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https://mathoverflow.net/questions/404907
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1
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Let $\boldsymbol{X} = (X\_1,X\_2)^{\rm T}\sim \mathcal{N}\_2(\boldsymbol{\mu}, \mathrm{\Sigma})$, where
\begin{eqnarray\*}
\boldsymbol{\mu} = (\mu\_1, \mu\_2)^{\rm T}& = &(\sqrt{\xi\_1\xi\_2/(\xi\_1+\xi\_2)}, 0)^{\rm T}\\
\mathrm{\Sigma} & = &\begin{pmatrix} 1 & -\rho\\
-\rho & 1\end{pmatrix}\\
\rho & = & \sqrt{\xi\_1\xi\_3/(\xi\_1+\xi\_2)(\xi\_2+\xi\_3)}.
\end{eqnarray\*}
It is given that $\xi\_1\leq\xi\_2\leq\xi\_3$, $\xi\_i\geq 0$ and $\sum\_{i=1}^3\xi\_i = 1$. I have a function
\begin{equation\*}
\pi(\boldsymbol{\mu};\boldsymbol{\xi}) = 1-\mathbb{P}(\boldsymbol{X}\leq \boldsymbol{t}) = 1-\mathbb{P}(X\_1\leq t \cap X\_2\leq t),
\end{equation\*}
where $t>0$. Under the constraints $\xi\_1\leq\xi\_2\leq\xi\_3$, $\xi\_i\geq 0$ and $\sum\_{i=1}^3\xi\_i = 1$, numerically we are getting the maximum of $\pi(\boldsymbol{\mu};\boldsymbol{\xi})$ where the non-zero mean $\mu\_1$ is maximized. This occurs when $\xi\_1 = \xi\_2 = \xi\_3 = 1/3$.
Can we use some kind of stochastic ordering arguments to prove the result theoretically?
|
https://mathoverflow.net/users/120111
|
A problem related to stochastic ordering
|
$\newcommand{\der}{\mathrm{der}}\newcommand{\tdert}{\mathrm{dert}}\newcommand{\erf}{\operatorname{erf}}\newcommand{\eqs}{\overset{\text{sign}}=}\newcommand{\tder}{\widetilde\der}$The answer is no.
Indeed, let $x:=\xi\_1$ and $y:=\xi\_2$, so that $\xi\_3=1-x-y$, $0\le x\le y\le1-x-y$, whence $x\in[0,1/3]$.
Consider further the case $y=x\in(0,1/3]$, so that
$$\mu\_1=M(x):=\sqrt{x/2},$$
$$\rho=-R(x),\quad R(x):=\sqrt{\frac{1-2x}{2(1-x)}}.$$
Let also
$$p\_t(x):=1-P(X\_1\le t,X\_2\le t).$$
Your desired result would imply that
\begin{equation\*}
p\_t(x)\le p\_t(1/3) \tag{0}
\end{equation\*}
for all $x\in(0,1/3]$ and all real $t>0$.
Note that
\begin{equation\*}
p\_t(x)=1-P(X\_1\le t,X\_2\le t)=P(M(x),-R(x)), \tag{1}
\end{equation\*}
where
\begin{equation\*}
P(m,r):=1-\int\_{-\infty}^t du \int\_{-\infty}^t dv\, f\_{m,r}(u,v)
\end{equation\*}
and $f\_{m,r}$ is the density function of the bivariate normal distribution with means $m,0$, variances $1,1$, and correlation $r$.
The key is Plackett's observation ([formula (3)](https://www.jstor.org/stable/2332716)) that
\begin{equation\*}
D\_r f\_{m,r}(u,v)=D\_v D\_u f\_{m,r}(u,v),
\end{equation\*}
where $D\_w$ denote the partial derivative with respect to a variable $w$. It follows that
\begin{equation\*}
\begin{aligned}
D\_r P(m,r)&:=-\int\_{-\infty}^t du \int\_{-\infty}^t dv \, D\_r f\_{m,r}(u,v) \\
&=-\int\_{-\infty}^t du \int\_{-\infty}^t dv \, D\_v D\_u f\_{m,r}(u,v) \\
&=-f\_{m,r}(t,t).
\end{aligned}
\tag{2}
\end{equation\*}
Next,
\begin{equation\*}
\begin{aligned}
P(m,r)&=1-\int\_{-\infty}^t du \int\_{-\infty}^t dv\, f\_{0,r}(u-m,v) \\
&=1-\int\_{-\infty}^{t-m} dw \int\_{-\infty}^t dv\, f\_{0,r}(w,v)
\end{aligned}
\end{equation\*}
and hence
\begin{equation\*}
\begin{aligned}
D\_m P(m,r)=\int\_{-\infty}^t dv\, f\_{0,r}(t-m,v)=\int\_{-\infty}^t dv\, f\_{m,r}(t,v);
\end{aligned}
\tag{3}
\end{equation\*}
the latter integral can be easily expressed in terms of the error function $\erf$ and elementary functions.
By (1) and a chain rule of differentiation,
\begin{equation\*}
D\_x p\_t(x)=D\_m P(M(x),-R(x))M'(x)-D\_r P(M(x),-R(x))R'(x).
\end{equation\*}
In particular,
\begin{equation}
D\_x p\_{1/10}(x)\big|\_{x=1/3}=\erf\left(\frac{1}{60} \left(3 \sqrt{6}-10\right)\right)+1-\frac{3
}{\sqrt{\pi }}\,e^{(30 \sqrt{6}-77)/1800}
=-0.739\ldots<0.
\end{equation}
So, inequality (0) fails to hold for $t=1/10$ and all $x$ in a left neighborhood of $1/3$. $\quad\Box$
|
1
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https://mathoverflow.net/users/36721
|
404955
| 166,070 |
https://mathoverflow.net/questions/404948
|
1
|
Toom's rule is a 2-dimensional cellular automaton which is known to have two distinct stationary measures in the thermodynamic limit, even after small perturbations to a probabilistic cellular automaton by introducing bit-flip or biased noise. What about more general local, but non-on-site noise. E.g., imagine after each step of the deterministic Toom's rule, flipping each pair of neighboring bits together, with a small constant probability. (This is well-defined since double-bit-flips commute with each other; if we want more general noise like low-probability CNOTs one would have to apply this noise in some kind of brick-layer-type manner.)
Is Toom's rule still stable under such more general noise? I assume there's nothing proven, but maybe there are numerical studies on this?
|
https://mathoverflow.net/users/115363
|
Is Toom's rule robust under local but non-on-site noise?
|
Toom's CA is robust under such noise. In fact, the class of noise distributions he considers is quite general. For a fixed CA $f$ on $\{0,1\}^{\mathbb{Z}^2}$ and a parameter $0 \leq \epsilon \leq 1$, let $M\_\epsilon$ be the set of Borel probability measures $\mu$ on $(\{0,1\}^{\mathbb{Z}^2})^ {\mathbb{N}}$ such that every finite set $V \subset \mathbb{Z}^2 \times \mathbb{N}$ satisfies
$$
\mu(\forall (\vec v, n) \in V : x(n+1)\_{\vec v} \neq f(x(n))\_{\vec v}) < \epsilon^{|V|}
$$
for a $\mu$-random itinerary $x = (x(n))\_{n \in \mathbb{N}} \in (\{0,1\}^{\mathbb{Z}^2})^ {\mathbb{N}}$.
For $y \in \{0,1\}^{\mathbb{Z}^2}$, let $M\_\epsilon(y)$ be the subset of those $\mu \in M\_\epsilon$ with $\mu(x(0) = y) = 1$ (we fix the initial configuration instead of allowing it to be random).
In Theorem 1 and Example 1 of [1], Toom proves that the north-east-self majority CA satisfies
$$
\lim\_{\epsilon \to 0} \sup\_{\mu \in M\_\epsilon(a^{\mathbb{Z}^2}), (\vec v, n) \in \mathbb{Z}^2 \times \mathbb{N}} \mu(x(n)\_{\vec v} \neq a) = 0
$$
for $a = 0$ and $a = 1$.
This means that the itinerary "remembers" its initial state if we store it in every cell, since every spacetime point has the same high probability of being in that state.
Intuitively, the marginals of the noise don't have to be independent or identically distributed at different spacetime points, as long as its error rate on any finite set of spacetime points is dominated by independent $\epsilon$-noise.
In particular, choosing to independently flip adjacent pairs, or even larger local patterns, with a small enough probability on every time step results in a robust CA.
[1] A. Toom. Stable and Attractive Trajectories in Multicomponent Systems. Multicomponent Random Systems, Dekker, 1980, v. 6, pp.549-576. Originally published in Russian. Available at [Toom's website](http://toomandre.com/my-articles/engmat/index.htm).
|
4
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https://mathoverflow.net/users/66104
|
404973
| 166,074 |
https://mathoverflow.net/questions/404986
|
2
|
For any point $x \in \mathbb R^n$, denote by $\delta\_x$ the Dirac Delta measure centered at $x$.
Let $a\_n$ be an sequence of positive numbers with $\lim\_{n \to \infty} a\_n = 0$, and let $d\_i$ be a countable dense subset of $\mathbb R^n$.
Given a rearrangement $a\_{n\_k}$ of $a\_n$, consider the space $\mathbb R^n$ under the Euclidean topology and the measure $\sum\_k a\_{n\_k} \delta\_{d\_k}$.
>
> **Question:** For any given $a\_n, d\_i$ satisfying the above consitions, does there exist a rearrangement such that the above space admits nowhere locally constant continuous functions that are integrable under the given measure?
>
>
>
*Note: A function is said to be nowhere locally constant if there exists no open set $S$ for which $f(x) = c$ for all $x$ in $S$.*
|
https://mathoverflow.net/users/173490
|
Does $\mathbb R^n$ equipped with a sum of Dirac delta measures admit nowhere locally constant continuous integrable functions?
|
Partition $(a\_n)$ into two subsequences $(a\_n')$ and $(a\_n'')$ with $\sum a\_n' < \infty$, and partition $(d\_i)$ into two subsequences $(d\_i')$ and $(d\_i'')$ such that $d\_i'' \to \infty$. Pair the $a\_i'$s with the $d\_i'$s and the $a\_i''$s with the $d\_i''$s. Then any bounded continuous function that vanishes at every $d\_i''$ will be integrable.
|
3
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https://mathoverflow.net/users/23141
|
404987
| 166,079 |
https://mathoverflow.net/questions/404990
|
8
|
Two morphisms of category $\ \mathbf C\ $ are isomorphic to one
another $\ \Leftarrow:\Rightarrow\ $ they are the opposite edges that are drawn horizontally (aimed East) of a commutative square that has the vertical edges (aimed North) being isomorphisms of $\ \mathbf C$.
**Problem** What is the minimum total number of morphisms
of a category such that there are isomorphic morphisms
$\ f\ $ and $\ u,\ $ and another isomorphic pair $\ g\ $ and $\ v,\ $ and the compositions $\ g\circ f\ $ and
$\ v\circ u\ $ exist but are not isomorphic?
I have an example of a category, as described above, that has a total number of $27$ morphisms. (*No, I've touched NO computer* :) ).
---
AN ADDITIONAL NOTE:
Now, that @HenrikRüping has provided his excellent [example](https://mathoverflow.net/a/404994) (most likely minimal), let me mention that my example was a monoid too (but of course) of all maps into itself of a 3-element set.
|
https://mathoverflow.net/users/110389
|
Isomorphic morphisms. A 27-morphism category
|
I see one example with 7 morphisms. It is a subcategory of the category of groups. The only object is the the Klein 4-group $(\mathbb{Z}/2)^2$, and the morphisms are generated by the two projections and the flip. That monoid has 7 elements, and the two projections are conjugate, and hence they are isomorphic. However $pr\_1\circ pr\_1=pr\_1$ is not isomorphic to $pr\_1\circ pr\_2=0$.
Edit: I miscounted the number of elements. Originally I thought of it as a semidirect product (which it is not). However in terms of matrices, it should consist exactly of the elements
$\pmatrix{1&0\\0&1}$,$\pmatrix{1&0\\0&0}$,$\pmatrix{0&0\\0&1}$,$\pmatrix{0&0\\0&0}$,$\pmatrix{0&1\\1&0}$,$\pmatrix{0&0\\1&0}$,$\pmatrix{0&1\\0&0}$.
|
11
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https://mathoverflow.net/users/3969
|
404994
| 166,081 |
https://mathoverflow.net/questions/404953
|
3
|
The $p$-adic field $\mathbb{Q}\_p$ has topological basis of open sets of the form $a+p^N \mathbb{Z}\_p$ for $0 \leq a \leq p^N-1$ and $N \in \mathbb{Z}$. These are indeed compact open sets. One can define Bernoulli distributions by $$\mu\_{B,k}(a+p^N \mathbb{Z}\_p)=p^{N(k-1)}B\_k \left(\frac{a}{p^N}\right), $$ where $B\_k(x)$ are Bernoulli polynomials and $B\_k=B\_k(0)$ are Bernoulli numbers.
These $\mu\_{B,k}$ extends to a distribution on $\mathbb{Z}\_p$.
But these Bernoulli distributions $\mu\_{B,k}$ do not define measure on $\mathbb{Z}\_p$.
To integrate $p$-adic continuous functions over the compact open subsets $\mathbb{Z}\_p$ or $\mathbb{Z}\_p^{\*}$.of $\mathbb{Q}\_p$, one defines the regularized Bernoulli distribution by $$\mu\_{k,\alpha}(U)=\mu\_{B,k}(U)-\alpha^{-k}\mu\_{B,k}(\alpha U)$$ for $\alpha \in \mathbb{Z}\_p^{\*}$, $k \in \{0\} \cup \mathbb{N}$ and $U=a+p^N \mathbb{Z}\_p$.
For $k=1,2, \cdots$, we have the measures $ \mu\_{1, \alpha}, \mu\_{2,\alpha}, \cdots, \mu\_{k, \alpha}, \cdots$. These measure or regularized Bernoulli distributions are related by the following relation $$\int\_{\mathbb{Z}\_p} f(x) d \mu\_{k,\alpha}=\int\_{\mathbb{Z}\_p} f(x) \cdot kx^{k-1} d \mu\_{1, \alpha},$$ for any continuous function $f: \mathbb{Z}\_p \to \mathbb{Z}\_p$.
This is all the story on $\mathbb{Q}\_p$.
Now consider a finite extension $K$ of $\mathbb{Q}\_p$. It has similar
compact-open subsets $\mathcal{O}\_K$, the ring of integers and topological
basis consisting of open sets $a+\pi O\_K$, where $\pi$ is the
uniformizer in the ring $\mathcal{O}\_K$.
**My question-**
Can we extend $\mu\_{B,k}$ and $\mu\_{k,\alpha}$ from $\mathbb{Z}\_p$ to $\mathcal{O}\_K$?
$$-------------------------------$$
Clearly, we can define Haar measure normalized with $\mu(O\_K)=1$ and
so $\mu(m\_K)=1/q$, where $m\_K$ is the maximal ideal and $q=\lvert\mathcal{O}\_K/m\_K\rvert$.
Also we know that Haar measure $\mu$ coincide with the Bernoulli distribution $\mu\_{B,k}$ for $k=0$.
So it appears trivially that we can define the Bernoulli distribution $\mu\_{B,0}$ on the finite extension $K$ as it equals to Haar measure and which can be defined on any locally compact Hausdorff space $K$.
I appreciate any help. Thanks
|
https://mathoverflow.net/users/122445
|
Bernoulli distributions and $p$-adic measure on $K$
|
Here is why I doubt that the question has been looked at and why it is not well-formulated, but I am really not confident in saying this. If it has been considered, I'd hope someone corrects me here.
$\newcommand{\ZZ}{\mathbb{Z}}\newcommand{\QQ}{\mathbb{Q}}$ Let $\mu$ be a measure taking open subsets of a group $\Gamma$ to a ring $A$. In your case $\Gamma$ is isomorphic to $\ZZ\_p^\times$ and $A$ is $\ZZ\_p$ or $\QQ\_p$ or $\mathbb{C}\_p$. Conceptually for the Bernoulli distribution, $\Gamma$ is the Galois group of the global extension of $\QQ$ defined by adjoining all $p$-power roots of unity. This $\mu$ corresponds to an element of the Iwasawa Algebra $A[\![\Gamma]\!]$. In your case the measures using Bernoulli numbers link to the $p$-adic $L$-functions and zeta-functions, interpolating values of the classical complex $L$-functions and the Riemann zeta function.
A natural change of the group would be a surjection $\Gamma'\to \Gamma$ corresponding to a further extension of the number field. You could then ask to interpolate $L$-values for more characters of that group. That has been done to some extend and conjecturally we have a good picture of what the right extensions of $\mu$ should be.
In your situation, however, you have a larger group $\Gamma'=\mathcal{O}\_K^\times$ that contains $\Gamma=\ZZ^\times\_p$ where $K/\QQ\_p$ is an extension. I changed the notation to emphasis that the maps $\mu$ really only see the group $\Gamma'$ rather than the ring of integers $\mathcal{O}\_K$. As a group $\Gamma'$ is a direct product of a finite group and $\ZZ\_p^{[K:\QQ\_p]}$; up to the finite bit it won't depend much on $K$, but only on its degree. Since there is no subextension of $\QQ$, this $\Gamma\to\Gamma'$ cannot correspond to a global Galois extension and to $L$-values.
When you are asking for an extension, the most natural thing would be to take the surjection $\mu\circ N\_{K/\QQ\_p}$. This would now take $N=N\_{K/\QQ\_p}:\Gamma'\to \Gamma$ and the natural map $A[\![\Gamma]\!]\to A[\![\Gamma']\!]$. It still could not correspond to a larger extension of $\QQ$ as there are no such extensions. We would have to start with a larger number field $L$, but then the Bernoulli measures even for $\Gamma$ have to be changed depending on $L$.
Up to the finite bit, your group $\Gamma'$ is the direct sum of $\Gamma$ and the group $\ker (N\_{K/\QQ\_p})$. That natural extension is then rather boring on this second group. Still ignoring the finite bits, you could more or less randomly choose a different measure on this second group as an extension, but I would not know what meaning to give to it.
(I said ignoring finite subgroups. There is a little interesting question as there are cases when $\Gamma'$ is not the direct product of $\Gamma$ and $\ker(N)$, which I have not thought about.)
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2
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https://mathoverflow.net/users/5015
|
405005
| 166,084 |
https://mathoverflow.net/questions/405011
|
8
|
In a [recent answer to an old MO question](https://mathoverflow.net/a/404872), I made a distinction between a "definition" of a mathematical object in the sense of axioms that characterize it, and a "definition" that explicitly constructs the object in question. For a concrete example, consider the "definition" of the real numbers as an ordered field with the least upper bound property, and the "definition" of the real numbers via Dedekind cuts or Cauchy sequences.
In the comments, David Roberts suggested that the words [synthetic and analytic](https://mathoverflow.net/questions/128016/why-do-mathematicians-prefer-one-definition-over-the-other-when-they-both-define/404872?noredirect=1#comment1037698_404872) be used to describe this distinction. He also [commented](https://mathoverflow.net/questions/128016/why-do-mathematicians-prefer-one-definition-over-the-other-when-they-both-define/404872?noredirect=1#comment1037810_404872) that the word "analytic" in this sense came from philosophy. That is, it is not to be confused with the use of the term "analytic" to mean involving calculus or real/complex/functional analysis.
The distinction between "synthetic geometry" and "analytic geometry" is well known, and "synthetic geometry" certainly is analogous to the development of the theory of the real numbers from the axioms for an ordered field with the l.u.b. property. But I was a little surprised at David Roberts's suggestion that the words are used more widely in mathematics.
>
> Are the words *synthetic X* and *analytic X* commonly used to describe this distinction in mathematics for any value of *X* other than geometry?
>
>
>
I have some background in philosophy and I am pretty sure that the terminology does not "come from philosophy" in any obvious sense. There is a famous analytic/synthetic distinction in philosophy, but Kant was the first to make a big deal about it, and the term "analytic geometry" dates back to Descartes at least. Moreover, the analytic/synthetic distinction in the sense of Kant (or later philosophers) does not really line up with the distinction between analytic geometry and synthetic geometry; for example, Kant thought that *all* of mathematics was "synthetic a priori," and Quine famously questioned whether the distinction even made sense.
Just to be clear, I am not asking for additional mathematical examples of distinctions that are analogous to the distinction between the two different "definitions" of the reals; they are easy to come up with. I'm just asking whether the terminology is already in use, and/or would be readily understood by people, and not confused with the other notion of "analytic".
(Come to think of it, I'm not even sure that "analytic geometry" is quite analogous to Dedekind cuts; I think of analytic geometry as referring to the study of Euclidean geometry using coordinates, rather than the explicit construction of a model of Hilbert's axioms. But never mind this quibble for now.)
|
https://mathoverflow.net/users/3106
|
Analytic/synthetic distinction in mathematics besides geometry?
|
The answer is here:
<https://ncatlab.org/nlab/show/synthetic+mathematics>
As you can see, there are several flavors available, such as synthetic topology, probability, domain theory, etc. This effort is by no mean new, but it is true that the categorical approach has once again emphasized the synthetic over the analytical
---
SOME THOUGHTS AND A BIT OF BACKGROUND
As usual your questions cut through to the very bone. I shall try to articulate an answer (or at least a sketch thereof), knowing all too well that this is impossible in full.
The two terms, Analysis and Synthesis, go back well before Kant, to the very beginning of western thought ( they somehow appeared in Aristoteles, for instance in his **Analytica Priora**, ie the first formalization of logic, but he probably incorporated previous knowledge from various sources).
It is worth visiting etimological dictionary:
Analysis, circa 1580s, "resolution of anything complex into simple elements" (opposite of synthesis), from Medieval Latin analysis (15c.), from Greek analysis "solution of a problem by analysis," literally "a breaking up, a loosening, releasing," noun of action from analyein "unloose, release, set free; to loose a ship from its moorings," in Aristotle, "**to analyze," from ana "up, back, throughout" (see ana-) + lysis "a loosening," from lyein "to unfasten"** (from PIE root \*leu- "to loosen, divide, cut apart").
```
**So, Analysis basically means to solve some concept into its simpler
constituents, whereas Synthesis is the opposite direction.**
```
In modern day we could say that math is like an acrobat balancing itself between two (yet entwined) sides, the analytic one, and the synthetic (in Lawvere's terminology the second is called **CONCEPTUAL**).
One would be tempted to replace them with constructive vs platonic, and perhaps one would not go too far off, but with many caveats.
The examples abound: for instance, one can define a smooth manifold intrinsically, or present it with explicit charts. A group is an instance of an abstract group, but Combinatorial Group Theory studies concrete groups generated by a set of elements and its relations. You can pick your field, and you almost invariably see that it straddles between an analytical approach and a synthetic one (though the mixture varies...)
Now, what I find extremely interesting is the interplay between these two sides: how to pass from the synthetic to the analytic viewpoint and conversely?
*I believe the method is universal: suppose you have several analytic expressions and you surmise they are in fact the same, then IF you can prove explicitly that this is indeed the case, you form an equivalence class, and you say that these two versions are different instances of presentation of one "conceptual" object.*
PS Descartes is the official father of analytic geometry, even though some Greek (forgot his name) anticipated him.
What is intriguing is that Descartes is in a way the father of both modern philosophy and mathematics. In his **Discourse of the Method** he elaborated his universal way of reasoning, which distinguished 4 phases, 2 of them being Analysis and Synthesis. It is not too much of a stretch to think that his analytic geometry is an application of his own philosophical method (of which he was very proud, and rightly so) . But, not knowing enough of this story, I could be wrong: it could be that the mathematician arrived before the philosopher. Either way, there is no doubt that this question carries a great relevance in both philosophy and mathematics.
PPS it is not by chance that Bill Lawvere has written (with Schanuel) a book called Conceptual Mathematics. Category Theory (including higher cats) is so far perhaps the best tool we have to treat Synthetic Math in full generality, though by no means the only one. That is both its greatness and its inherent limit: it captures the invariant side of math, but of course it hides the other one (think of groups for instance: to study the cat of Groups tells you all you need to know of general groups, but if I want to study say SU(5) I need to calculate, there is no way round it....
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7
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https://mathoverflow.net/users/15293
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405024
| 166,089 |
https://mathoverflow.net/questions/404993
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0
|
The following is a recursion for one point monotone Hurwitz numbers
$$
d \, m\_g(d) = 2(2d-3) \, m\_g(d-1) + d(d-1)^2 \, m\_{g-1}(d)\label{1}\tag{$\*$}
$$
with initial condition $m\_0 (1) =1$ and some of the other numbers are $ m\_0 (2) = 1, m\_1 (3) =10$.
Let denote the generating function by
$$F\_{g}(x) := \sum\_{g\geq 1} m\_g (d) x^d$$
$$
\begin{split}
x{\frac {\rm d}{{\rm d}x}}F\_g \left( x \right) &-4\,{x}^{2}{\frac {\rm d}
{{\rm d}x}}F\_g \left( x \right) +2\,xF\_g \left( x \right) \\
&= \left(x{\frac {\rm d}{{\rm d}x}}\right)^3F\_{g-1} \left( x \right) -2\,\left({x}{\frac {\rm d}
{{\rm d}x}}\right)^2F\_{g-1} \left( x \right) +\,xF\_{g-1} \left( x \right)
\end{split}\label{2}\tag{$\*\*$}
$$
Now we put the condition that $F\_g (x) =0 $ for $g<0$ hence
using \eqref{2} we get
$$
x{\frac {\rm d}{{\rm d}x}}F\_0 \left( x \right) -4\,{x}^{2}{\frac {\rm d}
{{\rm d}x}}F\_0 \left( x \right) +2\,xF\_0 \left( x \right)=0\label{3}\tag{$\*\*\*$}
$$
We get $F\_0 = C\sqrt{(1-4x)}$.
Now I can do a change of coordinate
$$x(z) = z-z^2 $$
The solution become rational. That is solution is $(1-2z)$.
From the equation \eqref{3} we can determine that the rational solution can have poles at $z=1/2$.
My question can we determine all other $F\_g$ in coordinates z and it's rational, and determine the poles or $F\_g$ and it's order? How much information we can conclude about $F\_g$ for $g>0$ having the solution of $F\_0 (z)$ and is rational. Any reference will be very helpful.
|
https://mathoverflow.net/users/45170
|
Rational solution of differential equation
|
The recurrence $(\*)$ gives the correct form of $(\*\*)$ as $$(1 - 4x) F'\_g(x) + 2 F\_g(x) = x^2 F'''\_{g-1}(x) + x F''\_{g-1}(x) \tag{\*\*}$$
The boundary condition which gives the the correct form of $F\_0$ is $$(1 - 4x) F'\_0(x) + 2 F\_0(x) = 1 \tag{\*\*\*}$$ yielding $F\_0 = \frac{1 + 2C\_0 \sqrt{1 - 4x}}2$ and we further require $2C\_0 = -1$ to give the offset Catalan numbers.
Then we get $$F\_1 = \frac{x^2 + C\_1(1-4x)^3}{(1-4x)^{5/2}}$$ where we desire $C\_1 = 0$; $$F\_2 = \frac{x^2(9x^2 + 20x + 1) + C\_2(1-4x)^6}{(1-4x)^{11/2}}$$ where we desire $C\_2 = 0$; $$F\_3 = \frac{x^2(450x^4 + 3080x^3 + 1770x^2 + 136x + 1) + C\_3(1-4x)^9}{(1-4x)^{17/2}}$$ where we desire $C\_3 = 0$, and I'm sure you can take it from there.
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1
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https://mathoverflow.net/users/46140
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405033
| 166,090 |
https://mathoverflow.net/questions/375409
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2
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I believe someone working in an alternative set theory called $$\{x|a\in x\}$$ the *essence* of *a*. Does someone recall a reference?
|
https://mathoverflow.net/users/37385
|
The essence of type free set theory?
|
Here:
"In [1944] Hailperin gave the first of a number of finite axiomatisations of NF now known. Many of them exploit the function $x\mapsto\{y:x∈y\}$ which is injective and total and is an $\in$-isomorphism. This function was known to Whitehead, who suggested to Quine that $\{y:x∈y\}$ should be called the “essence” of x (a terminology clearly suggested by a view of sets as properties-in-extension)."
Forster, Thomas, "Quine’s New Foundations", The Stanford Encyclopedia of Philosophy (Summer 2019 Edition), Edward N. Zalta (ed.), URL = <https://plato.stanford.edu/archives/sum2019/entries/quine-nf/>.
|
3
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https://mathoverflow.net/users/37385
|
405051
| 166,095 |
https://mathoverflow.net/questions/405057
|
0
|
Let the pdf of a multivariate normal distribution be
\begin{equation}
p\_{Z}(\mathbf{z})=\frac{1}{\left(2\pi \sigma^2 \right)^{k/2}}\exp(-{\mathbf{z}}^{\text{T}}\mathbf{z}/2\sigma^2).
\end{equation}
Consider a continuous function $f(\mathbf{s})$ supported on a compact set $\mathcal{S}$. Then let
\begin{equation}
I\_{\sigma}(\mathbf{s})=\int\_{\mathbb{R}^{k}} p\_{Z}(\mathbf{z}-\mathbf{s}) f(\mathbf{z}) d\Omega, \mathbf{s}\in \mathcal{S}.
\end{equation}
My question is whether $I\_{\sigma}(\mathbf{s})$ converges to $f(\mathbf{s})$ uniformly on $\mathcal{S}$ as $\sigma \to 0^{+}$. Can the condition of the compactness of $\mathcal{S}$ be relaxed? In addition, tight nonasymptotic lower and upper bounds on $I\_{\sigma}(\mathbf{s})$ are more welcome.
|
https://mathoverflow.net/users/149696
|
Asymptotic moment of a multivariate normal distribution
|
$\newcommand\si\sigma\newcommand\om\omega\newcommand\z{\mathbf z}\newcommand\R{\mathbb R}$Let us assume that your $d\Omega$ means $d\z$, so that
$$I\_\si(s):=\int\_{\R^k} p\_{Z}(z-s) f(z)\, dz,$$
and we shall assume that this definition holds for all $s\in\R^k$.
So,
$$I\_\si(s)=Ef(s+\si Z),$$
where $Z$ is a standard normal random vector in $\R^k$.
Let $\si\downarrow0$. Then for each $s\in\R^k$ we have $s+\si Z\to s$ in distribution, and [hence](https://en.wikipedia.org/wiki/Convergence_of_measures#Weak_convergence_of_measures) $I\_\si(s)=Ef(s+\si Z)\to f(s)$ for any bounded continuous function $f$.
Suppose now that for some bounded function $\om\colon[0,\infty)\to[0,\infty)$ we have $\om(0+)=0$ and $|f(x)-f(y)|\le\om(h)$ for any $x$ and $y$ in $\R^k$ such that $|x-y|\le h$; so, $\om$ is a modulus of continuity of $f$; here $|\cdot|$ denotes the Euclidean norm on $\R^k$ and on $\R$. (In particular, $f$ has such a modulus of continuity if $f$ is a continuous function with a compact support, or if $f$ is bounded and [Hölder continuous](https://en.wikipedia.org/wiki/H%C3%B6lder_condition).) Then for all $s\in\R^k$
$$|I\_\si(s)-f(s)|=|Ef(s+\si Z)-f(s)|\le E\om(\si Z)\to\om(0+)=0,$$
by dominated convergence, since $\si Z\to0$ in probability. It follows that, whenever a bounded modulus of continuity exists, $I\_\si(s)\to f(s)$ uniformly in all $s\in\R^k$.
Suppose now that $f$ is twice differentiable, with the second derivative bounded from above in the sense that for some real $b$ and all $s$ and $z$ in $\R^k$ we have $f''(s)(z,z)\le b|z|^2$. Then
$$f(s+\si Z)-f(s)\le\si f'(s)(Z)+ b\si^2|Z|^2/2.$$
Taking now the expectations, we get
$$I\_\si(s)-f(s)=Ef(s+\si Z)-f(s)\le bk\si^2/2.$$
Similarly, if $f$ is twice differentiable, with the second derivative bounded from below in the sense that for some real $a$ and all $s$ and $z$ in $\R^k$ we have $f''(s)(z,z)\ge a|z|^2$, then
$$I\_\si(s)-f(s)\ge ak\si^2/2.$$
|
2
|
https://mathoverflow.net/users/36721
|
405062
| 166,099 |
https://mathoverflow.net/questions/405060
|
0
|
Is there a term for the operation $A B A^T$? In colloquial terms, I might call this a "sandwich" of a matrix between another matrix and the transpose of that other matrix.
How about for the special case where $B$ (as well as $A B A^T$) is symmetric? This shows up often in Kalman filtering.
I would like to look up some properties of this kind of operation but I'm not sure what to search for.
|
https://mathoverflow.net/users/391616
|
Is there a term for the operation of multiplying the product of two matrices by the transpose of the first matrix?
|
If $B=XAX^T$ classically one says that A and B are congruent.
|
2
|
https://mathoverflow.net/users/56010
|
405073
| 166,100 |
https://mathoverflow.net/questions/405064
|
2
|
Is there a name for the following property of a commutative ring $R$:
its Jacobson radical $J$ is nilpotent, and $R/J$ is semi-simple?
(It is easily equivalent to: $R$ is a finite product of commutative local rings with nilpotent radical.)
|
https://mathoverflow.net/users/76506
|
Terminology for commutative ring whose Jacobson radical $J$ is nilpotent with semisimple quotient $R/J$
|
The word for a ring $R$ whose Jacobson radical $J$ is nilpotent and such that $R/J$ is semisimple is **semiprimary**. I don’t know if there is a more special word for the commutative case.
|
5
|
https://mathoverflow.net/users/19965
|
405076
| 166,101 |
https://mathoverflow.net/questions/405077
|
2
|
I am confused by the two conclusions in this paper ([DOI link behind paywall at Springerlink](https://link.springer.com/content/pdf/10.1007/s10878-019-00461-7.pdf)).
It shows that the equitable tree-coloring problem is $W[1]$-hard when parameterized
by treewidth.
However, it also shows that the equitable tree-coloring problem is polynomial solvable in the class
of graphs of bounded treewidth.
Does this mean $W[1]=FPT$?
|
https://mathoverflow.net/users/178444
|
$W[1]$-hard and FPT about the equitable tree-coloring problem
|
No. Fix $k \in \mathbb{N}$ and let $G$ be an $n$-vertex graph of treewidth at most $k$. If you analyze their polynomial-time algorithm that decides if $G$ has an equitable tree-colouring, you'll notice that the degree of the polynomial depends on $k$. For example, in the proof of Theorem 11, there is a bound of the form $O(n B^2(k+1) \lceil n/r \rceil^{2c(k+1)})$, where $r$ and $c$ are constants and $B(k+1)$ is the $(k+1)$-th Bell number. An FPT algorithm must run in time at most $f(k)n^d$, where $d$ is a constant independent of $k$. Thus, their algorithm is not an FPT algorithm.
Of course, it is widely believed (but not proved) that FPT $\neq$ W[1].
|
2
|
https://mathoverflow.net/users/2233
|
405082
| 166,102 |
https://mathoverflow.net/questions/404965
|
10
|
$\DeclareMathOperator\Hom{Hom}$It is well-known (see [Breen, Mikhailov, Touzé - Derived functors of the divided power functors](https://arxiv.org/abs/1312.5676) for example) that for $A$ a free abelian group we have
$$ H\_i(K(A,1); \mathbb{Z}) \cong {\bigwedge}^i A$$
the exterior powers of $A$
and that for $B$ an abelian group we have
$$ H\_2(K(B,2); \mathbb{Z}) = B, \quad H\_3(K(B,2); \mathbb{Z}) = 0, \quad H\_4(K(B,2); \mathbb{Z}) = \Gamma(B)$$
where $\Gamma(B)$ is Whitehead's gamma group. It is the universal recipient of a quadratic map. It can be defined as the free group generated by the symbols $\gamma(b)$ for $b \in B$ subject to two relations:
1. $\gamma(-b) = \gamma(b)$;
2. $\gamma(x + y + z) + \gamma(x) + \gamma(y)+ \gamma(z) = \gamma(x+y) + \gamma(y+z) + \gamma(z+x)$.
A map $f:K(A, 1) \to K(B,2)$ is given by a degree 2 cohomology class. We can use universal coefficients to compute this, and since $A$ is free the ext term vanishes, giving us:
$H^2(K(A,1); B) = \Hom( \wedge^2 A, B)$.
Thus given $f: {\bigwedge}^2 A \to B$, we have a map $f:K(A, 1) \to K(B,2)$, and thus an induced map on degree 4 homology: $f\_\*: {\bigwedge}^4 A \to \Gamma(B)$.
**Question**: Given $f$, what is this map $f\_\*: {\bigwedge}^4 A \to \Gamma(B)$?
Another way to phrase this is if I have a quadratic function on $B$ and a homomorphism $f$, then I should be able to combine these into a linear function on ${\bigwedge}^4 A$. How to do this?
|
https://mathoverflow.net/users/184
|
Induced map on $H_4$ of Eilenberg–MacLane spaces
|
Let me summarize comments above. Functoriality of the resulting map $f\_\*:H\_4(K(A,1),\mathbb{Z})\to H\_4(K(B,2),\mathbb{Z})$ translates into a request for the universal map $\wedge^4 A\to \Gamma(A)$ which in turn, being functorial in $A$, allows to guess the answer assuming $A$ is free and
$rk\ A=4$.
Note that $\Gamma(A)$ identifies with the second divided powers $\Gamma^2(A)$
under the map $\gamma(x)\to \frac{x^2}{2}=\gamma^2(x)$. Thus we can write the product in divided powers as $x\cdot y=\gamma(x+y)-\gamma(x)-\gamma(y)$.
Consider a map given by
$$a\wedge b \wedge c \wedge d\to a\wedge b\cdot c\wedge d - a\wedge c\cdot b\wedge d+a\wedge d\cdot b\wedge c$$
Under assumption $rk\ A=4$ it defines a functorial map $\wedge^4 A\to \Gamma(A)$ which is unique up
to a scalar. To ensure that the universal $f\_\*$ is given by the above formula, we have to check that the scalar is indeed equal to $1$. For a prime $p$ we have the reduction inclusion $H\_4(K(\wedge^2 A,2),\mathbb{Z})/p\to H\_4(K(\wedge^2 A,2),\mathbb{Z}/p)$, thus it is enough to consider similar maps
$\bar{f\_\*}:H\_4(K(A,1),\mathbb{Z}/p)\to H\_4(K(\wedge^2 A,2),\mathbb{Z}/p)$ for all primes $p$, which are compatible with $f\_\*$ under the reduction. Then, following
the definitions and Serre's description of EM-space cohomology $\mod p$, one can see that
its dual is given by the usual multiplication
$Sym^2(\wedge^2 (A/p)^\*)\overset{\wedge}{\to} \wedge^4 (A/p)^\*$. Dualizing again we see that our universal formula is simply the lifting of this comultiplication and there are no unexpected multipliers.
|
3
|
https://mathoverflow.net/users/8906
|
405083
| 166,103 |
https://mathoverflow.net/questions/405023
|
8
|
Are there two non-isotopic knots $K,K'$ in $S^3$ with the same $\mathrm{SL}\_2(\mathbb C)$ $A$-polynomials? If it's an open problem, has anyone suggested a method for finding them, or a reason why no such pair should exist (i.e., why the $A$-polynomial should be a complete knot invariant)?
My first guess for how to find such a pair $K, K'$ would be to try mutation, but I wasn't able to find much about (or figure out on my own) how mutation affects the $A$-polynomial.
|
https://mathoverflow.net/users/113402
|
Distinct knots with same $A$-polynomial
|
The torus knots $T\_{7,15}$ and $T\_{3,35}$ have the same A-polynomials. In general, if $p,q>1$ are coprime and odd then $T\_{p,q}$ has A-polynomial $(L-1)(LM^{pq}+1)(LM^{pq}-1)$, which only depends on the product $pq$. This follows from a few observations:
1. The [original paper](https://link.springer.com/article/10.1007%2FBF01231526) on the A-polynomial by Cooper-Culler-Gillet-Long-Shalen asserts (Proposition 2.7) that the A-polynomial of $T\_{p,q}$ is a multiple of $LM^{pq}+1$.
2. That factor comes from a family of representations $\pi\_1(S^3\setminus T\_{p,q})\to SL\_2(\mathbb{C})$ sending $\mu^{pq}\lambda$ to $-I$, and if we multiply all of these by the nontrivial character $\pi\_1(S^3\setminus T\_{p,q}) \to \{\pm1\}$ then (since $pq$ is odd) we get a different family for which $\mu^{pq}\lambda \mapsto +I$, contributing another factor of $LM^{pq}-1$.
3. Any irreducible representation $\rho: \pi\_1(S^3\setminus T\_{p,q}) \to SL\_2(\mathbb{C})$ with $\rho(\mu),\rho(\lambda)$ diagonal has to send $\mu^{pq}\lambda$ to $\pm I$, so every irreducible is accounted for by these two factors. This is because the complement $S^3\setminus T\_{p,q}$ is Seifert fibered, with $\mu^{pq}\lambda$ the class of a generic fiber, and so $\mu^{pq}\lambda$ is central in the knot group.
So in fact you can get arbitrarily many odd torus knots with the same A-polynomial. Namely, if $p\_1,\dots,p\_n$ are distinct odd primes, then you can get $2^{n-1}-1$ different such knots of the form $T\_{r,s}$ by letting $r$ and $s$ be the products of any two complementary, nonempty subsets of $\{p\_1,\dots,p\_n\}$. (This gives $2^n-2$ pairs $(r,s)$, but only half as many torus knots because $T\_{r,s}$ is isotopic to $T\_{s,r}$.) These all have the same A-polynomial, determined completely by the product $p\_1\dots p\_n$.
|
5
|
https://mathoverflow.net/users/428
|
405091
| 166,107 |
https://mathoverflow.net/questions/405097
|
2
|
Suppose that $(X,\leq )$ is an ordered set, we can define the maximum and the infimum of this set,now let $(X,A)$ be a measurable space and let $M(X,A)$ be the set of all measures on $(X,A)$, we now define an order $\leq$ by declaring that $\mu \_{1}\leqslant \mu \_{2}$ iff $\mu \_{1}\left ( A\_{0} \right )\leq \mu \_{2}\left ( A\_{0} \right )$ for each $A\_{0}\in A$ my questions are as follow.
(a) Let $\phi \neq Y\subseteq M(X,A)$ and for each $B$ in $A$, define $$\mu \left ( B \right )=\sup\left \{ \sum\_{n\in\mathbb{N} }^{}\varphi \left ( n \right )\left ( B\_{n} \right ):\varphi :\mathbb{N}\rightarrow Y,B\_{n}\mbox{ is a partition of }B \mbox{ with } B\_{n}\in A\mbox{ for each }n\right \}$$My question is that how to show that $\mu$ is a measure and it is a supremum of $Y$.
(b) is it true that every $\phi \neq Y\subseteq M(X,A)$ admits a lower bound?
|
https://mathoverflow.net/users/206621
|
Supremum with respect to the order of measures on $(X,A)$
|
$\newcommand\C{\mathscr C}\newcommand\ep{\varepsilon}\newcommand\A{\mathscr A}\newcommand\N{\mathbb N}$Yes, $\mu$ is a measure.
Indeed, let $\A:=A$. Say that a partition of a set $B\in\A$ is measurable if all its pieces are in $\A$.
For any $B\in\A$, let $P(B)$ denote the set of all countable measurable partitions of $B$. Then
$$\mu(B)=\mu\_Y(B):=\sup\Big\{\sum\_{n\in\N}f(n)(C\_n)\colon f\colon\N\to Y, (C\_n)\_{n\in\N}\in P(B)\Big\}. \tag{0}$$
Take any pairwise disjoint $B\_1,B\_2,\dots$ in $\A$ and let $B:=B\_1\cup B\_2\cup\cdots$.
Take any $(C\_n)\_{n\in\N}\in P(B)$ and any $f\colon\N\to Y$. Then $(C\_n\cap B\_j)\_{n\in\N}\in P(B\_j)$ for each $j\in\N$. So,
$$\sum\_{n\in\N}f(n)(C\_n)=\sum\_{n\in\N}\sum\_{j\in\N}f(n)(C\_n\cap B\_j)
=\sum\_{j\in\N}\sum\_{n\in\N}f(n)(C\_n\cap B\_j)
\le\sum\_{j\in\N}\mu(B\_j),$$
whence
$$\mu(B)\le\sum\_{j\in\N}\mu(B\_j). \tag{1}$$
Let us now show that
$$\mu(B)\ge\sum\_{j\in\N}\mu(B\_j). \tag{2}$$
Since $\mu\ge0$, here we may and will assume that $\mu(B\_j)<\infty$ for all $j\in\N$ (otherwise, (2) is trivial). Take now any real $\ep>0$. Then for each $j\in\N$ there is a partition $(C\_{j,n})\_{n\in\N}\in P(B\_j)$ and a function $f\_j\colon\N\to Y$ such that $\sum\_{n\in\N}f\_j(n)(C\_{j,n})\ge\mu(B\_j)-\ep/2^j$. Moreover, $(C\_{j,n})\_{(j,n)\in\N^2}\in P(B)$. So,
$$\mu(B)\ge\sum\_{j\in\N}\sum\_{n\in\N}f\_j(n)(C\_{j,n})
\ge\sum\_{j\in\N}(\mu(B\_j)-\ep/2^j)
=\sum\_{j\in\N}\mu(B\_j)-\ep.$$
So, (2) follows. By (1) and (2), $\mu$ is a measure.
---
Moreover, for any $B\in\A$, we have $(B,\emptyset,\emptyset,\dots)\in P(B)$. Taking now any measure $y\in Y$ and, say, letting $f(n)=y$ for all $n\in\N$, we see that
$\mu\ge y$ for any $y\in Y$. However, it is easy to see that in general $\mu$ will not be in $Y$.
It is also clear that any $y\in Y$ is a lower bound on $\mu$. Moreover, for any $Y\_0\subseteq Y$, the measure $\mu\_{Y\_0}$ is a lower bound on $\mu\_Y$.
|
1
|
https://mathoverflow.net/users/36721
|
405103
| 166,109 |
https://mathoverflow.net/questions/405069
|
9
|
A [classical result](http://www-users.math.umd.edu/%7Ewmg//fricke.pdf) of Fricke--Klein--Vogt from the late 1800s implies that the character variety $\chi\_\mathbb{C}$ associated to the free group $F\_2$ and the algebraic group $\mathrm{SL}\_2(\mathbb{C})$ is isomorphic to $\mathbb{C}^3$.
Question: What happens if we change $\mathbb{C}$ to a different (commutative unital) ring $R$? I.e., can one explicitly describe the character variety $\chi\_R$ associated to $F\_2$ and $\mathrm{SL}\_2$ over an arbitrary ring $R$? For which rings $R$, do we have $\chi\_R\simeq R^3$?
I suspect that $\chi\_\mathbb{R}$ is not isomorphic to $\mathbb{R}^3$ while $\chi\_{\mathbb{F}\_q}\simeq \mathbb{F}\_q^3$.
|
https://mathoverflow.net/users/41301
|
Character variety of the free group
|
Let $\pi$ be a free group of rank 2. The character variety of $SL\_{2,k}$-representations of $\pi$ is always isomorphic to affine 3 space $\mathbb{A}^3\_k$ *for any ring $k$*.
Let $A[\pi] = A[\pi]\_k$ denote the coordinate ring of $SL\_{2,k}\times SL\_{2,k}$, which we identify with $Hom(\pi, SL\_{2,k})$ by picking a basis of $\pi$. Thus
$$A[\pi] \cong k[\{a\_i,b\_i,c\_i,d\_i\}\_{i=1,2}]/(a\_id\_i-b\_ic\_i-1)\_{i=1,2}$$
Let $X\_i$ be the matrix variable $[[a\_i,b\_i],[c\_i,d\_i]]$. I claim that if we let $GL\_{2,k}$ act on $A[\pi]$ by simultaneous conjugation on matrices, then $A[\pi]^{GL\_{2,k}}$ is a polynomial ring in 3 variables.
$\newcommand{\tr}{\operatorname{tr}}$
$\newcommand{\bZ}{\mathbb{Z}}$
This is treated in Brumfiel-Hilden's book "SL(2)-representations of finitely presented groups" (combine their Propositions 9.1(ii) and Proposition 3.5). Specifically 9.1(ii) says that for any ring $k$, the ring of invariants $A[\pi]^{GL\_{2,k}}$ is isomorphic to a certain ring $TH[\pi]$ (a subalgebra generated by "traces"), and Proposition 3.5 says that $TH[\pi]$ is in fact a polynomial ring over $k$ in the variables $A = \tr(X\_1)$, $B = \tr(X\_2)$, and $C = \tr(X\_1X\_2)$.
However, there is an issue with their statement - they actually claim that $k[A,B,C] = TH[\pi] = A[\pi]^{GL\_2(k)}$ (taking invariants by $GL\_2(k)$ instead of $GL\_{2,k}$). This cannot be true, since when $k$ is a finite field, $A[\pi]^{GL\_2(k)}$ is the invariant ring of a 6-dimensional algebra by a finite group, which must also be 6-dimensional, whereas $k[A,B,C]$ is visibly 3-dimensional. However, this is the only issue in their exposition. If you replace all instances of $GL\_2(k)$ by $GL\_{2,k}$, then their proof is fine.
I recently encountered this error when I tried to use their results in a paper of mine. Here are two ways to address their gap:
1. When $k = \bZ$ or $k$ is an infinite field, $GL\_2(k)$ is Zariski-dense in $GL\_{2,k}$, and hence in these cases their argument essentially works as is. You can bootstrap from these cases to the general case as follows. Since you know the case for $k = \mathbb{Z}$, you have $\bZ[A,B,C] = A[\pi]\_\bZ^{GL\_{2,\bZ}}$, so it suffices to check that taking invariants commutes with base change to any ring $k$. For any ring $k$, the universal coefficients theorem (Jantzen, I Proposition 4.18) gives an exact sequence
$$0\longrightarrow A[\pi]\_\bZ^{GL\_{2,\bZ}}\otimes\_\bZ k\longrightarrow A[\pi]\_k^{GL\_{2,k}}\longrightarrow Tor\_1^\bZ(H^1(GL\_{2,\bZ},A[\pi]\_\bZ),k)\longrightarrow 0$$
Thus it suffices to show that $H^1(GL\_{2,\bZ},A[\pi]\_\bZ)$ is $\bZ$-flat (in fact it is 0, but we don't need this). For this, it suffices to check vanishing of the Tor group when $k = \mathbb{F}\_p$ (for all $p$), and since $\overline{\mathbb{F}\_p}$ is faithfully flat over $\mathbb{F}\_p$, it suffices to check this when $k = \overline{\mathbb{F}\_p}$, but since we are assuming Brumfiel-Hilden's result over infinite fields, the exact sequence above gives us this Tor vanishing for $k = \overline{\mathbb{F}\_p}$, as desired.
2. Another approach is to put together pieces of their argument to make your own direct argument. Firstly, Brumfiel-Hilden's arguments for Proposition 3.5 are correct. Specifically, you can check that for any ring $k$, the map $k[x,y,z] \rightarrow A[\pi]^{GL\_{2,k}}$ gives an isomorphism onto the subalgebra $T[\pi]\subset A[\pi]$ generated by traces $\tr(X\_{i\_1}X\_{i\_2}\cdots X\_{i\_r})$ where each $i\_j\in\{1,2\}$ and $r\ge 1$ (injectivity is Prop 3.2, surjectivity is prop 1.7). The next piece you need is the correct statement of "The first fundamental theorem of invariant theory". Namely, you need to replace $GL\_2(k)$-invariants in their Proposition 9.2 (first part) with $GL\_{2,k}$-invariants. This is due to Donkin (*Invariants of several matrices*) when $k = \bZ$ or $k$ is an algebraically closed field, but again the same argument as above using the universal coefficient theorem extends this to all rings. This theorem says that if $A[2]$ denotes the polynomial ring on 8 variables representing two 2x2 matrices, then $A[2]^{GL\_{2,k}}$ ($GL\_2$ acting by simultaneous conjugation) is generated as a $k$-subalgebra by traces and determinants of arbitrary products of the universal matrices. Thus, to complete the argument it suffices to check that the surjective map $A[2]\rightarrow A[\pi]$ induces a surjection on $GL\_{2,k}$-invariants (note the determinants map to $1\in A[\pi]$). For this one can use a piece of Brumfiel-Hilden's argument (specifically in $\S9$, p97-98, "surjectivity of $\rho\_n$"), which is correct and holds over arbitrary rings $k$.
|
5
|
https://mathoverflow.net/users/15242
|
405109
| 166,113 |
https://mathoverflow.net/questions/405099
|
11
|
For the statement of Carlson's theorem please see,
<https://en.wikipedia.org/wiki/Carlson%27s_theorem>.
There is an extension of Carlson's theorem that says that the condition that $f$ needs to vanish on integers can be replaced with $f$ vanishing on a subset A of integers provided that $A$ has upper density 1. This is a necessary and sufficient condition.
Now my question is as follows: Assuming that the condition on the growth of $f$ on the $y$-axis is more stringent, say for example that $f$ is uniformly bounded on the entire $y$-axis, can one obtain an extension of Carlson's theorem with $f$ vanishing on a monotone divergent sequence $a\_1<a\_2<\ldots$ with upper density strictly less than one?
|
https://mathoverflow.net/users/50438
|
An extension of the Carlson's theorem in complex analysis
|
If $f$ is bounded on the imaginary line, (and has exponential type) then
$f$ has completely regular growth in the sense of Levin-Pfluger, with indicator
$c|\cos\theta|$. This implies that density of zeros on the positive ray must
be zero. Moreover, density of zeros in any angle $|\arg z|<\pi/2-\epsilon$
and in the vertical angle is zero. Boundedness on the imaginary line condition can be relaxed to
$$\int\_{-\infty}^{\infty}\frac{\log^+|f(it)|}{1+t^2}dt<\infty.$$
See any of the two books of Levin (Distribution of zeros of entire functions, or
Lectures on entire functions) for terminology and proofs.
For weaker conditions on the imaginary line, implying some upper estimates on the density of real zeros, see, for example Theorem 7 (Chap IV, $\S$ 2 of the first book of Levin):
the lower density of zeros is at most
$$\frac{1}{2\pi}\left(h\left(\frac{\pi}{2}\right)+h\left(-\frac{\pi}{2}\right)\right),$$
where $h$ is the indicator.
See also MR0207986 and <https://arxiv.org/abs/0807.2054>
|
10
|
https://mathoverflow.net/users/25510
|
405112
| 166,114 |
https://mathoverflow.net/questions/405105
|
9
|
Is it known when
$2^{2n}-2^n+1$
is prime?
It seems to be only when n is 1,2,4 or 32.
|
https://mathoverflow.net/users/45242
|
When is $2^{2n}-2^n+1$ prime?
|
Expanding on the remarks by Pace Nielsen (with an additional result that we need only consider $n>2$ a multiple of $4.$):
$x^n-1=\prod\_{d \mid n}\phi\_d(x)$. The factors are irreducible over the rationals.
In particular $$x^6-1=\phi\_1(x)\phi\_2(x)\phi\_3(x)\phi\_6(x)=(x-1)(x+1)(x^2+x+1)(x^2-x+1)$$
And $$\phi\_6(x^n)=x^{2n}-x^n+1$$ is a factor of $x^{6n}-1.$ It is an irreducible rational polynomial exactly if $n$ is of the form $n=2^i3^j,$ then $$\phi\_6(x^n)=\phi\_{6n}(x)$$ .
But if $p$ is prime to $6$ then $$\phi\_6(u^p)=\phi\_p(u)\phi\_{6p}(u)$$ so for $n=pm$ $$\phi\_{6}(x^n)=\phi\_p(x^m)\phi\_{6p}(x^m).$$
Accordingly, a necessary condition for $b^{2n}-b^n+1$ to be prime is that $n$ is of the form $2^i3^j.$
In the event that $b \bmod 3=-1$ we also need $n$ even lest $b^{2n}-b^n+1 \bmod 3=0.$
Finally, in the given case that $b=2,$ we have that for $n=4k-2>2$ the integer $2^{2n}-2^n+1$ has two factors of roughly equal size.
For example $2^{36}-2^{18}+1=68719214593=246241\cdot 279073.$
This is due to the following interesting [Aurifeuillean factorization](https://en.wikipedia.org/wiki/Aurifeuillean_factorization)
$$\left(x^{4 k -2}+x^{2 k -1}+1+(x^{3 k -1}+x^{k})\right) \left(x^{4 k -2}+x^{2 k -1}+1-(x^{3 k -1}+x^{k})\right)\\ =x^{8 k -4}+(2 x^{6 k -3}-x^{6 k -2})-(x^{2 k}- 2 x^{2 k -1})+(3 x^{4 k -2}-2 x^{4 k -1})+1$$
Putting $x=2,$ the first two pairs become $0$ and the third $2^{4k-2}.$
So the question might be phrased "when does the Mersenne number $2^n-1$ have factors other than those suggested by algebra?" Here restricted to $n=6m.$ That is the subject of the [Cunningham Project](https://homes.cerias.purdue.edu/%7Essw/cun/) where much information (in very condensed form) can be found.
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13
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https://mathoverflow.net/users/8008
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405124
| 166,116 |
https://mathoverflow.net/questions/105926
|
7
|
I am trying to derive the [classic paper](http://alexandria.tue.nl/repository/freearticles/597601.pdf) in the title only by elementary means (no generating functions, no complex analysis, no Fourier analysis) although with much less precision. In short, I "only" want to prove that the average height $h\_n$ of a tree with $n$ nodes (that is, the maximum number of nodes from the root to a leaf) satisfies $h\_n \sim \sqrt{\pi n}$.
The outline is as follows. Let $A\_{nh}$ be the number of trees with height less than or equal to $h$ (with the convention $A\_{nh} = A\_{nn}$ for all $h \geqslant n$) and $B\_{nh}$ the number of trees of $n$ nodes with height greater than or equal to $h+1$ (that is, $B\_{nh} = A\_{nn} - A\_{nh}$). Then $h\_n = S\_n/A\_{nn}$, where $S\_n$ is the finite sum
$$
S\_n = \sum\_{h \geqslant 1} h(A\_{nh} - A\_{n,h-1}) = \sum\_{h \geqslant 1} h(B\_{n,h-1} - B\_{nh}) = \sum\_{h \geqslant 0} B\_{nh}.
$$
It is well known that $A\_{nn} = \frac{1}{n}\binom{2n-2}{n-1}$, for the set of general trees with $n$ nodes is in bijection with the set of binary trees with $n-1$ nodes, counted by the Catalan numbers.
Therefore, the first step is to find $B\_{nh}$ and then the main term in the asymptotic expansion of $S\_n$.
At this point the authors use analytical combinatorics (three pages) to derive
$$
B\_{n+1,h-1} = \sum\_{k \geqslant 1} \left[\binom{2n}{n+1-kh} - 2\binom{2n}{n-kh} + \binom{2n}{n-1-kh}\right].
$$
>
> My own attempt is as follows. I consider the bijection between trees with $n$ nodes
> and monotonic paths on a square grid $(n-1) \times (n-1)$ from $(0,0)$ to $(n-1,n-1)$ which do not cross the diagonal (and are made of two kinds of steps: $\uparrow$ and $\rightarrow$). These paths are sometimes called *Dyck paths* or *excursions*. I can express now $B\_{nh}$ in terms of lattice paths: it is the number of Dyck paths of length 2(n-1) and height greater than or equal to $h$. (Note: a tree of height $h$ is in bijection with a Dyck path of height $h-1$.)
>
>
> Without loss of generality, I assume that they start with $\uparrow$ (hence stay above the diagonal). For each path, I consider the first step crossing the line $y = x + h - 1$, if any. From the point above, all the way back to the origin, I change $\uparrow$ into $\rightarrow$ and vice versa (this is a *reflection* wrt the line $y=x+h$). It becomes apparent that the paths I want to count ($B\_{nh}$) are in bijection with the monotonic paths from $(-h,h)$ to $(n-1,n-1)$ which avoid the boundaries $y=x+2h+1$ and $y=x-1$. (See [figure](http://www.filedropper.com/lattice).)
>
>
>
In the classic book *Lattice Path Counting and Applications* by Mohanty (1979, page 6) the formula
$$
\sum\_{k \in \mathbb{Z}} \left[\binom{m+n}{m-k(t+s)} - \binom{m+n}{n+k(t+s)+t}\right],
$$
counts the number of monotonic paths in a lattice from $(0,0)$ to $(m,n)$, which avoid the boundaries $y = x - t$ and $y = x + s$, with $t > 0$ and $s > 0$. (This result was first established by Russian statisticians in the 50s.) Therefore, by considering a new origin at $(-h,h)$, we satisfy the conditions of the formula: $s=1$, $t=2h+1$ and the destination point (the upper right corner) is now $(n+h-1,n-h-1)$. Then
$$
B\_{nh} = \sum\_{k \in \mathbb{Z}} \left[\binom{2n-2}{n+h-1-k(2h+2)} - \binom{2n-2}{n-h-1+k(2h+2) + 2h+1}\right].
$$
This can be simplified in
$$
B\_{n+1,h-1} = \sum\_{k \in \mathbb{Z}} \left[\binom{2n}{n+1-(2k+1)h} - \binom{2n}{n-(2k+1)h}\right],
$$
which, in turn, is equivalent to
$$
B\_{n+1,h-1} = \sum\_{k \geqslant 0} \left[\binom{2n}{n+1-(2k+1)h} - 2\binom{2n}{n-(2k+1)h} + \binom{2n}{n-1-(2k+1)h}\right].
$$
The difference with the expected formula is that I sum over the odd numbers ($2k+1$), instead of all positive integers ($k$). First, I hoped that the even terms would cancel out, but that does not seem to be the case.
Any idea where is the problem?
[**Edit:**
Starting from the expected result, by the same elementary binomial manipulations, we have
$$
B\_{n,h} = A\_{nn} + \sum\_{k \in \mathbb{Z}}\left[\binom{2n-2}{n-k(h+1)} - \binom{2n-2}{n-1+k(h+1)}\right].
$$
But if we try to find a combinatorial interpretation to this sum in terms of bounded lattice paths, we fail: the destination point has coordinates $(n,n-2)$, which is below the inferior boundary $y = x - 1$, so the number of paths is $0$, but Mohanty's formula gives negative numbers in this kind of situation (although he does not mention this). Therefore, if we find a combinatorial interpretation for the absolute value of these numbers, we can understand the result in the same terms.]
[**Edit:** In response to a comment below, here are all the details in slow motion.
\begin{align}
B\_{n,h} &= \sum\_{k \in \mathbb{Z}}\left[\binom{2n-2}{n-(2k+1)(h+1)} - \binom{2n-2}{n-1+(2k+1)(h+1)}\right]\cr
B\_{n+1,h-1} &= \sum\_{k \in \mathbb{Z}}\left[\binom{2n}{n+1-(2k+1)h} - \binom{2n}{n+(2k+1)h}\right]\cr
&= \sum\_{k \in \mathbb{Z}}\left[\binom{2n}{n+1-(2k+1)h} - \binom{2n}{n-(2k+1)h}\right]\cr
&= \sum\_{k \geqslant 0}\left[\binom{2n}{n+1-(2k+1)h} - \binom{2n}{n-(2k+1)h}\right]\cr
&+ \sum\_{k > 0}\left[\binom{2n}{n+1+(2k-1)h} - \binom{2n}{n+(2k-1)h)}\right]\cr
&= \sum\_{k \geqslant 0}\left[\binom{2n}{n+1-(2k+1)h} - \binom{2n}{n-(2k+1)h}\right]\cr
&+ \sum\_{k \geqslant 0}\left[\binom{2n}{n+1+(2k+1)h} -\binom{2n}{n+(2k+1)h}\right]\cr
&= \sum\_{k \geqslant 0}\left[\binom{2n}{n+1-(2k+1)h} - \binom{2n}{n-(2k+1)h}\right]\cr
&+ \sum\_{k \geqslant 0}\left[\binom{2n}{n-1-(2k+1)h} -\binom{2n}{n-(2k+1)h}\right]\cr
&= \sum\_{k \geqslant 0}\left[\binom{2n}{n+1-(2k+1)h} - 2\binom{2n}{n-(2k+1)h} + \binom{2n}{n-1-(2k+1)h}\right].
\end{align}
]
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https://mathoverflow.net/users/26078
|
On "The Average Height of Planted Plane Trees" by Knuth, de Bruijn and Rice (1972)
|
I think the problem may be in the bijection. Consider instead the following bijection between (planted plane) trees with $n$ vertices and Dyck paths from $(0,0)$ to $(n-1,n-1)$: perform a depth-first search on the tree, moving $\uparrow$ in the Dyck path when you follow an edge away from the root, and $\rightarrow$ in the Dyck path when you follow an edge towards the root. Then a tree of height $h$ corresponds to a Dyck path which reaches $y=x+h-1$ but not $y=x+h$, so $A\_{n,h}$ counts the Dyck paths from $(0,0)$ to $(n-1,n-1)$ which avoid the boundaries $y=x+h$ and $y=x-1$. By the referenced formula of Mohanty,
$$A\_{n,h} = \sum\_{k \in \mathbb{Z}} \left[\binom{2n-2}{n-1-k(h+1)} - \binom{2n-2}{n+k(h+1)}\right]$$
Then
$$\begin{eqnarray\*}
B\_{n+1,h-1} &=& A\_{n+1,n+1} - A\_{n+1,h-1} \\
&=& \frac{1}{n+1} \binom{2n}{n} + \sum\_{k \in \mathbb{Z}} \left[ \binom{2n}{n+1+kh} - \binom{2n}{n-kh} \right] \\
&=& \frac{1}{n+1} \binom{2n}{n} + \sum\_{k \geqslant 0} \left[ \binom{2n}{n+1+kh} - \binom{2n}{n-kh} \right] + \\&& \sum\_{k > 0} \left[ \binom{2n}{n+1-kh} - \binom{2n}{n+kh} \right] \\
&=& \frac{1}{n+1} \binom{2n}{n} + \binom{2n}{n+1} - \binom{2n}{n} + \\&& \sum\_{k \geqslant 1} \left[ \binom{2n}{n+1+kh} - \binom{2n}{n-kh} - \binom{2n}{n+kh} + \binom{2n}{n+1-kh} \right] \\
&=& \sum\_{k \geqslant 1} \left[ \binom{2n}{n+1+kh} - \binom{2n}{n-kh} - \binom{2n}{n+kh} + \binom{2n}{n+1-kh} \right] \\
&=& \sum\_{k \geqslant 1} \left[ \binom{2n}{2n-(n+1+kh)} - \binom{2n}{n-kh} - \binom{2n}{2n-(n+kh)} + \binom{2n}{n+1-kh} \right] \\
&=& \sum\_{k \geqslant 1} \left[ \binom{2n}{n-1-kh} - 2\binom{2n}{n-kh} + \binom{2n}{n+1-kh} \right] \\
\end{eqnarray\*}$$
as desired.
|
1
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https://mathoverflow.net/users/46140
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405130
| 166,117 |
https://mathoverflow.net/questions/405117
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7
|
EDIT: After a little prompting by Mark Grant, I answered the first question in the comments. The second question remains open.
---
Let $M$ be a compact $3$-manifold with $\pi\_2(M) \neq 0$. The sphere theorem says that there is an embedded 2-sphere in $M$ that is nontrivial in $\pi\_2$.
In his book "Group theory and 3-dimensional topology", Stallings has a proof of this using his theorem about groups with infinitely many ends. He first has an easy reduction to the case where none of the boundary components of $M$ are $2$-spheres and all of them are incompressible. The next step is to prove that the universal cover $\tilde{M}$ has at least 2 ends, and I'm having trouble understanding his proof of this. I have two questions. The first one is probably pretty easy (I bet there is something silly I'm missing), but the second one seems fundamental.
What he says is as follows. Using Poincare duality and the fact that all the boundary components of the universal cover of $\tilde{M}$ are contractible, we have
$$\pi\_2(\tilde{M}) = H\_2(\tilde{M}) \cong H\_2(\tilde{M},\partial \tilde{M}) \cong H^1\_c(\tilde{M}),$$
so
$$H^1\_c(\tilde{M}) \neq 0.$$
Let $H^{k}\_e(\tilde{M})$ denote the homology of the cochain complex obtained by quotienting the using simplicial cochain complex of $\tilde{M}$ by the complex of cochains with compact (or, equivalently, finite) support. We then have a long exact sequence containing the segment
$$H^0\_c(\tilde{M}) \rightarrow H^0(\tilde{M}) \rightarrow H^0\_e(\tilde{M}) \rightarrow H^1\_c(\tilde{M}) \rightarrow H^1(\tilde{M}).$$
We have $H^1(\tilde{M})=0$ since $\tilde{M}$ is simply-connected. He then says that $H^0\_c(\tilde{M}) = 0$ since $\tilde{M}$ is noncompact. This is the first thing I don't understand:
* **Question 1**: Why can't $\tilde{M}$ be compact? This is equivalent to asking why the hypotheses imply that $\pi\_1(M)$ is infinite. I know that geometrization implies that $\pi\_1(M)$ being finite would imply that $M$ has universal cover $S^3$ and hence $\pi\_2(M) = 0$, but obviously this should not depend on that!
Let's grant that $\tilde{M}$ is noncompact. Stallings then says that $H^1(\tilde{M})=0$ since $\tilde{M}$ is $1$-connected. We thus have a short exact sequence
$$0 \rightarrow \mathbb{Z} \rightarrow H^0\_e(\tilde{M}) \rightarrow H^1\_c(\tilde{M}) \rightarrow 0.$$
Since the cokernel of this is nontrivial $H^0\_e(\tilde{M})$ has rank at least $2$. Stallings then asserts that this means that $\tilde{M}$ has at least $2$ ends.
* **Question 2**: Why is that? The number of ends of $\tilde{M}$ should equal the dimension of $H^0\_e(\tilde{M};\mathbb{F}\_2)$, but all we know is that the integral cohomology group $H^0\_e(\tilde{M})$ has rank at least $2$. If $\pi\_2(M) = H^1\_c(\tilde{M})$ consisted entirely of $p$-torsion with $p \neq 2$, then its contribution would disappear when we work mod-$2$.
|
https://mathoverflow.net/users/392655
|
Two details from Stallings's proof of the sphere theorem
|
What you need for your second question is that $H^1\_c(\tilde{M})$ is a free abelian group. As you noted, this is isomorphic to $H\_2(\tilde{M})$. Now, $\tilde{M}$ is a connected non-compact 3-manifold. A basic theorem of Whitehead says that a connected noncompact PL $n$-manifold is homotopy equivalent to an $(n-1)$-dimensional simplicial complex. See my note [here](https://www3.nd.edu/%7Eandyp/notes/NoncompactSurfaceFree.pdf) for a proof and references. Applying this to $\tilde{M}$, we find that $\tilde{M}$ is homotopy equivalent to a $2$-dimensional simplicial complex $X$, so $H\_2(\tilde{M}) \cong H\_2(X)$. Letting $C\_{\bullet}(X)$ be the simplicial chain complex, we have $C\_3(X) = 0$, so $$H\_2(X) = \text{ker}(C\_2(X) \stackrel{\partial}{\rightarrow} C\_1(X)).$$
In particular, $H\_2(X)$ is a subgroup of the free abelian group $C\_2(X)$, and thus is itself free abelian.
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4
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https://mathoverflow.net/users/317
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405140
| 166,120 |
https://mathoverflow.net/questions/405080
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3
|
Let $C
\_p$ be the Schatten-p-classes on a separable Hilbert spaces, $p\ge 1$.
Let ${\rm Tr}$ be the standard trace.
Let $y\in C\_p$ be a self-adjoint operator (or even a positive operator) and let $x\_n \in C\_{p'}$, $\frac1p+\frac1{p'}=1$, be a sequence of self-adjoint operators with $\|x\_n\|\_{p'}=1$ such that ${\rm Tr}(x\_n^\*y)\to \|y\|\_p$.
Let $u\_n |x\_n|=x\_n $ be the polar decomposition.
Do we have
$$\|u\_n^\* y - |y|\| \_p \to 0, ~{
\rm as~}n\to \infty?$$
|
https://mathoverflow.net/users/91769
|
norm estimates for Schatten class
|
I understand that you are happy with the case $y\geq 0$, so let's assume that for simplicity (I expect that small modifications should deal with arbitrary self-adjoint $y$). In that case, **and if** $1<p<\infty$, it is true that $\| u\_n^\* y - y\|\_p \to 0$. I am not sure about the extreme cases $p=1,\infty$, where the uniform convexity argument breaks down.
Here is the argument. It should work more generally in the non-commutative $L\_p$ space of an arbitrary von Neumann algebra. Without loss of generality, we can assume that $\|y\|\_p=1$. In that case, the unique $x \in C\_{p'}$ of norm one such that $Tr(x^\* y) = 1$ is $y^{\frac{p}{p'}}$. By the uniform convexity of $C\_{p'}$, the assumption that $\|x\_n\|\_{p'} \leq 1$ and $Tr(x\_n^\* y) \to 1$ implies that $\| x\_n - y^{\frac{p}{p'}}\|\_{p'} \to 0$.
Now consider the Mazur map $z=u|z| \in C\_{p'} \mapsto M\_{p',p}(z) := u |z|^{\frac{p'}{p}} \in C\_p$. It is known that it is continuous (and even uniformly continuous when restricted to the unit ball, see for example the papers by [Yves Raynaud](https://jot.theta.ro/jot/archive/2002-048-001/2002-048-001-003.pdf) or [Eric Ricard](https://arxiv.org/pdf/1407.8334.pdf) for the optimal estimates). Therefore, we obtain
$$\| u\_n |x\_n|^{\frac{p}{p'}} - y \|\_p = \| M\_{p',p}(x\_n) - M\_{p',p}(y^{\frac{p}{p'}})\|\_p \to 0.$$
$x\_n$ being self-adjoint, we have $u\_n = P\_n^+-P\_n^-$ for orthogonal projections $P\_n^+$ and $P\_n^-$ commuting with $x\_n$ and $|x\_n|$. Multiplying on both sides by $P\_n^-$ the preceding inequality, we get
$$ \| P\_n^- |x\_n|^{\frac{p}{p'}} + P\_n^- y P\_n^-\|\_p \to 0.$$
But since $0 \leq P\_n^- |x\_n|^{\frac{p}{p'}} \leq P\_n^- |x\_n|^{\frac{p}{p'}} + P\_n^- y P\_n^-$ ($y$ is positive), we get
$$\||x\_n|^{\frac{p}{p'}}- u\_n |x\_n|^{\frac{p}{p'}}\|\_p =2\|P\_n^- |x\_n|^{\frac{p}{p'}} \|\_p \leq 2\| P\_n^- |x\_n| + P\_n^- y P\_n^-\|\_p \to 0.$$
This allows to conclude:
$$ \|u\_n^\* y - y\|\_p \leq \|u\_n^\*(y - u\_n |x\_n|^{\frac{p}{p'}})\|\_p + \| |x\_n|^{\frac{p}{p'}} - u\_n |x\_n|^{\frac{p}{p'}}\|\_p + \| u\_n |x\_n|^{\frac{p}{p'}}-y\|\_p$$
goes to zero because each term does.
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2
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https://mathoverflow.net/users/10265
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405146
| 166,124 |
https://mathoverflow.net/questions/405095
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3
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Let $R$ be a Noetherian commutative ring. A complex of $R$-modules $P^{\bullet}$ is K-projective if for any acyclic complex $A^{\bullet}$, the complex of abelian groups $ Hom(P^{\bullet}, A^{\bullet})$ is acyclic. K-projective complexes were defined by Spaltenstein:
<http://www.numdam.org/article/CM_1988__65_2_121_0.pdf>.
Suppose now that $R$ has finite homological dimension. I've heard it stated that a (possibly unbounded) complex of projective modules $P^\bullet$ is K-projective. Is there a (preferably elementary) reference for this or proof?
|
https://mathoverflow.net/users/375103
|
K-projectivity for rings of finite homological dimension
|
There's a nice, short proof in
*Positselski, Leonid; Schnürer, Olaf M.*, [**Unbounded derived categories of small and big modules: is the natural functor fully faithful?**](http://dx.doi.org/10.1016/j.jpaa.2021.106722), J. Pure Appl. Algebra 225, No. 11, Article ID 106722, 23 p. (2021). [ZBL1464.18015](https://zbmath.org/?q=an:1464.18015).
where this is Proposition 4.1(b).
Suppose $R$ has global dimension $d<\infty$. Let $P^\bullet$ be a complex of projectives, and let $\alpha:P\_K^\bullet\to P^\bullet$ be a $K$-projective resolution. To prove that $P^\bullet$ is $K$-projective, it suffices to prove that $\alpha$ is a homotopy equivalence, or equivalently that the mapping cone of $\alpha$ is contractible.
So we just need to prove that an acyclic complex of projectives is contractible.
Let $Q^\bullet$ be an acyclic complex of projectives. Then the truncation
$$\dots\to Q^{-2}\to Q^{-1}\to Q^0\to 0\to\dots$$
is a projective resolution of some module, which has projective dimension at most $d$ by the assumption on the global dimension of $R$. Therefore the image of the differential $Q^{-d}\to Q^{-d+1}$ is projective.
Applying the same argument to shifts of $Q^\bullet$, every differential of $Q^\bullet$ has projective image, which (together with the fact that $Q^\bullet$ is acyclic) implies that $Q^\bullet$ is contractible.
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6
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https://mathoverflow.net/users/22989
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405151
| 166,126 |
https://mathoverflow.net/questions/405150
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6
|
Let $X$ be a complex smooth projective variety of dimension $d$. Let $K(X) := K(\text{Coh}(X))$ denote the Grothendieck group of coherent sheaves on $X$. For two coherent sheaves $E$ and $F$ on $X$, define their Euler pairing by
$$\chi(E,F) = \sum\_{i=0}^{d} (-1)^i \text{dim Ext}^i(E,F).$$
The Euler pairing descends to $K(X)$ by its additivity on short exact sequences in $\text{Coh}(X)$. Let $\ker\_L \subset K(X)$ denote the left radical of $\chi$, that is, $E \in \ker\_L$ if and only if $\chi(E,F) = 0$ for all $F \in K(X)$. By Serre duality, the left and right radicals of $\chi$ agree with each other. Define the numerical Grothendieck group of $X$ by
$$N(X) = K(X)/\ker\_L.$$
In [Bridgeland's notes on derived categories](http://www.tom-bridgeland.staff.shef.ac.uk/talks/warwick.pdf), he said that it's not clear whether the Chern character descends to $N(X)$, but it's true when $X$ is of dimension $\leq 2$.
Why does the Chern character descend to $N(X)$ when $d \leq 2$?
Suppose $d \leq 2$. I want to prove that if $\chi(E,F) = 0$ for all $F \in \text{Coh}(X)$, then $\text{ch}(E) = 0$. This should be a consequence of the Hirzebruch-Riemann-Roch (HRR) theorem, which says
$$\chi(E,F) = \int\_X \text{ch}(E)^\vee \text{ch}(F) \text{td}(X),$$
where $$\text{ch}(E)^\vee = \sum\_{i=0}^d (-1)^i \text{ch}\_i(E),$$ and $$\text{td}(X) = 1 + \frac{1}{2}c\_1(X) + \frac{1}{12}(c\_1(X)^2+c\_2(X)).$$Here I assume the Chern character takes values in the rational cohomology $H^\*(X) \otimes \mathbb{Q}$. If $d = 1$, i.e., $X$ is a curve, then the third term in $\text{td}(X)$ is zero. In this case, let $g$ be the genus of $X$, let $(r\_1, d\_1)$ be the rank and degree of the coherent sheaf $E$ and $(r\_2, d\_2)$ be those of $F$. Then by the HRR theorem, one obtains the following
$$\chi(E,F) = r\_1d\_2 - r\_2d\_1 + r\_1r\_2(1-g).$$
So choosing $(r\_2, d\_2) = (0, 1)$ kills $r\_1$, and further choosing $(r\_2, d\_2) = (1, 0)$ kills $d\_1$. So the case $d = 1$ is settled.
Now let $X$ be a surface. The question is to choose suitable $F$ to kill each $\text{ch}\_i(E)$. But I don't see how this is always possible.
|
https://mathoverflow.net/users/146366
|
Why does the Chern character descend to the numerical Grothendieck group for surfaces?
|
First, taking $F$ to be the structure sheaf of a point you "kill" $\mathrm{ch}\_0(E)$.
Next, since intersection pairing is non-degenerate on
$$
\mathrm{Im}(\mathrm{Pic}(X) \to H^2(X,\mathbb{Q})) = \mathrm{NS}(X)
$$
you can choose $F\_i$ to be a collection of the structure sheaves of curves on $X$ to "kill" $\mathrm{ch}\_1(E)$.
Finally, taking $F$ to be the structure sheaf of $X$ you "kill" $\mathrm{ch}\_2(E)$.
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5
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https://mathoverflow.net/users/4428
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405157
| 166,128 |
https://mathoverflow.net/questions/405160
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6
|
Let $p$ be a prime and let $n\geq 2$ be an integer.
The set $\mathbb{A}^n(\mathbb{F}\_p)$ has $p^n$ elements so it has $2^{p^n}$ subsets. How many of those subsets are of the form $V(\mathbb{F}\_p)$ with $V\subset \mathbb{A}^n$ a closed geometrically irreducible subvariety?
|
https://mathoverflow.net/users/393396
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Sets of $\mathbb{F}_p$-points of closed subvarieties of $\mathbb{A}^n$
|
All of them. One can even take $V$ to be a smooth hypersurface whose extension to projective space is smooth, by Poonen's Bertini theorem. This guarantees irreducibility.
Poonen's Bertini theorem states that there exists a smooth hypersurface in $\mathbb P^n$ of degree $d$ for all $d$ sufficiently large satisfying any finite number of local conditions at closed points, as long as those local conditions don't manifestly force singularities at those points.
For each point in $\mathbb A^n(\mathbb F\_p)$, both the condition that it lie in $V$ and that it not lie in $V$ are local conditions of just the sort considered by Poonen, and neither one forces smoothness.
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13
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https://mathoverflow.net/users/18060
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405165
| 166,130 |
https://mathoverflow.net/questions/405163
|
3
|
What is an example of a rigid object $A$ in an abelian monoidal category $\mathcal{M}$ that is not projective as an object in $\mathcal{M}$? (Since $\mathcal{M}$ is abelian projective just means that all exact sequences of the form $ 0 \to R \to Q \to P \to 0$ split.) What are sufficient conditions for such a rigid object to be projective?
|
https://mathoverflow.net/users/371382
|
A non-projective rigid object in an abelian monoidal category
|
If by "rigid" you mean the same thing as dualizable, an example is given by $\mathcal M = Mod\_{R[G]}$ for some commutative ring $R$ and group $G$, with monoidal structure given by $\otimes\_R$ and the diagonal action.
Then the object $R$ with a trivial $G$-action is the unit, hence it's rigid, but it's rarely projective: $\hom(R,-)$ is isomorphic to $(-)^G$, so it's projective if and only if the latter is exact, if and only if $H^\*(G,-)$ vanishes on $R$-modules - counterexamples are given by $R= \mathbb F\_p, G= $ any finite group with order divisible by $p$.
More generally, note the following : if $P$ is dualizable, with dual $P^\vee$ then $\hom(P,-)\cong \hom(\mathbf 1, P^\vee \otimes -)$, where $\mathbf 1$ is the unit. So there are two ways exactness can fail: $\hom(\mathbf 1,-)$ might not be exact (in which case, the unit is an example anyway), or $P^\vee\otimes -$ might not be exact (these are not necessary/sufficient conditions, because $\hom(\mathbf 1,-)$ might not be conservative either).
Note that in fact, the latter cannot happen if $\mathcal M$ is symmetric monoidal : indeed $P^\vee\otimes -$ is then both a left adjoint (hence right exact) and a right adjoint (hence left exact). I don't know what happens if $\mathcal M$ is just monoidal, because then $P^\vee$ might itself not be dualizable (say I fixed a side and am only referring to "left dualizable")
So in deciding if $P$ is projective, you really only have to check whether the unit $\mathbf 1$ is projective.
Note that it might still happen that the unit is not projective, yet $P$ is. Consider for instance the following : take a counterexample $\mathcal M$ as above, and an ordinary category of modules $\mathcal M'$ over a commutative ring. Then the unit of $\mathcal M\times\mathcal M'$ is not projective (it has the unit of $\mathcal M$ as a summand, and that one is not projective), yet any projective module in $\mathcal M'$ is projective in the product.
|
2
|
https://mathoverflow.net/users/102343
|
405166
| 166,131 |
https://mathoverflow.net/questions/405139
|
6
|
$\DeclareMathOperator\holim{holim}\DeclareMathOperator\hocolim{hocolim}$Let $\mathcal{F}$ be a simplicial (pre)sheaf on some site $\mathcal{C}$ (assume the site has enough stalks; if you like also assume every representable functor on $\mathcal{C}$ is a sheaf). Suppose $\mathcal{G}$ is a (pre)sheaf of groups acting on $\mathcal{F}$. Then we can form a homotopy quotient $\mathcal{F}\_{h\mathcal{G}}$ and a homotopy fixed point sheaf $\mathcal{F}^{h\mathcal{G}}$ via a section wise prescription (this is my naive 'on the spot' construction - I don't have a reference and would like one if it exists for the 'correct' construction):
$\mathcal{F}\_{h\mathcal{G}}(U) = \mathcal{F}(U)\_{h\mathcal{G}} := \hocolim\_{\mathcal{G}(U)} F(U)$,
$\mathcal{F}^{h\mathcal{G}}(U) = \mathcal{F}(U)^{h\mathcal{G}} := \holim\_{\mathcal{G}(U)} F(U)$,
for $U \in \mathcal{C}$.
If $\mathcal{F}\_x$ is a stalk, then we can also form:
$(\mathcal{F}\_x)\_{h\mathcal{G}} := \hocolim\_{\mathcal{G}\_x}\mathcal{F}\_x$,
$(\mathcal{F}\_x)^{h\mathcal{G}} := \holim\_{\mathcal{G}\_x}\mathcal{F}\_x$.
**Question 1**: Is $(\mathcal{F}\_x)\_{h\mathcal{G}} \simeq (\mathcal{F}\_{h\mathcal{G}})\_x$?
**Question 2**: Is $(\mathcal{F}\_x)^{h\mathcal{G}} \simeq (\mathcal{F}^{h\mathcal{G}})\_x$?
In both cases I am worried about commuting colimits (albeit filtrant) with homotopy limits/colimits.
Naively, 1) seems to be ok to me: using the explicit Borel model for the homotopy colimit, it's just the diagonal for a bisimplicial complex and stalks are more or less by definition taken level wise. Regardless, I worry I am missing some subtlety, and am quite lost with the dual homotopy fixed point model in terms of a totalization of a cosimplicial set.
**Question 3**: Is there a decent reference for these constructions for a neophyte for simplicial (pre)sheaves (I am not particularly versed with the subtleties of simplicial methods - my training is in representation theory)? I have Jardine's 'Local Homotopy Theory' which is a godsend for a lot of stuff, but seems to not quite have much along these lines.
**Added later:** In my questions I am implicitly assuming that stalks are given by a filtrant colimit. This is unnecessary for Question 1, but probably is required for Question 2 to have any hope of having an affirmative answer — along with the group being finite (see Dmitri's answer and the comments underneath it).
**Added even later:** Question 1 has been sorted in the affirmative by Dmitri's answer below. Question 2 also has an affirmative answer assuming that the diagram is finite (so say $G$ is a constant presheaf with stalk a finite group). This is Lemma 1.20 in Morel-Voevodsky "$\mathbb{A}^1$-homotopy of schemes". They state it without proof, so I give an elementary one here (leaving the details of a final check out).
**Please see Maxime Ramzi's counterexample in the comments to the accepted answer. Either I am missing something, or the claim in Morel-Voevodsky needs refinement**.
~~Do note, it has absolutely zilch to do with filtrant limits.:~~
The homotopy limit over a finite diagram can be expressed as a limit, over a different, but still finite diagram. Stalks, by definition, commute with finite limits. Hence, all that remains to be done is to check that when we do this commutation the resulting limit (in simplicial sets) is precisely the homotopy limit of the stalk (or that the canonical map between the two simplicial sets obtained is a weak equivalence).
In fact, essentially the same proof should also work for an arbitrary site using pullback to a Boolean localization (but I don’t really understand those, so…).
P.S. In all honesty I have not checked the last `check’. However, at this point I am impatient enough to give Morel-Voevodsky the benefit of the doubt (given that they use the cited result everywhere and it is a landmark paper).
|
https://mathoverflow.net/users/392998
|
Homotopy quotients, fixed points and stalks of simplicial (pre)sheaves
|
Taking stalks always commutes with taking homotopy orbits,
since filtered colimits of simplicial sets are also filtered homotopy colimits, and homotopy colimits commute with homotopy colimits.
Taking stalks commutes with taking homotopy fixed points
with respect to a finite group
only if additional fibrancy conditions are satisfied,
as explained in [another answer](https://mathoverflow.net/questions/287091/why-do-we-need-model-categories/287234#287234).
Without additional conditions, there are counterexamples.
Concerning references, there are not so many accessible ones.
Perhaps Dugger's [A primer on homotopy colimits](https://pages.uoregon.edu/ddugger/hocolim.pdf) may be helpful.
Bousfield and Kan in their book “Homotopy Limits,
Completions and Localizations” prove in §XII.3.5 that filtered colimits of simplicial sets are homotopy colimits.
|
3
|
https://mathoverflow.net/users/402
|
405175
| 166,132 |
https://mathoverflow.net/questions/316614
|
5
|
Let $f = (f\_1, \dotso, f\_n):\mathbb{R}^n \to \mathbb{R}^n$ be a smooth map and let $J$ be its Jacobian (determinant of the matrix with $ij$-th entry $\partial\_i f\_j$). We introduce the zero sets of $J$ and its derivatives
$$Z\_0 = J^{-1}(0), \quad Z\_1 = Z\_0 \cap (\nabla J)^{-1}(0), \quad \dotso$$
Here we define $Z\_k = Z\_{k - 1} \cap (\nabla^k J)^{-1}(0)$ for $k \geq 1$; $\nabla^k$ denotes the operation of taking $k$-th order derivatives. My question is: what are the generic properties of $J$? Do we have $Z\_1 = \emptyset$, $Z\_2 = \emptyset$ or $Z\_k = \emptyset$ for some $k$, generically, and what are the relations between $k$ and $n$? Do we have for all $n$, that $Z\_k = \emptyset$ generically for some $k$? It is fine to work on a compact set and consider generic properties on it, so $f: K \subset \mathbb{R}^n \to \mathbb{R}^n$ for some compact $K$.
A certain property is generic, if it holds on an open and dense set of $C^\infty(\mathbb{R}^n; \mathbb{R}^n)$ of maps between $\mathbb{R}^n \to \mathbb{R}^n$.
For example, for $n = 2$, Whitney [1] showed that generically we have $Z\_1 = \emptyset$. His idea was to perturb $f$ by polynomials in parameter families and prove that the relation on $Z\_k$ is large enough codimension. For $n = 3$, I believe we have the same result (computation). For $n \geq 4$ the set $Z\_1$ is too big and we need to include $Z\_2$, but I don't know counterexamples to this; the calculations also become increasingly difficult. In [2, section 1], there are some results related to stability/instability of maps by Thom, which need sufficient dimension.
[1] Whitney, Hassler
On singularities of mappings of euclidean spaces. I. Mappings of the plane into the plane.
Ann. of Math. (2) 62 (1955), 374–410.
[2] Arnolʹd, V. I. Singularity theory. Selected papers. Translated from the Russian. With an introduction by C. T. C. Wall. London Mathematical Society Lecture Note Series, 53. Cambridge University Press, Cambridge-New York, 1981.
|
https://mathoverflow.net/users/84963
|
Generic properties of Jacobians of smooth functions
|
This is answered in the paper <https://link.springer.com/content/pdf/10.1007/s00526-020-01740-6.pdf> (OA), Lemma 6.13; this Lemma is proved in the Appendix of the paper. The final answer is that one needs to take $k(n) = \left \lceil{n + 1}\right \rceil $ derivatives to obtain that generically $Z\_{k(n) - 1} = \emptyset$. The ideas used are the ones from the Whitney [1] reference, so the perturbations are by polynomials of certain degree; there is however some non-trivial linear algebra to deal with.
|
1
|
https://mathoverflow.net/users/84963
|
405204
| 166,141 |
https://mathoverflow.net/questions/405174
|
1
|
Let $m\geq 2$ be a fixed integer.
Let
$$f(n):=\begin{cases}
mf\left(\frac{n}{m}\right),&\text{if $n\mod m = 0$;}\\
1,&\text{otherwise}
\end{cases}$$
then if we have
$$a(n):=\begin{cases}
1,&\text{if $n=0$;}\\
a\left(\frac{n}{m}\right)+a\left(n-f\left(\frac{n}{m}\right)\right),&\text{if $n\mod m = 0$;}\\
a\left(\left\lfloor\frac{n}{m}\right\rfloor\right),&\text{otherwise}
\end{cases}$$
and also
$$s(n):=\sum\limits\_{k=0}^{m^n-1}a(k).$$
In particular, $s(0) = 1$ and $s(1) = m$.
I conjecture that for $m > 2$
$$s(n)=(m+3)s(n-1)-(2m+1)s(n-2).$$
For a case $m=2$ we get Bell numbers.
Is there a way to prove it?
Similar questions:
* [Sum with Stirling numbers of the second kind](https://mathoverflow.net/questions/406738/sum-with-stirling-numbers-of-the-second-kind)
* [Pair of recurrence relations with $a(2n+1)=a(2f(n))$](https://mathoverflow.net/questions/406902/pair-of-recurrence-relations-with-a2n1-a2fn)
* [Sequence that sums up to INVERTi transform applied to the ordered Bell numbers](https://mathoverflow.net/questions/407290/sequence-that-sums-up-to-inverti-transform-applied-to-the-ordered-bell-numbers)
* [Sequences that sums up to second differences of Bell and Catalan numbers](https://mathoverflow.net/questions/407758/sequences-that-sums-up-to-second-differences-of-bell-and-catalan-numbers)
|
https://mathoverflow.net/users/231922
|
Recurrence for the sum
|
Let $n=m^tk$ where $m\nmid k$. Then $f(n)=m^t$.
Furthermore, if $t>0$, then $f(n/m)=m^{t-1}$ and $n-f(n/m)=m^{t-1}(mk-1)$. It follows that $a(n)=a(m^{t-1}k)+a(m^{t-1}(mk-1))$ and further by induction on $t$,
$$
(\star)\qquad a(n)=\sum\_{i=0}^t \binom{t}{i} a\big(m^ik-\frac{m^i-1}{m-1}\big).
$$
---
**CASE $m>2$.** In this case, formula $(\star)$ reduces to
$$a(n) = a(k)+(2^t-1)a(k-1).$$
Let's analyze $s(n)$.
It is clear that for any $\ell\not\equiv 0\pmod{m}$, we have
$$\sum\_{k=0\atop k\equiv \ell\pmod{m}}^{m^n-1} a(k) = s(n-1)$$
and correspondingly
$$\sum\_{k=0\atop k\equiv 0\pmod{m}}^{m^n-1} a(k) = s(n) - (m-1)s(n-1)$$
Now we are ready to derive a recurrence for $s(n)$ by grouping the summation indices based on the power $m^t$ they contain:
\begin{split}
s(n) &= 1 + \sum\_{t=0}^{n-1} \sum\_{k=1\atop k\not\equiv 0\pmod{m}}^{m^{n-t}-1} \left(a(k) + (2^t-1)a(k-1)\right) \\
&=1 +\sum\_{t=0}^{n-1} \left((m-1)s(n-t-1) + (2^t-1)((m-2)s(n-t-1) + s(n-t)-(m-1)s(n-t-1))\right) \\
&=1 +\sum\_{t=0}^{n-1} \left((m-2^t)s(n-t-1) + (2^t-1)s(n-t)\right) \\
&=2 - 2^n + \sum\_{t=1}^n (2^{t-1}+m-1)s(n-t).
\end{split}
Restating the above recurrence in terms of the generating function $S(x):=\sum\_{n\geq 0} s(n)x^n$, we have
$$S(x) = \frac{2}{1-x} - \frac{1}{1-2x} + \left(\frac{x}{1-2x} + \frac{(m-1)x}{1-x}\right)S(x).$$
That is,
$$S(x) = \frac{1-3x}{1 - (m+3)x + (2m+1)x^2},$$
from where the required recurrence follows instantly.
---
**CASE $m=2$.** In this case, formula $(\star)$ takes form:
$$a(n)=a(k)+\sum\_{i=1}^t \binom{t}{i} a(2^{i-1}(k-1)).$$
It further follows that for $n=2^{t\_1}(1+2^{1+t\_2}(1+\dots(1+2^{1+t\_\ell}))\dots)$ with $t\_j\geq 0$, we have
\begin{split}
a(n) &= \sum\_{i\_1=0}^{t\_1} \binom{t\_1}{i\_1} \sum\_{i\_2=0}^{t\_2+i\_1} \binom{t\_2+i\_1}{i\_2} \sum\_{i\_3=0}^{t\_3+i\_2} \dots \sum\_{i\_\ell=0}^{t\_\ell+i\_{\ell-1}} \binom{t\_\ell+i\_{\ell-1}}{i\_\ell} \\
&=\prod\_{j=1}^\ell (\ell+2-j)^{t\_j}.
\end{split}
Grouping the summands in $s(n)$ by the number of unit bits, we have
\begin{split}
s(n) &= \sum\_{\ell=0}^n\ \sum\_{t\_1+t\_2+\dots+t\_{\ell}\leq n-\ell}\ \prod\_{j=1}^\ell (\ell+2-j)^{t\_j}\\
&= \sum\_{\ell=0}^n\ \sum\_{t\_1+t\_2+\dots+t\_{\ell}+t\_{\ell+1} = n-\ell}\ \prod\_{j=1}^{\ell+1} (\ell+2-j)^{t\_j}\\
&=\sum\_{\ell=0}^n [x^{n-\ell}]\ \prod\_{j=1}^{\ell+1} \frac1{1-jx} \\
&=\sum\_{\ell=0}^n S(n+1,\ell+1) \\
&=B\_{n+1},
\end{split}
where $S(n+1,\ell+1)$ are Stirling numbers of second kind, and $B\_{n+1}$ is Bell number.
|
4
|
https://mathoverflow.net/users/7076
|
405210
| 166,145 |
https://mathoverflow.net/questions/405098
|
1
|
Let $\boldsymbol{X} = (X\_1,X\_2)^{\rm T}\sim \mathcal{N}\_2(\boldsymbol{\mu}, \mathrm{\Sigma})$, where
\begin{eqnarray\*}
\boldsymbol{\mu} = (\mu\_1, \mu\_2)^{\rm T}& = &(\sqrt{\xi\_1\xi\_2/(\xi\_1+\xi\_2)}, 0)^{\rm T}\\
\mathrm{\Sigma} & = &\begin{pmatrix} 1 & -\rho\\
-\rho & 1\end{pmatrix}\\
\rho & = & \sqrt{\xi\_1\xi\_3/(\xi\_1+\xi\_2)(\xi\_2+\xi\_3)}.
\end{eqnarray\*}
It is given that $\xi\_1\leq\xi\_2\leq\xi\_3$, $\xi\_i\geq 0$ and $\sum\_{i=1}^3\xi\_i = 1$. I have a function
\begin{equation\*}
\pi(\boldsymbol{\mu};\boldsymbol{\xi}) = 1-\mathbb{P}(\boldsymbol{X}\leq \boldsymbol{t}) = 1-\mathbb{P}(X\_1\leq t \cap X\_2\leq t),
\end{equation\*}
where $t>1$. Under the constraints $\xi\_1\leq\xi\_2\leq\xi\_3$, $\xi\_i\geq 0$ and $\sum\_{i=1}^3\xi\_i = 1$, numerically we are getting the maximum of $\pi(\boldsymbol{\mu};\boldsymbol{\xi})$ where the non-zero mean $\mu\_1$ is maximized. This occurs when $\xi\_1 = \xi\_2 = \xi\_3 = 1/3$.
Can we use some kind of stochastic ordering arguments to prove the result theoretically?
|
https://mathoverflow.net/users/120111
|
A problem related to bivariate normal stochastic order
|
$\newcommand{\der}{\mathrm{der}}\newcommand{\tdert}{\mathrm{dert}}\newcommand{\erf}{\operatorname{erf}}\newcommand{\eqs}{\overset{\text{sign}}=}\newcommand{\tder}{\widetilde\der}$The statement about the maximum is true, but the proof has hardly anything to do with stochastic ordering arguments. Rather, the problem is reduced to certain problems of real algebraic geometry -- that is, to solving (rather complicated) systems of algebraic inequalities over $\mathbb R$.
Let $x:=\xi\_1$ and $y:=\xi\_2$, so that $\xi\_3=1-x-y$, $0\le x\le y\le1-x-y$,
$$\mu\_1=M(x,y):=\sqrt{\frac{x y}{x+y}},$$
$$\rho=-R(x,y),\quad R(x,y):=\sqrt{\frac{x(1-x-y)}{(x+y)(1-x)}}.$$
Let also
$$p\_t(x,y):=1-P(X\_1\le t,X\_2\le t)$$
and
\begin{equation\*}
G:=\{(x,y)\colon 0\le x\le y\le1-x-y\}=\{(x,y)\colon 0\le x\le1/3,x\le y\le(1-x)/2\}. \tag{-1}
\end{equation\*}
We want to show that
\begin{equation\*}
p\_t(x,y)\le p\_t(1/3,1/3) \tag{0}
\end{equation\*}
for all $(x,y)\in G$ and all real $t\ge1$.
Note that
\begin{equation\*}
p\_t(x,y):=1-P(X\_1\le t,X\_2\le t)=P(M(x,y),-R(x,y)), \tag{1}
\end{equation\*}
where
\begin{equation\*}
P(m,r):=1-\int\_{-\infty}^t du \int\_{-\infty}^t dv\, f\_{m,r}(u,v)
\end{equation\*}
and $f\_{m,r}$ is the density function of the bivariate normal distribution with means $m,0$, variances $1,1$, and correlation $r$.
The key is Plackett's observation ([formula (3)](https://www.jstor.org/stable/2332716)) that
\begin{equation\*}
D\_r f\_{m,r}(u,v)=D\_v D\_u f\_{m,r}(u,v),
\end{equation\*}
where $D\_w$ denote the partial derivative with respect to a variable $w$. It follows that
\begin{equation\*}
\begin{aligned}
D\_r P(m,r)&:=-\int\_{-\infty}^t du \int\_{-\infty}^t dv \, D\_r f\_{m,r}(u,v) \\
&=-\int\_{-\infty}^t du \int\_{-\infty}^t dv \, D\_v D\_u f\_{m,r}(u,v) \\
&=-f\_{m,r}(t,t)<0.
\end{aligned}
\tag{2}
\end{equation\*}
Next,
\begin{equation\*}
\begin{aligned}
P(m,r)&=1-\int\_{-\infty}^t du \int\_{-\infty}^t dv\, f\_{0,r}(u-m,v) \\
&=1-\int\_{-\infty}^{t-m} dw \int\_{-\infty}^t dv\, f\_{0,r}(w,v)
\end{aligned}
\end{equation\*}
and hence
\begin{equation\*}
\begin{aligned}
D\_m P(m,r)=\int\_{-\infty}^t dv\, f\_{0,r}(t-m,v)=\int\_{-\infty}^t dv\, f\_{m,r}(t,v)>0.
\end{aligned}
\tag{3}
\end{equation\*}
Also, for $(x,y)\in G\setminus\{(0,0)\}$,
\begin{equation\*}
(D\_x M(x,y),D\_y M(x,y))=\frac{(y^2,x^2)}{(x + y)^2},
\end{equation\*}
\begin{equation\*}
(D\_x R(x,y),D\_y R(x,y))=\frac{((1-2 x-y)y,-(1-x) x)}{(1-x)^2 (x + y)^2},
\end{equation\*}
so that $D\_x M(x,y)\ge0$ and $D\_x R(x,y)\ge0$.
So, by (1), a chain rule of differentiation, and the inequalities in (2) and (3),
\begin{equation\*}
D\_x p\_t(x,y)=D\_m P(M(x,y),-R(x,y))D\_x M(x,y)-D\_r P(M(x,y),-R(x,y))D\_x R(x,y)\ge0.
\end{equation\*}
So, the maximum of $p\_t(x,y)$ over $(x,y)\in G$ occurs at one of right boundaries of $G$, that is, where either $y=x$ and $y=(1-x)/2$.
Next, for $x\in[0,1/3]$ (cf. (-1)),
\begin{equation\*}
\frac d{dx}\,M(x,(1-x)/2)=\frac{1-2x-x^2}{(1+x)^2}>0,
\end{equation\*}
\begin{equation\*}
\frac d{dx}\,R(x,(1-x)/2)=\frac1{(1+x)^2}>0.
\end{equation\*}
So, by (1) and the inequalities in (2) and (3), $p\_t(x,(1-x)/2)$ is increasing in $x\in[0,1/3]$.
It remains to show that
\begin{equation\*}
\der(t,x)\overset{\text{(?)}}\ge0 \text{ for $x\in[0,1/3]$ and $t\ge1$,} \tag{4}
\end{equation\*}
where
\begin{equation\*}
\begin{aligned}
\der(t,x)&:=8 \sqrt{\pi } \sqrt{x}\, e^{\left(\sqrt{2} \sqrt{x}-2 t\right)^2/8}\, D\_x p\_t(x,x) \\
&=1+\erf\left(v(t,x)/2\right)-\frac{2
}{\sqrt{\pi } (1-x)}\,\sqrt{\frac{x}{1-2 x}}\,e^{u(t,x)/4}, \\
u(t,x)&:=t^2 \left(-4 \sqrt{2} \sqrt{\frac{1-2x}{1-x}} (1-x)+8 x-6\right) \\
&+2 t \left(-2
\sqrt{2} x^{3/2}+2 (1-x) \sqrt{\frac{1-2x}{1-x}} \sqrt{x}+\sqrt{2}
\sqrt{x}\right)+x (2 x-1), \\
v(t,x)&:=\sqrt{1-x} \left(\sqrt{\frac{1-2x}{1-x}} \left(\sqrt{2} t-\sqrt{x}\right)+2
t\right).
\end{aligned}
\end{equation\*}
By a chain rule of differentiation,
\begin{equation\*}
D\_x p\_t(x,x)=D\_m P(M(x),-R(x))M'(x)-D\_r P(M(x),-R(x))R'(x),
\end{equation\*}
where
\begin{equation\*}
M(x):=M(x,x),\quad R(x):=R(x,x).
\end{equation\*}
Using (2) and (3), we get
\begin{equation\*}
\der(t,x)=A\_1(x)e^{A\_2(t,x)}+1+\erf(A\_3(t,x));\tag{4.5}
\end{equation\*}
here and in what follows, $A\_1,A\_2,\dots$ are certain algebraic functions and $\erf$ is the error function.
So,
\begin{equation\*}
D\_t\der(t,x)
=A\_4(t,x)e^{A\_2(t,x)}+A\_5(t,x)e^{-A\_3(t,x)^2/2}. \tag{5}
\end{equation\*}
thus getting rid of the function $\erf$. Moreover,
\begin{equation\*}
A\_5(t,x)>0
\end{equation\*}
(for $t$ and $x$ as in (4)).
Letting now
\begin{equation\*}
\tdert(t,x):=\frac{D\_t\der(t,x)}{A\_5(t,x)e^{-A\_3(t,x)^2/2}},
\end{equation\*}
we get
\begin{equation\*}
D\_t\der(t,x)\eqs\tdert(t,x)=A\_6(t,x)e^{A\_7(t,x)}+1, \tag{6}
\end{equation\*}
where $a\eqs b$ means that $a$ and $b$ are of the same sign; this reduces the two exponential expressions $e^{A\_2(t,x)}$ and $e^{-A\_3(t,x)^2/2}$ in (5) to one such expression.
Now we have
\begin{multline\*}
D\_t\tdert(t,x)=e^{A\_7(t,x)}(1-x)
\sqrt{\frac{(1-2x)(1-x)}{x}} \Big(2+\sqrt2\,\sqrt{\frac{1-2x}{1-x}}\,\Big) \\
\times \big(3-4 x+2 \sqrt{2} \sqrt{(1-2 x) (1-x)}\,\big)>0
\end{multline\*}
(for $x\in(0,1/3]$).
Also,
\begin{equation\*}
\tdert(0,x)
=\frac{(1-2 x) \left(2-2 x+\sqrt{2} \sqrt{(1-2 x) (1-x)}\right)}{\left(2+\sqrt{2}
\sqrt{2+\dfrac{1-2 x}{1-x}}\right) (1-x)^2}>0.
\end{equation\*}
So, $\tdert(t,x)>0$ for all $t\ge0$ and all $x\in(0,1/3]$.
So, by (6), it is enough to show that
\begin{equation\*}
\der\_1(x):=\der(1,x)\overset{\text{(?)}}\ge0 \tag{7}
\end{equation\*}
for all $x\in(0,1/3]$.
Using (4.5) and some algebra, we get
\begin{equation\*}
\der\_1(x)=1-\frac{2/\sqrt\pi}{1-x}\,\sqrt{\frac x{1-2x}}\,e^{p(x)/4}+\erf(q(x)),
\end{equation\*}
where
\begin{equation\*}
p(x):=-6+2 \sqrt{2} \sqrt{x}+7 x-4 \sqrt{2} x^{3/2}+2 x^2+\sqrt{\frac{1-2 x}{1-x}} \left(-4 \sqrt{2}+4
\sqrt{x}+4 \sqrt{2} x-4 x^{3/2}\right),
\end{equation\*}
\begin{equation\*}
q(x):=\frac{1}{2} \sqrt{1-2 x} \left(\sqrt{2}-\sqrt{x}\right)+\sqrt{1-x}.
\end{equation\*}
Let
\begin{equation\*}
\der\_{11}(x):=\der\_1'(x)2\sqrt\pi\,
\frac{(1-2 x) (1-x)^2 \sqrt{(1-2x)x} }
{p\_1(x)e^{p(x)/4}},
\end{equation\*}
where
\begin{equation\*}
\begin{aligned}
p\_1(x)&:=(1-2x)(1-x) x p'(x)+2+2x-8 x^2 \\
&=2 + 2 x - 8 x^2 +
x (1 - 2 x) ((7 + \sqrt2/\sqrt{x} - 6 \sqrt2 \sqrt{x} + 4 x) (1 -
x) +p\_2(x)),
\end{aligned}
\end{equation\*}
\begin{equation\*}
p\_2(x):=2 (\sqrt2 - \sqrt{x})/
s(x) + (4 \sqrt2 + 2/\sqrt{x} - 6 \sqrt{x}) (1 - x) s(x),
\end{equation\*}
\begin{equation\*}
s(x):=\sqrt{\frac{1-2 x}{1-x}}.
\end{equation\*}
It is elementary to show that $p\_2(x)\ge5$ (for $x\in(0,1/3]$), and hence
\begin{equation\*}
\begin{aligned}
p\_1(x)&\ge2 + 2 x - 8 x^2 +
x (1 - 2 x) ((7 + \sqrt2/\sqrt{x} - 6 \sqrt2 \sqrt{x} + 4 x) (1 -
x) +5)>0.
\end{aligned}
\end{equation\*}
So,
\begin{equation\*}
\der\_{11}(x)\eqs\der\_1'(x).
\end{equation\*}
Next, let
\begin{equation\*}
\der\_{111}(x):=\der\_{11}'(x)
\sqrt{(1-2x)x} \, e^{p(x)/4+q(x)^2} \, p\_1(x)^2 =\tder\_{111}(\sqrt x),
\end{equation\*}
where
\begin{equation\*}
\begin{aligned}
\tder\_{111}(z)&:=
-3 \sqrt{2}+50 z+56 \sqrt{2} z^2-282 z^3-61 \sqrt{2} z^4 \\
&+392 z^5-284 \sqrt{2} z^6+248 z^7+540
\sqrt{2} z^8 \\
&-832 z^9-160 \sqrt{2} z^{10}+384 z^{11}-64 \sqrt{2} z^{12} \\
&-2 \sqrt{\frac{1-2z^2}{1-z^2}} \\
&\times\big(1-6 \sqrt{2} z-32 z^2+48 \sqrt{2} z^3 +33 z^4 \\
& -106 \sqrt{2} z^5 +136 z^6+48\sqrt{2} z^7-230 z^8 \\
&+80 \sqrt{2} z^9+48 z^{10}-64 \sqrt{2} z^{11}+32 z^{12}\big).
\end{aligned}
\end{equation\*}
The function $\tder\_{111}$ switches its sign only once on the interval $(0,1/\sqrt3]$, from $-$ to $+$, and hence $\der\_{11}'$ has the same sign pattern on the interval $(0,1/3]$. So, for some $x\_\*\in(0,1/3)$, the function $\der\_{11}$ is decreasing on $(0,x\_\*]$ and increasing on $[x\_\*,1/3]$. Also, $\der\_{11}(0+)=-3/2<0$ and $\der\_{11}(1/3)=-1.11\ldots<0$. So, $\der\_{11}<0$ on $(0,1/3]$. So, $\der\_1$ is decreasing on $(0,1/3]$, to $\der\_1(0)=1.31\ldots>0$.
Thus, (7) is proved. $\quad\Box$
|
2
|
https://mathoverflow.net/users/36721
|
405215
| 166,146 |
https://mathoverflow.net/questions/405212
|
4
|
Let $Y$ be a space, and $G$ a group. For simplicity we can take $G$ to be finite, $Y$ to be a point, if this avoids technical issues. Then for $X/Y$ a $G$ torsor, so a sheaf of sets with $G$ action on $X$, locally isomorphic to the sections of $G\times U \xrightarrow{\pi} U$, we can consider the automorphisms of this map, in the category of $G$ torsors.
This will be a sheaf of groups on $Y$, and if we choose a section over $U$, then we get base points of all of our fibres, so we get a local isomorphism with the constant sheaf of groups with value $G$. Then changing our section changes our isomorphism, by conjugation.
My question is then, given a sheaf of groups $G$, how can we tell whether it comes from a $G$ torsor in this manner? Or, what gadget parametrises families of these objects, where we know these groups are locally isomorphic to $G$, only canonically up to conjugation. For instance, if $G=A$ is abelian, then this automorphism sheaf will always be a constant sheaf with group $A$.
|
https://mathoverflow.net/users/128502
|
What extra structure does the group of automorphisms of a torsor carry?
|
I am not sure what kind of criterion you are after. You can give an answer in terms of cocycles or an answer in terms of killing an obstruction. Briefly, if $\mathcal{G}$ is a bundle of groups on $Y$ which is locally isomorphic to the trivial bundle of groups $\underline{G}$ with fiber $G$, then the first obstruction to $\mathcal{G}$ being the bundle of automorphisms of a $G$ torsor is that $\mathcal{G}$ is an inner twist of $\underline{G}$. Concretely this means the following. Consider the frame bundle of
$\mathcal{G}$, i.e. the bundle of fiberwise group isomorphisms
$Isom(\mathcal{G},\underline{G})$. This frame bundle is manifestly a right $Aut(G)$-torsor (via postcomposition) and the condition is that the structure group of $Isom(\mathcal{G},\underline{G})$ reduces from $Aut(G)$ to the subgroup $Inn(G) \subset Aut(G)$ of inner automorphisms of $G$. Concretely this means that if you look at the assoiciated $Out(G)$-torsor $Isom(\mathcal{G},\underline{G})/Inn(G)$, then this $Out(G)$-torsor is trivializable. Choosing a trivialization gives you an $Inn(G)$ torsor $P$, such that $Isom(\mathcal{G},\underline{G})$ is associated with $P$ via the injective homomorphism $Inn(G) \to Aut(G)$. Once this necessary condition is satisfied and $P$ is chosen ($P$ is well defined up to isomorphism), then you have a second condition which says that the $Inn(G)$-torsor $P$ lifts to a $G$-torsor $X$. From the short exact sequence
$$
1 \to Z(G) \to G \to Inn(G) \to 1,
$$
you see that the obstruction to the existence of such a lift is a degree two cohomology class $\alpha(P) \in H^{2}(Y,Z(G))$ with values in the center $Z(G)$ of $G$. The class $\alpha(P)$ is sometimes called the Brauer class of $P$ and if you choose e.g. a fine enough trivializing cover of $P$ and bounding cochain for the Cech cocycle corresponding to $\alpha(P)$ you can construct a $G$-torsor lifting $P$ explicitly.
|
4
|
https://mathoverflow.net/users/439
|
405218
| 166,147 |
https://mathoverflow.net/questions/405227
|
4
|
I am from physics background so I apologize in advance if my question is trivial.
Kojima proves for every finite group $G$, there is a hyperbolic 3-manifold such that its mapping class group equals $G$ ([here](https://core.ac.uk/download/pdf/82773183.pdf)). I wonder if it is possible to work out an ideal triangulation for the manifold he constructs.
Some examples of hyperbolic 3-manifold with explicit ideal triangulation and with some simple non-trivial mapping class group would also be very helpful.
|
https://mathoverflow.net/users/394173
|
Ideal triangulation of hyperbolic 3-manifold with generic mapping class group
|
The hyperbolic manifolds considered by Kojima are closed, so do not admit ideal triangulations in the "usual sense". (They admit "partially flat, spun, ideal triangulations" but I am not sure that is what you are interested in.)
It is an open question whether or not every finite volume cusped hyperbolic three-manifold admits an ideal triangulation (with all tetrahedra positively oriented).
Finally, the standard example of a manifold admitting an ideal triangulation (with all tetrahedra positively oriented) is the figure-eight knot complement. This manifold, called m004 in the snappy census, has mapping class group isomorphic to $D\_4$. Taking normal covers of m004 gives examples with more symmetry.
See the book [Hyperbolic knot theory](https://bookstore.ams.org/cdn-1598077516917/gsm-209), by Jessica Purcell, as reference on this material.
|
3
|
https://mathoverflow.net/users/1650
|
405230
| 166,152 |
https://mathoverflow.net/questions/405196
|
20
|
Recall that $A(X)$, the K-theory of a connected, pointed space X, is defined as the K-theory spectrum of the ring spectrum $\Sigma^\infty\_+ \Omega X$ (or via a plethora of alternative definitions). Is it known if the homotopy type of $A(X)$ determines the homotopy type of $X$? If not, what is the best one can hope for?
Of course, since $X$ is connected the space $\Omega X$ with its loop space structure determines the homotopy type of $X$, but I am not sure if this is still true when we take $\Sigma^\infty\_+$, I am worried we get might get $X$ only up to $\Sigma^n \Omega^n $. Then there is the question of if ring spectra of this type can have the same K-theory, perhaps we should assume $X$ simply connected to get a positive answer?
|
https://mathoverflow.net/users/134512
|
Does Waldhausen K-theory detect homotopy type?
|
The answer to the question
>
> Does the homotopy type of $()$ determine the homotopy type of $$?
>
>
>
is No in general. As you say, $A(X)$ is determined by the homotopy type of $\Sigma^\infty \Omega X\_+$ as an associative (or $A\_\infty$) ring spectrum, and this ring spectrum does not uniquely determine $X$, even if $X$ is simply-connected.
For example, suppose $Z$ is a pointed space. Let $T(Z)$ be the free associative $S$-algebra generated by $Z$. I.e.,
$$T(Z)=\Sigma^\infty S^0\vee \Sigma^\infty Z \vee (\Sigma^\infty Z)^{\wedge 2} \vee \cdots .$$
If $Z$ is connected, there is an equivalence of associative ring spectra
$$ T(Z) \simeq \Sigma^\infty \Omega\Sigma Z\_+.$$
The equivalence is a version of the classical James splitting. It is induced by a map of spectra $\Sigma^\infty Z \to \Sigma^\infty \Omega\Sigma Z\_+$, extended to a map of ring spectra $T(Z)\to \Sigma^\infty \Omega\Sigma Z\_+$ using freeness.
It follows that if $X$ and $Y$ are connected spaces such that $\Sigma X$ and $\Sigma Y$ are not equivalent, but $\Sigma^\infty X$ and $\Sigma^\infty Y$ are equivalent, then there is an equivalence $A(\Sigma X)\simeq A(\Sigma Y)$ providing a counterexample.
A couple of comments:
* It is well-known that there exist non-isomorphic groups $G$ and $H$ such that the group rings $\mathbb Z[G]$ and $\mathbb Z[H]$ are isomorphic. There are even examples with finite $G$ and $H$. One may wonder if for some of these examples the spherical group rings $\Sigma^\infty G\_+$ and $\Sigma^\infty H\_+$ are equivalent as associative ring spectra. If yes, then $BG$ and $BH$ would provide another counterexample.
* In general one can have non-equivalent ring spectra that have equivalent $K$-theories. For example, I believe that if $P$ and $Q$ are Morita equivalent in a suitable sense, then $K(P)\simeq K(Q)$. Can there be two spaces $X$ and $Y$ such that $\Sigma^\infty \Omega X\_+$ and $\Sigma^\infty \Omega Y\_+$ are not equivalent as ring spectra, but have equivalent categories of modules (in a strong enough sense to induce equivalence of $K$-theories)? It seems far fetched, but I don't know how to exclude this possibility. **Added later**: [A paper by Roggenkamp and Zimmerman](https://pdf.sciencedirectassets.com/271593/1-s2.0-S0022404900X00230/1-s2.0-002240499500121C/main.pdf?X-Amz-Security-Token=IQoJb3JpZ2luX2VjEIP%2F%2F%2F%2F%2F%2F%2F%2F%2F%2FwEaCXVzLWVhc3QtMSJHMEUCIQDPvFmhYCSH3uV39DkXovveN%2BOyDaQDB6Z%2F1kR3sS000gIgWfYmxwaYC3yWTlyGjTWWxBqI584z8DzOVDzIJUJcaEkqgwQI6%2F%2F%2F%2F%2F%2F%2F%2F%2F%2F%2FARAEGgwwNTkwMDM1NDY4NjUiDBSq2vFuLcecgiZDLirXAxDdL61ZrE4CZZcDfZvm6EeIjuIa1QUcLQpVO%2FnmGP4LboOLwHrmdToaRfYdRJsZKz1HYhCQ7vIcxrVZs7qfObCO9tCadIM8ROh1oIDyhbLkNRW3C6qwBnGQkOKTVGv9DLNCbszGXpbIDGk3%2FcPbPjo1n61XEGdjxcghedRSvYRAB2RRiGxX3ztHT42JLHZ0OJQPE2Wajhu7lI%2FYZeaLOXrZpKtTEVxUuaYa5CkHYaGJz590AaiAU1ayM%2FfrxjmJ%2FW9Xl2ninYRjDhz6Um7DTQPL1jL4Jr10gmSW6Sq9nWJSIhTKrN7GW49zKpJ%2BoVdD0ADeQB2DoE9Wh%2BgrEU%2BTZUKSRiDTRG8Hb7dy5dQO5Q4WsGqBGqkkrsmhU9u2vA6gVrKmhhkoeGA5PZZcHH32%2FWJJJr8K4IAfoDsqlZ21ycgEbb%2BncWPh9a8bGWLrU9i31vorMCaCvXtd21IEs8E8LX7qGXfZIs1L58ujJf8%2F0PyC8L%2FtxP%2FxB1DbgfxQU%2FJGAMuAdzTZ1mwxuBjgkPOSGmrerMyaOl8PyDWLlQghHOcOjHwRvNApCWmBM6SndIApBMoGaja16zQCUuZj4dvw9R8yyNt8Nt6JcK2eVai65y29JERsXqr3GDD7utuKBjqlAevKM%2Budwrs6ydAxjMs72xYyZttDS0qr4tyI0u7b8Pldqv2v1MqbxlbogF1m7Fj%2FoPhQ0r8QbHfCYP1YHGBCZzMiBzHUIGKWQT2bkVevTkSWhdueB80ZcVAVe5rIWd8%2FBHBu1jRpHOXi2uJQRWO3lFHkHnQvVpoE%2BrMvSS9%2Fut%2B75YFa31jvvM30fN5p68pVEeUP%2F84Bc0gHIiiXiiI8gfEs35g%2BKg%3D%3D&X-Amz-Algorithm=AWS4-HMAC-SHA256&X-Amz-Date=20211001T102935Z&X-Amz-SignedHeaders=host&X-Amz-Expires=300&X-Amz-Credential=ASIAQ3PHCVTYYFQ4EQ6V%2F20211001%2Fus-east-1%2Fs3%2Faws4_request&X-Amz-Signature=24c11fc3ebd73feb971192d31deb20e90feac3f06e41b8a1184a48754f9c0a91&hash=44d93e5ecbbfbf4421a06fd1b82235a6df599e452d5f327646c6c2b115f5f87d&host=68042c943591013ac2b2430a89b270f6af2c76d8dfd086a07176afe7c76c2c61&pii=002240499500121C&tid=spdf-f84bc980-2baf-4830-a450-6a9fdfd04fa3&sid=912d2868500e4540d55b51c-0b6669052cf9gxrqb&type=client) gives an example of two groups $G$ and $H$ for which the rings $\mathbb Z[G]$ and $\mathbb Z[H]$ are not isomorphic, but Morita equivalent. It follows that the Quillen $K$-theory of these rings is isomorphic. One may ask whether the $K$-theory spectra of $\Sigma^\infty G\_+$ and $\Sigma^\infty H\_+$ are equivalent as well.
|
23
|
https://mathoverflow.net/users/6668
|
405235
| 166,154 |
https://mathoverflow.net/questions/405238
|
0
|
Let $E$ be a Banach space, $\mathfrak{M}\_E$ indicate the family of all nonempty bounded subset of $E$, $\mathfrak{N}\_E$ the family of all relatively compact sets, and $Ker \mu=\{X\in \mathfrak{M}\_E$ such that: $\mu(X)=0\}$.
>
> **"Standard" Definition:**
> A mapping $\mu:\mathfrak{M}\_E\rightarrow \mathbb R^+$ is said to be a [**measure of noncompactness**](https://dml.cz/bitstream/handle/10338.dmlcz/105982/CommentatMathUnivCarol_021-1980-1_10.pdf) (shortly, MNC) in $E$ if it satisfies the following conditions (for any $X,Y,X\_n \in \mathfrak{M}\_E$ and $a\in E$):
>
>
> 1. The family $\operatorname{ker} \mu $ is nonempty and $\operatorname{ker} \mu \subset \mathfrak{N}\_E$.
> 2. $X\subseteq Y\Rightarrow \mu (X)\leq \mu (Y)$.
> 3. $\mu (\bar{X})=\mu (X)$.
> 4. $\mu (\operatorname{Conv}(X))=\mu (X)$.
> 5. $\mu (\lambda X+(1-\lambda) Y)\leq \lambda \mu (X)+(1-\lambda) \mu(Y)$ for $\lambda\in [0,1]$.
> 6. If $\{X\_n\}\subset \mathfrak{M}\_E^c$, such that $X\_{n+1}\subset X\_n$ for $n=1,2,...$ and if $\displaystyle\lim\_{n\rightarrow \infty} \mu(X\_n)=0$ then $X\_{\infty}=\bigcap\_{n=1}^{\infty}X\_n\neq \emptyset$.
>
>
>
---
Now, an MNC is said to be:
7. With **maximum property** if: $\mu(X \cup Y)=\max \{\mu(X), \mu(Y)\}$.
8. **Nonsingular** if: $\mu(\{a\} \cup X)=\mu(X)$.
---
In this question, I want to see if there is an MNC that is With maximum property but **Not** nonsingular.
|
https://mathoverflow.net/users/102228
|
A MNC with maximum property but not singular
|
1. First, according to [paper you cited](https://dml.cz/bitstream/handle/10338.dmlcz/105982/CommentatMathUnivCarol_021-1980-1_10.pdf), $\mathfrak{M}\_E$ is **the** [not "a"] family of all nonempty bounded subset**s** of $E$.
2. The examples of MNC's given in the [paper you cited](https://dml.cz/bitstream/handle/10338.dmlcz/105982/CommentatMathUnivCarol_021-1980-1_10.pdf) -- namely, the Kuratowski MNC $\alpha$, the ball measure $\chi$, and the MNC $\mu$ defined in Theorem 1 of that paper -- are all with the maximum property and nonsingular.
3. Perhaps, you rather wanted an MNC with the maximum property but not **non**singular. Such an MNC, say $\nu$, is given e.g. by the formula $\nu(X):=\sup\{|x|\colon x\in X\}$ for $E=\mathbb R$ and all $X\in\mathfrak{M}\_E$.
|
1
|
https://mathoverflow.net/users/36721
|
405255
| 166,162 |
https://mathoverflow.net/questions/405256
|
22
|
I apologize for this question which is obviously not research-level. I've been teaching to master students the standard generating sets of the symmetric and alternating groups and I wasn't able to give a simple, convincing example where it's useful to use the two-generating set $\{(1,2),(1,2,...,n)\}$. (I always find it annoying when we teach something and we're not able to convince the students that it's useful.) I asked a couple of colleagues and no simple answer came out -- let me stress that I'd like to find something simple enough, like a remark I could do in passing or an exercise that I could leave to the reader without cheating him/her. Do you know such examples ?
|
https://mathoverflow.net/users/17988
|
What is the standard 2-generating set of the symmetric group good for?
|
Let $f\in\mathbb{Q}[x]$ be an irreducible polynomial of prime degree $p$, with exactly $2$ non-real roots.
You can view the Galois group of $f$ (i.e., the Galois group of the splitting $f$) as a subgroup of $S\_p$. Complex conjugation shows that the Galois group contains a transposition. You can use Cauchy's theorem from group theory to show that the Galois group contains a $p$-cycle.
Then $f$ has Galois group $S\_p$. This uses the slightly stronger fact that $S\_p$ is generated by any transposition and $p$-cycle (which can be proved from the standard two-generating set).
In turn, constructing a polynomial with Galois group $S\_5$ is useful for proving insolvability of the quintic.
|
37
|
https://mathoverflow.net/users/95685
|
405257
| 166,163 |
https://mathoverflow.net/questions/405253
|
1
|
Can every polynomial $P(x)$ with integer coefficients we represented in the form
$$
P(x) = Q^2(x)R(x),
$$
where $Q(x)$ and $R(x)$ are polynomials with integer coefficients such that $R(x)$ has no repeated (complex) roots?
If this is true it should be well-known, then a reference would be helpful.
Intuition: if $\alpha$ is any complex root of $P(x)$ of even multiplicity $2k$, then factor $(x-\alpha)^k$ goes to $Q(x)$. Conversely, if $\alpha$ has odd multiplicity $2k+1$, then $(x-\alpha)^k$ goes to $Q(x)$ while $(x-\alpha)$ goes to $R(x)$. However, will $Q(x)$ and $R(x)$ have integer coefficients?
|
https://mathoverflow.net/users/89064
|
Square-free representation of polynomials with integer coefficients
|
Yes. Recall that $\mathbb{Z}[x]$ is a UFD and recall that, if $p(x)$ is irreducible in $\mathbb{Z}[x]$, then the roots of $p(x)$ are distinct.
So write $P(x) = \prod p\_i(x)^{a\_i}$, a product of irreducible polynomials in $\mathbb{Z}[x]$, and put $Q(x) = \prod p\_i(x)^{\lfloor a\_i/2 \rfloor}$ and $R(x) = \prod\_{a\_i \equiv 1 \bmod 2} p\_i(x)$.
|
3
|
https://mathoverflow.net/users/297
|
405258
| 166,164 |
https://mathoverflow.net/questions/405214
|
3
|
In the book [Number Theory IV](https://www.springer.com/gp/book/9783540614678) from Parshin, one can find this statement (precisely at p. 215) with the comment "*it is easy to see that...*":
>
> Let $a\_{ij}$ ($1\le i,j\le m$) be complex numbers with $|a\_{i,j}|\le H\_i$ ($1\le j\le m$).
> One assumes that the linear forms $L\_i(\underline X)=\sum\_{i=1}^ma\_ix\_i$ are linearly independent. Then, for any $\overline w=(w\_1,\cdots,w\_m)\in\mathbb C^m$, one has
> $$\sum\_{i=1}^m\frac{|L\_i(\overline w)|}{H\_i}\ge c\_1\frac{|\Delta|}{H\_1\cdots H\_m}$$
> where $\Delta=\det(a\_{i,j})$ and $c\_1=\frac1{m!}\max\_{1\le j\le m}|w\_j|$.
>
>
>
Unfortunately, for me it is not easy. Can one have hints to prove that? Thanks in advance
|
https://mathoverflow.net/users/33128
|
Minoration of linear forms
|
We claim that
$$
\sum\_{i=1}^m \frac{|L\_i(w)|}{H\_i} \geqslant \frac{|w|\_\infty}{(m-1)!}\frac{|\Delta|}{H\_1 \cdots H\_m},
$$
where $|w|\_\infty = \max\_{1 \leqslant j \leqslant m} |w\_j|$.
Indeed, first assume that $|A|\_\infty = \max\_{1\leqslant i,j \leqslant m} |a\_{ij}|\leqslant 1$, and $H\_i = 1$ for all $i$. Because $|Bw|\_\infty \leqslant |B|\_\infty |w|\_1$ for any matrix $B$ and $w \in \mathbb{C}^m$ (where $|w| = \sum\_{j=1}^m |w\_j|$), we have
$$
|w|\_\infty \leqslant |A^{-1}|\_\infty|Aw|\_1.
$$
Now $A^{-1} = \frac{1}{|\Delta|} \mathrm{Com}(A)^\top$ where $\mathrm{Com}(A)$ is the comatrix of $A$. Note that $|\mathrm{Com}(A)|\_\infty \leqslant (m-1)!$ by the Leibniz formula for determinants, since $|A|\_\infty \leqslant 1$. Thus
$$
|A^{-1}|\_\infty \leqslant \frac{(m-1)!}{|\Delta|}.
$$
Therefore, by definition of $L\_i$ $(i = 1, \dots, m)$ we get
$$
\sum\_{i=1}^m |L\_i(w)| = |Aw|\_1 \geqslant \frac{|w|\_\infty |\Delta|}{(m-1)!},
$$
which is the sought result in the case $|A|\_\infty \leqslant 1$ and $H\_1 = \cdots = H\_m = 1$.
Finally take a general $A$ and assume that $\max\_{1 \leqslant j \leqslant m}|a\_{ij}| \leqslant H\_i$ for $i = 1, \dots, m$. Writing $A = \begin{pmatrix} L\_1 \\ \vdots \\ L\_m \end{pmatrix}$ and applying the above estimate to the matrix
$
\tilde A = \begin{pmatrix} L\_1 / H\_1 \\ \vdots \\ L\_m / H\_m \end{pmatrix},
$
we get the claim.
|
2
|
https://mathoverflow.net/users/393977
|
405265
| 166,166 |
https://mathoverflow.net/questions/405252
|
4
|
I cannot find the exact same question asked anywhere in this site. I know the related Green-Tao theorem but the gaps between consecutive primes can grow unbounded so it does not seem helpful to answer this question.
What I have tried: assume the largest gap is D and without loss of generality it appears infinitely often. Then I try to apply the pigeonhole principle but don't know how.
Thanks in advance.
Added thought: will there always be a subsequence forming an infinitely long arithmetic progression in the original sequence? I think the answer is NO.
|
https://mathoverflow.net/users/393027
|
Are there arbitrarily long arithmetic progressions in every increasing sequence of positive integers with bounded gaps between consecutive terms?
|
You may of course use Szemeredi theorem, as suggested by Alexander Kalmynin.
If you need a more elementary argument, you may apply Van der Waerden theorem as follows: assuming that the gaps are bounded by $T$, color every positive integer $n$ to the color $i\in \{1,\dots,T\}$ if $nT+i$ belongs to your set (so each large enough integer gets at least one color), and find a large monochromatic arithmetic progression. It corresonds to a large progression in the initial set.
|
9
|
https://mathoverflow.net/users/4312
|
405274
| 166,171 |
https://mathoverflow.net/questions/405270
|
2
|
Let $X$ be an random $n \times d$ matrix with entries drawn iid from $N(0,1/d)$ and let $w$ be a unit-vector in $\mathbb R^d$. With $\lambda>0$, and define $G:=X^\top(XX^\top + \lambda I\_n)^{-1}X$. Finally, defined $\alpha := w^\top G^2 w$.
>
> **Question.** *In the limit $n,d \to \infty$ with $n/d \to \rho \in (0,\infty)$, what is the limitting value of $\alpha$ as a function of $\lambda$ and $\rho$ ?*
>
>
>
A useful subcase is when $\lambda \to 0^+$.
>
> **Question.** *What is the value of $\lim\_{\lambda \to 0^+}\lim\_{n,d \to \infty \\ n/d \to \rho}\alpha$ as a function of $\rho$ ?*
>
>
>
|
https://mathoverflow.net/users/78539
|
Asymptotics of $w^\top G^2 w$, where $w$ is a unit-vector, $G:=X^T(XX^T+t I_n)^{-1}X$, $t > 0$, and $X$ is an $n\times d$ gaussian random matrix
|
Let me calculate the expectation value of $\alpha$. The probability distribution of $X$ is invariant under orthogonal transformations, so without loss of generality I can orient the unit vector $w$ along one of the axes, $w\_i=\delta\_{ip}$, $p\in\{1,2,\ldots d\}$. Then
$$\mathbb{E}[\alpha]=\mathbb{E}\left(X^T(XX^T+\lambda I)^{-1}XX^T(XX^T+\lambda I)^{-1}X\right)\_{pp}.$$
Again because of orthogonal invariance the answer cannot depend on the value of the index $p$, hence we can sum over $p$ and divide by $d$, which gives the trace,
$$\mathbb{E}[\alpha]=\frac{1}{d}\mathbb{E}\,{\rm tr}\,\left(X^T(XX^T+\lambda I)^{-1}XX^T(XX^T+\lambda I)^{-1}X\right)$$
$$\qquad=\frac{1}{d}\mathbb{E}\,{\rm tr}\,\frac{W^2}{(W+\lambda I)^2},\;\;W=XX^{T}.$$
For the subcase $\lambda\rightarrow 0$ in the OP we thus find $\mathbb{E}[\alpha]=n/d \to \rho$.
For nonzero $\lambda$ and in the large-$n$ limit the result for $\mathbb{E}[\alpha]$ is the integral of $\mu^2(\mu+\lambda)^{-2}$ weighted by the [Marchenko-Pastur distribution](https://en.wikipedia.org/wiki/Marchenko%E2%80%93Pastur_distribution) $\rho(\mu)$ for the eigenvalues $\mu$ of $W$.
|
2
|
https://mathoverflow.net/users/11260
|
405278
| 166,173 |
https://mathoverflow.net/questions/405284
|
2
|
Prove that there exists five matrices $B\_i \in \mathbb{F}\_2^{5\times 10}$, $i\in \{1,2,3,4,5\}$, such that any two $B\_i$'s form an invertible matrix in $\mathbb{F}\_2^{10\times 10}$.
I am interested in a proof of existence that could be generalized to matrices with other dimensions.
Follow-up question when there are constraints on the $B\_i$'s: [Existence of matrices in $\mathbb{F}\_2$ with some invertibility properties](https://mathoverflow.net/questions/405296/existence-of-matrices-in-mathbbf-2-with-some-invertibility-properties)
|
https://mathoverflow.net/users/nan
|
Existence of matrices with some invertibility properties
|
If $B$, $B'$ are $5\times 10$ matrices of rank $5$ over $\mathbb{F}\_2$, the condition that the matrix formed by stacking $B$ on top of $B'$ is invertible is equivalent to the condition that the row span of $B$ and the row span of $B'$ have trivial intersection. The row spans are $5$-dimensional subspaces of $\mathbb{F}\_2^{10}$, so given five $5$-dimensional subspaces of $\mathbb{F}\_2^{10}$ with pairwise trivial intersection, we can choose bases for them to construct matrices $B\_1,\ldots,B\_5$.
Note that $\mathbb{F}\_2^{10}$ is isomorphic to $\mathbb{F}\_{2^5}^2$ as an $\mathbb{F}\_2$ vector space. A $1$-dimensional $\mathbb{F}\_{2^5}$-subspace of $\mathbb{F}\_{2^5}^2$ can be identified with a $5$-dimensional $\mathbb{F}\_2$ subspace of $\mathbb{F}\_2^{10}$, and every pair of $1$-dimensional $\mathbb{F}\_{2^5}$-subpsaces of $\mathbb{F}\_{2^5}^2$ have trivial intersection. There are 33 such subspaces, so we can find even find $B\_1,\ldots,B\_{33}\in\mathbb{F}\_2^{5\times 10}$ with the desired property.
|
3
|
https://mathoverflow.net/users/5263
|
405288
| 166,178 |
https://mathoverflow.net/questions/405295
|
37
|
Does there exist a map $f:\Bbb R^n \rightarrow \Bbb R^m$, where $n<m$ and $ n,m \in\Bbb N^+$ such that $f$ is surjective and differentiable?
|
https://mathoverflow.net/users/369324
|
Is there a differentiable map surjective from low to high dimension？
|
$\DeclareMathOperator\R{\mathbf{R}}$It's easy to check that the image of any locally Lipschitz map $f:\R^n\to\R^m$ has measure zero when $n<m$ (this encompasses the case of class-$\text{C}^1$ maps, but not the case of differentiable maps).
Indeed, extend $f$ to $F:\R^m\to\R^m$ by $F(x,y)=f(x)$. This is still locally Lipschitz. So it maps the subset $\R^n$ of measure zero to a subset of measure zero, see this [MathSE post](https://math.stackexchange.com/questions/139883/why-does-a-lipschitz-function-f-mathbbrd-to-mathbbrd-map-measure-zero-s) (it assumes Lipschitz, but the argument is local and $\R^m$ is a countable union of subsets on which $F$ is Lipschitz).
---
Taking into accounts the comments: here is a setting encompassing both the cases when $f$ is locally Lipschitz, and when $f$ is differentiable.
Suppose that for every $x\in\mathbf{R}^n$, we have
$$(\*)\qquad F\_f(x)=\limsup\_{y\to x,\;y\neq x}\frac{\|f(y)-f(x)\|}{\|y-x\|}<\infty.$$
Define, for $p$ positive integer
$$X\_p=\{x\in\mathbf{R}^n:\forall y\in\mathbf{R}^n:\|y-x\|\le 1/p \Rightarrow \|f(y)-f(x)\|\le p\|y-x\|\}.$$
Then $\mathbf{R}^n$ is the (countable) union of all $X\_p$, and $X\_p$ is a countable union of subsets $X\_{p,i}$ of diameter $\le 1/p$. And $f$ is $p$-Lipschitz on $X\_{p,i}$ (and also on its closure, in case one wishes to get closed subsets).
So the result indeed follows, not of the Lipschitz case as strictly said, but of the same statement replacing $\mathbf{R}^n$ with a subset of $\mathbf{R}^n$ with the restriction of the Euclidean distance (namely: for $n<m$ and $Y$ subset of $\mathbf{R}^n$, every Lipschitz function $Y\to\mathbf{R}^m$ has image of measure zero). The argument for the latter seems unchanged.
PS: for a reference, it is mentioned by @Kosh that Lemma 7.25 in Rudin's *Real and complex analysis* (initially published in 1966) does all the job: it asserts that any map $f:\mathbf{R}^m\to\mathbf{R}$ satisfying $(\*)$ maps measure zero subsets to measure zero subsets. The proof given here actually seems to roughly be the same as the one written (concisely) in Rudin's book.
|
25
|
https://mathoverflow.net/users/14094
|
405303
| 166,184 |
https://mathoverflow.net/questions/405318
|
-1
|
Let $M\in \mathbb{R}^{N\times N}$ be a given matrix and $k\ge 2$ be a given integer. Then my question is the following optimization problem:
Is there a polynomial-time solution to the following problem: $$S^\star = \arg\max\_{\substack{S\subset [N]:\\ |S|\le k}} \sum\_{i,j\in S}M\_{ij}?$$
This seems to be hard in general (that is, it requires exponential time-complexity), but I could not find a direct link to any known problem. I first thought that this problem is related to the maximum subset problem, but I am not sure. It will be really helpful if somebody can provide any reference to any related problem. It seems that approximate solutions can be found for this problem, but I was unable to find that too after a Google search. It will be really great if someone can give any reference.
|
https://mathoverflow.net/users/64194
|
Finding a $k$-subset which maximizes a matrix sum
|
This is the 0-1 [quadratic knapsack problem](https://en.wikipedia.org/wiki/Quadratic_knapsack_problem), which is NP-hard. The binary decision variables $x\_i$ indicate whether $i\in S$, and the knapsack capacity is $k$.
You can solve it via integer linear programming as follows. For $i<j$, let binary decision variable $y\_{i,j}$ represent $x\_i x\_j$. The problem is to maximize $\sum\_i M\_{i,i} x\_i + \sum\_{i<j} (M\_{i,j}+M\_{j,i}) y\_{i,j}$ subject to
\begin{align}
y\_{i,j} &\le x\_i &\text{for $i<j$} \tag1 \\
y\_{i,j} &\le x\_j &\text{for $i<j$} \tag2 \\
y\_{i,j} &\ge x\_i + x\_j - 1 &\text{for $i<j$} \tag3 \\
\sum\_i x\_i &\le k \tag4
\end{align}
Constraint $(1)$ enforces $y\_{i,j} \implies x\_i$.
Constraint $(2)$ enforces $y\_{i,j} \implies x\_j$.
Constraint $(3)$ enforces $(x\_i \land x\_j) \implies y\_{i,j}$.
Constraint $(4)$ enforces $|S| \le k$.
If $M\_{i,j} \ge 0$, you can omit $(3)$, which will naturally be satisfied because of the objective.
|
3
|
https://mathoverflow.net/users/141766
|
405320
| 166,190 |
https://mathoverflow.net/questions/405316
|
0
|
The following is a recursion for one point monotone Hurwitz numbers
$$
d \, m\_g(d) = 2(2d-3) \, m\_g(d-1) + d(d-1)^2 \, m\_{g-1}(d)\label{1}\tag{$\*$}
$$
with initial condition $m\_0 (1) =1$ and some of the other numbers are $ m\_0 (2) = 1, m\_1 (3) =10$.
Let denote the generating function by
$$F\_{g}(x) := \sum\_{d\geq 1} m\_g (d) x^d$$
$$
\begin{split}
x{\frac {\rm d}{{\rm d}x}}F\_g \left( x \right) &-4\,{x}^{2}{\frac {\rm d}
{{\rm d}x}}F\_g \left( x \right) +2\,xF\_g \left( x \right) \\
&= \left(x{\frac {\rm d}{{\rm d}x}}\right)^3F\_{g-1} \left( x \right) -2\,\left({x}{\frac {\rm d}
{{\rm d}x}}\right)^2F\_{g-1} \left( x \right) +\,xF\_{g-1} \left( x \right)
\end{split}\label{2}\tag{$\*\*$}
$$
Now we put the condition that $F\_g (x) =0 $ for $g<0$ hence
using \eqref{2} we get
$$
x{\frac {\rm d}{{\rm d}x}}F\_0 \left( x \right) -4\,{x}^{2}{\frac {\rm d}
{{\rm d}x}}F\_0 \left( x \right) +2\,xF\_0 \left( x \right)=0\label{3}\tag{$\*\*\*$}
$$
We get $F\_0 = C\sqrt{(1-4x)}$.
With the change of coordinates $x(z) = z -z^2$ the we get $F\_0 (z)$ to be a rational solution hence in coordinate $z$ for this particular equation we get all the solution $F\_g (z)$ to be rational. I hope I am not wrong here.
My question is the following given a general one point recursion of type (\*) that is
say recursion of type
$$\sum\_{i,j}^{i\_{max}, j\_{max}} p\_{ij} (d)n\_{g-i}(d-j) $$
From this one point recursion we can get differential equation of type (\*\*), for the generating function
$$F\_{g}(x) := \sum\_{d\geq 1} n\_g (d) x^d$$
So what constraints do we need to put on the polynomials $p\_{ij}(d)$ such that the solution will be rational for the differential equations.
In the example I gave is the differential equation is Linear ode, so is there any reference regarding rational solutions and their relation to the poles of solution.
|
https://mathoverflow.net/users/45170
|
Rational solution for linear differential equation
|
I must say that I don't much understand the motivation coming from number theory, but your question about rational solutions of ODEs has a definite answer, provided the equation has polynomial or rational coefficients. A standard reference is
>
> *Abramov, S. A.*, [**Rational solutions of linear differential and difference equations with polynomial coefficients**](http://dx.doi.org/10.1016/S0041-5553(89)80002-3), U.S.S.R. Comput. Math. Math. Phys. 29, No. 6, 7-12 (1989). [ZBL0719.65063](https://zbmath.org/?q=an:0719.65063).
>
>
>
The main idea is straight forward and I will summarize it here. The solution of a homogeneous linear ODE with rational coefficients (which can always be made polynomial by clearing denominators) can have a pole (including the poles at infinity) only at a [regular singular point](https://en.wikipedia.org/wiki/Regular_singular_point), and only if the [Frobenius method](https://en.wikipedia.org/wiki/Frobenius_method) reveals integer negative critical exponents. Taking the most negative integer critical exponents at each regular singular point gives you a universal polynomial denominator $Q$, so that any rational solution can be written as $P/Q$, with $P$ polynomial. The Frobenius analysis at infinity also gives you an upper bound on the degree of $P$, which makes $P/Q$ an ansatz for the most general rational solution with finitely many parameters. Now, it is only a matter of plugging in the ansatz into the ODE and expanding in a power series. The preceding analysis converts the equation into a finite dimensional linear algebra problem on the parameters of the $P/Q$ ansatz.
|
2
|
https://mathoverflow.net/users/2622
|
405330
| 166,194 |
https://mathoverflow.net/questions/345008
|
2
|
Let $P\subset\Bbb R^d$ be a *vertex-transitive polytope* aka. an *orbit polytope*.
Can there be a matrix $T\in\mathrm{SO}(\Bbb R^d)$ that commutes with all symmetries in $\mathrm{Aut}(P)\subset\mathrm O(\Bbb R^d)$?
Probably one approach to the question is as follows: can there be a vertex-transitive polytope $P\subset\Bbb R^d$ for which $\mathrm{Aut}(P)$ is (real) irreducible, but **not** absolutely irreducible (that is, not irreducible over $\Bbb C$).
Vertex-transitivity is necessary for all these questions. For example, there is a polytope (not vertex-transitive) whose symmetry group is a finite subgroup of $\mathrm{SO}(\Bbb R^2)$, which is real irreducible, but reducible over $\Bbb C$.
Since $\mathrm{SO}(\Bbb R^2)$ is commutative, every element of that group would then commute with $\mathrm{Aut}(P)$.
It is known that most commutative groups cannot be symmetry groups of vertex-transitive polytopes (only exceptions are elementary 2-abelian groups).
|
https://mathoverflow.net/users/108884
|
A matrix that commutes with all symmetries of a vertex-transitive polytope
|
There are examples of vertex-transitive polytopes with **irreducible** symmetry groups for which there are still a **non-scalar** transformations $T\in\mathrm O(\Bbb R^d)$ that commutes with all the symmetries of the polytopes.
Fix a group $G$ with the following properties:
* $G$ is neither abelian nor generalized dicyclic.
* the centralizer of every irreducible representation of $G$ is isomorphic to either $\Bbb C$ or $\Bbb H$.
Examples are contained among the finite subgroups of $\mathrm U(n)$ (for $n\ge 2$) or $\mathrm{SU}(2)$. The first property ensures that $G$ is not among the groups that are excluded as symmetry groups of vertex-transitive polytopes as determined in [1]. So there is a vertex-transitive polytope $P$ whose automorphism group can be considered as a matrix group representation of $G$. By the second assumption, the centralizer of this matrix group containst then more than just scalar matrices.
---
[1] E. Friese, F. Ladisch. *"Classification of Affine Symmetry Groups of Orbit Polytopes"*
|
0
|
https://mathoverflow.net/users/108884
|
405332
| 166,195 |
https://mathoverflow.net/questions/405294
|
3
|
**Motivation**
The notion of uniform integrability is important for formulating the Vitali convergence theorem. Unfortunately, different authors define uniform integrability differently, which causes quite a lot of confusion, as evident in the number of questions about uniform integrability and Vitali convergence theorem in [MSE](https://math.stackexchange.com/search?q=uniform+integrability+vitali+convergence+theorem). However, after reading many of those questions and the answers, I feel that many students are still confused. In this post, I want to clear all the confusion once and for all.
**Definitions**
Throughout this post, I would use the following definitions.
Let $(X, \mathcal{F}, \mu)$ be a measure space and $\Phi$ a collection of measurable functions on $(X, \mathcal{F})$, taking values in the extended real line.
1. We say $\Phi$ is **uniformly bounded in $L^1$** if
$$ \sup\_{f \in \Phi} \int\_X |f| d\mu < \infty $$
2. We say $\Phi$ **does not escape to vertical infinity** if for any $\epsilon > 0$, there exists $M > 0$ such that
$$\sup\_{f \in \Phi} \int\_{|f| \geq M} |f| d\mu < \epsilon $$
3. We say $\Phi$ **does not escape to width infinity** if for any $\epsilon > 0$, there exists $m > 0$ such that
$$\sup\_{f \in \Phi} \int\_{|f| \leq m} |f| d\mu < \epsilon $$
4. We call $\Phi$ **equi-integrable** if for any $\epsilon > 0$, there exists $\delta > 0$ such that whenever $A \in \mathcal{F}$ is a measurable set with $\mu(A) < \delta$, we have
$$\sup\_{f \in \Phi} \int\_A |f| d\mu < \epsilon $$
5. We call $\Phi$ **tight** if for any $\epsilon > 0$, there exists a measurable set $X\_0 \in \mathcal{F}$ such that $\mu(X\_0) < \infty$ and
$$\sup\_{f \in \Phi} \int\_{X \setminus X\_0} |f| d\mu < \epsilon $$
**Confusion**
Here comes the confusion about the definitions of uniform integrability:
1. Measure theory textbooks usually define uniform integrability as being equi-integrable.
2. Probability theory textbooks usually define uniform integrability as not escaping to vertical infinity.
3. In [Tao's blog post](https://terrytao.wordpress.com/2010/10/02/245a-notes-4-modes-of-convergence/), uniform integrability was defined as being uniformly bounded in $L^1$, not escaping to vertical infinity and not escaping to width infinity.
4. Yet some other authors define uniform integrability as being uniformly bounded in $L^1$ and equi-integrable.
What adds to more confusion is, the different definitions of uniform integrability are only equivalent under certain assumptions, while in general they are not equivalent. Moreover, different authors formulate the Vitali convergence theorem under different definitions of uniform integrability.
**Claim**
It is well-known that if $\mu$ is a finite measure, then $\Phi$ does not escape to vertical infinity if and only if it is uniformly bounded in $L^1$ and equi-integrable. For the general case, I would like to propose the following claim. To avoid confusion, I would avoid the term "uniformly integrable" altogether.
>
> Let $(X, \mathcal{F}, \mu)$ be a measure space and $\Phi$ a collection
> of measurable functions on $(X, \mathcal{F})$, taking values in the
> extended real line. We do not make any other assumption on $\mu$ or $\Phi$.
> Then the following conditions are equivalent:
>
>
> 1. $\Phi$ is uniformly bounded in $L^1$, does not escape to vertical infinity and does not escape to width infinity
> 2. $\Phi$ does not escape to vertical infinity and is tight
> 3. $\Phi$ is uniformly bounded in $L^1$, equi-integrable and tight
>
>
> Now, let $(f\_n)$ be a sequence of measurable functions and $f$ another
> measurable function on $(X, \mathcal{F})$. Suppose that:
>
>
> 1. the collection $\Phi = \{f\_n : n \in \mathbb{N}\}$ satisfies any one of the equivalent conditions above,
> 2. the sequence $(f\_n)$ converges to $f$ either almost everywhere or in measure,
>
>
> then we have $f \in L^1(\mu)$, and $(f\_n)$ converges to $f$ in
> $L^1(\mu)$.
>
>
>
**Questions**
1. Is my claim correct?
2. Is there any book or paper that makes a concerted effort to clear the confusion around uniform integrability and Vitali convergence theorem?
*Edit:*
3. [Iosif Pinelis' answer](https://mathoverflow.net/questions/405294/confusion-around-uniform-integrability-and-vitali-convergence-theorem/405336#405336) gives an counter-example in which a sequence of functions is uniformly bounded in $L^1$, does not escape to vertical infinity, and does not escape to width infinity. However, the sequence is neither equi-integrable nor tight. So how to connect [Tao's definition of uniform integrability](https://math.stackexchange.com/questions/4042711/different-definition-of-uniform-integrability) with equi-integrability and tightness?
|
https://mathoverflow.net/users/49284
|
Confusion around uniform integrability and Vitali convergence theorem
|
$\newcommand\R{\mathbb R}\newcommand{\ep}{\epsilon}\newcommand{\de}{\delta}$
Your claim is not quite correct.
E.g., let $\mu$ be the Lebesgue measure over $X:=\R$. Let
$$\Phi:=\{f\_n\colon\, n\in\mathbb N\},$$
where
$$f\_n(x):=\frac1{1+(x-n)^2}$$
for real $x$. Then
$\Phi$ is uniformly bounded in $L^1$, does not escape to vertical infinity, and does not escape to width infinity; however, $\Phi$
is not tight.
Also, $f\_n\to0$ pointwise and hence almost everywhere, but $f\_n\not\to0$ in $L^1(\mu)$.
---
On the the other hand, if $\Phi$ does not escape to vertical infinity, then $\Phi$ is equi-integrable (the condition that $\Phi$ does not escape to width infinity is not needed here). Indeed, suppose that $\Phi$ does not escape to vertical infinity, and take any real $\ep>0$. Then there is a real $M>0$ such that
$\sup\_{f\in\Phi}\int\_{|f|\ge M}|f|\,d\mu<\ep/2$. Let now $\de:=\ep/(2M)$. Then for any $f\in\Phi$ and any $A\in\mathcal F$ such that $\mu(A)<\de$ we have
\begin{equation}
\int\_A|f|\,d\mu\le\int\_{|f|\ge M}|f|\,d\mu+\int\_A M\,d\mu<\ep/2+M\de=\ep.
\end{equation}
So, $\Phi$ is equi-integrable, as claimed.
Also, if $\Phi=\{f\_n\colon\, n\in\mathbb N\}$ does not escape to vertical infinity and is tight, and if $f\_n\to f$ almost everywhere, then $f\_n\to f$ in $L^1(\mu)$.
|
1
|
https://mathoverflow.net/users/36721
|
405336
| 166,197 |
https://mathoverflow.net/questions/405321
|
1
|
Let $\Gamma\_0,\Gamma\_1,...$ be regeneration epochs.
If $(X\_n)\_{n \in \mathbb{N}}$ is a $\lambda$ biased random walk on a Galton-Watson tree, than the regeneration epochs are defined as:
$\Gamma\_0:=\inf\{\iota \ | X\_i\neq X\_{\iota} \ \forall i\leq \iota \ \text{and} \ X\_{j}\neq (X\_{\iota})\_\* \ \forall j\geq \iota \}$,
$\Gamma\_k:=\inf\{\iota \ | \iota>\Gamma\_{k-1}: X\_i\neq X\_{\iota} \ \forall i\leq \iota \ \text{and} \ X\_{j}\neq (X\_{\iota})\_\* \ \forall j\geq \iota \}$
---
I want to show that
$\{\Gamma\_0 < \infty \}=\mathcal{S}$, where $\mathcal{S}$ is the survival set.
---
The statement makes sense, of course. But i do not know how to start the proof.
Maybe someone has experience in the subject area and could help me.
Greetings : Fynn
|
https://mathoverflow.net/users/396065
|
Random walks on GW-trees (regeneration epochs/survival set)
|
You need to assume that the bias satisfies $\lambda<m$ where $m$ is the mean offspring. Then you can find the proof in Lemma 3.3 page 253 of [1].
[1] Lyons, Russell, Robin Pemantle, and Yuval Peres. "Biased random walks on Galton–Watson trees." Probability theory and related fields 106, no. 2 (1996): 249-264.
<https://link.springer.com/content/pdf/10.1007/s004400050064.pdf>
|
1
|
https://mathoverflow.net/users/7691
|
405346
| 166,201 |
https://mathoverflow.net/questions/405323
|
9
|
The "pants" bordism in dimension n is a bordism which goes from $S^n \sqcup S^n$ to $S^n$ witnessing the connected sum operation - equivalently by attaching 1-handle to the trivial bordism, equivalently doing surgery on a 0-sphere.
The "copants" bordism is the same manifold, but thought of as a bordism from $S^n$ to the disjoint union $S^n \sqcup S^n$. It can be understood as doing surgery on an (n-1)-sphere (the equator).
Now I want to understand when these manifolds can be understood as framed manifolds (both tangential framing and stable framing are relevant) and if they are framed what constraints they put on the framings of the boundaries.
Now we can embed $S^n$ into $\mathbb{R}^{n+1}$ in the standard way and choose the outward normal to identify $\underline{\mathbb{R}}^{n+1} \cong \tau\_{S^n} \oplus \underline{\mathbb{R}}$, giving $S^n$ a *standard* (n+1)-framing. This extends over the interior of the solid ball in $\mathbb{R}^{n+1}$.
Now the pants bordism can always be realized as an $(n+1)$-framed manifold because it too can be viewed as a subset of $\mathbb{R}^{n+1}$. Think of two spheres inside a larger sphere and take the intervening space between them. This gives a framed bordism from the disjoint union of two "standard" spheres to one "standard" sphere.
Now the framings on the spheres will be a torsor for $\pi\_n(SO(n+1)$. The standard framing gives us a basepoint (which we identify with the identity element in $\pi\_n(SO(n+1)$). The framings on the pants bordism are uniquely determined by the two framings on the incoming spheres. There are $\pi\_n(SO(n+1) \times \pi\_n(SO(n+1)$ many and if the framings of the two incoming spheres are (under our identification) $x, y \in \pi\_n(SO(n+1)$, then the framing of the outgoing sphere will be $x + y$.
Now when $n=1$, we can also frame the copants bordism. However for that bordism if the outgoing framings of the two outgoing spheres are $x$ and $y$, then the incoming sphere has framing $x + y + 2 \in \pi\_1 SO(2) = \mathbb{Z}$.
**Question:** My question is what happens in higher dimensions? Can the copants bordism be framed? If so what are the constraints of the bourdary framings? Is this question easier if we use stable framings instead?
I checked Kervaire-Milnor's "groups of homotopy spheres: I" paper. In this case the obstruction for extending the framing over the pants bordism always is trivial, as expected. However for the copants bordism in general there is the possibility of a non-trivial obstruction. Furthermore their result that the surgery sphere can always be reframed to eliminate the obstruction doesn't apply in this case (because the dimension of the surgery sphere is too high).
On the otherhand, I think we can just view the pants bordism as a bordism the otherway around to get some (tangential) framing on the copants. It is then a matter of pinning down how the framings on the boundaries change when we do this.
|
https://mathoverflow.net/users/184
|
Framed version of the "copants bordism"?
|
The key point is the identification $\tau(S^n)+\mathbb{R}$ with the restriction of tangent bundle of the bordism. I will read bordism from bottom to top.
Even to obtain pants bordism between standard spheres you need to choose the $\textit{inward}$ normal at bottom and $\textit{outward}$ normal at top for the above identification.
Let us examine the difference between pants and copants. In the first case we take a cartoon pants with wide sphere at top and pair of small spheres at bottom in $\mathbb{R}^{n+1}\times I$ s.t. the projection onto $\mathbb{R}^{n+1}$ is the described above picture with two spheres inside the bigger one. Note that induced framing from $\mathbb{R}^{n+1}$ is perfectly compatible with our identification.
Let us write $(s\_1,s\_2,s\_0)$ for the framing on the boundary spheres of this pants bordism.
Now reverse the picture (or simply read from top to bottom). Projection is the same and we can induce the framing from $\mathbb{R}^{n+1}$ as above. Note that each inward normal becomes outward and vice versa. Hence induced framing is $(R(s\_1),R(s\_2),R(s\_0))$, where $R$ is reflecting of a framing $\tau+\mathbb{R}$ over $S^n$ in normal direction $\mathbb{R}$. I claim that $R(s)=s-t$, where $t\in \pi\_n(SO(n+1))$ corresponds to a gluing function for tangent bundle over $S^{n+1}$: it is clear after considering filled disk $D^{n+1}$ bounding $S^n$ as upper and respectively lower semi-spheres of $S^{n+1}$.
Now, as was noted framed pants/copants respect an action of $\pi\_n(SO(n+1))\times \pi\_n(SO(n+1))$ by changing framings on the boundary components: $a\times b\cdot (s\_1,s\_2,s\_0)=(s\_1+a,s\_2+b,s\_0+a+b)$. In particular we have a copants framing corresponding to $(R(s\_1)+t,R(s\_2)+t,R(s\_0)+2t)$ which is $(s\_1,s\_2,s\_0+t)$.
In other words, given a copants bordism with the standard framing over output spheres we unavoidably (unless $S^{n+1}$ is parallelizable) get a non standard framing over input sphere. The difference with the standard framing is $t\in \pi\_n(SO(n+1))$ corresponding to $\tau(S^{n+1})$.
On the other hand standard copants bordism fall into stable range very quickly and the obstruction disappear: if you add $\mathbb{R}$ everywhere, then the same argument gives the difference in $\pi\_n(SO(n+2))$ which is the image of $\tau(S^{n+1})\in \pi\_n(SO(n+1))$ under inclusion $SO(n+1)\to SO(n+2)$. This is zero since $\tau(S^{n+1})+\mathbb{R}$ is trivial.
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5
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https://mathoverflow.net/users/8906
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405354
| 166,203 |
https://mathoverflow.net/questions/405358
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1
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The following theorem is well-known:
>
> **Theorem.** Let $M$ be a smooth manifold. A smooth covector field $\omega \in \Omega^1(M)$ is conservative, that is, $$\int\_{\mathbb{S}^1}f^\*\omega = 0 \qquad \forall f \in C^\infty(\mathbb{S}^1,M),$$ if and only if $\omega$ is exact.
>
>
>
For a proof, see for example Theorem 11.42 in the brilliant book on smooth manifolds by @JackLee. I wonder if one can generalise this result as follows:
Let $\omega \in \Omega^k(M)$ such that $$\int\_{\mathbb{S}^k}f^\*\omega = 0 \qquad \forall f \in C^\infty(\mathbb{S}^k,M).$$ Does then also follow that $\omega$ is exact?
|
https://mathoverflow.net/users/98139
|
Generalisation of conservative covector fields
|
Consider a 2-torus. The second homotopy is trivial, so the spheres can't feel the homology, or cohomology. The same for any manifold obtained by quotienting a simply connected manifold by a cocompact discrete group action, so all surfaces of genus 2 or more.
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3
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https://mathoverflow.net/users/13268
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405360
| 166,205 |
https://mathoverflow.net/questions/405349
|
7
|
Let $T:H\to H$ be a continuous operator on a Hilbert space.
Assume there exists an orthonormal base $(e\_j)\_{j\in\mathbb N}$, such that the sequence $Te\_j$ tends to zero.
Must $T$ be compact?
|
https://mathoverflow.net/users/nan
|
Criterion for compactness
|
$T$ is not necessarily compact. Let me produce a counterexample.
Let $H$ be any infinite dimensional real or complex separable Hilbert space. Let $(f\_{j,k})\_{1\leq k\leq j},(e\_{j})\_{j=1}^{\infty}$ be orthonormal bases for $H$.
Then let $T:H\rightarrow H$ be the bounded linear operator defined by letting
$T(f\_{j,k})=\frac{1}{\sqrt{j}}\cdot e\_{j}$ whenever $1\leq k\leq j$. We observe that $T(f\_{j,k})\rightarrow 0$ as $j\rightarrow\infty$.
We observe that $$\biggl\|\sum\_{j=1}^{\infty}\sum\_{k=1}^{j}a\_{j,k}f\_{j,k}\biggr\|^{2}=\sum\_{j=1}^{\infty}\sum\_{k=1}^{j}a\_{j,k}^{2}
=\sum\_{j=1}^{\infty}\sum\_{k=1}^{j}a\_{j,k}^{2}
\geq\sum\_{j=1}^{\infty}\frac{1}{j}\biggl(\sum\_{k=1}^{j}a\_{j,k}\biggr)^{2}$$
$$=
\sum\_{j=1}^{\infty}\biggl(\frac{1}{\sqrt{j}}\sum\_{k=1}^{j}a\_{j,k}\biggr)^{2}
=\biggl\|\sum\_{j=1}^{\infty}\biggl(\frac{1}{\sqrt{j}}\sum\_{k=1}^{j}a\_{j,k}\biggr)e\_{j}\biggr\|^{2}=\biggl\|T\biggl(\sum\_{j=1}^{\infty}\sum\_{k=1}^{j}a\_{j,k}f\_{j,k}\biggr)\biggr\|^2.$$
Therefore, $T:H\rightarrow H$ is a bounded linear operator with norm $1$.
However, if $y\in H$, then $y=\sum\_{j}a\_{j}e\_{j}$, then let
$x=\sum\_{j=1}^{\infty}\sum\_{k=1}^{j}\frac{1}{\sqrt{j}}\cdot a\_{j}f\_{j,k}$.
Then $$\|x\|^{2}=\sum\_{j=1}^{\infty}\sum\_{k=1}^{j}(\frac{1}{\sqrt{j}}\cdot a\_{j})^{2}=\sum\_{j=1}^{\infty}\sum\_{k=1}^{j}\frac{1}{j}a\_{j}^{2}=\sum\_{j=1}^{\infty}a\_{j}^{2}=\|y\|^{2}.$$
However, $$T(x)=\sum\_{1\leq k\leq j}\frac{1}{j}a\_{j}e\_{j}=\sum\_{j=1}^{\infty}a\_{j}e\_{k}.$$
Therefore, $T$ cannot be compact since $T$ maps the unit ball of $H$ surjectively onto the unit ball of $H$.
In fact, $T$ maps the orthonormal set $(\frac{1}{\sqrt{j}}\sum\_{k=1}^{j}f\_{j,k})\_{j=1}^{\infty}$ to the orthonormal basis $(e\_{j})\_{j=1}^{\infty}$.
|
12
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https://mathoverflow.net/users/22277
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405362
| 166,206 |
https://mathoverflow.net/questions/404734
|
4
|
I found this local isometric immersion from $\mathbb H^{n}$ into $\mathbb R^{2n-1}$, given by Schur (1886) in [Über die Deformation der Räume constanten Riemannschen Krümmungsmaasses](https://eudml.org/doc/157211) as follows, $(1\leq k\leq n-1)$:
\begin{align\*}
x\_{2k-1}&=\frac{a^2}{z\_n}\cos \frac{z\_k}{a}\\
x\_{2k}&=\frac{a^2}{z\_n}\sin \frac{z\_k}{a}\\
x\_{2n-1}&=a\int^{z\_n}\frac{\sqrt{z\_n^2-(n-1)a^2}}{z\_n^2}dz\_n
\end{align\*}
I'm trying to prove the following statements:
1. It is a local isometric immersion.
>
> Here, taking $\phi:\mathbb H^n\to \mathbb R^{2n-1}$ given by $\phi(z\_1,\dots,z\_n)=(x\_1,\dots,x\_{2n-1})$ I imagine that $\phi^\*g\_{\mathbb R^{2n-1}}=g\_{\mathbb H^n}$ which would prove that is a isometric immersion, but the conditions for $x\_{2n-1}$ to be well defined make it only a local immersion.
>
>
>
2. It has a constant curvature $K\equiv -1/a^2$.
>
> This is where I have some problems: is this a consequence of the above result? I'm trying with Christoffel's symbols.
>
>
>
3. Any ideas to prove that image $\phi(z\_1,\dots,z\_n)$ is not a complete surface?
I started to see this example as a coincidence but I was thinking a bit about what happens in $\mathbb R^3$: there are $3$ types of smooth surfaces of revolution with negative constant curvature given by $x(u,v)=(f(v)\cos u,f(v)\sin u,g(v))$, this is clear when solving
$$K=-\frac{f''(v)}{f(v)}.$$
Is there something similar in $\mathbb R^{2n-1}$, how many surfaces with these characteristics exist? is there a differential equation as in $\mathbb R^3$?
|
https://mathoverflow.net/users/171387
|
A local isometric immersion from $\mathbb H^{n}$ into $\mathbb R^{2n-1}$
|
The isometric immersion that you describe above is the higher dimensional pseudosphere. Now, concerning your final question, I presume that you need to search about isometric immersions of the hyperbolic space $\mathbb H^n$ by means of a warped product representation (of $\mathbb H^n$) into the Euclidean space.
Now, some additional things that you might be interested to:
1. There are many (explicit in some cases) local isometric immersions from $\mathbb H^n$ to $\mathbb R^{2n-1}$. These can be constructed by using either the Ribaucour or the Bäcklund transformation (for instance, see the papers by [Dajczer-Tojeiro](https://www.cambridge.org/core/journals/proceedings-of-the-london-mathematical-society/article/abs/an-extension-of-the-classical-ribaucour-transformation/DCF1A5B953590BBACC770B8BDB738177) and [Tenenblat-Terng](https://annals.math.princeton.edu/1980/111-3/p04)).
2. Local isometric immersions of the hyperbolic plane $\mathbb H^2$ into $\mathbb R^3$ imply "local" solutions, that is, solutions that are not defined on the whole $\mathbb R^2$, of the sine-Gordon equation and vice versa. Therefore, it follows from Hilbert's theorem that there is no "global" solution, that is, a solution defined on the whole plane $\mathbb R^2$, of the sine-Gordon equation. Just like in the case of dimension two, the same also happens in the higher dimensional case where now you will end up with a system of PDES (see for instance [Dajczer-Tojeiro](https://eudml.org/doc/153758)). We can have local solutions to this system but we don't know if there exists any global. The existence of a global solution would imply the existence of a global isometric immersion of $\mathbb H^n$ into $\mathbb R^{2n-1}$, which would give a non affirmative answer to the major still open problem (in submanifolds) up to this day, which is the following conjectured extension of Hilbert's theorem:
>
> There is no global isometric immersion from $\mathbb H^n$ to $\mathbb R^{2n-1}$
>
>
>
However, the above holds true in some very special cases. For instance:
* If the immersion is also minimal (the mean curvature vanishes) (see [Moore](https://www.google.com/url?sa=t&rct=j&q=&esrc=s&source=web&cd=&ved=2ahUKEwjjvoWztq7zAhUp_7sIHatZBGIQFnoECAIQAQ&url=https%3A%2F%2Fprojecteuclid.org%2Fjournals%2Fpacific-journal-of-mathematics%2Fvolume-40%2Fissue-1%2FIsometric-immersions-of-space-forms-in-space-)).
*I should also mention here that $\mathbb H^2$ admits no minimal immersion in any Euclidean space. (for a proof of this fact see either "*Lectures on minimal submanifolds*" by Lawson, or [Bryant](https://www.jstor.org/stable/1999793), or [Di Scala](https://academic.oup.com/blms/article-abstract/35/6/825/305609?redirectedFrom=fulltext)).*
* (weaker) If the immersion has also bounded mean curvature (see [here](https://link.springer.com/article/10.1134%2FS0001434607070024))
* (even weaker) If also the length of the mean curvature of the immersion does not go to infinity too fast, that is, exponentially fast (see [here](https://link.springer.com/article/10.1007%2Fs00013-020-01565-x))
I also recommend the following:
* The survey of Borisenko [here](https://www.ime.usp.br/%7Elymber/files/textos/RMS_56_3_R01.pdf)
* Chapter 5 of the book of Dajczer and Tojeiro *[Submanifold Theory
"Beyond an Introduction"](https://www.springer.com/gp/book/9781493996421)*
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4
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https://mathoverflow.net/users/85181
|
405373
| 166,208 |
https://mathoverflow.net/questions/405353
|
1
|
I am dealing with a knapsack-like problem with one difference from the conventional problem: the “weights” can be positive or negative and the constraint is $\sum w\_i x\_i \ge 0$ instead of $\sum w\_i x\_i \le W$. The "values" can also be positive or negative.
Can this be transformed to a knapsack problem or is it some other type of combinatorial optimization problem?
|
https://mathoverflow.net/users/396728
|
Knapsack like problem with nonnegative weight constraint
|
Assuming $x\_i$ is binary, perform a change of variables $\bar{x}\_i:=1-x\_i$ where $w\_i>0$:
\begin{align}
\sum\_i w\_i x\_i
&= \sum\_{i:w\_i>0} w\_i x\_i + \sum\_{i:w\_i<0} w\_i x\_i \\
&= \sum\_{i:w\_i>0} w\_i(1-\bar{x}\_i) + \sum\_{i:w\_i<0} w\_i x\_i \\
&= \sum\_{i:w\_i>0} w\_i + \sum\_{i:w\_i>0} (-w\_i)\bar{x}\_i + \sum\_{i:w\_i<0} w\_i x\_i
\end{align}
So $\sum\_i w\_i x\_i \ge 0$ is equivalent to
$$\sum\_{i:w\_i>0} w\_i\bar{x}\_i + \sum\_{i:w\_i<0} (-w\_i) x\_i \le \sum\_{i:w\_i>0} w\_i$$
|
1
|
https://mathoverflow.net/users/141766
|
405378
| 166,209 |
https://mathoverflow.net/questions/405021
|
2
|
This question is inspired by [Characterization of functors whose right adjoint is monadic?](https://mathoverflow.net/questions/385363/characterization-of-functors-whose-right-adjoint-is-monadic).
Let $F : \mathbf C \rightleftarrows \mathbf D : U$ be an adjunction, and suppose that we want to establish when the canonical comparison functor $\mathbf{Kl}(UF) \to \mathbf D$ is an equivalence. A necessary and sufficient condition is that [$F$ be essentially surjective on objects](https://mathoverflow.net/questions/26075/characterization-of-kleisli-adjunctions).
Is it possible to characterise this condition in terms only of the right adjoint $U : \mathbf D \to \mathbf C$?
|
https://mathoverflow.net/users/152679
|
Characterisation of functors whose left adjoint is Kleisli
|
Try this. I confess I haven't written out the proofs yet, but will do so if you (and the community) think this is an appropriate answer. Please excuse my renaming your categories according to my personal convention.
First, $U:{\mathcal A}\to{\mathcal S}$ must be faithful and reflect invertibility, cf Beck's theorem.
We rewrite essential surjectivity of $F:{\mathcal S}\to{\mathcal A}$ in terms of univeral properties:
For every "algebra" $A$, ie object of $\mathcal A$,
there are a "set" $X$ and a "function" $e:X\to U A$ in $\mathcal S$ that is universal in the sense that
for every other "algebra" $B$ and "function" $f:X\to U B$ in $\mathcal S$ there is a unique "homomorphism" $h:A\to B$ in $\mathcal A$ that extends $f$ in the sense that $f=e;U h$.
How do we find the object $X$? This is going to invoke $F$ (if that's allowed by the question!)
It need not be unique (up to iso), but the construction of the replacement for $\mathcal S$ is essentially the one in [this paper of mine](http://www.paultaylor.eu/ASD/sobsc) and the ones by Hayo Thielecke and Peter Selinger that it cites.
If appropriate equalisers exist in $\mathcal S$ then the canonical $e:X\to U A$ (in fact the terminal one) is given by the equaliser of $U A\rightrightarrows U F U A$.
In terms of $X$, the two maps are $\eta U F X=\eta U A$ and $U F\eta X$. We can derive the second of these from the same problem starting with $F U A$ in place of $A$, in which case we get a split equaliser.
|
2
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https://mathoverflow.net/users/2733
|
405379
| 166,210 |
https://mathoverflow.net/questions/405370
|
4
|
Let $A$ be a finite-dimensional (not necessarily unital) associative algebra over the field of complex numbers $\mathbb{C}$ (but I'm also interested in more general fields). Assume the multiplication on $A$ is non-degenerate, which means that $A= AA$ and if $a \in A$ satisfies $aA = 0$ or $Aa = 0$, then $a=0$. Is it true that $A$ is unital? If not, what is a counterexample?
---
Some easy observations:
* If $A$ is also simple, then it can be shown that the answer is positive. This follows for instance by the argument [here](https://mathoverflow.net/q/181839).
* If $A$ is a $C^\*$-algebra (which is automatically non-degenerate), then a finite-dimensional $C^\*$-algebra is automatically unital.
|
https://mathoverflow.net/users/nan
|
Is a non-degenerate finite-dimensional algebra unital?
|
There's a four-dimensional counterexample over any field.
$A$ has basis $\{e,a,b,c\}$, with all products of basis elements zero except for
$$e^2=e,\quad ab=c,\quad ea=a,\quad ec=c,\quad be=b,\quad ce=c.$$
(This is a codimension one ideal in the path algebra of the quiver with two vertices, two arrows $a$ and $b$ in opposite directions between the two vertices, modulo the relation $ba=0$.)
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5
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https://mathoverflow.net/users/22989
|
405381
| 166,211 |
https://mathoverflow.net/questions/405385
|
10
|
Let $P(z) = \prod\_{i = 1}^n (z - z\_i) \in \mathbb{C}[z]$ be a monic polynomial having all roots $z\_1, \dots, z\_n$ on the unit circle $\mathbb{T} := \{z \in \mathbb{C} : |z| = 1\}$.
What is known about upper bounds for
$$M(P) :=\max\_{|z| = 1} |P(z)|$$
in terms of the distribution of $z\_1, \dots, z\_n$ in $\mathbb{T}$ ?
On the one hand, if $z\_1, \dots, z\_n$ are equally spaced on $\mathbb{T}$ then we have that $|P(z)| = |z^n - 1|$ and so $M(P) \leq 2$. In other words, $M(P)$ is small.
On the other hand, if $z\_1, \dots, z\_n$ all belong to one half of $\mathbb{T}$, say $\Im(z) < 0$, then it follows easily that $M(P) \geq P(1) \geq (\sqrt{2})^n$, and so $M(P)$ is large.
My guess is that knowing that $z\_1, \dots, z\_n$ are "well distributed", in some precise quantitative way, would provide a nontrivial upper bound for $M(P)$.
I searched in the literature but I could not find anything like that. Thanks for any suggestion.
*Note.* [This](https://mathoverflow.net/questions/63525/polynomials-on-the-unit-circle) is a somehow similar question but regarding a lower bound for $M(P)$ and for "random" $P$.
|
https://mathoverflow.net/users/388007
|
Upper bound for maximum modulus of polynomial on unit circle in term of the distribution of its roots
|
Let me first start with the other side: does the maximum being small guarantee that the roots are equidistributed? This is indeed the case, and is a beautiful theorem of Erdos and Turan. For a recent exposition see [Equidistribution of zeros of polynomials](https://arxiv.org/abs/1802.06506) (published paper in the Amer. Math. Monthly). This shows that
$$
{\mathcal D}(P) \le C \sqrt{N \log M(P)},
$$
with $C= 8/\pi$, and ${\mathcal D}(P)$ denotes the discrepancy of the angles of the roots. That is, ${\mathcal D}(P)$ is the maximum over all intervals $(\alpha, \beta)$ of $(0, 2\pi]$ of the absolute value of the difference between $N(\beta-\alpha)/(2\pi)$ and the number of roots with argument between $\alpha$ and $\beta$. In fact the result is a bit more precise, and recently there's been progress in improving the constant $C$ by [Shu and Wang](https://arxiv.org/pdf/2109.11006.pdf) (see also [Carneiro et al](https://arxiv.org/abs/2104.00105)).
Your question is about the other direction: to bound $M(P)$ in terms of the discrepancy ${\mathcal D}(P)$. This can also be done, using a result of [Carneiro and Vaaler](https://www.ams.org/journals/tran/2010-362-11/S0002-9947-2010-04886-X/S0002-9947-2010-04886-X.pdf) for example. Their Theorem 8.1 (with small changes of notation) shows that for any $1\le K\le N$
$$
\log M(P) \le \frac{N\log 2}{K+1} + \sum\_{k=1}^{K} \frac 1k \Big| \sum\_{j=1}^{N} z\_j^k\Big|.
$$
If the roots are equidistributed then the power sums $\sum\_{j=1}^{N} z\_j^k$ are small, and the desired bound would follow. In fact, with ${\mathcal D}(P)$ being the discrepancy, partial summation shows that
$$
\Big| \sum\_{j=1}^{N} z\_j^k \Big| \le 2\pi k {\mathcal D}(P),
$$
so that from Carneiro--Vaaler we would get
$$
\log M(P) \le \frac{N\log 2}{K+1} + 2\pi K {\mathcal D}(P).
$$
Choosing $K$ to minimize this, we obtain the complementary bound
$$
\log M(P) \le C \sqrt{N {\mathcal D}(P)},
$$
for a suitable constant $C$.
|
11
|
https://mathoverflow.net/users/38624
|
405397
| 166,220 |
https://mathoverflow.net/questions/405395
|
-1
|
If $B\_k(x)$ are the Bernoulli polynomials, then (by definition, if you like) we get that
$$\sum\_{k=0}^{\infty}B\_k(x)\frac{t^k}{k!}=\frac{te^{tx}}{e^t-1}$$
My question is whether or not there is a known formula for
$$\mathcal{G}(x;t):=\sum\_{k=0}^{\infty}B\_{2k+1}(x)\frac{t^{2k+1}}{(2k+1)!}.$$
The motivation for this question is that while studying modular forms the formula
$$\sum\_{k=1}^{\frac{m-1}{2}}k\mathcal{G}\left(\frac{k}{m},t\right)=\frac{mt}{8\sinh\left(\frac{t}{2}\right)\sinh\left(\frac{t}{2m}\right)}-\frac{t}{8\sinh^{2}\left(\frac{t}{2m}\right)}$$
appeared for every odd integer $m$. This seems to imply that there is something going on with $\mathcal{G}(x;t)$, at least for rational values of $x$.
|
https://mathoverflow.net/users/159298
|
Closed form for odd part of Bernoulli Polynomial generating function, $\sum_{k=0}^{\infty}B_{2k+1}(x)\frac{t^{2k+1}}{(2k+1)!}$
|
Using Pietro Majer's bisection formula we find by a straightforward computation (I did it with Maple, but I'm sure it could be done without too much difficulty by hand) that the OP's formula for
$$\sum\_{k=1}^{\frac{m-1}{2}}k\mathcal{G}\left(\frac{k}{m},t\right)$$ is indeed true.
|
2
|
https://mathoverflow.net/users/10744
|
405399
| 166,222 |
https://mathoverflow.net/questions/405048
|
3
|
John Isbell defined a notion of a median algebra (although the original idea is due to Birkhoff). A median algebra is a set $S$ with a ternary operation $[x,y,z]$ satisfying a set of [axioms](https://en.wikipedia.org/wiki/Median_algebra). The example to have in mind comes from lattice theory. Suppose $(L, \wedge, \vee, 0, 1)$ is a lattice. Then, we can define $[x, y, z] = (x \vee y) \wedge (y \vee z) \wedge (x \vee z)$ (with a little work, you can show it satisfies said axioms). The intuition is clear: if, for example $x \preceq y \preceq z$, then $[x,y,z] = y$. More interesting things can happen if $x, y, z$ do not lie in a chain.
My question is this: is there a nice notion of median of a) a finite subset of elements of a lattice or b) and arbitrary subset of a complete lattice. In order for said notion to be considered a median, it must, at the bare minimum, satisfy the property that if $C$ is a chain, then $\mathrm{median} (C)$ is a median (maybe there are two) of the chain.
|
https://mathoverflow.net/users/128639
|
Medians in lattice theory
|
Yes. There is a nice notion of an $n$-ary median term in a distributive lattice and in a median algebra as long as $n$ is odd.
Proposition: Suppose that $1\leq k\leq n$.
1. There is a term $t$ in the language of lattices such that if $x\_{1}\leq\dots\leq x\_{n}$, then $t(x\_{1},\dots,x\_{n})=x\_{k}$ and where $t$ satisfies the identity $t(z\_{1},\dots,z\_{n})=t(z\_{\sigma(1)},\dots,z\_{\sigma(n)})$ for each permutation $\sigma\in S\_{n}$.
2. If $s,t$ are terms in the language of lattices such that if
$x\_{1}\leq\dots\leq x\_{n}$, then $s(x\_{1},\dots,x\_{n})=t(x\_{1},\dots,x\_{n})=x\_{k}$ and where $s(z\_{1},\dots,z\_{n})=s(z\_{\sigma(1)},\dots,z\_{\sigma(n)})$ and
$t(z\_{1},\dots,z\_{n})=t(z\_{\sigma(1)},\dots,z\_{\sigma(n)})$ for each permutation $\sigma\in S\_{n}$, then
$s(u\_{1},\dots,u\_{n})=t(u\_{1},\dots,u\_{n})$ whenever $u\_{1},\dots,u\_{n}$ are elements of some distributive lattice.
For example, suppose that $m\_{n,k}^{+}(z\_{1},\dots,z\_{n})=\bigwedge\_{R\in[n]^{k}}(\bigvee\_{r\in R}z\_{r})$, and suppose that
$m\_{n,k}^{-}(z\_{1},\dots,z\_{n})=\bigvee\_{R\in[n]^{n+1-k}}(\bigwedge\_{r\in R}z\_{r})$. Then the terms $m\_{n,k}^{-},m\_{n,k}^{+}$ satisfy $1$ in the above proposition.
To prove $2$, we observe that $2$ holds for the $2$ element distributive lattice, and that the variety of all distributive lattices is generated by the $2$ element distributive lattice.
For distributive lattices, we may just write $m\_{n,k}$ for $m\_{n,k}^{-}$ or $m\_{n,k}^{+}$ since $m\_{n,k}^{-}$ and $m\_{n,k}^{+}$ are logically equivalent for distributive lattices.
By the next proposition, we see that the for distributive lattice, the medians $m\_{n,k}$ are precisely the terms which are symmetric with respect to all permutations in $S\_{n}$.
Proposition: If $t$ is a term in the language of distributive lattices such that $t(z\_{1},\dots,z\_{n})=t(z\_{\sigma(1)},\dots,z\_{\sigma(n)})$ for each permutation $\sigma$, then there is some $k$ with $1\leq k\leq n$ such that
$t(z\_{1},\dots,z\_{n})=m\_{n,k}(z\_{1},\dots,z\_{n})$ whenever $z\_{1},\dots,z\_{n}$ belong to some distributive lattice.
If $t$ is a term in the language of all distributive lattices, then let $t^{d}$ be the term obtained from $t$ by replacing every instance of $\wedge$ with $\vee$ and every instance of $\vee$ with $\wedge$.
Proposition: If $n=2r-1$, then the median $m\_{2r-1,r}$ is (up-to-logical equivalence) the only $2r-1$-ary term $t$ in the language of Boolean algebras where
$t(z\_{1},\dots,z\_{n})=t(z\_{\sigma(1)},\dots,z\_{\sigma(n)})$ for each permutation $\sigma$ and where $t=t^{d}$.
The term $m\_{2r-1,r}(x\_{1},\dots,x\_{2r-1})$ can actually be defined in terms of the ternary median operator. The theory of $2r-1$ median operations has been studied in the paper "The algebra of majority consensus" by Hans-Jorgen Bandelt and Gerasimos C. Meletiou.
Define terms $$(w\_{1},\dots,w\_{i+1},x^{i})=m(m(m(w\_{1},w\_{2},x),w\_{3},x),\dots,w\_{i+1},x).$$
The following proposition reconstructs the $2r-1$-ary median $m\_{2r-1,r}$ from
the $2r-3$-ary median $m\_{2r-3,r-1}$ and the ternary median $m$.
Proposition: In a distributive lattice, we have
$$m\_{2r-1,r}(x\_{1},\dots,x\_{2r-1})$$
$$=m\_{2r-3,r-1}(x\_{1},\dots,x\_{r-2},(x\_{r-1},x\_{r},x\_{r+1}),(x\_{r-1},x\_{r},x\_{r+1},x\_{r+2}^{2}))\dots(x\_{r-1}\dots,x\_{2r-2},x\_{2r-1}^{r-1})).$$
There are other ways of constructing the $2r-1$-median operator from the ternary median. The $2r-1$-median operator is the only $2r-1$-ary term $t$ up to logical equivalence in the language of ternary median algebras such that $t(z\_{1},\dots,z\_{2r-1})=t(z\_{\sigma(1)},\dots,z\_{\sigma(2r-1)})$ for all permutations $\sigma\in S\_{2r-1}$.
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4
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https://mathoverflow.net/users/22277
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405400
| 166,223 |
https://mathoverflow.net/questions/405361
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6
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Let $B \subset \mathbb R^n$ be the unit ball.
Consider a Borel measurable set $E \subset B$ with positive Lebesgue measure $|E|>0$ (say $|E| = |B|/2$).
Then, Lebesgue's density theorem, says that for a.e. $x\in E$
$$
\lim\_{r \downarrow 0} \frac{|B(x,r)\backslash E|}{|B(x,r)|} = 0.
$$
We can restate it as follows: for a.e. $x\in E$, for all $\epsilon>0$ there exists $r\_0 = r\_0(x, \epsilon)>0$ such that
$$
|B(x,r)\backslash E| \leq \epsilon |B(x,r)|, \quad 0<r<r\_0(x,\epsilon) .
$$
I am particularly interested in the dependence $\epsilon(r, x)$.
I have a question about this. Probably it has been studied but I have not been able to find any reference.
Given $E$, can we prove some uniformity for $\epsilon$ in a positive measure set (maybe of measure smaller than $|E|$)? That is, can we find some $r\_\*>0$ and $\phi$ continuous with $\phi(0)=0$ such that
$$
\epsilon(r,x) \leq \phi(r), \quad 0<r<r\_\*
$$
for all $x \in \tilde E$ for some Borel set $\tilde E\subset E$ with $0<|\tilde E|\leq |E|$.
**Edit:**
Initially I had two questions but I have decided to delete one.
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https://mathoverflow.net/users/173610
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Set where the speed of convergence is uniform in Lebesgue's density theorem
|
Let $$f\_n(x) = \sup\_{r \in {\mathbb Q} \cap [\frac{1}{n+1},\frac{1}{n})} \frac{|B(x,r)\setminus E|}{|B(x,r)|}\,,$$ so that $f\_n(x) \to 0$ for a.e. $x \in E$. By Egorov's theorem [1], for every $\epsilon>0$ there is a subset
$\tilde{E} \subset E$ with $|E \setminus \tilde{E}| <\epsilon$, such that
$f\_n(x) \to 0$ uniformly on $\tilde{E}$. It follows that
$$
\lim\_{r \downarrow 0} \frac{|B(x,r)\backslash E|}{|B(x,r)|} = 0
$$
uniformly in $\tilde{E}$, even when the limit is considered for real $r>0$.
[1] <https://en.wikipedia.org/wiki/Egorov%27s_theorem>
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4
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https://mathoverflow.net/users/7691
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405405
| 166,224 |
https://mathoverflow.net/questions/405363
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3
|
Let $0\neq \beta\in\overline{\mathbb{Z}}$ and let $n$ be a positive integer coprime to $N\_{\mathbb{Q}(\beta)/\mathbb{Q}}(\beta)$. Say that $n$ is a Fermat pseudoprime to base $\beta$ if
$$\beta^{n^{[\mathbb{Q}(\beta):\mathbb{Q}]}-1}\equiv 1\pmod n.$$
If $p$ is a prime which does not divide $N\_{\mathbb{Q}(\beta)/\mathbb{Q}}(\beta)$ and $\mathcal{O}$ is the ring of integers of $\mathbb{Q}(\beta)$, then $\bar{\beta}\in\mathcal{O}/p\mathcal{O}$ is a unit, and $\mathcal{O}/p\mathcal{O}$ is a product of residue fields, each with degree dividing $[\mathbb{Q}(\beta):\mathbb{Q}]$, so $p$ passes the test. This justifies the use of the word 'pseudoprime'.
If $\beta\in\mathbb{Z}\_{\ge 2}$, there are many ways to construct composite $n$ which are Fermat pseudoprimes to base $\beta$. For example, Cipolla's method: Let $p\nmid \beta(\beta^2-1)$ be an odd prime, then $(\beta^{2p}-1)/(\beta^2-1)$ is composite and a pseudoprime to base $\beta$.
**Questions:**
* For which $\beta$ do we have a way of constructing infinitely many composite base $\beta$ pseudoprimes?
* For which $\beta$ do we know there exist infinitely many composite base $\beta$ pseudoprimes?
**Edit:** Just [read](https://mathoverflow.net/a/49966) that if $n$ is a Carmichael number coprime to the discriminant of $\mathbb{Q}(\beta)$, such that the minimal polynomial of $\beta$ splits completely modulo $p$ for all $p\mid n$, then $\beta^n\equiv \beta\pmod n$. This suggests the answer to the second question is affirmative.
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https://mathoverflow.net/users/165478
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Can we construct composite Fermat pseudoprimes to integral algebraic bases?
|
The answer to both questions is yes. Just like Alford, Granville and Pomerance's proof that there are infinitely many Carmichael numbers, my proof that there are infinitely many Carmichael numbers where the minimal polynomial splits completely is essentially a constructive proof.
Both proofs need to jump through some hoops to get provability that you would not need in practice.
Grantham, Jon. There are infinitely many Perrin pseudoprimes. *J. Number Theory* **130** (2010), no. 5, 1117–1128.
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1
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https://mathoverflow.net/users/11722
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405406
| 166,225 |
https://mathoverflow.net/questions/405434
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4
|
$\DeclareMathOperator\SL{SL}$Let $d$ be a metric on the upper-half plane $\mathbb H = \{(x,y) : y > 0\}$ which is invariant with respect to the action of $\SL(2, \mathbb R)$ to $\mathbb H$ which is defined by
$$A \cdot z = \frac{az+b}{cz+d}, \ \ A = \begin{pmatrix} a & b \\ c & d \end{pmatrix} \in \SL(2, \mathbb R), \ z \in \mathbb H. $$
Specifically,
$$ d(A \cdot z, A \cdot w) = d(z,w), \ \ z, w \in \mathbb H, A \in \SL(2, \mathbb R).$$
Then, without assuming that $(\mathbb H,d)$ is geodesic, can we say that $d$ is Gromov-hyperbolic? Our original motivation comes from the square roots of some $f$-divergences between univariate Cauchy distributions which are non-geodesic metrics on $\mathbb H$.
|
https://mathoverflow.net/users/116429
|
Gromov hyperbolicity for (non-geodesic) metrics on the upper-half plane invariant with respect to SL(2, R) action
|
(Recall that an arbitrary metric space is Gromov-hyperbolic if
$$\sup\_{a,b,c,d}\Big((ab+cd)-\max(ac+bd,ad+bc)\Big)<\infty;$$
where $ab$ is the distance between $a$ and $b$. Note that this passes to subspaces.)
The answer is no: a counterexample being the square root of the usual distance on $\mathbf{H}^2$, which is invariant under the whole isometry group. For $\mathbf{R}\_{\ge 0}$ embeds isometrically into $\mathbf{H}^2$, and just the square root of the usual distance on $\mathbf{R}\_{\ge 0}$ is not hyperbolic:
Choose $a=0$, $b=2t$, $c=t$, $d=3t$. Then $ab+cd=2\sqrt{2}t$, $ac+bd\le ad+bc=(\sqrt{3}+1)\sqrt{t}$, so $(ab+cd)-\max(ac+bd,ad+bc)=(2\sqrt{2}-\sqrt{3}-1)\sqrt{t}$ is unbounded when $t\to\infty$.
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7
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https://mathoverflow.net/users/14094
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405441
| 166,234 |
https://mathoverflow.net/questions/405439
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0
|
Let $(X\_n)\_{n\in\mathbb{N}\_0}$ be a biased Random Walk on Galton-Watson tree with $\lambda\in(\lambda\_c,m)$.
How can I obtain the following equation:
$\sum\_{k=0}^{n-1}\mathbb{E}\_{e\_\*}[|X\_{k+1}|-|X\_k| \ | \mathcal{S}]=\sum\_{k=0}^{n-1}\mathbb{E}\_{e\_\*}[\frac{\nu(X\_k)-\lambda}{\nu(X\_k)+\lambda} \ | \mathcal{S}]$
$\mathcal{S}$ is here the survival set and $\mathbb{E}$ the expectation regarding the annealed probab. $\mathbb{P}$.
I thought that the sum on the left side is a sum of -1 and 1 and don’t see the link to the right side.
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https://mathoverflow.net/users/396065
|
Random walks on GW-trees (transformation)
|
It would be more useful if you specify exactly where you are reading this and what is $e\_\*$. Is the walk on a GW-tree or an augmented GW tree?
The underlying reason for this identity is that if a vertex $v$ has $b$ children with weight 1 each, and one parent with weight $\lambda$, then the net drift (=expected increment) of the biased RW from $v$ toward the children is $b/(b+\lambda)-\lambda/(b+\lambda)$.
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1
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https://mathoverflow.net/users/7691
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405445
| 166,236 |
https://mathoverflow.net/questions/405435
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4
|
Let $Y$ be a Stein manifold and $D\subset\subset Y$ be a Stein domain. I think $\overline D$ has connected boundary, and it should be somewhere, but I cannot find a reference for this. Thanks
|
https://mathoverflow.net/users/70148
|
Connectedness of boundary of a Stein domain
|
This is a consequence of the fact that any Stein manifold with complex dimension at least 2 has one end. This follows from Hartogs extension across compact sets in Stein manifolds. You can find a proof of this Hartogs
extension in the paper of Serre on Serre duality Commentarii Mathematici Helvetici volume 29 1955 pages 9-26 on page 22 .
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7
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https://mathoverflow.net/users/4696
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405448
| 166,239 |
https://mathoverflow.net/questions/405170
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3
|
Let $(\mathcal{M},\otimes)$ be a symmetric monoidal model category; I'll assume for simplicity that every object is fibrant. Suppose that the unit $I$ is NOT cofibrant. I'm interested in whether the derived tensor product with the unit is oplax/strong monoidal.
On the one hand, since $-\otimes^L I$ is the left derived functor of $-\otimes Q$ where $Q$ is a cofibrant replacement of $I$, an oplax structure on the derived functor could come from a monoidal Quillen adjunction with left adjoint $-\otimes Q$. This requires to find a "natural" diagonal map $Q\to Q\otimes Q$, and that the map $Q\otimes Q \to I$ be a weak equivalence.
On the other hand, the unit axiom seems to indicate that $-\otimes^L I$ is strong monoidal anyway, since $I\otimes^L I\simeq I$; but this isomorphism can be presented by two maps: $Q\otimes Q \to Q\otimes I$ and $Q\otimes Q \to I\otimes Q$. Does this really induce a strong monoidal structure, and are the two structures "the same" ?
|
https://mathoverflow.net/users/138396
|
On the derived functor of the tensor product in a monoidal category
|
Yes, the tensor product with a cofibrant replacement can be turned into a lax monoidal functor, where the lax structure maps are weak equivalences.
Consider the model category $\def\Mon{{\rm Mon}} \Mon(M)$ of monoids in $M$.
This model structure exists if $M$ satisfies the monoid axiom,
which is almost always true in practice.
Consider a cofibrant replacement $Q$ of the monoid $1$ in $\Mon(M)$.
The underlying object of $Q$ is a cofibrant object in $M$.
(See, for example, Theorem 6.7 in [arXiv:1410.5675](https://arxiv.org/abs/1410.5675), but earlier references probably exist.)
Now, the monoid structure of $Q$ equips the functor $Q⊗-$ with a structure of a lax monoidal functor whose lax structure maps are weak equivalences
because of the unit axiom for $C$.
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2
|
https://mathoverflow.net/users/402
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405453
| 166,240 |
https://mathoverflow.net/questions/405425
|
3
|
This question comes from book [Ju-Yi Yen and Marc Yor](https://link.springer.com/book/10.1007/978-3-319-01270-4) [P59](https://i.stack.imgur.com/DJGja.png) and [P60](https://i.stack.imgur.com/3C4qD.png),
On page 59, "Define $\mathcal{Z}\_\omega=\{t:B\_t(\omega)=0\},$ and $\tau\_l$ is the inverse local time. The complement of $\mathcal{Z}\_\omega$is shown to be $\mathcal{Z}\_\omega^c=\bigcup\_{s>0}\left]\tau\_{s-},\tau\_s\right[$, where $]\tau\_{s-},\tau\_s[$ are the maximal intervals of constancy of the local time $L$. This defines the excursion process as $e\_l(\omega)=B(\tau\_{l-}+t) 1\_{(t\leq \tau\_s-\tau\_{s-})}$"
On page 60, "$(e\_s, s\geq 0)$ is a Poisson point process."
My question is:
1. how to understand the definition of excursion process $e\_l(\omega)=B(\tau\_{l-}+t) 1\_{(t\leq \tau\_s-\tau\_{s-})}$? what is the picture?
2. how to prove this excursion process is a Poisson Point Process?
|
https://mathoverflow.net/users/147009
|
How to prove excursion process is a Poisson point process?
|
The standard reference for Brownian excursions is Chapter 12 in the classic book [1]. Other developments of excursion theory can be found in [2]-[4] and many other sources.
To develop the intuition, a good approach is to start with random walks.
(See, e.g., the exposition of local time in [6]).
The counting measure on the zero set $\{Z\_0=0, Z\_1, Z\_2,\ldots \}$ of the symmetric simple random walk (SRW) on the integers,
suitably normalized, converges to the Brownian local time at zero. Hence it is natural to understand the excursions of SRW of duration at least $2\ell$ that start in one of $Z\_{an},\dots, Z\_{bn}$. Using the reflection principle and Stirling, see e.g. [5] Thm 9.3, we know that the chance a SRW excursion will be "long" (=have duration greater than $2\ell$) is asymptotic to $p\_\ell=(\pi \ell)^{-1/2}$ as $\ell$ grows. Moreover, these events are independent for each of the excursions started at $Z\_{an},\dots, Z\_{bn}$, so the total number of "long" excursions among these has a Binomial Bin$(bn-an,p\_\ell)$ distribution. To get a nontrivial limit as $n,\ell \to \infty$, take $\ell=\ell(n)=tn^2$ so that $p\_\ell=(\pi t)^{-1/2}/n$. We deduce that the number of "long" excursions (of duration $\ge 2tn^2$) started in $Z\_{an},\dots, Z\_{bn}$ has a limit distribution $$\lim\_n { \rm Bin}(bn-an,(\pi t)^{-1/2}/n)={\rm Poisson}((b-a) \cdot (\pi t)^{-1/2})\,.$$ These counts of long excursions are independent for disjoint intervals
$[a\_1,b\_1], [a\_2,b\_2],\ldots$, so the counting measure of long excursions will indeed converge to a Poisson process as $n \to \infty$.
[1] Revuz, Daniel, and Marc Yor. Continuous martingales and Brownian motion. Vol. 293. Springer Science & Business Media, 2013.
[2] Pitman, J., and M. Yor. "Itô's excursion theory and its applications." Japanese Journal of mathematics 2, no. 1 (2007): 83.
[3] Watanabe, S. (2010). Itô’s theory of excursion point processes and its developments. Stochastic processes and their applications, 120(5), 653-677.
<https://core.ac.uk/download/pdf/82811223.pdf>
[4] K. L. Chung, Excursions in Brownian motion, Ark. Mat., 14 (1976), 155–177.
[5] Révész, Pál. Random walk in random and non-random environments. World Scientific, 2013.
[6] Mörters, Peter, and Yuval Peres. Brownian motion. Vol. 30. Cambridge University Press, 2010.
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3
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https://mathoverflow.net/users/7691
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405455
| 166,241 |
https://mathoverflow.net/questions/405436
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2
|
Let $D\_1$ be a domain with smooth boundary and assume that $D\_1$ is a proper subset of $D\_2$ which is itself a bounded domain in $\mathbb R^n$ with a smooth boundary. Assume also that $D\_2\setminus D\_1$ is connected. We write $L^2(D\_2\setminus D\_1)$ for the set of functions in the space
$$\{f \in L^2(D\_2)\,:\,\textrm{supp}(f)\subset D\_2\setminus \overline{D\_1}\}$$
let us define the mapping
$$S: L^2(D\_2\setminus \overline{D\_1})\mapsto H^{\frac{3}{2}}(\partial D\_1),$$
through
$$ Sf:= u|\_{\partial D\_1},$$
where $u \in H^2(D\_2)$ is the unique solution to the equation
$$ \Delta u =f \quad \text{on $D\_2$},$$
subject to $u|\_{\partial D\_2}=0$. Is it true that the image of $S$ is dense in $H^{\frac{1}{2}}(\partial D\_1)$?
|
https://mathoverflow.net/users/50438
|
Density of traces of solutions to an elliptic equation
|
The answer is yes: take any smooth function $g\_0$ on $\partial D\_1$ and solve the Dirichlet problem
$$
\begin{cases}
\Delta g = 0 & \text{ on } D\_1\\
g = g\_0 & \text{ on } \partial D\_1.
\end{cases}
$$
Now extend $g$ to a smooth function on $\mathbb{R}^n$. Multiply by a smooth cutoff function $\eta$ which is $1$ on $D\_1$ and compactly supported on $D\_2$.
Then $f = \Delta (\eta g)$ is smooth and supported on $D\_2 \setminus \bar{D}\_1$, so in particular lies in the given $L^2$ space. This shows that $g\_0$ is in the image of your operator $S$, and smooth functions are dense in $H^s$.
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1
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https://mathoverflow.net/users/378654
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405457
| 166,243 |
https://mathoverflow.net/questions/405268
|
5
|
Short version: what are some interesting hyperdoctrines for classical (not intuitionistic) first-order logic, that are not models in the traditional sense? (Where the terminal and initial hyperdoctrines are "uninteresting".)
Long version:
The categorical semantics of first-order logic are given by [hyperdoctrines](https://ncatlab.org/nlab/show/predicate+logic). This is in contrast with the traditional semantics in terms of model theory.
In brief, in the traditional picture, we think of the semantics of a first-order theory as being some set $U$ ("the universe") plus an interpretation of every constant symbol / function symbol / predicate symbol in that universe (so, eg, for the unary function symbol $f$, the model provides $[\![f]\!] : U \to U$, and so on and so forth.) We can then interpret every proposition in the theory, interpreting (for example) $f x = x$ as the subset of $U$ where $[\![f]\!]$ is fixed.
By contrast, in the categorical style, we think of the semantics of a first-order theory T as being a [hyperdoctrine](https://ncatlab.org/nlab/show/hyperdoctrine) plus some interpretations for the symbols and etc, which I'll call a "hyperdoctrine for T". This is more-or-less a stream of lattices, equipped with some strucures relating the lattices to one another, plus interpretations for the various symbols in terms of the lattices. Roughly speaking, the nth lattice is thought of as the possible interpretations of a proposition with n variables free, and the structures relating the lattices are about substitution and quantification.
The latter framework is more general. For instance, we can turn a traditional model for a theory T into a hyperdoctrine for T by letting the nth lattice be the lattice of all subsets of $U^n$. But we also have new hyperdoctrines: most notably a terminal one, corresponding to the choice where every lattice in the stream is the trivial (one-object) lattice; and the initial one, corresponding to the syntax.
And presumably, the categorical semantics also add a whole host of more interesting "new models". Like, presumably there are hyperdoctrines (for, say, 1st order arithmetic) that assert some combination of sentences, that no traditional model asserts. (Such combinations must necessarily be infinite, on account of the traditional completeness theorem, but still. (ETA: Not quite; see the comments below.)) And, like, yes, the terminal and initial hyperdoctrines show us some boring ways that this is true, but surely this newfound generality does more than just bolt a new "initial" and "terminal" model onto the traditional models. So, what are some of these new models?
(Ideally in the classical setting; I know we can get topological models and stuff if we consider intuitionistic logic, but it still seems to me that even classically we must have additional interesting hyperdoctrines, and I'd like to know what they are.)
(Ideally I'm looking for hyperdoctrines that feel motivated in their own right, more like "the lattices contain only the propositions that satisfy the following natural property" than "well given any hyperdoctrine we can generate a new one by bolting on a spandrel; just do that to the initial model". My apologies for the vagueness of this constraint. What I'm really after here are intuitions about how hyperdoctrines expand the space of models.)
(If I'm wrong in my assumption that there are interesting hyperdoctrines aside from the initial and terminal one, I'd also be happy to hear about that.)
|
https://mathoverflow.net/users/15201
|
What are some interesting hyperdoctrines that are not classical models?
|
Maybe I am wrong, but it seems to me that the other answers are misunderstanding the question. The emphasis on syntactic hyperdoctrines seems to me beside the point.
A (classical, first-order) hyperdoctrine is a categorical (semantic) structure, consisting of a category $C$ and a functor $P: C \to \rm BoolAlg$ together with adjoints and a Beck-Chevalley condition. It seems to me that the OP wants to consider a particular construction of such a hyperdoctrine as follows: given a set $U$, let $C$ be the full subcategory of $\rm Set$ determined by the objects $U^n$, and let $P(U^n)$ be the powerset of $U^n$. Let's call this hyperdoctrine ${\rm Set}|\_U$.
For a particular theory $T$ (which, in general, should be multi-sorted) and hyperdoctrine $C$, one can then consider the notion of a "model of $T$ in $C$". This consists of an interpretation function assigning an object of $C$ to each sort of $T$, a morphism of $C$ to each function symbol of $T$, and a predicate in some $P(X)$ to each relation symbol of $T$, such that the axioms of $T$ are "satisfied". I think this is what the OP means by a "hyperdoctrine for $T$". One can rephrase this in a more highbrow way by building a "syntactic hyperdoctrine" $S\_T$ out of $T$ and saying that a model of $T$ in $C$ is a morphism of hyperdoctrines $S\_T \to C$, but that isn't necessary.
It seems to me that the OP is asking whether there are any interesting models of $T$ in hyperdoctrines $C$ not of the form ${\rm Set}|\_U$ (and also where $C$ is not terminal and not the initial $T$-hyperdoctrine $S\_T$).
There are some fairly trivial answers to that question. One, which I think appears in the other answers, is that for any theory $T'$ extending $T$, there is a canonical model of $T$ in $S\_{T'}$. Another is that instead of models in ${\rm Set}|\_U$, we can consider models in $\rm Set$ itself; if $T$ has only one (base) sort then any such model factors uniquely through some ${\rm Set}|\_U$, so it is not very different (but if $T$ has more than one sort, then this more general kind of model is the "correct" one to think about).
However, I think the most satisfying answer is that
**You can replace $\rm Set$ by any (Boolean) category.**
In other words, given any category $C$, you can define a hyperdoctrine over $C$ where $P(X)$ is the poset of subobjects of $X\in C$. If $C$ is a Boolean category, this will be a classical first-order hyperdoctrine, call it ${\rm Sub}(C)$. Then you can consider models of any theory $T$ in ${\rm Sub}(C)$, and they will be new, different, and interesting, and can satisfy principles that don't hold in $\rm Set$.
You get many more interesting models if you generalize to intuitionistic logic, in which case you use Heyting categories instead of Boolean categories and every elementary topos is an example. This leads to the whole field of topos theory and the categorical semantics of internal languages. But there are also interesting Boolean categories other than $\rm Set$, such as ${\rm Set}^X$ for any set $X$, or even for any groupoid $X$.
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4
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https://mathoverflow.net/users/49
|
405465
| 166,245 |
https://mathoverflow.net/questions/382462
|
0
|
Is there a discrete topological dynamical system $(X,f)$, where $X$ is a compact metric space (with distance $d$), which is transitive but not minimal, such that $h(f)>0$ and every point is a full entropy point?
By transitive I mean that there is a point with a dense orbit. By minimal I mean that every orbit is dense. By full entropy point I mean the concept defined in [1](https://www.ams.org/journals/tran/2007-359-12/S0002-9947-07-04357-7/S0002-9947-07-04357-7.pdf), that is a point $x$ such that the topological entropy, restricted to any of its closed neighborhoods, coincides with the entropy $h(f)$ of the system.
Finally, for every closed neighbor $K(x)$ of $x$, let $N(\epsilon, n, K)$ be the largest cardinality of an $(n,\epsilon)$-separated subset $A\subset K$ in the metric $d\_n(x,y)=\max\_{i=0,\dots,n}d(f^i(x),f^i(y))$. Then the entropy restricted to $K(x)$, in symbols $h(f,K)$, is defined as the usual limit
$$\lim\_{\epsilon\to 0}\lim\_{n\to\infty} \frac 1n\log{N(\epsilon, n, K)}.$$
Notice that $h(f)=\sup\_K h(f,K)$, where the supremum is taken over all compact subsets of $X$.
|
https://mathoverflow.net/users/167834
|
Non-minimal system in which every point is a full entropy point
|
I guess this is not hard to do with a subshift. Take $X$ to be your favorite minimal positive entropy subshift on $\{0,1\}$ (such examples are constructed in Hahn-Katznelson, among other works).
Now, choose any $x \in X$ and define $y$ on $\{0,1,\*\}$ as
$y = .\* \ x\_1 \* x\_2 x\_3 \* x\_4 x\_5 x\_6 \* x\_7 x\_8 x\_9 x\_{10} \* \ldots$
Define $Y$ to be the orbit closure of $y$. Clearly $Y$ is transitive by definition, and is not minimal since it strictly contains $X$. It's not hard to check that $h(Y) = h(X)$; by the ergodic theorem, any ergodic measure on $Y$ gives $\*$ zero measure, and so is supported on $X$. Then $h(Y) = h(X)$ by the variational principle.
Finally, every point of $Y$ should have full entropy, since every point $z \in Y$ contains arbitrarily long words of $X$, and so all words in the language of $X$ by minimality. So if I understand your definition correctly, $z$ should have full entropy.
I don't see how any of this could be carried over to $[0,1]$, but as we discussed in the comments, it may be impossible to construct such an example on the interval.
|
1
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https://mathoverflow.net/users/116357
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405471
| 166,249 |
https://mathoverflow.net/questions/405470
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4
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Two players take turns coloring edges on an $n$-by-$n$ grid. Both players use the same color. Every time a player surrounds a square of the grid, they mark that square with their name and go again. A player cannot pass; they must move if it is their turn. The goal is to make as many squares as possible. What is the score with optimal play as a function of $n$? Note that ties are only possible if $n$ is odd so that the total number of squares is even.
To state it more formally: there is a graph with vertices labelled by pairs $(i,j)$ for $i,j \in \{1,\ldots,n\}$. There is an edge between a pair of vertices $(i,j)$ and $(k,l)\ $ iff $\ |i-k|+|j-l|=1.\ $ The state of the game is described by specifying the player whose move it is and specifying a bit for each edge, saying whether that edge is "colored" or "uncolored". One player moves first. Initially all edges are uncolored. On a turn, a player picks an uncolored edge and colors it. If the edge that a player colors makes at least one elementary square colored (i.e. that edge is in a cycle of length $4$ of colored edges), then the player scores one point for each such elementary square, and it is that player's turn again. Otherwise, it is the other players turn. Game terminates when there are no uncolored edges. Each player tries to maximize their own score.
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https://mathoverflow.net/users/83174
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Who wins this two player game of making squares?
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This just the game of [Dots and Boxes](https://en.wikipedia.org/wiki/Dots_and_Boxes). There is a huge literature on this game. In particular, Berlekamp's book referenced in the above link shows how difficult this game is.
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13
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https://mathoverflow.net/users/2807
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405472
| 166,250 |
https://mathoverflow.net/questions/405468
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15
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Belyi's theorem establishes a correspondence between smooth projective curves defined over number fields and the so called *dessins d'enfants* which are bipartite graphs embedded on an oriented surface so that each component of their complement is a topological disk. To be more precise, given a smooth projective curve $X$ of genus $g$ over $\overline{\Bbb{Q}}$, Belyi's theorem guarantees the existence of a branched cover $f:X(\Bbb{C})\rightarrow\Bbb{P}^1(\Bbb{C})$ unramified outside $\{0,1,\infty\}$. Then, $f^{-1}([0,1])$ is an embedded graph that can be considered as the $1$-skeleton of a CW structure on $X(\Bbb{C})$. Conversely, such a graph on an oriented compact surface of genus $g$ equips it with a complex structure and such a branched cover.
**My Question:** Can dessins be used to defined a "reasonable" (e.g. comparable to a Weil height) height function on the moduli space $\mathcal{M}\_g(\overline{\Bbb{Q}})$? For instance, is the minimum possible number of edges of a dessin (i.e. the minimum possible degree of a Belyi map on the complex curve) be used as a height function? Any reference to the literature on this question is highly appreciated.
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https://mathoverflow.net/users/128556
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Height functions on $\mathcal{M}_g(\overline{\Bbb{Q}})$ defined via dessins d'enfants?
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If $X$ is a (smooth projective) curve over $\overline{\mathbb{Q}}$, we define
>
>
> >
> > The Belyi degree $\deg\_B(X)$ of $X$ to be the minimum degree of a Belyi map $X\to \mathbb{P}^1\_{\overline{\mathbb{Q}}}$.
> >
> >
> >
>
>
>
The Belyi degree is a function on $\mathcal{M}\_g(\overline{\mathbb{Q}})$ which satisfies the following Northcott-type finiteness property.
**Proposition. (Strong Northcott)** For every integer $d$, the set of $\overline{\mathbb{Q}}$-isomorphism classes of curves $X$ over $\overline{\mathbb{Q}}$ with $\deg\_B(X)\leq d$ is finite.
*Proof.* Like all finiteness statements, this one also boils down to some "general" finiteness statements. In this case, the statement (seemingly arithmetic in nature) is a consequence of a (topological) finiteness property of the fundamental group of $\mathbb{P}^1\setminus \{0,1,\infty\}$. Indeed, the proposition can be proven using the fact that the fundamental group of $\mathbb{P}^1\setminus \{0,1,\infty\}$ is finitely generated, and that a finite generated group has only finitely many finite index subgroups of index at most $d$. QED
Note that this Proposition can be used to enumerate all curves over $\overline{\mathbb{Q}}$. Simply "write" down the curves of Belyi degree at most $3$, then $4$, then $5$, etc.
Note that the Northcott property satisfied by the Belyi degree is much stronger than that of any Weil height $h$. The Northcott property for a Weil height usually requires in addition a bound on the *degree* of the point.
The Strong Northcott property implies that, given a Weil height $h$ (or any function!) on $\overline{\mathbb{Q}}$, there is a function $f(\deg\_B(-))$ such that
$$ h(X) \leq f(\deg\_B(X)).$$
Thus, any function on $\ {\mathcal{M}\_g}(\overline{\mathbb{Q}})$ is bounded by a function in the Belyi degree (simply because of the above Proposition). For example, the genus of $X$ is bounded by $\deg\_B(X)$. This follows from the Riemann-Hurwitz formula.
There are a few natural (arithmetic) invariants on $\ {\mathcal{M}\_g}(\overline{\mathbb{Q}})$ such as the Faltings height for which one can write down explicit bounds. For example:
**Theorem.** If $X$ is a curve over $\overline{\mathbb{Q}}$ with Faltings height $h\_F(X)$, then $$h\_F(X) \leq 10^8 \deg\_B(X)^6.$$
This (with many more explicit inequalities) is proven in [1]. The motivation for proving such inequalities is that they can be used to control the running time of certain algorithms computing coefficients of modular forms.
The question of actually computing the Belyi degree of a curve is an interesting one. An algorithm (which I would not recommend trying to implement) for doing so is given in [2].
[1] A. Javanpeykar. *Polynomial bounds for Arakelov invariants of Belyi curves, with an appendix by Peter Bruin.* Algebra and Number Theory, Vol. 8 (2014), No. 1, 89–140.
[2] A. Javanpeykar and J. Voight. *The Belyi degree of a curve is computable*
Contemp. Math., 2019, 722, p. 43-57.
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11
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https://mathoverflow.net/users/4333
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405480
| 166,253 |
https://mathoverflow.net/questions/405262
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18
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This question was previously posted [here](https://math.stackexchange.com/questions/4185940/is-the-p-adic-density-of-the-image-of-a-polynomial-always-rational) on MSE.
Let $P(x)$ be a polynomial with integer coefficients, and let $p$ be a prime number. For $n\in\mathbb N$, let $I\_n$ be the number of integers $i\in\{1,\dotsc,p^n\}$ such that there is an integer $x$ for which $P(x)\equiv i\bmod p^n$. Now define
$$\delta:=\lim\_{n\rightarrow\infty}\frac{I\_n}{p^n}.$$
Remark that this limit exists since $\frac{I\_n}{p^n}\geq \frac{I\_{n+1}}{p^{n+1}}$ for all $n$. One could say that $\delta$ is ‘the $p$-adic density of the image of $P$’.
Now I have the following question: is $\delta$ a rational number for all polynomials $P$ and primes $p$?
This question is connected to another question [Cardinality of the image of a polynomial modulo $p^n$](https://mathoverflow.net/questions/295365/cardinality-of-the-image-of-a-polynomial-modulo-pn) on MathOverflow, which asks for general information on the behavior of $I\_n$ as $n\rightarrow\infty$.
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https://mathoverflow.net/users/394725
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Is the p-adic density of the image of a polynomial always rational?
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Using the strategies suggested by @Merosity on MSE and @Gro-Tsen and @RP\_ on MO, I have found a proof that the density is indeed always rational.
Let $P$ be a polynomial with integer coefficients, and let $p$ be a prime number. If $P$ is a constant polynomial, then we obtain $\delta=0$ which is rational. So assume that $P$ is nonconstant. We shall prove that $\delta$ is rational by means of a chain of lemmas. Each lemma only uses the previous lemma.
**Lemma**
Let $g:\mathbb Z\_p\rightarrow\mathbb Z\_p$ be a power series
$g(x)=\sum\_{i=0}^{\infty}g\_ix^i$ with $g\_i\in\mathbb Z\_p$ for all $i$ and
$g\_i\rightarrow0$ as $i\rightarrow\infty$. Suppose that $g'(0)=1$. Then the
restriction of $g$ to $p\mathbb Z\_p$ has image $g(0)+p\mathbb Z\_p$.
**Proof**
The proof is analoguous to that of Hensel's lemma.$\tag\*{$\blacksquare$}$
Now let $v\in\mathbb Z\_p$ for which $P'(v)=0$, and let
$n\_v\geq2$ be the largest integer such that $(x-v)^{n\_v}$
divides $P(x)-P(v)$. Let $Q\_v(x)\in\mathbb Z\_p[x]$ be the polynomial such that $P(x)=P(v)+(x-v)^{n\_v}Q\_v(x-v)$. Now $Q\_v(0)\neq0$.
**Lemma**
For all $v\in\mathbb Z\_p$ such that $P'(v)=0$, there is an integer $N\_v\in\mathbb N$ such that $$P(v+p^{N\_v}\mathbb Z\_p)=P(v)+p^{n\_vN\_v}Q\_v(0)f\_v(\mathbb Z\_p),$$ where $f\_v:\mathbb Z\_p\rightarrow\mathbb Z\_p$ is defined by $f\_v(z):=z^{n\_v}$.
**Proof**
Define the function $R:\mathbb Z\_p\rightarrow\mathbb Z\_p$ by $$R(x)=\sum\_{i=0}^{\infty}r\_ix^i:=\sum\_{i=0}^{\infty}p^{2i}n\_v^{i}\binom{\frac{1}{n\_v}}{i}x^i.$$ For all $i$, we have $v\_p(r\_i)=v\_p(p^{2i}n\_v^{i}\binom{\frac{1}{n\_v}}{i})\geq v\_p(p^{2i})-v\_p(i!)>i$, so all $r\_i$ are $p$-adic numbers and $\lim\_{i\rightarrow\infty}r\_i=0$. Therefore, $R(x)$ is well-defined.\
It follows from the definition of $R$ that $R(x)^{n\_v}=1+p^2n\_vx$ for all $x\in\mathbb Z\_p$. Now define the function $K:\mathbb Z\_p\rightarrow \mathbb Z\_p$ by $$K(x):=R\left(\frac{Q\_v(p^2n\_vQ\_v(0)x)-Q\_v(0)}{p^2n\_vQ\_v(0)}\right).$$ Then $K(x)=\sum\_{i=0}^{\infty}k\_ix^i$ for coefficients $k\_i\in\mathbb Z\_p$ with $\lim\_{i\rightarrow\infty}k\_i=0$.
It follows for all $x\in\mathbb Z\_p$ that $$Q\_v(0)K(x)^{n\_v}=Q\_v(0)\left(1+p^2n\_v\frac{Q\_v(p^2n\_vQ\_v(0)x)-Q\_v(0)}{p^2n\_vQ\_v(0)}\right)=Q\_v(p^2n\_vQ\_v(0)x).$$
Therefore, we see that \begin{align\*}P(v+p^2n\_vQ\_v(0)x)&=P(v)+(p^2n\_vQ\_v(0)x)^{n\_v}Q\_v(p^2n\_vQ\_v(0)x)\\
&=P(v)+(p^2n\_vQ\_v(0)x)^{n\_v}Q\_v(0)K(x)^{n\_v}\\
&=P(v)+Q\_v(0)(p^2n\_vQ\_v(0)xK(x))^{n\_v}\end{align\*}
for all $x\in\mathbb Z\_p$. Since $\frac{d(xK(x))}{dx}\big|\_{x=0}=K(0)=R(0)=r\_0=1$, we can use the previous lemma to see that the image of $xK(x)$ restricted to $p\mathbb Z\_p$ is $p\mathbb Z\_p$. Therefore, the image of $P$ restricted to $v+p^3n\_vQ\_v(0)\mathbb Z\_p$ is $P(v)+Q\_v(0)(p^3n\_vQ\_v(0))^{n\_v}f\_v(\mathbb Z\_p)$. So the lemma holds for $N\_v:=v\_p(p^3n\_vQ\_v(0))$.
$\tag\*{$\blacksquare$}$
**Lemma** For all $v\in\mathbb Z\_p$ such that $P'(v)=0$, there is a finite set $S\_v\subset\mathbb Z$ such that $$P(v+p^{N\_v}\mathbb Z\_p)=P(v)+p^{n\_vN\_v}Q\_v(0)\left(\{0\}\cup\left(\bigcup\_{s\in S\_v}\bigcup\_{i=0}^{\infty}p^{in\_v}(s+p^{2v\_p(n\_v)+1}\mathbb Z\_p)\right)\right).$$
**Proof**
Define $S\_v:=\{a^{n\_v}\mid 0<a<p^{v\_p(n\_v)+1},p\nmid a\}$. Then it follows from arguments similar to the proof of Hensel's lemma that for all $i\geq0$, the image of $f\_v$ restricted to $p^i\mathbb Z\_p\backslash p^{i+1}\mathbb Z\_p$ is equal to $p^{in\_v}\bigcup\_{s\in S\_v}(s+p^{2v\_p(n\_v)+1}\mathbb Z\_p)$. Taking the union over all $i$, we see that the image of $f\_v$ equals $\{0\}\cup\left(\bigcup\_{s\in S\_v}\bigcup\_{i=0}^{\infty}p^{in\_v}(s+p^{2v\_p(n\_v)+1}\mathbb Z\_p)\right)$. Therefore, this lemma follows from the previous lemma.
$\tag\*{$\blacksquare$}$
**Lemma**
For all $\sigma\in P(\mathbb Z\_p)$, there is an integer $M\_{\sigma}\geq0$ such that $$P(\mathbb Z\_p)\cap(\sigma+p^{M\_{\sigma}}\mathbb Z\_p)$$ has rational $p$-adic density.
**Proof**
First, suppose that there is an $x\in\mathbb Z\_p$ such that $P(x)=\sigma$ and $P'(x)\neq0$. Then it follows from arguments similar to Hensel's lemma that $P(\mathbb Z\_p)$ contains a neighbourhood of $\sigma$. This immediately proves the lemma.\
Now suppose that for all $x\in\mathbb Z\_p$ such that $P(x)=\sigma$, we have $P'(x)=0$. Let $V\_{\sigma}:=P^{-1}(\sigma)$, then $V\_{\sigma}$ is a finite set since $P$ is nonconstant. Since $P$ is continuous, we can choose $M\_{\sigma}\in\mathbb Z\_{\geq0}$ such that $P^{-1}(\sigma+p^{M\_{\sigma}}\mathbb Z\_p)$ is contained in $\bigcup\_{v\in V\_{\sigma}}(v+p^{N\_v}\mathbb Z\_p)$. Now it follows that $$P(\mathbb Z\_p)\cap(\sigma+p^{M\_{\sigma}}\mathbb Z\_p)=\bigcup\_{v\in V\_{\sigma}}P(v+p^{N\_v}\mathbb Z\_p)\cap(\sigma+p^{M\_{\sigma}}\mathbb Z\_p).$$
Using the previous lemma, we see that this set equals
$$\bigcup\_{v\in V\_{\sigma}}\left(\sigma+p^{n\_vN\_v}Q\_v(0)\left(\{0\}\cup\left(\bigcup\_{s\in S\_v}\bigcup\_{i=0}^{\infty}p^{in\_v}(s+p^{2v\_p(n\_v)+1}\mathbb Z\_p)\right)\right)\right)\cap(\sigma+p^{M\_{\sigma}}\mathbb Z\_p).$$
Let $n:=\mathrm{lcm}\_{v\in V\_{\sigma}}(n\_v)$. Then there exists an integer $C>0$ and a finite collection of numbers $a\_l\in \mathbb Z\_p\backslash p^{C}\mathbb Z\_p$, $1\leq l\leq L$, such that our set can be written as
$$P(\mathbb Z\_p)\cap (\sigma+p^{M\_{\sigma}}\mathbb Z\_p)=
\{\sigma\}\cup\bigcup\_{l=1}^{L}\left(\sigma+\bigcup\_{i=0}^{\infty}p^{in}(a\_l+p^{C}\mathbb Z\_p)\right).$$
The $p$-adic density of this set is $$\lvert\{a\_l\bmod p^{C}\mid 1\leq l\leq L\}\rvert\cdot \frac{p^n}{p^n-1}\cdot p^{-C}$$ which is a rational number.
$\tag\*{$\blacksquare$}$
**Theorem**
The $p$-adic density of $P(\mathbb Z\_p)$ is rational.
**Proof**
When we define $B\_{\sigma}:=\sigma+p^{M\_{\sigma}}\mathbb Z\_p$ for all $\sigma\in P(\mathbb Z\_p)$, then $\{B\_{\sigma}\mid \sigma\in P(\mathbb Z\_p)\}$ is an open cover of $P(\mathbb Z\_p)$. Since $\mathbb Z\_p$ is a compact set and $P$ is continuous, the image $P(\mathbb Z\_p)$ is also a compact set. Therefore, the open cover has a finite subcover $\{B\_{\sigma\_1},\dots,B\_{\sigma\_q}\}$ which is minimal. The sets in this subcover must be pairwise disjoint, so it follows that $P(\mathbb Z\_p)$ is the disjoint union of the sets $B\_{\sigma\_i}\cap P(\mathbb Z\_p)$ for $1\leq i\leq q$. Therefore, the $p$-adic density of $P(\mathbb Z\_p)$ is the sum of the densities of these sets. By the last lemma, all those densities are rational. Therefore their sum is also rational.$\tag\*{$\blacksquare$}$
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7
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https://mathoverflow.net/users/394725
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405487
| 166,257 |
https://mathoverflow.net/questions/405479
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1
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Poincaré inequality is stated as follows:
>
> Let $ \Omega $ be bounded, connected, open subset pf $ \mathbb{R}^n $, with a $ C^1 $ boundary $ \partial \Omega $. Assume that $ 1\leq p<\infty $ and $ u\in W^{1,p}(\Omega) $. Then there exists a constant $ C $, depending only on $ n, p $ and $ \Omega $, such that
> $$
> \left\|u-\frac{1}{|\Omega|}\int\_{\Omega}u\right\|\_{L^p(\Omega)}\leq C\left\|Du\right\|\_{L^p(\Omega)}
> $$
> for each function $ u\in W^{1,p}(\Omega) $.
>
>
>
I have already know the proof from compactness argument. Now I want to prove this inequality by direct computation. First, I consider the condition that $ \Omega $ is convex and prove the inequality. Now I want to deal with the general case by using the extension theorem of Sobolev space. However, I can not deal with the error terms when using the theorem. Can you give me some hints or references?
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https://mathoverflow.net/users/241460
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How to prove Poincaré inequality by using extension theorem?
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It seems to me that some useful information you can find in paragraph 1.5 of
"Differentiable Functions on Bad Domains" by V. G. Mazia, S. Pobozchi.
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1
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https://mathoverflow.net/users/362006
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405492
| 166,259 |
https://mathoverflow.net/questions/405476
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5
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I am interested in a "dynamical" modification of the cardinals $\mathfrak r$ and $\mathfrak r\_\sigma$, well-known in the theory of cardinal characteristics of the continuum.
For a compact metrizable space $X$, let $\mathfrak r\_X$ be the smallest cardinality of a family $\mathcal R$ of infinite subsets of $\omega$ such that for every sequence $(x\_n)\_{n\in\omega}\in X^\omega$ there exists $R\in\mathcal R$ such that the subsequence $(x\_n)\_{n\in R}$ converges in $X$.
It is easy to see that $\mathfrak r\_2\le \mathfrak r\_X\le\mathfrak r\_{2^\omega}$ for every compact metric space $X$ containing more than one point. Moreover, it can be shown that
$$\mathfrak r\_X=\begin{cases}\mathfrak r\_2&\mbox{if $|X|\le\omega$};\\
\mathfrak r\_{2^\omega}&\mbox{if $|X|>\omega$}.
\end{cases}
$$
Here $2$ is the doubleton $\{0,1\}$, endowed with the discrete topology.
In the theory of cardinal characteristics of the continuum, the cardinals $\mathfrak r\_2$ and $\mathfrak r\_{2^\omega}$ are denoted by $\mathfrak r$ and $\mathfrak r\_\sigma$, respectively.
As written in [the survey of Blass](http://www.math.lsa.umich.edu/%7Eablass/hbk.pdf), it is not known whether $\mathfrak r=\mathfrak r\_\sigma$ in ZFC.
Now I introduce a dynamical modification of the cardinal $\mathfrak r\_X$.
Let us recall that a *dynamical system* is a pair $(X,f)$ consisting of a compact metric space $X$ and a continuous function $f:X\to X$. Define the iterations $f^n$ of $f$ by the recursive formula: $f^0$ is the identity map of $X$ and $f^{n+1}=f\circ f^n$ for $n\in\omega$.
**Definition.** For a dynamical system $(X,f)$, let $\mathfrak r\_{(X,f)}$ be the smallest cardinality of a family $\mathcal R$ of infinite subsets of $\omega$ such that for every $x\in X$ there exists $R\in\mathcal R$ such that the sequence $(f^n(x))\_{n\in R}$ coverges in $X$.
It is clear that $\mathfrak r\_{(X,f)}\le\mathfrak r\_X$. If $f$ is the identity map of $X$, then $\mathfrak r\_{(X,f)}=1$, so the inequality can be strict.
>
> **Problem.** Is it consistent that $\mathfrak r\_{(2^\omega,f)}<\mathfrak r\_{2^\omega}$ for the shift map $f:2^\omega\to 2^\omega$, $f:(x\_n)\_{n\in\omega}\mapsto (x\_{n+1})\_{n\in\omega}$?
>
>
>
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https://mathoverflow.net/users/61536
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The cardinal characteristic $\mathfrak r_{(X,f)}$ of a dynamical system
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Unfortunately, $\mathfrak r\_{(2^\omega,f)}\ge\mathfrak r$. Indeed, let $\mathcal R$ be a family of infinite subsets of $\omega$ such that $|\mathcal R|=\mathfrak r\_{(2^\omega,f)}$ and for any $x=(x\_n)\_{n\in\omega}\in 2^\omega$ there exists $R\in\mathcal R$ such that the sequence $(f^n(x))\_{n\in R}$ converges in $2^\omega$. Then the sequence $(x\_n)\_{n\in R}$ of the first coordinates of the sequence $(f^n(x))\_{n\in R}$ stabilizes, so the family $\mathcal R$ witness that $\mathfrak r\le \mathfrak r\_{(2^\omega,f)}$.
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https://mathoverflow.net/users/43954
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405503
| 166,262 |
https://mathoverflow.net/questions/405477
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11
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In a [comment](https://mathoverflow.net/questions/405256/what-is-the-standard-2-generating-set-of-the-symmetric-group-good-for#comment1038737_405273) at the recent question [What is the standard 2-generating set of the symmetric group good for?](https://mathoverflow.net/questions/405256/what-is-the-standard-2-generating-set-of-the-symmetric-group-good-for), it was remarked that the symmetric groups $S\_n$ for $n\gt 2$, $n\neq 5,6,8$, can be generated by an element of order 2 and an element of order 3 (G. A. Miller, Bull. Amer. Math. Soc. **7** (1901), 424-426 doi:[10.1090/S0002-9904-1901-00826-9](https://doi.org/10.1090/S0002-9904-1901-00826-9)). The remaining three nonabelian cases can of course be generated by a pair of elements, but these are cycles of length $5,6,8$ respectively. What is the best that can be done in these cases, and is there a conceptual reason why these are exceptional? (eg the presence of the nontrivial outer automorphism of $S\_6$? Or some action on an exceptional combinatorial object?)
---
**ADDED** By 'conceptual' proof I mean something more like 'structural', or the analogue of what in combinatorics is a 'bijective proof'. There should be some actual construction for the generic case that clearly breaks down for the small cases, due to a lack of space. Compare for instance the deep understanding of what goes wrong with the sort of handle moves that happen in high-dimensional topology, when we go down to dimension 4, and then why the replacement there will not work in lower dimensions. Simply counting two sets and noticing they have the same number of elements isn't the sort of thing I want. Nor do I want a proof that just writes down a pair of generators and checks they work, but of course I do want to see said generators.
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https://mathoverflow.net/users/4177
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Low-order symmetric group 2-generation: n=5,6,8
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It is also possible to use the (exceptional) outer automorphism of order $2$ of $S\_{6}$ to give an "explanation" of why $S\_{6}$ is not $\{2,3\}$-generated, along the lines I used in comments for $S\_{5}$ above. Take a $6$-cycle $\sigma \in S\_{6}.$ Then $\sigma^{2}$ is a product of two disjoint three cycles and $\sigma^{3}$ is a product of three disjoint $2$-cycles. These clearly commute. Now take an outer automorphism $\tau$ of $S\_{6}$ which sends products of two disjoint three cycles to three cycles. Then $\tau$ must also send products of three disjoint transpositions to transpositions, since $\tau(\sigma^{2})$ and $\tau(\sigma^{3})$ must commute.
Now suppose that $S\_{6}$ is $\{2,3\}$-generated say $S\_{6} = \langle \alpha, \beta : \alpha^{2} = \beta^{3} = 1 \rangle.$ Then we may apply $\tau$ if necessary, and assume that $\beta$ is a $3$-cycle. Then $\alpha$ is an odd permutation, so is either a transposition, or a product of three disjoint transpositions.
In the former case, we have a contradiction since there is a point fixed by both $\alpha$ and $\beta$. In the latter case, none of the transpositions in $\alpha$ can fix all points moved by $\beta$, for otherwise that transposition would be central in $\langle \alpha, \beta \rangle.$ It follows that $\alpha$ sends each point moved by $\beta$ to a point fixed by $\beta$ and conversely. It follows that $\beta$ and $\beta^{\alpha}$ commute. Now $\alpha$ normalizes the Abelian subgroup $\langle \beta, \beta^{\alpha}\rangle $, so that $\langle \alpha, \beta \rangle = \langle \alpha \rangle \langle \beta^{\alpha} ,\beta \rangle$ has order dividing $18$, a contradiction.
I do not know if there is an argument using the fact that $S\_{8}$ is isomorphic to ${\rm GL}(4,2)\langle \gamma \rangle$, where $\gamma$ is the transpose inverse automorphism, to "explain" that $S\_{8}$ is not $\{2,3\}$-generated.
Later edit: It would have been better perhaps to use the outer automorphism of $S\_{6}$ to reduce to the case that $\alpha$ is a transposition,(in which case, generation requires that $\beta$ is a product of two disjoint $3$-cycles), and then note the general fact that when $n >1$, $S\_{2n}$ is never generated by a transposition $\alpha$ and an element $\beta$ which
is a product of two disjoint $n$-cycles. For if it were, we may conjugate the pair and assume that $\alpha = (12).$ If either of the $n$-cycles in $\beta$ were disjoint from $\alpha$, then that $n$-cycle would be central in $\langle \alpha, \beta \rangle = S\_{2n}$, a contradiction. Hence both $n$-cycles of $\beta$ contain a point moved by $\alpha$.
We may conjugate $\beta$ by a permutation fixing both $1$ and $2$ and assume that $\beta = (1357 \ldots 2n-1)(2468 \ldots 2n)$ without disturbing the generation property. Then $\langle \alpha, \alpha^{\beta}, \ldots, \alpha^{\beta^{n-1}}\rangle$ = $\langle (12),(34), \ldots , (2n-1 2n) \rangle$ is Abelian and normal in $\langle \alpha, \beta \rangle = S\_{2n},$ a contradiction.
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12
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https://mathoverflow.net/users/14450
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405504
| 166,263 |
https://mathoverflow.net/questions/405498
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2
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Suppose $\Omega\subset\mathbb{R}^n$ is a regular open set, $f\in L^2(\Omega)$ and consider the following elliptic problem.
$$-\Delta u + u=f'(u) , \;\;u\_{|\partial \Omega}=0,$$
where $f'$ is the derivative of a function $f:\mathbb{R}\to\mathbb{R}$ and $f'(u)\in L^2(\Omega)$ for all $u\in H^1\_0(\Omega)$ (choose any assumptions you like on $f'$ to obtain this). Suppose also that that $f''$ exists.
It is well known that $u$ minimizes the energy functional $J:H^1\_0(\Omega)\to \mathbb{R}$ given by
$$J(u)=\frac{1}{2}\int\_\Omega|\nabla u|^2 +\frac{1}{2}\int\_\Omega u^2-\int\_\Omega f(u).$$
Now this also means that $u$ is a root of the Frechet (equiv to Gâteaux in this case) derivative $J':H^1\_0(\Omega)\to H^{-1}(\Omega)$. In particular, this means that
$$J'(u)v=\int\_\Omega\nabla u\cdot\nabla v+\int\_\Omega u v-\int\_\Omega f'(u)v=0, \;\;\;\text{ for all }v\in H^1\_0(\Omega).$$
(Notice that this is nothing but the variational formulation of the pde)
Suppose now that we wish to apply the [Newton method in Banach spaces](https://www.ams.org/journals/proc/1955-006-05/S0002-9939-1955-0071730-1/S0002-9939-1955-0071730-1.pdf) to minimize the function $J'$. I will assume that $J'':H^1\_0(\Omega)\to L(H^1\_0(\Omega),H^{-1}(\Omega)) $ exists and that $J''(u)$ be identified with the continuous bilinear form $J''(u)(\cdot,\cdot)\in \mathcal{B}(H^1\_0(\Omega))$ as
$$J''(u)(v,w)=\int\_\Omega\nabla v\cdot\nabla w+\int\_\Omega vw-\int\_\Omega f''(u)vw, \;\;\;\text{ for all }v,w\in H^1\_0(\Omega).$$
Starting with a guess $u\_0\in H^1\_0(\Omega)$, the usual Newton method iterative procedure to minimize $J'$ is then to solve to following "abstract" equation
$$J''(u\_n)(u\_{n+1}-u\_n)=-J'(u\_n). \;\;\;\text{(Equality is in $H^{-1}(\Omega)$)}.\tag{1}\label{1}$$
1. I would like to check whether or not \eqref{1} is equivalent to $$J''(u\_n)(u\_{n+1},v)-J''(u\_n)(u\_n,v)=-J'(u\_n)v, \;\;\;\text{ for all }v\in H^1\_0(\Omega).$$
2. If the above is true, to apply Newton method, is is necessary that that $$J''(u)^{-1}:L(H^{-1}(\Omega),H^1\_0(\Omega))\simeq \mathcal{B}(H^1\_0(\Omega))\to H^1\_0(\Omega)$$ exists is bounded in the operator norm. How does one even start thinking about such a problem?
---
**Important Fix**: Question 2 is wrong. What we need to show is that $J''(u)$ is invertible for every $u$, ie, $J''(u)^{-1}:H^{-1}(\Omega)\to H^1\_0(\Omega)$ exists.
|
https://mathoverflow.net/users/105925
|
Semi-linear elliptic problem, energy functionals, Fréchet derivatives and the Newton method in Banach spaces
|
Let me write out the equation $J''(u)w = g$ for $g\in H^{-1}$ and $w\in H^1\_0(\Omega)$. This is equivalent to
$$
\int\_\Omega \nabla w \cdot \nabla v - f''(u)wv = g(v) \quad \forall v\in H^1\_0(\Omega).
$$
This is the weak formulation of
$$
-\Delta w - f''(u)w = g$$
plus boundary conditions. To show existence of solutions, you need some assumptions on $f''$ to be able to invoke Lax-Milgram theorem or other existence theorems. The easiest one would be to require $f''(u) \le0$ for all $u$.
In order to apply the convergence theory of Newton's method, you will need that $J''(u)^{-1}$ is a bounded linear operator, and that the inverses $J''(u)^{-1}$ are bounded on a neightborhood of the reference solution.
|
2
|
https://mathoverflow.net/users/48485
|
405506
| 166,264 |
https://mathoverflow.net/questions/405460
|
5
|
Consider a continuous ODE,
$$\dot x = f(x), f \in C^1$$
$\dot x = 0$ for all $x \in K \subset \mathbb{R}^n$, where we assume that $K$ is a closed but unbounded set of non-isolated equilibrium. For example, $K$ could be a line, or a half space, or a ray, etc.
>
> Suppose that this ODE admits an energy function $E$ such that $E(x) > 0$
> for all $x \neq K$, $E(x) = 0$ for all $x \in K$, and, $\dot E(x) < 0$ for
> all $x \neq K$, $\dot E(x) = 0$ for all $x\in K$.
>
>
>
Does there exist any theorem or result that allow us to conclude the converge of $x$ towards $K$ or some point in $K$ starting from any point $x(0) \in \mathbb{R}^n \backslash K$?
The challenge here is, unlike what has been suggested here: [Conditions for convergence to non-isolated fixed points](https://mathoverflow.net/questions/244449/conditions-for-convergence-to-non-isolated-fixed-points) $K$ is assumed to be unbounded. This eliminates the existence of some compact set which contains $\{x| \dot E = 0\}$, hence the [Krasovsky-LaSalle theorem](https://math.libretexts.org/Bookshelves/Differential_Equations/Book%3A_Ordinary_Differential_Equations_(Wiggins)/07%3A_Lyapunov%E2%80%99s_Method_and_the_LaSalle_Invariance_Principle/7.01%3A_Lyapunov%E2%80%99s_Method_and_the_LaSalle_Invariance_Principle) does not apply.
|
https://mathoverflow.net/users/145401
|
Which result guarantees convergence of solution of an ODE to a set of non-compact, non-isolated equilibrium?
|
The extension to the unbounded setting is due to [Hale 1969: Dynamical systems and stability](https://www.sciencedirect.com/science/article/pii/0022247X69901759), where he states your statement provided that the positive orbit $\mathscr{O}^+(x\_0)$ of the initial point $x\_0$ is contained in a compact set. See [Theorem 1](https://www.sciencedirect.com/science/article/pii/0022247X69901759).
This might sounds a bit unsatisfactory, but I don't think that some assumption of compactness, which might depend on the initial point, can be avoided in this generality without using any other special structure of the equations.
Note, that the missing compactness of the orbit is also the essential bit in the counter-example of AVK.
|
3
|
https://mathoverflow.net/users/13400
|
405515
| 166,267 |
https://mathoverflow.net/questions/404099
|
1
|
Let $X$ be an algebraic variety over $\mathbb{C}$ (the ground field is not important but this makes things easier I think) and $G$ an algebraic group acting over it. Let's say we know that there's a normal subgroup $K \cong \mathbb{G}\_m$ such that $K$ acts trivially on $X$ and $G/K$ acts in a free way on $X$.
I think one could put less restrictive hypothesis on $K$ actually: maybe reductive should be enough.
I'm denoting as $\mathcal{X}=[X/G]$ the quotient stack. Is it always true that $$H^\*(\mathcal{X}) \cong H^\*(X/(G/K)) \otimes H^\*(BK) $$ and $$H^\*\_c(\mathcal{X}) \cong H^\*\_c(X/(G/K)) \otimes H^\*\_c(BK) $$
(everything is meant to be with $\overline{\mathbb{Q}\_{\ell}}$ coefficients.
What should be a possible counterexample?
|
https://mathoverflow.net/users/146464
|
Cohomology of quotient stack
|
Let $G$ be a linear algebraic group, and $K\subset G$ a closed normal subgroup with quotient $L = G/K$. So all three groups are linear algebraic. Further, assume $L$ is connected.
Let $X$ be a variety with $G$-action such that $K$ acts trivially, $L$ acts freely and we have a quotient $X/L$.
**Step 1:** The canonical map $BG \to BL$ is a fiber bundle with fiber $BK$. As $L$ is connected the mondromy action on the cohomology of the fibers, namely $H^\*(BK)$, is trivial. Further, the Leray-Serre spectral sequence associated to this fiber bundle degenerates at $E\_2$ because both $BK$ and $BL$ have cohomology concentrated in even degrees.
Note: The claim about odd vanishing in cohomology holds because $K$ and $L$ are linear algebraic groups and we are using torsion free coefficients (it is ultimately due to the Bruhat decomposition). To emphasize: no connected assumptions are needed for this vanishing, just torsion free coefficients and linear algebraic groups.
**Step 2:** We have a pullback square (this **claim** should be checked - I have not worked out the details carefully):
$\require{AMScd}$
\begin{CD}
[X/G] @>{}>> BG\\
@VVV @VVV\\
X/L @>{}>> BL
\end{CD}
The monodromy action on the cohomology of the fibers $H^\*(BK)$ for the bundle $[X/G] \to X/L$ is the pullback of the monodromy action in $BG \to BL$. Hence, it is trivial.
Further, the elements of $H^\*(BK)$ are permanent cycles in the Leray-Serre spectral sequence of $[X/G] \to X/L$ because they are so for that of $BG \to BL$. This gives the required claim:
$H^\*([X/G]) \cong H^\*(BK) \otimes H^\*(X/L)$.
Note: even assuming I haven't done something completely dumb above, this says nothing about ring structure. Further, the requirement that $L$ act freely on $X$ should be unnecessary - replace $X/L$ by $[X/L]$ (again, modulo me not being delusional).
The point (assuming the pullback square is correct) at the heart of the matter is that $BG \to BL$ is the **universal** example of a fiber bundle with structure group $G$ in which the $G$-action on the fibers factors through $L$ (i.e., $K$ acts trivially).
|
1
|
https://mathoverflow.net/users/392998
|
405516
| 166,268 |
https://mathoverflow.net/questions/405512
|
3
|
I am looking for a reference for the **Stallings-Epstein-Waldhausen construction** (constructing an incompressible surface in a 3-manifold from a nontrivial splitting of the fundamental group).
I know of Proposition 2.3.1 in the following article, but was hoping for a more comprehensive / textbook-like treatment.
Marc Culler and Peter B. Shalen: Varieties of Group Representations and Splittings of 3-Manifolds. Annals of Mathematics, Vol. 117, No. 1 (Jan., 1983), pp. 109-146.
My goal is to find out under which conditions the resulting incompressible surface is orientable.
|
https://mathoverflow.net/users/129446
|
Reference request: Stallings-Epstein-Waldhausen construction
|
As an answer to your first question: You can find an exposition of the pull-back construction in Scott's notes "[An introduction to 3-manifolds](https://www.math.cuhk.edu.hk/course_builder/1617/math6071b/scott_An%20Introduction%20to%203-manifolds-1974.pdf)".
As an answer to your second question: If the group we are splitting over is not free, then we can detect the orientability of the surface by abelianising the group. If the group is free, then we are obtaining a surface with boundary. As HJRW notes, the surface obtained is always two-sided. Thus it suffices to know that $M^3$, the ambient three-manifold, is orientable. If $M^3$ is not orientable, there is still a way to detect the orientation of the splitting surface, by lifting to the orientation double cover and seeing how much of the edge group lifts.
|
3
|
https://mathoverflow.net/users/1650
|
405532
| 166,270 |
https://mathoverflow.net/questions/405482
|
0
|
Any element in a boolean extension field $a\in GF(2^n)$ can be presented by a boolean vector $a\_{(2)} \in GF(2)^n$. For any element $a\in GF(2^n)$, there exists a boolean matrix $M\_a\in GF(2)^{n\times n}$ such that $(ab)\_{(2)} = M\_a \cdot (b)\_{(2)}$.
**The question:**
For any non-zero $\vec v\in GF(2)^n$,
is the mapping $a\mapsto \vec v^T M\_a$ a bijection?
In other words, if $a\in GF(2^n)$ is a uniform random field element, is $\vec v^T M\_a$ uniform random as well?
Note that, the mapping $a \mapsto M\_a \vec v$ is a bijection for any non-zero $\vec v\in GF(2)^n$.
Because there exists non-zero $v\in GF(2^n)$ that $v\_{(2)} = \vec v$,
then $M\_a \vec v = M\_a \cdot v\_{(2)} = (av)\_{(2)}$.
In other words, if $a$ if random,
so every column vector in $M\_a$ is random, every linear combination of column vectors in $M\_a$ is random. I wonder if the same holds for its row vectors.
|
https://mathoverflow.net/users/165642
|
Does binary extension field multiplication matrix have random rows?
|
The function $a \mapsto M\_a$ is a linear transformation from $GF(2)^n$ to $GF(2)^{n \times n}$, so $a \mapsto \vec{v}^{T} M\_a$ is a linear transformation from $GF(2)^n$ to the dual vector space $(GF(2)^n)^\*$. Both of these vector spaces are $n$-dimensional. To see that the linear transformation $a \mapsto \vec{v}^{T} M\_a$ is a bijection, we will show that its kernel is zero-dimensional, i.e. the only $a \in GF(2)^n$ such that $\vec{v}^{T} M\_a=0$ is $a=0$.
By assumption $\vec{v} \neq 0$, so there exists $\vec{w} \in GF(2)^n$ such that $\vec{v}^{T} \vec{w} \neq 0$. Let $b \in GF(2^n)$ be the field element such that $b\_{(2)} = \vec{w}$. If $a \neq 0$, we may define $c = a^{-1} b \in GF(2^n)$ and $\vec{x} = c\_{(2)} \in GF(2)^n$. Then we have $M\_a \vec{x} = (a c)\_{(2)} = (a a^{-1} b)\_{(2)} = b\_{(2)} = \vec{w}$ so $\vec{v}^{T} M\_a \vec{x} = \vec{v}^{T} \vec{w} \neq 0$. Hence, when $a \neq 0$ it follows that $\vec{v}^{T} M\_a \neq 0$, so there are no non-zero vectors in the kernel of the linear transformation $a \mapsto \vec{v}^{T} M\_a$, as claimed.
|
2
|
https://mathoverflow.net/users/8049
|
405540
| 166,272 |
https://mathoverflow.net/questions/405545
|
1
|
I need to solve the following equation for $P \in\mathbb{R}^{r\times d}$
$$P - G\_1G\_2(\lambda P^\top(PAP^\top)^{-1}P + A^{-1} ) = 0,$$
where the other quantities are known: $A\in\mathbb{R}^{d\times d}$, $G\_1 \in\mathbb{R}^{r\times d}$, $G\_{2} \in\mathbb{R}^{d\times d}$, $\lambda\in\mathbb{R}$.
I have already tried everything I know. If the equation had involved scalar quantities the solution would simply be
$$
p = g\_1g\_2(a^{-1}(1+\lambda)),
$$
but unfortunately in the matrix case I don't know how to proceed. Maybe there should be some way to decompose it using SVD or some trace trick?
|
https://mathoverflow.net/users/156139
|
Matrix equation with projection matrix
|
The solution for $P$ to
$$P - G\_1G\_2(\lambda P^\top(PAP^\top)^{-1}P + A^{-1} ) = 0$$
is
$$P=(1 +\lambda )G\_1 G\_2 A^{-1},$$
as one can check by substitution into
$$G\_1G\_2P^\top(PAP^\top)^{-1}P=G\_1 G\_2(G\_1G\_2A^{-1})^\top\bigl(G\_1G\_2(G\_1G\_2A^{-1})^\top\bigr)^{-1}G\_1G\_2A^{-1}=G\_1G\_2A^{-1}.$$
|
5
|
https://mathoverflow.net/users/11260
|
405548
| 166,274 |
https://mathoverflow.net/questions/405511
|
4
|
Let $X$ be a smooth scheme over a field $k$ and let $\mathsf{D}\_{\text{qc}}(\mathcal{D}\_X)$ be the full subcategory of $\mathsf{D}(\mathcal{D}\_X\mathsf{-Mod})$ composed of the complexes of $\mathcal{D}\_X$-modules with quasi-coherent cohomology.
If $f:X\to Y$ is a morphism between such schemes, we have natural functors
$$f\_\*:\mathsf{D}\_{\text{qc}}(\mathcal{D}\_X)\leftrightarrows \mathsf{D}\_{\text{qc}}(\mathcal{D}\_Y):f^!.$$
If we restrict to the subcategory of complexes with *holonomic* cohomology, then the Verdier duality functor allows us to find left adjoints $f^\*$ and $f\_!$ of $f\_\*$ and $f^!$, respectively.
**I wonder if those adjoints already exist in $\mathsf{D}\_{\text{qc}}(\mathcal{D}\_X)$.**
Perhaps we can use Brown representability for left adjoints (as in Neeman's book about triangulated categories) or some other adjoint functor theorem...
In Neeman's paper *The Grothendieck Duality Theorem via Bousfield's Techniques*, it is proven that $f\_\*$ has a **right** adjoint (even though in the text it is said that it's a left adjoint). Perhaps we can use this functor to construct our left adjoints?
|
https://mathoverflow.net/users/131975
|
Six functor formalism for quasi-coherent $D$-modules
|
They do not exist in general. The simplest example is maybe to take $X\rightarrow Y$ to be the closed embedding of the origin inside $\mathbb{A}^1.$ Then $f\_\*$ sends a vector space $V$ to the $\mathcal{D}\_{\mathbb{A}^1}$-module $V\otimes\delta\_0$, where $\delta\_0$ is the irreducible $\mathcal{D}\_{\mathbb{A}^1}$-module set-theoretically supported at $0$. The point is that because $\delta\_0$ is infinite-dimensional as a vector space, this functor does not commute with products and hence cannot have a left adjoint.
|
3
|
https://mathoverflow.net/users/51424
|
405559
| 166,278 |
https://mathoverflow.net/questions/405566
|
4
|
It is known that in a stable $\infty$-category $\mathcal{C}$ and $X \in \mathcal{C}$, the suspension $X[1]$ is defined by the pushout of $0\leftarrow X \rightarrow 0$. However this does not make sense in usual category since the pushout is the zero object $0$.
Now all we can do is following the formal definition of limit/colimit for an $\infty$-category. But the cost is that we lose the intuition. What (intuition) should one keep in mind when thinking about pushout/pullback in a stable $\infty$-category.
|
https://mathoverflow.net/users/133871
|
How to understand pushout/pullback in a stable $\infty$-category
|
Let me try to give some intuition by examining two important examples. One should start from the definition: the suspension $ΣX$ is the universal choice of $Y$ filling of a square
$$\require{AMScd}
\begin{CD}
X @>{p}>> \ast\\
@V{p}VV @VVV \\
\ast @>>> Y
\end{CD}\,.$$
To understand the suspension we need to understand what's the *datum* of such a square. This is given by two maps $y\_0:\ast\to Y$ and $y\_1:\ast\to Y$ (two ``points'' of $Y$) and a homotopy $H:y\_0p\simeq y\_1p$ as maps $X\to Y$. That is, the suspension is the universal recipient of two points and a homotopy between the two constant maps at $y\_0$ and $y\_1$.
Let us go now in the ∞-category of spaces (or, if you prefer the name, animæ). Then a homotopy of maps $H:X\to Y$ is exactly a map $H:X\times [0,1]\to Y$. Since we require the homotopy to be between two constant maps we see that a commuting square as above is exactly the datum of a map
$$\ast\amalg\_{X\times\{0\}}X\times [0,1]\amalg\_{X\times\{1\}} \ast\to Y$$
One needs to check the universal property more carefully, but as one would expect the left hand side is exactly the universal recipient. That is
$$ ΣX=\ast\amalg\_{X\times\{0\}}X\times [0,1]\amalg\_{X\times\{1\}} \ast$$
Note that the right hand side is exactly the classical suspension of $X$, thus justifying the name $ΣX$.
---
Now for a more algebraic (and stable) example: the derived category of a ring. In this case, when representing everything by chain complexes, the two points are no data (since there's only one possible map $0\to Y$), and a homotopy is the same as a collection of maps $H\_n:X\_{n+1}\to Y\_n$ such that $dH\_n+H\_nd=0$. But this is exactly the same as a map of chain complexes $X[1]\to Y$ (remember than in $X[1]$ the differential inherits a sign to make the formulas canonical -- this is exactly its origin!). Therefore in $\mathscr{D}(R)$, the suspension is given by the usual shift.
---
So what's the reason for these new phenomena? The point is that a commuting square in an ∞-category in general contains *more* information than the square in a 1-category: it's not enough that you say that the two composition are equal, you also have to provide a *reason* for them to be equal (i.e. a homotopy). Therefore your pushout needs to account for this additional data, and this is why it is nontrivial.
|
8
|
https://mathoverflow.net/users/43054
|
405571
| 166,282 |
https://mathoverflow.net/questions/405573
|
5
|
Let $X$ be a smooth projective variety, $\mathscr T$ a torsion sheaf with irreducible support of codimension $1$, say $Z$. Then the first Chern class $c\_1(\mathscr T)$ is of form $r[Z]$. Is there anything we can say about the positivity of $r$?
Any help is appreciated.
|
https://mathoverflow.net/users/313627
|
First Chern class of torsion sheaves
|
The coefficient $r$ is equal to the length of $\mathcal{T}$ at the generic point of $Z$, so it is positive.
|
7
|
https://mathoverflow.net/users/4428
|
405575
| 166,283 |
https://mathoverflow.net/questions/405580
|
5
|
At the end of
<https://encyclopediaofmath.org/index.php?title=Cubic_hypersurface#References>
it is stated that the diagonal cubic hypersurface
$$
\sum\_{i=0}^{2m+1} a\_i x\_i^3 = 0, m\ge 2
$$
(and presumably $a\_i\not=0$) is rational. Is this true over the complex numbers, or any field of characteristic zero? Where can I find a reference of this result? Thanks!
|
https://mathoverflow.net/users/66397
|
diagonal cubic hypersurfaces
|
Yes, this is true over $\mathbb{C}$, and rather easy. You can assume your equation is $\sum x\_i^3=0$. For convenience, let me call the coordinates $x\_0,\ldots ,x\_m;y\_0,\ldots ,y\_m$. Then your hypersurface $X$ contains the $m$-planes $P\_1: x\_i=-y\_i$ and $P\_2: x\_i=-\rho y\_i$, with $\rho =e^{2\pi i/3}$. Note that $P\_1\cap P\_2=\varnothing$. Now consider the rational map $\varphi :P\_1\times P\_2 -\!--\!\!> X$ defined as follows: given general points $p\_1\in P\_1$ and $p\_2\in P\_2$, the line $\langle p\_1,p\_2\rangle$ intersects $X$ in a third point $\varphi (p\_1,p\_2)$. It is easy to see that $\varphi $ is birational.
|
12
|
https://mathoverflow.net/users/40297
|
405584
| 166,285 |
https://mathoverflow.net/questions/405452
|
2
|
Let $G$ be an affine algebraic group (let's say over $\mathbb{C}$). If necessary one can assume $G$ to be reductive. Imagine one has $X$ over which $G$ acts freely: moreover, we have a locally closed subvariety $Y$ such that $X=G \cdot Y$.
Moreover, one has a subgroup $H \subseteq G$ such that $H$ stabilizes $Y$. We know also that given any two points $y\_1,y\_2 \in Y$ and $g \in G$ such that $g y\_1=y\_2$ we have $g \in H$.
As everything acts freely one can make up two quotient varieties $Y/H$ and $X/G$ with the natural morphism $$f:Y/H \to X/G .$$
The hypotheses imply that $f$ is bijective. Is it also an isomorphism? I'm not assuming anything on normality or smoothness
|
https://mathoverflow.net/users/146464
|
Quotient variety and subgroups
|
I don't think that this is true without extra hypotheses like normality of $X$ (if $X$ is normal then it follows as explained by Damian Rossler in the comments). Here is a possible example.
We will take $G = GL\_2$ and $H=1$ (the latter is just the trivial group). Let $C$ be the cuspidal curve $C = Spec(k[x,y]/(y^2-x^3))$.
Let's take $X = GL\_2 \times C$, equipped with the natural action of $GL\_2$ by right multiplication on the first factor (this is just the trivial $GL\_2$-bundle over $C$).
Inside $GL\_2$ we have the additive group $\mathbb{G}\_a$ of strictly upper triangular matrices. So we have a closed subvariety $Z = \mathbb{G}\_a \times C$ inside $X$. We can write $Z = Spec(k[x,y,t](y^2-x^3))$. Inside $Z$ we can define $Y$ to be the vanishing locus of the polynomial $xt-y$. So we get a closed subvariety $Y \hookrightarrow X$ given by
$$ Y = Spec(k[x,y,t]/(y^2-x^3, xt-y))$$
Notice that $X/GL\_2 =C$ and the projection $Y \to C$ onto the second component is the map $Y/H \to X/GL\_2$ in this case.
$Y$ is actually isomorphic to $\mathbb{A}^1$, and the projection $Y \to C$ exhibits $Y$ as the normalization of the cuspidal curve $C$. In particular $Y \to C$ is not an isomorphism.
Note that the projection $Y \to C$ is bijective on points, so $Y$ contains exactly one point in each fiber of the $GL\_2$-bundle $X \to C$. This implies that $GL\_2 \cdot Y = X$ and also that there are no two closed points of $Y$ that belong to the same $GL\_2$-orbit. In other words, the hypotheses are satisfied for $Y \hookrightarrow X$ (with $H = 1$) and the map $Y/H \to X/G$ is not an isomorphism.
One can probably do variations of this to get examples with $H$ nontrivial.
|
4
|
https://mathoverflow.net/users/339730
|
405588
| 166,286 |
https://mathoverflow.net/questions/405589
|
2
|
Let X and Y be separable Banach spaces.
Let $f:X\rightarrow Y$ be a Baire-1 function, which is the pointwise limit of a sequence of continuous functions $f\_n:X\rightarrow Y$.
Define $E$ as the set of $x$ such that $f\_n(x\_n)\rightarrow f(x)$ fails to hold for some sequence $\{x\_n\}$ approaching x.
Is $E$ empty?
If not, if $\mu$ is a countably additive measure on $X$, defined on the Borel $\sigma$-algebra, is $\mu(E)=0$?
|
https://mathoverflow.net/users/401135
|
Convergence of sequences for Baire-1 functions
|
Recall:
**1.** Given $f\_n\in C(X,Y)$ point-wise convergent to $f$, the above set $E$ always contains the discontinuity set of $f$ (starting from any $x\_n\to x$ with $ f(x\_n)\not\to f(x)$ one has for a subsequence $ f\_{k\_n}(x\_n)-f(x\_n) \to0$, and re-naming the sequences one can also assume $ f\_{n}(x\_n)-f(x\_n) \to0$ so $ f\_n(x\_n)\not\to f(x)$).
**2.** The characteristic function of any closed set $C$ of a metric space is always of class Baire-$1$: $f\_n(x):=\big(1-n\,d(x,C)\big)\_+$ is $0$ for all $n$ if $x\in C$, and it is eventually $0$ if $x\notin C$.
**3.** Assuming $X$ *separable and of positive dimension,* for any non-zero finite Borel measure $\mu$ on $X$, there is a closed set with empty interior and $\mu(C)>0$, thus a Baire-$1$ function $\chi\_C$ with discontinuity set $\partial C=C$ of positive measure $\mu(C)$ . (Indeed, since $X$ has at most a countable set of points with positive measure, which has empty interior because $\dim X>0$, and since $X$ is separable, $X$ also has a countable dense set $\{s\_n\}\_{n\in\mathbb N}$ of points of measure $0$; since $\mu$ is finite, for each $n\ge1$ there is a ball $B(s\_n,\epsilon\_n)$ of measure less than $3^{-n}\mu(X)$. So $X\setminus \cup\_{n\in\mathbb N}B(s\_n,\epsilon\_n)$ is a closed set with empty interior and measure $\mu\big(X\setminus \cup\_{n\in\mathbb N}B(s\_n,\epsilon\_n)\big)\ge \mu(X)-
\sum\_{n\in\mathbb N}\mu(B(s\_n,\epsilon\_n))\ge\mu(X)/2>0\;$ ).
|
2
|
https://mathoverflow.net/users/6101
|
405592
| 166,287 |
https://mathoverflow.net/questions/405543
|
7
|
$\newcommand{\oUConf}{\widehat{\mathrm{UConf}}}\newcommand{\UConf}{\mathrm{UConf}}\newcommand{\oGr}{\widehat{\mathrm{Gr}}}\newcommand{\Gr}{\mathrm{Gr}}\newcommand{\SO}{\mathrm{SO}}\newcommand{\Spin}{\mathrm{Spin}}\newcommand{\String}{\mathrm{String}}\newcommand{\U}{\mathrm{U}}\newcommand{\SU}{\mathrm{SU}}\newcommand{\O}{\mathrm{O}}\newcommand{\Sp}{\mathrm{Sp}}\newcommand{\Mp}{\mathrm{Mp}}$We can describe the classifying spaces of the groups $\O\_n$, $\SO\_n$, $\U\_n$, $\Sp\_n$, $\Sigma\_n$, and $A\_n$ as follows:
* $\mathrm{B}\O\_n$, $\mathrm{B}\U\_n$, $\mathrm{B}\Sp\_n$ are the Grassmanians of $n$-planes $\Gr\_n(\mathbb{R}^\infty)$, $\Gr\_n(\mathbb{C}^\infty)$, and $\Gr\_n(\mathbb{H}^\infty)$;
* $\mathrm{B}\SO\_n$ is the Grassmanians of *oriented* $n$-planes $\oGr\_n(\mathbb{R}^\infty)$;
* $\mathrm{B}\Sigma\_n$ is the unordered configuration space $\UConf\_n(\mathbb{R}^\infty)$ of $n$ points on $\mathbb{R}^\infty$;
* $\mathrm{B}A\_n$ is the space $\oUConf\_n(\mathbb{R}^\infty)$ whose points are [$n$ points on $\mathbb{R}^\infty$ together with an orientation of their spanned $n$-plane](https://mathoverflow.net/questions/30113);
Are there similar "geometric" descriptions for $\mathrm{B}\Spin\_n$ and $\mathrm{B}\String\_n$?
What about $\mathrm{B}\SU\_n$ (is there a [more explicit description than "the $3$-connected cover of $\mathrm{B}\U\_n$"](https://math.stackexchange.com/a/2850463)?), [$\mathrm{B}\Mp\_n$](https://en.wikipedia.org/wiki/Metaplectic_group), [$\mathrm{B}\widetilde{A}\_n$](https://groupprops.subwiki.org/wiki/Double_cover_of_alternating_group), and [$\mathrm{B}\mathcal{A}\_n$](https://arxiv.org/abs/1605.09192)?
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https://mathoverflow.net/users/130058
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Geometric models for the classifying spaces of the spin and string covers of the orthogonal, symplectic, and symmetric groups
|
For $\def\B{{\rm B}} \def\bB{{\bf B}} \def\Spin{{\rm Spin}} \def\String{{\rm String}} \B\Spin(n)$, simply equip the $n$-planes with a [spin structure](https://mathoverflow.net/questions/122748/what-is-a-spinor-structure/122798#122798), as originally proposed by Stolz and Teichner.
For $\B\String(n)$, equip the $n$-planes with a string structure,
as described in a [paper by Douglas and Henriques](http://andreghenriques.com/PDF/TringWP.pdf).
Also, I am not sure what the intended difference between $\B$ and $\bB$ is,
but if $\bB$ does refer to the stack version of these classifying spaces,
then the above constructions continue to work perfectly well
for stacks: to a smooth manifold $S$ assign the groupoid respectively 2-groupoid of vector bundles with base $S$ and whose fibers are equipped with a spin respectively string structure.
This yields the stacks $\bB\Spin(n)$ and $\bB\String(n)$, and applying the shape functor to these stacks yields the classifying spaces $\B\Spin(n)$ and $\B\String(n)$.
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5
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https://mathoverflow.net/users/402
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405603
| 166,289 |
https://mathoverflow.net/questions/405578
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2
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$\DeclareMathOperator\FSym{FSym}$Let $p$ be a prime and $S$ be a transitive Sylow $p$-subgroup of $\FSym(\mathbb{N})$, the finitary symmetric group of the set of all natural numbers.
Question: Is $S$ totally imprimitive and uniserial (i.e. for every $i\in \mathbb{N}$, there exists a unique system of imprimitivity on $\mathbb{N}$ with blocks of size $p^i$—this concept is introduced by P.M. Neumann in [*The structure of finitary permutation groups, Arch. Math. 27 (1976), 3-7* [DOI link](https://doi.org/10.1007/BF01224634)])?
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https://mathoverflow.net/users/98061
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On Sylow subgroups of finitary symmetric groups
|
Yes: all transitive Sylow $p$-subgroups of $\mathrm{FSym}(\mathbb{N})$ are permutation isomorphic to the infinite wreath product $C\_p \wr C\_p \wr \ldots $. This permutation group is uniserial and has a unique system of blocks of size $p^i$ for each $i \in \mathbb{N}$; one such block is an orbit of the canonical subgroup isomorphic to $C\_p \wr \ldots \wr C\_p$ (with $i$ factors) used to construct the wreath product.
For a reference, see (2) on page 422 of [this paper](https://www.sciencedirect.com/science/article/pii/S002186931600017X) by Agnieszka Bier, Yuriy Leshchenko and Vitaliy Sushchanskyy.
(Just to be clear, here $C\_p \wr C\_p \wr \ldots $ is the restricted wreath product, with the left-most $C\_p$ corresponding to blocks of size $p$, not the profinite wreath product $\ldots \wr C\_p \wr C\_p$. The latter is the automorphism group of the infinite $p$-ary rooted tree in which the right-most $C\_p$ corresponds to the action on the $p$ branches below the root. It does not contain any finitary permutation except the identity.)
**Edit.** Since the linked paper proves far more than we need (it's really about automorphism groups of directed limits of trees), let me add a proof using only Wielandt's theorem that a finitary primitive permutation group on an infinite set $\Omega$ is either $\mathrm{FSym}(\Omega)$ or its index $2$ subgroup $\mathrm{Alt}(\Omega)$. Since neither is a $p$-group, it follows that a maximal $p$-subgroup $P$ of $\mathrm{FSym}(\mathbb{N})$ is imprimitive.
It is well known that every block of a imprimitive finitary permutation group is finite. (*Proof.* If $\Gamma$ is a block then $\Gamma g \cap \Gamma = \varnothing$ for any $g$ moving $\omega \in \Gamma$ to $\omega g \not\in \Gamma$, so $\Gamma \subseteq \mathrm{supp}\ g$.) Moreover, if $P$ has a maximal proper block $\Delta$ then $P$ acts primitively on the set $\{\Delta g : g \in P\}$, again contradicting that $P$ is a $p$-group. Therefore there is an infinite chain of finite blocks $\Gamma\_1 \subset \Gamma\_2 \subset \ldots $ and $P$ embeds in the restricted iterated wreath product $S\_{\Gamma\_1} \wr S\_{\Gamma\_2} \wr \ldots$ where, since $P$ is a $p$-group, each $S\_{\Gamma\_i}$ is a finite $p$-group. By maximality, each $S\_{\Gamma\_i}$ is a Sylow $p$-subgroup of a finite symmetric group, so of the expected form $C\_p \wr \ldots \wr C\_p$. In particular $P$ is totally imprimitive and uniserial.
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3
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https://mathoverflow.net/users/7709
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405611
| 166,290 |
https://mathoverflow.net/questions/405015
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2
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A function $q(x)$ is said to be completely monotonic on an interval $I$ if $q(x)$ has derivatives of all orders on $I$ and $(-1)^{n}q^{(n)}(x)\ge0$ for $x\in I$ and $n\ge0$. See Chapter 1 in the monograph [1] below.
A positive function $q(x)$ is said to be logarithmically completely monotonic on an interval $I\subseteq\mathbb{R}$ if it has derivatives of all orders on $I$ and its logarithm $\ln q(x)$ satisfies $(-1)^k[\ln q(x)]^{(k)}\ge0$ for $k\in\mathbb{N}=\{1,2,\dotsc\}$ on $I$. See Definition 1 in th article [2] below.
A logarithmically completely monotonic function on $I$ must be completely monotonic on $I$, but not conversely. See Theorem 1 in [2] and related texts in the references [1, 3, 4] below.
The famous Bernstein-Widder's theorem (on page 161 Theorem 12b in the book [5]) reads that a necessary and sufficient condition that $q(x)$ should be completely monotonic for $0<x<\infty$ is that
\begin{equation} \label{berstein-1}\tag{w}
q(x)=\int\_0^\infty \textrm{e}^{-xt}\textrm{d}\,\alpha(t),
\end{equation}
where $\alpha(t)$ is non-decreasing and the integral \eqref{berstein-1} converges for $0<x<\infty$.
It is trivial that the exponential function $\textrm{e}^{1/x}$ is logarithmically completely monotonic on $(0,\infty)$. Consequently, by the above-mentioned Theorem 1 in [2], we conclude that the function $\textrm{e}^{1/x}$ is completely monotonic on $(0,\infty)$.
Motivated by the Bernstein-Widder's theorem mentioned above, we pose a question:
**What is the explicit expression of the measure $\alpha(t)$ such that
\begin{equation} \label{exp-frac1x}\tag{+}
\textrm{e}^{1/x}=\int\_0^\infty \textrm{e}^{-xt}\textrm{d}\,\alpha(t)
\end{equation}
converges for $0<x<\infty$?** See Section 4 in the paper [6] below.
References
1. R. L. Schilling, R. Song, and Z. Vondracek, *Bernstein Functions*, 2nd ed., de Gruyter Studies in Mathematics **37**, Walter de Gruyter, Berlin, Germany, 2012; available online at <https://doi.org/10.1515/9783110269338>.
2. F. Qi and C.-P. Chen, *A complete monotonicity property of the gamma function*, J. Math. Anal. Appl. **296** (2004), 603--607; available online at <https://doi.org/10.1016/j.jmaa.2004.04.026>.
3. C. Berg, *Integral representation of some functions related to the gamma function*, Mediterr. J. Math. **1** (2004), no. 4, 433--439; available online at <https://doi.org/10.1007/s00009-004-0022-6>.
4. B.-N. Guo and F. Qi, *A property of logarithmically absolutely monotonic functions and the logarithmically complete monotonicity of a power-exponential function*, Politehn. Univ. Bucharest Sci. Bull. Ser. A Appl. Math. Phys. **72** (2010), no. 2, 21--30.
5. D. V. Widder, *The Laplace Transform*, Princeton University Press, Princeton, 1946.
6. Xiao-Jing Zhang, Feng Qi, and Wen-Hui Li, *Properties of three functions relating to the exponential function and the existence of partitions of unity*, International Journal of Open Problems in Computer Science and Mathematics **5** (2012), no. 3, 122--127; available online at <https://doi.org/10.12816/0006128>.
7. <https://math.stackexchange.com/a/4262516/945479>
8. <https://math.stackexchange.com/a/4262498/945479>
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https://mathoverflow.net/users/147732
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What is the integral representation of the exponential function $e^{1/t}$ on $(0,\infty)$?
|
For $k\in\mathbb{N}\_0=\{0,1,2,\dotsc\}$ and $z\ne0$, let
\begin{equation}\label{exp=k=sum-eq-degree=k+1}
H\_k(z)=\textrm{e}^{1/z}-\sum\_{m=0}^k\frac{1}{m!}\frac1{z^m}.
\end{equation}
For $\Re(z)>0$, the function $H\_k(z)$ has the integral representations
\begin{equation}\label{exp=k=degree=k+1-int}
H\_k(z)=\frac1{k!(k+1)!}\int\_0^\infty {}\_1F\_2(1;k+1,k+2;t)t^k \textrm{e}^{-zt}\textrm{d}\,t
\end{equation}
and
\begin{equation}\label{exp=k=degree=k+1-int-bes}
H\_k(z)=\frac1{z^{k+1}}\biggl[\frac1{(k+1)!}+\int\_0^\infty \frac{I\_{k+2} \bigl(2\sqrt{t}\,\bigr)}{t^{(k+2)/2}} \textrm{e}^{-zt}\textrm{d}\,t\biggr],
\end{equation}
where the hypergeometric series
\begin{equation}
{}\_pF\_q(a\_1,\dotsc,a\_p;b\_1,\dotsc,b\_q;x)=\sum\_{n=0}^\infty\frac{(a\_1)\_n\dotsm(a\_p)\_n} {(b\_1)\_n\dotsm(b\_q)\_n}\frac{x^n}{n!}
\end{equation}
for $b\_i\notin\{0,-1,-2,\dotsc\}$, the shifted factorial $(a)\_0=1$ and
\begin{equation}
(a)\_n=a(a+1)\dotsm(a+n-1)
\end{equation}
for $n>0$ and any real or complex number $a$, and the modified Bessel function of the first kind
\begin{equation}\label{I=nu(z)-eq}
I\_\nu(z)= \sum\_{k=0}^\infty\frac1{k!\Gamma(\nu+k+1)}\biggl(\frac{z}2\biggr)^{2k+\nu}
\end{equation}
for $\nu\in\mathbb{R}$ and $z\in\mathbb{C}$.
References
1. Feng Qi and Shu-Hong Wang, *Complete monotonicity, completely monotonic degree, integral representations, and an inequality related to the exponential, trigamma, and modified Bessel functions*, Global Journal of Mathematical Analysis **2** (2014), no. 3, 91--97; available online at <https://doi.org/10.14419/gjma.v2i3.2919>.
2. Feng Qi, *Properties of modified Bessel functions and completely monotonic degrees of differences between exponential and trigamma functions*, Mathematical Inequalities & Applications **18** (2015), no. 2, 493--518; available online at <https://doi.org/10.7153/mia-18-37>.
3. Feng Qi and Christian Berg, *Complete monotonicity of a difference between the exponential and trigamma functions and properties related to a modified Bessel function*, Mediterranean Journal of Mathematics **10** (2013), no. 4, 1685--1696; available online at <https://doi.org/10.1007/s00009-013-0272-2>.
4. Bai-Ni Guo and Feng Qi, *Some integral representations and properties of Lah numbers*, Journal for Algebra and Number Theory Academia **4** (2014), no. 3, 77--87.
5. Xiao-Jing Zhang, Feng Qi, and Wen-Hui Li, *Properties of three functions relating to the exponential function and the existence of partitions of unity*, International Journal of Open Problems in Computer Science and Mathematics **5** (2012), no. 3, 122--127; available online at <https://doi.org/10.12816/0006128>.
6. Bai-Ni Guo and Feng Qi, *An explicit formula for Bell numbers in terms of Stirling numbers and hypergeometric functions*, Global Journal of Mathematical Analysis **2** (2014), no. 4, 243--248; available online at <https://doi.org/10.14419/gjma.v2i4.3310>.
7. Feng Qi and Xiao-Jing Zhang, *Complete monotonicity of a difference between the exponential and trigamma functions*, Journal of the Korea Society of Mathematical Education Series B: The Pure and Applied Mathematics **21** (2014), no. 2, 141--145; available online at <https://doi.org/10.7468/jksmeb.2014.21.2.141>.
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1
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https://mathoverflow.net/users/147732
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405612
| 166,291 |
https://mathoverflow.net/questions/405610
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5
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Consider two diffusions given by
$$X\_j(t)=\int\_0^t a\_j(s,X\_j(s))\,dW\_s$$
for $j=1,2$ and $t\ge 0$, where $W\_\cdot$ is a standard Wiener process/Brownian motion and the $a\_j$'s are smooth enough functions such that $0\le a\_1\le a\_2$.
Does it then follow that $P(|X\_1(1)|>x)\le P(|X\_2(1)|>x)$ for all real $x$?
At least, does this comparison hold when $a\_2$ is a constant?
---
There are a number of results comparing two diffusions with the same diffusion coefficients but with the drift of one of the two diffusions greater than the drift of the other one -- see e.g. [Nakao](https://projecteuclid.org/download/pdf_1/euclid.ojm/1200775827). However, I have been unable to find a comparison of the desired kind.
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https://mathoverflow.net/users/36721
|
A comparison of diffusions
|
The inequality is not true in general — additional assumptions are needed. I think some kind of monotonicity of $a\_1$ and $a\_2$ should help, but this is merely a guess.
Here is a counterexample. Consider $a\_2 = 1$, so that $X\_2(1) = W(1)$. Let $a\_1(s, x) = 1$ when $|x| < 1$ and $a\_1(s, x) = 0$ otherwise. Then $X\_2(1) = W(1 \wedge \tau)$, where $\tau$ is the hitting time of $\{-1, 1\}$ for $W$. It is fairly straightforward to see that if $|X\_2(1)| \geqslant 1$, then $|X\_1(1)| = 1$, while if $|X\_2(1)| < 1$, then still $|X\_1(1)| = 1$ with positive probability. Thus, the desired inequality fails to hold for $x = 1 - \epsilon$ with $\epsilon > 0$ small enough.
The question asks for smooth $a\_1$ and $a\_2$, so one should mollify them appropriately, but this is just a technicality, I hope the idea is clear.
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7
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https://mathoverflow.net/users/108637
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405618
| 166,292 |
https://mathoverflow.net/questions/405632
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4
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Chapter 2, Exercise 25 of R. Stanley's "Enumerative Combinatorics" Vol. 1 asserts that
$$ \sum\_{m,n \geq 0} \left(\sum\_{t \geq 0} f\_i(m,n)t^i\right)\frac{x^m}{m!}\frac{y^n}{n!} = e^{-x-y}\sum\_{m,n \geq 0} (1+t)^{mn} \frac{x^m}{m!}\frac{x^n}{n!},$$
where $f\_i(m,n)$ is the number of $m\times n$ $0,1$-matrices with at least one $1$ in every row and every column, and with $i$ total $1$'s.
I know how to prove this, following the method suggested in the exercise of using the principle of inclusion exclusion to get a formula for $\sum\_{t \geq 0} f\_i(m,n)t^i$ and then carrying out some standard but slightly tricky generating function manipulations.
**Question**: Is there a "bijective" proof of this identity? Note that $(1+t)^{mn}$ is evidently the generating function of *all* $m\times n$ $0,1$-matrices (according to number of $1$'s).
I guess it's a little hard to imagine what a bijective proof could be considering the appearance of exponential generating functions and the factor $e^{-x-y}$, but what I really want to know is:
**Question, reformulated**: Is there any more conceptual explanation for this identity? I am especially interested in an explanation for the way that the generating functions $\sum\_{t \geq 0} f\_i(m,n)t^i$ and $(1+t)^{mn}$ for $m\times n$ $0,1$-matrices with/without restrictions enter into the expression.
|
https://mathoverflow.net/users/25028
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$0,1$-matrices with $1$ in every row/column vs. all $0,1$-matrices
|
If you multiply both sides by $e^{x+y}$ there's a simple bijective proof: In a nutshell, every 0-1 matrix consists of a matrix with a 1 in every row and column together with some all-zero rows and columns.
If you restate this in terms of bipartite (or more precisely, bicolored—bipartite with a specified bipartition) graphs you get something more familiar from the point of view of exponential generating functions: every bicolored graph consists of a bicolored graph with no isolated vertices together with a set of isolated vertices.
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4
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https://mathoverflow.net/users/10744
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405633
| 166,293 |
https://mathoverflow.net/questions/405186
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6
|
Let $p:X\to S$ be the unique map from a (locally compact) topological space $X$ to a point. Since $\underline{\hom}(\underline{\mathbb{Z}},-)$ is the identity functor, we have that $\Gamma(X,-)=\hom(\underline{\mathbb{Z}},-)$ and so
$$H^i(X,-)=\operatorname{Ext}^i(\underline{\mathbb{Z}},-)=\hom\_{\mathsf{D}(X)}(\underline{\mathbb{Z}},-[i]).$$
We can then use the adjunction to write this as $H^i(X,-)=\hom\_{\mathsf{D}(S)}(\underline{\mathbb{Z}},p\_\*- [i])$. In particular, we have that $H^i(X,\underline{\mathbb{Z}})=\hom\_{\mathsf{D}(S)}(\underline{\mathbb{Z}},p\_\*p^\*\underline{\mathbb{Z}} [i])$. The same kind of reasoning shows that $H^i\_c(X,\underline{\mathbb{Z}})=\hom\_{\mathsf{D}(S)}(\underline{\mathbb{Z}},p\_!p^\*\underline{\mathbb{Z}} [i])$.
Now, most references on Verdier duality would call $\hom\_{\mathsf{D}(S)}(\underline{\mathbb{Z}},p\_\*p^!\underline{\mathbb{Z}} [-i])$ the $i$-th homology of $X$. I get that this is defined precisely in a way that recovers Poincaré duality in its general form. **Is there more to it? Why does this deserve to be called a homology group?**
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https://mathoverflow.net/users/131975
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Why does $p_*p^! A$ deserve to be called homology with coefficients in $A$?
|
One nice geometric way to view homology is a measurement of your space $X$ given by probing $X$ with other, nicer spaces, for instance, singular homology probes with simplices. One good reason to call this abstract thing homology is that it too admits elements corresponding to nice enough maps from test objects into $X$.
Lets work in a general six functor formalism, but I'm secretly thinking of constructible sheaves on nice spaces. Then we have our unit object $\mathbf{1}$ ($\underline{\mathbb{Z}}$ in our case) and a dualising object $p^!\mathbf{1}:=\omega$, in our case, this is the usual dualising sheaf $p^!\underline{\mathbb{Z}}$.
Then we can define a "smooth" object $Z$ to be one which admits an isomorphism $\gamma:\mathbf{1}\_Z\rightarrow \omega\_Z[-d\_Z]$, for some integer $d\_Z$. In the constructible/sheafy setting, manifolds give plenty of examples of "smooth" objects, and $d\_Z$ is the dimension.
Then if we have $Z$ a closed subset of $X$, or more generally, if $i\_\*\cong i\_!$ for $i:Z\rightarrow X$, then we have the following map, where all arrows are coming from adjunctions in our formalism, except the second one which says $Z$ is "smooth".
$$\mathbf{1}\_X\rightarrow i\_\*\mathbf{1\_Z}\xrightarrow{i\_\*\gamma} i\_\*\omega\_Z[-d\_Z]\xrightarrow{\cong}i\_!\omega\_Z[-d\_Z]\rightarrow\omega\_X[-d\_Z]$$
So the formalism gives us elements of this group for each "smooth" closed subset $Z$ in $X$, so in my mind, this deserves the label of homology, it admits a direct link to the geometry of nice subsets of your space!
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3
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https://mathoverflow.net/users/128502
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405640
| 166,295 |
https://mathoverflow.net/questions/405066
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5
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Let $P$ be a parabolic subgroup of $\mathrm{GL}\_n$ and $u\in P$ a unipotent element. The parabolic Springer fibre associated to $(P,u)$ can be defined by
$$
\mathcal{P}\_u:=\{gP\in G/P \mathrel\vert g^{-1}u g\in P\}\subseteq G/P.
$$
It is known that these varieties admit affine pavings; see for instance, [Fresse - Existence of affine pavings for varieties of partial flags associated to nilpotent elements](https://arxiv.org/abs/1305.3355). It follows that $\lvert\mathcal{P}\_U(\mathbb{F}\_q)\rvert$ is a polynomial in $q$. This polynomial depends only on the partitions $\lambda$ and $\nu$ associated to $P$ and $U$. Thus, we can denote it by $f\_{\lambda,\nu}$.
Question: Is there an explicit formula for $f\_{\lambda,\nu}$?
For usual Springer fibres (i.e. when $P=B$) an answer is provided at [Fresse - Betti numbers of Springer fibers in type A](https://arxiv.org/abs/0706.3656). Note that the usual Springer fibres are pure; thus, the polynomial counting the number of points is the same as the Poincaré polynomial.
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https://mathoverflow.net/users/41301
|
Number of points of parabolic Springer fibres
|
The earliest reference I could find that works out the polynomial $f\_{\lambda, \nu}(q)$ is the paper
>
> R. Hotta, N. Shimomura ["The Fixed Point Subvarieties of Unipotent Transformations
> on Generalized Flag Varieties and the Green Functions"](https://eudml.org/doc/163262) Math. Ann. 241, 193-208 (1979)
>
>
>
where the authors provide a recursive description of a certain statistic on tableaux which gives the desired polynomial. I also like the calculation in
>
> A. Lascoux, B. Leclerc, J.-Y. Thibon, "Ribbon Tableaux, Hall-Littlewood Functions, Quantum Affine Algebras and Unipotent Varieties" J. Math. Phys. 38(2), 1041-1068 (1997) ([arxiv](https://arxiv.org/abs/q-alg/9512031))
>
>
>
where things are stated a little more explicitly: Recall the Kostka polynomials $K\_{\lambda,\mu}(q)$, and their modified anaolgues $\tilde{K}\_{\lambda, \nu}(q)$ (the generating functions of the charge and cocharge statistic, respectively, see [here](https://www.symmetricfunctions.com/kostkaFoulkes.htm)). If you define the polynomial $$\tilde{Q}'\_{\lambda}(X,q)=\sum\_{\mu}\tilde{K}\_{\lambda, \mu}(q)s\_{\mu}(X)$$
then our $f\_{\lambda,\nu}(q)$ is the coefficient of the monomial symmetric function $m\_{\nu}$ when $\tilde{Q}'\_{\lambda}$ is expressed in the monomial symmetric function basis. Thus, you can write
$$f\_{\lambda,\nu}(q)=\sum\_{\mu}\tilde{K}\_{\lambda, \mu}(q)K\_{\mu,\nu}.$$
As a side note, calculating the cohomology of these parabolic Springer fibers (sometimes referred to as Spaltenstein varieties) used to be the only known proof for the nonnegativity of the coefficients of the Kostka polynomials. The combinatorial understanding of charge/cocharge came later. :)
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2
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https://mathoverflow.net/users/2384
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405647
| 166,298 |
https://mathoverflow.net/questions/405659
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2
|
I want to find a condition on $\delta(G)$ (ex. $\delta(G) \geq an$) that guarantees $\kappa(G)=\delta(G)$ where $\kappa(G)$ is the vertex-connectivity of a bipartite graph $G$, and $\delta(G)$ is the minimum degree of $G$.
In other words, I want to prove that the statement
>
> If $\delta(G) \geq an$, then $\kappa(G)=\delta(G)$
>
>
>
holds for bipartite graphs.
I know that $\delta(G) \geq \frac{n+1}{4}$ implies $\kappa'(G)=\delta(G)$ where $\kappa'(G)$ is the edge-connectivity of $G$, and it is sharp.
Also, I proved that bipartite $G$ is connected if $\delta(G) \geq \frac{n+1}{4}$.
But I have no idea with the **vertex-connectivity**.
Here are my ideas to find such boundary $an$:
1. (Contrapositive) Prove the statement "If $\kappa(G)<\delta(G)$, then $\delta(G)<an$."
2. Use the statement $\delta(G) \geq \frac{n+1}{4}\Rightarrow \kappa'(G)=\delta(G)$. Find the boundary of $\delta(G)$ that implies $\kappa(G)=\kappa'(G)$, and then show this boundary also satisfies $\delta(G) \geq \frac{n+1}{4}$.
Would you help me?
|
https://mathoverflow.net/users/384338
|
Connectivity and the minimum degree of bipartite graph
|
This becomes true at $a = \frac{1}{3}$.
**Claim.** If $G$ is an $n$-vertex bipartite graph such that $\delta(G) \geq \frac{1}{3}n$, then $\delta(G)=\kappa(G)$.
*Proof.* Let $(A,B)$ be the bipartition of $G$. Suppose $\kappa(G)<\delta(G)$ and let $X \subseteq V(G)$ be such that $G-X$ is disconnected and $|X|=\kappa(G)$. Let $A=A\_1 \sqcup A\_2 \sqcup A\_3$ and $B=B\_1 \sqcup B\_2 \sqcup B\_3$ be such that $A\_1=X \cap A$, $B\_1=X \cap B$, and $G[A\_2 \cup B\_2]$ and $G[A\_3 \cup B\_3]$ are both a union of connected components of $G-X$. Since $\kappa(G) < \delta(G)$, none of $A\_2, A\_3, B\_2, B\_3$ are empty. Observe that the degree in $G$ of each vertex in $A\_2, A\_3, B\_2, B\_3$ are at most $|B\_2|+|B\_1|, |B\_3|+|B\_1|, |A\_2|+|A\_1|, |A\_3|+|A\_1|$, respectively. By taking the average of these four numbers, it follows that for some $Y \in \{A\_2, A\_3,B\_2,B\_3\}$ every vertex in $Y$ has degree at most $(n+\kappa(G))/4$. Thus $\delta(G) \leq (n+\kappa(G))/4 < (n+\delta(G))/4$, which implies $\delta(G) < \frac{n}{3}$.
On the other hand, the following claim shows that $\frac{1}{3}$ is best possible.
**Claim.** For all $\ell \in \mathbb{N}$ exists a bipartite graph $G$ on $3\ell+1$ vertices such that $\delta(G)=\ell$, but $\kappa(G) < \delta(G)$.
*Proof.* Let $G\_1$ and $G\_2$ be copies of $K\_{\ell, \ell}$ with bipartitions $(A\_1,B\_1)$ and $(A\_2,B\_2)$, respectively. Let $X\_1 \subseteq A\_1$ and $X\_2 \subseteq A\_2$ both be of size $k$, and let $G$ be the graph obtained from $G\_1$ and $G\_2$ by identifying $X\_1$ and $X\_2$. Note that $G$ is a bipartite graph with $4\ell-k$ vertices, minimum degree $\ell$, and vertex-connectivity $k$. Setting $k=\ell-1$ proves the claim.
|
5
|
https://mathoverflow.net/users/2233
|
405664
| 166,300 |
https://mathoverflow.net/questions/405663
|
2
|
Does there exist a terminal $3$-fold $X$ with a curve $C\subset X$ such that $K\_X\cdot C < 0$ admitting a Mori flip $X\dashrightarrow Y$, flipping $C$ to a curve $C'\subset Y$, where the singular locus of $Y$ consists of a single point of type $\frac{1}{2}(1,1,1)$, $C'$ passes through the singular point of $Y$, and $-K\_Y\cdot C' < -\frac{1}{2}$?
Is there any classification result of $3$-fold terminal flips that could answer to this question?
Thank you.
|
https://mathoverflow.net/users/14514
|
Existence of terminal $3$-fold flips
|
Yes - there are very many such examples, and you can cook up examples by a procedure called 'Mori's algorithm'.
A k2A flipping neighbourhood is a 3-fold flipping contraction $f\colon(C\subset X)\to (P\in Y)$, where the exceptional curve $C$ is irreducible and $X$ has two $cA\_{d\_i-1}/\tfrac1{r\_i}(1,-1,a\_i,0)$ hyperquotient singularities along $C$. You can construct such neighbourhoods as a 1-parameter $\mathbb Q$-Gorenstein smoothing of the contraction of a rational curve in a surface $f\colon (C\subset H\_X)\to (P\in H\_Y)$, where $H\_X$ has two $T$-singularities of type $\tfrac{1}{r\_i^2d\_i}(1,r\_id\_ia\_i-1)$ for $i=1,2$ along $C$.
Mori's algorithm says that if $\delta=r\_1a\_2+r\_2a\_1-r\_1r\_2>0$ (which is the integer such that $K\_XC=-\frac{\delta}{r\_1r\_2}$), then set $d\_1=d\_3=d\_5=\ldots$ and $d\_2=d\_4=d\_6=\ldots$ and write down a table
$$ \begin{array}{ccccccc}
d\_1\delta & d\_2\delta & d\_1\delta & d\_2\delta & \ldots & d\_n\delta & d\_{n+1}\delta \\ \hline
r\_1 & r\_2 & r\_3 & r\_4 & \cdots & r\_n & r\_{n+1} \\
r\_1-a\_1 & a\_2 & a\_3 & a\_4 & \cdots & a\_n & a\_{n+1}
\end{array} $$
where $r\_{i+1} = \delta d\_ir\_i - r\_{i-1}$ and $a\_{i+1} = \delta d\_ia\_i - a\_{i-1}$. If $r\_1>r\_2$ then eventually the sign of $r\_i$ will become negative. Let $n$ be the first $n$ such that $r\_n>0$ and $r\_{n+1}<0$. Then the flip $(C^+\subset X^+)$ will have singularities of type $cA\_{d\_i-1}/\tfrac1{r\_i}(1,-1,a\_i,0)$ along $C^+$ for $i=n,n+1$. Moreover $K\_{X^+}C^+=\frac{\delta}{r\_n(-r\_{n+1})}>0$.
Note that if $d=1$ then a $cA\_{d-1}/\tfrac1{r}(1,-1,a,0)$ singularity is simply a $\tfrac1{r}(1,-1,a)$ quotient singularity, and if $r=1$ then this quotient singularity is actually just a smooth point. Therefore, to give a fairly arbitrary example which address your question, we can take $(r\_1,a\_1,d\_1)=(34,25,1)$ and $(r\_2,a\_2,d\_2)=(7,2,1)$ with $\delta=5$ and write down the table
$$ \begin{array}{cccc}
5 & 5 & 5 & 5 \\ \hline
34 & 7 & 1 & -2 \\
9 & 2 & 1 & 3
\end{array} $$
which tells me that $(C\subset X)$ has $\tfrac1{34}(1,9,25)$ and $\tfrac17(1,2,5)$ singularities with $K\_XC=-\frac{5}{238}$ and the flip $(C^+\subset X^+)$ has one $\tfrac12(1,1,1)$ singularity along $C^+$ and $K\_{X^+}C^+=\frac{5}{2}$. If you look at the resolution of the surface singularity $(P\in H\_Y)$ downstairs, you will find that it is a $\tfrac1{15}(1,4)$ singularity whose resolution is a chain of two rational curves of self-intersection $-4$. The flip $X^+$ is obtained by resolving one of the $-4$ curves and taking a $\mathbb Q$-Gorenstein smoothing of the $\tfrac14(1,1)$ point on the resulting surface.
You make arbitrarily complicated examples by cooking up the two rightmost columns of the table correctly so that $\delta>1$, and then extending the table to the left as far as you want.
|
4
|
https://mathoverflow.net/users/104695
|
405675
| 166,302 |
https://mathoverflow.net/questions/405624
|
4
|
Consider a diffusion given by
$$X\_t=\int\_0^t a(s,X\_s)\,dW\_s$$
for $t\ge 0$, where $W\_\cdot$ is a standard Wiener process/Brownian motion and $a$ is a smooth enough positive function bounded away from $0$.
Does then $X\_1$ have a bounded pdf?
---
This [interesting answer by James Martin](https://mathoverflow.net/a/84289/36721) shows that, without assuming that $a$ is bounded away from $0$ **and** replacing $a(s,X\_s)$ by $a\big(s,(X\_u\colon 0\le u\le s)\big)$, it is possible that $P(X\_1=0)>0$. See also comments to that answer.
|
https://mathoverflow.net/users/36721
|
Bounded density for diffusions with diffusion coefficients bounded away from $0$
|
The "yes" answer follows immediately from [Theorem 2.5 in this paper by Kusuoka](https://reader.elsevier.com/reader/sd/pii/S0304414916300850?token=83AF411073D65047F42CC4F378ABD1D841ADA751E9CEE4DCE66CD933F7D9311AC9D96047079D3456564E461FAA7C60B7&originRegion=us-east-1&originCreation=20211007170037), which implies that $X\_1$ has a normal-like pdf $p$, such that
$$c\_1 e^{-b\_1x^2}\le p(x)\le c\_2 e^{-b\_2x^2}$$
for all real $x$, where $c\_1,b\_1,c\_2,b\_2$ are positive real constants depending only on $\inf\_{t,x} a(t,x)>0$, $\sup\_{t,x} a(t,x)<\infty$, and $\sup\_{t,x,y\ne x}|(a(t,x)-a(t,y)|/|x-y|<\infty$.
|
1
|
https://mathoverflow.net/users/36721
|
405686
| 166,307 |
https://mathoverflow.net/questions/405687
|
8
|
I want to characterize Hausdorffness of a locally convex space only using categorical terms of the additive category LCS of locally convex spaces and continuous linear maps, i.e., terms like mono- or epimorphisms, categorical limits or colimits, or images and kernels are allowed but the toplological definition *Distinct points have disjoint neighbourhoods* is forbidden.
Using the field $\mathbb K$ (either real or complex) as a special object, two characterizations of Hausdorffness are
* Every morphism $f:\mathbb K\to X$ is strict (i.e., its canonical factorization $\dot f:$ coimage$(f) \to$ im$(f)$ is an isomorphism)
* There is a monomorphism $X\to \mathbb K^I$ for some set $I$ (where $\mathbb K^I$ is a categorical product, this characterization uses Hahn-Banach.)
These two characterizations would fit the bill if $\mathbb K$ is characterized in categorical terms.
The questions are thus:
* Is there a characterization of Hausdorffness in terms of LCS without using the field $\mathbb K$?
* Is there a characterization of $\mathbb K$ in LCS?
A similar question could of course be asked for the categories of all topological spaces or (to have enough morphisms) all completely regular spaces. Mayby a reference in this direction would help for the questions in LCS.
|
https://mathoverflow.net/users/21051
|
Is Hausdorffness a categorical property in the category of locally convex spaces?
|
For a category $\mathcal{C}$, let $\mathcal{C}'$ denote the full subcategory of $\mathcal{C}$ whose objects are the non-terminal objects of $\mathcal{C}$.
In a category, say that an object $Y$ is final if for every object $X$ there exists an epimorphism $X\to Y$.
In turn, say that an object of $\mathcal{C}$ is pre-final if it is a final object of $\mathcal{C}'$.
Then say that an object $Y$ of $\mathcal{C}$ is pseudo-Hausdorff if $\mathrm{Hom}(X,Y)$ is reduced to a singleton for every pre-final $X$.
---
Then in the category $\mathcal{C}$ of locally convex spaces (and also topological vector spaces over an arbitrary Hausdorff field), the terminal objects are those spaces reduced to $\{0\}$. In both $\mathcal{C}$ and $\mathcal{C}'$, epimorphisms are just surjective maps (this uses the existence of non-Hausdorff objects). In turn, the pre-final objects are those 1-dimensional non-Hausdorff spaces. And the pseudo-Hausdorff objects are then the Hausdorff spaces.
|
6
|
https://mathoverflow.net/users/14094
|
405690
| 166,308 |
https://mathoverflow.net/questions/404668
|
12
|
Suppose $E$ is a complex-oriented ring spectrum whose formal group law is isomorphic to the additive one. As the title suggests, we might as well change the complex orientation so that the formal group law is literally the additive one. Is $E$ an $H\mathbb{Z}$-algebra?
|
https://mathoverflow.net/users/163893
|
Is every complex oriented ring spectrum with additive FGL an Eilenberg-Maclane spectrum?
|
This is an answer to the question in the title, which is what I had meant to ask: is an $E$ as in the question body an $H\mathbb{Z}$-module? (the last sentence of the question body is stronger and likely has a negative answer)
The answer is yes. Here is an outline of the proof. The vast majority of the following was outlined to me by Robert Burklund and Eric Peterson, but as usual all the errors are mine, etc, etc.
The first step is to reduce to the connective case by noting that $E$ being an $H\mathbb{Z}$-module is equivalent to asking that the unit map $\mathbb{S}\rightarrow E$ factor through $H\mathbb{Z}$, as a map of spectra, and the unit map of $E$ always factors through its connective cover.
The next step is to show that the coation of the mod $p$ dual Steenrod algebra $(H\mathbb{F}\_p)\_\*H\mathbb{F}\_p$ on $(H\mathbb{F}\_p)\_\*E$ (and also $(H\mathbb{F}\_p)\_\*(E/p)$ which is needed later) on is cofree for all $p$. There are probably a lot of ways to do that, my favorite being to take the "formal geometry" perspective by considering the action-of-a-group-scheme-on-another-scheme that the coaction map in question corepresents, and constructing an equivariant map to a scheme on which the group scheme clearly acts freely. This is where you use the fact that you can choose a complex orientation of $E$ in which the formal group law of $E$ is equal to the additive formal group law.
The next step is to make some arguments using the Adams spectral sequence: the previous step implies that the $H\mathbb{F}\_p$-ASS for $E/p$ is concentrated on the zero line. Since $E$ is connective, $E/p$ is $H\mathbb{F}\_p$-nilpotent which implies that the Hurewicz map $E/p\rightarrow H\mathbb{F}\_p\wedge E/p$ is an injection on homotopy groups. Choosing a basis of $\pi\_\*(E/p)$ and extending it to a basis of $\pi\_\*(H\mathbb{F}\_p\wedge E/p)$ defines an equivalence from a wedge of shifts of $H\mathbb{F}\_p$ $X\_p$ to $H\mathbb{F}\_p\wedge E/p$. Let $Y\_p$ be the summand of $X\_p$ corresponding to basis elements of $\pi\_\*(E/p)$. Then the map $E/p\rightarrow H\mathbb{F}\_p\wedge E/p\simeq X\_p\rightarrow Y\_p$ is a weak equivalence, and $Y\_p$ is an $H\mathbb{F}\_p$-module.
The final step is to invoke a lemma that says that if $E/p$ is an $H\mathbb{F}\_p$-module for all $p$, then $E$ is an $H\mathbb{Z}$-module, which involves an argument about $k$-invariants being visibile mod $p$ for some $p$.
|
4
|
https://mathoverflow.net/users/163893
|
405695
| 166,310 |
https://mathoverflow.net/questions/405542
|
1
|
This question can be seen as a continuation of [Lipschitz continuity of $\mathbb P[\tau>t]$ with respect to $t$](https://mathoverflow.net/questions/396424/lipschitz-continuity-of-mathbb-p-taut-with-respect-to-t)
Consider the martingale given as
$$X\_t=1+\int\_0^t a(s,X\_s)dW\_s,\quad \forall t\ge 0.$$
Denote $\tau:=\inf\{t\ge 0: X\_t\le0\}$. My question is whether $t\mapsto \mathbb P[\tau>t]$ is Holder continuous, i.e. $\exists C>0, \alpha\in (0,1]$ s.t.
$$0\le \mathbb P[\tau>t]-\mathbb P[\tau>t+\Delta t]\le C\Delta t^{\alpha}.$$
Any solution, references or comments are appreciated. Here we assume that $0<\underline a \le \inf\_{(t,x)} a(t,x)\le \sup\_{(t,x)} a(t,x)\le \overline a$, $t\mapsto a(t,x)$ is continuous and $|a(t,x)-a(t,y)|\le L|x-y|$ for some $L>0$.
PS : My idea is as follows :
\begin{eqnarray}
\mathbb P[\tau>t]-\mathbb P[\tau>t+\Delta t] &=& \mathbb P\left[\inf\_{0\le s\le t}X\_s>0, X\_t+\inf\_{t\le u\le t+\Delta t}\int\_t^ua(s,X\_s)dW\_s\le 0\right] \\
&=& \mathbb P\left[\inf\_{0\le s\le t}X\_s>0, X\_t>0, X\_t+\inf\_{t\le u\le t+\Delta t}\int\_t^ua(s,X\_s)dW\_s\le 0\right] \\
&\le &\mathbb P\left[X\_t>0, X\_t+\inf\_{t\le u\le t+\Delta t}\int\_t^ua(s,X\_s)dW\_s\le 0\right] \\
&=& \int\_{(0,\infty)}\mathbb P\left[x+\inf\_{t\le u\le t+\Delta t}\int\_t^ua(s,X\_s)dW\_s\le 0\Big|X\_t=x\right]\mathbb P[X\_t\in dx].
\end{eqnarray}
Therefore, it suffices to estimate the conditional probability
$$\mathbb P\left[x+\inf\_{t\le u\le t+\Delta t}\int\_t^ua(s,X\_s)dW\_s\le 0\Big|X\_t=x\right]$$
|
https://mathoverflow.net/users/261243
|
First hitting time for non-homogeneous diffusion martingale
|
For $h:=\Delta t>0$, you had
$$P(\tau>t)-P(\tau>t+h)=\int\_{(0,\infty)}
P(M\le-x|X\_t=x)\,P(X\_t\in dx),$$
where
$$M:=\inf\_{t\le u\le t+h}J\_u,\quad J\_u:=\int\_t^u a(s,X\_s)\,dW\_s.$$
By [Doob's martingale inequality](https://en.wikipedia.org/wiki/Doob%27s_martingale_inequality#Statement_of_the_inequality),
$$P(M\le-x|X\_t=x)\le x^{-2}\,E(J\_{t+h}^2|X\_t=x)
\le x^{-2}\,ha\_2^2,$$
where $a\_2:=\overline a$ and $a\_1:=\underline a$.
The crucial point is that the pdf $p\_t$ of $X\_t$ for $t>0$ is [bounded](https://mathoverflow.net/a/405686/36721) so that
$$p\_t(x)\le\frac c{\sqrt t}\,e^{-bx^2/t}\le\frac c{\sqrt t}$$
for some positive real constants $c,b$ depending only on $a\_1,a\_2,L$ and for all real $x$. So,
$$
\begin{align}
P(\tau>t)-P(\tau>t+h)&\le
\int\_{(0,\infty)}\min(1,x^{-2}\,ha\_2^2)\,\frac c{\sqrt t}\,dx \\
&=
\sqrt h\int\_{(0,\infty)}\min(1,u^{-2}\,a\_2^2)\,\frac c{\sqrt t}\,du \\
&=\frac C{\sqrt t}\,\sqrt h,
\end{align}
$$
where $C>0$ is a real constant depending only on $a\_1,a\_2,L$.
---
To get now a uniform Hölder continuity, we can reason as follows:
$$P(\tau>t)-P(\tau>t+h)\le1-P(\tau>t+h)=P(\tau\le t+h)\le a\_2^2(t+h)/1^2,$$
again by Doob's martingale inequality. So,
$$P(\tau>t)-P(\tau>t+h)\le\min\Big(1,\frac C{\sqrt t}\,\sqrt h,a\_2^2(t+h)\Big)\le C\_1h^{1/3},$$
where $C\_1>0$ is a real constant depending only on $a\_1,a\_2,L$ (to verify the latter inequality, consider separately the three cases when (i) $h\ge1$, (ii) $t\le h^{1/3}<1$, or (iii) $t>h^{1/3}$).
Using here an exponential inequality (see e.g. [Theorem 3.1](https://projecteuclid.org/download/pdf_1/euclid.aop/1176988477)) instead of Doob's one, one can improve the factor $h^{1/3}$ to $h^{1/2}\ln\frac1h.$
|
3
|
https://mathoverflow.net/users/36721
|
405700
| 166,313 |
https://mathoverflow.net/questions/405706
|
2
|
I want to prove the following statement:
>
> Let $u$ be a vertex in a $2$-connected graph $G$. Then $G$ has two spanning trees such that for every vertex $v$, the $u,v$-paths in the trees are independent.
>
>
>
I tried to show this, but surprisingly, I have proved another statement.
>
> A graph with $\vert V(G) \vert \geq 3$ is $2$-connected iff for any two vertices $u$ and $v$ in $G$, there exist at least two independent $u,v$-paths.
>
>
>
And I can assure that it is true, since I could find it from other papers.
I think this one may help me proving the desired statement, but I have no idea how to use it properly.
Would you help me find a such way, or suggest another proof of the first statement?
|
https://mathoverflow.net/users/384338
|
Two independent spanning trees of $2$-connected graph
|
Yes, this is true. We will prove the following stronger lemma.
**Lemma.** Let $G$ be a $2$-connected graph and $u \in V(G)$. Then $G$ contains two spannings trees $T\_1$ and $T\_2$ such that for all $a,b \in V(G) \setminus \{u\}$ (possibly $a=b$), either
* the $ua$-path in $T\_1$ and the $ub$-path in $T\_2$ are internally-disjoint, or
* the $ub$-path in $T\_1$ and the $ua$-path in $T\_2$ are internally disjoint.
*Proof.* We proceed by induction on $|V(G)|+|E(G)|$. Since $G$ is $2$-connected, $G$ has an [ear decomposition](https://en.wikipedia.org/wiki/Ear_decomposition) $(C, P\_1, \dots, P\_k)$, with $u \in V(C)$. If $G=C$, let $e\_1$ and $e\_2$ be the two edges incident to $u$. It is easy to check that we may take $T\_1=C \setminus e\_1$ and $T\_2=C \setminus e\_2$. Thus, we may assume $k \geq 1$.
If the last ear $P\_k$ is just an edge $e$, then $G \setminus e$ is $2$-connected, and we can apply induction. Thus, we may assume $|V(P\_k)| \geq 3$. Suppose $P\_k=x\_1x\_2 \dots x\_\ell$. Observe that $G'=G-\{x\_2, \dots, x\_{\ell-1}\}$ is $2$-connected. By induction, $G'$ contains two spanning trees $T\_1'$ and $T\_2'$ such that for all $a,b \in V(G') \setminus \{u\}$, either
* the $ua$-path in $T\_1'$ and the $ub$-path in $T\_2'$ are internally-disjoint, or
* the $ub$-path in $T\_1'$ and the $ua$-path in $T\_2'$ are internally disjoint.
In particular, either
* the $ux\_1$-path in $T\_1'$ and the $ux\_\ell$-path in $T\_2'$ are internally-disjoint, or
* the $ux\_\ell$-path in $T\_1'$ and the $ux\_1$-path in $T\_2'$ are internally disjoint.
In the first case, $T\_1:=T\_1' \cup x\_1x\_2 \dots x\_{\ell-1}$ and $T\_2:=T\_2' \cup x\_2x\_3 \dots x\_\ell$ are the required spanning trees of $G$. In the second case, $T\_1:=T\_1' \cup x\_2x\_3 \dots x\_\ell$ and $T\_2:=T\_2' \cup x\_1x\_2 \dots x\_{\ell-1}$ are the required spanning trees of $G$.
|
1
|
https://mathoverflow.net/users/2233
|
405710
| 166,319 |
https://mathoverflow.net/questions/405707
|
5
|
It would lead me too far to explain how I stumbled upon the somewhat obscure identities
$$\sum\_{m=0}^n \binom{n}{m} (1-m)^m m^{n-m}=(-1)^n d\_n, \quad \sum\_{m=0}^n \binom{n}{m} (1-m)^{m-1} m^{n-m}=1,$$
where $d\_n=n!\sum^{n}\_{k=0} (-1)^k /k!$ is the $n$-th de Montmort number, when doing some algebro-geometric considerations. No doubt they are well-known to professional combinatorialists. As stated, they look rather unmotivated; surely they are special cases of a much broader class of identities. Ideally I would like to get a reference for the latter.
|
https://mathoverflow.net/users/104669
|
A certain type of combinatorial identity, involving de Montmort numbers
|
In John Riordan's book *Combinatorial Identities*, page 21, is the formula
$$\sum\_{k=0}^n \binom{n}{k}(x+k)^k (y+n-k)^{n-k}=
\sum\_{k=0}^n \binom{n}{k} k!\, (x+y+n)^{n-k}.$$
(There is a typo in the formula given in the book—this is the correct version.) Riordan writes, "This is usually called Cauchy's formula," but I don't know of a reference to Cauchy's work. The case $x=-1, y=-n$ is the OP's first formula.
Sections 1.5 and 1.6 of Riordan's book are devoted to Abel-type formulas.
|
5
|
https://mathoverflow.net/users/10744
|
405716
| 166,323 |
https://mathoverflow.net/questions/405717
|
6
|
How many sublattices does the powerset lattice $2^n$ contain for $n$ finite? (up to equality, not isomorphism)
I thought for sure this would be easy to find on OEIS, but so far I am coming up empty.
I really am interested in seeing a list of small examples, say up to $n=5$ or $6$ maybe, although perhaps already things blow up too much at that point for this to be feasible. Ideally there would be a systematic way to write down examples, but just as an entryway to the literature, I thought I'd ask this as a counting question.
|
https://mathoverflow.net/users/2362
|
How many sublattices are contained in the powerset lattice of a finite set?
|
[OEIS A306445:](https://oeis.org/A306445) 2, 4, 13, 74, 732, 12085, 319988, 13170652, 822378267, 76359798228, 10367879036456, 2029160621690295, 565446501943834078, 221972785233309046708, 121632215040070175606989, 92294021880898055590522262, 96307116899378725213365550192, 137362837456925278519331211455157, 266379254536998812281897840071155592
Number of collections of subsets of $\{1, 2,\ldots,n\}$ that are closed under union and intersection (starting at $n=0$).
This count includes the empty sublattice, so if you don't want that you should subtract one from every term.
|
14
|
https://mathoverflow.net/users/75735
|
405719
| 166,324 |
https://mathoverflow.net/questions/405631
|
11
|
It's well known that $1$ is the sum of three cubes infinitely many different ways but is it true for perhaps the tetrahedral numbers as well? Let $T\_n = (1/6)n(n+1)(n+2)$. Then the following are the first solutions less than 6000 of the form $T\_a - T\_b - T\_c = 1$.
$$
\begin{array}{c c c}
a & b & c\\
6 & 5 & 4\\
8 & 7 & 5\\
11 & 9 & 8\\
23 & 19 & 17\\
33 & 32 & 14\\
45 & 43 & 22\\
51 & 49 & 24\\
76 & 75 & 25\\
85 & 84 & 27\\
209 & 207 & 63\\
238 & 228 & 117\\
323 & 304 & 177\\
340 & 323 & 177\\
369 & 318 & 262\\
380 & 317 & 284\\
449 & 422 & 248\\
715 & 707 & 229\\
1105 & 1022 & 655\\
1493 & 1438 & 707\\
2319 & 2173 & 1302\\
2406 & 2405 & 258\\
3183 & 2982 & 1789\\
5950 & 5093 & 4282\\
5985 & 5904 & 2047\\
\end{array}
$$
Are there infinitely many solutions?
|
https://mathoverflow.net/users/265714
|
Prove that $1$ is the sum of three tetrahedral numbers infinitely many different ways
|
There are infinitely many solutions. I'll show below that there are infinitely many positive integers $k$ for which $93k^{2} - 288k + 276 = z^{2}$ for some positive integer $z$. From such a $z$, we get a solution by setting
$a = \frac{z+3k}{6} - 1$, $b = 2k-3$, and $c = \frac{z-3k}{6} - 1$. Here's a table of solutions in this family:
| $k$ | $z$ | $a$ | $b$ | $c$ |
| --- | --- | --- | --- | --- |
| $1$ | $9$ | $1$ | $-1$ | $0$ |
| $23$ | $207$ | $45$ | $43$ | $22$ |
| $163$ | $1557$ | $340$ | $323$ | $177$ |
| $18005$ | $173619$ | $37938$ | $36007$ | $19933$ |
| $135460$ | $1306314$ | $285448$ | $270917$ | $149988$ |
| $15104876$ | $145666134$ | $31830126$ | $30209749$ | $16725250$ |
The choice of variables here is related to Wojowu's comment that setting $a = c+k$ yields an elliptic curve. This elliptic curve is isomorphic to $y^{2} = x^{3} - 144k^{2}x + (-432k^{6} + 1728k^{4} + 10368k^{3})$. This elliptic curve always has the point $(12k^{2} - 24k, 36(k-4)k^{2})$ on it, but this corresponds to the "trivial solution" $a = b = k-3$ and $c = -3$. I observed that setting $x = dk^{2} - 24k$ yields $y^{2} = k^{4} \cdot (\text{quadratic in } k)$. Setting $d = 24$ gives $y^{2} = 144k^{4} (93k^{2} - 288k + 276)$.
The equation $93k^{2} - 288k + 276 = z^{2}$ is equivalent after completing the square to $w^{2} - 93z^{2} = -4932$ with the restriction that $w \equiv 42 \pmod{93}$. (We have $w = 93k-144$.) We can rewrite $w^{2} - 93z^{2} = -4932$ as $N\_{\mathbb{Q}(\sqrt{93})/\mathbb{Q}}(w + z \sqrt{93}) = -4932$. One solution is $-51 + 9 \sqrt{93}$. To find more solutions, note that a fundamental unit is $u = \frac{29 + 3 \sqrt{93}}{2}$ and $u^{6} = 295293601 + 30620520 \sqrt{93} \equiv 1 \pmod{93}$. So an infinite family of such $z$'s can be found by looking at the coefficient of $\sqrt{93}$ in $(-51+9\sqrt{93}) (295293601 + 30620520 \sqrt{93})^{r}$ for $r \geq 1$.
This family of solutions is quite sparse, but other families can be found by choosing a different value of $d$ in $x = dk^{2} - 24k$.
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16
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https://mathoverflow.net/users/48142
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405721
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