parent_url
stringlengths 37
41
| parent_score
stringlengths 1
3
| parent_body
stringlengths 19
30.2k
| parent_user
stringlengths 32
37
| parent_title
stringlengths 15
248
| body
stringlengths 8
29.9k
| score
stringlengths 1
3
| user
stringlengths 32
37
| answer_id
stringlengths 2
6
| __index_level_0__
int64 1
182k
|
|---|---|---|---|---|---|---|---|---|---|
https://mathoverflow.net/questions/409170
|
6
|
Working in $\mathsf{ZFC}$ + "There is a weakly compact cardinal" and letting $\kappa$ be the least weakly compact cardinal, say that a logic $\mathcal{L}$ is **loraxian** iff every $\mathcal{L}$-definable-over-$V\_\kappa$ subtree of $2^{<\kappa}$ has an $\mathcal{L}$-definable-over-$V\_\kappa$ branch. [Enayat and Hamkins showed](https://arxiv.org/pdf/1610.02729.pdf) (among other things) in $\mathsf{ZFC}$ that first-order logic is not loraxian, but I don't see how to use their techniques to address other logics - the key point being that everything leading up to case $2$ of Theorem $2.6$ is highly $\mathsf{FOL}$-specific.
As is my wont, I'm specifically curious about the situation with respect to second-order logic:
>
> Is second-order logic consistently loraxian?
>
>
>
All I've been able to see is the following two basic observations:
* In contrast with case $2$, case $1$ of E/H's Theorem $2.6$ is fairly coarse: roughly speaking, by considering a tree whose nodes code choice functions for the $V\_\alpha$s with $\alpha<\kappa$, we have that $\mathcal{L}$ is loraxian only if there is some weakly compact $\kappa$ such that $V\_\kappa$ has an $\mathcal{L}$-definable-over-$V\_\kappa$ well-ordering.
* Meanwhile, there is a silly red herring here. My initial guess was that $\mathsf{SOL}$ would **obviously** be loraxian since we can define the nodes which belong to some path. However, since $\kappa$ is weakly compact *in reality* this is actually no power whatsoever. What posed a problem for $\mathsf{FOL}$ isn't an inability to detect when a node should extend to a path, but rather (very roughly) the issue of piecing together a *single* path in a consistent way - and I don't see that this goes away for $\mathsf{SOL}$.
More generally I'm interested in *anything* on loraxian logics, but $\mathsf{SOL}$ seems like a natural starting point.
|
https://mathoverflow.net/users/8133
|
Can $\mathsf{Ord}$ be weakly compact from a second-order perspective?
|
If $\kappa$ is weakly compact and there is a wellorder of $V\_{\kappa+1}$ definable over $V\_{\kappa+1}$ without parameters, then second order logic is Loraxian for $V\_\kappa$: the least branch through a definable tree is definable. In $L$ (or in fact any of the known canonical inner models), there is such a wellorder.
You didn't ask, but I think it is consistent for second order logic to fail to be Loraxian. After a preparatory forcing, one can add a Cohen subset $G$ of $\kappa$ without destroying its weak compactness. The forcing can be factored as adding a Suslin tree $T$ and then adding $G$ as a branch through $T$. I am not an expert here, but I think one can probably force over $V[G]$ to make $T$ definable over $V\_{\kappa+1}$, arranging that $G$ remains undefinable and $\kappa$ remains weakly compact. Probably one can arrange that any subset of $V\_\kappa$ definable over $V\_{\kappa+1}^{V[G,H]}$ is in fact be definable over $V\_{\kappa+1}^{V[T]}$ from $T$. This is an (admittedly sketchy) local version of the arguments from Cheng-Friedman-Hamkins's ["Large cardinals need not be large in HOD."](https://arxiv.org/abs/1407.6335)
|
4
|
https://mathoverflow.net/users/102684
|
409308
| 167,557 |
https://mathoverflow.net/questions/409321
|
3
|
My construction is as follows: Let $X\_k$ be a real-valued continuous random variable centered at $k$ (an integer), having distribution $F\_k(x,s)$ where $k$ is the location parameter and $s$, a strictly positive real number, is the scale (fixed, not depending on $k$). Thus $s$ is typically a monotonic increasing function of the variance of $X\_k$. We assume that the support domain of $F\_k$ is the set of all real numbers $x$, and that the $X\_k$'s are independently distributed, with $k \in \mathbb{Z}$ . The points of my point process are the $X\_k$'s, and this is how my point process is defined.
Obviously if the support domain of $F\_k$ is compact, say equal to $[k-s, k+s]$, I end up with a non-Poisson point process. But let's focus on processes meeting the criteria mentioned above. In addition, add these constraints: if $s=0$ then $X\_k=k$ (the variance is zero). If $s\rightarrow\infty$ then the variance of $X\_k$ tends to infinity.
What is weird, and counter-intuitive to me, is that the resulting point process, however small or large $s$ is, is always stationary Poisson with intensity equal to $1$. I will illustrate this in the case when $F\_k$ is the logistic distribution as computations are somewhat simple, but if you look at my computations in the example section, it seems that it would also be true whether $F\_k$ has a Cauchy, Gaussian, Laplace, or other symmetric standard continuous distribution on the full real line.
**Question**:
Can you confirm (by checking my computations in the example below) whether this is true or not? Are there any exceptions? What happens if $F\_k$ is not a symmetric distribution centered at $k$? Or is there something wrong in my computations?
Generalization to two dimensions, with $k$ replaced by $(k,l)\in \mathbb{Z}^2$ and $F\_{k,l}$ being the bivariate distribution attached to $X\_{k,l}$, would be nice to discuss, but this is not part of my question. In particular, what if the joint distribution $F\_{k,l}(x,y)$ is not equal to the product of its marginals, and the parameter $s$ is replaced by a covariance matrix? Also, by increasing the granularity of the underlying lattice by a factor $\lambda$, the resulting process is expected to be Poisson with intensity $\lambda$, but this is not part of my question either.
**Example**:
Let $B=[a, b]$ with $a<b$ be an interval on the real line, and $N(B)$ be the random variable that denotes the number of points from the point process defined above, that are in $B$. Let $F\_k(x)$ be the logistic distribution (CDF) defined by
$$F\_k(x) = \frac{1}{2}+\frac{1}{2}\tanh \Big(\frac{x-k}{2s}\Big).$$
Let $p\_k=p\_k(B)$ be the probability that $X\_k \in B$, with $B=[a,b]$. That is,
$$p\_k=\frac{1}{2}\Big[\tanh \Big(\frac{b-k}{2s}\Big)-\tanh \Big(\frac{a-k}{2s}\Big)\Big].$$
The expected number of points in $B$ is
$$E[N(B)]=\sum\_{k=-\infty}^{\infty} p\_k = b-a.$$
I used Mathematica to compute the value of the above infinite series, and surprisingly, it does not depend on $s$. Of course it is supposed, based on intuition, to be proportional to $b-a$ but the proportionality factor is always $1$. Also, one would guess that the number of points in non-overlapping Borel sets, to be independent. Then I decided to compute the distribution of $N(B)$, with $B=[a, b]$. Let $q\_n=P(N(B)=n)$, for $n=0,1,2$ and so on.
$$q\_0 = \prod\_{k=-\infty}^\infty (1-p\_k)$$
$$q\_1 = q\_0 \prod\_{k=-\infty}^\infty \frac{p\_k}{1-p\_k},$$
*Note: the remaining of this section is wrong. I decided to leave it as is so that readers can relate to the comments / answer provided by other authors. The fixed version can be found in the last section called update, added recently at the bottom of my question.*
$$\begin{align}
q\_2 = &\frac{q\_0}{2!} \prod\_{i\neq j} \frac{p\_i p\_j}{(1-p\_i)(1-p\_j)} \\
= &
\frac{q\_0}{2!}\Big[\Big(\prod\_i \frac{p\_i}{1-p\_i}\Big)\Big(\prod\_j\frac{p\_j}{1-p\_j}\Big) (1+o(1))\Big]\\
= & \frac{q\_0}{2!}\cdot\Big(\frac{q\_1}{q\_0}\Big)^2,
\end{align}$$
$$\begin{align}
q\_3 = & \frac{q\_0}{3!} \prod\_{i\neq j\neq l} \frac{p\_i p\_j p\_l}{(1-p\_i)(1-p\_j)(1-p\_l)}\\
= &
\frac{q\_0}{3!}\Big[\Big(\prod\_i \frac{p\_i}{1-p\_i}\Big)\Big(\prod\_j\frac{p\_j}{1-p\_j}\Big) \Big(\prod\_l\frac{p\_l}{1-p\_l}\Big)(1+o(1))\Big]\\
= & \frac{q\_0}{3!}\cdot\Big(\frac{q\_1}{q\_0}\Big)^3,
\end{align}$$
and continuing iteratively, one finds that
$$q\_n=\frac{q\_0}{n!}\cdot\Big(\frac{q\_1}{q\_0}\Big)^n.$$
So we are dealing with a Poisson distribution, and we showed that its expectation is $b-a = \mu(B)$. With the assumed independence assumptions between disjoint Borel sets, we meet all the criteria to conclude that we are dealing with a Poisson process of intensity $1$. I used the notation $o(1)$ assuming the indices in the products for $q\_2$ and $q\_3$ were varying from $-k$ to $+k$, with $k\rightarrow\infty$.
**Background**:
Initially, I wanted to study the behavior of series such as $\zeta(z)=\sum\_{k=1}^\infty k^{-z}$ when replacing the index $k$ by $X\_k$, where $(X\_k)$ are the points of a stochastic process as defined in the introduction. The idea was to have a distribution $F\_k$ attached to $X\_k$, centered at $k$, and such that when $s\rightarrow 0$, $X\_k \rightarrow k$, as for the logistic distribution. In other words, replacing $\zeta(z)$ by a random function, with a limiting case being $\zeta(z)$ itself, and studying the properties, especially as $s\rightarrow 0$.
For this to work, I needed a point Process that is non-Poisson regardless of $s$ of course. So far, I haven't found a solution yet since I ended up with pure Poisson processes. While I failed, I've found it interesting enough to further explore these original constructions of Poisson processes, discovered by accident.
**Update**
Regardless of the distribution $F\_k$, the distribution of $N(B)$ is not Poisson as claimed, but rather Poisson-Binomial of parameters $p\_k$, with $k\in \mathbb{Z}$ and $p\_k=P(X\_k\in B)$. The only difference with a standard Poisson-binomial distribution (see [here](https://en.wikipedia.org/wiki/Poisson_binomial_distribution)) is that in my case, the number of parameters is infinite. In particular, my values for $E[N(B)], q\_0, q\_1$ are correct but those for $q\_n$ with $n>1$ are wrong. Also,
$$\mbox{Var}[N(B)]= \sum\_{k=-\infty}^\infty (1-p\_k)p\_k.$$
I'll check if I can get a closed form when $F\_k$ is the logistic distribution.
An example when Poisson-binomial becomes Poisson at the limit, is as follows. Pick up randomly an integer between $0$ and $n-1$, another one between $0$ and $n$, then another one between $0$ and $n+1$, and so on, and stop after picking a last one between $0$ and $\lambda n$ (here $\lambda > 1$). This problem arises in a Sieve-like algorithm, where the numbers you pick up are residues of a number massively larger than $\lambda n$, modulo $n, n+1, n+2$ and so on. When $n\rightarrow\infty$, the probability that exactly $k$ of the numbers selected are zero, is $(\log \lambda)^k \cdot \lambda^{-1}/k!$. Does this mean that the chance that a number much larger than $\lambda n$ has exactly $k$ divisors between $n$ and $\lambda n$ is $(\log \lambda)^k \cdot \lambda^{-1}/k!$, as $n\rightarrow\infty$, with the average number of divisors in that interval being $\log\lambda$?
|
https://mathoverflow.net/users/140356
|
Does my construction always result in a stationary Poisson point process of intensity $1$? How so?
|
This is not true. Indeed, suppose that $X\_k=X\_{s;k}=k+sZ\_k$, where $s\downarrow0$ and the $Z\_k$'s are **any** iid random variables (r.v.'s).
To obtain a contradiction, suppose that, for the random Borel measure $\mu\_s$ over $\mathbb R$ defined by $\mu\_s(B):=\sum\_{k\in\mathbb Z}1(X\_{s;k}\in B)$, the distribution of the random variable (r.v.) $\mu\_s(B)$ is Poisson with parameter $\lambda(s)|B|$ for some $\lambda(s)>0$ and all Borel $B$, where $|B|$ is the Lebesgue measure of $B$.
Note that
\begin{equation}
\mu\_s((-1/2,1/2))\to1 \tag{1}
\end{equation}
in probability (see details on (1) below). Therefore and because the r.v. $\mu\_s((-1/2,1/2))$ has the Poisson distribution with parameter $\lambda(s)$,
necessarily $\lambda(s)\to0$ and hence $\mu\_s((-1/2,3/2))\to1$ in probability. However, similarly to (1) we have $\mu\_s((-1/2,3/2))\to2$ in probability, a contradiction.
So, the random measure $\mu\_s$ cannot be Poisson for all $s>0$.
---
*Proof of (1):* Note that $\mu\_s(B)=\sum\_{k\in\mathbb Z}1(Z\_k\in\frac{B-k}s)$ and hence
\begin{equation}
1-\mu\_s((-1/2,1/2))=s\_1-s\_2,
\end{equation}
where
\begin{equation}
s\_1:=1-1\Big(Z\_0\in\Big(\frac{-1/2}s,\frac{1/2}s\Big)\Big),
\end{equation}
and
\begin{equation}
s\_2:=\sum\_{k\in\mathbb Z\setminus\{0\}}1\Big(Z\_k\in\Big(\frac{-1/2-k}s,\frac{1/2-k}s\Big)\Big).
\end{equation}
Next,
\begin{equation}
Es\_1=1-P\Big(Z\_0\in\Big(\frac{-1/2}s,\frac{1/2}s\Big)\Big)\to0
\end{equation}
and
\begin{equation}
Es\_2=\sum\_{k\in\mathbb Z\setminus\{0\}}P\Big(Z\_0\in\Big(\frac{-1/2-k}s,\frac{1/2-k}s\Big)\Big)\le Es\_1.
\end{equation}
Therefore and because $s\_1,s\_2\ge0$, we have
\begin{equation}
E|\mu\_s((-1/2,1/2))-1|\le Es\_1+Es\_2\to0.
\end{equation}
So, by Markov's inequality, (1) follows.
|
2
|
https://mathoverflow.net/users/36721
|
409326
| 167,565 |
https://mathoverflow.net/questions/408960
|
17
|
Here are some examples of compact homogeneous 3 manifolds for different Thurston geometries:
(1) Spherical: $\mathbb{S}^3 \cong \mathrm{SU}\_2$ modulo any finite subgroup
(2) Euclidean: 3 torus $\mathbb{R}^3/\mathbb{Z}^3$
(3) $\mathbb{S}^2 \times \mathbb{R}$: 2 sphere times a circle $ S^2\times S^1 \cong (SO\_3/SO\_2) \times SO\_2 $ or projective plane times a circle $ \mathbb{RP}^2 \times S^1 \cong (SO\_3/O\_2) \times SO\_2 $
(4) Nil: Take the three dimensional real Heisenberg group, which is just the group of upper triangular $3\times 3$ real matrices with all 1s on the diagonal, and mod out by the subgroup with integer entries. This quotient is called the "Heisenberg nilmanifold." This is an iterated principal circle bundle like all compact nilmanifolds in any dimension (<https://arxiv.org/abs/1805.06585>).
(5) $\widetilde{\mathrm{SL}\_2(\mathbb{R})}$: $ SL\_2(\mathbb{R}) $ mod a Fuchsian surface group gives a unit tangent bundle (a type of circle bundle) over a hyperbolic surface.
Now I want to find examples for the other 3 geometries of a matrix group $ G $ mod some cocompact subgroup $ H $ whose quotient is a 3 manifold with the desired geometry.
(6) Sol: Could someone suggest a $ G $ and an $ H $ such that $ G/H $ is a compact 3 dimensional manifold with Sol geometry? (EDIT: this is answered here <https://math.stackexchange.com/questions/4317139/bianchi-classification-of-solvable-lie-groups-and-cocompact-subgroups> )
(7) $ \mathbb{H}^2 \times R $: Could someone suggest a $ G $ and an $ H $ such that $ G/H $ is a compact 3 manifold with $ \mathbb{H}^2 \times R $ geometry ?
(8) Hyperbolic $ \mathbb{H}^3 $: Could someone suggest a $ G $ and an $ H $ such that $ G/H $ is a compact hyperbolic 3 manifold?
(EDIT: As Moishe Kohan states in his answer every compact homogeneous 3 manifold is either $ G/H $ for $ G $ 3 dimensional and $ H $ discrete (which covers my examples here for all geometries except $ S^2 \times R $) or if it is not of this form then it must be a fiber bundle of the following sort: either a bundle of circles over a sphere/projective plane/torus/klein bottle or a bundle of sphere/projective plane/torus/klein bottle over a circle)
|
https://mathoverflow.net/users/387190
|
Examples of the Thurston geometries with transitive Lie group action
|
This is an answer to questions 7 and 8 (I have to say, having 8 questions in one post is way too much for my taste):
>
> Suppose that $M$ is a finite-volume quotient of $H^3$ or a compact quotient of $H^2\times {\mathbb R}$ by a discrete group of isometries. Then $M$ cannot be homogeneous (in the non-Riemannian sense!), meaning that there is no (finite-dimensional) connected Lie group $G$ and its closed subgroup $H$ such that $M$ is diffeomorphic to $G/H$.
>
>
>
This can be derived, for instance, from Theorem C in
G. D. Mostow, A structure theorem for homogeneous spaces.
Geom. Dedicata 114 (2005), 87–102.
Mostow's theorem provides the most complete, to my knowledge, description of smooth connected homogeneous manifolds: Mostow describes them as iterated fiber bundles with homogeneous fibers where, with exception of one of the fibrations, the fibers of the bundles are diffeomorphic to quotients of connected Lie groups by discrete subgroups. When you translate this into the setting of 3-dimensional homogeneous manifolds, this becomes the statement that a homogeneous connected 3-manifold $M$ is either diffeomorphic to $G/\Gamma$ where $G$ is a connected Lie group and $\Gamma< G$ is discrete, or $M$ is a fiber bundle whose base and fiber are either circle and a surface of Euler characteristic $\ge 0$ or vice-versa. In the case of hyperbolic 3-manifolds (of finite volume) and compact $H^2\times {\mathbb R}$-manifolds neither type is possible. For instance, a fibration of one of the above types is ruled out by the fact that $\pi\_1(M)$ is not (virtually) solvable. To rule out a representation of $M$ as a quotient $G/\Gamma$, where $G$ is a 3-dimensional connected Lie group and $\Gamma< G$ is discrete, one observes that, for the same algebraic reason, $G$ has to be locally isomorphic to $SL(2, {\mathbb R})$. But in this case, $M$ is a manifold which admits the $\widetilde{SL}(2, {\mathbb R})$-geometry, which is known to be incompatible with a hyperbolic structure of finite volume and with a $H^2\times {\mathbb R}$-structure in the case of compact manifolds.
For the sake of completeness:
1. The classes of **noncompact** manifolds of finite volume of the types $H^2\times {\mathbb R}$ and $\widetilde{SL}(2, {\mathbb R})$ are indistinguishable topologically.)
2. "Most" quotients of ${\mathbb H}^3$ by torsion-free discrete subgroups cannot be diffeomorphic to quotients of $\widetilde{SL}(2, {\mathbb R})$ by discrete subgroups. The exceptions all have free fundamental groups or fundamental groups isomorphic to ${\mathbb Z}^2$ or to the fundamental group of the Klein bottle.
|
9
|
https://mathoverflow.net/users/39654
|
409329
| 167,566 |
https://mathoverflow.net/questions/409334
|
4
|
Recently, I want to know how well can a $\ell\_1$ ball be approximated by the image of a $\ell\_2$ ball under a linear transform. I formulate this problem as the following optimization problem.
\begin{aligned}
&\min\_{\mathbb{H}\in \mathcal{M}\_{n}} \max\_{\left\| \mathbf{x}\right\|\_2 \le 1} &&\left\|\mathbb{H}\mathbf{x} \right\|\_1 \\
&\quad\quad\text{s.t.} &&|\det(\mathbb{H})|=1
\end{aligned}
where $\mathbf{x}\in \mathbb{R}^n$ and $\mathcal{M}\_{n}$ denotes the set of $n\times n$ square matrices in $\mathbb{R}$.
Asymptotic analysis (in terms of $n \to \infty$) and numerical algorithms are also appreciated.
|
https://mathoverflow.net/users/149696
|
How to solve this minimax matrix optimization problem?
|
$\newcommand{\1}{\mathbf 1}\newcommand{\ep}{\varepsilon}\newcommand{\tr}{\operatorname{tr}}$The min-max value is $\sqrt n$.
Indeed, take any real $n\times n$ matrix $H$ with $|\det H|=1$. By the singular value decomposition,
\begin{equation}
H=U^TDV,
\end{equation}
where $U$ and $V$ are some orthogonal matrices and $D$ is the diagonal matrix with diagonal entries $d\_1,\dots,d\_n$ such that $d\_1\cdots d\_n=1$. Then
\begin{equation}
\max\_{\|x\|\_2\le1}\|Hx\|\_1=\max\_{\|z\|\_2\le1}\|U^TDz\|\_1.
\end{equation}
So,
\begin{equation}
\begin{aligned}
&\min\_{H\colon\,|\det H|=1}\max\_{\|x\|\_2\le1}\|Hx\|\_1 \\
&=\min\_{U,D}\max\_{\|z\|\_2\le1}\,\max\_{\ep\in\{-1,1\}^n}\sum\_{i=1}^n e\_i^T D\_\ep U^TDz \\
&=\min\_{U,D}\max\_{\ep\in\{-1,1\}^n}\,\max\_{\|z\|\_2\le1}\,\sum\_{i=1}^n e\_i^T D\_\ep U^TDz \\
&=\min\_{U,D}\max\_{\ep\in\{-1,1\}^n}\,\|DUD\_\ep\1\|\_2,
\end{aligned}
\end{equation}
where (i) $\min\_{U,D}$ denotes the minimum over all orthogonal matrices $U$ and all diagonal matrices $D$ with diagonal entries $d\_1,\dots,d\_n$ such that $d\_1\cdots d\_n=1$; (ii) the $e\_i$'s are the standard basis vectors; (iii) $\ep:=(\ep\_1,\dots,\ep\_n)\in\{-1,1\}^n$ and $D\_\ep$ is the diagonal matrix with diagonal entries $\ep\_1,\dots,\ep\_n$; and (iv) $\1:=\sum\_{i=1}^n e\_i$.
Now the crucial point: considering $\ep$ as a random point uniformly distributed on $\{-1,1\}^n$, we get the expected value of $E\|DUD\_\ep\1\|\_2^2$:
\begin{equation}
E\|DUD\_\ep\1\|\_2^2=\1^T ED\_\ep U^T D^2 U D\_\ep \1
=\tr U^T D^2 U=\tr D^2=\sum\_{i=1}^n d\_i^2\ge n,
\end{equation}
since $d\_1\cdots d\_n=1$. Hence,
\begin{equation}
\max\_{\ep\in\{-1,1\}^n}\,\|DUD\_\ep\1\|\ge
E\|DUD\_\ep\1\|\_2\ge\sqrt n.
\end{equation}
So,
\begin{equation}
\min\_{H\colon\,|\det H|=1}\max\_{\|x\|\_2\le1}\|Hx\|\_1\ge\sqrt n.
\end{equation}
On the other hand,
\begin{equation}
\min\_{H\colon\,|\det H|=1}\max\_{\|x\|\_2\le1}\|Hx\|\_1\le
\max\_{\|x\|\_2\le1}\|x\|\_1=\sqrt n.
\end{equation}
Thus,
\begin{equation}
\min\_{H\colon\,|\det H|=1}\max\_{\|x\|\_2\le1}\|Hx\|\_1=\sqrt n,
\end{equation}
as claimed.
|
6
|
https://mathoverflow.net/users/36721
|
409335
| 167,568 |
https://mathoverflow.net/questions/409167
|
8
|
Let $P$ be an irreducible Markov matrix, and $\pi$ its stationary distribution. Let $D$ be a perturbation matrix which is zero except for two entries in row $r$:
$$D\_{rg}=+1 \qquad D\_{r\ell}=-1.$$
Let $\widetilde{P}(\delta)=P+\delta D$ and let $\widetilde{\pi}(\delta)$ be its stationary distribution. The matrix $\widetilde{P}(\delta)$ is a perturbed version of $P$ where transitions from $r$ happen a little more to the gaining state $g$ and a little less to the losing state $\ell$.
In [1], Conlisk derived, based on identities in [2], the following remarkable formula for the effect on the stationary probability of the gaining state:
$$ \frac{d}{d\delta} \widetilde{\pi}\_g(\delta) = \pi\_r\frac{m\_{\ell g}}{m\_{gg}}.$$
Here $M$ is the mean first-passage time matrix, with $m\_{ij}$ defined to be the expected number of steps, when the chain starts at $i$, that it takes to first visit $j$ after that.
Conlisk's proof [1], and Schweitzer's calculations that underlie it, use the fundamental matrix of the chain. **It seems that such a simple formula should have a fairly simple, and probabilistic, proof.** Is there one out there?
[1] Conlisk, J., 1985. Comparative statics for Markov chains. Journal of Economic Dynamics and Control, 9(2), pp.139-151.
[2] Schweitzer, P.J., 1968. Perturbation theory and finite Markov chains. Journal of Applied Probability, 5(2), pp.401-413.
|
https://mathoverflow.net/users/7967
|
Probabilistic proof for derivative of invariant distribution of a Markov chain
|
Here's another, perhaps more probabilistic, approach. It's known that $\pi$ can be represented in terms of the mean occupation times as follows. Fix a state, for convenience $r$. Then, for any state $j$,
$$
\pi\_j=m\_{rr}^{-1} E^r\left[ \sum\_{n=0}^{T\_r-1}1\_{\{X\_n=j\}}\right],
$$
where $E^r$ denotes expectation for the chain started in $r$, and $T\_r$ is the first return time to $r$. (This dates back to work of Chung and Harris from the '50s; see the extension below at (\*).) The "redirection" you have in mind will only be (possibly) enacted in the first transition, so conditioning on that transition will allow you easily to develop a formula for $\tilde m\_{rr}(\delta)\cdot\tilde\pi\_j(\delta)$: For $j\not=r$,
$$
\tilde m\_{rr}\tilde\pi\_j
=
\tilde P\_{r\ell}\cdot E^\ell\left[ \sum\_{n=0}^{T\_r-1}1\_{\{X\_n=j\}}\right]
+\tilde P\_{rg}\cdot E^g\left[ \sum\_{n=0}^{T\_r-1}1\_{\{X\_n=j\}}\right]
+\sum\_{i\not=\ell,g,r} P\_{ri}\cdot E^i\left[ \sum\_{n=0}^{T\_r-1}1\_{\{X\_n=j\}}\right]
$$
and likewise
$$
m\_{rr}\pi\_j
=
P\_{r\ell}\cdot E^\ell\left[ \sum\_{n=0}^{T\_r-1}1\_{\{X\_n=j\}}\right]
+ P\_{rg}\cdot E^g\left[ \sum\_{n=0}^{T\_r-1}1\_{\{X\_n=j\}}\right]
+\sum\_{i\not=\ell,g,r} P\_{ri}\cdot E^i\left[ \sum\_{n=0}^{T\_r-1}1\_{\{X\_n=j\}}\right].
$$
(I have abbreviated $\tilde\pi\_j(\delta)$ to $\tilde \pi\_j$, etc.) Subtracting
$$
\tilde m\_{rr}\tilde\pi\_j-m\_{rr}\pi\_j = \delta\left\{ E^g\left[ \sum\_{n=0}^{T\_r-1}1\_{\{X\_n=j\}}\right]- E^\ell\left[ \sum\_{n=0}^{T\_r-1}1\_{\{X\_n=j\}}\right]\right\}.
$$
Similarly,
$$
\tilde m\_{rr} =m\_{rr}+\delta(m\_{gr}-m\_{\ell r}).
$$
The derivative of $\tilde\pi\_g(\delta)$ at $\delta=0$ therefore satisfies
$$
{\tilde\pi'\_g(0)\over \pi\_r}= E^g\left[ \sum\_{n=0}^{T\_r-1}1\_{\{X\_n=g\}}\right]- E^\ell\left[ \sum\_{n=0}^{T\_r-1}1\_{\{X\_n=g\}}\right]+\pi\_g\cdot(m\_{\ell r}-m\_{gr}).
$$
By the strong Markov property at time $T\_g$,
$$
E^g\left[ \sum\_{n=0}^{T\_r-1}1\_{\{X\_n=g\}}\right]- E^\ell\left[ \sum\_{n=0}^{T\_r-1}1\_{\{X\_n=g\}}\right]=E^g\left[ \sum\_{n=0}^{T\_r-1}1\_{\{X\_n=g\}}\right]\cdot P^\ell[T\_g>T\_r].
$$
It's known (see Lemma 7 in Section 2.2 of the unpublished book of Aldous and Fill on Markov Chains: <https://www.stat.berkeley.edu/users/aldous/RWG/book.html>) that
$$
E^g \left[\sum\_{n=0}^{T\_r-1}1\_{\{X\_n=g\}}\right]=\pi\_g\cdot(m\_{gr}+m\_{rg}).
$$
To verify Conlisk's identity we thus need to show that
$$
m\_{\ell g}=(m\_{rg}+m\_{gr})\cdot P^\ell[T\_r<T\_g]+m\_{\ell r}-m\_{gr}.
$$
This follows immediately from Corollary 10 in section 2.2 of Aldous and Fill.
The discussion there is probabilistic and based on the following observation (Proposition 3 of Section 2.1): If $S$ is a (possibly randomized) stopping time with $X\_S=i$ and $E^i[S]<\infty$, then
$$
E^i\left[\sum\_{n=0}^{S-1} 1\_{\{X\_n=j\}}\right]=\pi\_j\cdot E^i[S].\qquad\qquad(\*)
$$
It may be noted that A&F develop several formulas involving the fundamental matrix using this fact, and further probabilistic arguments.
|
1
|
https://mathoverflow.net/users/42851
|
409341
| 167,569 |
https://mathoverflow.net/questions/408726
|
3
|
(This question is related to [Splitting a space into positive and negative parts](https://mathoverflow.net/questions/7709/splitting-a-space-into-positive-and-negative-parts) but different.)
Given a finite-dimensional vector space $V$ over $\mathbb{R}$, what I call a "positive-negative splitting" for a symmetric bilinear form $\langle\cdot,\cdot\rangle$ on $V$ is a splitting $V=V\_+\oplus V\_-$ (not necessarily $\langle\cdot,\cdot\rangle$-orthogonal) such that the restrictions of $\langle\cdot,\cdot\rangle$ to $V\_+$ and $V\_-$ are positive definite and negative definite, respectively.
If there exists such a splitting, then $\langle\cdot,\cdot\rangle$ is written in matrix form as
$$
\langle x,y\rangle=
x^\mathsf{T}
\begin{pmatrix}
A\_+&B\\
B^\mathsf{T}&A\_-
\end{pmatrix}
y,
$$
where $A\_+$ and $A\_-$ are positive and negative definite symmetric matrices, respectively. We can calculate the determinant of the above block matrix by Gauss elimination and get
$$
\det
\begin{pmatrix}
A\_+&B\\
B^\mathsf{T}&A\_-
\end{pmatrix}=\det(A\_+)\det(A\_-\!-B^\mathsf{T}A\_+^{-1}B)\neq0
$$
(note that $A\_-\!-B^\mathsf{T}A\_+^{-1}B$ is negative definite). So $\langle\cdot,\cdot\rangle$ is non-degenerate in this case.
My question is:
**For a Hilbert space $(\mathcal{H},(\cdot|\cdot))$ and a bilinear form $\langle\cdot,\cdot\rangle:=(A\cdot|\cdot)$ on $\mathcal{H}$ given by a self-adjoint operator $A:\mathcal{H}\to\mathcal{H}$, does the existence of a positive-negative splitting still imply that $\langle\cdot,\cdot\rangle$ is non-degenerate?**
|
https://mathoverflow.net/users/17294
|
Infinite-dimensional analogue of "positive-negative splitting implies non-degeneracy"
|
A simple proof:
Assume $\mathcal{H}=\mathcal{H}\_+\oplus\mathcal{H}\_-$, with $( A u| u)>0$ (resp. $<0$) for all nonzero $u\in\mathcal{H}\_+$ (resp. $u\in\mathcal{H}\_-$). We want to show $\ker A=0$. To this end, assume $Au=0$ and write $u=u\_++u\_-$ with $u\_\pm\in\mathcal{H}\_\pm$. If either $u\_+$ or $u\_-$ is zero, the assumption easily implies $u=0$. Otherwise, we have
$$
0=(Au|u\_+)=(Au\_+|u\_+)+(Au\_-|u\_+)>(Au\_-|u\_+),
$$
$$
0=(Au|u\_-)=(Au\_+|u\_-)+(Au\_-|u\_-)<(Au\_+|u\_-).
$$
In view of the self-adjointness, we get a contradiction.
|
1
|
https://mathoverflow.net/users/17294
|
409353
| 167,570 |
https://mathoverflow.net/questions/409249
|
4
|
Let $\mathbb{K}$ be a field (not assumed to be algebraically closed, but we can assume characteristic 0 if necessary), and let $\mathfrak{g}$ be a semisimple Lie algebra.
A Lie subalgebra $\mathfrak{p} \leq \mathfrak{g}$ is said to be **parabolic** if
$$\mathfrak{p}^\perp = \operatorname{nil} \mathfrak{p},$$
where $\mathfrak{p}^\perp$ is the Killing perp, and $\operatorname{nil} \mathfrak{p}$ is the nilradical of $\mathfrak{p}$ (the unique maximal nilpotent ideal in $\mathfrak{p}$).
The nilradical gives us a short exact sequence
$$0 \to \operatorname{nil} \mathfrak{p} \to \mathfrak{p} \to \mathfrak{p}^{\text{red}} \to 0$$
Where $\mathfrak{p}^{\text{red}} = \mathfrak{p} / \operatorname{nil} \mathfrak{p} $ is reductive.
I have seen it mentioned in various places that this short exact admits a splitting on the right in the special case that $\mathfrak{p}$ is parabolic. Just naively, it seems like this should have to do with using the Killing form, since $\mathfrak{p}^\perp = \operatorname{nil} \mathfrak{p}$. But the Killing form is in general degenerate on $\mathfrak{p}$, so I'm not quite sure how to go about this.
Is there a way to construct this splitting (in a manner as choice-free as possible) without extending to an algebraically closed field $\overline{\mathbb{K}}$?
|
https://mathoverflow.net/users/56938
|
Reductive Levi decomposition of a parabolic subalgebra
|
There are many such splittings. Any complementary (aka opposite) parabolic subalgebra (i.e. $\mathfrak{q}$ such that $\mathfrak{p} \oplus \mathfrak{q}^\perp = \mathfrak{g}$ and so on) provides a unique splitting $\mathfrak{p} = \mathfrak{p} \cap \mathfrak{q} \oplus \mathfrak{p}^\perp$. The space of complementary parabolic subalgebras is a $\exp \mathfrak{p}^\perp$-torsor or in other words an affine space modelled on $\mathfrak{p}^\perp$.
We can also equate these to a particular adjoint orbit in $\mathfrak{p}$. There is a particular element $\xi^\mathfrak{p}\_\mathfrak{q}$ of $\mathfrak{p}$ called the grading element (aka canonical element) for which $ \mathfrak{q}^\perp \oplus \mathfrak{p} \cap \mathfrak{q} \oplus \mathfrak{p}^\perp$ are the $-1,0,1$ eigenspaces for $\mathrm{ad}\ \xi^\mathfrak{p}\_\mathfrak{q}$. This uniquely identifies the complementary subalgebra and the splitting.
Check out [this paper](https://arxiv.org/abs/1607.00370) on arxiv for an exhaustive treatment of this only assuming the basics of Lie algebras.
Also [this paper](https://scholarworks.umass.edu/cgi/viewcontent.cgi?article=2159&context=math_faculty_pubs), covers the filtrations and gradings we are seeing here quite succintly in the preliminary chapter.
Edit: an additional point is that the span of $\xi^\mathfrak{p}\_\mathfrak{q}$ is exactly the 1-dimensional centre of $\mathfrak{p} \cap \mathfrak{q}$ and its orthocomplement is semisimple.
|
3
|
https://mathoverflow.net/users/163024
|
409360
| 167,572 |
https://mathoverflow.net/questions/409146
|
1
|
I'm studying a paper and in the introduction appears the following:
It is well known that existence of critical points and solvability of Euler-Lagrange equations are related, and there is and extensive literature about critical points which are minimizers, specially for functionals defined on the Sobolev space $W\_{0}^{1,p}(\Omega),\; p>1,$ by
$$J(u)=\int\_{\Omega}\mathcal{F}(x,u,Du) dx,$$
where $\Omega$ is bounded, open subset of $\mathbb{R}^N.$
**DOUBT:** However, I'm struggling to find this extensive literature, and I also would like to find definition and properties of Euler-Lagrange equations.
Thanks in advance.
I appreciate any help.
|
https://mathoverflow.net/users/156922
|
Definition of Euler-Lagrange equation and properties, where can I find?
|
These [lecture notes](https://sites.pitt.edu/~hajlasz/Notatki/hel-97.pdf) by Piotr Hajłasz might have the introductory level you are looking for:
>
> The lectures will be divided into two almost independent streams. One
> of them is the theory of Sobolev spaces with numerous aspects which go
> far beyond the calculus of variations. The second stream is just
> calculus of variations. The theory of Sobolev spaces is a basic
> technical tool for the calculus of variations, however it suffices to
> know only basic results for most of the applications in that context.
>
>
>
|
0
|
https://mathoverflow.net/users/11260
|
409363
| 167,574 |
https://mathoverflow.net/questions/409367
|
3
|
$\newcommand{\dmod}{\text{-}\mathrm{mod}}$
Let $A$ be a finite-dimensional $k$-algebra, $A\dmod$ be a category of finite-dimensional A-modules and $\mathrm{U}\_A:A\dmod \to \textbf{Vect}\_k$ be a forgetful functor. We can reconstruct $A$ as $\mathrm{End}(\mathrm{U}\_A)$ by using Tannaka reconstruction thorem.
**Question** : Does the claim hold even if the assumption of "finite-dimensional" is excluded?
|
https://mathoverflow.net/users/49781
|
Is a smallness condition necessary in the Tannaka reconstruction theorem?
|
Yes, of course. You still have a natural homomorphism $A\rightarrow END(U\_A)$. Since ${}\_AA$ is a free $A$-module, an endomorphism $x\in END(U\_A)$ is determined by its value $x\_A$ on ${}\_AA$. This proves that the natural homomorphism is an isomorphism:
$$x\_A \in End (A\_{End\_{{}\_AA}})= End (A\_{{A}})=A.$$
|
3
|
https://mathoverflow.net/users/5301
|
409369
| 167,576 |
https://mathoverflow.net/questions/406880
|
3
|
Let $A$ be an ADE-hypersurface singularity in dimension one.
For example in Dynkin type $A\_n$, A is given by $K[[x,y]]/(x^2+y^{n+1})$.
Then $A$ is CM-finite and let $M$ be the direct sum of all indecomposable maximal CM-modules of $A$ and $B=\underline{End\_A}(M)$ the stable endomorphism ring of $M$.
>
> Question: Is there a description by quiver and relations of $B$ depending on the Dynkin type somwhere?
>
>
>
It should be a finite dimensional quiver algebra but I was not able to find a reference for the explicit description, while in dimension two it leads to the famous preprojective algebras. Im especially interested in the cases of $A\_n$ (there the algebra is a preprojective algebra when $n$ is odd?) and $E\_6$.
|
https://mathoverflow.net/users/61949
|
Quiver and relations for ADE singularities in dimension one
|
I found the answer in theorem 8.7 in the survey article on periodic algebras by Erdmann and Skowronski. They are certain (twisted) mesh algebras of Dynkin type.
|
0
|
https://mathoverflow.net/users/61949
|
409376
| 167,577 |
https://mathoverflow.net/questions/409327
|
3
|
In the process of computing [Shapley values](https://en.wikipedia.org/wiki/Shapley_value), I observed an interesting combinatorial constant. I am not exactly sure where such behavior comes. And here is the conjecture.
**Notations**
For any finite non-empty sequence of number $Q \subset \mathbb{R}$, I can construct a vector $C = (c\_0, c\_1, c\_2, \cdots, c\_{|Q|})$ using such a mapping $C(Q)$:
$$
\sum\_{i=0}^{|Q|} c\_i x^ i = \prod\_{q\_i \in Q} (1+q\_ix)
$$
Given a sequence $A = a\_1, a\_2, \cdots, a\_m $, denote $A\_{\neg i}$ as $A \setminus \{a\_i\}$. $|A| = m$ denotes size/length of $A$ is $m$.
With size $m \in \mathbb{N}$, denote a constant vector $N(m)$ as $(1/{m-1 \choose 0}, 1/{m-1 \choose 1}, \cdots, 1/{m-1 \choose m-1 } )$. And $\langle\cdot, \cdot \rangle$ is the inner product operation.
**Conjecture**
For any finite non-empty sequence $A \subset \mathbb{R}$ with $|A| = m$, i conjecture if $0 \in A$ then:
$$
\sum\_{i \in A} (1 - i) \langle C(A\_{\neg i}), N(m) \rangle = m
$$
I'm able to obtain consistent results from simulations, but I don't know how to prove it. Any help on a proof/dis-proof would be appreciated!
**Example**
For $A = 1, 2, 3$, the size $m$ is clearly 3.
$A\_{\neg 1} = 2, 3$, $A\_{\neg 2} = 1, 3$, $A\_{\neg 3} = 1, 2$
$C(A\_{\neg 1}) = (1, 5, 6)$, $C(A\_{\neg 2}) = (1, 4, 3)$, $C(A\_{\neg 3}) = (1, 3, 2)$.
$N(3) = (1, 0.5, 1)$
$$
(1-1)\* \langle (1, 5, 6), (1, 0.5, 1) \rangle \\
+ (1-2)\* \langle (1, 4, 3), (1, 0.5, 1) \rangle \\
+ (1-3)\* \langle (1, 3, 2), (1, 0.5, 1) \rangle \\
= -15
$$
which is not $m$.
On the other hand:
For $A=0, 2, 3$, the size $m$ is still 3.
$A\_{\neg 0} = 2, 3$, $A\_{\neg 2} = 0, 3$, $A\_{\neg 3} = 0, 2$
$C(A\_{\neg 0}) = (1, 5, 6)$, $C(A\_{\neg 2}) = (1, 3, 0)$, $C(A\_{\neg 3}) = (1, 2, 0)$.
$N(3) = (1, 0.5, 1)$
$$
(1-0)\* \langle (1, 5, 6), (1, 0.5, 1) \rangle \\
+ (1-2)\* \langle (1, 3, 0), (1, 0.5, 1) \rangle \\
+ (1-3)\* \langle (1, 2, 0), (1, 0.5, 1) \rangle \\
= 3
$$
which is $m$.
|
https://mathoverflow.net/users/136067
|
Conjecture on some combinatorial constant
|
We can rewrite $$\sum\_{i=0}^{|Q|} c\_i x^ i = \prod\_{q\_i \in Q} (1+q\_ix)$$ as $$c\_i = \sum\_{\substack{S \subseteq Q \\ |S| = i}} \prod\_{q \in S} q$$
Then $$\sum\_{a \in A} (1 - a) \langle C(A\_{\neg a}), N(m) \rangle$$ can be split into two sums: $$\left(\sum\_{a \in A} \langle C(A\_{\neg a}), N(m) \rangle \right) - \left(\sum\_{a \in A} a \langle C(A\_{\neg a}), N(m) \rangle \right)$$ Call the first sum $L$ (for lower) and the second sum $U$ (for upper). Note that $L$ only contains terms with products of $0$ to $m-1$ elements of $A$, and $U$ only contains terms with products of $1$ to $m$ elements of $A$.
Consider a subset $S \subset A$ of size $0 < k < m$. It occurs in $L$ once for every $a \in A \setminus S$ and each time with weight $\binom{m-1}{k}^{-1}$ from the dot product, for a total weight of $(m-k)\binom{m-1}{k}^{-1} = \frac{(m-k)!k!}{(m-1)!}$; it occurs in $U$ once for every $a \in S$ and each time with weight $\binom{m-1}{k-1}^{-1}$ from the dot product, for a total weight of $k\binom{m-1}{k-1}^{-1} = \frac{(m-k)!k!}{(m-1)!}$; and so all of these terms cancel.
Therefore the only terms which survive are the term from $L$ corresponding to the subset of size $0$ and the term from $U$ corresponding to the subset of size $m$. The former occurs with weight $\frac{(m-0)!0!}{(m-1)!} = m$; the latter with weight $\frac{(m-m)!m!}{(m-1)!} = m$ (but negated because of the subtraction).
Therefore the sum evaluates to $$m\left(1 - \prod\_{a \in A}a\right)$$
|
2
|
https://mathoverflow.net/users/46140
|
409382
| 167,579 |
https://mathoverflow.net/questions/409344
|
4
|
A Cauchy-Euler operator is an operator that leaves homogeneous polynomial of a certain degree invariant, named after the [Cauchy-Euler differential equations](https://en.wikipedia.org/wiki/Cauchy%E2%80%93Euler_equation)
We consider the operator
$$(Lf)(x) = \langle Ax,\nabla \rangle \langle \nabla,Ax \rangle f(x),$$
where $A \in \mathbf C^{d \times d}$ is a constant matrix and $\nabla$ the gradient.
I am asking what the spectrum of $L$ on the space of all homogeneous polynomial is?
Let me illustrate this for the case $d=1:$
Then $$Lf(x) = \vert a \vert^2 x \partial\_x^2( x f(x)) =\vert a \vert^2 (x^2 f''(x)+2x f'(x)). $$
If we then assume that $f(x)=x^m$ with $m \ge 0$ we see that
$$Lf(x) = \vert a \vert^2 (m(m-1)+2m) f(x) = \vert a \vert^2 m(m+1) f(x).$$
Thus, the spectrum is precisely $\vert a \vert^2 m(m+1)$ where $m \in \mathbb N\_0$ is arbitrary.
|
https://mathoverflow.net/users/119875
|
Spectrum Cauchy-Euler operator
|
Miscellaneous results.
* If $A$ is strictly upper triangular, then $x\cdot\nabla$ consists only is terms $x\_j\partial\_k$ with $j<k$. The action of $L$ over homogenous polynomials of degree $d$ is described, in the basis of monomials written in lexicographic order, by a strictly upper triangular matrix. hence the only eigenvalue is $\lambda=0$.
In general
\begin{eqnarray\*}
Lf & = & (Ax\cdot\nabla){\rm div}(fAx)=(Ax\cdot\nabla)(\nabla f\cdot Ax+f{\rm Tr}A) \\
& = & (Ax)^T\nabla^2f(Ax)+\nabla f\cdot(A^2+A{\rm Tr A})x.
\end{eqnarray\*}
* Degree one (linear forms). One has
$$L[v\cdot x]=v\cdot(A^2+A{\rm Tr A})x.$$
The eigenforms correspond to eigenvectors of $(A^2+A{\rm Tr A})^T$. The eigenvalues are the numbers $a(a+{\rm Tr}A)$ where $a\in\sigma(A)$.
* Degree two (quadratic forms). One has
$$L[\frac12x^TSx]=(Ax)^TS(Ax)+(Sx)\cdot(A^2+A{\rm Tr A})x.$$
The eigenpairs correspond to the symmetric matrices $S$ solutions of
$$2A^TSA+S(A^2+A{\rm Tr A})+(A^2+A{\rm Tr A})^TS=\lambda S,$$
where $\lambda$ is the eigenvalue.
If $a\in\sigma(A)$ and $A^Tv=av$, then $S=vv^T$ is a solution, with
$$\lambda=4a^2+2a{\rm Tr}A.$$
More generally, if $a,b\in\sigma(A)$, $A^Tv=av$ and $A^Tw=bw$, then $S=vw^T+wv^T$ is a solution, with
$$\lambda=(a+b)^2+(a+b){\rm Tr}A.$$
**Edit**. The examples above path the way to a general solution. Consider the action of $L$ over forms (homogeneous polynomials) of degree $d$. Then we obtain eigenforms as follows. Let $a\_1,\ldots,a\_d$ be a list of numbers, all of them being eigenvalues of $A$ (repetition is allowed). Let $v\_j$ be an eigenvector of $A^T$ associated with $a\_j$, that is $A^Tv\_j=a\_jv\_j$. Then
$$f(x):=\prod\_jv\_j\cdot x$$
satisfies
$$Lf=\lambda f,\qquad\lambda=\left(\sum\_ja\_j\right)^2+\left(\sum\_ja\_j\right){\rm Tr} A.$$
To see this, remark that
$$\nabla f=f\sum\_j\frac{v\_j}{v\_j\cdot x},\qquad\nabla^2 f=f\sum\_{j\ne k}\frac{v\_j}{v\_j\cdot x}\otimes\frac{v\_k}{v\_k\cdot x}.$$
If $A$ has simple eigenvalues, ordered in some way $\mu\_1,\cdots,\mu\_n$, then you obtain an eigenform for every $n$-uplet $(m\_1,\ldots,m\_n)$ such that $m\_1+\cdots+m\_n=d$, where you take $m\_1$ times $\mu\_1$ in the list $(a\_j)\_j$, etc... The number of $n$-uplets, summing up to $d$, being equal to the dimension of forms of degree $d$ in $n$ variables, this gives a diagonalisation of $L$. Actually, one can see the products as monomials in the system of coordinates $(v\_1\cdot x,\ldots,v\_n\cdot x)$.
By a continuity argument, the spectrum of $L$ over $d$-forms is always the set of numbers
$$\left(\sum\_ja\_j\right)^2+\left(\sum\_ja\_j\right){\rm Tr} A,$$
counted with multiplicities.
|
3
|
https://mathoverflow.net/users/8799
|
409385
| 167,581 |
https://mathoverflow.net/questions/409387
|
4
|
Let $\mathcal{C}\_{\mathrm{aut}}(G, F)$ be the category of automorphic representations of a connected reductive group $G$ over a number field $F$.
If this is a Tannakian category, it has an associated Tannakian fundamental group $G^{\mathrm{aut}}\_F$. What is it and how is it related to $G$ itself (for example, for $G = \operatorname{GL}\_n$)? In other words, what does Tannakian duality recover in this case?
EDIT: In light of Bugs Bunny's answer below, if we consider the group of endomorphisms of the forgetful functor from $\mathcal{C}\_{\mathrm{aut}}(G, F)$ (with completed tensor product) to the category of Hilbert spaces, how is the group related to $G$?
|
https://mathoverflow.net/users/469664
|
Tannakian fundamental group of automorphic representations
|
It is not a tannakian category. The issue is the tensor product. Let $V$ and $W$ be automorphic representations. The algebraic tensor product $V\otimes W$ is no longer automorphic. You need some kind of completion $V\widehat{\otimes}W$ but, according to David Loeffler, the completion is too big to be automorphic.
|
3
|
https://mathoverflow.net/users/5301
|
409397
| 167,585 |
https://mathoverflow.net/questions/409388
|
3
|
Suppose that $A\subseteq R^{m}$, let $B(R^{m})$ be the set of all Borel subsets of $R^{m}$. We say that $B\subseteq R^{m}$ is a **Borel envelope** of $A$ if $B\in B(R^{m})$ and for every $L^{m}$ measurable set (here $L^{m}$ means Lebesgue outer measure) $F$ in $R^{m}$ one has $L^{m}(A\cap F)=L^{m}(B\cap F)$. Then I want to show the following: Let $B\in B(R^{m})$ and $A\subseteqq B$. Prove that $B$ is a Borel envelope of A iff $L^{m}(S)=0$ whenever $S$ is an $L^{m}$-measurable subset of $B \setminus A$. I want to use the fact that there is a Borel set $A'$ such that $L^{m}(A)=L^{m}(A')$ but I don't know what to do next.
|
https://mathoverflow.net/users/206621
|
$B$ is a Borel envelope of $A$ iff any measurable subset of $B\setminus A$ has Lebesgue measure 0
|
$\newcommand{\de}{\delta}\newcommand\R{\mathbb R}\newcommand{\Z}{\mathbb{Z}}$To simplify the writing, let us first work locally, say with subsets of $(0,1]^m$, instead of $\R^m$, so that to avoid infinite values of $L^m$. (Going back to $\R^m$ is then straightforward: using [Carathéodory's criterion](https://en.wikipedia.org/wiki/Carath%C3%A9odory%27s_criterion), for any set $A\subseteq\R^m$ we "collect" its "finite pieces" by the formula $L^m(A)=\sum\_{z\in\Z^m}L^m(A\cap(z+(0,1]^m))$.)
Then it is easy to show (see Lemma 1 at the end of this answer) that the definition of a Borel envelope simplifies as follows: a Borel set $B$ is a Borel envelope of $A$ if $B\supseteq A$ and $L^{m}(A)=L^{m}(B)$.
We want to show the following: Suppose that $B$ is Borel and $A\subseteq B$; then $B$ is a Borel envelope of $A$ iff $L^{m}(S)=0$ whenever $S$ is an $L^{m}$-measurable subset of $B\setminus A$.
**The only-if part:** Suppose that $B$ is a Borel envelope of $A$. To obtain a contradiction, suppose that $L^{m}(S)>0$ for some $L^{m}$-measurable $S\subseteq B\setminus A$. Then $S\subseteq B$ and for the $L^{m}$-measurable set $S\_1:=B\setminus S$ we have $A\subseteq S\_1$ and hence $L^{m}(A)\le L^{m}(S\_1)=L^{m}(B)-L^{m}(S)<L^{m}(B)$, which contradicts the condition $L^{m}(A)=L^{m}(B)$, as desired.
**The if part:** Suppose that $L^{m}(S)=0$ whenever $S$ is an $L^{m}$-measurable subset of $B\setminus A$. To obtain a contradiction, suppose that $B$ is not a Borel envelope of $A$ -- that is, $L^{m}(A)<L^{m}(B)$.
Note that $L^{m}(A)=\inf\{L^{m}(C)\colon C\supseteq A, C\text{ Borel}\}$. So,
$L^{m}(B)>L^{m}(C)$ for some Borel set $C$ such that $C\supseteq A$, whence $L^{m}(B)> L^{m}(C\cap B)$ and $L^{m}(B\setminus C)=L^{m}(B)-L^{m}(C\cap B)>0$, whereas the set $S:=B\setminus C$ is $L^m$-measurable and contained in $B\setminus A$ -- a desired contradiction.
To complete the proof, we only to need to state and prove
>
> **Lemma 1:** A Borel set $B\subseteq(0,1]^m$ is a Borel envelope of a set $A\subseteq(0,1]^m$ iff $B\supseteq A$ and $L^{m}(A)=L^{m}(B)$.
>
>
>
*Proof of Lemma 1:* The only-if part of Lemma 1 is trivial: just take, for instance, $F=(0,1]^m$ or $F=\R^m$.
Let us now consider the if part of Lemma 1. Here we suppose that $A\subseteq B\subseteq(0,1]^m$, $B$ is Borel, and $L^{m}(A)=L^{m}(B)$. To obtain a contradiction, suppose that $B$ is not a Borel envelope of $A$. Then $L^{m}(A\cap F)\ne L^{m}(B\cap F)$ and hence $L^{m}(A\cap F)<L^{m}(B\cap F)$ for some $L^{m}$-measurable set $F\subseteq\R^m$. Also, obviously, $L^{m}(A\cap F^c)\le L^{m}(B\cap F^c)$, where $F^c:=\R^m\setminus F$. Adding the latter two inequalities and using [Carathéodory's criterion](https://en.wikipedia.org/wiki/Carath%C3%A9odory%27s_criterion) again (together with the upper bound $L^{m}(A)\le L^{m}((0,1]^m)\le1<\infty$), we get
$L^{m}(A)=L^{m}(A\cap F)+L^{m}(A\cap F^c)<L^{m}(B\cap F)+L^{m}(B\cap F^c)
=L^{m}(B)$, which does contradict the condition $L^{m}(A)=L^{m}(B)$.
This completes the proof of Lemma 1 and thus the entire proof of the desired characterization of the Borel envelope.
|
2
|
https://mathoverflow.net/users/36721
|
409401
| 167,586 |
https://mathoverflow.net/questions/409375
|
10
|
If we add the following axiom schema to ZF-Reg., would the resulting theory prove $\sf AC$?
**Definable sets Choice:** if $\phi$ is a formula in which only the symbol $``y"$ occurs free, then:
$$\forall X (X=\{y \mid \phi\} \to \\\exists f (f:X \setminus \{\emptyset\} \to \bigcup X \land \forall x (f(x) \in x)))$$
If not, then which form of choice this is equivalent to, over axioms of ZF-Reg.?
The same question but over axioms of ZF was [answered](https://math.stackexchange.com/a/4314918/489784) to the positive, and since ZF-Reg. is consistent with a [principle](https://math.stackexchange.com/questions/4315049/is-it-consistent-with-sf-zf-fnd-that-no-class-whose-union-is-the-whole-univer?answertab=active#tab-top) stating that every class whose union is the universe won't be well-orderable, then this would kill the *least* fashioned argument used above.
It was [noted](https://math.stackexchange.com/questions/4314897/is-choice-over-definable-sets-equivalent-to-ac?answertab=active#comment8991608_4314918) that this matter is complicated, and that the answer might be to the negative, hence this question!
|
https://mathoverflow.net/users/95347
|
Is choice over definable sets equivalent to AC over axioms of ZF-Reg.?
|
Yes, indeed this kind of choice in general doesn't imply $\mathsf{AC}$ over $\mathsf{ZF}-\mathsf{Reg}$.
I will reason in $\mathsf{ZFC}$ and construct an interpretation of $\mathsf{ZF}-\mathsf{Reg}$, where $\mathsf{AC}$ fails, but choice for definable sets holds. The idea is to define a modified permutation model $M$ of $\mathsf{ZFU}+\lnot\mathsf{AC}$, where internally the collection of all urelements is a proper class and from external perspective any two urelements could be swapped by an automorphism. Definable choice holds in $M$, since due to the presence of automorphisms and the fact that urelements form a proper class, the only definable sets are sets whose transitive closure contains no urelements. Finally we convert $M$ to the desired model $M'$ of $(\mathsf{ZF}-\mathsf{Reg})$ by transforming each urelement $a$ into the set $\{a\}$.
Now let me outline the construction of $M$ in more details. We fix a countable set of urelements $U$ together with a function $r\colon U\to \mathbb{Q}$ such that $|r^{-1}(q)|=\aleph\_0$, for all $q\in \mathbb{Q}$. We consider a group $G$ of permutations of $U$ consisting of all permutations $\pi\colon U\to U$ such that $r\circ \pi=r$. Let filter $\mathcal{F}$ on the lattice of subgroups of $G$ be generated by all the subgroups $\mathsf{Fix}\_S=\{\pi\in G\mid \forall a\in S(\pi(a)=a)\}$, where $S$ is a finite subset of $U$.
We naturally define the universe $V\_U$ of sets with urelements from $U$. Naturally we define the support function $\mathsf{supp}\colon V\_U\to \mathcal{P}(U)$ and the action of $G$ on $V\_U$. We say that $x\in V\_U$ is symmetric if $\{\pi \in G\mid \pi x= x\}\in \mathcal{F}$. And we say that $x\in V\_U$ is hereditarily symmetric if it is symmetric and all elements of the transitive closure $\mathsf{TC}(x)$ are symmetric. We say that $x\in V\_U$ is bounded if the set of rationals $r[\mathsf{supp}(x\cup\mathsf{TC}(x))]$ is bounded from above. The universe $M$ is a subuniverse of $V\_U$ consisting of all $x\in V\_U$ that are hereditarily symmetric and bounded.
Since $U$ itself isn't bounded, the urelements form a proper class in $M$. The proof that $M$ satisfies $\mathsf{ZFU}$ is essentially the standard proof that permutation models satisfy $\mathsf{ZFU}$. The proof that $M$ doesn't satisfy $\mathsf{AC}$ is also essentially standard: there are no well ordering of the set $r^{-1}(0)$, since otherwise we would be able to split $r^{-1}(0)$ into two disjoint infinite subsets and they wouldn't be symmetric. For any two $a,b\in A$ we easily find a permutaion $\sigma$ of $A$ such that $\sigma(a)=b$, $\sigma(b)=a$, $\forall c,d\in A(r(c)=r(d)\iff r(\sigma(c))=r(\sigma(d)))$, and for any $S\subseteq A$ the set $r[S]$ is bounded from above iff $r[\sigma[S]]$ is bounded from above. Obviously, such a permutation naturally extends to an automorphism of $M$ swapping $a$ with $b$.
Note that in fact I choose $M$ in such a manner that it satisfies even collection (and not merely replacement). Namely, consider the models $M\_q\subseteq M$, for $q\in\mathbb{Q}$ that consist of all $x\in M$ such that $r[\mathsf{supp}(x)]$ is bounded from above by some $q'<q$. Clearly each $M\_q$ is isomorphic to $M$ but furthermore $M\_q$ is an elementary submodel to $M$, since we have isomorphisms fixing any given finite family of elements of $M\_q$. Thus we always could find the result for an instance of collection as follows. We fix $q\in \mathbb{Q}$ such that all parameters used in this instance are from $M\_q$. Then we find least $\alpha$ such that all desired witnesses could be found in $\alpha$-th level $V\_U(\alpha)$ of $V\_U$. And finally we observe that due to elementary submodel of $M$ and $M\_q$ all the necessary witnesses could be found already in $V\_U(\alpha)\cap M\_q$, which in fact in fact is a bounded and hereditarily symmetric set.
|
11
|
https://mathoverflow.net/users/36385
|
409405
| 167,587 |
https://mathoverflow.net/questions/409406
|
7
|
Let $k=\mathbb{F}\_q$ be a finite field, and let $X$ be a smooth projective variety over $k$. Suppose that $X\_{\overline{k}}$ is birational to $\mathbb{P}^n\_{\overline{k}}$, do we know
(1)If $X$ is necessarily birational to $\mathbb{P}^n\_k$?
(2)If $X$ necessarily has a $k$-point?
|
https://mathoverflow.net/users/nan
|
Geometrically rational variety over a finite field
|
(1) **No**: There exist minimal cubic surfaces over finite fields (see for example <https://arxiv.org/abs/1611.02475>). Such surfaces are non-rational over the ground field.
(2) **Yes**: This is a special case of a more general result of Esnault: <https://arxiv.org/abs/math/0207022>
This proves the congruence $\#X(\mathbb{F}\_q) \equiv 1 \bmod q$, which clearly implies the existence of a rational point.
|
6
|
https://mathoverflow.net/users/5101
|
409410
| 167,588 |
https://mathoverflow.net/questions/409407
|
12
|
From my understanding, mathematics sometimes gives rise to new physical/tangible laws and the converse is also true. In particular, physical phenomena give rise to new mathematics.
In all of the cases that I have seen, the mathematics is usually formalized. That is, definitions, lemmas, theorems and their proofs are developed in either of the two cases mentioned above.
Are there ever cases where formally defining physical phenomena in mathematical language is unnecessary?
For example, [deep learning has recently been formalized](https://arxiv.org/abs/2106.10165). Is this formalism necessary for developing new techniques in the field? I can see how it would make sense if the formal theory was developed first.
|
https://mathoverflow.net/users/469551
|
Is it ever unnecessary to mathematically formalize a concept?
|
Your question is
>
> Are there ever cases where formally defining physical phenomena in mathematical language is unnecessary?
>
>
>
It is **never possible** to define physical phenomena **directly** in mathematical language. Mathematics (and even sciences) can deal with real phenomena only through models.
To deal with a model mathematically, it must be a mathematical model, described in mathematical terms.
I think all mathematical models go back to real needs -- even if very indirectly, through many layers of abstraction and generalization.
So, at least in applications of mathematics, an important goal is to faithfully and efficiently model real phenomena of interest, so as to be able to make accurate enough predictions about the real phenomena.
The clarity and effectiveness of mathematical presentation are of course very important. Therefore, formalization can hardly ever hurt, unless it is taken to such an extent as to actually hamper readers' understanding. The power of formalization is to code very rich, deep, and complex mathematical content into very compact, unambiguously defined, and easy to grasp at once (after some practice) mathematical symbols and formulas.
It is usually helpful if a formal presentation is complemented by appropriate illustrations and imagery. In particular, if a new term is introduced, it helps to give it a descriptive name, easy to remember.
However, if the presentation is very informal and mainly consists of many images each with multiple possible interpretations, then it is easy for the uninitiated reader to quickly get lost (as happened with me many times).
---
To me, the presentation of deep learning theory [linked in your post](https://arxiv.org/abs/2106.10165) does not look mathematical -- see e.g. formulas (2.30) and (2.32) on p. 51 there.
|
15
|
https://mathoverflow.net/users/36721
|
409413
| 167,590 |
https://mathoverflow.net/questions/409379
|
7
|
This question is an extension of my previous question last year (see [2020]) in which I asked about the (consensus of a) definition of a weak $n$-category.
Here are some background: while strict $n$-categories are easily defined, they are not sufficient for $n>2$. Therefore weak $n$-categories need to be defined. What a definition of a weak $n$-category should satisfy was proposed in [BD1995]. However, many proposals have since been given (see [Lei2001] or [2020]). And as David White pointed out in [2020], we had not reached to a consensus yet.
This question focuses on a smaller part of the problem.
**Question:** In order to prove that different models of $n$-categories to be equivalent, there must be a well-defined notion of a $(n+1)$-category to start with. So how is it possible to really prove the equivalence?
I guess this relates to a philosophical problem that in order to justify (anything) one needs to justify the setting in which we justify. Before entering the realm of formalized arguments, we can postulate some desired results (as in [BD1995], or called "specification" in compsci's term), but nothing can stop people from building different settings (or called "implementation" in compsci's term). How could such problem be resolute without brute-force [translating](https://math.stackexchange.com/questions/3869884/translating-developments-over-different-foundations) results from different foundations?
* [2020]: [Definition of a $n$-category](https://mathoverflow.net/questions/374783/definition-of-an-n-category/374795)
* [BD1995]: Higher-dimensional Algebra and Topological Quantum Field Theory-[John C. Baez and James Dolan]-[arXiv:q-alg--9503002]
* [Lei2001]: [A Survey of Definitions of n-Category](http://www.tac.mta.ca/tac/volumes/10/1/10-01abs.html)-[Tom Leinster]-[Theory and Applications of Categories, Vol. 10, 2002, No. 1, pp 1-70]
|
https://mathoverflow.net/users/124549
|
Equivalences of $n$-categories
|
As Marc Hoyois indicates in the comments, historically this *was* a major obstruction, past tense "was". My feeling is that these days, there is a nice perspective that whatever weak $n$-categories are, they are the objects of some $(\infty,1)$-category $n\mathrm{Cat}$. (Of course, weak $n$-categories are the objects of more than an $(\infty,1)$-category. But the extra $n$ dimensions in $n\mathrm{Cat}$ should be recoverable from looking at exponential objects.) It was a substantial feat to develop a theory of $(\infty,1)$-categories, but it has been more or less done.
Moreover, surely $n\mathrm{Cat}$ will be not just some weird $(\infty,1)$-category, but in fact a *presentable* $(\infty,1)$-category, and these can be "presented" by model categories. So the question is "just" one of finding the correct Quillen equivalence class of model categories, where "correct" means that it should match your intuition about weak $n$-categories.
Note that I am not saying that there is complete consensus about which presentable $(\infty,1)$-category deserves the name $n\mathrm{Cat}$. I certainly have opinions on the matter (namely: set $0\mathrm{Cat}:= \mathrm{Set}$, and define inductively $n\mathrm{Cat}$ to be the $(\infty,1)$-category of $(\infty,1)$-categories enriched in $(n{-}1)\mathrm{Cat}$), but I don't have the sociological data to conclude that my opinions are shared by the majority, let alone that there exists a consensus.
I should also emphasize that, although I do believe there to be a single correct answer to the question of which presentable $(\infty,1)$-category should be called "$n\mathrm{Cat}$", this answer doesn't satisfy, or at least doesn't obviously satisfy, all natural desiderata. Most notably, it is not very algebraic. I mean, it isn't very *non*-algebraic — it is about as algebraic as is the notion of "Kan simplicial set" — but it isn't very algebraic either. However, I counter that the search for a highly algebraic theory of $n$-categories is most likely a fool's errand. Any such theory will in particular include a highly algebraic description of homotopy $n$-types. Postnikov and Whitehead provide a "lowly algebraic" description of homotopy $n$-types, and I am pessimistic about there being anything better.
|
7
|
https://mathoverflow.net/users/78
|
409420
| 167,591 |
https://mathoverflow.net/questions/409421
|
40
|
That is, is there an open cover of $\mathbb{R}P^n$ by $n$ sets homeomorphic to $\mathbb{R}^n$?
I came up with this question a few years ago and I´ve thought about it from time to time, but I haven´t been able to solve it. I suspect the answer is negative but I´m not very sure. Also, is there an area of topology which studies questions like this one?
|
https://mathoverflow.net/users/172802
|
Can the nth projective space be covered by n charts?
|
Expanding on the comment by @user127776, the key reference is Palais, "Lusternik-Schnirelman Theory on Banach Manifolds", Topology 5 (1966),
where it is proved that if $X$ can be covered by $n$ contractible closed sets, then the cup-length of $X$ is strictly less than $n$.
(Here the cup-length is the largest $n$ such that for some field $F$ and some elements $c\_1,\ldots,c\_n$ in $H^\*(X,F)$, we have $c\_1\cup\ldots\cup c\_n\neq 0$.)
This rules out covering ${\mathbb RP}^n$ with $n$ closed contractible sets, which should suffice here (after slightly shrinking the given $n$ copies of ${\mathbb R}^n$).
**Editing to add:**
More generally, suppose $X$ is a compact Hausdorff space covered by $n$ closed sets $X\_1,\ldots, X\_n$ with all $H^1(X\_i,{\mathbb Z}/2{\mathbb Z})=0 $. (Equivalently, any (real) line bundle on $X\_i$ is trivial.)
**Theorem.** Any line bundle on $X$ can be generated by $n$ sections.
**Proof.** Let $\hat{X}= Spec(C(X,{\mathbb R}))$, so that $X$ imbeds in $\hat{X}$. Note that:
1. Because $X$ is normal, each $X\_i$ is defined by the vanishing of a continuous function, so the $\hat{X}\_i$ form a closed covering of $\hat{X}$.
2. By Swan's theorem, the map that takes a vector bundle over $\hat{X}$ to its pullback over $X$ is an equivalence of categories (and likewise with $X$ replaced by $X\_i$).
Now because every line bundle on $X\_i$ is trivial, so is every line bundle on $\hat{X}\_i$.
Because $\hat{X}$ is an affine scheme, a line bundle corresponds to a projective module, which in turn is the image of an idempotent matrix with entries in $C(X,{\mathbb R})$. A little thought reveals that this matrix can be taken to be $n\times n$. It follows that any line bundle on $\hat{X}$ is generated by $n$ sections. Therefore (by the Swan correspondence) so is any line bundle on $X$, as advertised.
**Corollary.** For any $c\in H^1(X,{\mathbb Z}/2{\mathbb Z})$, the $n$-fold cup product $c^n\in H^n(X,{\mathbb Z}/2{\mathbb Z})$ is zero.
**Proof.** $c$ is the first Stiefel-Whitney class of some line bundle $\xi$. Let $\phi\_\xi:X\rightarrow {\mathbb RP}^\infty$ be the classifying map of $\xi$. The $n$ sections guaranteed by the theorem provide a factorization of $\phi\_\xi$ through ${\mathbb RP}^{n-1}$. But $H^n({\mathbb RP}^{n-1},{\mathbb Z}/2{\mathbb Z})=0$.
|
37
|
https://mathoverflow.net/users/10503
|
409422
| 167,592 |
https://mathoverflow.net/questions/409424
|
4
|
Let $f: \mathbb R^n \to \mathbb R$ be a Lipschitz continuous function with *strict* Lipschitz constant $L > 0$.
That is, $|f(x) - f(y)| < L|x - y|$ for all $x \neq y$ in $\mathbb R^d$.
**Question:** What is the maximal Hausdorff dimension of the set on which $f$ is differentiable and $|Df| = L$?
**Remarks:**
1. $\mathcal H^{n-1}$ is trivially achievable.
2. By sticking together a maximising sequence, we may achieve the supremal Hausdorff dimension, so we are justified in speaking of the maximum.
|
https://mathoverflow.net/users/173490
|
On the set on which $|Df|$ is maximal for Lipschitz $f$
|
The maximal dimension is $n$, and it can be even of positive Lebesgue measure.
For $n=1$, consider a fat cantor set $K$. Then the primitive
$$
f(x):=\int\_0^x \chi\_K(t)dt
$$
is a maximizer. Indeed, since $K$ is nowhere dense, we get $|f(x)-f(y)|< |x-y|$. On the other hand, by the fundamental theorem of calculus (i.e., some form of Lebesgue differentiation theorem) we obtain that $f$ is differentiable with derivative 1 at all density points of $K$, which form a positive measure set.
In general dimension it is sufficient to consider the same function of one coordinate, say $f(x\_1)$.
|
5
|
https://mathoverflow.net/users/140505
|
409432
| 167,594 |
https://mathoverflow.net/questions/409431
|
3
|
Let $X$ be a random variable following a $\mathrm{Binomial}(n,p)$ distribution, and let $$Y=\min\{X,n-X\}.$$ Ispired by the problem posed by C. Clement on <https://math.stackexchange.com/questions/1696256/expectation-and-concentration-for-minx-n-x-when-x-is-a-binomial>, I want to ask whether there exists some constant $c>0$ such that $\mathbb{E}(Y)\geq c\cdot\min\{p,1-p\}\cdot n$ for all $0<p<1$. If this is not true, can we find some $p\_0$ with $\frac{1+\sqrt{5}}{4}\leq p\_0<1$ such that if $X\sim \mathrm{Binomial}(n,p\_0)$ then $\mathbb{E}(Y)\geq (1-p)[(1+4p)n-8(1+p)]$?
|
https://mathoverflow.net/users/75264
|
A lower bound for the expectation of $\min\{X,n-X\}$ when $X$ follows a $\mathrm{Binomial}(n,p)$ distribution
|
Since $\min(a,b)=(a+b-|a-b|)/2$, your question is really about upper bounding $E|2X-n|$, or, equivalently, $E|X-n/2|$:
$$E\min(X,n-X)=n/2-2E|X-n/2|.$$
You can upper bound $E|X-n/2|$ using Jensen's inequality:
$E|X-n/2|\le\sqrt{E(X-n/2)^2}$.
The latter, if I'm not mistaken, evaluates to
$$ n\sqrt{
p(1-p)/n+p^2-p+1/4
}
=:nF(p).$$
For $n$ sufficiently large and $p$ sufficiently small (certainly, $p\le 0.65$; the exact value can be easily computed), we have
$2F(p)\le c(1-p)$
for some universal $c>0$.
That means
that
$n/2-2F(p)n\ge cnp$, so your conjecture holds for this range of $p$. [I'm confident that with a bit more care you can extend the result to all $p$, perhaps with a worse constant. Update: this "confidence" has proven misplaced, see Dmitry Krachun's answer below!]
|
3
|
https://mathoverflow.net/users/12518
|
409434
| 167,595 |
https://mathoverflow.net/questions/408964
|
7
|
*Note: This is a generalisation of [an earlier problem](https://mathoverflow.net/questions/397726/on-equibounded-sequences-in-l-infty) as suggested by user Jochen Glueck in the comments.*
Let $1 \leq p < q \leq \infty$, and $f\_n: [0, 1] \to \mathbb R$ be a sequence of functions in the closed unit ball of $L^q$.
**Question:** Is it true that there exists a constant $C < 2$, depending only on $p$ and $q$ such that
$$\inf\_{n\_k} \sup\_{i,j \in\mathbb N} \|f\_{n\_i} - f\_{n\_j}\|\_{L^p} \leq C?$$
Where the infimum is taken over all increasing sequences of natural numbers $n\_k$.
If so, what is the sharpest such constant for each $p, q$?
*Remark: In the linked problem, the sharp constant $C = 1$ for $p = 1, q = \infty$ is obtained in the answer by Yuval Peres.*
|
https://mathoverflow.net/users/173490
|
$L^p$ bounds on tails of bounded $L^q$ sequences
|
Let me prove that such constant $C$ always exists.
It is not hard to find such $\alpha$, $\beta$ that the inequality $$x^p\leqslant \alpha x^q+\beta$$
holds for all positive $x$ and turns into equality if and only if $x=2$. Then $$|f-g|^p+|g-h|^p+|f-h|^p\leqslant \alpha (|f-g|^q+|g-h|^q+|f-h|^q)+3\beta.$$
Since all three differences $|f-g|$, $|g-h|$, $|f-h|$ can not be equal to $\pm 2$, we actually have
$$|f-g|^p+|g-h|^p+|f-h|^p\leqslant \alpha (|f-g|^q+|g-h|^q+|f-h|^q)+3\tilde{\beta}, \quad \text{with some}\,\,\tilde{\beta}<\beta$$
Then integrating against $[0,1]$ we get $$\|f-g\|\_p^p+\|g-h\|\_p^p+\|f-h\|\_p^p\leqslant \alpha (\|f-g\|\_q^q+\|g-h\|\_q^q+\|f-h\|\_q^q)+3\tilde{\beta}.$$
If all $f,g,h$ are in the closed unit ball in $L^q$, this yields
$$\|f-g\|\_p^p+\|g-h\|\_p^p+\|f-h\|\_p^p\leqslant 3(2^q\alpha+\tilde{\beta}).$$
Since $2^q\alpha+\beta=2^p$, we see that at least one of expressions $\|f-g\|\_p, \|g-h\|\_p, \|f-h\|\_p$ is at most $(2^q\alpha+\tilde{\beta})^{1/p}=:C<2$.
Now join $n$ and $m$ by a red edge if $\|f\_n-f\_m\|\leqslant C$ and by a blue edge otherwise. By infinite Ramsey theorem, there exists either a blue triangle or an infinite red clique. The second case is impossible, the first case is what we need.
|
6
|
https://mathoverflow.net/users/4312
|
409437
| 167,597 |
https://mathoverflow.net/questions/195060
|
9
|
$\DeclareMathOperator\Spin{Spin}\DeclareMathOperator\Sp{Sp}\DeclareMathOperator\SL{SL}$The exceptional isomorphism $\Spin(5,1)\simeq \SL(2,\mathbb{H})$ is well-known, and I can find references that say the maximal compact of $\Spin(5,1)$ is $\Spin(5) \simeq \Sp(2)$. So I know the answer to the question, but not the how or why. In particular, is there a proof that $\Sp(2)$ is maximal compact in $\SL(2,\mathbb{H})$ not via the exceptional isomorphisms? Perhaps some sort of analogue of Gram-Schmidt or other explicit factorisation?
|
https://mathoverflow.net/users/4177
|
Maximal compact subgroup of $\mathrm{SL}(2,\mathbb{H})$
|
$\DeclareMathOperator\GL{GL}\DeclareMathOperator\SL{SL}\DeclareMathOperator\SO{SO}\DeclareMathOperator\Sp{Sp}$YCor's [comment](https://mathoverflow.net/questions/195060/maximal-compact-subgroup-of-mathrmsl2-mathbbh#comment485760_195060) contains the essential idea needed for the proof, but maybe a few more details would be helpful. (If YCor does provide something similar later, feel free to award YCor's answer the bounty.)
Consider the mapping $\sigma:\GL(2,\mathbb{H})\to M\_2(\mathbb{H})$ given by
$$
\sigma(A) = A^\* A
$$
where $A^\*$ is the conjugate transpose of $A$ in $M\_2(\mathbb{H})$, the $2$-by-$2$ matrices with entries in $\mathbb{H}$. Then $\sigma(A)=\sigma(A)^\*$ (using the fact that $\overline{pq}= \overline{q}\,\overline{p}$ for $p,q\in\mathbb{H}$). Consequently, the image of $\sigma$ lies in the $6$-dimensional real subspace $S\_2(\mathbb{H})$, consisting of the matrices $s\in M\_2(\mathbb{H})$ that satisfy $s = s^\*$. Due to the associativity of multiplication in $\mathbb{H}$ and the above-mentioned conjugation identity, we have
$$
(AB)^\*sAB = B^\*(A^\*sA)B,
$$
so it follows that the mapping $\rho(s,A)=A^\*sA$ defines a (right) representation of $\GL\_2(\mathbb{H})$ on $S\_2(\mathbb{H})$.
Now, define the quadratic form $Q:S\_2(\mathbb{H})\to\mathbb{R}$ by
$$
Q\left(\begin{pmatrix}a&x\\\overline{x}&b\end{pmatrix}\right) = ab-x\overline{x}.
$$
when $a,b\in\mathbb{R}$ and $x\in\mathbb{H}$. Note that $Q$ has signature type $(1,5)$ as a real quadratic form.
The crucial identity (which can be proved by hand just by writing it out) is that
$$
Q(A^\*sA) = Q(s)\,Q(A^\*A).
$$
It follows that $Q(A^\*A) = 1$ defines $\SL(2,\mathbb{H})$ as a codimension $1$ closed subgroup of $\GL(2,\mathbb{H})$. In particular, the representation $\rho$ sends $\SL(2,\mathbb{H})$ into $\SO(Q)\simeq\SO(1,5)$, and it is easy to show that the kernel of this homomorphism is $\{\pm I\_2\}\subset\SL(2,\mathbb{H})$.
By definition, the stabilizer of $I\_2$ under the right representation $\rho$ is the 10-dimensional Lie group usually denoted in differential geometry by $\Sp(2)$, and its orbit under this right action by $\SL(2,\mathbb{H})$ must thus have dimension $5$ and hence be the nappe of the hyperboloid $Q(s)=1$ consisting of those $s$ with positive trace, i.e., hyperbolic $5$-space.
The connectedness of $\Sp(2)$ follows from the well-known fibration $\Sp(1)\to \Sp(2)\to S^7$, so it follows that $\Sp(2)$ is a nontrivial double cover of the identity component of $\SO(Q)$. Since $\SL(2,\mathbb{H})$ is simple and not compact, the signature type of its Killing form cannot be $(0,15)$, and since its Lie algebra splits as a module over ${\mathfrak{sp}}(2)$ into two irreducible pieces of dimension $10$ and $5$ corresponding to ${\mathfrak{sp}}(2)$ (on which it is negative definite) and its orthogonal complement, the type must be $(5,10)$. Thus, a maximal compact in $\SL(2,\mathbb{H})$ must have dimension $10$, so $\Sp(2)$ must be a maximal compact.
Finally, if you want a ‘factorization’ (analogous to the QR decomposition in the real case), you can show that every $A\in\SL(2,\mathbb{H})$ can be factored uniquely in the form
$$
A = Q\begin{pmatrix} a & 0\\ 0& a^{-1}\end{pmatrix}\begin{pmatrix} 1 & x\\ 0& 1\end{pmatrix},
$$
with $Q\in\Sp(2)$, $a\in\mathbb{R}^+$, and $x\in\mathbb{H}$, which also shows that $\SL(2,\mathbb{H})$ is diffeomorphic to $\Sp(2)\times\mathbb{R}^5$. (This is just the KAN decomposition for $\SL(2,\mathbb{H})$.)
|
11
|
https://mathoverflow.net/users/13972
|
409454
| 167,602 |
https://mathoverflow.net/questions/409463
|
5
|
Let $(z\_i)$ be a square-summable sequence which is even summable but not absolute summable, i.e. $\sum\_{i=1}^{\infty} \vert z\_i \vert = \infty$,$\sum\_{i=1}^{\infty} \vert z\_i \vert^2 < \infty$ and $\sum\_{i=1}^{\infty} z\_i$ exists. I would like to ask if the following function $$f(\mu):=\prod\_{i=1}^{\infty}(1+\mu^2 \vert z\_i \vert^2 - 2 \mu \Re(z\_i))$$
is necessarily entire?
I must say that I do not even know if this product exists away from the real axis. On the real axis it is clear that it exists by using that $(1+x) \le e^x$ such that
$\vert f(\mu) \vert \le e^{\mu^2 \sum\_i \vert z\_i \vert^2 -2\mu \Re \sum\_i z\_i }.$
Assuming it was entire, does there exist a similar growth bound on $\vert f(\mu) \vert$ as the one I obtained on the real axis?
|
https://mathoverflow.net/users/457901
|
Is this infinite product entire?
|
This function is (on the real line, at least) the product of
$$ \exp( \mu^2 \sum\_{i=1}^\infty |z\_i|^2 - 2 \mu \Re(\sum\_{i=1}^\infty z\_i)) \quad (1)$$
and the [Hadamard type product](https://en.wikipedia.org/wiki/Weierstrass_factorization_theorem)
$$ \prod\_{i=1}^\infty E\_1( 2 \mu\Re z\_i - \mu^2 |z\_i|^2) \quad(2)$$
where $E\_1$ is the first elementary factor
$$ E\_1(z) := (1-z) \exp(z).$$
The expression (1) is clearly entire in $\mu$; the product (2) is locally uniformly convergent (from the standard bound $E\_1(z) = 1+O(|z|^2)$ when $|z| \leq 1$) and so is also entire. A refinement of this analysis (using for instance the upper bound $|E\_1(z)| \leq \exp(O(|z|^2))$ for all $z$) also gives growth bounds comparable to the ones you already located in the real case.
An alternate factorisation is
$$ \exp( - 2 \mu \Re \sum\_{i=1}^\infty z\_i ) \prod\_{i=1}^\infty E\_1(\mu z\_i) E\_1(\mu \overline{z\_i}).$$
Thus this function is an order two entire function with zeroes precisely at $1/z\_i, 1/\overline{z\_i}$ (counting multiplicity), which specifies the function uniquely up to quadratic exponential factors (such as (1)) by the [Hadamard factorisation theorem](https://en.wikipedia.org/wiki/Weierstrass_factorization_theorem#Hadamard_factorization_theorem).
|
11
|
https://mathoverflow.net/users/766
|
409465
| 167,604 |
https://mathoverflow.net/questions/409426
|
4
|
I imagine this to be a very classical question in complex analysis:
Consider the Hadamard product
$$g(\mu) = \prod\_{n=1}^{\infty}E\_1(\mu z\_n),$$
where $E\_1(z):=(1-z)e^z$ is the first elementary factor
for some sequence $z\_n \to 0$ fast enough, such that $g$ is entire. In fact, choosing $ (z\_m)\_{m \in \mathbb N}$ to be square-summable is sufficient.
I wonder if there are any estimates on the Taylor coefficients of $g$ known?
I can only think of the following very pedestrian approach:
Using the elementary inequality $$\left\lvert (1-w)e^w \right\rvert \le e^{\vert w \vert^2/2},$$
we can estimate
$$\vert g(\mu) \vert \le e^{\frac{\vert \mu^2\vert}{2} \sum\_{m=0}^{\infty} \vert z\_m \vert^2}=:h(\mu).$$
This implies for the power series $g(\mu) = \sum\_n a\_n \mu^n$ that
$\vert a\_n \vert \le r^{-n} \sup\_{\mu = r} h(\mu)$ for any $r.$
Optimizing over $r$ we find that
$$\vert a\_n \vert \le \frac{ e^{n/2}\Vert (z\_m)\_{m \in \mathbb N} \Vert^n\_{\ell^2}}{n^{n/2}}.$$
I am wondering if there is a systematic way to improve these estimates of $\vert a\_n\vert$ (apart from some obvious improvements such as $g'(0)=0$.
|
https://mathoverflow.net/users/457901
|
Taylor coefficients of Hadamard product
|
I doubt this bound can be improved much, at least for even $n$. Indeed: set $z\_1 = z\_2 = \ldots = z\_k = k^{-1/2}$ and $z\_{k+1} = z\_{k+2} = \ldots = 0$, so that the $\ell^2$ norm of $(z\_n)$ is $1$. (Intuitively, this is the worst-case scenario.) Then
$$ g(\mu) = (E\_1(k^{-1/2} \mu))^k = (1 - k^{-1/2} z)^k e^{z \sqrt k} .$$
As $k \to \infty$, the above functions converge to $e^{-z^2/2}$ locally uniformly on $\mathbb C$, and so the $n$th Taylor coefficient of $g$ converges to the $n$th Taylor coefficient of $e^{-z^2/2}$, which is equal to
$$ \frac{1}{2^{n/2} (n/2)!} \approx \frac{e^{n/2}}{n^{(n+1)/2} \sqrt{\pi}} $$
when $n$ is even. Compared to your "pedestrian" bound, this is off by a factor $n^{-1/2}$.
My guess would be that things are not different for odd $n$, but I did not think about it.
|
3
|
https://mathoverflow.net/users/108637
|
409472
| 167,608 |
https://mathoverflow.net/questions/409418
|
13
|
Would've been a better question for Christmas than Thanksgiving, but alas...
Let $t\_n$ denote the number of rooted, unlabeled trees on $n$ vertices ([OEIS A000081](http://oeis.org/A000081)). These are the isomorphism classes of rooted trees under root-preserving isomorphisms. Let $T(z) = \sum\_{n\geq 1} t\_n z^n$ be the corresponding generating function. In 1937, using his [enumeration under symmetry theorem](https://en.wikipedia.org/wiki/P%C3%B3lya_enumeration_theorem), Pólya showed that
$$ T(z) = z \prod\_{i=1}^{\infty}e^{\frac{T(z^i)}{i}}.$$
By differentiating this identity one obtains the recurrence
$$ (n-1)\cdot t\_n = \sum\_{i=1}^{n-1}t\_{n-i}\sum\_{m \mid i}mt\_{m}$$
for $n> 1$. This is such a nice recurrence that I wonder:
**Question**: Is there a bijective proof of this recurrence for $t\_n$?
|
https://mathoverflow.net/users/25028
|
Bijective proof of recurrence for rooted unlabeled trees
|
**Late edit**: having now read through the OP comments, I can see that my proof is essentially a carbon copy of @darij.grinberg's approach (although my derivation was independent). I'm okay to delete this answer once/if darij chooses to post theirs.
Pick a canonical ordering of unlabelled rooted trees, say, with lexicographical comparison of tuples $(n, T\_1, \ldots, T\_k)$, where $n$ is the number of vertices, $T\_1, \ldots, T\_k$ is the non-descending sequence of children subtrees. Throughout $T, T\_1, T\_2$ are unlabelled rooted trees in canonical form (that is, subtrees of any vertex are ordered as above).
Let $A\_n$ be the set of pairs $(T, v)$ with $|T| = n$, $v$ is a non-root vertex of $T$. Also, let $B\_n$ be the set of tuples $(T\_1, T\_2, k, u)$ such that $|T\_1| + k|T\_2| = n$, $u$ is a vertex of $T\_2$. Observe that $|A\_n|$ and $|B\_n|$ are LHS and RHS of the recurrence in OP, more readily seen by rewriting $(n - 1)t\_n = \sum\_{m = 1}^{n - 1}mt\_m \sum\_{0 < km < n} t\_{n - km}$.
The bijection between $A\_n$ and $B\_n$ is as follows:
* we associate $(T\_1, T\_2, k, u)$ to $(T, v)$ by:
+ $T\_2$ = the subtree of $T$ containing $v$,
+ $u$ = the respective vertex of $T\_2$,
+ $k$ = the number of children subtrees of $T$ isomorphic to $T\_2$ not later than the copy containing $v$ in the ordered sequence of children subtrees,
+ $T\_1$ = the result of removing the $k$ children subtrees from $T$.
* we associate $(T, v)$ to $(T\_1, T\_2, k, u)$ by inserting $k$ copies of $T\_2$ as children subtrees of the root of $T\_1$ (and naming the result $T$), and picking the respective vertex $u$ in the $k$-th copy as $v$.
|
6
|
https://mathoverflow.net/users/106512
|
409496
| 167,614 |
https://mathoverflow.net/questions/409499
|
10
|
Let $ G $ be a linear algebraic group. Is it true that a subgroup $ H $ of $ G $ is Zariski closed if and only if there exists a representation $ \pi: G \to \mathrm{GL}(V) $ and a vector $ v \in V $ such that the stabilizer $ G\_v:=\{g \in G: \pi(g)v=v \} $ is equal to $ H $?
I think one implication is clear since $ G\_v $ is certainly a subgroup and the equation $ \pi(g)v=v $ is polynomial in the matrix entries. Thus any stabilizer must be Zariski closed.
I am not sure of the reverse implication. Is it really true that every Zariski closed subgroup of $ G $ arises as the stabilizer of some vector in some representation?
|
https://mathoverflow.net/users/387190
|
Is every Zariski closed subgroup a stabilizer?
|
Chevalley's theorem (see Theorem 4.19 in [Milne](https://www.jmilne.org/math/CourseNotes/iAG200.pdf)) is very close to this. It says
>
> Let $G$ be a linear algebraic group and let $H$ be a Zariski closed subgroup. Then there is a representation $V$ of $G$ and a one dimensional subspace $L$ of $V$ such that $H$ is the stabilizer of $\mathbb{P}(L)$ in the action of $G$ on $\mathbb{P}(V)$.
>
>
>
So this is close to what you asked for, but uses projective rather than linear representations.
To see that you can't get exactly what you asked for, work over a field of characteristic zero, take $G = \text{SL}\_2$ and let $H$ be the Borel subgroup $B:=\left[ \begin{smallmatrix} \ast & \ast \\ 0 & \ast \end{smallmatrix} \right]$. The classification of $\text{SL}\_2$-representations is well known, and we see that $V^B = V^{\text{SL}\_2}$ for any $\text{SL}\_2$-representation $V$.
|
23
|
https://mathoverflow.net/users/297
|
409501
| 167,615 |
https://mathoverflow.net/questions/409500
|
3
|
Grothendieck once asked "What is a meter?" (<https://golem.ph.utexas.edu/category/2006/08/letter_from_grothendieck.html>). This innocent sounding question, made me to think about how coordinate systems are defined in physics.
How are coordinate systems in physics defined, for example in special relativity where the coordinate system is assumed to be given.
**I have tried a possibility, and was asking myself, if there are other mathematical ways to define a coordinate system.** (Thanks for your help):
Here is a possible
For the definition of a physical affine basis $B = (v\_1 = p\_1-p\_0, v\_2 = p\_2-p\_0, v\_3=p\_3-p\_0)$ of the spatial space $\mathbb{R}^3$, there
has to be an a priori defined basis $\hat{B}$, so that $B$ can refer to this basis $\hat{B}$.
Observer $A$ at point $p=p\_0$ can do the following:
* Choose nearby points $p\_1,p\_2,p\_3$ in spatial space,
without actually to give those points any a priori coordinates and define "affine vectors": $$v\_1 = p\_1 p\_0, v\_2 = p\_2 p\_0, v\_3 = p\_3 p\_0$$
* It is possible for the observer $A$ to measure the distance between two (nearby) points: $$d\_{ij} = d(p\_i,p\_j), 0\le i,j \le 3$$
* Oberser $A$ then can compute the gram matrix:
$$G = (g\_{ij}), g\_{ij} = \frac{1}{2}(d\_{0i}^2+d\_{0j}^2-d\_{ij}^2), 1\le i,j \le 3$$
we assume by the Schönberg criterion, that this matrix is a positive definite matrix, and hence especially a Gram matrix.
* Observer $A$ then can do Cholesky decomposition of the Gram matrix : $$G = C C^T, C = (x\_1,x\_2,x\_3)$$
* Observer $A$ can do the Gram-Schmidt procedure to get an orthonormal basis: $e\_1,e\_2,e\_3$ given $x\_1,x\_2,x\_3$.
* Hence a posteriori observer $A$ can choose the points $p\_i$ in such a way that $$G=\mathbf{1}$$,
since $\left< e\_i,e\_j \right> = \delta\_{ij},$ where $\delta$ is the Kronecker $\delta$.
Observer $B$ at point $q$ can do the same procedure, to get the orthonormal Basis $\mathbf{1} = G\_q = C\_q C\_q^T$.
By the [unitary freedom of square roots](https://en.wikipedia.org/wiki/Square_root_of_a_matrix#Unitary_freedom_of_square_roots), there exists an orthogonal matrix $O\_{pq}$ such that $C\_p O\_{pq} = C\_q$
hence:
$$O\_{pq}=C\_q C\_p^{-1} = C\_q C\_p^T$$
From this last equation we get:
* $O\_{pq}^T = O\_{pq}^{-1} = C\_p C\_q^{-1}= O\_{qp}$
* $O\_{pp} = \mathbf{1}$
* $O\_{pr} O\_{rq} = O\_{pq}$
Setting $w\_{pq} := \log(O\_{pq})$ we get
* $w\_{pq} = -w\_{qp}$
* $w\_{pp} = \mathbf{0}$
* $w\_{pr} + w\_{rq} = w\_{pq}$
Hence one might define the affine vectors between the points $p,q$ as $w\_{pq}$. The distance between $p,q$ might be given as:
$$d(p,q) = |w\_{pq}|\_F = |\log(O\_{pq})|\_F = |\log(C\_q C\_p^T)|\_F$$
where $|.|\_F$ denotes the Frobenius norm.
Also asked here, in case it is not appropriate for this forum:
<https://physics.stackexchange.com/questions/679409/how-are-spatial-coordinate-systems-in-physics-defined>
|
https://mathoverflow.net/users/165920
|
How are spatial coordinate systems in physics defined?
|
This question has been explored in the context of global positioning systems, which need to account for general relativity. The traditional Minkowski coordinates $(t,x,y,z)$ of flat space-time do not allow for an immediate positioning in an unknown gravitational field.
[Tarantola](https://en.wikipedia.org/wiki/Albert_Tarantola) and colleagues propose a symmetric coordinate system with four times, see [Gravimetry, Relativity, and the Global Navigation Satellite Systems](https://arxiv.org/abs/0905.3798) and this [talk.](http://www.ipgp.fr/~tarantola/Files/Professional/Teaching/Seminar/Lessons/Coll/Coordinates-RG.pdf) If four satellite clocks – having an arbitrary space-time trajectory – broadcast their proper time – using electromagnetic signals,– then, any observer receives, at any point along his personal space-time trajectory, four times, corresponding to the four signals arriving at that space-time point. These four times, $\tau\_1,\tau\_2,\tau\_3,\tau\_4$, are, by definition, the coordinates of the space-time point.
In [Using pulsars to define space-time coordinates](https://arxiv.org/abs/0905.4121) Coll and Tarantola propose to replace the satellite clocks by pulsars, to obtain a relativistic coordinate system valid in a domain larger than our Solar system.
---
.
"What is a meter?" Albert Tarantola in front of the original meter.
|
6
|
https://mathoverflow.net/users/11260
|
409506
| 167,617 |
https://mathoverflow.net/questions/409504
|
20
|
In classical probability theory, the (multivariate) Gaussian is in some sense the "nicest quadratic" random variable, i.e. with second moment a specified positive-definite matrix. I do not know how to make this precise, but non-precisely what I mean is that 1. Gaussian shows up everywhere, and 2. it is universal/canonical/... in some sense, e.g. as in the central limit theorem.
My **question** is whether for many noncommutative probability spaces (an algebra $A$ over $\mathbf{C}$ and a map $E:A\to\mathbf{C}$, with conditons), there *also* exists a "nicest quadratic" random variable $X\in A$, satisfying analogous properties to the Gaussian.
|
https://mathoverflow.net/users/119012
|
Is there a noncommutative Gaussian?
|
The theory of classical independence and classical convolution can be generalised to noncommutative settings in several ways. The most famous one is that of [free independence and free convolution](https://en.wikipedia.org/wiki/Free_probability) (introduced by Voiculescu), but there is also boolean independence and boolean convolution (introduced by Speicher and Woroudi in [Boolean convolution](https://mathscinet.ams.org/mathscinet-getitem?mr=1426845)); monotone independence and monotone convolution (introduced by Muraki in [Monotonic independence, monotonic central limit theorem and monotonic law of small numbers](https://mathscinet.ams.org/mathscinet-getitem?mr=1824472)); and anti-monotone independence and anti-monotone convolution (the order-reversal of the previous notion). There are classification results of Speicher ([On universal products](https://mathscinet.ams.org/mathscinet-getitem?mr=1426844)) and Muraki ([The five independences as natural products](https://mathscinet.ams.org/mathscinet-getitem?mr=2016316)) that show that these are the only notions of independence (or convolution) that obey some natural set of axioms. (Speicher's classification assumed that convolution is commutative, so omitted the monotone and anti-monotone cases that were later discovered by Muraki.)
For each such concept of independence, there is a central limit theorem. Classically, the limiting distribution is the gaussian; in free probability it is the semicircular law; in the boolean case it is the Bernoulli distribution; and in the monotone and anti-monotone cases it is the arcsine law. See Section 9.2.1 of the recent thesis [Evolution equations in non-commutative probability](https://escholarship.org/content/qt8n39f7mt/qt8n39f7mt.pdf?t=qbrc2a) of David Jekel (and Chapter 5 of that thesis contains a more detailed history of the development of these notions of independence). For the classical and free independence concepts, at least, there is also an associated notion of entropy, and these distributions extremise the entropy amongst all distributions of a fixed mean and variance; again, Jekel's thesis has further information. (For the free case, of course, pretty much any introduction to free probability will contain these facts.)
EDIT: There is also finite free convolution (see [Marcus, Spielman, and Srivastava - Finite free convolutions of polynomials](https://arxiv.org/abs/1504.00350)), in which the analogue of the gaussian is the distribution of zeroes of Hermite polynomials.
|
28
|
https://mathoverflow.net/users/766
|
409514
| 167,620 |
https://mathoverflow.net/questions/409512
|
5
|
(This question may be too elementary for this site — I'm fine if it needs to be moved to math.stackexchange.)
If I approximate a nice planar curve by a straight line, the tangent, then the second derivative tells me which side of the line the curve lies on locally. If instead I approximate the curve by a circle — the osculating circle — what determines the answer to the analogous question? Which side of the osculating circle does the curve lie on locally?
---
Update: I'm accepting one of the [answers](https://mathoverflow.net/a/409538) below, but it's surprising and with my poor geometric intuition not particularly easy. This situation is the exact opposite of the tangent line analogy I made in the question. In that case the graph of a function lies locally on one side or the other of the tangent line at generic points, and only crosses at inflection points. But for the osculating circle it is the reverse: the curve crosses the circle at generic points, and only lies (locally) inside or outside the circle at maxima or minima of the curvature.
The can be deduced from the Tait–Kneser theorem, as the [answer](https://mathoverflow.net/a/409538) below claims. A nice explication by Ghys, Tabachnikov and Timorin can be found at [Osculating curves: around the Tait–Kneser Theorem](https://arxiv.org/abs/1207.5662).
The theorem itself says wherever the curvature is monotonic (e.g. when parametrized by arclength), the osculating circles are pairwise disjoint and nested. It takes some geometric visualization to realize this means the circles can not lie on one side of the curve as the point of contact moves.
This can be a challenge to see on actual graphics, because the curve and circle agree up to second order terms. A *Mathematica* demonstration may be found [here](https://demonstrations.wolfram.com/InvertingAPointInTheOsculatingCirclesOfACurve/#more) which allows one to select in particular the example of an ellipse which was mentioned in the comments.
|
https://mathoverflow.net/users/6756
|
Osculating circle
|
[Tait–Kneser theorem](https://en.wikipedia.org/wiki/Tait%E2%80%93Kneser_theorem) says that generically curves crosses its osculating circle. If not (that is, if the curve is locally supported by its osculating circle), then the point is a vertex of the curve, but the supporting condition is stronger a bit.
If it is a point of local minimim/maximum of the curvature, then the curve lies locally outside/inside of its osculating circle [see 6.4 [here](https://arxiv.org/abs/2012.11814v2)].
|
4
|
https://mathoverflow.net/users/1441
|
409538
| 167,627 |
https://mathoverflow.net/questions/409516
|
8
|
Let $X\subset \mathbb{P}^n\_k$ be a smooth projective variety, a point $p\in \mathbb{P}^{n,\vee}\_k$ gives rise to a hyperplane $H\_p\subset \mathbb{P}^n$, hence an intersection $X\_p:=H\_p\cap X$.
We say a line $L\subset\mathbb{P}\_k^{n,\vee}$ is a Lefschetz pencil if
(1) There exists $0,\infty \in L(k)$ such that $H\_0, H\_\infty, X$ intersect transversally.
(2) For $p\in L-\{s\_1,...,s\_r\}$, the section $X\_p$ is smooth, $H\_{s\_i}\cap X$ has exactly one quadratic ordinary singularity.
When $k$ is an algebraically closed field of $\mathrm{char}(k)=p$, do we know if Lefschetz pencils always exist? If not, can we relax the assumption by allowing $L$ be a smooth curve in $\mathbb{P}^{n,\vee}\_k$?
(From Theorem 3 [here](https://arxiv.org/pdf/0911.1470.pdf), Lefschetz pencils always exist over any field after $d$-uple embedding of $X$)
|
https://mathoverflow.net/users/nan
|
Does Lefschetz pencil always exist in char $p$?
|
I think you are asking whether a Lefschetz pencil exists without re-embedding. Then the answer is no. In cor. 3.5.0 of expose XVII of SGA 7, Katz gives a necessary and sufficient condition for a Lefschetz pencil to exist. In the case of a hypersurface $X=V(F)$ in $\mathbb{P}^n$ and $char k=p\not=2$, the condition amounts to the Gauss map $\phi:X\to (\mathbb{P}^n)^\vee$ defined by $\phi(x\_0,\ldots, x\_n) = (\partial F/\partial x\_i)$ being seperable.
But he shows this condition fails for $F= \sum x\_i^{p^n+1}$.
|
8
|
https://mathoverflow.net/users/4144
|
409540
| 167,628 |
https://mathoverflow.net/questions/409304
|
2
|
Let $m \geqslant 1$ be a fixed integer.
Let $f(n)$ be [A007814](https://oeis.org/A007814), exponent of highest power of $2$ dividing $n$, a.k.a. the binary carry sequence, the ruler sequence, or the $2$-adic valuation of $n$.
Then we have an integer sequence given by
\begin{align}
a\_1(0)& = 1\\
a\_1(2n+1)& = a\_1(n)\\
a\_1(2n)& = a\_1(n-2^{f(n)})+a\_1(2n-2^{f(n)})
\end{align}
Here $a\_1(n)$ is [A243499](https://oeis.org/A243499), product of parts of integer partitions as enumerated in the table [A125106](https://oeis.org/A125106).
Let
$$a\_m(n) = \sum\limits\_{k=0}^{n}(\binom{n}{k}\operatorname{mod} 2)a\_{m-1}(k)$$
Also
$$s\_m(n)=\sum\limits\_{k=0}^{2^n-1}a\_m(k)$$
I conjecture that $s\_m(n)$ is Stirling transform of
$$1, m, m^2, m^3, \cdots$$
In other words
$$s\_m(n)=\exp(-m)\sum\limits\_{k=0}^{\infty}(k + 1)^n\frac{m^k}{k!}=\sum\limits\_{k=0}^{n}{n+1\brace k+1}m^k$$
Is there a way to prove it?
|
https://mathoverflow.net/users/231922
|
Modulo $2$ binomial transform of A243499 applied $k$ times
|
The definition of $a\_1$ given in OEIS is based on a bijection between integer partitions and natural numbers. A partition $\lambda\_1\geq\lambda\_2\geq\dots\geq\lambda\_m>0$ with exactly $m$ parts corresponds to the number
$$2^{\lambda\_1+m-2}+2^{\lambda\_2+m-1}+\dots+2^{\lambda\_m-1}.$$
The definition of $a\_1$ can then be written as
\begin{equation}\label{a1d}(1)\qquad a\_1\left(2^{\lambda\_1+m-2}+2^{\lambda\_2+m-1}+\dots+2^{\lambda\_m-1}\right)=\lambda\_1\dotsm\lambda\_m.\end{equation}
When $n$ is a natural number, I will write $\|n\|$ for the number of ones in the binary expansion of $n$, and $n\preceq m$ if all binary digits in $n$ are smaller than or equal to those in $m$. It is well-known that $\binom nk\,\operatorname{mod}\,2=1$ if and only if $k\preceq n$. It is easy to see from this that
$$a\_m(n)=\sum\_{k\preceq n}(m-1)^{\|n-k\|}a\_1(k).$$
Roughly speaking, we go from $k$ to $n$ by changing zeroes to ones, and for each of the $\|n-k\|$ zeroes we can choose to change it in any one of $m-1$ binomial transforms.
The final ingredient is the identity
$$\left\{\array{k+l\\l}\right\}=\sum\_{l\geq \lambda\_1\geq\lambda\_2\geq\dots\geq\lambda\_k> 0}\lambda\_1\lambda\_2\dotsm\lambda\_k.$$
Probably this is well-known. I verified it using induction; just let me know if you need more explanation. By (1), this can be written
\begin{equation}\label{sa}(2)\qquad\left\{\array{k+l\\l}\right\}=\sum\_{0\leq j<2^{l+k-1},\,\|j\|=k}a\_1(j).\end{equation}
We now have all ingredients we need. We have
$$s\_m(n)=\sum\_{0\leq k<2^n}a\_m(k)=\sum\_{0\leq k<2^n}\sum\_{l\preceq k}(m-1)^{\|k-l\|}a\_1(l).$$
For fixed $l$ and $j=\|k-l\|$, we obtain $k$ by choosing $j$ from $n-\|l\|$ zeroes in $l$ and changing them to ones. Thus, we can write
$$s\_m(n)=\sum\_{0\leq l<2^n}a\_1(l)\sum\_{j}\binom{n-\|l\|}{j}(m-1)^{j}
=\sum\_{0\leq l<2^n}a\_1(l)m^{n-\|l\|}.$$
Writing this as a sum over $k=n-\|l\|$ and using (2) gives indeed
$$s\_m(n)=\sum\_{k=0}^n m^k\left\{\array{n+1\\k+1}\right\}.
$$
|
2
|
https://mathoverflow.net/users/10846
|
409557
| 167,634 |
https://mathoverflow.net/questions/365897
|
8
|
There are many well-known excellent blogs like the ones by T. Tao, G. Kalai, J. Baez, etc. Many of them use the WordPress engine.
I have been surprised to find that there are some excellent blogs on some unconventional platforms like [Telegram](https://en.wikipedia.org/wiki/Telegram_(software)) (created in particular by mathematician [Nikolai Durov](https://en.wikipedia.org/wiki/Nikolai_Durov)) or Twitter. Telegram blogs ("channels") seem to be "invisible" outside Telegram.
**Question:** what are some not so well-known, but noteworthy math blogs on some "non-standard" platforms, especially invisible from the outside?
Let me give some examples. Some of the following blogs by MathOverflow participants:
Twitter:
* [@littmath](https://twitter.com/littmath) (in English) by [Daniel Litt](https://mathoverflow.net/users/6950/daniel-litt)
Telegram:
* [Graphs and Machine Learning](https://t.me/graphML) (in English) by Sergei Ivanov
* [Математические байки](https://t.me/mathtabletalks) (in Russian) by [Victor Kleptsyn](https://mathoverflow.net/users/31371/victor-kleptsyn)
* [tropical saint petersburg](https://t.me/tropicalgeometry) (in Russian) by [Nikita Kalinin](https://mathoverflow.net/users/4298/nikita-kalinin)
* [fpmath](https://t.me/fedyap) ( English/Russian ) by [Fedor Petrov](https://mathoverflow.net/users/4312/fedor-petrov)
* [Математическая свалка Сепы](https://t.me/math_dump_of_sepa) (in Russian) by Sergei O. Ivanov
|
https://mathoverflow.net/users/10446
|
Mathematical blogs on "non-standard" platforms (Telegram, Twitter, Dzen , ... )
|
44 Telegram math channels are [here](https://t.me/math_channels) (5 in English).
|
0
|
https://mathoverflow.net/users/469930
|
409566
| 167,639 |
https://mathoverflow.net/questions/333831
|
4
|
Let $L$ be a restricted Lie algebra over a field $F$ of characteristic $p>0$. Assume that the following condition holds:
For every restricted ideal $I$ of $L$, the minimal restricted subalgebras of $L/I$ are pairwise non-isomorphic.
>
> **QUESTION**: Is $L$ necessarily abelian?
>
>
>
I already know that the answer is affirmative if one assumes that $L$ is nilpotent.
|
https://mathoverflow.net/users/17582
|
A condition on minimal restricted subalgebras of a restricted Lie algebra
|
The answer is positive if $L$ is finite-dimensional and $F$ is algebraically closed.
Indeed, suppose first that $L$ is $p$-nilpotent. Then the Frattini restricted subalgebra of $L$ is given by $\Phi(L)=[L,L]+L^{[p]}$, and $L/\Phi(L)$ is abelian with trivial $p$-map. Thus the hypothesis forces that $L/\Phi(L)$ is 1-dimensional, so that $L$ is cyclic and we are done.
On the other hand, if $L$ has an element $x$ that is not $p$-nilpotent, then consider the Jordan-Chevalley decomposition $x=x\_n+x\_s$ of $x$, where $x\_n$ is $p$-nilpotent, $x\_s$ is semisimple, and $[x\_n,x\_s]=0$. Then the restricted subalgebra $T$ generated by $x\_s$ is a torus and so, as $F$ is algebraically closed, $T$ has an $F$-basis consisting of elements $t$ such that $t^{[p]}=t$. By the hypothesis, we deduce that $L$ contains a unique (1-dimensional) torus $T$. Using again the fact that $F$ is algebraically closed, it follows that $L$ is nilpotent and so $T$ is contained in the center $Z(L)$ of $L$. In particular, $L/Z(L)$ is $p$-nilpotent and from the first part we infer that $L/Z(L)$ is cyclic. This implies that $L$ is abelian, yielding the claim.
|
2
|
https://mathoverflow.net/users/14653
|
409568
| 167,640 |
https://mathoverflow.net/questions/409526
|
2
|
I have been reading the paper ['Improved Bounds for the Sunflower Lemma' (Ann. of Math., Vol. 194(3), pp. 795-815)](https://annals.math.princeton.edu/2021/194-3/p05), and have not managed to understand the following:
1. I would like a formalization for what a $p$-biased distribution (pg. 796) is, since $R$ is treated both as a set and a random-variable in Definition 1.5 (pg. 796) and I am not certain how this is done.
What I have tried: Consider the measure space $(X, \mathcal{P}(X), \mathbb{P})$ and the measurable space $(X, \mathcal{P}(X))$. We define the random-variable $\mathcal{R}: X \to X$ such that $\mathcal{R}: t \mapsto t$, and the probability-measure $\mathbb{P}: \mathcal{P}(X) \to [0, \infty]$, such that $\forall \, t \in X: \mathbb{P}(\{t\}) = p$, and $\forall \, S \subseteq X:$ $\mathbb{P}(S) := \prod\_{s \in S} \mathbb{P}(\{s\}).$ Then, the distribution function is as follows- $\forall \, A \in \mathcal{P}(X):$ $$\mathbb{P}^{X}(A) = \mathbb{P}(\{w \in X: \mathcal{R}(w) \in A\}) = \mathbb{P}(A).$$ Is this indeed the formal definition of a $p$-biased distribution? If so, what does Definition 1.5 mean (, since $R$ is considered both as a r.v. and a set in the definition)? If not, what is a $p$-biased distribution formally?
2. Assuming a heuristic understanding of $U(X,p)$ as being as stated by the authors on pg. 796 before Definition 1.5, we consider Lemma 1.6. On pg. 797, the paper claims that by 'the union bound', we have that $$\textrm{Pr} \: [\forall \, i \in [r], \, \exists \, S \in \mathcal{F}: S \subseteq Y\_{i}] > 0$$, where $Y\_{i}$ is the set of all points in $X$ coloured with colour $i$. I am not able to follow this, since I am not able to say anything non-trivial about $\textrm{Pr} (\cup\_{i \in [r]} \varepsilon\_{i})$.
3. When the authors define the weight function $\sigma$ on pg. 799, they restrict its domain to $\mathcal{F}$, but on pg. 800, they refer to $\sigma(\mathcal{F}\_T)$, which doesn't make sense to me, since it doesn't appear that $\forall \, T \subseteq X, \, T \neq \emptyset: \mathcal{F}\_T \subset \mathcal{F}.$ Therefore, I was wondering if we require that the domain of $\sigma$ be $\mathcal{P}(X)$ instead.
4. On pg. 800, after Definition 2.1, the authors state that if $(\mathcal{F},\sigma)$ is s-spread, where $s= (s\_{0}; s\_{1}, \dots, s\_{w})$ is a weight-profile, in particular, $\mathcal{F}$ is a $w$-set-system.
What I am able to show is that if indeed $\mathcal{F}$ contains a set $S$ with $|S| > w$, then, we have $$\forall R \in \mathcal{F}, S \subset R: \sigma (R \, \backslash \, S) = 0.$$ What I thought I could try showing was that $\sigma$ would be $0$ everywhere, which is not allowed, by definition of a weight function on pg. 799.
Any help on any or all of these questions is appreciated.
|
https://mathoverflow.net/users/nan
|
On 'Improved Bounds for the Sunflower Lemma' [Alweiss, Lovett, Wu, Zhang]
|
1. The random variable $R$ here takes values in the power set ${\mathcal P}(X)$ of $X$, not in $X$: it's a random *set* in $X$, not a random *point*. For the purposes of this argument, the only important features of $R$ are that the events $(x \in R)$ for each $x \in X$ are independent events of probability $p$. But if you really wanted to formally model this random variable by a measurable map on a probability space as you attempted to do above, one could do so by taking the probability space $({\mathcal P}(X), {\mathcal P}({\mathcal P}(X)), {\mathbb P})$ with the probablity measure
$$ {\mathbb P}(\{R\_0\}) = p^{\# R\_0} (1-p)^{\# (X \backslash R\_0)}$$
for all $R\_0 \subset X$ (i.e., $R\_0 \in {\mathcal P}(X))$), and take $R: {\mathcal P}(X) \to {\mathcal P}(X)$ to be the identity map $R: R\_0 \mapsto R\_0$. However, the precise probability space model for this random set $R$ is not of particular relevance; see [this post of mine](https://terrytao.wordpress.com/2010/01/01/254a-notes-0-a-review-of-probability-theory/) for some further discussion.
2. Once one can prove that ${\mathbb P}(E\_i) > 1-1/r$ for all $i=1,\dots,r$, the union bound (applied to the complements of $E\_i$) shows that ${\mathbb P}(\bigwedge\_{i=1}^r E\_i) > 0$, thus the $E\_i$ simultaneously hold with positive probability.
3. Presumably this will become clearer at the later stages of the argument where the quantity $\sigma({\mathcal F}\_T)$ is actually used, but I would guess that here the authors are implicitly abusing notation by identifying the link ${\mathcal F}\_T = \{ S \backslash T: S \in {\mathcal F}, T \subset S \}$ with the associated subset $\{ S: S \in {\mathcal F}, T \subset S \}$ of ${\mathcal F}$ for the purpose of computing weights.
4. Apply Definition 2.1(ii) with $T$ replaced by your set $S$ with $|S| > w$ and with the conventions indicated in my response to item 3.
More generally, you might like to look at my [general advice on how to read mathematical papers](https://terrytao.wordpress.com/advice-on-writing-papers/on-compilation-errors-in-mathematical-reading-and-how-to-resolve-them/) (in particular how to resolve issues like 3, 4 that are often coming from some typo or implicit convention on the author's part).
|
7
|
https://mathoverflow.net/users/766
|
409574
| 167,642 |
https://mathoverflow.net/questions/409553
|
32
|
Consider the hierarchy of relative geometric constructibility by
straightedge and compass. Namely, given a geometric figure $B$, a
set of points in the plane, we define that geometric figure $A$ is
*constructible from* $B$, written as $$A\leq B,$$ if from points in
$B$ using straightedge and compass we may construct every point in
$A$.
This is a partial preorder on geometric figures, considered as sets of
points in the Euclidean plane. The order gives rise to a
corresponding strict order notion $A<B$, which holds when from $B$
we may construct $A$ but not conversely. That is, $$A<B\qquad\text{
if and only if }\qquad A\leq B\quad\text{ and }\quad B\not\leq A.$$
**Question.** Is the hierarchy of relative geometric constructibility a
dense order? That is, if $A<B$ for figures in the plane, is there a figure $C$
such that $A<C<B$?
I would also be interested to know whether the answer depends on
considering only finite sets of points or infinite sets.
[**Update.** The main question is now answered by the maximality argument of Pace Nielsen below, but remains open in the case of finite figures. That is, if $A$ and $B$ are finite geometric figures with $A<B$, must there be some figure $C$ strictly between?]
Let me make a few observations. First, relative constructibility is symmetric on line segments. That is, if a line segment $AB$ is constructible from $CD$, then $CD$ is
constructible from $AB$. Thus, the relation of relative constructibility on line segments is an equivalence relation, with no occurences of the strict order. The argument is given on [my blog
post](http://jdh.hamkins.org/the-hierarchy-of-geometric-constructibility-can-we-go-back/).
For this reason, there will be no counterexamples to density using
segments only.
The symmetry claim is not true, however, for triangles. For
example, from a unit length and $\sqrt[3]{2}$, we can construct a
unit isosceles right triangle, but from such a triangle, we cannot
construct the length $\sqrt[3]{2}$. This instance is also discussed on my blog post.
Since this case has the feeling of being perhaps a minimal extension, it might be considered as a natural candidate counterexample, and several people (including David Madore and Nikolay Nikolov) have proposed various arguments that make definite progress. But the matter is currently still open.
A constructively closed set $S$ of points in the plane is
determined by any two of them and the set of points on the
corresponding line $\ell$. The reason is that if point $p\in S$,
then the distances between $p$ and two points on $S\cap\ell$ will
be a distance witnessed on $S\cap\ell$, and so we can construct $p$
via circles from points on $S\cap\ell$.
Therefore, a constructively closed set of points in the plane is
determined by a unit segment and the real field of distances
realized by that set. (one can also simply view the points as
complex numbers, as determined by a unit segment from the set.) For
this reason, the question is equivalent to the following:
**Question.** If $K< L$ are quadratically closed field
extensions of the rationals, must there be a quadratically closed
field $F$ strictly between them?
$$K< F< L$$
Such a field would correspond to a strictly intermediate set of points between $K$ and $L$ in the hierarchy of relative constructibility.
Regarding the natural candidate counterexample, the question would be: is there a quadratically closed field strictly between the quadratic closure of $\mathbb{Q}$ and the quadratic closure of $\mathbb{Q}(\sqrt[3]{2})$?
|
https://mathoverflow.net/users/1946
|
Is the hierarchy of relative geometric constructibility by straightedge and compass a dense order?
|
There are three answers. Throughout let $qcl(F)$ be the quadratic closure of a field $F$ inside $\mathbb{C}$.
**Part 1:** Yes there is a quadratically closed field strictly between $qcl(\mathbb{Q})$ and $qcl(\mathbb{Q}(2^{1/3}))$. First, find an $S\_4$-extension $K/\mathbb{Q}$ containing $\mathbb{Q}(2^{1/3})$. (My friend Darrin Doud tells me that the Galois closure of adjoining a root of $x^6 + 3x^4 + 3x^2 + 3$ will be such an $S\_4$ extension.) Note that $K$ is a subfield of $qcl(\mathbb{Q}(2^{1/3}))$, being constructible from $2^{1/3}$ by a sequence of square roots.
Now, $K$ has a unique subfield $\mathbb{Q}(\sqrt{-3})$ quadratic over $\mathbb{Q}$. Then $K/\mathbb{Q}(\sqrt{-3})$ is an $A\_4$-extension, and hence there exists a non-Galois subfield $F$ with $[F:\mathbb{Q}(\sqrt{-3})]=4$. Note, in particular, that the Galois closure of $F$ requires a degree $3$ extension. We know that $F$ is not contained in $qcl(\mathbb{Q})$ because $qcl(\mathbb{Q})$ is Galois but contains no cubic extension of $\mathbb{Q}$. Also, $qcl(F)$ is proper in $qcl(\mathbb{Q}(2^{1/3}))$, since $qcl(F)$ is formed only using power-of-2 extensions of $\mathbb{Q}$.
**Part 2:** The order is not dense. Just use Zorn's lemma (or transfinite induction, if you want to avoid the axiom of choice) to construct a quadratically closed subfield of $qcl(\mathbb{Q}(2^{1/3}))$ that is maximal with respect to not containing $2^{1/3}$.
**Part 3:** If by "geometrical figure" you mean a *finite* set of points, then I don't know the answer to the first question anymore. (Edited to add: I've now also answered this question in the negative. I've put it as a separate answer.)
|
19
|
https://mathoverflow.net/users/3199
|
409580
| 167,645 |
https://mathoverflow.net/questions/409045
|
3
|
$\DeclareMathOperator\ora{ora}$Let $A\_0$ be the adjacency matrix of graph $G$ and $P\_0$
permutation matrix of multiplicative order $\rho$.
Let $X$ be positive integer and $B\_0=P\_0^X A\_0 P\_0^{-X}$.
>
> Q1 Given $A\_0,P\_0,B\_0$ can we find $X$ efficiently?
>
>
>
Positive answer need not mean graph isomorphism is efficient,
since in GI we don't know $P\_0$.
Related problem:
Given $P\_0$ and $C\_0=P\_0^Y$ we can find $Y$ efficiently since
$\rho$ is divisor of factorial $n$ and $n$-smooth.
Another relation:
Assume $\ora(U)$ is an oracle which computes $P\_0^{U X} A\_0 P\_0^{-U X}$
given integer $U$.
Assume $X=d d'$ where $d$ is factor of $\rho$.
Then $\ora(\frac{\rho}{d})=P\_0^{\rho d'} A\_0 P\_0^{-\rho d'}=A\_0$ and we have information about $X$.
Repeat the same idea for other factors of $\rho$ and multiplying
by small powers of $P\_0$ to get congruences and $X$ via crt.
In an [answer](https://mathoverflow.net/questions/408757/diffie-hellman-cryptography-based-on-graph-isomorphism) Joseph Van Name appears to claim the following
which is very close to computational diffie-hellman problem (CDH).
Given $P\_0^X A\_0 P\_0^{-X}$ and $P\_0^Y A\_0 P\_0^{-Y}$, he can efficiently
compute $P\_0^{X+Y} A\_0 P\_0^{-X-Y}$
Setting $X=Y$, he can double $P\_0^X, P\_0^{2X},P\_0^{4X} \ldots P\_0^{2^kX}$ and by repeated doublings,
he can compute $\ora(U)$.
>
> Q2 Does Joseph's attack works for Q1?
>
>
>
>
> Q3 Can we drop knowing $P\_0$ and find $P\_0^X$ efficiently from only $A\_0,B\_0$?
>
>
>
|
https://mathoverflow.net/users/12481
|
Relation graph isomorphism to discrete logarithm
|
Yes. We can efficiently find all integer solutions $X$ to the equation $B\_{0}=P\_{0}^{X}A\_{0}P\_{0}^{-X}$ since we can either conclude that there is no integer $X$ with $B\_{0}=P\_{0}^{X}A\_{0}P\_{0}^{-X}$ or we can show that $B\_{0}=P\_{0}^{X}A\_{0}P\_{0}^{-X}$ precisely when $X$ satisfies a system of linear congruence equations.
Suppose that the permutation corresponding to $P\_{0}$ can be written as a product of $r$ disjoint cycles of lengths $n\_{1},\dots,n\_{r}$. Without loss of generality, we may therefore assume that $P\_{0}$ is a block matrix $(P\_{i,j})\_{1\leq i\leq r,1\leq j\leq r}$ where each entry $P\_{i,j}$ is the $n\_{i}\times n\_{j}$-zero matrix whenever $i\neq j$ and where each $P\_{i,i}$ is an $n\_{i}\times n\_{i}$-permutation matrix corresponding to the cycle permutation.
Let us make $A\_{0},B\_{0}$ as well into block matrices. Suppose that $A\_{0}=(A\_{i,j})\_{1\leq i\leq r,1\leq j\leq r}$ and
$B\_{0}=(B\_{i,j})\_{1\leq i\leq r,1\leq j\leq r}$ where each $A\_{i,j},B\_{i,j}$ is an $n\_{i}\times n\_{j}$-matrix.
Then $$P\_{0}^{X}A\_{0}P\_{0}^{-X}=(P\_{i,i}^{X}A\_{i,j}P\_{j,j}^{-X})\_{1\leq i\leq r,1\leq j\leq r}.$$ Therefore, $P\_{0}^{X}A\_{0}P\_{0}^{-X}=B\_{0}$ if and only if
$B\_{i,j}=P\_{i,i}^{X}A\_{i,j}P\_{j,j}^{-X}$ for each $i,j$.
If there exists $i,j$ where $B\_{i,j}\neq P\_{i,i}^{X}A\_{i,j}P\_{j,j}^{-X}$ for all integers $X$ with $0\leq X<\text{Lcm}(m\_{i},m\_{j})$, then there does not exist an integer $X$ with $B\_{i,j}\neq P\_{i,i}^{X}A\_{i,j}P\_{j,j}^{-X}$, and therefore the equation $B\_{0}=P\_{0}^{X}A\_{0}P\_{0}^{-X}$ has no solution.
On the other hand, if for each pair $i,j$, there exists an integer $X\_{0}$ with $0\leq X\_{0}<\text{Lcm}(m\_{i},m\_{j})$ where $B\_{i,j}=P\_{i,i}^{X\_{0}}A\_{i,j}P\_{j,j}^{-X\_{0}}$, then
$$H=\{Y\in\mathbb{Z}\mid B\_{i,j}=P\_{i,i}^{X}B\_{i,j}P\_{j,j}^{-X}\}$$ is a subgroup of $\mathbb{Z}$ of index at most $\text{Lcm}(m\_{i},m\_{j})$. If $v\_{i,j}$ is the index of $H$, and $u\_{i,j}=X\_{0}$, then
$B\_{i,j}=P\_{i,i}^{X}A\_{i,j}P\_{j,j}^{-X}$ if and only if $X=u\_{i,j}\mod v\_{i,j}.$ Therefore, $P\_{0}^{X}A\_{0}P\_{0}^{-X}=B\_{0}$ if and only if
$X=u\_{i,j}\mod v\_{i,j}$ for $1\leq i\leq r,1\leq j\leq r.$ One can easily find the (possibly empty) set of all integer solutions $X$ to the equation $B\_{0}=P\_{0}^{X}A\_{0}P\_{0}^{-X}$.
|
4
|
https://mathoverflow.net/users/22277
|
409594
| 167,649 |
https://mathoverflow.net/questions/409604
|
4
|
I hope to ask what the outer automorphism group of the Lie group $\text{SL}\_2(\mathbb{R})$ is, just as an abstract group. It seems like Dieudonné's paper *On the automorphisms of the classical groups* explicitly left out the case $\text{SL}\_n(K)$ for $n=2$. Hua's appendix solves it for $\text{SL}^\pm\_2(\mathbb{R})$ which according to Hua is not the same as the case $\text{SL}\_2(\mathbb{R})$, since -1 is not in the commutator subgroup of $\mathbb{R}$. I will appreciate any help or any reference.
related: [What is the outer automorphism group of SL(2,Fq)?](https://mathoverflow.net/questions/348440/what-is-the-outer-automorphism-group-of-operatornamesl2-mathbbf-q)
|
https://mathoverflow.net/users/141136
|
What is the outer automorphism group of the Lie group $\text{SL}_2(\mathbb{R})$ as an abstract group?
|
It is a standard fact that the automorphism group of $G=\mathrm{SL}\_2(\mathbf{R})$ as topological group equals $\mathrm{PGL}\_2(\mathbf{R})$, which is also the automorphism group of the Lie algebra viewed as $\mathbf{R}$-algebra. In particular, the outer automorphism group as topological group is cyclic of order 2.
>
> I claim that this is the same as abstract group. Namely, every automorphism is continuous. So, as an abstract group, the automorphism group of $G=\mathrm{SL}\_2(\mathbf{R})$ is reduced to $\mathrm{PGL}\_2(\mathbf{R})$, and hence its outer automorphism group is cyclic of order 2.
>
>
>
Note: this is highly false for $\mathrm{SL}\_2(\mathbf{C})$: while its outer automorphism group as topological group is cyclic of order 2 (induced by complex conjugation), its automorphism group as abstract group has cardinal $2^{\mathfrak{c}}$ (use automorphisms of the abstract field $\mathbf{C}$).
---
Notation: for $h$ in a group $H$, let $C\_h$ be the centralizer of $h$ in $H$, and $h$ the normalizer of $h$. Write $X\_H$ as the set of $h\in H$ such that $C\_h$ is maximal among solvable subgroups of $H$.
**Lemma 1:** for $g\in G=\mathrm{SL}\_2(\mathbf{R})$, we have $g\in X\_G$ iff $g$ is non-central elliptic (i.e. $|\mathrm{Trace}(g)|<2$).
Proof: direct verification. If $g$ is central then its centralizer is not solvable. If $g$ is non-central diagonalizable, or $\pm g$ is unipotent, then its centralizer is contained in an upper triangular group. Conversely if $g$ is non-central elliptic, its centralizer is conjugate to $\mathrm{SO}(2)$, which is maximal solvable in $G$.
**Lemma 2.** Let $T$ be the upper triangular group. Let $H$ be a maximal solvable subgroup of $G$. Suppose that $H$ is infinite and not virtually abelian. Then $H$ is conjugate to $T$.
Proof: then $H$ is Zariski closed. So its unit component has finite index, hence is non-abelian and infinite. So it is conjugate to $T$.
**Lemma 3:** all maximal abelian subgroups of $T$, except $[T,T]$, are conjugate.
Proof: easy standard elementary verification (also follows by conjugation of maximal tori in solvable algebraic groups).
**Lemma 4:** let $F$ be a subset of $G$. Then $F$ is bounded if and only if there exist $g,h\in X\_G$ and $n$ such that $F\subset (C\_gC\_h)^n$.
Proof: if $g\in X\_G$ then $C\_g$ is compact. So the condition implies that $F$ is bounded. Also, if $g,h\in X\_G$ don't commute, $C\_g\subset C\_h$ is a compact generating subset of $G$ and Baire's theorem implies the result.
Now let $f$ be an automorphism. Let us prove that $f$ is continuous. It is enough to prove that $f$ is continuous at $1$. Suppose $g\_n\to 1$. By Lemma 4, $f$ preserves boundedness, so $(f(g\_n))$ is bounded. Supposing by contradiction that $f(g\_n)$ doesn't converge to $1$, we can suppose, after extraction, that it converges to $g\neq 1$. There exist $i\_n\to\infty$ such that $g\_n^{i\_n}$ also tends to $1$. So $f(g\_n)^{i\_n}$ is bounded as well. This already implies that $|\mathrm{Trace}(g)|\le 2$.
Let $L$ be the set of $g$ that occur as limit of $f(g\_n)$ for $g\_n$ tending to $1$. This is a normal subgroup. By the previous paragraph, it consists of matrices of $|\mathrm{Trace}|\le 2$. Hence it is either identity, or $\{\pm 1\}$. In the first case, we are done.
In the other case, this already implies that the $f$ induces a continuous automorphism of $\mathrm{PSL}\_2(\mathbf{R})$. Hence, after composing by a continuous automorphism, we can suppose that $f$ induces the identity on $\mathrm{PSL}\_2(\mathbf{R})$. Hence $f(g)=e(g)g$ for all $g$ with $e(g)\in\{\pm 1\}$. Then $e$ is a homomorphism $\mathrm{SL}\_2(\mathbf{R})\to\{\pm 1\}$. Since $\mathrm{SL}\_2(\mathbf{R})$ is abstractly a perfect group, we have $e=1$. So $f$ is the identity and the proof is completed.
|
6
|
https://mathoverflow.net/users/14094
|
409608
| 167,654 |
https://mathoverflow.net/questions/409613
|
1
|
Let $\rho\_1, \rho\_2 \in L^1(\Omega;\mathbb R\_+)$ such that $\int \rho\_i|\ln \rho\_i| < \infty$. Is it true that there exists a constant $C>0$ such that
\begin{align\*}
\int\_\Omega \left(\rho\_{1} \ln \frac{\rho\_{1}}{\rho\_{2}}\right) d x d y \leq C\int\_\Omega |\rho\_1 - \rho\_2| d x d y
\end{align\*}
holds?
|
https://mathoverflow.net/users/nan
|
Prove $\int_\Omega \left(\rho_{1} \ln \frac{\rho_{1}}{\rho_{2}}\right)dx dy \leq C\int_\Omega |\rho_1-\rho_2|dxdy$ for $0 \le \rho_1, \rho_2 \in L^1$
|
Certainly not, if $C$ is supposed to be real. For instance, suppose that $\rho\_1$ and $\rho\_2$ are probability densities such that $\rho\_1\rho\_2=0$. Then the left-hand side of your inequality is $\infty$, whereas its right-hand side is $2C$.
|
1
|
https://mathoverflow.net/users/36721
|
409614
| 167,655 |
https://mathoverflow.net/questions/406249
|
20
|
Let $f(x)=\log|2\sin(x/2)|$ (the normalizing factor $2$ is chosen to have the average over the period equal to $0$). Does there exist $a>0$ such that all sums $\sum\_{k=1}^n f(ak)\ge 0$? The computations (run up to the values of $n$ where I could not rely on the floating point precision any more) show that it may be the case even for $a=2\pi\sqrt 2$ but I'll be happy with any $a$ (or with a proof that no such $a$ exists).
This question came up in our joint study of Leja interpolation points with Volodymyr Andriyevskyy but the connection is a bit too convoluted to be explained here :-)
|
https://mathoverflow.net/users/1131
|
Can all partial sums $\sum_{k=1}^n f(ka)$ where $f(x)=\log|2\sin(x/2)|$ be non-negative?
|
The answer to this question is yes, which is proved in the following [paper](https://arxiv.org/abs/2111.12974) that appeared on arXiv today. Specifically, part $(ii)$ of Theorem 2 of this paper, in the case $b = 1$, says that for $\beta = \frac{\sqrt{5}-1}{2}$ the smallest term of the sequence
$$P\_N(\beta) = \prod\limits\_{r = 1}^N 2|\sin(\pi r\beta)|$$
is the first term $P\_1(\beta)$ (this is the same as the sequence in OP up to denoting $a = 2\pi \beta$ and taking the exponent). Since
$$P\_1(\beta) = 2|\sin(\pi \beta)| = 1,86\ldots > 1,$$
the sequence in the OP for $a = 2\pi \beta$ is strictly positive and even has a uniform poisitve lower bound $\log(1,86\ldots) = 0,62\ldots$ .
Note also that the case $b = 2$ of the same Theorem gives us $\beta = \sqrt{2}-1$. By periodicity of sine, it's the same as $\beta' = \sqrt{2}$, which translates to $a = 2\pi \sqrt{2}$ from the OP. So, for this $a$ infimum is also the first term, which is also strictly positive.
|
7
|
https://mathoverflow.net/users/104330
|
409618
| 167,657 |
https://mathoverflow.net/questions/409586
|
1
|
I would like to understand the irreducible components of a projective algebraic set.
Given an irreducible and homogeneous polynomial $H(w,x,y)\in \mathbb{C}[w,x,y]$ we define
$H\_i(w,x\_0,x\_i):=H(w,x\_0,x\_i)\in \mathbb{C}[w,x\_0,x\_1,\dotsc,x\_n]$ and the projective algebraic set $Z(H\_1,\dotsc,H\_n)\subseteq \mathbb{P}^{n+1}$.
How many irreducible components of dimension one does this set have? Are all of them isomorphic?
Does $H$ give some informations about the function field of those curves?
What I suppose is that there should be some symmetries between the these curves, but I don't know how to attack this problem.
You may know some references dealing with the same kind of questions (maybe some intersection theory).
This problem arises from the following:
I'm interested in finding explicitly non trivial embeddings of curves in a higher dimensional projective space.
(By trivial embedding I mean $Z(H(w,x\_0,x\_1),x\_2−x\_1,\dotsc,x\_n−x\_1)$, I would like the curve to "spread" among all coordinates.)
|
https://mathoverflow.net/users/155306
|
Irreducible components of a projective variety
|
We have either $H(0,0,1)=0$ or $H(0,0,1) \neq 0$. In the first case, the locus $w=x\_0=0$ is contained in $Z(H\_1,\dots, H\_n)$ and is either an irreducible component of dimension $n-1$ or contained in an irreducible component of dimension $n$, and in the second case the locus $w=x\_0=$ does not intersect $Z(H\_1,\dots, H\_n)$, so in either case we can ignore this locus, at least as long as $n \geq 3$. (If $n=2$, it might provide an irreducible component of dimension 1).
After removing this locus, $Z(H\_1,\dots, H\_n)$ maps to $\mathbb P^1$ by the projection with coordinates $(w:x\_0)$. This projection is the fiber product of $n$ copies of the map $Z(H) \to \mathbb P^1$ with coordinates $(w:x)$. So we are looking at an $n$-fold fiber product of a map of curves.
If any fiber of the map $Z(H) \to \mathbb P^1$ has positive dimension, i.e. if $Z(H)$ contains a line of the form $aw+bx=0$, then the fiber product will contain the $n$'th power of this line, an irreducible component of dimension $n$.
Ignoring these components, we are taking the $n$-fold fiber product of a flat morphism of curves, which is a scheme of dimension $1$, so every irreducible component has dimension $1$. To calculate the irreducible components, we may restrict to an open set where the morphism is finite étale. Then we may calculate components by Grothendieck's Galois theory - the étale morphism $Z(H) \to \mathbb P^1$ corresponds to a finite set with an action of the fundamental group of this open subset, and each irreducible component corresponds to an orbit of this group action on the $n$'th power of the finite set.
So the exact number of components depends on the degree of this covering and its monodromy group (inside the symmetric group of that degree).
There certainly are symmetries among the components, and it depends entirely on the combinatorics of this monodromy group action. For example, if the degree is $d$ and the monodromy group is the full symmetric group $S\_d$, then components are governed by $S\_d$-orbits on $n$-tuples of choices from a set of $d$ letters, which are the same thing as partitions of $\{1,\dots,n\}$ into at most $d$ parts, and two components are birational if their partitions have the same numbers of parts.
|
3
|
https://mathoverflow.net/users/18060
|
409622
| 167,660 |
https://mathoverflow.net/questions/409633
|
3
|
Let $\pi:X\rightarrow W$ be a morphism of smooth projective varieties over a field $k$ whose generic fiber is a smooth quadric, and let $r$ be the dimension of the fibers of $\pi$.
Does there always exists a rank $r+2$ vector bundle $f:\mathcal{E}\rightarrow W$ on $W$ such that $X$ can be embedded in $\mathbb{P}(\mathcal{E})$ as a divisor and $X\cap f^{-1}(w) = \pi^{-1}(w)$ for all $w\in W$?
When $r=1$ this is true. Indeed, $\omega\_{X}^{-1}$ is relatively very ample and we can take $\mathcal{E} = \pi\_{\*}\omega\_{X}^{-1}$. Does there exist a similar construction for $r\geq 2$?
In this paper
ARNAUD BEAUVILLE, Variétés de Prym et jacobiennes intermédiaires,
Annales scientifiques de l’É.N.S. 4e série, tome 10, no 3 (1977), p. 309-391
A quadric bundle over $\mathbb{P}^2$ is defined as a smooth projective variety $X$ with a morphism $\pi:X\rightarrow\mathbb{P}^2$ whose fibers are isomorphic to $r$-dimensional quadrics.
According to Proposition $1.2$ there exists a vector bundle $\mathcal{E}\rightarrow\mathbb{P}^2$ of rank $r+2$ and a form $q\in H^0(\mathbb{P}^2,Sym^2\mathcal{E}(h))$ such that $X$ identifies with the zero locus of $q$ in the projective bundle $\mathbb{P}(\mathcal{E})\rightarrow\mathbb{P}^2$.
|
https://mathoverflow.net/users/14514
|
Embedding quadric bundles
|
No. For instance, take your favorite $\mathbb{P}^1$-bundle $Y \to W$ which is not a projectivization of a rank 2 vector bundle and set
$$
X := Y \times \mathbb{P}^1.
$$
This is a quadric surface bundle over $W$, but if there is an embedding $X \hookrightarrow \mathbb{P}(\mathcal{E})$, then its restriction to any fiber of $X$ over $\mathbb{P}^1$ would give a rational section of $Y \to W$, which is impossible.
|
5
|
https://mathoverflow.net/users/4428
|
409635
| 167,664 |
https://mathoverflow.net/questions/409643
|
4
|
Suppose we have an $(\infty,1)$-category $\mathcal{C}$. There are two ways I can think of to produce an $(\infty,0)$-category from $\mathcal{C}$, and I'm wondering if they're equivalent.
The first way is as follows. Let $\operatorname{Cat}$ be the $\infty$-category of $\infty$-categories, and let $\operatorname{Grpd}$ be the $\infty$-category of $\infty$-groupoids. There is a forgetful functor $F: \operatorname{Grpd} \to \operatorname{Cat}$ which admits a left adjoint $Str: \operatorname{Cat} \to \operatorname{Grpd}$, which one could call the "groupoidification functor". Roughly, $Str(\mathcal{C})$ inverts all the arrows in $\mathcal{C}$.
The second method starts by viewing $\mathcal{C}$ as a category object in the infinity category $\operatorname{Grpd}$. More precisely, there is a fully faithful functor $s: \operatorname{Cat} \to Fun(\Delta^{op},\operatorname{Grpd})$, which assigns to $\mathcal{C}$ the simplicial object
$$s \mathcal{C}: \Delta^{op} \longrightarrow \operatorname{Grpd}$$
whose space of zero simplices is the core of $\mathcal{C}$, its $1$-simplices is the space of morphisms in $\mathcal{C}$, and so on. By taking the geometric realization of $s\mathcal{C}$ we get an $\infty$-groupoid $|s\mathcal{C}|$.
Question: Is the composition $|-| \circ s$ equivalent to the groupoidification functor? This seems intuitively obvious to me; e.g. two $0$-simplices of $s\mathcal{C}$ are identified in $|s\mathcal{C}|$ whenever they're joined by a $1$-simplex.
EDIT: I replaced all instances of "strictification" with "groupoidification," in order to avoid any confusion for future readers.
|
https://mathoverflow.net/users/101861
|
Groupoidification of infinity categories and geometric realization
|
Yes, they are equivalent, and this is why people sometimes use $|C|$ to denote $Str(C)$.
Consider the following composite $Fun(\Delta^{op},\mathrm{Grpd}) \to Fun^{cpl, Segal}(\Delta^{op}, \mathrm{Grpd}) \to Cat\_\infty \to \mathrm{Grpd}$ where the first map is the left adjoint to the inclusion of complete Segal spaces, the second is the equivalence between complete Segal spaces and $\infty$-categories, and the last one is $Str$.
I claim that this map is given by geometric realization, i.e. $\mathrm{colim}\_{\Delta^{op}}$. For this, because all the maps appearing are left adjoints, it suffices to show that the composite of right adjoints is equivalent to the constant functor.
But the string of right adjoints sends an $\infty$-groupoid $X$ to the $[n]\mapsto map([n],X)$, which is constant equivalent to $X$ - indeed, because $|\Delta^{op}|$ is contractible, to show that a simplicial groupoid is constant it suffices to show that it sends all maps $[0]\to [n]$ in $\Delta^{op}$ to equivalences, but $map([0],X)\to map([n],X)$ is an equivalence, as the map of $\infty$-categories $[n]\to [0]$ induces an equivalence $Str([n])\to Str([0])$.
In particular this shows that the equivalence $map([\bullet ], X)\simeq X$ is natural in $X$ as well, as it is induced by a natural transformation in $\Delta^{op}$.
It follows that if you precompose it with the inclusion $Fun^{cpl,Segal}(\Delta^{op},\mathrm{Grpd})\to Fun(\Delta^{op},\mathrm{Grpd})$, you get exactly $\mathrm{colim}\_{\Delta^{op}}$ , but because the composite $Fun^{cpl,Segal}(\Delta^{op},\mathrm{Grpd})\to Fun(\Delta^{op},\mathrm{Grpd})\to Fun^{cpl, Segal}(\Delta^{op},\mathrm{Grpd})$ is equivalent to the identity, this composite is also equivalent to $Str$, which proves the claim.
|
6
|
https://mathoverflow.net/users/102343
|
409644
| 167,666 |
https://mathoverflow.net/questions/409491
|
4
|
Let $A$ be a non-separable reflexive Banach algebra. Every separable subspace of $A$ is contained in a separable 1-complemented subspace [[Lindenstrauss,1966](https://doi.org/10.1090/S0002-9904-1966-11606-3)]. It is straightforward to show that every *separable subalgebra* is contained in a separable subalgebra $W$ of the form $$W=\bigcup\_{n=1}^{\infty} E\_n$$ where $(E\_n)$ is an increasing sequence of separable 1-complemented subspaces of $A$.
**Question:** Is it true that every separable subalgebra of $A$ is contained in a separable complemented subalgebra?
|
https://mathoverflow.net/users/164350
|
Separable subalgebras of non-separable reflexive Banach algebras
|
I am inclined to say *yes*. This is because reflexive Banach spaces have [projectional resolutions of the identity](https://www.cambridge.org/core/services/aop-cambridge-core/content/view/80C18D3F91097CD50DE5EA763C6BEE08/S0004972700013344a.pdf/on-projectional-resolution-of-identity-on-the-duals-of-certain-banach-spaces.pdf), which are increasing ordinal-indexed nets of contractive, commuting projections that exhaust the whole space and satisfy certain compatibility conditions. From this, it follows that separable subspaces of reflexive Banach space are contained in 1-complemented separable subspaces as you mentioned, but we got actually more.
Now, let $A$ be a separable subalgebra of a reflexive Banach algebra $B$. Denote by $\langle S \rangle$ denote the closed subalgebra generated by $S\subset B$. Take $A^1$ to be a separable subspace of $B$ containing $A$ that is 1-complemented by a projection $P^1$ from some fixed PRI and set $A\_1 = \langle A^1 \rangle$. Let $A^2$ be a separable subspace of $B$ containing $A\_1$ that is 1-complemented by a projection $P^2$ from the same PRI and set $A\_2 = \langle A^2 \rangle$. Continue this process recursively so that you get intertwined sequences $A^1 \subset A\_1 \subset A^2 \subset A\_2 \ldots$. In particular, the unions of $(A\_n)$ and $(A^n)$ have the same closure, call it $A\_\omega$. Readily, $A\_\omega$ is a closed subalgebra being the closure of an increasing chain of subalgebras. Since projections from a PRI commute, $(P^n)$ converge in the weak operator topology to a projection, which looks like a projection onto $A\_\omega$.
|
4
|
https://mathoverflow.net/users/15129
|
409646
| 167,667 |
https://mathoverflow.net/questions/409649
|
1
|
I have a smooth projective surface $X$, and two flat family of elliptic curves on it: $E\_{1,t}$ and $E\_{2,t}$, (I don't know what either $t$ runs through!) such that
(1), for any i={1，2}, the closed points of $X$ are the disjoint union of closed points of all $E\_{i,t}$.
(2), the intersection number of $E\_{1,t}$ and $E\_{2,t}$ is always 1.
Can we conclude that $X$ is an abelian surface?
|
https://mathoverflow.net/users/177957
|
Characterization of an Abelian surface
|
I think that the answer is yes, at least if you are working over an algebraically closed field of char $0$. Let me try to give an argument, I hope that there are no mistakes.
You have two flat families $E\_1 \to C\_1$ and $E\_2 \to C\_2$ of elliptic curves over some unknown bases $C\_1, C\_2$.
The first assumption says that the map $E\_1 \to X$ induces a bijection on closed points. If the field is of characteristic 0, one can use Zariski's main theorem to conclude that $E\_1 \to X$ is an isomorphism of varieties (see e.g. [Bijection implies isomorphism for algebraic varieties](https://mathoverflow.net/questions/264204/bijection-implies-isomorphism-for-algebraic-varieties)). The same holds for $E\_2 \to X$. Using these identifications $X \cong E\_1$ and $X \cong E\_2$, we get two flat projections $X \to C\_1$ and $X \to C\_2$, where the fibers are elliptic curves. In particular $C\_1$ and $C\_2$ must be smooth (by flat descent of smoothness).
I am interpreting the assumption that the intersection numbers are always 1 to mean that the fibers of the induced morphism $X \to C\_1 \times C\_2$ are singleton points, i.e. any two fibers of the families $E\_1$ and $E\_2$ respectively intersect in a single point (is this the precise condition you have in mind?). Since $C\_1 \times C\_2$ is smooth, this implies again that $X \to C\_1 \times C\_2$ is an isomorphism. But now by choosing a closed point $x\_1 \in C\_1$, we see that $C\_2 = x\_1 \times C\_2$ is isomorphic to the fiber of $E\_1 \to C\_1$ at $x\_1$. This means that $C\_1$ is an elliptic curve. The same holds for $C\_2$. Therefore $X \cong C\_1 \times C\_2$ is a product of elliptic curves.
|
1
|
https://mathoverflow.net/users/339730
|
409652
| 167,668 |
https://mathoverflow.net/questions/393898
|
25
|
I remember having read, about 15 years ago, a transcript of a lecture given by Richard K. Guy, titled "How not to be a graduate student". He gave lots of advice, mostly humorous, concealing sharp and deep observations.
Unfortunately, I haven't been able to locate this text on the Internet; I have the feeling I had read it in a newsgroup list, and these things don't seem to exist anymore.
Does anyone perchance have a copy of this?
|
https://mathoverflow.net/users/10481
|
Looking for source: "How not to be a graduate student"
|
I was able to locate it, \TeX ed it, and uploaded it to <https://www.math.uni-sb.de/ag/bartholdi/pub/Guy-NotGrad.pdf>
|
25
|
https://mathoverflow.net/users/10481
|
409664
| 167,670 |
https://mathoverflow.net/questions/409660
|
0
|
A Banach algebra $A$ is a *dual Banach algebra* if it is a dual Banach space with a (not necessarily unique) predual $A\_{\ast}$, and the multiplication on $A$ is separately weak\*-continuous. Dual Banach algebras are naturally analogous to von Neumann algebras [[Daws2007](https://doi.org/10.4064/sm178-3-3), [Daws2011](https://doi.org/10.3318/PRIA.2011.111.1.3)]; thus, one may expect a given dual Banach algebra to possess a rich structured subset of idempotents. (e.g., structured as explained in [[Laustsen2003](https://doi.org/10.1017/S0017089502008947), [DawsHorvath2021](https://doi.org/10.4153/S0008414X20000565)] )
**Question:** Let $E$ be a reflexive Banach space and $A\subseteq B(E)$ be a dual Banach algebra. Does $A$ contain *projectional skeletons* (e.g. [[Kubis2009](https://doi.org/10.1016/j.jmaa.2008.07.006)] ) or *projectional resolutions of identity* ?
|
https://mathoverflow.net/users/164350
|
Projectional skeletons in dual Banach algebras
|
No. $A = \ell\_\infty$ is a dual Banach algebra. Every separable complemented subspace of $A$ is finite-dimensional, so there is no way to exhaust $A$ by nicely complemented separable subspaces.
|
4
|
https://mathoverflow.net/users/15129
|
409666
| 167,671 |
https://mathoverflow.net/questions/409663
|
4
|
Excuse my naive question and please let me explain it:
In everyday life we experience 3 spatial "dimensions" + time etc.
Usually the 3 dimensions are represented by a coordinate system and mathematically as the vector space $R^3$.
In constrast, the word "dimension" in dimensional analysis has a more objective realism which is based on physical measurements. For instance let $l$ be a transcendental number.
Then $1,l,l^2,l^3,\cdots$ will be *linearly independent* over the rational numbers, and hence can serve as a basis in a vector space. For instance, instead of saying the point in spatial dimension has coordinates $(x,y,z)$ we might as well write:
$$x+y \cdot L + z \cdot L^2$$
where $L$ represents a "transcendental number over the rationals" and is associated with meter or kilometer etc.
Take for instance the derived unit $s/t^k$ for $k=0,1,2$ where $s$ denotes length and $t$ denotes time.
Then $s$=length, $s/t =$ velocity, $s/t^2=$ acceleration. If we view these as a basis of a vector space, then this vector space has 3 dimensions. But no one would count subjectively these as dimensions.
On the other hand the seven basic SI-units (or 6 if you do not want to count $mol$), could be seen as a transcendental numbers over the rationals or reals, and hence powers of those transcendental numbers could give a basis for vector spaces. Monomials of these transcendent numbers would correspond, as is done in dimensional analysis, to derived units. Adding and subtracting for instance $LT^{-1}+M$ would give a point in the "space of physical quantities" (**Edit**: and indicate the measured velocity and mass of an object). As a vector space over the reals this space has infinite dimension but has transcendence degree of $7$. Every derived / basic physical quantity measured by SI-Units would correspond to a point in this space / field of transcendence degree $6$ (or $7$ if you count $mol$ which I will not do):
$$\mathbb{Q}(T,L,M,I,\Theta,J)$$
Excuse my naive question: Is there any reason from physics to discard this point of view?
**Edit**:
It seems that the main idea of this question can be implemented through the ring (Laurent polynomial ring) $L:=\mathbb{R}[T,T^{-1},L,L^{-1},M,M^{-1},I,I^{-1},\Theta, \Theta^{-1},J,J^{-1}]$
**second Edit**:
I was asked to give at least one application of this idea, which I would like to do:
Application:
Let $k(x,y) = \frac{xy}{x^2+y^2-xy}$ for $x \neq 0,y \neq 0, x,y \in \mathbb{R}$ be a Jaccard-Similarity / positive definite kernel defined on $\mathbb{R}$.
We can define a similarity and positive definite kernel on the Laurent polynomial ring $L$ as :
$$K(x,y) := \frac{1}{N\_x + N\_y - N\_{xy}} \sum\_{X\_i^{\alpha\_i}=Y\_j^{\beta\_j}} k(a\_i,b\_j)$$
for $x = \sum\_{i} a\_i X^{\alpha\_i},y = \sum\_{j} b\_j X^{\beta\_j}$ and $X = (T , L, M, I, \Theta,J)$, $\alpha\_i, \beta\_j \in \mathbb{Z}^6$ and $X^{\alpha\_i},X^{\beta\_j}$ are multinomials, and $N\_x = $ number of nonzero $a\_i$, $N\_y =$ number of nonzero $b\_j$, $N\_{xy} =$ number of $(X^{\alpha\_i} = X^{\beta\_j})$.
Since $k(ca,cb) = k(a,b)$ for all $a,b,c \neq 0$, we deduce that $K(c \cdot x,c \cdot y) = K(x,y)$ for all $x,y \in L$, $c \neq 0, c \in \mathbb{R}$.
Hence this (or any other similarity and positive definite kernel $k$ with $k(ca,cb) = k(a,b)$. This is to make sure, that rescaling of physical units, does not change the similarity between objects.) gives us a possibility to measure the similarity / inner product
of two physical objects $A,B$ each of which is defined through measurements $x = \sum\_{i} a\_i X^{\alpha\_i}$ and $y = \sum\_{j} b\_j X^{\beta\_j}$.
Since there are different possibilities to measure similarites / define positive definite kernels, there should be different possibilities to define
equality / similarity between two physical objects $A$ and $B$.
Hence aposteriori $L$ is a Hilbert space by the Aaronszajn-Kolmogorov theorem.
Example:
The meaning of this kernel is to compare two physical objects. For instance $A = 10 m/s + 1 kg$, $B = 9 m/s + 2kg$, $C = 1 m/s^2+10kg$. Then $K(A,B) = 226/276 = 0.8278$, $K(A,C) = 10/91= 0.10989$, $K(B,C) = 5/21 = 0.2381$. Hence $A$ is most similar to $B$, $B$ is most similar to $A$, $C$ is most similar to $B$, and $A,C$ are the most dissimilar physical objects in this list.
|
https://mathoverflow.net/users/165920
|
Is the "space of physical quantities" a field of transcendence degree $6$ or $7$ over the rationals?
|
I don't think there are many meaningful situations where physical quantities of different dimension are added together (in fact, this is widely regarded as a taboo, and precisely the sort of mistake that dimensional analysis is supposed to prevent us from doing), so I don't think viewing the space of physical quantities as a field is the most fecund or useful description.
Instead, let me offer the following alternative point of view, which I think better matches what physics does and *why* it does it, and which is, at any rate, how I think of physical dimensions:
Consider the “**group of homogeneities**” $G = (\mathbb{R}\_+^\times)^k$ which is the product of $k$ copies of the positive real numbers, where $k$ is the number of “base units” (so $k=7$ in the SI system, but this depends on the sort of theory we wish to study: economists might have reasons to consider currencies or commodities as different units, whereas relativists will not consider time and length to be different units): the point is that $G$ acts on physical quantities by multiplying each one of the $k$ base quantities by the corresponding positive real number (e.g., $g=(2,3,1,\ldots)$ might multiply lengths by $2$, times by $3$ and preserve masses).
The crucial physics point is that: *insofar as the choice of units is arbitrary, $G$ acts as a group of physics-preserving symmetries* (e.g., multiplying all lengths by $2$ and times by $3$ while preserving masses should keep physics the same). So we seek quantities that are covariant under $G$. This is the whole point of dimensional analysis.
Now the group $G^\*$ of characters of $G$, i.e., the group of continuous morphisms $\lambda\colon G\to \mathbb{R}\_+^\times$ is the group $\mathbb{Z}^k$ with $(d\_1,\ldots,d\_k) \in \mathbb{Z}^k$ sending $g := (g\_1,\ldots,g\_k) \in G$ to $g\_1^{d\_1}\cdots g\_k^{d\_k}$. Really we should forget about $k$ and care about the group $G$ which might not have a clear “basis of dimensions” (e.g., is it electrical charge or electrical current which is the “base unit”? this is meaningless) and might even incorporate other transformations than positive-real-homogeneities (e.g., changes of sign on certain quantities might be physics-preserving).
So I propose that: **a “physical dimension” is an element $\lambda$ of $G^\*$ and a quantity of that dimension is an element of the corresponding representation $V\_\lambda$ of $G$** (which is $1$-dimensional because $G$ is abelian), i.e., $G$ acts on $V\_\lambda$ by $g \mapsto v \mapsto g^\lambda\cdot v$.
This $V\_\lambda$ is $1$-dimensional, so it is $\mathbb{R}$ as an $\mathbb{R}$-vector space, but it is so in a non-canonical way, and **a unit of measure for the physical dimension $\lambda$ is a basis for $V\_\lambda$** (probably a *positive* basis since $V\_\lambda$ still has a natural orientation).
Taking the **product** of two physical quantities of dimensions $\lambda$ and $\mu$ consists of taking the tensor product $V\_\lambda \otimes V\_\mu \buildrel\sim\over\to V\_{\lambda+\mu}$, while the inverse of a physical quantity $\lambda$ should be seen as living in the dual space $V\_\lambda^\vee \buildrel\sim\over\to V\_{-\lambda}$. The **sum**, on the other hand, is meaningful only insofar as it compatible under the action of $\lambda$, i.e., adding elements of the same $V\_\lambda$. (Of course, you are welcome to consider $V\_\lambda \oplus V\_{\lambda'}$, in some cases it might be meaningful, e.g., adding different currencies in economics, but the direct sum is then canonically split.)
What you propose to do, by taking arbitrary rational combinations of the base units and considering them as transcendental quantities, would be germane if the physically relevant acting group were the full Cremona group of automorphisms of rational fractions. I don't think we can make any physical sense of such an action, so this point of view is bound to remain very artificial.
**PS:** I'd like to add the following remark showing that the group of symmetries (so, in my description, $k$) might very well depend on the kind of theory being considered: Americans like to measure heights in *feet* and plane distances in *miles*, which makes some kind of sense in a situation where one might apply different homotheties on both; the “fundamental slope” of $5280\,\mathrm{ft}/\mathrm{mi}$ breaks this symmetry, exactly in the same way the “fundamental speed” known as the speed of light of $299\,792\,458\,\mathrm{m}/\mathrm{s}$ breaks the possibility of applying different homotheties on times and lengths: there is no fundamental difference between $5280\,\mathrm{ft}/\mathrm{mi}$ and $299\,792\,458\,\mathrm{m}/\mathrm{s}$ from the dimensional point of view: they are conversion factors breaking the possibility of choosing two units independently which, from the pre-breaking point of view, appears as a fundamental constant of nature (which can be measured experimentally: if the foot and the mile had been defined separately using a reference height and a reference plane length, the former would be an experimental quantity until redefinition of one of the units, exactly like the speed of light in the SI system).
|
14
|
https://mathoverflow.net/users/17064
|
409669
| 167,674 |
https://mathoverflow.net/questions/409639
|
4
|
Given any $n\in\mathbb N$, consider the [the Sylvester-Hadamard-Walsh matrix](https://en.wikipedia.org/wiki/Walsh_matrix) $M=(a\_{i,j})\_{i,j\in 2^n}$ of size $2^n\times 2^n$ and for a number $p\in[1,\infty)$, let
$$\nu\_{n,p}=\max\_{F\subseteq 2^n}\Big(\sum\_{j\in 2^n}\big|\sum\_{i\in F}a\_{i,j}\big|^p\Big)^{1/p}\quad\mbox{and}\quad \tilde \nu\_{n,p}=\frac1{2^{2^n}}\sum\_{F\subseteq 2^n}\Big(\sum\_{j\in 2^n}\big|\sum\_{i\in F}a\_{i,j}\big|^p\Big)^{1/p}.$$
For $p=2$, the Pithagoras Theorem and the orthogonality of the rows of the matrix $M$ imply that $\nu\_{n,2}=2^n$. Using this equality, it is easy to show that $\nu\_{n,p}=2^n$ for all $p\in[2,\infty)$.
If $p\in[1,2]$, then by the Holder inequality, we obtain
$$2^n\le\nu\_{n,p}\le 2^{n(\frac1p+\frac12)}.$$ In particular, $2^n\le\nu\_{n,1}\le 2^{3n/2}$. On the other hand, computer calculations show that $\tilde \nu\_{n,1}$ and $\nu\_{n,1}$ are much smaller than $2^{3n/2}$ (the values of $\tilde\nu\_{n,1}$ are calculated using the formula
$$\tilde\nu\_{n,1}=\frac1{2^{2^n}}\Big(\sum\_{i=0}^{2^n}{2^n\choose i}i+2(2^n-1)\sum\_{0\le i<j\le 2^n}{2^n\choose i}{2^n\choose j}(j-i)\Big)$$
of Alex Ravsky suggested in his comment):
$$
\begin{array}{c|c|c|c|c|c|c}
n&2^n&\tilde \nu\_{n,1}&\nu\_{n,1}&\lfloor 2^{3n/2}\rfloor&\tilde\nu\_{n,1}/2^{3n/2}\\
\hline
0&1&1&1&1&1\\
1&2&1.5&2&2&0.53...\\
2&4&4.25&6&8&0.53...\\
3&8&11.65...&14&22&0.51...\\
4&16&31.56...&40&64&0.49...\\
5&32&85.41...&\ge 96&181&0.47...\\
6&64&232.28..&??&512&0.45...\\
7&128&636.09...&??&1448&0.43...\\
8&256&1754.09...&??&4096&0.42...\\
9&512&4866.56...&??&11585&0.42...\\
\end{array}
$$
>
> **Problem 1.** Is $\nu\_{n,1}\ge\frac12 2^{3n/2}$? Is $\tilde\nu\_{n,1}\ge \varepsilon 2^{3n/2}$ for some $\varepsilon>0$?
>
>
> **Problem 2.** Is $\tilde\nu\_{n,1}=o(2^{3n/2})$? Is $\nu\_{n,1}=o(2^{3n/2})$?
>
>
> **Problem 3.** Find nontrivial lower and upper bounds on the number $$\lambda\_1=\limsup\_{n\to\infty}\frac1n\log\_2(\nu\_{n,1}).$$ Is $1<\lambda\_1<\frac32$?
>
>
>
|
https://mathoverflow.net/users/61536
|
The largest $\ell_p$-norm of a sum of rows of a Sylvester-Hadamard-Walsh matrix
|
$\newcommand{\tnu}{\tilde\nu}$Continuing Alex Ravsky's comment, we have
\begin{equation\*}
2^{2^n}\tnu\_{n,1}=\sum\_{j=0}^{2^n-1}S\_j=S\_0+(2^n-1)S\_1, \tag{1}
\end{equation\*}
where
\begin{equation\*}
S\_j:=\sum\_{F\subseteq[2^n]}\Big|\sum\_{i\in F}a\_{i,j}\Big|,
\end{equation\*}
and for each $j\ne0$
\begin{equation\*}
S\_j=S\_1=\sum\_{k,l=0}^M\binom Mk \binom Ml |k-l|,
\end{equation\*}
where
\begin{equation\*}
M=M\_n:=2^{n-1}. \tag{2}
\end{equation\*}
So,
\begin{equation\*}
S\_1=2^{2M} E|K-K'|, \tag{3}
\end{equation\*}
where, for each natural $n$, $K=K\_n$ and $K'=K'\_n$ are independent random variables (r.v.'s) each with the binomial distribution with parameters $M,1/2$. By the central limit theorem, the distribution of $(K-M/2)/\sqrt{M/4}$ converges weakly to the standard normal distribution (as $n\to\infty$), and hence $V\_n:=(K-K')/\sqrt{M/4}$ converges in distribution to $Z-Z'$, where $Z$ and $Z'$ are independent standard normal r.v.'s.
Also, $EV\_n^2=2<\infty$ for all $n$ and hence, by the [de la Vallée-Poussin theorem](https://en.wikipedia.org/wiki/Uniform_integrability#Relevant_theorems), the $V\_n$'s are uniformly integrable.
Therefore (see e.g. [Theorem 3.5, p. 31](http://cermics.enpc.fr/%7Emonneau/Billingsley-2eme-edition.pdf)), $E|V\_n|\to E|Z-Z'|$ and hence, by (3),
\begin{equation\*}
S\_1\sim 2^{2M} \sqrt{\frac M4}\;E|Z-Z'|=2^{2M} \sqrt{\frac M4}\;\frac2{\sqrt\pi}. \tag{4}
\end{equation\*}
Also,
\begin{equation\*}
S\_0=\sum\_{F\subseteq[2^n]}\sum\_{i\in F}1=\sum\_{k=0}^{2M}\binom{2M}k k=2^{2M}M. \tag{5}
\end{equation\*}
Collecting the pieces (1), (5), (4), and (2), we finally get
\begin{equation\*}
\tnu\_{n,1}\sim\frac1{\sqrt{2\pi}}\,2^{3n/2}.
\end{equation\*}
|
2
|
https://mathoverflow.net/users/36721
|
409686
| 167,675 |
https://mathoverflow.net/questions/409688
|
1
|
If $H = S\_n$ then then the [*fundamental symmetric polynomials*](https://en.wikipedia.org/wiki/Elementary_symmetric_polynomial) allow to write any $S\_n$-invariant polynomial $f$ as a polynomial expression of these elementary symmetric functions. In other words, $\mathbb{C}[x\_1, \dots ,x\_n]^{S\_n} = \mathbb{C}[e\_1, \dots ,e\_n]$.
Now, given $H \le S\_n$ a subgroup of the symmetric group, is there a general way to compute a system of invariants for $\mathbb{C}[x\_1, \dots , x\_n]^H$ ?
EDIT:
A brute-force approach to find invariants for $\mathbb{C}[x\_1, \dots,x\_n]^H$ might be the Reynolds Operator?
It is defined as:
$$R : \mathbb{R}[x\_1, \dots, x\_n] \rightarrow \mathbb{R}[x\_1, \dots, x\_n]^H$$
$R(f):= \frac{1}{|H|}\sum\_{g \in H}f(g \cdot \textbf{x})$.
So just for simplicity consider $\mathbb{Z}\_3 \le S\_3$ and we define a group action
\begin{align}
\mathbb{Z}\_3 \times V &\rightarrow V\\
(g, \textbf{x}) &\mapsto g \cdot \textbf{x} := (x\_1+g, \dots, x\_n +g)
\end{align}
where $V$ is a real vector space of dimension $d$.
So for example if $d=2$ and we have $f(x,y)=x+y$ then the $\mathbb{Z}\_3$ invariant version of $f$ might be
$$f\_{inv}(x,y)= f(x,y)+f(x+1,y+1)+f(x+2,y+2) = 3x+3y$$
Is this correct?
|
https://mathoverflow.net/users/470162
|
A question about finding a system of invariants for a subgroup $H$ of the symmetric group $S_n$
|
It depends what you mean by "compute. The ring $R$ of invariants is spanned as a vector space by symmetrized monomials $\sum\_{h\in H} h\cdot m$, where $m$ is a monomial.
$R$ is generated as a $\mathbb{C}$-algebra by those symmetrized monomials of degree at most $h$. However, there is no nice description of a minimal generating set or the relations (syzygies) among these generators, etc. Various algebra packages can make these computations for reasonably small groups. Some additional information is [here](https://math.mit.edu/~rstan/pubs/pubfiles/38.pdf). See Theorem 1.2 in particular. This paper deals with more general group actions than $H\leq S\_n$, but this extra condition does not affect the difficulty in finding a minimal set of generators.
|
3
|
https://mathoverflow.net/users/2807
|
409693
| 167,676 |
https://mathoverflow.net/questions/409676
|
1
|
Let $A$ and $B$ be two $n\times n$ hermitian matrices. Does $U^{\*}AU+B \prec\_{w} A+B$ for any unitary matrix $U$? Here the notation $``\prec\_{w}"$ stands for the weak majorization, that is, $x\prec\_{w} y$ if and only if $\sum\limits\_{j=1}^{k}\lambda\_{j}^{\downarrow}(x)\leq \sum\limits\_{j=1}^{k}\lambda\_{j}^{\downarrow}(y)$, for each $1\leq k\leq n$, and $\{\lambda\_{j}^{\downarrow}(x)\}\_{j=1}^{n}$ and $\{\lambda\_{j}^{\downarrow}(y)\}\_{j=1}^{n}$ are eigenvalues of $x$ and $y$ ordering in the non-increasing order respectively.
|
https://mathoverflow.net/users/129593
|
Weak majorizations for sum of two hermitian matrices
|
Consider $A = \left[\begin{matrix} 1&0 \\ 0&0\end{matrix}\right]$, $B = \left[\begin{matrix} 0&0 \\ 0&1\end{matrix}\right]$ and $U = \left[\begin{matrix} 0&1 \\ 1&0\end{matrix}\right]$
Then $A+B = \left[\begin{matrix} 1&0 \\ 0&1\end{matrix}\right]$ and $U^\*AU + B = \left[\begin{matrix} 0&0 \\ 0&2\end{matrix}\right]$.
Therefore, $U^\*AU + B \succ A+B$ and $U^\*AU + B \nprec\_w A+B$.
I guess the easy answer to your question is that when $U=I$ everything is fine. In general, most unitaries will not satisfy this weak majorization order.
|
0
|
https://mathoverflow.net/users/76593
|
409697
| 167,677 |
https://mathoverflow.net/questions/409645
|
25
|
I have a question that arose while reading Milnor's "Characteristic Classes". I will use the notation from that book.
Let $M$ be a smooth manifold and let $\zeta$ be a complex vector bundle on $M$. Milnor defines a connection on $M$ to be a map $\nabla\colon C^{\infty}(\zeta) \rightarrow C^{\infty}(\tau\_{\mathbb{C}}^{\ast} \otimes \zeta)$ satisfying the Leibniz identity, where $\tau\_{\mathbb{C}}$ is the complexified tangent bundle of $M$.
In Lemma 4 of Appendix C, he proves that such a connection can be extended to a map $\hat{\nabla}\colon C^{\infty}(\tau\_{\mathbb{C}}^{\ast} \otimes \zeta) \rightarrow C^{\infty}(\wedge^2 \tau\_{\mathbb{C}}^{\ast} \otimes \zeta)$ satisfying an appropriate Leibniz rule. However, his proof is just a definition in local coordinates, with the details left to the reader. I verified these details, though they were a bit of a pain.
However, I feel like there must be a more conceptual definition of $\hat{\nabla}$ that makes no reference to local coordinates. Does anyone know one?
|
https://mathoverflow.net/users/470094
|
Conceptual definition of the extension of a connection to 1-forms
|
If we denote by $\nabla$ the connection on $E\to M$, then we can define an exterior differential $d^\nabla:\Gamma(\Lambda^pM\otimes E)\to\Gamma(\Lambda^{p+1} M\otimes E) $ by
$$ d^\nabla \alpha (X\_0,\dots, X\_p) =
\sum\_i (-1)^i \nabla\_{X\_i}(\alpha(\tilde{X\_0}, \dots , \hat {\tilde{X\_i}}, \dots, \tilde X\_p))
+ \sum\_{i\neq j} -(1)^{i+j} \alpha ([\tilde X\_i, \tilde X\_j], X\_0, \dots, \hat X\_i,\dots, \hat X\_j, \dots, X\_p).$$
where $X\_i\in T\_x M$; $\tilde X\_i$ denotes an extension of $X\_i$ to a neighbourhood of $x\in M$, and the hat above something denotes that that argument has been omitted.
This formula can be found in Besse's book "Einstein Manifolds" pg 24, beware I recall there are a few typos in that part of the book.
The pattern for this definition is the usual one used to extend the covariant derivative to the tensor algebra, modified by alternating the result and using the covariant derivative.
|
19
|
https://mathoverflow.net/users/99042
|
409703
| 167,679 |
https://mathoverflow.net/questions/408676
|
6
|
This may be a naive question, but I have been unable to find a reference that answers it directly, at least at a level that I can understand. My intuition from physics is that non-ergodicity is typically associated with conserved quantities, which should be atypical in generic systems without special symmetries, but I'm curious if there is a more rigorous way to arrive at this result (or a counterargument!)
|
https://mathoverflow.net/users/151658
|
Are almost all measure-preserving flows on compact manifolds ergodic?
|
As it was mentioned before, KAM tells you that in the $C^r$-topology, for $r$ sufficiently large, ergodicity is not "typical".
For homeomorphism Oxtoby-Ulam proved [here](https://www.jstor.org/stable/1968772) that ergodicity is $C^0$ typical.
For the $C^1$-topology there is a nice recent result by Avila-Crovisier-Wilkinson [here](https://link.springer.com/article/10.1007/s10240-016-0086-4) which proves that for a $C^1$-typical volume preserving diffeomorphisms either you have zero (metric) entropy or you are ergodic. In this second case, they actually obtain something called non-uniformly Anosov (which gives more dynamical information).
In higher regularity ($r>1$) with more dynamical information something can be said. You can take a look at Pugh-Shub's conjecture for partially hyperbolic systems.
|
6
|
https://mathoverflow.net/users/117630
|
409706
| 167,680 |
https://mathoverflow.net/questions/409681
|
5
|
I keep running into the statement that "the generic k3 surface has Picard rank 1".
For instance the answer of [this question](https://mathoverflow.net/questions/124880/picard-group-of-a-k3-surface-generated-by-a-curve?newreg=99ae833537454a0eb215109d82c0525c) (end) and [this paper](https://chrome-extension://efaidnbmnnnibpcajpcglclefindmkaj/viewer.html?pdfurl=https%3A%2F%2Fwww.math.nyu.edu%2F%7Etschinke%2Fbooks%2Ffinite-fields%2Ffinal%2F02_elsenhans.pdf&clen=225439&chunk=true) (following Example 1.1) or [this paper](https://chrome-extension://efaidnbmnnnibpcajpcglclefindmkaj/viewer.html?pdfurl=https%3A%2F%2Farxiv.org%2Fpdf%2Falg-geom%2F9711031.pdf&clen=299222&chunk=true) (proof of Theorem 3.5) all state this fact without reference.
I simply cannot track down a reference for this (I tried e.g. Huybrechts) -- perhaps because it is very simple.
I would greatly appreciate a reference, or if very simple an argument.
|
https://mathoverflow.net/users/470144
|
Reference request: Generic k3 surface has Picard number 1
|
Welcome new contributor. I am just writing my comment as an answer, and expanding on the observation of Prof. Arapura. For a smooth, projective scheme $X$ over a field $k$, the space of first order deformations of $X$ as a $k$-scheme is naturally isomorphic to the $k$-vector space $H^1(X,T\_X)$.
For a locally free sheaf $E$ on $X$ of rank $r>0$, the "Atiyah extension" is an element in $\text{Ext}^1\_{\mathcal{O}\_X}(E,E\otimes\_{\mathcal{O}\_X}\Omega^1\_{X/k})$ whose "characteristic polynomial" has coefficients in $\text{Ext}^r\_{\mathcal{O}\_X}(\mathcal{O}\_X,\Omega^r\_{X/k}) = H^r(X,\Omega^r\_{X/k})$ that are equal to the images of the degree $r$ Chern classes of $E$ under the cycle class map from $\text{CH}^r(X)$ to the "de Rham cohomology groups" $H^r(X,\Omega^r\_{X/k})$. In particular, the "pairing" of the Atiyah extension with an element of $H^1(X,T\_X)=\text{Ext}^1\_{\mathcal{O}\_X}(\Omega^1\_{X/k},\mathcal{O}\_X)$ gives an element in $\text{Ext}^2\_{\mathcal{O}\_X}(E,E)$. As proved in references on deformation theory (e.g., Illusie's book on the cotangent comples or Grothendieck's earlier LNM on the good 2-term truncation of the cotangent complex), this element is the obstruction to lifting $E$ to a locally free sheaf on the corresponding first-order deformation of $X$. Altogether, this defines a $k$-linear map,
$$
H^1(X,T\_X) \to \text{Hom}\_k(\text{Ext}^1\_{\mathcal{O}\_X}(E,E\otimes\_{\mathcal{O}\_X}\Omega^1\_{X/k}), \text{Ext}^2\_{\mathcal{O}\_X}(E,E)).
$$
Now consider the case when $E$ is an invertible sheaf. The $k$-linear map above reduces to the form,
$$
H^1(X,T\_X) \to \text{Hom}\_k(H^1(X,\Omega^1\_{X/k}),H^2(X,\mathcal{O}\_X)).
$$
For a polarized K3 surface $X$ over $\mathbb{C}$, using the trivialization of $\Omega^2\_{X/\mathbb{C}}$, this map is "equivalent to" the usual cup-product pairing,
$$
H^1(X,\Omega^1\_{X/\mathbb{C}}) \to \text{Hom}\_{\mathbb{C}}(H^1(X,\Omega^1\_{X/\mathbb{C}}),H^2(X,\Omega^2\_{X/\mathbb{C}}).
$$
The cup-product pairing is a perfect pairing. Thus, this $\mathbb{C}$-linear map is an isomorphism of $\mathbb{C}$-vector spaces.
Finally, since $H^1(X,\mathcal{O}\_X)$ vanishes for a K3 surface, the cycle class map from the Picard group to $H^1(X,\Omega^1\_{X/\mathbb{C}})$ is injective (in characteristic $p$, the kernel of this cycle class map obviously contains the $p$-power image of the Picard group, so this is one place characteristic $0$ is crucial). Thus the induced $\mathbb{C}$-linear map,
$$
H^1(X,T\_X) \to \text{Hom}\_{\mathbb{Z}}(\text{Pic}(X),H^2(X,\mathcal{O}\_X)),
$$
is surjective. In particular, if the Picard rank is at least $1$, if we fix a saturated rank $1$ sublattice, there are first order deformations such that the corresponding "obstruction map" on $\text{Pic}(X)$ has kernel precisely equal to this saturated rank $1$ sublattice. Since also $H^2(X,T\_X)$ vanishes, these first order deformations extend. Therefore, there are deformations of the K3 surface over a $1$-dimensional base such that the Picard lattice of a very general member of the family equals the saturated rank $1$ sublattice, i.e., the Picard rank is $1$. In applications, typically we choose the saturated rank $1$ sublattice to be generated by an ample divisor class, so that the deformation is a family of projective K3 surfaces (rather than just Kaehler K3 surfaces that are not projective).
Some part of this is included in the proof of Claim 3.5 of my note about Artin's axioms.
Artin's axioms, composition and moduli spaces
[http://www.math.stonybrook.edu/~jstarr/papers/moduli4.pdf](http://www.math.stonybrook.edu/%7Ejstarr/papers/moduli4.pdf)
|
11
|
https://mathoverflow.net/users/13265
|
409719
| 167,684 |
https://mathoverflow.net/questions/409710
|
1
|
Let $X$ be an (algebraic) K3 surface, then we have $H^{2,0}(X)=\langle \omega\_X\rangle$, where $\omega\_X$ is the period. Suppose $G=\langle g\rangle$ is a finite group acting on $X$ and $g$ as an automorphism of $X$ doesn't fix $\omega\_X$(e.g. $g$ is purely non-symplectic), then why is $h^{2,0}(X/G)=0$?
What's in my head is
$$
H^k(X,\mathbb{Q})^G\cong H^k(X/G,\mathbb{Q}).
$$ by the Leray-Cartan-Serre spectral sequence. But it requires free action and it's of coefficient $\mathbb Q$. Do we have
$$
H^{i,j}(X)^G\cong H^{i,j}(X/G)
$$ in general?
Also, I am not sure if we always have Hodge structure on normal varieties in general. Here $X/G$ is a normal variety, I guess we can define $H^{i,j}(X/G):=H^j(X/G,\omega\_{X/G}^i)$? I know the canonical divisor is well-defined for normal varieties, so for my case that $X$ is K3, $H^{2,0}(X/G)$ is well defined but is $H^j(X/G,\omega\_{X/G}^i)$ well-defined in general?
|
https://mathoverflow.net/users/104837
|
The Hodge number $h^{2,0}$ of (finite) quotient variety of a K3 surface
|
Your variety $X/G$ is an orbifold; on
a singular variety, the Hodge decomposition
does not work, but on an orbifold, it works
just as well. Then $G$ acts on $H^\*(X,{\Bbb Q})$, and
$H^\*(X/G,{\Bbb Q})$ is the space of $G$-invariants
(this is more or less a definition of
$H^\*(X/G)$ for an orbifold).
Similarly, $H^{p,q}(X/G)$ is the space
of $G$-invariants in $H^{p,q}(X)$.
Then $H^{2,0}(X/G)= H^{2,0}(X)^G=0$ because your $G$-action
is not holomorphically symplectic.
To use this argument, you would probably
need to compare the orbifold cohomology
with the usual cohomology of $X/G$.
This is not very hard to do, see
<https://www.pnas.org/content/42/6/359>
I. Satake,
"ON A GENERALIZATION OF THE NOTION OF MANIFOLD",
PNAS June 1, 1956 42 (6) 359-363.
|
3
|
https://mathoverflow.net/users/3377
|
409721
| 167,686 |
https://mathoverflow.net/questions/409724
|
5
|
Let $B\_t$ be a standard Brownian motion and let $M\_t:=\sup \_{s\le t}B\_s$ be the maximum process. What is the distribution of $2M\_1-B\_1$? is it elementary?
|
https://mathoverflow.net/users/161778
|
What is the distribution of $2M_1-B_1$ where $M_t$ is the maximum process of the the Brownian motion $B_t$
|
Yes, the pdf of this distribution is
\begin{equation}
u\mapsto 2u^2 f(u)\,1(u>0) \tag{1}
\end{equation}
where $f$ is the standard normal pdf.
Indeed, by [Proposition 2](https://ocw.mit.edu/courses/sloan-school-of-management/15-070j-advanced-stochastic-processes-fall-2013/lecture-notes/MIT15_070JF13_Lec7.pdf),
\begin{equation}
G(m,b):=P(M\_1>m,B\_1\le b)=P(B\_1>2m-b)=1-F(2m-b)
\end{equation}
for real $m,b$ such that $m>b\_+:=\max(0,b)$, where $F$ and $f$ are the standard normal cdf and pdf, respectively. So, for the joint pdf $g$ of $(M\_1,B\_1)$ we have
\begin{equation}
g(m,b)=-\frac{\partial^2 G(m,b)}{\partial m\,\partial b}
=\frac{\partial^2 F(2m-b)}{\partial m\,\partial b}
=2(2m-b)f(2m-b)
\end{equation}
if $m>b\_+$, with $g(m,b)=0$ otherwise. So, for
\begin{equation}
U:=2M\_1-B\_1
\end{equation}
and all real $u>0$ we have
\begin{equation}
\begin{aligned}
&P(U\le u) \\
&=\iint\_{\mathbb R^2}dm\,db\,g(m,b)\,1(m>b\_+,\,2m-b<u) \\
&=\int\_0^u dm\,\int\_{2m-u}^m db\,g(m,b) \\
&=2 F(u)-2 u f(u)-1,
\end{aligned}
\end{equation}
with $P(U\le u)=0$ for $u\le0$. Differentiating $P(U\le u)$ in $u$, we confirm that the function given by (1) is the pdf of the distribution of $U=2M\_1-B\_1$.
|
4
|
https://mathoverflow.net/users/36721
|
409729
| 167,688 |
https://mathoverflow.net/questions/409705
|
3
|
I'm not entirely sure what I'm trying to ask.
According to my understanding of the Erlangen programme, each "geometry" (in the sense of Euclidean or hyperbolic or elliptic geometry) is defined in some sense by its abstract group of congruences. My question is whether there's a way to go from the abstract group to the "salient" objects of the geometry.
Examples, because I'm not sure what I mean:
The "salient" objects in Euclidean plane geometry are the points, lines and circles (I suppose). The lines and points correspond in some sense to the reflections, because the set of fixed points of a reflection is either a point or a line. While circles aren't fixed points of any Euclidean congruence transformation, they are the result of rotating a point around.
The Euclidean example suggests we can get the "salient" objects either via fixed points or via orbits of subgroups.
In the case of the elliptic plane, I know that the isometry group is $SO(3)$. When a transformation in $SO(3)$ acts on $\mathbb R^3$, its set of fixed points is a line through the origin. These are the points of the elliptic plane. But elliptic geometry has also got lines and circles. Circles are the result of spinning a point around.
Question: Is there a systematic way of deriving the "salient" objects from an abstract group?
|
https://mathoverflow.net/users/75761
|
Getting the "salient" geometric objects out of an abstract congruence group
|
If $G$ is a reductive Lie group, then it may be viewed as a group of symmetries for a geometry whose salient objects are the coset spaces $G/P$ with respect to parabolic subgroups $P$. This point of view is alluded to in [TWF249](https://math.ucr.edu/home/baez/week249.html), worked out in detail for $G = \operatorname{PGL}(3)$ (as mentioned in [TWF250](https://math.ucr.edu/home/baez/week250.html), not the isogenous group $\operatorname{SL}(3)$, despite what's claimed in TWF178) in [TWF178](https://math.ucr.edu/home/baez/week178.html), and sketched for some more exotic groups in [TWF187](https://math.ucr.edu/home/baez/week187.html).
In TWF249, Baez refers to double coset spaces, but my understanding of the surrounding discussion seems actually to emphasise single coset spaces (which are anyway nicer, because $G$ acts on them!). Anyway, since the question is open ended, perhaps an open ended answer is also acceptable.
|
4
|
https://mathoverflow.net/users/2383
|
409733
| 167,690 |
https://mathoverflow.net/questions/409511
|
2
|
A corollary of the Mostow-Palais theorem is that every homogeneous space for a compact group is a linear group orbit. In other words, if $ H $ is a closed subgroup of a compact group $ K $ then there exists some representation $ \pi: K \to GL(V) $ and $ v \in V $ such that the orbit
$$
\mathcal{O}\_v=\{ \pi(k)v: k \in K\}
$$
is diffeomorphic to $ K/H $ (the stabilizer of $ v $ is exactly $ H $).
Is the converse true; does every compact linear group orbit admit a transitive action by a compact group? In other words, if $ G $ is a group, $ \pi: G \to GL(V) $ a representation, $ v \in V $, and the stabilizer
$$
\mathcal{O}\_v=\{ \pi(g)v: g \in G \}
$$
is compact, then must there exist some compact group $ K $ acting transitively on the manifold $ \mathcal{O}\_v $?
|
https://mathoverflow.net/users/387190
|
Compact linear group orbit equivalent to linear compact group orbit
|
Answer is yes. By theorem of Mostow mentioned in this question
[Homogeneous manifold deformation retracts onto compact submanifold](https://mathoverflow.net/questions/345905/homogeneous-manifold-deformation-retracts-onto-compact-submanifold)
["Covariant Fiberings of Klein spaces" Mostow 1955]
If G and G' both have finitely many connected components (for example if they are algebraic groups) then G/G' is a vector bundle over K/K' where K and K' are maximal compacts.
This is the case here because the image of a representation is always an algebraic group and the stabilizer of a vector is always Zariski closed and thus always an algebraic group.
So if G/G' is a linear group orbit which is compact then the vector bundle part is trivial and we just have that linear group orbit is K/K'.
Just a note here that a similar result to the Mostow result is corollary 2 of "Simply Connected Homogeneous Spaces" 1950 by Montgomery which states that if G is connected and G' has finitely many connected components and G/G' compact then maximal compact K acts transitively.
|
2
|
https://mathoverflow.net/users/387190
|
409737
| 167,691 |
https://mathoverflow.net/questions/409736
|
10
|
Let $f: \mathbb Q \to \mathbb R$ be a continuous function.
An *extension* of $f$ is a function $\tilde f: \mathbb R \to \mathbb R$ such that $\tilde f = f$ on $\mathbb Q$.
We say an extension $\tilde f$ of $f$ is *maximally continuous* if for any other extension $g$ of $f$, we have that if $g$ is continuous at $x \in \mathbb R$, then so is $\tilde f$.
>
> **Question:** For any continuous function $f: \mathbb Q \to \mathbb R$, does there exist a maximally continuous extension of $f$?
>
>
>
**Remark:** One can always obtain an extension that is continuous at every point in $\mathbb Q$, see for example, the answer in [this post](https://mathoverflow.net/questions/394565/extending-continuous-functions-from-mathbb-q-to-mathbb-r) by Fedor Petrov.
|
https://mathoverflow.net/users/173490
|
Maximally continuous extension of continuous functions from $\mathbb Q$ to $\mathbb R$
|
There even exists a largest set $X$ to which $f$ can be continuously extended.
The trick is the following result (which I state here in more generality, to point out which topological assumptions one needs):
**Theorem.** Let $X,Y$ be topological spaces, where $Y$ is $T\_3$, and let $D \subseteq X$ be dense. Let $f: D \to Y$ be continuous and assume that the following property is satisfied for each point $x \in X \setminus D$ (equivalently, each point $x \in X$):
$(\*)$ There exists $y\_x \in Y$ such that, for each net $(x\_j)$ in $D$ that converges to $x$, the net $(f(x\_j))$ converges to $y\_j$.
Then $f$ has a (obviously, unique) continuous extension to $X$.
(I gave a proof of this in a topology course a year ago, but unfortunately I don't know a reference - maybe somebody else can help out with a reference?)
**Application to your situation.** For $D = \mathbb{Q}$, let $X$ be the set of all $x \in \mathbb{R}$ which satisfy property $(\*)$. Then $f$ can be continuously extended to $X$, and this extension is clearly the largest one.
|
8
|
https://mathoverflow.net/users/102946
|
409739
| 167,693 |
https://mathoverflow.net/questions/409746
|
3
|
I have been playing around with interesting integer sequences and came across [Schröder number](https://en.wikipedia.org/wiki/Schr%C3%B6der_number) which defines the number of lattice paths of n x n grid.
The recurrence formula to calculate these numbers is as follows:
$$
S\_n = 3S\_{n-1} + \sum\_{k=1}^{n-2}S\_kS\_{n-k-1} \text{ for }n\ge2 \text{ with } S\_0 =1, S\_2=2
$$
The first few numbers in the sequence are:
$$
1, 2, 6, 22, 90, 394, 1806, 8558
$$
I was curious about the following ratio and its limit:
$$
\lim\_{n\rightarrow\infty}\frac{S\_{n-1}}{S\_{n}} = ?
$$
So I wrote a small C++ program with mixed-precision to handle the large numbers and surprisingly there appears to be an asymptote that is not zero. So far my little program has reached n = 20,000, but it's slowing down significantly as the numbers are getting truly large but it has given a preliminary result of
$$
\lim\_{n\rightarrow\infty}\frac{S\_{n-1}}{S\_{n}} = 0.17158...
$$
My question now is, is this simply a ghost limit for not having gone far enough, or does a limit exist that is not zero? My math expertise is lacking somewhat to attempt to find that limit analytically (can it even be done?).
If a limit exists would it imply that, for large n, the number of additional lattice paths of a n x n grid only increases by about 17% by expanding the grid by 1?
Code for anyone who wants to try it for themselves:
```
#include <iostream>
#include <gmp.h>
#include <gmpxx.h>
#include <vector>
void schroder(std::vector<mpz_class>& n_vec);
int main(int argc, char **argv)
{
int n=2;
int n_max;
std::cout << "Enter n_max: ";
std::cin >> n_max;
std::vector<mpz_class> s = {mpz_class(1), mpz_class(2)};
while (n<n_max){
mpq_class div (s[s.size()-2],s.back());
gmp_printf ("%d\t %.*Ff\n", n, 100, mpf_class(div,500));
schroder(s);
n++;
}
return 0;
}
void schroder(std::vector<mpz_class>& n_vec){
int n = n_vec.size();
mpz_class sum = mpz_class(0);
for(int k = 1; k<=(n-2); k++){
sum += n_vec[k]*n_vec[n-k-1];
}
sum += 3*n_vec[n-1];
n_vec.push_back(sum);
}
```
|
https://mathoverflow.net/users/470225
|
Limit of the Schröder numbers ratio
|
The g.f. of these numbers (see the link) is $\sum\_{n=0}^\infty S\_nx^n=\frac{1-x-\sqrt{1-6x+x^2}}{2x}$. Thus radius of convergence is the same of the radical, that is the modulus of the smaller root of $1-6x+x^2$, which is $3-\sqrt{8}=\lim\_{n\to\infty}\frac{S\_{n-1}}{S\_n}$.
**(edit 12/3/21).** I feel obliged to improve a little this answer. The above argument shows that if the limit of the ratio $S\_{n-1}/S\_{n}$ exists, its value is $3-\sqrt{8}$. However, it is not immediately clear that there should be a limit (e.g. the ratio of coefficients of $\frac {1+ax}{1-x^2}$ alternates between $a $ and $1/a$). Here is an elementary existence argument. From the g.f. one gets the two-term recurrence
$$S\_n=\frac{6n-3}{n+1}S\_{n-1}-\frac{n-2}{n+1}S\_{n-2}$$
So $\rho\_n:=S\_n/S\_{n-1}$ satisfies
$$\cases{\rho\_1=2\\\\\rho\_n=\frac{6n-3}{n+1}-\frac{n-2}{n+1}\frac1{\rho\_{n-1}}.}$$
It follows easily by induction e.g. $1\le \rho\_n\le 6$. Since $x\mapsto-\frac1x$ is increasing we have $ \liminf\_{n\to\infty}(-1/\rho\_{n-1})=-1/{\liminf\_{n\to\infty}\rho\_{n}}$ and $ \limsup\_{n\to\infty}(-1/\rho\_{n-1})=-1/{\limsup\_{n\to\infty}\rho\_{n}}$. Therefore both $\liminf\_{n\to\infty}\rho\_n$ and $\limsup\_{n\to\infty}\rho\_n$ solve the fixed point equation $\lambda=6-\frac1\lambda$; lying in the interval $[1,6]$ they both coincide with $3+\sqrt{8}$, which is therefore the limit. (By a little more computation or more thinking one should be able to prove that $\rho\_n$ in fact increasing).
|
5
|
https://mathoverflow.net/users/6101
|
409753
| 167,697 |
https://mathoverflow.net/questions/409755
|
4
|
Let $f:M \to M$ be a $C^{1}$ diffeomorphism on a compact Riemannian manifold with a normalized Riemannian volume $\mathrm{Leb}$. Given an $f$-invariant Borel probability $\mu$ in $M$, we call the *basin of attraction of $\mu$* the set $B(\mu)$ of the points $x \in M$ such that the averages of Dirac measures along the orbit of $x$ converge to $\mu$ in the weak\* sense:
$$\lim \_{n \rightarrow+\infty} \frac{1}{n} \sum\_{j=0}^{n-1} \varphi\left(f^{j}(x)\right)=\int \varphi d \mu$$ for any continuous $\varphi: M \rightarrow \mathbb{R}$. Then we say that $\mu$ is a *physical measure* for $f$ if the basin of attraction $B(\mu)$ has positive Lebesgue measure in $M$.
A particular type of physical measures are the so-called Sinai–Ruelle–Bowen, or SRB,
measures which have the property of having nonzero Lyapunov exponents $\mu$-almost
everywhere and admitting a system of conditional measures such that the conditional measures on unstable manifolds are absolutely continuous with respect to the Lebesgue measures $\mathrm{Leb}$ on these manifolds induced by the restriction of the Riemannian structure.
An ergodic SRB measure is physical. The figure-eight attractor has
a physical measure with a positive Lyapunov exponent which is not an
SRB measure. I want to know whether there is an example of an SRB measure which is not physical.
|
https://mathoverflow.net/users/127839
|
An example of an SRB measure which is not a physical measure
|
You can just take an Anosov map on $T^2$ and multiply by identity on the circle. Then, you will have SRB measures supported on $T^2 \times pt$ which are not physical.
|
4
|
https://mathoverflow.net/users/5753
|
409761
| 167,698 |
https://mathoverflow.net/questions/409760
|
7
|
Wolfram alpha calculates the integral
$$\int\limits\_0^\infty \frac{x^2\ln{x}}{e^x-1}dx=2\zeta^\prime(3)+3\zeta(3)-2\gamma\zeta(3).$$
However, I need to cite the source of this identity (the table of integrals, or the article where this integral was calculated). Could you indicate any?
|
https://mathoverflow.net/users/32389
|
The source of the Integral
|
One way to get the claimed value of the given integral, $J$ in notation below, is by starting from the standard relation
$$
\begin{aligned}
\zeta(s)
&= \frac 1{\Gamma(s)}\int\_0^\infty \frac {x^{s-1}}{e^x-1}\; dx\ ,\qquad
\text{ so }\\
\zeta'(s)
&=
\frac\partial{\partial s}
\left(\
\frac 1{\Gamma(s)}\int\_0^\infty \frac {x^{s-1}}{e^x-1}\; dx
\ \right)
\\
&=
\frac 1{\Gamma(s)}\int\_0^\infty \frac {x^{s-1}\; \ln x}{e^x-1}\; dx
-
\underbrace{\frac{\Gamma'(s)}{\Gamma(s)}}\_{=\psi(s)}\zeta(s)\ ,\qquad\text{leading to }
\\
\zeta'(3)&=\frac 1{\Gamma(3)}\underbrace{\int\_0^\infty \frac {x^2\; \ln x}{e^x-1}\; dx}\_{\text{our integral }J}
-
\psi(3)\zeta(3)\ .
\\[2mm]
&\qquad\text{Extracting $J$ from above,}\\[2mm]
J &=\Gamma(3)\; \zeta'(3) \ +\ \Gamma(3)\; \psi(3)\; \zeta(3)
\\
&=2\zeta'(3)+(3-2\gamma)\zeta(3)\ ,
\end{aligned}
$$
where we have used the relations $\displaystyle\psi(x+1)=\frac 1x+\psi(x)$ and $\psi(1)=-\gamma$, e.g. from [here](https://www.nbi.dk/%7Epolesen/borel/node5.html). Explicitly, this gives the value for $\displaystyle\psi(3)=\frac 12+\psi(2)=\frac 12+1+\psi(1)=\frac 12(3-2\gamma)$.
$\square$
---
The above shows how to get similar relations when there is an other power of $x$ instead of $x^2$ in the numerator of then integrand, and gives an interpretation of the involved coefficients.
|
18
|
https://mathoverflow.net/users/122945
|
409769
| 167,701 |
https://mathoverflow.net/questions/409754
|
12
|
This is somewhat inspired by [Factoring a function from a finite set to itself](https://mathoverflow.net/questions/408116/factoring-a-function-from-a-finite-set-to-itself).
Fix natural number $n$ and let $[n] := \{1,2,\ldots,n\}$. Set $g\_0 \colon [n]\to [n]$ to be the identity, and for $i \geq 1$ define $g\_i := f\_i \circ g\_{i-1}$ where the $f\_i\colon [n] \to [n]$ are chosen (independently and) uniformly at random among all functions from $[n]$ to $[n]$.
What is the expected value of the smallest $t$ for which $g\_t$ is a constant function? (More generally, what is the distribution of this random variable $t$?)
**EDIT**: As Peter Taylor explained, it is easy to view this also as a Markov chain on $[n]$ where at time $t$ our state $a\_t$ is the size of the image of $g\_t$. And as I mentioned in the comments then the trajectory of this Markov chain $(a\_1-1,a\_2-1,\ldots)$ gives a random partition with part sizes $\leq n-1$; the expected time to a constant function is the expected length of this partition.
There is also a natural $q$-analog of this problem, where instead of random functions $[n]\to [n]$ we look at random linear functions $\mathbb{F}\_q^n\to \mathbb{F}\_q^n$. This gives a Markov chain on $\{0,1,\ldots,n\}$ where our state $a\_t$ is the dimension of the image of $g\_t$. Of course now the transition probabilities of the Markov chain involve the parameter $q$ (and should recover the previous case with $q=1$).
|
https://mathoverflow.net/users/25028
|
Expected number of compositions needed to get constant function
|
This question was completely settled by J.A. Fill here:
<https://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.8.641>
|
5
|
https://mathoverflow.net/users/48831
|
409780
| 167,704 |
https://mathoverflow.net/questions/409543
|
14
|
There are familiar analytic equiconsistency proofs for Euclidean and hyperbolic geometry. Those proofs are so robustly geometric that it seems like they must have synthetic analogues.
Looking into the literature, though, I wonder if I am too optimistic about this. The most common rigorous axiomatizations for synthetic Euclidean and hyperbolic geometry, so far as I can tell, are Hilbert's from his [Foundations of Geometry](https://math.berkeley.edu/%7Ewodzicki/160/Hilbert.pdf), with two changes. They omit the non-elementary axiom of continuity, and they add something to assure that circles will actually have points of intersections with lines and other circles that they cross. The only difference, in these axiomatizations, between Euclidean and hyperbolic geometry is in the axiom of parallels. But this synthetic hyperbolic geometry is incapable of some constructions that we take pretty much for granted both in Euclidean geometry and also (more to my point) in analytic hyperbolic geometry. Notably $n$-section of lines, see [trisection-of-a-hyperbolic-line-segment](https://math.stackexchange.com/questions/1184542/trisection-of-a-hyperbolic-line-segment). Maybe the analytic equiconsistency proofs really do not transfer well to synthetic geometry.
I know Greenberg's discussion of axiomatic issues in [Greenberg](https://www.maa.org/sites/default/files/pdf/upload_library/22/Ford/Greenberg2011.pdf). But I do not have all his references at hand. Judging from the ones I do have (including [Hartshorne](https://rads.stackoverflow.com/amzn/click/com/0387986502)) it seems likely that his discussion of equiconsistency for the two geometries (pp.213-214) refers to analytic presentations of geometry.
Can I find mutually interpretable axioms for synthetic Euclidean and hyperbolic geometry?
Edit: Erik Walsberg's comment about Tarski's axioms answers my title question, even though not in the way I had in mind when I wrote the text. My text was ambiguous about "synthetic" methods, in just the way that Tarski [What is elementary geometry?](http://www.geographicknowledge.de/SeminarLFG/Tarski_Whatiselementarygeometry.pdf) means when he says "In colloquial language the term elementary geometry is used loosely [...with] no well determined meaning."
Hilbert, Tarski, and Greenberg all show that the important logical distinction characterizing elementary methods is not between using or not using coordinates in some field. It is between using or not using higher order notions like point-set continuity and limits. First order algebraic considerations on fields (notably pythagorean fields) are already implicit in Euclid and central to successful first-order elementary geometry.
The work that led me to this question is about interpretation in first order logic, and not about compass (or horocompass) and straightedge or other such construction methods, and really not about avoiding coordinates. So my title was true to my actual concern. Some of my text concerning synthetic methods was less relevant (though those questions too intrigue me). Walsberg's comment is really an answer both to the question as titled and to my actual concern, though he correctly saw I had another issue about methods also in mind.
|
https://mathoverflow.net/users/38783
|
Are there good mutually interpretable axioms for synthetic Euclidean and hyperbolic geometry?
|
Here is an expanded version of my previous comments. There are a lot of things to check here and I haven't.
In my opinion the right axioms for Euclidean/Hyperbolic geometry are the Tarski axioms. Tarski works in a system where the domain is $\mathbb{R}^2$ and you have two relations, a ternary betweenness relation and an equidistance relation $E(x,y;x',y')$ which says that the line segments $\overline{xy}$ and $\overline{x',y'}$ are congruent. In this system all the axioms are phrased in terms of points, not in terms of both points and lines like in Euclid. Lines emerge as definable sets.
The Tarski axioms for Euclidean geometry are on [Wikipedia](https://en.wikipedia.org/wiki/Tarski%27s_axioms), and you just change the parallel postulate to get axioms for Hyperbolic geometry. The usual proof of bi-interpretability goes by showing that both theories are bi-interpretable with the theory of $(\mathbb{R},+,×,<)$, i.e. the theory of real closed fields. I'm not sure where the hyperbolic case was written down, maybe by Szmielew. I don't know of a synthetic version of the proof.
But it should be possible to get a synthetic proof. There are well-known Euclidean models of the hyperbolic plane, like the Klein model, so it should be enough to make a Hyperbolic model of the Euclidean plane.
**Edit:** I think that my initial attempt at the Hyperbolic model of the Euclidean plane fails, because equidistance isn't definable. Following a suggestion of @ColinMcLarty, I will describe a different model that I think works. This model is from Greenberg's book "Euclidean and Non-euclidean geometries".
Let $\mathbb{H}$ be the Hyperbolic plane. In this model the points are just the points in $\mathbb{H}$ and the lines all all lines in $\mathbb{H}$ through the origin together with the curves in $\mathbb{H}$ that are equidistant from a line through the origin.
To get a model of Euclidean geometry we need a betweenness and equidistance relationship, and these relationships need to be definable in $\mathbb{H}$. The betweenness relationship is easy, you just need to observe that the lines form a uniformly definable family of sets. Equidistance is more complicated.
Greenberg describes a map $\rho : \mathbb{H} \to \mathbb{R}^2$ and says that the equidistance relation on $\mathbb{H}$ is the pull-back of the equidistance relation on $\mathbb{R}^2$ by $\rho$. We can put polar coordinates on $\mathbb{H}$ in the same way as on the Euclidean plane. We fix a ray $\ell$ through the origin and let the polar coordinates of $p \in \mathbb{H}$ be $(r,\theta)$ where $r$ is the Hyperbolic distance from the origin to $p$ and $\theta$ is the angle that $p$ makes with $\ell$. If $p \in \mathbb{H}$ is the point with polar coordinates $(r,\theta)$ then $\rho(p)$ is the point
$$
\rho(p) = (\sinh r \sin \theta, \sinh r \cos \theta) = \sinh r (\cos \theta, \sin \theta).
$$
So $\rho(p)$ is the point in $\mathbb{R}^2$ whose (euclidean) polar coordinates are $(\sinh r, \theta)$.
Now let's suppose that $\mathbb{H}$ is the Poincare disc model. I think it is enough to show that $\rho$ is semialgebraic, i.e. definable in $(\mathbb{R},+,\times,<)$. The Hyperbolic distance $r$ of $p$ from the origin is not a semialgebraic function of $p$, but it is the arcsinh of a semialgebraic functions, see [Wikipedia](https://en.wikipedia.org/wiki/Poincar%C3%A9_disk_model#Metric_and_curvature). (On Wikipedia there is a factor of $2$ in from the arcsinh, but that is just a scaling factor, so we can drop it). I think it should be pretty clear that $\rho$ is semialgebraic.
Once we know that $\rho$ is semialgebraic we know that the equidistance relation on our Hyperbolic model of Euclidean geometry is semialgebraic. This should imply that it is definable in Hyperbolic geometry, but one would need extra arguments to see it directly.
|
6
|
https://mathoverflow.net/users/152899
|
409785
| 167,706 |
https://mathoverflow.net/questions/409656
|
2
|
I have been reading section 7 of Serre's "Quelques applications du théorème de densité de Chebotarev" (<http://www.numdam.org/item/PMIHES_1981__54__123_0/>), and in particular have been trying to understand the proof of Theorem 17(i) (pg.60-62) which gives the order of magnitude of the count $M\_f(x)$ of non-zero Fourier coefficients upto $x$, of a non-CM modular form of weight $k \geq 2$ and level $\Gamma\_0(N)$ with Nebentypus $\omega$.
For context, the idea of the proof is to consider the sets $R\_0$ (respectively, $R\_{1/2}$) of $f \in M\_k(\Gamma\_0(N), \omega)$ for which the lower asymptotic density of $M\_f(x)/x$ (respectively, $M\_f(x)\sqrt{log x}/x$) is zero, and show that each of these two subsets is stable under the action of the $\mathbb C$-subalgebra $\mathfrak H$ of $End\_{\mathbb C} M\_k(\Gamma\_0(N), \omega)$ generated by the Hecke operators $T\_p$ $(p \nmid N)$ and $U\_p$ $(p|N)$, whereafter he shows by means of a contradiction argument that $R\_0$ is precisely the set of CM-cusp forms (in the sense of Ribet) of weight $k$ and level $\Gamma\_0(N)$, Nebentypus $\omega$ (henceforth denoted, $S\_k(\Gamma\_0(N), \omega)^{\text{cm}}$), while $R\_{1/2}=0$.
In order to achieve the aforementioned contradiction, he assumes that there is some form $f \not\in S\_k(\Gamma\_0(N), \omega)^{\text{cm}}$ which lies in $R\_0$, so that $f$ can be uniquely written as a sum $g+h$ of some nonzero $g \in S\_k(\Gamma\_0(N), \omega)^{\text{non-cm}} \oplus E\_k(\Gamma\_0(N), \omega)$ and some $h \in S\_k(\Gamma\_0(N), \omega)^{\text{cm}}$, with $S\_k(\Gamma\_0(N), \omega)^{\text{non-cm}}$ denoting the orthogonal complement of $S\_k(\Gamma\_0(N), \omega)^{\text{cm}}$ in $S\_k(\Gamma\_0(N), \omega)$ (i.e., the space of non-CM cusp forms) and with $E\_k(\Gamma\_0(N), \omega)$ denoting the space of Eisenstein series with usual parameters. He considers the $\mathfrak H$-module $\mathfrak Hg$ generated by $g$ and claims that it contains a simple submodule $\Sigma$, which must therefore be one-dimensional (as a complex vector space) since $\mathfrak H$ is a commutative algebra.
The last sentence is what I cannot seem to be able to see.
1. First, as far as I know, nonzero modules over *Artinian* rings have simple submodules. If the ring is not Artinian, then even cyclic modules need not have simple submodules (as happens in the ring $\mathbb Z$). Is it known that $\mathfrak H$ is Artinian? If so, I would really appreciate a easy-to-understand reference with minimal machinery used. If not, how do we get a simple submodule of $\mathfrak H g$ in this case?
2. I know that simple modules over *finitely generated* $\mathbb C$-algebras are one-dimensional complex vector spaces (and can see it as a consequence of the weak Hilbert Nullstellensatz). But our algebra $\mathfrak H$ here is clearly not finitely generated, right? So even we do end up with a simple submodule $\Sigma$ of $\mathfrak H g$, why is it one-dimensional?
I would really appreciate some help with these queries, they are the only places in the proof which I cannot fill in. I fear I might be missing something simple, in which case I apologize in advance. Thank you for your time.
|
https://mathoverflow.net/users/157984
|
On some claims on cyclic modules over Hecke algebra used in Serre's "Quelques applications du théorème de densité de Chebotarev"
|
The set $\mathfrak H g$ is a finite dimensional complex vector space on which the Hecke operators ($T\_p$ for $p\nmid N$ and $U\_p$ for $p\mid N$) act. The Hecke operators are normal and commute with each other, hence $\mathfrak H g$ has a basis consisting of simultaneous eigenfunctions of the Hecke operators. If $b$ is an element of such an eigenbasis, then $\Sigma=\mathfrak H b$ is a one-dimensional submodule that Serre talks about. This is straightforward linear algebra, no need to look at ring-theoretic properties of $\mathfrak H$.
|
1
|
https://mathoverflow.net/users/11919
|
409787
| 167,707 |
https://mathoverflow.net/questions/409782
|
6
|
Let $A$ and $B$ be unital $C^\*$-algebras, so we can view these as operator systems, and it makes sense to consider their injective envelopes $I(A)$ and $I(B)$. These injective envelopes become $C^\*$-algebras for the Choi-Effros product.
Given a unital $\*$-morphism $f: A \to B$, is it true that there exists a unique unital $\*$-morphism $\overline{f}: I(A) \to I(B)$ that extends $f$?
Once the above question is answered positively (if the answer is positive), the following will probably be easy:
Is this construction functorial? I.e. is $I(-)$ a functor from the category of unital $C^\*$-algebras to the category of unital $C^\*$-algebras (with morphisms unital $\*$-homomorphisms?
A reference is more than enough for me to be satisfied with an answer.
|
https://mathoverflow.net/users/216007
|
Is the injective envelope functorial?
|
One can view $A$ and $B$ as sitting completely isometrically inside their injective envelopes $I(A)$ and $I(B)$. Then by injectivity a unital \*-homomorphism (or more generally a unital completely positive map) $f:A\rightarrow B\subseteq I(B)$ extends to a unital completely positive map $\overline f:I(A) \rightarrow I(B)$.
[Edit: this should work]
Paulsen in [this paper](https://www.ams.org/journals/tran/2011-363-09/S0002-9947-2011-05203-7/S0002-9947-2011-05203-7.pdf), Proposition 3.5, points out that any C$^\*$-algebra containing $K(H)$ has injective envelope $B(H)$. Then $A = K(H) + \mathbb C I$ has $I(A) = B(H)$.
Consider the $\*$-homomorphism $f:A\rightarrow \mathbb C$ given by $f(k+\alpha I) = \alpha$. Note that $I(\mathbb C) = \mathbb C$ and that any state of the Calkin algebra $B(H)/K(H)$ precomposed with the quotient map $q:B(H)\rightarrow B(H)/K(H)$ extends the map $f$. Therefore, $\overline f$ need not be a $\*$-homomorphism or unique.
|
8
|
https://mathoverflow.net/users/76593
|
409793
| 167,710 |
https://mathoverflow.net/questions/409791
|
2
|
Let $p(x), q(x)$ be two p.d.f.s of distributions on $\mathbb{R}$.
I am interested in finding the subset $E$ that maximizes the quantity
$$\frac{\int\_{E}\min(p(x),q(x))\mathrm{d}x}{\int\_{E}\max(p(x),q(x))\mathrm{d}x}.$$
In fact, I'm mostly interested if this quantity has been studied before --- I have some current progress in characterizing the set $E$ (one should add points to it when $p(x)$ and $q(x)$ are close multiplicatively, meaning $p(x)/q(x)$ or its inverse are small), and don't mind on working on this more myself, but it definitely seems like the kind of thing that might be well-known since the 60's.
The context this comes up in (a cryptographic one) is probably not particularly useful in finding pointers to this, as I expect (if it has shown up before) it would be in a fairly different field.
|
https://mathoverflow.net/users/101207
|
Subset which maximizes $\frac{\int_E\min(p(x), q(x))}{\int_E\max(p(x), q(x))}$?
|
You want to maximize
\begin{equation\*}
R(E):=\frac{\int\_E\ m}{\int\_E\ M}
\end{equation\*}
over all admissible sets $E$, that is, over all Lebesgue-measurable sets $E$ such that $\int\_E\ M>0$, where
\begin{equation\*}
m(x):=\min(p(x),q(x)),\quad M(x):=\max(p(x),q(x)).
\end{equation\*}
For
\begin{equation\*}
r(x):=\frac{m(x)}{M(x)}
\end{equation\*}
and real $c$, let
\begin{equation\*}
E\_c:=\{x\colon r(x)\ge c\}
\end{equation\*}
and then
\begin{equation\*}
s:=\sup\Big\{c\colon E\_c\text{ is admissible}\Big\}
=\sup\Big\{c\colon \int\_{E\_c}\ M>0\Big\}.
\end{equation\*}
Note that $s$ is the essential supremum of $r$ with respect to the measure $\mu$ given by $\mu(dx)=M(x)dx$. In particular, we have $s\in[0,1]$, since $0\le r\le1$. Also, $r\le s$ $\mu$-a.e., that is, almost everywhere with respect to the measure $\mu$. So, $m\le sM$ $\mu$-a.e., and hence $\int\_E\ m\le s\int\_E\ M$ for any Lebesgue-measurable set $E$, which implies
\begin{equation\*}
R(E)\le s
\end{equation\*}
for all admissible sets $E$.
Take now any real $a<s$. Then the set $E\_c$ is admissible and
$\int\_{E\_a}m=\int\_{E\_a}rM\ge a\int\_{E\_a}M$, whence $R(E\_a)\ge a$, for any real $a<s$.
Thus,
\begin{equation\*}
\sup\{R(E)\colon E\text{ is admissible}\}=s. \tag{1}
\end{equation\*}
---
If the set $I:=\{x\colon M(x)>0\}$ is a (possibly infinite) interval and the pdf's $p,q$ are continuous on $I$, then $p(z)=q(z)>0$ for some $z\in I$ (otherwise, we would have $p<q$ on $I$ or $p>q$ on $I$, which would contradict $p$ and $q$ being pdf's). So, in this case $s=1$.
In general (and usually), the supremum in (1) will not be attained. For instance, consider $p(x)=2x\,1(0<x<1)$ and $q(x)=2(1-x)\,1(0<x<1)$. Then the supremum in (1) is $s=1$ and is not attained -- because then $r<1$ almost everywhere and hence $R(E)<1=s$ for all admissible $E$.
|
3
|
https://mathoverflow.net/users/36721
|
409798
| 167,711 |
https://mathoverflow.net/questions/409748
|
4
|
Let $ A\_{n}(F) $ be the collection of all skew-symmetric matrices over the field $ F $ ($\operatorname{char} F \neq 2 $). Let M be a subspace of $ A\_{n}(F) $ such that all non zero elements have rank $ 2 $ . Here we consider $ F = \mathbb{R} $ .
Then what will be the maximum dimension of $ M $ when $ n = 6 $ ? I have already got $ \dim M \geq 5 $ .As if consider this subsapce
$$\left[\begin{array}{cc}0 & a & b & c & d & e \\ -a & 0 & 0 & 0 & 0 & 0 \\-b & 0 & 0 & 0 & 0 & 0 \\-c & 0 & 0 & 0 & 0 & 0 \\-d & 0 & 0 & 0 & 0 & 0 \\-e & 0 & 0 & 0 & 0 & 0 \\ \end{array}\right]$$
where $ a,b,c,d,e \in \mathbb{R} $ or . So in this space any element of is of rank $ 2 $.
Does it possible that $ \dim M = 6 $ or $ \geq 6 $ . I have also interested in What will be the maximum dimension of $ M $ when $ n \geq 6 $ ( except $ n = 8 $).
|
https://mathoverflow.net/users/215016
|
The upper bounds on rank $ 2 $ real matrices
|
A skew-symmetric matrix of rank 2 is of the form $(uv^T-vu^T)$ for some column vectors $u$ and $v$. Let's denote it $u\wedge v$. Also, if $u, v, w, z$ are linearly independent then $u\wedge v+w\wedge z$ has rank 4. It follows that a subspace of ${\mathfrak{so}}(n)$ consisting of matrices of rank $\leq 2$ is of the form $u\wedge V\_0$ where $V\_0$ is a subspace of ${\mathbb R}^n$, which we can assume lies in the orthogonal complement to ${\mathbb R}u$. In particular, the maximum dimension is $(n-1)$.
|
3
|
https://mathoverflow.net/users/26635
|
409802
| 167,714 |
https://mathoverflow.net/questions/409722
|
1
|
Let
$$Y\_t:=1+\int\_0^t b(s)ds + W\_t,\quad\forall t\ge 0,$$
where $b:\mathbb R\_+\to[1,2]$ is continuous and $(W\_t)\_{t\ge 0}$ is a standard Brownian motion. Denote $\tau:=\{t\ge 0: Y\_t\le 0\}$ and $X\_t:=Y\_{t\wedge \tau}$. It is known from [On the marginal distributions of an absorbed diffusion](https://mathoverflow.net/questions/407535/on-the-marginal-distributions-of-an-absorbed-diffusion) that the law of $X\_t$, denoted by $\mu\_t$, has the following decomposition:
$$\mu\_t(dx) = \alpha(t)\delta\_0(dx) + p\_t(x)dx,\quad \forall t>0.$$
Can we show (under suitable conditions) $p\_t(0+):=\lim\_{x\to 0+}p\_t(x)=0$ for every $t>0$?
PS : When $b$ is constant, e.g. $b\equiv 1$, we have
$$\int\_x^{\infty}p\_t(y)dy = \mathbb P[X\_t>x] = \mathbb P[\inf\_{0\le s\le t}Y\_s>0, Y\_t>x]= \mathbb P[\sup\_{0\le s\le t}(-s+W\_s)<1, -t+W\_t<1-x],\quad \forall t,x>0.$$
Using the joint density of the drifted Brownian Motion and its running maximum, one has
$$\mathbb P[\sup\_{0\le s\le t}(-s+W\_s)<1, -t+B\_t<1-x]=\int\_0^1 dm \int\_{-\infty}^{1-x}{\bf 1}\_{\{y\le m\}} e^{-t/2-y}\frac{2(2m-y)}{\sqrt{2\pi t^3}}e^{-(2m-y)^2/2t}dy,$$
which yields by differentiating with respect to $x$
$$p\_t(0+)=-\lim\_{x\to 0+} \frac{\partial \mathbb P[X\_t>x]}{\partial x}=0.$$
Can we extend to the general function $b$? The key is to show the existence of the joint density of $(Y\_t, \inf\_{0\le s\le t}Y\_s)$ but I do not know how prove it.
|
https://mathoverflow.net/users/261243
|
Does the density of a stopped drifted Brownian motion vanish at zero?
|
We can re-write the problem in terms of $W(t)$ alone, or, even better, in terms of the drifted Brownian motion $\tilde W(t) = W(t) - M t$, where $M$ is the supremum of $|b(s)|$.
Define
$$ B(t) = -1 - \int\_0^t b(s) ds - M t ,$$
so that $X(t) = \tilde W(t) - B(t)$ up to time
$$ \tau = \inf \{ t > 0 : \tilde W(t) \leqslant B(t) \} . $$
Fix $t\_0 > 0$ and define
$$ \sigma = \inf \{ t \in (0, t\_0] : \tilde W(t) \leqslant B(t\_0) \} . $$
Since $B$ is a non-increasing function, we clearly have $\sigma \geqslant \tau$, and hence the measure
$$\mu(dx) = \mathbb P(t\_0 < \tau, \tilde W(t\_0) - B(t\_0) \in dx)$$
is dominated by the measure
$$\nu(dx) = \mathbb P(t\_0 < \sigma, \tilde W(t\_0) - B(t\_0) \in dx) .$$
The latter is, however, just the distribution at time $t\_0$ of the drifted Brownian motion $\tilde W(t) - B(t\_0)$, killed upon hitting $0$. As you write in the statement of the problem, this is known to have a density function continuously vanishing at zero, and hence $\mu(dx)$ also has a density function continuously vanishing at zero.
It remains to note that $\mu$ is precisely the distribution of $X(t\_0)$, up to an extra atom at $0$.
---
*Remark:* A more general approach to the problem, which seems to work also when $b(s)$ is an (adapted) stochastic process rather than a deterministic function, would involve showing first that the distribution of $Y(t)$ — and thus also that of $X(t)$ — has a bounded density function (save for an atom at $0$), and then using Chapman–Kolmogorov equation and a comparison argument similar to the one given above to conclude that the density function of the distribution of $X(t)$ goes to zero at $0$.
|
3
|
https://mathoverflow.net/users/108637
|
409808
| 167,716 |
https://mathoverflow.net/questions/409805
|
-2
|
It is conjectured that for any integer $k\not\equiv \pm 4\pmod 9$ there are infinitely many integer solutions to
$$
a^3+b^3+c^3=k.
$$
Some cases for integer $k$ becomes too hard like $42$ which it were presented as the following in 2019 by Bouker
$$
(−80538738812075974)^3 + 80435758145817515^3 + 12602123297335631^3=42,
$$ Some others $k$ less than $10^3$ are already known we cite $114,627,390,\ldots $. My question here is not to know how we can solve other cases for $k$, but my question is: what is the purpose behind the previous conjecture? Rather than that why do we need to represent integers as the sum of three cubes?
For example, spending a century of research to represent numbers like 114 or 390 or … as sum of three cubes is it just curiosity or will it provide a new addition to number theory?
|
https://mathoverflow.net/users/51189
|
Why do we need to represent integers as the sum of three cubes?
|
The computational and theoretical number theory necessary to actually *find* such expressions is nontrivial and interesting. You should read the papers, eg Booker and Sutherland, *[On a question of Mordell](https://arxiv.org/abs/2007.01209)*. The actual problem is merely an excuse that drives people to develop new mathematics.
You might as well ask why we need to know that $n$th powers aren't the sum of two nonzero $n$th powers—but it led to deep understanding of modularity of Galois representations.
|
9
|
https://mathoverflow.net/users/4177
|
409810
| 167,717 |
https://mathoverflow.net/questions/406827
|
4
|
Let $USp(2n)$ be the compact symplectic group of size $2n$, $dA$ its Haar measure
of total mass one, and $\det(1−A)$ being computed for the standard representation of
$A\in USp(2n)$ as a matrix of size $2n$. Also, $C\_n=\frac1{n+1}\binom{2n}n$ be Catalan number.
This paper entitled [A note on random matrix integrals, moment identities, and Catalan numbers](https://web.math.princeton.edu/%7Enmk/catalan11.pdf) (page 5) lists the equality
$$\int\_{USp(2n)}\det(1+A)^rdA=\det(C\_{r+1+i+j})\_{i,j=0}^{n-1} \tag1$$
whose justification is solely based on just **equality as numbers**.
>
> **QUESTION.** Is there a conceptual or combinatorial reason for (1)?
>
>
>
|
https://mathoverflow.net/users/66131
|
Moment integrals and determinants
|
Yes, there is a combinatorial way to see this.
Firstly, it is much easier to show a baby version of this type of phenomenon. A while ago I read a nice [blog post](https://qchu.wordpress.com/2010/03/07/walks-on-graphs-and-tensor-products/) by Qiaochu explaining the identity
$$\int\_{0}^1 (2\cos \pi x)^n (2\sin^2\pi x)dx=\begin{cases} 0 &\text{ if n is odd} \\
C\_{n/2} & \text{ if n is even} \end{cases}$$
by interpreting this integral as $\int\_{SU(2)}\chi\_V(g)^nd\mu$, with respect to the normalized Haar measure, where $V$ is the defining representation. This in turn is the multiplicity of the trivial representation in $V^{\otimes n}$. Next, you make a graph with vertices corresponding to the irreducible representations of $SU(2)$, and add as many edges from $A$ to $B$ as the multiplicity of $A$ in $B\otimes V$. The irreducible representations of $SU(2)$ are the symmetric powers $\operatorname{Sym}^n(V)$, thus they are in bijection with the natural numbers. As a corollary of the fact that
$$\operatorname{Sym}^n(V)\otimes V\cong \operatorname{Sym}^{n-1}(V)\otimes \operatorname{Sym}^{n+1}(V)$$
we obtain that the graph is the half infinite line
$$0\leftrightarrow 1\leftrightarrow 2\leftrightarrow \cdots$$
and the desired multiplicity is the number of walks from zero to itself of length $n$, thus giving the answer in terms of the Catalan numbers.
---
Now for your question, the representation of $USp(2n)$ under consideration is the exterior algebra of the standard representation, $W$. So $V=\bigoplus\_{k=0}^{2n}\Lambda^{k}(W)$ and your integral measures the multiplicity of the trivial representation in the representation $V^{\otimes r}$. A general combinatorial set up that keeps track of these types of multiplicities is that of [crystal bases](https://en.wikipedia.org/wiki/Crystal_base). It is much more technical but it is what ultimately brings the result in terms of enumerations of certain types of tableaux or non-intersecting lattice paths. Fortunately, the details of this specific problem are all worked out in the paper
>
> Se-jin Oh, Travis Scrimshaw "Identities from representation theory" Discrete Math., 342(9):2493–2541, 2019 ([arxiv](https://arxiv.org/abs/1805.00113))
>
>
>
|
3
|
https://mathoverflow.net/users/2384
|
409812
| 167,718 |
https://mathoverflow.net/questions/409767
|
2
|
I'm learning about Lie groupoids and was inspired (by Mackenzie's book) to consider the following problem.
Consider first a principal bundle $P\xrightarrow G M$; we can construct the quotient manifold
$$
\Omega=\frac{P\times P}{G},
$$
obtained as the orbit space of the pair action $g(u\_2,u\_1)=(gu\_2,gu\_1)$. In this way, we obtain a Lie groupoid (called the *gauge groupoid* and denoted $\Omega\rightrightarrows M$) with the source and target maps $s,t\colon \Omega\rightarrow M$ given by $s[u\_2,u\_1]=u\_1$ and $t[u\_2,u\_1]=u\_2$, and partial multiplication $[u\_3,u\_2'][u\_2,u\_1]=[u\_3,u\_1]$ iff $u\_2'=u\_2$.
In particular, the universal cover $\mathrm{SU}(2)\rightarrow \mathrm{SO}(3)$, given by the adjoint representation, may be observed as a principal bundle ${\mathrm{SU}(2)}\xrightarrow{\mathbb Z\_2}{\mathrm{SO}(3)}$. The map $\mathrm{SU}(2)\times \mathrm{SU}(2)\rightarrow \mathrm{SO}(4)$, given by $(p,q)\mapsto (x\mapsto pxq^{-1})$, where $p,q,x$ are viewed as quaternions, induces the isomorphism of Lie groups
$$
\frac{\mathrm{SU}(2)\times \mathrm{SU}(2)}{\mathbb Z\_2}\cong \mathrm{SO}(4),
$$
so we actually obtain a Lie groupoid $\mathrm{SO}(4)\rightrightarrows \mathrm{SO}(3)$, with the source and target fibres diffeomorphic to $\mathrm{SU}(2)$. Additionally, for any $x\in \mathrm{SO}(3)$, the vertex group $s^{-1}(x)\cap t^{-1}(x)\cong\mathbb Z\_2$, indicating that $\mathrm{SO}(4)\rightrightarrows \mathrm{SO}(3)$ should be isomorphic to the fundamental groupoid of $\mathrm{SO}(3)$.
Now consider the (physically more juicy) universal cover $\mathrm{SL}(2,\mathbb C)\xrightarrow{\mathbb Z\_2}{\mathrm{SO}^+(1,3)}$, where $\mathrm{SO}^+(1,3)$ denotes the component of the Lorentz group $\mathrm O(1,3)$ which contains the identity matrix -- this is a Lie subgroup (the proper ortochronous Lorentz group). By the general construction, we obtain a gauge groupoid
$$
\frac{\mathrm{SL}(2,\mathbb C)\times \mathrm{SL}(2,\mathbb C)}{\mathbb Z\_2}\rightrightarrows \mathrm{SO}^+(1,3).
$$
We would again like to identify the space of arrows with a known space, but it is now 12-dimensional. The only sensible candidate that comes to my mind is the complexification $\mathrm{SO}^+(1,3)\_{\mathbb C}$, but I don't know where to start confirming whether this is the case. So the question is:
$$
\frac{\mathrm{SL}(2,\mathbb C)\times \mathrm{SL}(2,\mathbb C)}{\mathbb Z\_2}\stackrel{?}\cong\mathrm{SO}^+(1,3)\_{\mathbb C}.
$$
Lastly, if anyone has any intuition regarding the obtained gauge groupoid over $\mathrm{SO}^+(1,3)$, feel free to make a hand-wavey explanation of what this object could physically represent.
|
https://mathoverflow.net/users/126003
|
Gauge groupoid of Lorentz group & complexification
|
After inspecting the general construction of the complexification of a Lie group, I've arrived to a positive answer to the question of whether $\mathrm{SO}^+(1,3)\_{\mathbb C}$ is isomorphic to $\frac{\mathrm{SL}(2,\mathbb C)\times \mathrm{SL}(2,\mathbb C)}{\mathbb Z\_2}$. The argument goes as follows.
Given a Lie group $G$, denote by $\pi\colon\widetilde G\rightarrow G$ the universal covering projection and by $\widetilde G\_{\mathbb C}$ the (unique up to iso.) simply connected complex Lie group with Lie algebra $\mathfrak g\_{\mathbb C}=\mathfrak g\otimes \mathbb C$. Let $\phi\colon \widetilde G\rightarrow \widetilde G\_{\mathbb C}$ be the unique homomorphism (by Lie's second theorem) such that $\mathrm d\phi\_e$ is the canonical inclusion $\mathfrak g\hookrightarrow \mathfrak g\_{\mathbb C}$. The complexification $G\_{\mathbb C}$ of $G$ is constructed abstractly as
$$
G\_{\mathbb C}=\frac{\widetilde G\_{\mathbb C}}{\phi(K)^\*},
$$
where $K=\ker\pi$ is the fundamental group of $G$, and $\phi(K)^\*$ is the smallest closed normal subgroup of $\widetilde G\_{\mathbb C}$ which contains $\phi(K)$ (by inspecting the adjoint representation of $\widetilde G\_{\mathbb C}$, one can show that $\phi(K)$ is in the centre of $\widetilde G\_{\mathbb C}$, hence so is $\phi(K)^\*$).
In our case, the following isomorphism of Lie algebras is crucial:
$$
\mathfrak{sl}(2,\mathbb C)\_{\mathbb C}\cong \mathfrak{sl}(2,\mathbb C)\oplus \mathfrak{sl}(2,\mathbb C).
$$
Note that since $\mathfrak{so}(1,3)\cong \mathfrak{sl}(2,\mathbb C)$, there holds (up to an isomorphism) $$\widetilde{\mathrm{SO}^+(1,3)}\_{\mathbb C}=\mathrm{SL}(2,\mathbb C)\times \mathrm{SL}(2,\mathbb C).$$ Since $\ker \pi=\{I,-I\}$ is the fundamental group of $\mathrm{SO}^+(1,3)$, and $-I=\exp(\mathrm{diag}(i\pi,-i\pi))$, we have (using the appropriate identifications) $$\phi(\{I,-I\})=\{(I,I),(-I,-I)\},$$
which is already closed and normal in $\mathrm{SL}(2,\mathbb C)\times \mathrm{SL}(2,\mathbb C)$.
We may now indeed conclude that the gauge groupoid of the Lorentz group is
$$
\mathrm{SO}^+(1,3)\_{\mathbb C}\rightrightarrows \mathrm{SO}^+(1,3)
$$
(Note that I was, however, unsuccessful in using the universal complexification property to establish the wanted isomorphism.)
|
0
|
https://mathoverflow.net/users/126003
|
409814
| 167,720 |
https://mathoverflow.net/questions/409817
|
3
|
$\DeclareMathOperator\Ch{Ch}$Let $M$ be a connected manifold of finite type. We denote $\Ch\_{\mathbb{Q}}(M),$ $\Ch\_{\mathbb{Z}}(M)$ and $\Ch\_{\mathbb{\pm}\mathbb{Z}}(M)$ by cohomological dimensions of $M$ over $\mathbb{Q},$ $\mathbb{Z}$ and $\pm\mathbb{Z}$ (coefficients
in the orientation sheaf $\mathbb{\pm}\mathbb{Z}$) respectively. Is it always true that $\Ch\_{\mathbb{Q}}(M)\leq \Ch\_{\mathbb{Z}}(M)\leq \Ch\_{\mathbb{\pm}\mathbb{Z}}(M)?$ I know that if $M$ is orientable then $\Ch\_{\mathbb{Z}}(M)= \Ch\_{\mathbb{\pm}\mathbb{Z}}(M).$ Can we say that $\Ch\_{\mathbb{Z}}(M)< \Ch\_{\mathbb{\pm}\mathbb{Z}}(M)$ if $M$ is non-orientable? $H^{\*}(M,\mathbb{\pm}\mathbb{Z})$ is the homology of $Hom\_{\mathbb{Z}[\pi\_{1}(X)]}(C\_{\*}(M^{c}),\mathbb{Z}),$ where $C\_{\*}(M^{c})$ is the singular chain complex of universal cover $M^{c}$ of $M,$ and where the action of (the class of) a loop on the $\mathbb{Z}$ is multiplication by $\pm1$ according to whether this loop preserves or reverses orientation. We denote by $Ch\_{\pm\mathbb{Z}}(M)$ the smallest integer with property that
$$H^{i}(M,\pm\mathbb{Z})=0,\quad\mbox{ for all $i>Ch\_{\pm\mathbb{Z}}(M)$}.$$
|
https://mathoverflow.net/users/126407
|
Relation between cohomological dimensions of manifolds
|
By Bredon's Sheaf Theory, Proposition II.16.15, if $X$ is locally paracompact then $\dim\_L X\leq \dim\_{\mathbb Z}X$ for any ring $L$ with unit. So that should answer the question about the relation between $\dim\_{\mathbb Q}$ and $\dim\_{\mathbb Z}$.
For the other part, I'm not completely sure what you mean by $\mathbb Z\_\pm$, but if it's something to do with twisted coefficients then my guess is that for nice spaces like this it would be possible to prove the $\mathbb Z$ and $\pm\mathbb Z$ dimensions are the same using covering spaces. I'll think some more about the details.
Update: looking more carefully at the definition of dimension in Bredon, for a family of supports $\Phi$ and ring with unit $L$, he defines $\dim\_{\Phi,L} X$ to be the least integer $n$ such that $H^k\_{\Phi}(X; A)=0$ for all sheaves $A$ of $L$-modules and all $k>n$. Since we're using manifolds, the system of supports can be taken to be the usual closed supports. So, assuming this is the same notion of cohomological dimension that you mean, I'm not sure then what it means to talk about dimension over something that's not a fixed ring. Can you say more about your definition?
|
2
|
https://mathoverflow.net/users/6646
|
409821
| 167,721 |
https://mathoverflow.net/questions/409832
|
10
|
A group is $d$-quasirandom if every nontrivial complex representation has dimension at least $d$. Gowers introduced quasirandomness in [this paper](https://arxiv.org/abs/0710.3877) and proved that every nonabelian finite simple group of order $n$ is $\sqrt{\log n}/2$-quasirandom.
**Question:** What is the correct (asymptotic) lower bound for the quasirandomness of nonabelian finite simple groups?
Gowers indicates that a theorem of Jordan (Theorem 14.12 [here](https://www.cefns.nau.edu/%7Efalk/classes/511/Isaacs_Character_theory.pdf)) implies a slightly better bound. Indeed, this theorem gives that a nonabelian simple group of order $n$ has a nontrivial representation of dimension $d$ only if $n\leq (d!)12^{d(\pi(d+1)+1)}$, which implies that these groups are $\Omega(\sqrt{\log n\log\log n})$-quasirandom.
Of course, the correct bound should be obtainable using the classification of finite simple groups. The sporadic groups are irrelevant to the asymptotics, and the alternating groups are $\Omega(\log n/\log\log n)$-quasirandom. It remains to study the groups of Lie type. What I want can be read off the first column of the generic character tables of these groups, but to my surprise, these aren't known yet (see LeechLattice's answer [here](https://mathoverflow.net/questions/391136/is-the-representation-of-finite-simple-groups-fully-understood/)). Interestingly, the answer [here](https://mathoverflow.net/questions/400864/is-operatornamepsl2-q-the-most-quasirandom-group) describes how to find an upper bound on the smallest nontrivial representation in these cases, whereas I want a lower bound.
|
https://mathoverflow.net/users/29873
|
How quasirandom are the nonabelian finite simple groups?
|
Suppose $G$ is a finite simple group of order $n$ with a nontrivial representation of degree $d$. Then $G$ is isomorphic to a subgroup of $U(d)$. By Collins's sharp version of Jordan's theorem (<https://www.degruyter.com/document/doi/10.1515/JGT.2007.032/html>), $G$ has an abelian normal subgroup of index at most $(d+1)!$, which must be trivial since $G$ is simple, so $|G| \leq (d+1)!$. Rearranging, $d \gtrsim \log n / \log\log n$.
Collins's work builds on work of Weisfeiler that I think was unfinished by the time of his disappearance.
**Edit:** Specializing to simple $G$ actually reduces Collins's paper to a reference to a reference to the paper of Seitz and Zalesskii (<https://www.sciencedirect.com/science/article/pii/S0021869383711324?via%3Dihub>) mentioned by David Craven in the comments, so that's really the heart of the matter. We thereby get the slightly stronger bound $n \leq (d+1)!/2$ (for sufficiently large $d$). Apart from alternating groups I think you can read out a much stronger bound like $n \leq \exp O((\log d)^2)$, or $d \geq \exp \Omega( (\log n)^{1/2})$ (I am guessing the next worst case is $\mathrm{SL}\_n(2)$).
|
12
|
https://mathoverflow.net/users/20598
|
409837
| 167,728 |
https://mathoverflow.net/questions/408778
|
10
|
As far as I know, there is a classification of all prime knots with less than 16 crossings.
It seems that there is already a fast enough algorithm to distinguish a knot from an unknot.
So in principle there is a huge amount of data to implement a deep learning machine which will recognize (and distinguish) knots up to some very good accuracy.
Is it something that mathematicians have tried to do? Any references?
|
https://mathoverflow.net/users/17895
|
Deep learning for knot theory. Classification
|
I saw two articles today (12/2/21) that reminded me of this post. I am mentioning them here to potentially help the OP:
1. *[Learning knot invariants across dimensions](https://arxiv.org/abs/2112.00016)* by Jessica Craven, Mark Hughes, Vishnu Jejjala, Arjun Kar (on arXiv)
2. *[DeepMind’s AI helps untangle the mathematics of knots](https://www.nature.com/articles/d41586-021-03593-1?fbclid=IwAR3mVpuTb-gL6WDBiP7coW1vGzfK2CosYfz92Fp8sZlfUPrhGhTjSPlbuhA)*, by Davide Castelvecchi (on nature.com).
|
8
|
https://mathoverflow.net/users/12218
|
409841
| 167,730 |
https://mathoverflow.net/questions/409845
|
2
|
$\DeclareMathOperator\CSS{CSS}$It is well known that for a set $A$ of integers, if $\gcd(A) = d$,
then the set of (integer) linear combinations of $A$ is $d\mathbb{Z}$.
I'm looking for a probability generalization of this, namely the following.
Let $\varepsilon>0$, a finite set $A$ of positive integers with $\gcd(A) = d$.
Let $N$ be large (depending on $A$, $\varepsilon$) and $\alpha\in A^N$ such that
the density of every $a\in A$ in $\alpha$ satisfies
$\lvert\alpha^{-1}(a)\rvert/N\geq \varepsilon$.
Let $\CSS(\alpha)$ (consecutive sum set) denote the set of $b\in\mathbb{N}$ such that for some $n$, $m$, $b = \alpha(n)+\alpha(n+1)+\dotsb+\alpha(n+m-1)$.
**Question**: Do we have
$\lvert\CSS(\alpha) \rvert/\sum\_n\alpha(n)\geq (1-\varepsilon)/d$?
|
https://mathoverflow.net/users/74918
|
Is the consecutive sum set large in general?
|
No. Consider for instance $A = \{3,5\}$ (so $d=1$) and take $\alpha = (3,5,3,5,3,5,\dots)$. The partial sums $\alpha(1)+\dots+\alpha(n)$ are always equal to $3$ or $0$ mod $4$, so the partial sums $\alpha(n)+\dots+\alpha(n+m-1)$ always avoid $2$ mod $4$. Hence $|CSS(\alpha)|/\sum\_n \alpha\_n$ cannot exceed $3/4-o(1)$ as $N \to \infty$, which will be less than $(1-\varepsilon)/d$ when $\varepsilon$ is small and $N$ is large.
|
3
|
https://mathoverflow.net/users/766
|
409851
| 167,732 |
https://mathoverflow.net/questions/409416
|
5
|
In general, if I understand correctly, the representation theory of the braid groups is quite complicated, and there's no classification of the irreducibles. However, the braid groups form a sort of system of groups just as the symmetric groups do, and so one can ask about representation stability for coherent systems of representations of braid groups. Editing in response to Andy's comments below: it may be that "representation stability" doesn't mean anything because we can't decompose braid group representations in the same way we can decompose symmetric group representations. So a more basic question would be: is there any sensible way to talk about systems of braid group representations stabilizing that doesn't involve decomposition into irreducibles?
This is a very basic question, mostly just a reference request.
Note that I am not asking about representation stability for the braid groups themselves.
|
https://mathoverflow.net/users/11546
|
Representation stability for systems of braid group representations
|
A good way to handle systems of braid group representations is to consider the category of functors $\mathcal{C}\to R\textrm{-Mod}$, where $\mathcal{C}$ is a category with the braid groups as automorphisms. The braid groupoid $\beta$ (ie the groupoid with natural numbers as objects and braid groups as automorphisms) is then a subcategory of such $\mathcal{C}$. Note that $\beta$ itself is not quite satisfactory for such $\mathcal{C}$ since a functor $\beta\to R\textrm{-Mod}$ encodes a family of representations where the representations of $B\_{n}$ is independent of the one of $B\_{n+1}$. In other words, we would like $\mathcal{C}$ to encode compatibilities between the representations. There already exist good candidates for such category:
* the partial braid category, see Section 2.3 of Palmer <https://arxiv.org/pdf/1308.4397.pdf;>
* the Quillen’s bracket construction applied to $\beta$, denoted by $\mathcal{U}\beta$, and applied by Randal-Williams and Wahl <https://arxiv.org/abs/1409.3541> [RWW].
The latter has the significant advantage that a large class of classical families of representations of the braid groups define functors $\mathcal{U}\beta\to R\textrm{-Mod}$:
1. the Burau representations; see Example 4.3 of [RWW].
2. the Tong-Yang-Ma and Lawrence-Krammer-Bigelow representations; see Section 1.2 of <https://arxiv.org/pdf/1702.08279.pdf> [S1].
3. the whole family of the Lawrence-Bigelow representations; see Section 5.2.1.1 of <https://arxiv.org/pdf/1910.13423.pdf> [PS].
Also, there are notions of polynomiality which allows us to characterise and prove more properties on these systems of representations:
* the notion of (strong) polynomiality, a.k.a finite degree coefficient systems: the Burau representation is of degree $1$ (see Example 4.15 of [RWW]), the Tong-Yang-Ma representation is of degree $1$ and Lawrence-Krammer-Bigelow representations is of degree $2$ (see Propositions 3.25 and 3.33 of [S1]). This is the appropriate notion to prove twisted homological stability result see [RWW].
* the notion of *weak* polynomial functors, which has originally been introduced for symmetric monoidal categories (for instance $FI$) by Djament and Vespa <https://arxiv.org/abs/1308.4106>, and generalised to categories of the same type as $\mathcal{U}\beta$ (namely pre-braided monoidal categories) in <https://arxiv.org/pdf/1709.04278.pdf> (see Section 4.2). An advantage of this notion is that it reflects more accurately than the strong polynomiality the behaviour of functors in the stable range. For instance, Church, Miller, Nagpal and Reinhold <https://arxiv.org/pdf/1706.03845.pdf> compute the weak polynomial degree (named "stable degree" in this paper) of some FI-modules. Moreover, denoting by $Pol\_{d}(\mathcal{U}\beta)$ the category of weak polynomial functor of degree less or equal to $d$, we can define the quotient category
$$Pol\_{d+1}(\mathcal{U}\beta)/Pol\_{d}(\mathcal{U}\beta).$$
These quotient categories provide a new tool to handle families of representations with a sensible way to classify them. In particular, it doesn’t involve decomposition into irreducibles. See also Palmer <https://arxiv.org/pdf/1712.06310.pdf> for a comparison of the various instances of the notions of twisted coefficient system and polynomial functor. Hence weak polynomiality might be viewed as a refinement of representation stability phenomena and a sensible to talk about system of braid group representations.
|
7
|
https://mathoverflow.net/users/469926
|
409853
| 167,733 |
https://mathoverflow.net/questions/409799
|
1
|
Suppose we have a distribution $u\in B\_{\infty,\infty}^\alpha$, the Besov space with regularity coefficient $\alpha>0$. How to prove the folowing inequality?
$$
\|u\|\_{L^\infty}\leqslant c\|u\|\_{B\_{\infty,\infty}^\alpha}
$$
for some constant $c$.
|
https://mathoverflow.net/users/69279
|
How to prove that the L-infinity norm is smaller than the Besov norm?
|
With $\sum\_{\nu \ge 0}\phi\_\nu(\xi)=1$ be a Littlewood-Paley partition of unity we find that $u=\sum\_{\nu \ge 0}\phi\_\nu(D)u$ and thus
since
$$
\Vert u\Vert\_{B^\alpha\_{\infty, \infty}}=\sup\_{\nu\in \mathbb N}
2^{\nu \alpha}\Vert\phi\_\nu(D)u\Vert\_{L^\infty},
$$
we get
$$
\Vert u\Vert\_{L^\infty}\le \sum\_{\nu \ge 0}2^{\nu \alpha}\Vert\phi\_\nu(D)u\Vert\_{L^\infty}2^{-\nu \alpha}
\le \Vert u\Vert\_{B^\alpha\_{\infty, \infty}}
\underbrace{\sum\_{\nu \ge 0}2^{-\nu \alpha}}\_{c\_\alpha}.
$$
|
1
|
https://mathoverflow.net/users/21907
|
409855
| 167,734 |
https://mathoverflow.net/questions/409859
|
3
|
Let us call a *cap* the intersection of the boundary of 3-dimensional convex compact set $K$ in $\mathbb{R}^3$ with a half-space bounded by a plane $H$ such that the orthogonal projection to $H$ of this intersection is contained in $K\cap H$.
**Question.** Is it true that a cap equipped with the intrinsic metric is an Alexandrov space (with boundary) with non-negative curvature.
**Remark.** In the case when $K$ has smooth boundary and $H$ intersects $K$ transversally, the question is equivalent to asking whether the second fundamental form of the boundary of the cap is non-negative.
|
https://mathoverflow.net/users/16183
|
Is a cap an Alexandrov space?
|
Note that the union of $K\cap H$ with its reflection is a convex set.
Therefore, its surface $\Sigma$ is an Alexandrov space.
Your space is a quotient $\Sigma/\mathbb{Z}\_2$ by isometric involution.
Therefore, it is an Alexandrov space as well.
(There is a closely related problem *Convex hat*, page 21 in my [PIGTIKAL](https://arxiv.org/abs/0906.0290).)
|
4
|
https://mathoverflow.net/users/1441
|
409861
| 167,736 |
https://mathoverflow.net/questions/409864
|
6
|
Let us call a *cap* the intersection of the boundary of 3-dimensional convex compact set $K$ in $\mathbb{R}^3$ with a half-space bounded by a plane $H$ such that the orthogonal projection to $H$ of this intersection is contained in $K\cap H$.
**QUESTION.** Given a metric on a closed 2-dimensional disk which has non-negative curvature in the sense of Alexandrov. Can the disk be isometrically imbedded into $\mathbb{R}^3$ as a cap?
A reference would be helpful.
|
https://mathoverflow.net/users/16183
|
Isometric imbedding of a 2-disk into Euclidean 3-space
|
Take doubling of the disc, we obtain a metric on the sphere.
By Perelman's theorem it had nonnegative curvature in the sense of Alexandrov.
Therefore, by Alexandrov's theorem, it is isometric to a convex surface in the Euclidean space.
This convex surface is unique up to congruence (Pogorelov's theorem). Therefore, the involution of our sphere extends to a reflection of the space.
In particular, the boundary of the disc lies in one plane.
**Postscript.** You may also proceed as suggested by Joseph O'Rourke --- approximate the disc by polyhedral space, apply Alexandrov's theorem for polyhedral surfaces (which has uniqueness for free) and pass to the limit.
|
7
|
https://mathoverflow.net/users/1441
|
409866
| 167,738 |
https://mathoverflow.net/questions/409857
|
16
|
>
> Prove that there exist infinitely many integers $x$ such that integer $P(x)=x^3-2$ is a sum of two squares of integers.
>
>
>
Ideally, I am looking for a proof method that also applies for other $P(x)$, such as, for example, $P(x)=x^3+x+1$.
For $x=4t+3$, $P=(4t+3)^3-2$ is $1$ modulo $4$. By a well-believed (but difficult) Bunyakovsky conjecture, $P$ is a prime infinitely often, and every prime that is $1$ modulo $4$ is a sum of two squares.
To find an unconditional proof, it suffices to find polynomials $A(t)$, $B(t)$ and $C(t)$ with rational coefficients such that $A(t)^2 + B(t)^2 = C(t)^3-2$. Are there any good heuristics to find such polynomials? Is there any computer algebra system that helps to guess a solution of a polynomial equation over $Q[t]$?
One way to guess $C(t)$ is to form a set $S$ of all integers up to (say) $10^5$ that are sums of two squares, and look for polynomials $C(t)$ such that $C(t)^3-2$ belong to $S$ for all small $t$. I then checked all polynomials of degree up to $4$ and coefficients up to $12$, and found, for example, a polynomial $D(t)=3 + 8 t + 12 t^2 + 8 t^3 + 4 t^4$ such that $D(t)^3-1$ is always a sum of two squares. But no $C(t)$ found in this range, and increasing the degree and/or coefficients makes the enumeration infeasible. Are there methods to find a polynomial with all values in the given set, that are better than enumeration?
|
https://mathoverflow.net/users/89064
|
Representing $x^3-2$ as a sum of two squares
|
The answer is similar to one provided [here.](https://artofproblemsolving.com/community/c6h2007167p14224750) The approach is elementary and proves stronger fact that it can be expressed as sum of two coprime squares.
We consider the product $(n+2)(n^3-2)$ which is equal to $n^{4}-2n+2n^3-4$. Now we observe that $$(n+2)(n^3-2)=(n^2+n-7)^2+(13n^2+12n-53).$$
Proceeding as shown in the link we can easily get that $13n^2+12n-53$ is a perfect square for infinitely many $n\equiv 3\pmod{4}$ with first solution as $13(3^2)+12(3)-53=10^2$. If one chooses such a $n$ then clearly $n^3-2$ doesn't have any prime divisor of form $4k+3$ as $\gcd(n^2+n-7,13n^2+12n-53)\mid 25\cdot 59$ and $59$ doesn't divide $n^3-2$ if one looks at the general solution of the equation.
Since, $n^3-2$ doesn't have any prime divisor of form $4k+3$ and $n\equiv 3\pmod{4}$ we can infer from the folklore result that it can be expressed as sum of two coprime squares.
|
16
|
https://mathoverflow.net/users/160943
|
409867
| 167,739 |
https://mathoverflow.net/questions/409873
|
1
|
Let $d$ be a large positive integer. Fix a unit-vector $v \in \mathbb R^d$, and scalars $b,c \in \mathbb R$ with $b > 0$. Let $X$ be a log-concave random vector in $\mathbb R^d$ normalized so that $\mathbb E[(1/d)\|X\|^2] = 1$, WLOG.
>
> **Question 1.** *Is there a nontrivial lower-bound for $\alpha:=\mathbb E[e^{-b(v^\top X - c)^2}]$ in terms of $b$ and $c$ ?*
>
>
>
For example, if $X$ is distributed according to $N(0,I\_d)$, then $Z:=v^\top X$ has distribution $N(0,1)$, and so direct integration gives
$$
\alpha = \mathbb E\_{Z \sim N(0,1)}[e^{-b(Z-c)^2}] = \sqrt{\frac{1}{1 + 2b}}e^{-bc^2/(1 + 2b)}.
$$
>
> **Question 2.** *Same question as **Question 1**, additional condition that $X$ is isotropic.*
>
>
>
|
https://mathoverflow.net/users/78539
|
Lower-bound for $\mathbb E[e^{-b(v^\top X - c)^2}]$, when $X$ is log-concave in high-dimensions
|
The answer to Question 1 is no. Indeed, let $X=\sqrt d\,V v$, where $V\sim N(0,1)$. Then
\begin{equation}
Ee^{-b(v\cdot X-c)^2} = \frac1{ \sqrt{1 + 2bd}} e^{-bc^2/(1 + 2bd)}\to0
\end{equation}
(as $d\to\infty$). So, the only lower bound on $Ee^{-b(v\cdot X-c)^2}$ in general is the trivial bound $0$.
---
The answer to Question 2 is yes, even without the log-concavity condition. Indeed, let
\begin{equation}
l\_{b,c}:=\inf\_{|y|<\sqrt2}\frac{e^{-b(y-c)^2}}{2-y^2}
=\min\_{|y|<\sqrt2}\frac{e^{-b(y-c)^2}}{2-y^2}>0.
\end{equation}
Then $e^{-b(y-c)^2}\le l\_{b,c}(2-y^2)$ for all real $y$ and hence
\begin{equation}
Ee^{-b(v\cdot X-c)^2}
\ge l\_{b,c}(2-E(v\cdot X)^2)=l\_{b,c}
\end{equation}
-- because, if $X$ is isotropic and $E(1/d)\|X\|^2=1$, then $E(v\cdot X)^2=\|v\|^2=1$.
So, $l\_{b,c}>0$ is a nontrivial lower bound on $Ee^{-b(v\cdot X-c)^2}$ in the "isotropic" case.
(It is not hard to see that $l\_{b,c}\ge e^{-2b-2bc^2}$. The latter lower bound is not far off. Indeed, using the Prékopa–Leindler inequality, one can see that $Ee^{-b(v\cdot X-c)^2}\le e^{-b-bc^2}$ if $EX=0$, $E(1/d)\|X\|^2=1$, $X$ is isotropic, and the pdf of $X$ is log-concave.)
|
1
|
https://mathoverflow.net/users/36721
|
409880
| 167,742 |
https://mathoverflow.net/questions/409876
|
3
|
Let $k$ be a field, $A$ a finite dimensional $k$-algebra, $X$ a finite dimensional indecomposable (left) $A$-module and $M$ an infinite dimensional (left) $A$-module. Further $X\subseteq M$ and for every finite dimensional $A$-module $N$ with $X\subseteq N\subseteq M$ the inclusion $X\hookrightarrow N$ splits. Is it true that $X \hookrightarrow M$ must split? If not, does somebody know a counterexample?
|
https://mathoverflow.net/users/145920
|
Split monomorphisms of modules - does the finite case imply the infinite case?
|
Since $A$ is a finite-dimensional $k$-algebra, $M$ is the direct limit of its finite-dimensional submodules $N$ which contain $X$. Now the sequence $0\to X\to M\to M/X\to 0$ is pure-exact, since it is the direct limit of the split exact sequences $0\to X\to N\to N/X\to 0$. Also $X$ is finite-dimensional over $k$, so it is pure-injective. Thus the sequence splits, and $X$ is a direct summand of $M$.
|
5
|
https://mathoverflow.net/users/425351
|
409909
| 167,746 |
https://mathoverflow.net/questions/373792
|
13
|
This is a "forcing-absolute" followup to [this question](https://math.stackexchange.com/questions/3568276/is-there-a-specific-infinitary-sentence-second-order-logic-cant-capture), whose answer was largely unsatisfying. The question is:
>
> Suppose $V=L$. Is there an $\mathcal{L}\_{\infty,\omega}$-sentence $\varphi$ such that **in no forcing extension** is $\varphi$ equivalent to a second-order sentence?
>
>
>
Throughout, by "forcing" I mean "*set* forcing," although the (tame) class forcing version also seems potentially interesting.
EDIT: I forgot to add that I'm restricting attention to **infinite** structures here (which is key to my comment below that every projectively-definable infinitary sentence is second-order expressible). As Fedor Pakhomov commented below, without this restriction the problem is trivial since second-order theories of finite structures can't be changed by forcing. I do *not*, however, want to restrict attention to countable structures.
---
I've decided to focus on models of $\mathsf{V=L}$ since that hypothesis seems to add interesting flavor to the question in a few ways:
* It implies that such a $\varphi$ must **not** be in $\mathcal{L}\_{\omega\_1,\omega}$. This is because $(1)$ every constructible real is projective in some forcing extension and $(2)$ every projectively definable $\mathcal{L}\_{\omega\_1,\omega}$-sentence is equivalent to a second-order sentence. So a positive answer would have to crucially rely on uncountable Boolean combinations - which seems a bit odd, because each specific $\mathcal{L}\_{\infty,\omega}$-sentence is equivalent to some $(\mathcal{L}\_{\omega\_1,\omega})^{V[G]}$-sentence in an appropriate forcing extension $V[G]$ (just collapse the size of the sentence), but isn't an obvious contradiction since the "potential projectivity" fact about $L$ doesn't seem to lift to arbitrary forcing extensions of $L$.
* It rules out the "silly" solution provided by large cardinals. If large cardinals exist - specifically, enough to guarantee **projective absoluteness** - then any infinitary sentence which is not equivalent to a second-order sentence in $V$ remains so in all forcing extensions. *(Note that this would give us an example in $\mathcal{L}\_{\omega\_1,\omega}$ for that matter.)* But $\mathsf{V=L}$ breaks this "hammer," so that we seem to be forced to do some actual work.
* If the answer is yes, there is in fact a *definable* example, namely the least such sentence with respect to the $L$-ordering. Of course this is silly, but it suggests that there might be canonical examples in a more interesting sense. By contrast I could imagine models of $\mathsf{ZFC+V\not=L}$ where the existence of such a sentence is guaranteed nonconstructively (e.g. by a more intricate counting argument), and so no canonical example need exist.
That said, since it seems plausible that the $\mathsf{V=L}$-situation is more difficult to attack than I'd hoped, I'm also interested in results for other extensions of $\mathsf{ZFC}$.
|
https://mathoverflow.net/users/8133
|
Is there an infinitary sentence which is absolutely not second-order expressible?
|
For any given finite signature $\Omega$ there is a second-order sentence $\varphi$ of the signature $\Omega$ such that $\mathsf{ZFC}+V=L$ proves that for any $\mathcal{L}\_{\infty,\omega}$-formula $\psi$ of the signature $\Omega$ there is a poset $P$ for which it is $\Vdash\_P$-forced that for any infinite model $\mathfrak{M}$ we have $\mathfrak{M}\models \varphi$ iff $\mathfrak{M}\models \psi$.
$\varphi$ should be a second-order sentence expressing in all infinite models $\mathfrak{M}$ that the following holds (see below for a more explicit construction of $\varphi$):
1. there exists the greatest ordinal $\alpha$ such that the value $\aleph\_\alpha^{L\_\beta}$ is the same for all large enough countable limit ordinals $\beta$;
2. $\alpha$ is the position in $<\_L$ of some $\mathcal{L}\_{\omega\_1,\omega}$-sentence $\psi$
3. $\psi$ is true in $\mathfrak{M}$.
Given $\mathcal{L}\_{\infty,\omega}$-formula $\psi$ we consider $\alpha\_0$ that is the position of $\psi$ in $<\_L$. Let $P$ be the Levy collapse of $\aleph\_{\alpha\_0}$ onto $\omega$. Let us check that for this $\varphi,\psi$, and $P$ we have the desired equivalence.
Indeed let us reason inside $\Vdash\_P$. Clearly, $\psi$ became an $\mathcal{L}\_{\omega\_1,\omega}$-formula.
Observe that $\alpha$ from 1. will always be equal to $\alpha\_0$. Indeed, let us choose countable limit $\beta\_0>\alpha\_0$ such that all ordinals $\gamma<\alpha\_0$ that aren't $L$-cardinals aren't cardinals in $L\_{\beta\_0}$. Clearly, for any limit $\beta>\beta\_0$, we have $\aleph\_{\alpha}^{L\_\beta}=\aleph\_\alpha^L$. For any $\alpha'>\alpha\_0$ since $\aleph\_{\alpha'}^L$ isn't countable and any countable $\gamma$ we could find limit countable $\beta'$ such that $\aleph\_{\alpha'}^{L\_{\beta'}}$ is either undefined or has the value $>\gamma$. Hence $\alpha\_0$ is the ordinal $\alpha$ from 1.
Using this the proof of semantical equivalence of $\varphi$ and $\psi$ in infinite models is trivial.
---
Now let me give more detailed description of $\varphi$.
Let $\mathsf{KPUL}\_2(\in,\in\_u)$ be a second-order formula depending on element variable $\alpha$ and binary predicate symbols $\in,\in\_u$ asserting that we have $(\mathsf{KPU}-\mathsf{Foundation})+\textsf{Second-Order-Foundation}+L[\mathfrak{M}]=V$ for the structure where each element of the underlying domain simultaniously represent itself (as an urelement) and some set, $\in$ gives membership of sets in sets, $\in\_u$ gives membership of urelements in sets. Formula $\mathsf{Emb}(R,\in,\in\_u,\in',\in\_u')$ expresses that $\mathsf{KPUL}\_2(\in,\in\_u)\land \mathsf{KPUL}\_2(\in',\in\_u')$ and the unary function $f$ gives an end embedding of the admissible set given by $(\in,\in\_u)$ into the admissible set $(\in',\in\_u')$. Formula $\mathsf{St}(\alpha,\beta,\kappa,\in,\in\_u)$ expresses that $\mathsf{KPUL}\_2(\in,\in\_u)$ and in the corresponding admissible set $\alpha$ is a countable ordinal, $\beta$ is a countable limit ordinal, $\kappa$ is the ordinal that is $\alpha$-th cardinal according to $L\_\beta$, and for any $f,\in',\in\_u',\beta'$ if $\mathsf{Emb}(f,\in,\in\_u,\in',\in\_u')$ and $\beta'>f(\beta)$ is a countable limit ordinal according to the admissible set given by $(\in',\in\_u')$, then there is $f(\alpha)$-th cardinal according to $L\_{\beta'}$ and it is equal to $f(\kappa)$. Formula $\mathsf{MSt}(\alpha,\in,\in\_u)$ expresses that $\mathsf{St}(\alpha,\in,\in\_u)$ and we don't have $\mathsf{St}(\alpha+1,\in,\in\_u)$. Formula $\mathsf{LC}(\alpha,\psi,\in,\in\_u)$ expresses that in the admissible set given by $(\in,\in\_u)$, $\alpha$ is an ordinal, $\psi$ is a constructible $\mathcal{L}\_{\omega\_1,\omega}$-sentence of the signature $\Omega$, and its place in $<\_L$-order is $\alpha$. Formula $\mathsf{Tr}(\psi,\in,\in\_u)$ expresses that in the admissible set given by $(\in,\in\_u)$, $\psi$ is an $\mathcal{L}\_{\omega\_1,\omega}$-sentence of the signature $\Omega$ that is true in the underlying model $\mathfrak{M}$.
We put $\varphi$ to be:
$$\exists \in,\in\_u,\alpha,\psi(\mathsf{KPUL}\_2(\in,\in')\land \mathsf{MSt}(\alpha,\in,\in')\land \mathsf{LC}(\alpha,\psi,\in,\in')\land \mathsf{Tr}(\psi,\in,\in')).$$
|
6
|
https://mathoverflow.net/users/36385
|
409919
| 167,749 |
https://mathoverflow.net/questions/409235
|
4
|
Let $G$ be a smooth affine group scheme over a base $S$. $G$ acts on a scheme $X$ over $S$. Let $x$ be an $S$-point in $X$. Then we have an orbit map $G\to X$. I wonder when the image (set-theoretically) of this map is locally closed, and the induced scheme structure (the minimal one) on the orbit is smooth over $S$.
|
https://mathoverflow.net/users/5082
|
Smoothness of orbit of group scheme
|
$\textbf{Edit by afh:}$ *Unfortunately this answer is not correct. I apologize, there is a small bug in one of the last steps in the argument below (the surjective morphism of flat schemes at the end does not need to be a closed immersion/isomorphism). In fact the statement of the proposition below is not true even if the base is a DVR and $G$ is etale and quasifinite. The proof below only shows that the set theoretic image is locally closed when the base is a DVR.*
Let me add another answer addressing the first comment of the OP above, where he asks for possible hypothesis that ensure the orbit is locally closed. I hope I am not making a lot of mistakes. In summary, there is a positive answer to the question if one assumes that the scheme theoretic stabilizer $G\_x$ of the section $x$ is flat.
Here is the setup: let $S$ be a Noetherian scheme (probably can be removed) and let $X$ be a scheme of finite type over $S$. Let $G$ be a smooth affine algebraic group over $S$, and fix an action of $G$ on $X$. Let $x: S \to X$ be a section. If the scheme theoretic stabilizer $G\_x$ is flat over $S$, then we can form the quotient algebraic space $G /G\_x$ over $S$. By construction, the action morphism
$$ G \to X, \; \; g \mapsto g \cdot x$$
factors through the quotient $G \to G/G\_x$. Therefore we get a natural morphism $G/G\_x \to X$. We shall denote $O= G/G\_x$ and call it the orbit.
$\textbf{Proposition}$ In the setup above (under the assumption that $G\_x$ is $S$-flat), the orbit $O$ is represented by a smooth scheme over $S$ and the natural morphism $O \to X$ is a locally closed immersion.
Proof: First, since smoothness can be checked flat locally and $G$ is smooth, the fppf quotient morphism $G \to G/G\_x$ shows that $O$ is smooth over $S$. It is not difficult to show that the orbit morphism $O \to X$ is a monomorphism of algebraic spaces (cf. Section 2.1 in <https://arxiv.org/abs/0804.2242>). Now, since $X$ is a scheme, we can apply <https://stacks.math.columbia.edu/tag/03XX> to the monomorphism $O \hookrightarrow X$ to conclude that $O$ is a scheme. It remains to check that $O \hookrightarrow X$ is a locally closed immersion.
We will use EGA IV (15.7.6). This Corollary in EGA says that if the valuative criterion for local properness (to be explained below) is satisfied for $ \phi: O \hookrightarrow X$, then the morphism $\phi$ factors as a composition $h \circ g$, where $h$ is an open immersion and $g$ is proper. In this case this would mean that $g$ is a proper monomorphism, and hence a closed immersion. In this (quasicompact) situation, this would in turn imply that $\phi: O \to X$ is a locally closed immersion. We are left to prove the valuative criteria mentioned above.
This is what we have to show. Let $R$ be a DVR with field of fractions $K$. Suppose that we are given a morphism $Spec(R) \to X$ that factors set theoretically through the set theoretic image $\phi(O)$. The local valuative criterion stipulates that any section $Spec(K) \to O \times\_{X} Spec(K)$ must extend uniquely to a section $Spec(R) \to O \times\_{X} Spec(R)$. In order to check this, we are free to base change using the morphism $Spec(R) \to X \to S$ in order to assume that the base $S$ is the spectrum of a DVR.
So we assume that $S$ is $Spec(R)$ with generic point $\eta$ and special point $s$. Take the scheme theoretic image of $Z \subset X$ of the morphism $\phi: O \hookrightarrow X$. Since $S$ is a DVR and $O$ is $S$-flat, it follows that $Z$ is automatically flat over $X$ (this is the crucial reason why we passed to a DVR). It can be checked that $G$ still acts on $Z$ (as the scheme theoretic image of a quasicompact $G$-equivariant morphism), so we might as well replace $X$ with $Z$ and assume that $X$ is flat and $O$ is scheme theoretically dense in $X$. Now taking scheme closure commutes with flat base change, so the generic fiber $O\_{\eta} \hookrightarrow X\_{\eta}$ is scheme theoretically dense. The usual argument for orbits over fields (notice that the construction of $O$ commutes with arbitrary base-change!) shows that $O\_{\eta} \hookrightarrow X\_{\eta}$ is an open immersion. Since $O$ is smooth, this shows that $X\_{\eta}$ is geometrically reduced with the same dimension as $O\_{\eta}$, and the boundary $B\_{\eta} = X\_{\eta} \setminus O\_{\eta}$ has strictly smaller dimension. By flatness, the dimension of $X\_{s}$ is the same as the dimension of $X\_{\eta}$, and so we must have that $X\_{s}$ and $O\_{s}$ have the same dimension. Again, the usual argument for fields implies that $O\_{s} \hookrightarrow X\_s$ is locally closed, and since it is full dimension and smooth this means that the image of $O\_s \subset X\_s$ is open (but note that $X\_s$ could be nonreduced, so we don't know yet that $O\_s \to X\_s$ is an open immersion).
We equip the boundary with $B\_{\eta}$ with its reduced subscheme structure. Since $G\_{\eta}$ is geometrically reduced, it acts on $B\_{\eta}$. Take the scheme theoretic closure in $X$ of the boundary $B\_{\eta} \to X\_{\eta}$, let's call it $B$. $B$ is $G$-stable, and the fibers have smaller dimension than the fibers of $O$. This means that $B\_s$ must be disjoint from the open orbit $O\_s$ of bigger dimension. Hence the closed subset $B$ is disjoint from the image of $\phi: O \hookrightarrow X$, and we can replace $X$ with $X \setminus B$. Hence we can assume that the generic fiber $\phi\_{\eta} : O\_{\eta} \to X\_{\eta}$ is an isomorphism. Finally, by removing the closed complement $X\_s \setminus O\_s$ inside the generic fiber $X\_s$, we can assume that the morphism of $R$-flat schemes $\phi: O \to X$ becomes an infinitesimal closed immersion when restricted to the special fiber ($O\_s$ will be the reduced subscheme of $X\_s$ if $X\_s$ is not reduced). Since the restriction to the generic fiber is also a closed immersion (isomorphism actually), this implies that $O \hookrightarrow X$ is a closed immersion. Since $O$ is schematically dense inside $X$, this means that $O \hookrightarrow X$ must be an isomorphism. In summary, we have shown that $O \to X$ induces an open immersion into its scheme theoretic closure. This implies that $O \to X$ is a locally closed immersion, and so by the same proposition in EGA IV (15.7.6), it satisfies the valuative criterion for local properness. QED
$\textbf{Last Remark/ Warning:}$ The scheme theoretic image $Z$ of the locally closed immersion $O \hookrightarrow X$ does not need to commute with passing to fibers. In other words, the fibers of the orbit $O$ do not need to be dense in the fibers of the closure $Z$. Therefore one cannot in principle conclude properties of the fibers of $Z$ in terms of properties of $O$. An example that I like is to let $R$ be a DVR with uniformizer $\pi$ and consider the scheme $X = Spec(R[s,t]/(st - \pi))$. We can let $\mathbb{G}\_m$ act on $X$ with weight $-1$ on $t$ and weight $1$ on $s$. Consider the section $(t,s) = (1,\pi)$. The stabilizer of this section is trivial, and the orbit $O$ is the open immersion $O \hookrightarrow X$ with closed complement the vanishing locus of $t$. At the generic fiber we have an isomorphism $O\_{\eta} \cong X\_{\eta}$ (there is a single orbit at the generic fiber), but at the special fiber the complement $X\_s \setminus O\_s$ is the vanishing locus of $t$, which contains two orbits: another open orbit of $(t,s) = (0,1)$, and the closed orbit at the origin $(t,s) = (0,0)$.
|
3
|
https://mathoverflow.net/users/339730
|
409924
| 167,752 |
https://mathoverflow.net/questions/409661
|
3
|
We refer to [1] for the notions used in this post.
The Grothendieck ring of a fusion category (over $\mathbb{C}$) is a fusion ring, but there are fusion rings which are not of this form (categorification problem), for example if there are only two simple objects $1,X$ (up to isomorphism) then the fusion ring is completely determined by a non-negative integer $n$ such that $X \otimes X \simeq 1 \oplus n X$, and it is a Grothendieck ring of a fusion category if and only if $n=0,1$; see [2].
Note that a fusion category is a finite semisimple tensor category. Now, if we drop out the semisimple assumption:
**Question 1**: Is there a fusion ring which is not the Grothendieck ring of a finite tensor category?
**Question 2**: Is there a fusion ring which is the Grothendieck ring of a finite tensor category, but not of a fusion category?
[1, Remark 4.9.2]: the Grothendieck ring of a finite tensor category is not always a fusion ring; for example if $X$ is the $2$-dimensional irreducible representation of $S\_3$ over a field of characterisitic two, then $[X \otimes X^\* : \boldsymbol{1}] >1$.
*Bonus question*: Is there a finite tensor category over $\mathbb{C}$ whose Grothendieck ring is not a fusion ring?
---
References
[1]: P. Etingof, S. Gelaki, D. Nikshych, V. Ostrik; Tensor categories. Mathematical Surveys and Monographs (2015) 205.
[2]: Ostrik, Viktor. Fusion categories of rank 2. Math. Res. Lett. 10 (2003), no. 2-3, 177--183
|
https://mathoverflow.net/users/34538
|
Non-semisimple categorification problem of fusion rings
|
Here are answers by Pavel Etingof (reproduced with his authorization):
**Question 1**: Is there a fusion ring which is not the Grothendieck ring of a finite tensor category?
*Answer*: Yes, for example the rings in Example 8.19 in
<https://arxiv.org/pdf/math/0203060.pdf>
given by $gX=Xg=X, X^2=X+\sum\_{g\in G}g$ do not admit a categorification (even non-semisimple) when $|G|>2$.
Indeed, assume the contrary. Since $gX=Xg=X$, we get that $X^2$ both projects to $g$ and contains $g$ for any $g\in G$. Thus each g and hence $\oplus\_{g\in G}g$ is a direct summand of $X^2$. So any category which categorifies this ring must be semisimple. But in Example 8.19 it is shown that this ring has no semisimple categorification.
**Question 2**: Is there a fusion ring which is the Grothendieck ring of a finite tensor category, but not of a fusion category?
*Answer*: I don’t know. In positive characteristic i think it is likely that such examples exist.
**Bonus question**: Is there a finite tensor category over $\mathbb{C}$ whose Grothendieck ring is not a fusion ring?
*Answer*: absolutely. For example see the book "Tensor categories",
[http://www-math.mit.edu/~etingof/egnobookfinal.pdf](http://www-math.mit.edu/%7Eetingof/egnobookfinal.pdf)
p.105 (small quantum $sl(2)$). In fact, this is the typical behavior. Moreover, if the category is pivotal factorizable and non-semisimple (as in the above example), this must happen by
<https://arxiv.org/pdf/1703.00150.pdf>
Namely, it has a simple projective object $X$, so $X\otimes X^\*$ is projective and contains projective cover $P(1)$ as a summand. If the Grothendieck ring is a fusion ring then $[X\otimes X^\*:1]=1$, so since the category is unimodular (as it is factorizable), we get $P(1)=1$, hence it is semisimple.
|
4
|
https://mathoverflow.net/users/34538
|
409926
| 167,753 |
https://mathoverflow.net/questions/408737
|
5
|
$\DeclareMathOperator\FSym{FSym}\DeclareMathOperator\Sym{Sym}$*Notation: for $X$ a set, $\Sym(X)$ the group of permutations of $X$, and let $\FSym(X)$ be the subgroup of finitely supported permutations of $X$ (it is generated by transpositions).*
Let $G$ be an infinite group. Let $P\_G$ be the subgroup of $\Sym(G)$ generated by left translations and $\FSym(G)$. Equivalently, these are permutations of $G$ that coincide with a left translation outside a finite subset.
Thus $P\_G$ is naturally a semidirect product $\FSym(G)\rtimes G$. One can prove various things about this group (e.g., if $G$ is finitely generated so is $P\_G$, the group $P\_G$ is never finitely presented, never Kazhdan, etc.
The case $G=\mathbf{Z}$ of this construction is particularly well-known (as far as I know it essentially appears in a 1937 paper of B.H. Neumann) and well-documented.
I remember reading a paper about groups $P\_G$ (at least 10 years ago), including these results (or so of them). Despite significant efforts to find the right keywords, I wasn't able to locate it. Does anybody identify this (or any paper referring to this construction in general, not just $G=\mathbf{Z}$)?
|
https://mathoverflow.net/users/14094
|
Permutations of a group that are eventually left translations
|
**Theorem** (Elek–Szabo)**.** Let $G$ be an infinite residually finite hyperbolic group with Property (T). Then $P\_G$ is a finitely generated sofic group that is not residually amenable.
That's Theorem 3 of *Elek, Gábor; Szabó, Endre*, [**On sofic groups.**](http://dx.doi.org/10.1515/JGT.2006.011), J. Group Theory 9, No. 2, 161-171 (2006). [ZBL1153.20040](https://zbmath.org/?q=an:1153.20040) [arXiv:math/0305352](https://arxiv.org/abs/math/0305352).
The construction is also somewhat reminiscent of Houghton's groups.
|
3
|
https://mathoverflow.net/users/24447
|
409929
| 167,754 |
https://mathoverflow.net/questions/409942
|
3
|
Let $X$ be a closed subspace of a Banach space Y. I have functionals $f\_0, f\_1, \ldots, f\_n\in X^\*$ such that $f\_0$ is in the span of the remaining ones. I fix an extension of $f\_0$ to $Y$; let me call it $F\_0$. Can I extend them to functionals $F\_1, \ldots, F\_n$ on $Y$ in a way that $F\_0$ is in the span of $F\_1, \ldots, F\_n$? I don't really care about preserving the norms of the original functionals.
|
https://mathoverflow.net/users/470412
|
A Hahn-Banach type extension problem for multiple functionals
|
If $f\_0\ne0$ or if $f\_0=0$ and the $f\_1,\dotsc,f\_n$ are not linearly independent, then the answer is trivial:
In this case there is another functional, say $f\_1$, which is in the span of the remaining ones, say $f\_1=\sum\_{k\ne1}\lambda\_kf\_k$ with $\lambda\_0\ne0$: Extend the $f\_k$ to $F\_k$ $(k=2,\dotsc,n)$ and put $F\_1=\sum\_{k\ne1}\lambda\_kF\_k$.
In the remaining case, $f\_0=0$ and $f\_1,\dotsc,f\_n$ being linearly independent, the answer to your question is obviously positive if and only if $F\_0=0$ (because $\sum\_{k=1}^n\lambda\_kF\_k=F\_0$ implies by restriction to the subspace $\lambda\_1=\dotsc=\lambda\_n=0$).
|
6
|
https://mathoverflow.net/users/165275
|
409948
| 167,759 |
https://mathoverflow.net/questions/409959
|
4
|
I've written a paper that a) demonstrates an equivalence between conditional complexity $K$($Y$|$X$) in information theory and the random component of an effect size estimate $r\_{xy}$, and then b) shows that certain metrics of conditional complexity related to Hamming distance can be interpreted as indicating the sign and magnitude of a certain random component of $r\_{xy}$. In other words, a function of the Hamming distance between two vectors of scores is indicative of the accuracy of the scores' sample correlation (not to be confused with the precision of $r\_{xy}$, its standard error.) A reasonable statistician (or editor of a statistics journal) might reject this claim out of hand because the conventional interpretation of estimation theory has no mechanism for accuracy estimation: it only recognizes one kind of information, parameter (Fisher) information, whereas the random component of a particular estimate from a particular sample isn't a parameter. Or they might just not be used to thinking of "information" in two different ways. So the paper would be very important if correct, but its correctness may be difficult to evaluate.
Indeed, I've submitted the paper to two journals, Psychometrika and Statistical Papers, and I've gotten two form-letter rejections. I fully accept that the respective editors may have given the paper keen consideration, bringing to bear both an open mind and the necessary interdisciplinary knowledge, and they simply found it to be wrong. This isn't me just complaining about rejections, but I don't want to just keep submitting it blindly either way. Assuming it isn't junk, can you recommend a journal that would be receptive to this kind of paper?
|
https://mathoverflow.net/users/170051
|
What journal(s) do you recommend for submitting a paper on a topic that spans information theory and estimation theory?
|
IEEE transactions on Information Theory comes to mind. Their specifications of topics is broad.
Fisher information regularly appears in papers there.
>
> The IEEE Transactions on Information Theory is a journal that publishes theoretical and experimental papers concerned with the transmission, processing, and utilization of information. The boundaries of acceptable subject matter are intentionally not sharply delimited. Rather, it is hoped that as the focus of research activity changes, a flexible policy will permit this Transactions to follow suit. Current appropriate topics are best reflected by recent Tables of Contents; they are summarized in the titles of editorial areas that appear on the inside front cover.
>
>
>
|
3
|
https://mathoverflow.net/users/17773
|
409965
| 167,765 |
https://mathoverflow.net/questions/409944
|
6
|
Normal compact linear operators on Hilbert spaces have infinitely many (counting multiplicities) eigenvalues by the spectral theorem.
For non-normal operators this no longer has to be true.
There exist even examples of compact operators without eigenvalues such as weighted shifts and the Volterra operator $Tf(t) = \int\_0^t f(s) \ ds$ on $L^2(0,1),$ which when applied to a polynomial basis can also be interpreted as a shift. So a perhaps bold guess would be that we want to avoid some generalized infinite Jordan blocks.
*I would therefore like to understand: Do there exist other criteria not assuming normality that imply the existence of infinitely many eigenvalues for compact operators on Hilbert spaces?*
|
https://mathoverflow.net/users/119875
|
Criteria for operators to have infinitely many eigenvalues
|
The only criterion I know is based on a Theorem in the second book of Dunford and Schwartz, see Theorem X1.6.29 and following. If the resolvent of an Hilbert-Schimdt operator satisfies some decay estimates on some rays dividing the complex plane, then the span of the generalized eigenfuntions is dense. The theorem generalizes to operators having trace-class powers and to closed operators with trace-class powers of the resolvent.
|
2
|
https://mathoverflow.net/users/150653
|
409966
| 167,766 |
https://mathoverflow.net/questions/409918
|
2
|
I heard that there is an $\infty$-category $\mathbf{Top}\_\infty$ whose objects are topological spaces, whose 1-morphisms are continuous maps, whose 2-morphisms are homotopies, whose 3-morphisms are homotopies between homotopies, and so on.
*Question 1:* What is a homotopy between homotopies? What is a homotopy between *such* homotopies? (I can't find a definition using Google.)
*Question 2:* Is there any definition of $\mathbf{Top}\_\infty$ in the literature if we model $\infty$-categories as *quasicategories*?
This $\infty$-category should have at least the property that its homotopy category $\mathrm{h}\mathbf{Top}\_\infty$ is ["the naive homotopy category"](https://en.wikipedia.org/wiki/Homotopy_category#The_naive_homotopy_category).
My motivation is the following: If we have defined $\mathbf{Top}\_\infty$, then we can consider the subcategory $\mathbf{Type}\_\infty\subseteq \mathbf{Top}\_\infty$ of all CW complexes. We need that category in order to formulate Grothendieck's homotopy hypothesis:
>
> There is an equivalence of $\infty$-categories $\mathbf{Type}\_\infty \to \infty\mathbf{Grp}\_\infty$.
>
>
>
Note that the $\infty$-category $\infty\mathbf{Grp}\_\infty$ of $\infty$-groupoids has already been defined in the literature. It is discussed in Chapter 3 of Lurie's *Higher Topos Theory* (consider the subcategory of the $\infty$-category of all $\infty$-categories consisting of Kan complexes).
*Question 3:* Can the homotopy hypothesis be proved in this setting?
|
https://mathoverflow.net/users/470393
|
Precise definition of the $\infty$-category of spaces, continuous maps, homotopies, homotopies between homotopies, and so on
|
Here's an answer for question 1. (This bothered me for a long time too, I also could never find a formal definition in the literature!)
---
If one uses 'nice' topological spaces (so that Top has an [internal hom](https://mathoverflow.net/questions/403714)), then one can define a **homotopy** from a map $f: X\to Y$ to another map $g:X\to Y$ to be a continuous map
$$H:[0,1]\to\mathbf{Map}(X,Y)$$
with $H(0)=f$ and $H(1)=g$.
---
Then, given two such homotopies $H\_1,H\_2:[0,1]\to\mathbf{Map}(X,Y)$, we can define a **homotopy of homotopies** or a **$2$-homotopy** $\eta: H\_1\to H\_2$ to be a continuous map
$$\eta:[0,1]\to\mathbf{Map}([0,1],\mathbf{Map}(X,Y))$$
with $\eta(0)=H\_1$ and $\eta(1)=H\_2$.
---
One can keep doing this, defining a notion of **$n$-homotopy** using that of an **$(n-1)$-homotopy**.
---
To get the usual definition (which works for arbitrary topological spaces!), one uses that the functors $(-)\times X$ and $\mathbf{Map}(X,-)$ are adjoint.
---
So a **homotopy**
$$H:[0,1]\to\mathbf{Map}(X,Y)$$
is the same as a continuous map
$$H^\dagger:X\times[0,1]\to Y,$$
where the endpoint conditions are now given by
\begin{align\*}
H^\dagger(x,0) &= f(x),\\
H^\dagger(x,1) &= g(x)
\end{align\*}
for all $x\in X$.
---
Similarly, a **homotopy between homotopies**
$$\eta:[0,1]\to\mathbf{Map}([0,1],\mathbf{Map}(X,Y))$$
is the same as a map
$$\eta^\dagger:[0,1]\times[0,1]\to\mathbf{Map}(X,Y)$$
such that
\begin{align\*}
\eta^\dagger(0,t) &= H\_1(t),\\
\eta^\dagger(1,t) &= H\_2(t)
\end{align\*}
for all $t\in[0,1]$, which is the same as a map
$$\eta^\ddagger:[0,1]\times[0,1]\times X\to Y$$
such that
\begin{align\*}
\eta^\ddagger(0,t,x) &= H^\dagger\_1(t,x),\\
\eta^\ddagger(1,t,x) &= H^\dagger\_2(t,x)
\end{align\*}
for all $t\in[0,1]$ and all $x\in X$.
---
One can keep going like this for $n$-homotopies, applying adjointness $0$ to $n-1$ times, leading to $n$ equivalent definitions of an $n$-homotopy!
|
4
|
https://mathoverflow.net/users/130058
|
409969
| 167,768 |
https://mathoverflow.net/questions/409797
|
8
|
I am looking for an example where a transferred model structure fails to exist, *even if one is willing to work with semi-model category.* But let me be more precise:
Let's say I have a combinatorial model category $C$, a locally presentable category $D$ and an adjunction :
$$ L: C \rightleftarrows D : U$$
A classical (at least - mentioned on the nLab) necessary and sufficient condition (in this case) for the existence of a transferred model structure on $D$ is that one has the following two:
**(A)** For every object $X \in D$ such that $U(X)$ is fibrant, there exists a "path object"
$X \overset{a}{\to} P \overset{p}{\to} X \times X$ such that $U(a)$ is a weak equivalence and $U(p)$ is a fibration.
**(B)** There exists a "fibrant replacement" functor and natural transformation $X \overset{a\_x}{\to} FX$ on $D$, such that $U(FX)$ is fibrant and $U(a\_x)$ is a weak equivalence.
I know examples where condition (B) fails, but I can't find an example where (A) fails. Do you know one ?
Some details and motivations:
In practice, it appears that condition (A) is often almost free and condition (B) is the hard one. For example, if $C$ is a simplicial model category, $D$ is simplicially enriched (with cotensor) and the adjunction is a simplicial adjunction, you can take $P$ to be the cotensor $P = X^{\Delta[1]}$. The same applies with other enrichement.
Now, it also appears that condition (A) is sufficient to build a "transferred model structure" on D, at least if one is willing to work with right semi-model category and slightly generalizing what one means by transferred model structure. So failure of condition (B) isn't really a deal breaker, but just an additional hassle.
This being said, I can't find a single example where condition (A) fails.
|
https://mathoverflow.net/users/22131
|
example of "really" non-existent transferred model structure
|
The usual example in operad theory is when $C$ is a combinatorial, monoidal model category and $D$ is the category of commutative monoids in $C$. Unless $C$ satisfies a strong condition (that in my thesis, I called the *commutative monoid axiom*) guaranteeing symmetric powers are homotopically well-behaved, $D$ won't even have a semi-model structure.
For example, if $C = Ch(\mathbb{F}\_p)$ is chain complexes over a field $k$ of characteristic p, then it is easy to show that $D$ can't have a transferred semi-model structure. You know that, if it did, then the generating trivial cofibrations would be of the form $Sym(J)$ where $J$ is the set of generating trivial cofibrations in $C$, and $Sym$ is the free commutative monoid function (L in your notation). Recall that maps in $J$ look like $0\to D(n)$ where $D(n)$ is the chain complex with one copy of $k$ in degrees $n$ and $n-1$, and identity boundary map.
Let's be quite explicit. Take $p=2$. Then $Sym(0)=k \to Sym(D(n))$ is not a weak equivalence, because if $y\in D(n)$ is non-zero then $y^2 \in Sym(D(n))$ is a cycle of degree $2n$ which is not a boundary. This is Example 3.7 in [Model Categories and Simplicial Methods](https://arxiv.org/abs/math/0609537).
|
4
|
https://mathoverflow.net/users/11540
|
409987
| 167,774 |
https://mathoverflow.net/questions/409973
|
10
|
A real/complex rational atlas on a smooth closed manifold $M$ is an atlas with charts homeomorphic to Euclidean open sets in $\Bbb{R}^n$/$\Bbb{C}^n$ covering $M$ and real/complex rational transition maps. A real/complex rational structure is a maximal collection of compatible mentioned real/complex atlases. Two rational structures are called isomorphic if there is a topological automorphism $\sigma$ of $M$ such that each coordinate representation of $\sigma$ under a rational chart pair (from each structure) is rational. The existence of a rational structure does not imply being algebraic because every complex torus admits one (the transition maps are translations, thus polynomial), so I want to ask about how rigid these structures are.
**Q$1$**: Are there topological invariants classifying whether a smooth manifold admits a real/complex rational structure? Or on the contrary every smooth manifold admits one?
**Q$2$**: Obviously a real/complex rational structure belongs to a smooth/holomorphic structure. Consider the converse problem: what is the moduli space of real/complex rational structures compatible with the smooth/holomorphic structure on $M$? Specifically, **is it true that the number of complex rational structures belonging to the holomorphic structure of $M$ is at most $1$?** Counterexamples are welcome.
**Q$3$**: A [related question](https://mathoverflow.net/questions/317125/defining-algebraic-manifold-without-referring-to-schemes) exists on this website, and the first answer gives a sheaf-sense definition of rational structures. However its conclusion -- the algebraic structure of $\Bbb{C}^n$ can be pullbacked onto $M$ -- seems to contradict my complex torus example. I hope someone can help.
|
https://mathoverflow.net/users/166298
|
Algebraic atlas on smooth manifolds
|
The answers to your question in the case $n=1$ are well-known. In higher dimensions, the answers are less complete, but something is known.
For example, in the real case when $n=1$, there is only one smooth, connected compact $1$-manifold, the circle, and, for each natural number $k\ge1$, there is a rational structure $\mathcal{R}\_k$, which is the rational structure induced on the $k$-fold connected cover of $\mathbb{RP}^1\simeq S^1$ (which is smoothly diffeomorphic to $S^1$, of course). Two rational structures $\mathcal{R}\_i$ and $\mathcal{R}\_j$ on the circle are isomorphic if and only if $i=j$, and every rational structure on the circle is isomorphic to some $\mathcal{R}\_i$.
Meanwhile, in the complex case when $n=1$, specifying a rational structure on a compact Riemann surface of genus $g$ (i.e., a topological surface of genus $g$ with a fixed underlying holomorphic structure) is easily seen to be equivalent to specifying a *projective structure* on the surface. For $g>1$, it is known that the moduli of projective structures on a given compact Riemann surface is equivalent to the moduli of quadratic holomorphic differentials, a complex vector space of dimension $3g{-}3$. For example, see R. C. Gunning's *On uniformization of complex manifolds: The role of connections*. In particular, these provide counterexamples sought by the OP.
In higher dimensions the situation is more complicated because the pseudogroup of rational maps with a rational inverse, i.e., the so-called *birational pseudogroup*, in either $\mathbb{R}^n$ or $\mathbb{C}^n$ is not well-understood when $n>1$.
However, when $n=2$, the question of when a given compact complex surface has a rational structure is well-understood since we have a classification of compact complex surfaces, by the work of Kodaira. Not all such surfaces have a rational structure; for example, a K3 surface does not.
|
15
|
https://mathoverflow.net/users/13972
|
409990
| 167,775 |
https://mathoverflow.net/questions/409819
|
7
|
Assuming that $x$ is a sequence of $l$ bits (i.e. a binary word of length $l$) and $0 \le m < l$, let $R(x, m)$ denote the result of the left bitwise rotation (i.e. the left [circular shift](https://en.wikipedia.org/wiki/Circular_shift)) of $x$ by $m$ bits. For example, if $x = 0100110001110000$, then $l = 16$ and $$\begin{array}{l}
R(x,0) = x = {\rm{0100110001110000}},\\
R(x,1) = {\rm{1001100011100000}},\\
R(x,2) = {\rm{0011000111000001}},\\
\ldots \\
R(x,15) = {\rm{0010011000111000}}.
\end{array}$$
Let $A \oplus B$ denote the result of the bitwise [“exclusive or”](https://en.wikipedia.org/wiki/Exclusive_or) operation for two sequences of $l$ bits. For example, $$0100110001110000 \oplus 1010010001000010 = 1110100000110010.$$
Let $H(x)$ denote the number of non-zero bits in $x$ (i.e. the [Hamming weight](https://en.wikipedia.org/wiki/Hamming_weight) of $x$).
Assuming that $x$ is an $l$-bit word, let $f(x)$ denote the minimal element (the smallest number) in the tuple $$\begin{array}{l}
(H(x \oplus R(x, 1)),\\
H(x \oplus R(x, 2)),\\
\ldots, \\
H(x \oplus R(x, l - 2)),\\
H(x \oplus R(x, l - 1))).
\end{array}$$
For example, $$\begin{array}{l}
f(10011110100010010011) = 10,\\
f(11111111111111111111111111111111) = 0,\\
f(10000000000000000000000000000000) = 2,\\
f(10011100001111100000011111110000) = 8,\\
f(01000110111111100011100100100101) = 16.
\end{array}$$
Question: assuming that for any even natural number $n \ge 2$ there exists a $2n$-bit word $x$ such that $f(x) = n$ and $H(x) = n$, what can be a *time- and space-efficient* algorithm which, given an arbitrary (even) $n \ge 2$, allows to find *at least one* such $x$? For example, if $n = 10$, it is easy to check all $20$-bit words and find (in the lexicographic order) $$\begin{array}{l}
x\_0 = 00000101011011001111,\\
x\_1 = 00000101011110011011,\\
x\_2 = 00000101101110011101,\\
x\_3 = 00000101110011101101,\\
\ldots,\\
x\_{718} = 11111010100001100100,\\
x\_{719} = 11111010100100110000.
\end{array}$$
But as the value of $n$ grows, it becomes infeasible to check the huge amount of elements, even if one skips any element $x$ such that $H(x) \neq n$. For example, how can one find a $256$-bit word $x$ such that $f(x) = H(x) = 128$?
Let $S\_n (n = 2, 4, 6, \ldots)$ denote cardinality of the set of solutions for the corresponding $n$. Note that each solution $x$ is an element of the *family* of solutions, and this family contains exactly $2n$ elements, namely, $x$ and $(2n-1)$ its rotations. So we can pick a single (e.g. lexicographically first) element of a family and call it a *canonical solution* (all other elements of the corresponding family can be generated from this solution in a trivial way). Then let $C\_n (n = 2, 4, 6, \ldots)$ denote cardinality of the set of canonical solutions for the corresponding $n$. If my computations are correct, we have
$$\begin{array}{l}
S\_2 = 4 = 2^2,\\
C\_2 = 1 = 2^0,\\
S\_4 = 48 = 2^5 + 2^4,\\
C\_4 = 6 = 2^2 + 2^1,\\
S\_6 = 144 = 2^7 + 2^4,\\
C\_6 = 12 = 2^3 + 2^2,\\
S\_8 = 768 = 2^9 + 2^8,\\
C\_8 = 48 = 2^5 + 2^4,\\
S\_{10} = 720 = 2^9 + 2^7 + 2^6 + 2^4,\\
C\_{10} = 36 = 2^5 + 2^2,\\
S\_{12} = 5376 = 2^{12} + 2^{10} + 2^8,\\
C\_{12} = 224 = 2^7 + 2^6 + 2^5,\\
S\_{14} = 3360 = 2^{11} + 2^{10} + 2^8 + 2^5,\\
C\_{14} = 120 = 2^6 + 2^5 + 2^4 + 2^3,\\
S\_{16} = 4096 = 2^{12},\\
C\_{16} = 128 = 2^7.
\end{array}$$
Interestingly, six of the first eight elements of the tuple $(C\_2, C\_4, \ldots)$ are [highly composite numbers](https://en.wikipedia.org/wiki/Highly_composite_number): $1, 6, 12, 48, 36, 120$. I cannot explain the fact that $S\_{16}$ and $C\_{16}$ are powers of two, but it does not look like a random coincidence.
|
https://mathoverflow.net/users/122796
|
Is there an efficient generalized algorithm to find at least one binary word with the maximum rotational imbalance and the full $\{0, 1\}$-balance?
|
The function $f(x)$ is closely related to the notion of *autocorrelation*, which for a binary sequence $x$ of length $|x|=N$ and shift $w$ can be expressed as
$$\textbf{AC}\_x(w) := N - 2H(x\oplus R(x,w)).$$
The values of $\textbf{AC}\_x(w)$ for various non-trivial shifts (i.e. $1\leq w\leq N-1$) are called *out-of-phase autocorrelation values*.
So,
$$f(x) = \min\_{1\leq w\leq N-1} H(x\oplus R(x,w)) = \frac{N}2 - \frac12\max\_{1\leq w\leq N-1} \textbf{AC}\_x(w).$$
For the known constructions for optimal binary sequence w.r.t. autocorrelation, I refer to the paper [Cai and Ding (2009)](https://doi.org/10.1016/j.tcs.2009.02.021).
In particular, for $N=q-1\equiv 0\pmod{4}$, where $q$ is a power of prime, the Sidelnikov–Lempel–Cohn–Eastman construction produces a sequence $x$ with $H(x)=\frac{N}2$ and out-of-phase autocorrelation values $\{0,-4\}$, thus giving $f(x)=\frac{N}2$ as requested in the question. The value $N=256$ well fits into this construction since $q=257$ is prime. Here is one such 256-bit word:
```
0011101101010000101000110101100110101001111011001111111100011111010110011111100000100000011011001011011001110001000001000011110100001011010001010001001001100011110011100101011101011010100110000010100100101010110011101001111110011000001101111101000000111011
```
|
4
|
https://mathoverflow.net/users/7076
|
410006
| 167,778 |
https://mathoverflow.net/questions/410002
|
4
|
Let $X=(X,\|\cdot\|)$ be a Banach space and suppose that $F\subset X$ is a finite-dimensional subspace. There is then an equivalent norm $|\cdot|$ on $F$ such that $|\cdot|$ is induced by an inner product on $F$ (i.e. $|\cdot|$ will satisfy the parallelogram law) and it follows that
\begin{equation} c\_{(F,|\cdot|)}|x|\leq\|x\|\leq C\_{(F,|\cdot|)}|x| \end{equation}
for some constants $c,C>0$ and for all $x\in F$. **Here is my main question:** is there a name for the following property?
There exists $M\geq 1$ such that for every finite-dimensional subspace $F\subset X$, there is an equivalent norm $|\cdot|$ on $F$ that is induced by an inner product on $F$ and is such that $1\leq \frac{C}{c}\leq M$.
Clearly, any Hilbert space has this property by taking $M=1$ and $|\cdot|=\|\cdot\|$. Are there examples of non-Hilbert spaces that have this property? Is this property related somehow to the type/cotype of $X$?
|
https://mathoverflow.net/users/165007
|
Name for certain property of equivalent norms on finite-dimensional subspaces of a Banach space
|
Your condition implies that $X$ is isomorphic to a Hilbert space with isomorphism constant at most $M^2$. The distance condition implies that both type 2 constant and cotype 2 constant of $X$ is bounded by $M$. By Kwapien theorem the Banach-Mazur distance of $X$ to a Hilbert space is bounded by type 2 constant times cotype 2 constant.
As a reference see e.g., Albiac-Kalton.
|
9
|
https://mathoverflow.net/users/3675
|
410008
| 167,779 |
https://mathoverflow.net/questions/409899
|
5
|
Let $A$ be a UFD, that is also a $k$-algebra, where $k$ is a field of characteristic $\not=2$ (for instance polynomials over $k$).
Is every involution in $\mathrm{GL}\_n(A)$ diagonalisable?
This is of course true over the field of fractions of $A$.
In this [question](https://math.stackexchange.com/questions/174488/diagonalizing-matrices-over-ufds)
it is explained that $$M=\left(\begin{array}{rr}
3&-1\\-1 & 3\end{array}\right)$$
is diagonalisable over $\mathbb{Q}$ but not over $\mathbb{Z}$. However, it is not an involution.
|
https://mathoverflow.net/users/23758
|
Is every matrix involution over a UFD diagonalisable?
|
Here is an answer based on my comment, and Geoff Robinson's earlier comment.
Let $A$ be a domain with $2\in A^\times$, and let $M$ be an $n\times n$ matrix with $M^2=I$. It is convenient to consider $M$ as an $A$-endomorphism of $A^n$ and $I$ as the identity endomorphism of $A^n$. Put $S\_+ =\{x\in A^n\,:\,Mx=x\}$ and $S\_-=\{x\in A^n\,:\,Mx=-x\}$. As noted in Geoff Robinson's comment, since $2\in A^\times$, we have
$A^n=S\_+\oplus S\_-$. (This follows readily from the fact that $x=\frac{1}{2}(x+Mx)+\frac{1}{2}(x-Mx)$ for all $x\in A^n$ and $M^2=I$.) In particular, both $S\_+$ and $S\_-$ are f.g. projective $A$-modules.
Suppose that every f.g. projective $A$-module is free. Then $S\_+$ and $S\_-$ are free. Let $E\_+$ and $E\_-$ be bases for these spaces. Then $E=E\_+\cup E\_-$ is a basis of $A^n$ consisting of eigenvalues of $M$, which means that $M$ is a diagonalizable.
On the other hand, if there exists a f.g. projective $A$-module $P$ which is not free, then we can construct a non-diagonalizable involution $M:A^n\to A^n$ as follows. There exists a f.g. $A$-module $Q$ such that $A^n\cong P\oplus Q$ for some $n\in\mathbb{N}$. Now take $M$ to be the matrix corresponding to $\mathrm{id}\_P\oplus (-\mathrm{id}\_Q)$. For this $M$, we have $S\_+=P$ and $S\_-=Q$. If $M$ were diagonalizable, then $A^n$ would have a basis $E$ consisting of vectors $x\in A^n$ with $Mx\in \{\pm x\}$ (here we need $A$ to be a domain). Consequently, $E = (E\cap S\_+)\cup (E\cap S\_-)$, which means that $E\_+:=E\cap S\_+$ is a basis of $S\_+=P$. This contradicts our choice of $P$, so $M$ is not diagonalizable.
There are examples of UFDs admitting f.g. projective modules which are not free --- see this [question](https://mathoverflow.net/questions/54356/nonfree-projective-module-over-a-regular-ufd), for instance. Quoting from this [answer](https://mathoverflow.net/a/54379/86006), one can take $A=\mathbb{C}[a,b,c,x,y,z]/(ax+by+cz-1)$, the module $P$ to be the kernel of $(\alpha,\beta,\gamma)\mapsto a\alpha+b\beta+c\gamma:A^3\to A$, and $Q=(x,y,z)A$. One has $P\oplus Q=A^3$, so it should be possible to write the resulting $M\in \mathrm{M}\_3(A)$ explicitly.
|
8
|
https://mathoverflow.net/users/86006
|
410012
| 167,781 |
https://mathoverflow.net/questions/409982
|
3
|
The motivation for the following is to convert the integro-differential equation
\begin{equation}
\kappa\ddot x+\dot x=-kx+\beta\int\_{-\infty}^t W'(x(t)-x(s))e^{s-t}ds,
\end{equation}
into a system of nonlinear ODEs
\begin{equation}
\begin{split}
\dot{x}&=v,\\
\dot{v}&=\frac{1}{\kappa}(\beta M\_0-v-kx),\\
\dot{M}\_n&=-M\_n+W^{(n+1)}(0)+vM\_{n+1}.
\end{split}
\end{equation}
Let
$$M\_n=\int\_{-\infty}^t W^{(n+1)}(x(t)-x(s))e^{s-t}ds,$$
for all $n\in\mathbb{N}$ where $W^{(n+1)}(z)$ is the $n+1$-th derivative of $W$ with respect to $z$. I can't see how the last line of the proof below is equivalent to $-M\_{n+1}+W^{(n+2)}(0)+vM\_{n+2}$? Perhaps I made the wrong substitution on line 3. Any help would be much appreciated.
Applying the Leibniz integral rule, observe that for $n=0$,
\begin{equation\*}
\begin{split}
\dot{M\_0}&=\frac{\partial}{\partial t}\int\_{-\infty}^t W^{(1)}(x(t)-x(s))e^{s-t}ds\\
&=W^{(1)}(0)+\int\_{-\infty}^t e^{s-t}\left(\dot{x}W^{(2)}(x(t)-x(s))-W^{(1)}(x(t)-x(s))\right)ds,\\
&=-M\_0+W^{(1)}(0)+\dot{x}M\_1.
\end{split}
\end{equation\*}
The inductive step is then
\begin{equation\*}
\begin{split}
\dot{M}\_{n+1}
&= \frac{\partial}{\partial t} \left(\frac{\dot{M}\_n+M\_n-W^{(n+1)}(0)}{v}\right),\\
&=\frac{v\frac{\partial}{\partial t}\left(\dot{M}\_n+M\_n-W^{(n+1)}(0)\right)-\dot{v}(\dot{M}\_n+M\_n-W^{(n+1)}(0))}{v^2},\\
&=\frac{v\ddot{M}\_n+v\dot{M}\_n-\dot{v}\dot{M}\_n-\dot{v}M\_n-vW^{(n+2)}(0)+\dot{v}W^{(n+1)}(0)}{v^2},\\
&=\frac{\ddot{M}\_n+\dot{M}\_n-\dot{v}M\_{n+1}-W^{(n+2)}(0)}{v}.\\
\end{split}
\end{equation\*}
|
https://mathoverflow.net/users/167822
|
Inductive proof that $\dot{M}_{n+1}=-M_{n+1}+W^{(n+2)}(0)+vM_{n+2}$
|
Just as you did for $n=0$, for any $n\ge0$ write
\begin{equation\*}
\begin{split}
\dot{M\_n}&=\frac{\partial}{\partial t}\int\_{-\infty}^t W^{(n+1)}(x(t)-x(s))e^{s-t}ds\\
&=W^{(1)}(0)+\int\_{-\infty}^t e^{s-t}\left(\dot{x}(t)W^{(n+2)}(x(t)-x(s))-W^{(n+1)}(x(t)-x(s))\right)ds \\
&=-M\_n+W^{(n+1)}(0)+\dot{x}M\_{n+1},
\end{split}
\end{equation\*}
to get the desired result.
Induction is not needed here. Also, your multiline differentiation display is incorrect. In particular, the derivative of the constant $W^{(n+1)}(0)$ is $0$, not $W^{(n+2)}(0)$.
|
2
|
https://mathoverflow.net/users/36721
|
410022
| 167,786 |
https://mathoverflow.net/questions/409758
|
2
|
I would like to know if there is some uniform construction out of a given category $\mathcal C$ that *freely* throws in all quotients,to form a new category $\mathcal C'$. Preferably $\mathcal C'$ has all (small?) quotients, but if it only contains the quotients out of $\mathcal C$ I can also live with it.
I know that the presheaf category $[\mathcal C^{\mathrm {op}} , \mathrm{Set}]$ is the free cocompletion of $\mathcal C$, which means that I get all the quotients, but I also get a whole bunch of other stuff. Maybe it has to do with $\mathrm{Set}$ contains all the colimits. In fact $\mathrm{Set} = [\mathbb 1^{\mathrm {op}}, \mathrm{Set}]$, which is the free cocompletion of the category $\mathbb 1$. so maybe we should consider $[\mathcal C^{\mathrm {op}}, \mathbb Q]$ where $\mathbb Q$ is something like the category that somehow contains all quotients generated from $\mathbb 1$? But then $\mathbb Q = \mathbb 1$ since there are no new quotients to add at all...
On the other hand, maybe this is related to some sort of categorified process of "setoid-ification".
Is there existing results concerning this question? Any help is appreciated.
|
https://mathoverflow.net/users/136535
|
Freely add all quotients to a category
|
In general, the way to construct a free completion of a category under only *some* colimits is to take a full subcategory of the presheaf category $[C^{\rm op},\rm Set]$ that's the closure of the representables under the colimits in question. In particular we can do this for quotients, although there are different meanings we might pick for "quotient".
1. If by "quotient" we mean simply a coequalizer, then the closure of the representables under coequalizers is the free cocompletion under coequalizers, with a universal property relative to mapping into other categories with coequalizers.
2. If by "quotient" we mean the quotient of an [internal equivalence relation](https://ncatlab.org/nlab/show/congruence), then the closure of the representables under such quotients is called the [ex/lex completion](https://ncatlab.org/nlab/show/regular+and+exact+completions). As long as the category $C$ already has finite limits, so that it makes sense to talk about internal equivalence relations therein, this has a universal property relative to finite-limit-preserving functors into exact categories. This can be generalized to other kinds of exact completions.
|
6
|
https://mathoverflow.net/users/49
|
410025
| 167,787 |
https://mathoverflow.net/questions/409986
|
-4
|
Call $L$-function any element of an L-rig (see [Are there infinitely many L-rigs?](https://mathoverflow.net/questions/372349/are-there-infinitely-many-l-rigs) for a definition). Suppose $F$ and $G$ are two primitive L-functions. Is $F\otimes G$ itself primitive? If yes, does the primitivity of $F\otimes G$ imply the primitivity of both $F$ and $G$?
Edit: perhaps I should add some motivation. I learnt that an irreducible representation of the direct product of two finite groups $G\_{1}$ and $G\_{2}$ is the tensor product of irreducible representations of the respective groups, so drawing and analogy between irreducible representation of a finite group and primitive L-function, I came to think about this question.
|
https://mathoverflow.net/users/13625
|
Does Rankin-Selberg convolution preserve primitivity?
|
Let $L(s,F)$ be the $L$-function of a self-dual $\mathrm{GL}(2)$ holomorphic cuspidal newform without complex multiplication and with trivial nebentypus. Let $G=\mathrm{Sym}^2 F$ be the symmetric square lift of $F$. Then $L(s,F)$ is a primitive $\mathrm{GL}(2)$ $L$-function (due to Hecke) and $L(s,G)$ is a primitive $\mathrm{GL}(3)$ $L$-function (due to Gelbart and Jacquet), but
$$L(s,F\otimes G) = L(s,F)L(s,\mathrm{Sym}^3 F),$$
which is not primitive. Simpler and more complicated examples abound; this seemed like a nice example from the middle ground.
|
4
|
https://mathoverflow.net/users/111215
|
410035
| 167,791 |
https://mathoverflow.net/questions/410003
|
1
|
Let $(X,d)$ be a complete and separable metric space and, for $1\leq p<\infty$, let $(\mathcal{P}\_p(X,d),W\_p)$ be the $p$-Wasserstein space on $(X,d)$. For which $p$ and $(X,d)$ is $(\mathcal{P}\_p(X,d),W\_p)$ a $CAT(\kappa)$ space?
---
I know that for $p=2$ and $(X,d)$ a Banach space, $(\mathcal{P}\_2(X,d),W\_2)$ is a Hadamard ($CAT(0)$) space; but there must be other cases...
|
https://mathoverflow.net/users/469470
|
When are Wasserstein spaces $CAT(\kappa)$?
|
Almost never...
Note that there is an isometric embedding $X\to W\_p(X)$, so $X$ has to be CAT(κ). Second the space $W\_p(X)$ contains symmetric $p$-product $S^n(X)=X^{\times n}/S\_n$ so $p=2$, or $X$ is one a point-space.
Now if $\dim X>1$, then you get into trouble with extending geodesic thru a $\delta$-measure in $S^2(X)$, so you get $\dim X\le1$.
|
4
|
https://mathoverflow.net/users/1441
|
410038
| 167,792 |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.