflow.net/users/4312
|
409271
| 167,546 |
https://mathoverflow.net/questions/409270
|
-3
|
Knowing the value of $S=\sum\_{k=1}^n s\_k$ with $s\_k\geq 0$,
is it possible to obtain an upper bound on $\sum\_{k=1}^n\sqrt{s\_k}$ better than $n \times \sqrt{\max\_{1\leq k\leq n} s\_k}$ ?
|
https://mathoverflow.net/users/148279
|
Bounding sum of square roots in function of the sum value
|
Yes: by the [generalized mean inequality](https://en.wikipedia.org/wiki/Generalized_mean#Generalized_mean_inequality) (or, more specifically, by the [AM--QM inequality](https://en.wikipedia.org/wiki/HM-GM-AM-QM_inequalities)), $\sqrt{nS}$ is an upper bound on $\sum\_{k=1}^n\sqrt{s\_k}$, which is better than
$n\sqrt{\max\_{1\le k\le n}s\_k}$.
|
2
|
https://mathoverflow.net/users/36721
|
409282
| 167,549 |
https://mathoverflow.net/questions/409292
|
3
|
Let $R$ be a commutative (noetherian, if needed) ring, let $f\_1，\ldots，f\_r\in R[x\_1，\ldots，x\_n]$ and $A=R[x\_1，…，x\_n]/(f\_1，\ldots，f\_r)$, when is $A$ flat over $R$?
I found a nice answer for the case $n=r=1$ [here](https://math.stackexchange.com/a/3667137/95662), but I don't know how to formulate a characterisation in general, or if it is solved. Any idea or reference is welcome.
|
https://mathoverflow.net/users/42571
|
Flatness of finitely presented algebras
|
There is a nice criterion that is usually applied in this situation.
A is flat over $R$ if the Krull dimension of the fiber ring $A\_{\mathfrak{P}}/\mathfrak{P}$ for all prime ideals $\mathfrak{P} \subset R$ is $n-r$ (or the fiber is empty). In scheme theoretic language, the fibers either have the expected dimension (are complete intersections) or are empty. This is what is called a relative global complete intersection here <https://stacks.math.columbia.edu/tag/00SK>. Flatness follows from <https://stacks.math.columbia.edu/tag/00SW>.
This is a generalization of the criterion you cited when $r=1$. In that case the fibers of $Spec(R[x\_1, \ldots, x\_n]) \to Spec(R)$ are affine spaces over a field (so integral) and the condition ensures that the restriction of the polynomial $f$ to every fiber over some closed and open subset $U \subset Spec(R)$ is nonzero, so it must cut a closed subset of the fiber of dimension $n-1$ (or empty if it restricts to a constant). The idempotent comes from the fact that if $Spec(R)$ is not connected, then we can allow the preimage over the complement of this open and closed to be just isomorphic to the whole affine space $\mathbb{A}^n\_{Spec(R) \setminus U}$, ($f$ is allowed to be identically $0$ on $Spec(R) \setminus U$).
|
3
|
https://mathoverflow.net/users/339730
|
409293
| 167,551 |
https://mathoverflow.net/questions/409287
|
11
|
Let $f(z)=\sum a\_nz^n$ be a Taylor series that converges for $|z|<1$ and satisfies
$$
|f(z)|\le \frac{1}{(1-|z|)^{k}}
$$
for some fixed $k>0$.
**Question:** What can I deduce about the growth of the Taylor coefficients $a\_n$?
**Partial result:** By judiciously selecting the location of the contour in the formula $a\_n=\oint z^{-n}f(z)\tfrac{dz}{2\pi i z}$, namely, by performing the integration over the contour $|z|=\tfrac{n}{n+k}$ [which is the minimum of $|z|^{-n}(1-|z|)^{-k}$], I can get the "trivial bound" $|a\_n|< c\cdot n^k$.
But I suspect that this is not sharp.
In particular, the growth of the Taylor coefficients of $(1-z)^{-k}$ is only $n^{k-1}$. Not $n^k$.
**More precise formulation of the question:**
What is the optimal $k'>0$ such that
$$
|f(z)|\le \frac{1}{(1-|z|)^{k}}\quad\Rightarrow\quad |a\_n|< c\cdot n^{k'}
$$
for all $f(z)=\sum a\_nz^n$.
From the above arguments, I know that $k-1\le k'\le k$.
|
https://mathoverflow.net/users/5690
|
Estimating the growth of the Taylor coefficients given the growth of the function at the boundary
|
The optimal exponent is $k$. Such examples are given by sparse power series. This is actually trivial in the case $k=0$ (which was not included in the OP). Then we can simply take $f(z)=\sum j^{-2} z^{N(j)}$, say. This is obviously bounded, and the coefficients $a\_n$ will not satisfy $|a\_n|\lesssim n^{-\epsilon}$ for any $\epsilon>0$ if $N(j)$ increases fast enough.
For positive $k$, we can similarly consider something like
$$
f(z)=\sum n^{-2} \left(n^n\right)^k z^{n^n} .
$$
Using calculus to find the maximum, we see that
$$
x^k(1-\delta)^x \le C\delta^{-k} .
$$
Thus $f$ satisfies the desired bound, but the coefficients do not satisfy $|a\_n|\lesssim n^{k-\epsilon}$ for any $\epsilon>0$.
|
12
|
https://mathoverflow.net/users/48839
|
409301
| 167,554 |
https://mathoverflow.net/questions/409295
|
3
|
I'd like to compute the derivative of an expected value w.r.t one of the parameters that define the mean of a Gaussian:
$ Z=\int \mathcal{N}(x;\mu,\Sigma)f(x) \, dx $, then $ \frac{dZ}{dK}=\text{??}$ when $\mu=g(K)\in\mathbb{R}^{n\times 1}$.
I have three problems with this: 1. The dimensions are not adding up for me. 2. How can I compute this also numerically if $g(K)$ gets too complicated? 3. How do I do this if $\Sigma=h(K)\in\mathbb{R}^{n\times n}$, too?
1. This is what I have so far, if I ignore the dependency of $\mu$ in $K$ for a minute:
$$ \frac{dZ}{d\mu} = \int \frac{d\mathcal{N}(x;\mu,\Sigma)}{d\mu}f(x)\,dx $$
and since $\mathcal{N}(x;\mu,\Sigma)=(2\pi)^{-n/2} \det(\Sigma)^{-1} \exp(-0.5(x-\mu)^T\Sigma^{-1}(x-\mu))$, we have that $\frac{d\mathcal{N}(x;\mu,\Sigma)}{d\mu}=\mathcal{N}(x;\mu,\Sigma)\frac{d\ln\mathcal{N}(x;\mu,\Sigma)}{d\mu}$
so now we can get:
$$ \frac{dZ}{d\mu} = \int \mathcal{N}(x;\mu,\Sigma)\frac{d\ln\mathcal{N}(x;\mu,\Sigma)}{d\mu} f(x) \, dx = \mathbb{E}\left[\frac{d\ln\mathcal{N}(x;\mu,\Sigma)}{d\mu}\right]$$
This means I can compute $$ \frac{dZ}{d\mu} = \mathbb{E}\left[\frac{d\ln\mathcal{N}(x;\mu,\Sigma)}{d\mu}\right] = \mathbb{E}\left[\Sigma^{-1}(x-\mu)\right]$$
Now, comes the problem - this expected value is going to be a vector $\in \mathbb{R}^{n\times 1}$. Now if I want to derive the gradient w.r.t $K$ then:
$$ \frac{dZ}{dK} = \frac{dZ}{d\mu}\frac{d\mu}{dK}$$ and lets say that $\mu$ is the result of calculations of $K\in\mathbb{R}^{p\times q}$, what are the dimensions of $\frac{d\mu}{dK}$? Even if I got the ordering or transpose wrong in the matrix derivations, I still don't see how that dimensions should work because I expect that $ \frac{dZ}{dK} \in\mathbb{R}^{p\times q}$.
An example for $\mu$ can be:
$$ \mu = \left[ AKv ; (AK)^2v; (AK)^3v ; \ldots \right] \in \mathbb{R}^{n\times 1} $$ where the constant $A\in\mathbb{R}^{q\times p}, v\in\mathbb{R}^{q\times 1}$.
2. Is there any software that can compute this symbolically? Or numerically? Matlab will not derive a vector by a matrix. I'm trying to do this with PyTorch in Python with the [autograd](https://pytorch.org/tutorials/beginner/basics/autogradqs_tutorial.html#optional-reading-tensor-gradients-and-jacobian-products), but the problem is again the dimensions, I'm not sure what to put in the vector $v$ that multiplies the gradient.
3. Now let's suppose that $K$ also affects $\Sigma$ too, how do I compute the total influence (gradient) of $K$ on $Z$? With the technique above I can also compute:
$$ \frac{dZ}{d\Sigma} = \mathbb{E}\left[\frac{d\ln\mathcal{N}(x;\mu,\Sigma)}{d\Sigma}\right] = \mathbb{E}\left[\Sigma^{-1}(x-\mu)(x-\mu)^T\Sigma^{-1}-\Sigma^{-1}\right] $$
|
https://mathoverflow.net/users/301078
|
Derivative of an integral of a Gaussian
|
**Q1:** $dZ/dK$ is a $p\times q$ matrix with elements
$$\bigl[dZ/dK\bigr]\_{ij}=\sum\_{k=1}^n\frac{\partial \mu\_k}{\partial K\_{ij}}\,\mathbb{E}[f(x)\,\Sigma^{-1}\cdot(x-\mu)]\_k.$$
**Q2:** To evaluate this you will have to specify the function $f(x)$. For some $f$ you will be able to evaluate the expectation value symbolically, but in general a numerical evaluation should be easy and accurate because of the Gaussian weight factor.
**Q3:** If $\Sigma$ also depends on $K$ you add the terms $\sum\_{kl}(\partial\Sigma\_{kl}/\partial K\_{ij})(\partial Z/\partial\Sigma\_{kl})$.
|
1
|
https://mathoverflow.net/users/11260
|
409307
| 167,556 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.