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https://mathoverflow.net/questions/442004 | 1 | Let $A$ be a normalized $N\times N$ GOE matrix. Let $\sigma^1<\sigma^2<\dots <\sigma^N$. We know that the largest eigenvalue converges $\sigma^N$ to 2 almost surely. Assume that $X\_1,\dots, X\_N$ are iid random variables with law $\mu$ so that $E\_{X\sim \mu}[e^{\alpha X}]<\infty$ for some $\alpha>0$.
Now, I try to prove that
$$
P(\frac{1}{N}\sum\_{i=1}^N \sigma^i X\_i^2\ge ax)\le \mbox{something like $(E[e^{\alpha X^2}])^N e^{-a\alpha xN}$}
$$
which goes to zero as $N\to \infty$.
---
My reasons for having this upper bound estimate are that if we consider the following thing: By Chebyshev's inequality
$$
P(\frac{1}{N}\sum\_{i=1}^N X\_i^2\ge ax)=P(\exp(\alpha\sum\_i X\_i^2)\ge e^{a\alpha x N})\le \frac{(E[e^{\alpha X\_0^2}])^N}{e^{a\alpha x N}} \to 0
$$
as $N\to \infty$.
The inequality I want to prove has one more random variable $\sigma^i$ that is not easy to handle.
| https://mathoverflow.net/users/168083 | How to bound $ P(\frac{1}{N}\sum_{i=1}^N \sigma^i X_i^2\ge ax)$ for eigenvalues of a normalized $N\times N$ GOE matrix? | Since we have the joint law [GOE eigenvalues](https://www.lpthe.jussieu.fr/%7Eleticia/TEACHING/Master2019/GOE-cuentas.pdf) density $\rho\_{N}$
$$P[\sigma\_{i}\in A\_{i}]=\iint\_{A\_{i}}\frac{1}{Z\_{N}}e^{-\frac{1}{4}\sum\_{i=1}x\_{i}^{2}}\prod\_{k>m}|x\_{k}-x\_{m}|\prod dx\_{i}$$
and so
$$P(\frac{1}{N}\sum\_{i=1}^N \sigma^i X\_i^2\ge at)=\iint\_{R^{N}} P(\frac{1}{N}\sum\_{i=1}^N x\_{i} X\_i^2\ge ax) \rho\_{N}(x)dx$$
$$\leq \iint\_{R^{N}} \frac{\prod\_{i}E[e^{\alpha x\_{i} X\_i^2}])}{e^{a\alpha t N}} \rho\_{N}(x)dx.$$
To proceed from here one needs to know explicity the MGF for $X^{2}$ so as to estimate. Do you have any estimates for it at least?
| 1 | https://mathoverflow.net/users/99863 | 452064 | 181,729 |
https://mathoverflow.net/questions/451961 | 3 | I was reading stuffs about Riesz energy which is defined for an open subset $U\in\mathbb{R}^d$ by $I\_s(U)=\int\_U\int\_U|x-y|^{-s}\ dx\ dy$ where $dx$ and $dy$ are Lebesgue measure in $U$. Now if I take for simplicity $U$ to be a ball I am not sure how to evaluate the integration. If I shift the integral to origin i.e like calculating the integral $|x|^{-s}$ over a ball is fine using polar change of coordinate. But here it's not only in one ball over another ball also, I am not sure how to evaluate it or I am missing something. Any help is very much appreciated for my understanding.
| https://mathoverflow.net/users/493060 | Calculation of Riesz energy for balls | This is just an extended comment to [Iosif Pinelis's answer](https://mathoverflow.net/a/452021/108637) above, which provides an answer in terms of an unknown constant $C\_{d,s}$. Here we evaluate this constant.
Let $B$ be the unit ball. If $f(x)=(1-|x|^2)^p$ if $x\in B$ and $f(x)=0$ otherwise, then
$$(-\Delta)^{\alpha/2}f(x)=2^\alpha\frac{\Gamma(p+1)\Gamma(\frac{d+\alpha}2)}{\Gamma(p+1-\frac\alpha2)\Gamma(\frac d2)}{\_2F\_1}(\tfrac{d+\alpha}2,\tfrac\alpha2-p;\tfrac d2;|x|^2).$$
Here $p > -1$, ${\_2F\_1}$ is the Gauss's hypergeometric function, and $(-\Delta)^{\alpha/2}$ is the fractional Laplace operator for $\alpha > 0$ and the Riesz potential operator for $\alpha \in (-d, 0)$.
I would not be surprised if the above formula actually appeared in some paper by Boris Rubin, Stefan Samko, or one of the other authors working on Riesz potentials, but as far as I can tell, for $\alpha \in (0, 2)$ this is due to Bartłomiej Dyda [DOI:10.2478/s13540-012-0038-8](https://doi.org/10.2478/s13540-012-0038-8), and extension to an arbitrary $\alpha > -d$ is given in my paper with Bartłomiej Dyda and Alexey Kuznetsov [DOI:10.1007/s00365-016-9336-4](https://doi.org/10.1007/s00365-016-9336-4); for further discussion, see my survey [DOI:10.1515/9783110571622-007](https://doi.org/10.1515/9783110571622-007).
We set $p=0$ and $\alpha=s-d$ to get
$$(-\Delta)^{-(d-s)/2}\mathbb 1\_B(x)=2^{s-d}\frac{\Gamma(\frac s2)}{\Gamma(1+\frac{d-s}2)\Gamma(\frac d2)}{\_2F\_1}(\tfrac s2,\tfrac{s-d}2;\tfrac d2;|x|^2).$$
Using the definition of the Riesz potential operator, this translates to
$$\int\_B|x-y|^{-s}dy=\frac{2\pi^{d/2}}{(d-s)\Gamma(\frac d2)}{\_2F\_1}(\tfrac s2,\tfrac{s-d}2;\tfrac d2;|x|^2).$$
Integrating this with respect to $x\in B$, we obtain
$$\int\_B\int\_B|x-y|^{-s}dydx=\frac{4\pi^d}{(d-s)(\Gamma(\frac d2))^2}\int\_0^1r^{d-1}{\_2F\_1}(\tfrac s2,\tfrac{s-d}2;\tfrac d2;r^2).$$
Using [http://functions.wolfram.com/07.23.21.0003.01], we find that
$$\int\_B\int\_B|x-y|^{-s}dydx=\frac{4\pi^d{\_2F\_1}(\tfrac s2,\tfrac{s-d}2;1+\tfrac d2;1)}{d(d-s)(\Gamma(\frac d2))^2},$$
and [http://functions.wolfram.com/07.23.03.0002.01] leads to
$$\int\_B\int\_B|x-y|^{-s}dydx=\frac{4\pi^d\Gamma(1+\tfrac d2)\Gamma(1+d-s)}{d(d-s)\Gamma(1+\frac{d-s}2)\Gamma(1+d-\frac s2)(\Gamma(\frac d2))^2}.$$
After some simplification, we arrive at
$$\int\_B\int\_B|x-y|^{-s}dydx=\frac{2^{d-s+1}\pi^{d-1/2}\Gamma(\frac{1+d-s}2)}{(d-s)(d-\frac s2)\Gamma(\tfrac d2)\Gamma(d-\frac s2)}.$$
(Watch out: this may be full of typos! I only checked the final answer for $d = 1$, and it seems to agree with the result of an explicit calculation.)
| 4 | https://mathoverflow.net/users/108637 | 452065 | 181,730 |
https://mathoverflow.net/questions/452073 | 1 | This is a follow up to [this question](https://math.stackexchange.com/q/4747105/1176963), where I wish to partition the reals into two sets $A$ and $B$ that are dense (with positive measure) in every non-empty sub-interval $(a,b)$ of $\mathbb{R}$.
(In this case, I want to find $\lim\_{t\to\infty} \lambda(A\cap [-t,t])/(2t)$ and $\lim\_{t\to\infty} \lambda(B\cap [-t,t])/(2t)$, where $\lambda$ is the Lebesgue measure restricting the Lebesgue outer measure $\lambda^{\*}$ on sets measurable in the [Caratheodory sense](https://en.m.wikipedia.org/wiki/Carath%C3%A9odory%27s_criterion).)
I have a rough understanding of the [answer](https://math.stackexchange.com/a/4747177/1176963); however, I'm unsure of the measures of $A$ and $B$.
**Question:** What is the measure of $A$ and $B$? If both measures are $1/2$, how do we change the [answer](https://math.stackexchange.com/a/4747177/1176963) so the measures are positive but unequal?
| https://mathoverflow.net/users/87856 | What is the measure of two sets which partition the reals into subsets of positive measure? | We can make those limits be any two positive numbers that add to $1$, or we can make them nonconvergent. The reason is that in any interval, we can construct $A$ and $B$ on that interval by using the iterated fat-Cantor set construction (as described [here](https://twitter.com/JDHamkins/status/1683929594175401984)), and we can make $A$ have arbitrarily small measure on that one interval, which is what bof refers to in his comment.
By repeating those small-measure sets on successive intervals, or by using large-measure sets as desired on successive intervals, we can control the limit value of your asyptotic measure so as to realize any desired value or a nonconvergent value.
| 3 | https://mathoverflow.net/users/1946 | 452078 | 181,733 |
https://mathoverflow.net/questions/452072 | 0 | Is there a formula to construct a Collatz (3x + 1) sequence of arbitrary length that is strictly increasing? Obviously one can do this with a strictly decreasing sequence by just taking $2^n$ but I haven't seen a way to do it with a strictly increasing sequence.
| https://mathoverflow.net/users/509865 | Finding a strictly increasing Collatz sequence of arbitrary length | Start with a number equal to $-1$ modulo $2^n$. Then, after one step, the number is $\frac{3(-1)+1}{2}=-1$ modulo $\frac{2^n}{2}=2^{n-1}$, so inductively it will increase $n$ steps.
| 3 | https://mathoverflow.net/users/35593 | 452080 | 181,734 |
https://mathoverflow.net/questions/452026 | 4 | I'm making a research on Galois theory, and found something interesting regarding the ring of symmetric polynomials:
At least up to 5 variables, we can rewrite the elementary symmetric polynomials with powers of symmetric polynomials of degree 1.
For the formulation of the question, i will introduce the index set $I\_q=\mathbb{N}\_{<q}$, where $\mathbb{N}$ are the natural numbers, including zero.
First, we define the subset $S\_k\subset I\_q^k$ as
$$S\_k=\{(i\_1;\dots;i\_k)\in I\_q^k\mid i\_1<i\_2<\dots<i\_k\}$$
This allows us to define the $k$-th elementary symmetric polynomial with $q$ variables $e\_k=e\_k(x\_0;\dots;x\_{q-1})$ as:
$$e\_k=\sum\_{(i\_1;\dots;i\_k)\in S\_k}x\_{i\_1}\dots x\_{i\_k}$$
We also define the symmetric $p$-th powers of first degree polynomials $s\_j^p=s\_j^p(x\_0;\dots;x\_{q-1})$ to be:
$$s\_j^p=\sum\_{(i\_1;\dots;i\_j)\in S\_j}(x\_{i\_1}+\dots+x\_{i\_j})^p$$
The question is then: can every elementary symmetric polynomial $e\_k$ be written as a linear combination of these symmetric $k$-th powers of first degree polynomials $s\_j^k$ for an arbitrary number of variables $q$?
As said earlier, i have made some progress, and found a pattern up to $q=5$ variables. I will leave my findings here:
[For $q=1$]
$$e\_1=s\_1^1$$
[For $q=2$]
$$\begin{align\*}
e\_1&=s\_1^1+0s\_2^1\\
e\_2&=\frac{1}{2}(-s\_1^2+s\_2^2)
\end{align\*}$$
[For $q=3$]
$$\begin{align\*}
e\_1&=s\_1^1+0s\_2^1+0s\_3^1\\
e\_2&=\frac{1}{2}(-s\_1^2+0s\_2^2+s\_3^2)\\
e\_3&=\frac{1}{6}(s\_1^3-s\_2^3+s\_3^3)
\end{align\*}$$
Continuing in a similar fashion, we can construct $q$-dimensional vector spaces $V\_k$ over $\mathbb{Q}$, spanned by the set of symmetric $k$-th powers $G\_k=\{s\_1^k; s\_2^k;\dots;s\_q^k\}$.
The $k$-th elementary symmetric polynomial is then a vector $\vec{e\_k}\in V\_k$ in this setting. Note that these representations are not unique, e.g., for $q=2$, $\vec{e\_1}=(1;0)=(0;1)$, and hence $G\_k$ is not a basis for $V\_k$. I'll add a few more values for $q$:
[For $q=4$]
$$\begin{align\*}
\vec{e\_1}&=(1;0;0;0)\\
\vec{e\_2}&=\frac{1}{2}(-1;0;0;1)\\
\vec{e\_3}&=\frac{1}{6}(4;2;-3;-1)\\
\vec{e\_4}&=\frac{1}{24}(-1;-1;1;1)
\end{align\*}$$
[For $q=5$]
$$\begin{align\*}
\vec{e\_1}&=(1;0;0;0;0)\\
\vec{e\_2}&=\frac{1}{2}(-1;0;0;0;1)\\
\vec{e\_3}&=\frac{1}{6}(-6;0;-1;3;0)\\
\vec{e\_4}&=\frac{1}{24}(-5;2;-3;4;-1)\\
\vec{e\_5}&=\frac{1}{120}(1;-1;1;-1;1)
\end{align\*}$$
etc.
| https://mathoverflow.net/users/504252 | Can the ring of symmetric polynomials be generated by powers of symmetric polynomials of degree 1? | Yes.
For variables $x\_1,\dots, x\_k$, we have
$$ \sum\_{J \subseteq \{1,\dots, k \} } (-1)^{ k- |J|} \left(\sum\_{j \in J} x\_j\right)^k = k! x\_1 \dots x\_k $$
so that
$$ e\_k = \sum\_{(i\_1; \dots; i\_k )\in S} x\_{i\_1} \dots x\_{s\_k} = \frac{1}{k!} \sum\_{(i\_1; \dots; i\_k )\in S} \sum\_{J \subseteq \{i\_1,\dots, i\_k \}}(-1)^{ k- |J|} \left(\sum\_{j \in J} x\_{j} \right)^k $$
and each set $J$ occurs $\binom{ q-|J|}{k- |J|} $ times, so
$$ e\_k = \sum\_{ \substack{ J\subseteq I\_q \\ |J| \leq k}} (-1)^{ k -|J|} \frac{ \binom{ q- |J|} {k-|J|}}{k!} \left(\sum\_{j \in J} x\_j\right)^k =\sum\_{j=0}^k (-1)^{ k -j} \frac{ \binom{ q- j} {k-j}}{k!} s\_j ^k.$$
| 9 | https://mathoverflow.net/users/18060 | 452081 | 181,735 |
https://mathoverflow.net/questions/451272 | 4 | One-Counter Nets (OCNs) are finite-state machines equipped with an integer counter that
cannot decrease below zero and cannot be explicitly tested for zero.
An OCN $A$ over alphabet $\sum$ accepts a word $w \in \sum^\*$ from initial counter value $c\in\mathbb{N}$ if there is a run of $A$ on $w$ from an initial state to an accepting state in which the counter,starting from value $c$, does not become negative. Thus, for every counter value $c\in\mathbb{N}$ the OCN $A$ defines a language $L(A,c) \subseteq \sum^\*$.
As is the case with many computational models, certain decision problems for deterministic OCNs (OCNs that admit a single legal transition for each state $q$ and letter $\sigma$), denoted DOCNs, are computationally easier than for nondeterministic OCNs.
A one-counter net (OCN) is a finite automaton whose transitions are labelled both by letters and by integer weights. Formally, an OCN is a tuple $A = \langle \sum, Q, s\_0, \delta, F\rangle$ where $\sum$ is a finite alphabet, $Q$ is a finite set of states, $s\_0 \in Q$ is the initial state, $\delta \subseteq Q × \sum × \mathbb{Z} × Q$ is the set of transitions, and $F \subseteq Q$ are the accepting states. We say that an OCN is **deterministic** if for every $s \in Q, \delta \in \sum,$ there is at most one transition $(s, \sigma, e, s′)$ for some $e ∈ \mathbb{Z}$ and $s′ \in Q$.
For a transition $t = (s, \sigma, e, s′)\in \delta$ we define ***eff***$(t) = e$ to be its (counter) effect.
A path in the OCN is a sequence $\pi= (s\_1, \sigma\_1, e\_1, s\_2)(s\_2, \sigma\_2, e\_2, s\_3)\dots(s\_k, \sigma\_k, e\_k, s\_{k+1})\delta^\*.$
Such a path $\pi$ is a cycle if $s\_1 = s\_{k+1}$, and is a simple cycle if no other cycle is a proper infix of it. We say that the path $\pi$ reads the word $\sigma\_1\sigma\_2\dots\sigma\_k \in \sum^∗.$
The effect of $\pi$ is ***eff***$(\pi) =\Sigma\_{i=1}^k e\_i$, and its *nadir*, denoted ***nadir***$(\pi)$, is the minimal effect of any prefix of $\pi$(note that the nadir is non-positive, since ***eff***$(\epsilon) = 0)$.
A configuration of an OCN is a pair $(s, v) \in Q × \mathbb{N}$ comprising a state and a non-negative integer. For a letter $\sigma \in \sum$ and configurations $(s, v),(s′, v′)$ we write $(s,v)\xrightarrow{\sigma}(s′, v′)$ if there exists $d\in\mathbb{Z}$ such that $v′ = v + d$ and $(s, \sigma, d, s′)\in\delta$.
A run of $A$ from initial counter $c$ on a word $w = \sigma\_1\sigma\_2\dots\sigma\_k\in\sum^\*$is a sequence of configurations $\rho = (q\_0, v\_0),(s\_1, v\_1)\dots(s\_k, v\_k)$ such that $v\_0 = c$ and for every $1 \leq i \leq k$ it holds that $(s\_{i−1}, v\_{i−1})\xrightarrow{\sigma\_i}(s\_i, v\_i)$. Since configurations may only have a non-negative counter, this enforces that the counter does not become negative.
**My question**:
Is the following problem decidable? (does there exist an algorithm that solves it for all instances)
* Input: Given an OCN $A$ and a DOCN $D$,
* Question: is $L(A,0)=L(D,0)$?
**References**
(<https://drops.dagstuhl.de/opus/volltexte/2022/17081/pdf/LIPIcs-CONCUR-2022-18.pdf>)
Piotr Hofman, Patrick Totzke "[Trace Inclusion for One-Counter Nets Revisited](https://arxiv.org/abs/1404.5157)".
Piotr Hofman, Richard Mayr, Patrick Totzke "[Decidability of Weak Simulation on One-counter Nets](https://arxiv.org/abs/1304.4104)".
| https://mathoverflow.net/users/377873 | Equivalence between deterministic and non-deterministic counter net | The short answer is that as far as I'm aware, this question is open.
It is however very close to ones that are settled. I provide some more detail below.
As you've correctly pointed out, the decidability status of decision problems relating to counter automata
often depends on the presence of non-determinism, but also on the acceptance criterion.
The initial counter value should not matter as long as it is fixed (you can always build another automaton in the same class that sets the initial value via some initial chain of states. This will change the language of the automaton but will not change the decidability status of most decision problems). I will assume below that the initial counter is $0$ always.
There are three natural ways to define accepting runs, and thus the language of the automaton:
1. **Trace semantics**: A run is accepting if the effect of every prefix is greater equal to $0$. So you simply don't starve. Let's write $T(A)$ for the set of words in $\Sigma^\*$ so that the automaton $A$ (with some fixed initial state) has some accepting run on them. This is the semantics your original is about.
2. **Coverability semantics**: A run is accepting if it does not (starve as above) and ends in a "final" state. This is a designated subset of the control states, fixed as part of the automaton's definition. Let's write $C(A)$ for the resulting coverability language of $A$.
3. **Reachability semantics**: A run is accepting if it satisfies 1), 2) and ends in counter value $0$. Let's write $R(A)$ for the corresponding language of $A$.
#### Let's consider the reachability semantics first.
Here, the language universality problem ($R(A) = \Sigma^\*$?) is already undecidable ([0], theorem 10).
Since you can easily build a *deterministic* finite automaton, and therefore OCN, with $\Sigma^\*$, the undecidability of universality transfers to both inclusion and equivalence.
#### For the coverability and trace semantics,
universality remains decidable [1,2], yet inclusion is not [3].
In fact, [2,3] are written solely in terms of trace semantics but the argument for the upper bounds (decidability and complexity) in [2] works the same for coverability semantics, and the undecidability in [3] trivially transfers from the easier trace inclusion ($T(A)\subseteq T(B)$?) to coverability language inclusion ($C(A)\subseteq C(B)$?) by making all states final.
The argument for the undecidability of trace inclusion also works if the system $A$ on the left is deterministic, but *not* if $B$ on the RHS is deterministic.
You can reduce inclusion to equivalence by a small trick:
to check if $T(A)\subseteq T(B)$ you can equivalently check if $T(A') = T(B')$ for the modified automata where
$A' \rightarrow A$,
$A' \rightarrow B$
and $B'\rightarrow B$.
(after one step, $B'$ continues as $B$ and $A'$ can continue either as $A$ or $B$).
This will *make* $A'$ non-deterministic. It shows that trace (and thus cover) equivalence $T(A)=T(B)$ is undecidable for two non-deterministic OCN.
#### Your question
Trace equivalence ($T(A) = T(B)$?) between a OCN $A$ and a DOCN $B$, is open as fact as I'm aware.
Notice that you **cannot** decide it by checking two-way inclusion
because, as argued above, the inclusion $T(B)\subseteq T(A)$ is undecidable.
**Remark 1:**
The inclusion $T(A)\subseteq T(B)$ is in fact decidable if $B$ is deterministic. This can be shown in several ways, for example by checking simulation preorder [4] (if the RHS is deterministic then language inclusion coincides with simulation).
Alternatively, you can complement $B$ into a one-counter *automaton* $B'$ (that has explicit zero-testing transitions) then checking emptiness for $A\times B'$ via [5].
This will also work for a slightly larger class of "history-deterministic" automaton the RHS [6].
**Remark 2:**
If you consider languages of infinite words,
these are significantly more complex, both topologically and in terms of decision problems.
The languages recognised by OCN accepting by infinitely often visiting final states,
are not even Borel [7] and have and undecidable universality problem [8].
So in this case again, the equivalence of infinite trace languages is undecidable.
I hope this is useful; thanks for the very interesting question! Best wishes, P.
---
[0]: *Petri nets and regular languages*. Rüdiger Valk, Guy Vidal-Naquet
<https://doi.org/10.1016/0022-0000(81)90067-2>
[1]: *Petri Nets and Regular Processes* Petr Jančar, Javier Esparza, Faron Moller; <https://doi.org/10.1006/jcss.1999.1643>
[2]: *Trace inclusion for one-counter nets revisited* Piotr Hofman, and Patrick Totzke <https://doi.org/10.1016/j.tcs.2017.05.009>
[3]: *Decidability of Weak Simulation on One-Counter Nets*. Piotr Hofman, Richard Mayr, and Patrick Totzke. <http://doi.org/10.1109/LICS.2013.26>
[4]: *Simulation Over One-Counter Nets*. Piotr Hofman, Sławomir Lasota, Richard Mayr, and Patrick Totzke
(The paper appeared in LMCS, I cannot post more than 7 links apparently, sorry)
[5]: *Reachability in Petri nets with inhibitor arcs.* Klaus Reinhardt. <https://doi.org/10.1016/j.entcs.2008.12.042>
[6]: *On History-Deterministic One-Counter Nets*. Adityaf Prakash & K. S. Thejaswini
<https://link.springer.com/chapter/10.1007/978-3-031-30829-1_11>
[7]: *Büchi VASS recognise w-languages that are $\Sigmaˆ1\_1$-complete*. Michal Skrzypczak. <https://arxiv.org/abs/1708.09658>
[8]: *On Büchi One-Counter Automata*.
Böhm, Stanislav; Göller, Stefan; Halfon, Simon ;Hofman, Piotr
<https://drops.dagstuhl.de/opus/volltexte/2017/7019/>
| 4 | https://mathoverflow.net/users/509800 | 452092 | 181,739 |
https://mathoverflow.net/questions/452037 | 5 | Let us say that a partial order is "countably closed with infima" if every descending $\omega$-sequence has an infimum.
Suppose $P$ and $Q$ are posets that are countably closed with infima, and for some dense $D \subseteq Q$, there is a projection $\pi : D \to P$, i.e. $\pi$ is an order-preserving map such that whenever $p \leq \pi(q)$, then there is $q' \leq q$ in $D$ such that $\pi(q') = p$. Assume also that $D$ is closed under descending $\omega$-sequences, and $\pi$ is continuous on such sequences.
Whenever $G \subseteq P$ is generic, we define the quotient $Q/G$ as $\{ q : (\exists d \in D) d \leq q \wedge \pi(d) \in G \}$.
It is a well-known fact that under these assumptions, if $D = Q$ (i.e. $\pi$ is defined everywhere), then $Q/G$ is forced to be countably closed with infima. Does this necessarily hold when $D$ is a proper subset of $Q$?
| https://mathoverflow.net/users/11145 | Countable closure of quotient forcing | This does (unfortunately?) not hold in general.
Consider the following (trivial) forcings $P$ and $Q$:
The only thing $Q$ does is generically pick out some $N\leq\omega$, it does this in the following way. $Q$ has two types of conditions:
* atoms deciding $N$, one atom $a\_m$ representing "$N=m$" for $m\leq\omega$,
* conditions $q\_m$ representing "$N\geq m$" for integers $m<\omega$.
$Q$ is ordered in the obvious way and is countably closed with infima: Essentially the only nontrivial decreasing chain is $(q\_m)\_{m<\omega}$ and it converges to $a\_\omega$. Let $D$ be the dense set of the first type, i.e. the atoms.
Now $P$ has only two (incompatible) conditions, $p\_{\mathrm{fin}}$ representing "$N$ is finite" and $p\_{\mathrm{inf}}$ for "$N$ is $\omega$" and this suggests how to define the projection $\pi\colon D\rightarrow P$. For $G=\{p\_{\mathrm{fin}}\}$, we get $Q/G=Q-\{a\_\omega\}$ which is not countably closed.
| 6 | https://mathoverflow.net/users/125703 | 452097 | 181,740 |
https://mathoverflow.net/questions/260561 | 5 | To have the composition of two monads be a monad itself, we need a
distributive law natural transformation satisfying certain coherence
laws.
I'm interested in the strict 2-monad case, i.e. a strict 2-functor
equipped with unit and counit natural transformations that satisfy the
zig-zag equations on the nose.
I presume in such a case it's still possible for the distributive law
to satisfy its coherence laws only up to a modification.
If so, what coherence laws do those modifications have to satisfy?
| https://mathoverflow.net/users/756 | Coherence laws when composing 2-monads | These are known as [pseudo-distributive laws](https://ncatlab.org/nlab/show/pseudo-distributive+law) and are the most common notion of distributive law of 2-dimensional monads, even when both 2-dimensional monads in question are strict (i.e. 2-monads rather than pseudomonads). These have been well studied, and there are many equivalent formulations of the coherence conditions. The most comprehensive account is Walker's [Distributive laws, pseudodistributive laws and decagons](https://arxiv.org/abs/2102.12468).
| 1 | https://mathoverflow.net/users/152679 | 452100 | 181,742 |
https://mathoverflow.net/questions/452095 | 9 | The framed Pontryagin Thom construction produces a graded ring isomorphism from the framed cobordism ring $\Omega^{fr}\_\*$ to the stable ring of homotopy groups of spheres $\pi\_\*(\mathbb{S})$. I have a few questions about geometric interpretations of algebraic facts under this correspondence.
1. Every framed manifold has vanishing Stiefel-Whitney numbers, which implies that it is unoriented nullbordant, i.e. the zero element in $\Omega\_\*^O$. Given a manifold and a framing, is there a way to abstractly construct an unoriented nullbordism?
2. By Serre's finiteness, every element of positive degree in $\pi\_\*(\mathbb{S})$ is torsion. Given a framed manifold $M$, can we prove "geometrically" that there exists an integer $k$ such that the disjoint union of $k$ copies of $M$ is framed null?
3. Similarly by Nishida's nilpotence, given a framed manifold $M$, can we prove "geometrically" that there exists an integer $k'$ such that the $k'$-fold product of $M$ is framed null?
4. A priori, the two integers $k$ and $k'$ defined in the questions above are framed bordism invariants. But do they really depend on the framing? In other words, can we find a manifold $M$ and two "non-bordant" framings such that $M^{k'}$ (or $\sqcup^k M$) is null with one framing but not with the other?
I am aware that, at first glance, I am basically asking for geometric proofs of very involved results by Serre and Nishida, so interesting special cases and examples to any of the questions, such as in [this thread](https://mathoverflow.net/questions/201587/nilpotence-of-the-stable-hopf-map-via-framed-cobordism), would suffice!
| https://mathoverflow.net/users/484277 | Torsion and nilpotence of framed manifolds under the Pontryagin-Thom map | 1. I don't know how satisfied you would be by the paper:
"AN ELEMENTARY GEOMETRIC PROOF OF TWO THEOREMS OF THOM"
SANDRO BUONCRISTIANO and DEREK HACON
4. I think this happens already with the framed circle. One framing is nullbordant, but the other one is not (so k=1 or k=2). But maybe I don't understand the question.
| 9 | https://mathoverflow.net/users/12156 | 452107 | 181,744 |
https://mathoverflow.net/questions/450520 | 4 | This is inspired by the [problem of the Hoffman-Singleton Decomposition of](https://mathoverflow.net/a/450325) [$K\_{50}$](https://mathoverflow.net/a/450325). I wanted to look at smaller variants of this kind of problem, and so naturally I started wondering:
>
> Can the (edges of the) complete graph $K\_{16}$ be partitioned into $10$ cube graphs $Q\_3$?
>
>
>
By brute force, this can probably be done (or refuted??) in seconds, but are there really no *more elegant* methods?
A related question: Start with the tesseract graph $Q\_4$ and remove a certain 1-factor (= a perfect matching, i.e. 8 isolated edges). Call the resulting 3-regular graph $G$. For which choices of the 1-factor is it possible to **cover $K\_{16}$ by $5$ copies of $G$**?
Note that if we remove 8 “parallel” edges of the hypercube, $G$ is a (disconnected) $2Q\_3$, and so the resulting question is more constrained than the initial question of covering $K\_{16}$ by $10$ copies of $Q\_3$.
FWIW: Intuitively, I would conjecture that *none of those decompositions exists*. In that case, a weaker question would be whether at least, $K\_{16}$ may be covered by $5$ (possibly different) graphs of the type "$Q\_4\backslash 8K\_2$". Or might there even be a kind of parity argument to disprove that altogether, knowing that all those $G$ are bipartite?
| https://mathoverflow.net/users/29783 | Are there decompositions of $K_{16}$ by certain 3-regular graphs? | Such a decomposition exists. In particular a decomposition of 5 copies of $2Q\_3$ exists.
We can label each vertex with a binary number from 0000 to 1111 (or more mathematically with elements of $C\_2^4$). Note that each three linearly independent numbers $a, b, c$ correspond to a copy of $2Q\_3$ by connecting $u$ and $v$ if and only if $u \oplus v \in \{a,b,c\}$.
Thus we need to partition $\{0001, \ldots, 1111\}$ into five linearly independent sets. You can easily do this by hand, but if you want to use more fancy math you might want to look into partitioning the multiplicative group $C\_{15}$ of the finite field of order $F\_{16}$. For example, $\{\{a^{3n}, a^{3n+1}, a^{3n+2}\} \mid n \in \{1,2,3,4,5\}\}$ where $a$ is a generator.
| 1 | https://mathoverflow.net/users/502833 | 452109 | 181,745 |
https://mathoverflow.net/questions/452105 | 4 | $\DeclareMathOperator\Hom{Hom}$Let $A$ be a finite abelian group and $\text{Sym}(A)$ the (abelian) group of symmetric bilinear forms over $A$ valued in $\mathbb{R}/\mathbb{Z}$.
A quadratic function on $A$ is $q:A\rightarrow \mathbb{R}/\mathbb{Z}$ such that $q(-a)=q(a)$ and
$$
\chi\_q(a,b)=q(a+b)-q(a)-q(b)
$$
is bilinear. Hence if $q$ is quadratic $\chi\_q\in \text{Sym}(A)$. The map $q\rightarrow \chi\_q$ is clearly linear and thus defines a homomorphism from the group $Q(A)$ of quadratic functions to $\text{Sym}(A)$.
1. Is this map surjective? In other words, is there always a quadratic refinement of a symmetric bilinear form valued in $\mathbb{R}/\mathbb{Z}$? I think the answer is yes but I am not sure about a concrete proof.
2. Assuming the answer to the first question is yes, what is the kernel of the map? I think it should be isomorphic to $\Hom(A,\mathbb{Z}\_2)$ by looking at examples (and from other abstract arguments, see [Is there a clear pattern for the degree $2n$ cohomology group of the $n$'th Eilenberg-MacLane space?](https://mathoverflow.net/questions/450594/is-there-a-clear-pattern-for-the-degree-2n-cohomology-group-of-the-nth-eile)), but It is not clear to me how to prove this, and even how to see $\Hom(A,\mathbb{Z}\_2)$ as a subgroup of $Q(A)$.
| https://mathoverflow.net/users/495347 | Quadratic refinements of a bilinear form on finite abelian groups | 1. By the classification of finite abelian groups we have $A \cong C\_1 \oplus \cdots \oplus C\_k$ for some cyclic groups $C\_i = \langle g\_i \rangle$ of prime power order $r\_i$. Let $\chi: A \times A \to \mathbb R / \mathbb Z$ be a symmetric bilinear map. Observe that $\chi(g\_i, g\_j)$ has order dividing $\gcd(r\_i, r\_j)$ for all $i,j$. For each $i = 1, \dots, k$, let $q\_i$ be a solution to $2q\_i = \chi(g\_i, g\_i)$ such that $q\_i$ has odd order if $r\_i$ is odd (there is a unique choice if $r\_i$ is odd and two choices if $r\_i$ is even). Now define $q : A \to \mathbb R / \mathbb Z$ by
$$q\left(\sum\_{i=1}^k n\_i g\_i\right) = \sum\_{i=1}^k n\_i^2 q\_i + \sum\_{1 \le i < j \le k} \chi(g\_i, g\_j) n\_i n\_j.$$
This is well-defined by our careful choice of $q\_i$ and it is easy to check that $\chi\_q = \chi$.
2. As mentioned in the comments, the kernel consists of the homomorphisms $q : A \to \mathbb R / \mathbb Z$ taking half-integer values, which can obviously be identified with $\mathrm{Hom}(A, \mathbb Z / 2 \mathbb Z)$.
| 7 | https://mathoverflow.net/users/20598 | 452113 | 181,747 |
https://mathoverflow.net/questions/452068 | 3 | Let $U\subset\mathbf{P}^n\_{\mathbf{Z}}$ be an open subscheme such that the smooth morphism $U\to\text{Spec}(\mathbf{Z})$ is surjective. Suppose $U(\mathbf{Q})\neq\varnothing$ and $U(\mathbf{Z}\_p)\neq\varnothing$ for all primes $p$.
>
> Do we have $U(\mathbf{Z})\neq\varnothing$?
>
>
>
The answer is probably "no", but I could benefit from related references.
>
> Does there exist a finite extension $K/\mathbf{Q}$ such that, if $R$ is the integral closure in $K$ of $\mathbf{Z}$, then $U(R)\neq\varnothing$, and what can one say about $R$? (for example, how ramified is it?)
>
>
>
This question came up while reading through this [paper](http://www.numdam.org/item/10.5802/aif.1863.pdf) of Autissier.
>
> Do we at least have $U(\overline{\mathbf{Z}})^{\text{Gal}(\overline{\mathbf{Q}}/\mathbf{Q})}\neq\varnothing$, and under what conditions on $U$ can we deduce $U(\mathbf{Z})\neq\varnothing$? (presumably some condition on the codimension/singularities of the complement).
>
>
>
A variant of the same question, with $U(\mathbf{Q})\neq \varnothing$ replaced by $U(\overline{\mathbf{Z}})\neq\varnothing$.
>
> Does there exist a finite extension $K/\mathbf{Q}$ such that, if $R$ is the integral closure in $K$ of $\mathbf{Z}$, then $U(R)\neq\varnothing$ if $U(\mathbf{Z}\_p)$ and $U(\overline{\mathbf{Z}})$ are nonempty for all $p$, and what can one say about $R$?
>
>
>
| https://mathoverflow.net/users/509694 | $\mathbf{Z}$-points of quasi-projective schemes | As Jason Starr points out, $U = \mathbf P^1{\setminus}\{[1{:}0],[1{:}4],[0{:}1]\}$ is a counterexample to the first question:
* It has a $\mathbf Z[1/3]$-point $[1{:}1]$ and a $\mathbf Z[1/2]$-point $[1{:}2]$, so $U(\mathbf Q) \neq \varnothing$ and $U(\mathbf Z\_p)\neq\varnothing$ for all primes $p$;
* But it does not have a $\mathbf Z$-point: such a point is of the form $[a{:}b]$ with $a,b \in \mathbf Z$ and $\gcd(a,b) = 1$. Then $a \neq 0$ since $[a{:}b] \neq [0{:}1]$, so it is not possible that both $b$ and $4a-b$ are a unit in $\mathbf Z$. Then there exists a prime $p$ dividing either $b$ or $4a-b$. If $p \mid b$, then $[a{:}b]$ intersects $[1{:}0]$ above $p$, and if $p \mid 4a-b$, then $[a{:}b]$ intersects $[1{:}4] = [a{:}4a]$ above $p$. So $[a{:}b]$ never lands entirely in $U$.
The answer to the second question is positive: following the references in Autissier's paper quickly leads to the following result of Skolem [Sko34] (see also [MB89, 1.1] — I was unable to access the original source [Sko34] at the moment).
**Theorem.** *Let $p \in \bar{\mathbf Z}[x\_1,\ldots,x\_n]$ be a polynomial. Then the following are equivalent:*
* *there exist $a\_1,\ldots,a\_n \in \bar{\mathbf Z}$ such that $p(a\_1,\ldots,a\_n)$ is a unit;*
* *$p$ is primitive.*
This means that a standard open $U = D(f) \subseteq \mathbf A^n\_{\mathbf Z}$ has an $\mathcal O\_K$-point for some number field $K$ if and only if $f$ is primitive, i.e. if and only if $U \to \operatorname{Spec} \mathbf Z$ is surjective. If $f$ is moreover homogeneous, then the standard open $D\_+(f) \subseteq \mathbf P^{n-1}\_{\mathbf Z}$ therefore also has an $\mathcal O\_K$-point.
Now if $U \subseteq \mathbf P^n\_{\mathbf Z}$ is any open such that $U \to \operatorname{Spec} \mathbf Z$ is surjective, then it contains a standard open with the same property: for each component $Z\_i$ of the complement $Z$ of $U$, pick a primitive polynomial $f\_i$ in the homogeneous ideal of $Z\_i$. Setting $f = \prod\_i f\_i$ and $V = D\_+(f) \subseteq U$ does the trick. Since $V$ contains an $\mathcal O\_K$-point for some number field $K$, so does $U$.
(This is all subsumed by Rumely's theorem, but in this case we really only need Skolem's version!)
If you want to make this more effective, you have to look into the proofs of Skolem and Rumely.
---
**References.**
[MB89] L. Moret-Bailly, [*Groupes de Picard et problèmes de Skolem, I*](https://doi.org/10.24033/asens.1581). Ann. Sci. Éc. Norm. Supér. (4) **22**.2 (1989), p. 161-179. [ZBL0704.14014](https://zbmath.org/?q=an:0704.14014).
[Sko34] Th. Skolem, *Lösung gewisser Gleichungen in ganzen algebraischen Zahlen, insbesondere in Einheiten*. Skr. Norske Vid.-Akad. **10** (1935), p. 1-19. [ZBL0011.19701](https://zbmath.org/?q=an:0011.19701).
| 3 | https://mathoverflow.net/users/82179 | 452122 | 181,750 |
https://mathoverflow.net/questions/452121 | 0 | Suppose the following situation: we have to buy $n$ goods $g\_1,\,\dots,\,g\_n$ starting at day 1 and we can't buy more than one good per day.
On day $d$ the prices are $p\_1^d,\,\dots,\,p\_n^d;\quad p\_i^d\lt p\_i^{d+1}$, i.e. the prices for the goods keep rising.
>
> **Question:**
>
> what is the optimal buying strategy if the only available about the goods we have is the price for the current day and the next?
>
> Under which conditions is always buying the good, whose price will increase the most, the optimal strategy?
>
>
>
The actual motivation for the question comes from combinatorial optimization, where one can't foresee all effects of exchanging elements that are subjected to topological constraints e.g. in graph theory.
| https://mathoverflow.net/users/31310 | Optimality of a "shopping" heuristic | With the information given, for any strategy you might pick, there exists an instance of the game that makes this strategy awful. For example, suppose $n=2$. Let $i$ be the index of the good you buy at day $1$, and suppose that $p\_i^3=p\_i^2+\epsilon$ and that for all $j\neq i$, $p\_j^3=p\_j^2+N$ for small $\epsilon$ and very large $N$.
If you want the notion of "optimal strategy" to make sense, you need to give more information. For example, you could ask that the daily price increments follow a given probability law on $\mathbb R\_+$.
Assuming you have done so, let $X^d$ be the price increment vector between days $d$ and $d+1$, seen as a random variable in $(\mathbb R\_+)^n$. If the $(X^d)\_{1\leq i\leq d}$ are independant, then your strategy is optimal. Without independence, this is not always true.
| 3 | https://mathoverflow.net/users/158721 | 452123 | 181,751 |
https://mathoverflow.net/questions/452116 | 5 | In quantum mechanics we have position and momentum operators $P$ and $Q$ acting on $L^2(\mathbb{R})$ in the usual way. I'm wondering what the von Neumann algebra generated by the bounded functions of $P$ and $Q$ is. I.e., what is
$$
\{f(Q), f(P)\,\vert\,f\in L^\infty(\mathbb{R})\}''?
$$
I have a hunch that the answer may have something to do with the [Wigner-Weyl transform](https://en.wikipedia.org/wiki/Wigner%E2%80%93Weyl_transform), but (to my surprise) I have not been able to find any reference treating this question.
| https://mathoverflow.net/users/480683 | von Neumann algebra of canonical commutation relations | The C$^\*$-algebra generated by the exponentials $e^{isQ}$ and $e^{itP}$, $s,t \in \mathbb{R}$, is the CCR algebra. The von Neumann algebra they generate in this representation is all of $B(L^2(\mathbb{R}))$. That is because $e^{isQ}$ is multiplication by $e^{isx}$ and $e^{itP}$ is translation by $t$. Any operator that commutes with multiplication by $e^{isx}$ for all $s$ is a multiplication operator, and the only multiplication operators that commute with all translations are scalar multiples of the identity. So the result follows from the double commutant theorem.
| 8 | https://mathoverflow.net/users/23141 | 452125 | 181,752 |
https://mathoverflow.net/questions/452118 | 4 | I believe there is no good notion of homotopy groups for an arbitrary simplicial set $S$. However, when $S$ is *fibrant* - meaning that $S\to \*$ is a fibration - there is a definition. The singular realization of a topological space is fibrant, and one recovers the homotopy groups this way. Given a small category $\mathcal{C}$, its nerve $N\mathcal{C}$ is a simplicial set which is almost never fibrant; it is, though, whenever $\mathcal{C}$ is a groupoid.
Given an exact category $\mathcal{E}$, Quillen defines another category $Q(\mathcal{E})$ whose objects are the same as $\mathcal{E}$ but morphisms are given by "roofs" $\bullet \twoheadleftarrow \bullet \hookrightarrow \bullet$ where the epi and the mono are parts of exact sequences in $\mathcal{E}$. Composition is given by pullbacks, and is well-defined due to the axioms of exact categories.
Then, Quillen defines the $K$-group $K\_i(\mathcal{E})$ as the $(i+1)$th homotopy group of the topological realization of the simplicial set given by the nerve $N\mathcal{E}$ of $\mathcal{E}$.
I am curious: **since one can already define homotopy groups of simplicial sets, is there a way to bypass the topological realization of $N\mathcal{E}$?**
Since the category $\mathcal{E}$ is almost never a groupoid (although all morphisms are mono), I believe that the simplicial set $N\mathcal{E}$ is not fibrant. Yet, can one replace $Q(\mathcal{E})$ by some groupoid $G(\mathcal{E})$ functorially attached to $\mathcal{E}$ and define $K\_{i}(\mathcal{E})$ directly as $\pi\_{i+1}(NG(\mathcal{E}))$?
Sorry if this all well-known.
Many thanks!
| https://mathoverflow.net/users/66686 | Can one bypass the geometric realization in the definition of algebraic $K$-theory? |
>
> I believe there is no good notion of homotopy groups for an arbitrary simplicial set S.
>
>
>
It depends on what “good” means.
Kan's original definition works for arbitrary pointed simplicial sets: $$\def\Exi{\mathop{\sf Ex^∞}}π\_k(S,\*):=[S^k,\Exi S],$$
where $\Exi$ was defined by Kan in 1950s and $S^k$ can be taken to be any simplicial model for the $n$-sphere, e.g., $S^k=Δ^k/∂Δ^k$. (A modern exposition is found in Goerss–Jardine, Section III.4.) If “good” means usable in practice, there are many situations where the above definition is good.
>
> is there a way to bypass the topological realization of NE?
>
>
>
Yes, we can use the above definition of homotopy groups.
>
> Yet, can one replace Q(E) by some groupoid G(E) functorially attached to E and define Ki(E) directly as πi+1(NG(E))?
>
>
>
No, because the higher homotopy groups of the nerve of any groupoid necessarily vanish: $π\_k(NG,\*)≅0$ for $k>1$ and any basepoint $\*$. (That is to say, the nerve is a disjoint union of Eilenberg–MacLane spaces.)
| 6 | https://mathoverflow.net/users/402 | 452130 | 181,753 |
https://mathoverflow.net/questions/452127 | 0 | I would like to pose a question regarding the distance between a geodesic $\gamma(t)$ and a perturbed geodesic $\gamma\_{\epsilon}(t)$ on a Riemannian manifold. Specifically, is the distance controlled by a bound on the associated Jacobi field?
Following @Deane and @Otis (With heartfelt thanks), I'd like to attempt this question from a different perspective.. It may not be as strict, but I will do my best to gradually refine it. Starting from the definition (**Actually, I want to point out that when $t$ is sufficiently large, the distance between the geodesic still has an upper bound**) :
Let $ M $ be a Riemannian manifold and there exists a smooth map
$\Gamma: [0,\epsilon] \times [0,t] \rightarrow M$,
such that
\begin{align\*}
\Gamma(0,t) = \gamma(t), \quad
\Gamma(\epsilon,t) = \gamma\_\epsilon(t),
\end{align\*}
Let $\sigma(s)$ denote a unit speed minimizing geodesic from $\gamma(0)$ to $\gamma\_\epsilon(0)$, and $\epsilon$ be the length of $\sigma$.
Suppose we have a bound on the Jacobi field $\partial\_s\Gamma(s,t)$ for each $s \in (0,\epsilon)$ of the form
\begin{equation}
|\partial\_s\Gamma| \leq C.
\end{equation}
Note: This is essential; see @Deane's answer for details.
Try to demonstrate that (It also holds when $t$ is sufficiently large.)
\begin{equation}
\frac{1}{2} d^2 (\gamma\_\epsilon(t),\gamma(t))\leq C \epsilon.
\end{equation}
**Attempts (Update)**
Let
\begin{equation\*}
D(\epsilon) = \frac{1}{2}\int\_0^\epsilon d^{{2}}(\gamma\_s(t),\gamma(t)) ds.
\end{equation\*}
For each $ s \in (0, \epsilon) $, the Jacobi field $ \partial\_s \Gamma(s, t) $ satisfies the bound
\begin{equation\*}
\left|\partial\_s \Gamma\right| \leqslant C,
\end{equation\*}
where $C>0$ is a constant.
\begin{align\*}
(D'(\epsilon)) & = \frac{1}{2} d^2 (\gamma(t),\gamma\_\epsilon(t))= \frac{1}{2}
\int\_0^\epsilon \langle \nabla d^{2}(\gamma(t),\gamma\_s(t)), \partial\_s\Gamma(s,t) \rangle ds\\
& \leq \int\_0^\epsilon \langle d(\gamma(t),\gamma\_s(t)) , \vert \partial\_s\Gamma(s,t) \vert \rangle ds \\
& \leq C \int\_0^\epsilon d(\gamma(t),\gamma\_s(t)) ds.
\end{align\*}
According to the inequality $\frac{1}{2} d^2 (\gamma(t),\gamma\_\epsilon(t)) \leq C \int\_0^\epsilon d(\gamma(t),\gamma\_s(t)) ds$, we can obtain the following bound \begin{equation\*}d(\gamma(t),\gamma\_\epsilon(t)) \leq C\epsilon. \end{equation\*}
| https://mathoverflow.net/users/127966 | Compute distance between geodesics and perturbed geodesics on a Riemannian manifold via Jacobi field $\vert J \vert$ | You should formulate your question in more detail and rigor. Let me try to do this here and elaborate on Otis's comments.
Let $M$ be a Riemannian manifold and $\gamma: [0,1] \rightarrow M$ be a constant speed geodesic. You say that $\gamma\_\epsilon: [0,1] \rightarrow M$ is a perturbed geodesic but do not say what that means. I'll assume that this means there is a smooth map
$$ \Gamma: [0,\epsilon] \times [0,1] \rightarrow M $$
such that for any $t\in [0,1]$,
\begin{align\*}
\Gamma(0,t) &= \gamma(t)\\
\Gamma(\epsilon,t) &= \gamma\_\epsilon(t)
\end{align\*}
and for each $s \in [0,\epsilon]$, the curve
$$t \mapsto \Gamma(s,t)$$ is a constant speed geodesic.
Your goal is to estimate the distance between $\gamma(t)$ and $\gamma\_\epsilon$ and propose to use the observation that
$$
d(\gamma(t),\gamma\_\epsilon(t)) \le \ell(\Gamma(\cdot,t)) = \int\_{0}^{\epsilon} |\partial\_s\Gamma(s,t)|\,ds.
$$
By the Cauchy-Schwarz inequality, the following bound also holds:
$$
(d(\gamma(t),\gamma\_\epsilon(t)))^2 \le \int\_{0}^{\epsilon} |\partial\_s\Gamma(s,t)|^2\,ds.
$$
Along each geodesic $\Gamma(s,\cdot)$, the vector field $\partial\_s\Gamma(s,\cdot)$ is a Jacobi field. You denote
$$ J = \partial\_s\Gamma(0,\cdot). $$
It is clear that a bound on $|J(t)|=|\partial\_s\Gamma(0,t)|$ does not imply any bound on $|\partial\_s\Gamma(s,t)|$ for $0 < s \le \epsilon$. As Otis observes, this implies that your assumptions are not enough to obtain a bound on $d(\gamma(t),\gamma\_\epsilon(t))$.
If, on the other hand, if you have a bound on the Jacobi field $\partial\_s\Gamma(s,\cdot)$ for each $s \in (0,\epsilon)$ of the form
$$ |\partial\_s\Gamma| \le C, $$
then you obtain trivially that
$$ d(\gamma(t),\gamma\_\epsilon(t)) \le C\epsilon. $$
If you assume a sectional curvature bound, then, as Otis observes, you still need to assume more. One possible assumption consists of (sufficiently small) bounds on $d(\gamma(0),\gamma\_\epsilon(0))$ and $d(\gamma(1),\gamma\_\epsilon(1))$. Standard comparison theorems then imply the bounds you want. There are other assumptions you could make that imply similar bounds.
Last, I want to say again that if you want to ask a question on math.stackexchange.com (where I think this question really belongs) or MathOverflow, you should write it as carefully as I have done above. Often, by doing only that, you can see what's going on more clearly.
| 3 | https://mathoverflow.net/users/613 | 452147 | 181,755 |
https://mathoverflow.net/questions/452161 | 4 | Projective representations of a group $G$ are classified by the second group cohomology $H^2(G,U(1))$. If $G$ is finite and abelian, it is isomorphic to the direct product of cyclic groups
$$
G\cong C\_{n\_1,...,n\_k}:=\mathbb{Z}\_{n\_1}\times ...\times \mathbb{Z}\_{n\_k}
$$
Hence I was wandering if there is any known result for
$$
H^2(C\_{n\_1,...,n\_k},U(1))
$$
I know that
$$
H^2(\mathbb{Z}\_n,U(1))=0 \ , \ \ \ \ \ \ H^2(\mathbb{Z}\_{n\_1}\times \mathbb{Z}\_{n\_2},U(1))\cong \mathbb{Z}\_{\text{gcd}(n\_1,n\_2)}
$$
but I don't know about are a general result or even a result for $k=3$.
| https://mathoverflow.net/users/495347 | Projective representations of a finite abelian group | The answer is $\ \displaystyle\bigoplus\_{i<j}\ \mathbb{Z}/\!\gcd(n\_i,n\_j)$. The reason is that for $G$ finite, $H^2(G,U(1))$ is the dual abelian group of $H\_2(G,\mathbb{Z})$. Now use the Künneth formula.
| 5 | https://mathoverflow.net/users/460592 | 452166 | 181,759 |
https://mathoverflow.net/questions/452168 | 7 | $\require{AMScd}$Given an endofunctor $F : C\to C$, its category of algebras is the inserter of $F$ and the identity functor. This means that there is a square
$$\begin{CD}
Alg(F) @>j>> C \\
@VjVV \Rightarrow@| \\
C @>>F> C
\end{CD}$$
filled by a 2-cell $Fj\Rightarrow j$ and 1- and 2-terminal among all such.
Given a monad $T : C\to C$, one can build a similar diagram
$$\begin{CD}
EM(T) @>U>> C \\
@VUVV \Rightarrow@| \\
C @>>T> C
\end{CD}$$
where $EM(T)$ is the Eilenberg-Moore category and $TU=UFU\Rightarrow U$ is $U\epsilon$, where $F\dashv U$ is the free-forgetful adjunction, and $\epsilon : FU\Rightarrow 1$ its counit.
However, this is not an inserter, but a subobject thereof: $EM(T)$ is smaller than the inserter of $T$ and $1$ because one has to require compatibility of endofunctor algebras with the monad structure.
Also, the usual presentation of $EM(T)$ as a limit is always described as more complicated: $EM(T)$ is the lax limit of the monad regarded as a lax functor from the point, and this is a 2-limit of the monad regarded as a simplicial object, as in [Street, p.178](https://www.sciencedirect.com/science/article/pii/002240497690013X), or [Flexible limits, p.4](https://core.ac.uk/download/pdf/82151786.pdf).
Hence my question:
>
> Is there any hope to present $EM(T)$ concisely, giving a universal property to the cell $U\epsilon$ above? Maybe an equalizer obtained from $Alg(T)$ (the endofunctor algebras of the underlying functor, i.e. $Ins(T,1)$)? Or an iterated inserter, or some other "simple" 2-limits combined together in a clever way that can be written down "finitely"?
>
>
>
| https://mathoverflow.net/users/7952 | Eilenberg-Moore category as a 2-dimensional limit | Yes, the Eilenberg–Moore object for a monad $T$ can be presented in terms of two [equifiers](https://ncatlab.org/nlab/show/equifier) of the inserter $\mathbf{Ins}(T, 1)$. Denoting by $\phi \colon TU \Rightarrow U$, we equify $1\_U$ and $\phi \cdot \eta U \colon U \Rightarrow U$, as well as $\phi \cdot T \phi$ and $\phi \cdot \mu U \colon T^2 U \Rightarrow U$. This serves to impose the two axioms of an algebra for a monad.
One explicit reference for this is §2.78 of Adámek–Rosický's *Locally presentable and accessible categories*. It is also described in [this MathOverflow answer](https://mathoverflow.net/a/352917) (see also [this question](https://mathoverflow.net/questions/413937/constructing-the-e-m-category-of-a-monad-out-of-inserters-and-equifiers)), for instance.
| 10 | https://mathoverflow.net/users/152679 | 452172 | 181,760 |
https://mathoverflow.net/questions/452171 | 8 | From the answer to another question I asked ([Projective representations of a finite abelian group](https://mathoverflow.net/questions/452161/projective-representations-of-a-finite-abelian-group/452166#452166)) and from the structure theorem of finite abelian groups it follows that if $A$ is a finite abelian group with $H^2(A,U(1))=0$ then for every subgroup $B\subset A$ we also have $H^2(B,U(1))=0$ (namely $U(1)$ is a cohomologically trivial $A$-module).
1. Is there a way to prove this without using the structure theorem and computing explicitly the groups? For instance, is it the case that the map $H^2(A,U(1))\rightarrow H^2(B,U(1))$ induced by $K(B,1)\rightarrow K(A,1)$ is surjective? If yes, why?
2. Depending on the answer to the previous question this might be trivial or not: is it true that for any $A$-module $M$ and any $n$ such that $H^n(A,M)=0$, then $H^n(B,M)=0$? If not, are there conditions on $M$ (and maybe $n$) for which this is true?
| https://mathoverflow.net/users/495347 | Trivial group cohomology induces trivial cohomology of subgroups | For any abelian group $A$ we have a canonical isomorphism $\bigwedge^2A\to H\_2(A,\mathbb{Z})$, given by the (anti-symmetric) Pontrjagin product $H\_1(A,\mathbb{Z})\times H\_1(A,\mathbb{Z}) \to H\_2(A,\mathbb{Z})$, see for example Section 6 of Breen, "[On the functorial homology of abelian groups](https://doi.org/10.1016/S0022-4049(98)00140-6)". It follows that if $A$ is finite and $B$ is a subgroup of $A$ then $H\_2(B,\mathbb{Z}) \to H\_2(A,\mathbb{Z})$ is injective. Now $H^2(A,U(1))$ is the Pontrjagin dual abelian group to $H\_2(A,\mathbb{Z})$, and so for $B$ a subgroup of $A$ the restriction map $H^2(A,U(1))\to H^2(B,U(1))$ is surjective.
**Edit.** And now I see that I was too fast in writing this. It is not true that if $B \to A$ is an inclusion of finite abelian groups then $\bigwedge^2 B \to \bigwedge^2 A$ is injective. Let $A=\mathbb{Z}/p \times \mathbb{Z}/p^2$, generated by elements $u$ and $v$, and let $B$ be the subgroup $\mathbb{Z}/p\times \mathbb{Z}/p$ generated by $u$ and $pv$. Then $u\wedge pv$ is non-zero in $B\wedge B$ but zero in $A \wedge A$ because it equals $pu\wedge v$. So the answer to question 1 is no, for this inclusion.
Nonetheless, it is true for an inclusion of finite abelian groups, that if $\bigwedge^2A=0$ then $\bigwedge^2B=0$. Thus $H^2(A,U(1))=0$ implies $H^2(B,U(1))=0$.
But beware that for infinite abelian groups, even this is false. Look for example, at the inclusion $\mathbb{Z}/p\times\mathbb{Z}/p\to\mathbb{Z}/p\times\mathbb{Q}/\mathbb{Z}$; in this case $\bigwedge^2A=0$ but $\bigwedge^2B=\mathbb{Z}/p$.
| 11 | https://mathoverflow.net/users/460592 | 452176 | 181,762 |
https://mathoverflow.net/questions/452160 | 2 | As usual, for an $r$-uniform hypergraph $G$, denote by $ex\_r(n,G)$ the maximum number of edges an $r$-uniform, $G$-free hypergraph on $n$ vertices can have, and let $\lim \frac{ex\_r(n,G)}{\binom nr}\stackrel{n\to\infty}\longrightarrow\pi\_r(G)$.
Define the maximum of this quantity over graphs with $m$ edges as $\pi\_r(m)=\max \{\pi\_r(G) : e(G)=m\}$.
>
> Is it true that $\lim \pi\_r(m)\stackrel{r\to\infty}\longrightarrow 0$ for every $m$?
>
>
>
For $m=1,2$ this follows from well-known statements for set families, but I couldn't find anything for higher values.
On one hand, this feels like it should be simple, on the other, it seems to have strong consequences.
| https://mathoverflow.net/users/955 | Turán density of hypergraphs with very few edges | Seems, the answer is negative for $m=3$: consider $H = \{AB, BC, AC\}$ for disjoint $k$-element sets $A,B,C$. Then for $r=2k$ the is an $r$-graph $F$ which density is close to 0.5: consider only $r$-tuples that have an odd intersection with the first half-part $V\_1$ of vertex set. Suppose there is a copy of $H$. Then at least two of sets $A,B,C$ have intersection with $V\_1$ of the same parity. Union of two such sets do not belong to $F$; a contradiction.
| 5 | https://mathoverflow.net/users/479618 | 452179 | 181,764 |
https://mathoverflow.net/questions/452030 | 7 | This is a less ambitious version of [Is the Lebesgue measure of the $x$ so that this exponential sum is $o(\sqrt{N})$ positive?](https://mathoverflow.net/questions/451975/is-the-lebesgue-measure-of-the-x-so-that-this-exponential-sum-is-o-sqrtn) .
Consider $$S\_N(x):=\sum\_{n=1}^N \exp\left(2\pi i\left(\frac12n^2x+\alpha n\right)\right)$$
where $\alpha$ is irrational. If $x$ is an integer then $S\_N(x)$ is bounded for all $N$. Finding rational $x$ shouldn't be hard either. However, is there some pair $x, \alpha$ of irrational numbers so that $S\_N(x)=o(\sqrt N)$?
| https://mathoverflow.net/users/479223 | Does there exist some irrational $x,\alpha$ so that this Weyl sum is $o(\sqrt N)$? | The estimate
$$ S\_N(x) = S\_N(x,\alpha) = o(N^{1/2}) \tag {1}$$
cannot hold when $x$ is irrational.
Heuristically, the reason comes from the (approximate) modularity properties of $S\_N(x,\alpha)$; this expression is periodic mod $1$ in both $x$ and $\alpha$, and from the [van der Corput B-process](https://en.wikipedia.org/wiki/Van_der_Corput%27s_method) (Poisson summation) one morally has a relation
$$ |S\_N(x,\alpha)|
\sim x^{-1/2} |S\_{Nx}( -1/x, \alpha/x)|$$
when $x$ is non-zero (and $Nx$ is not too small). From periodicity, we see heuristically that if (1) holds for some choice of $x,\alpha,N$, then $x$ (and $\alpha$) can be placed in the interval $[-1/2,1/2]$ while remaining irrational and hence non-zero; from the B-process we may then replace $x,\alpha$ with $-1/x, \alpha/x$ and lower $N$ by a factor of $x$. Iterating this, we morally arrive at the point where (1) would hold in the range $N \sim 1$, which is absurd because the left-hand side jumps by a unit phase when $N$ increments by $1$, while the right-hand side would now be much less than $1$ due to the $o()$ factor.
To make this argument rigorous we introduce the [Jacobi theta function](https://en.wikipedia.org/wiki/Theta_function)
$$ \vartheta(z; \tau) := \sum\_{n \in {\bf Z}} e^{\pi i \tau n^2 + 2\pi i z n}$$
defined for complex $z,\tau$ with $\mathrm{Im} \tau > 0$. Morally we have
$$ S\_N(x,\alpha) \approx \vartheta( \alpha; x + i N^{-1/2} ).$$
We also will need absolute constants
$$ 0 < \delta \lll \varepsilon\_0 \lll \frac{1}{C} \lll 1$$
chosen so that $C$ is sufficiently large, $\varepsilon\_0$ is sufficiently small depending on $C$, and $\delta$ is sufficiently small depending on both $C$ and $\varepsilon\_0$. For any $x,\alpha,\varepsilon$ with $0 < \varepsilon < 1/C$, let $P(x,\alpha,\varepsilon)$ denote the assertion that the estimate
$$ |\vartheta(z; \tau)| \leq \delta (\mathrm{Im} \tau)^{-1/4} (1 + C \mathrm{Im}(\tau))$$
holds whenever $z, \tau$ are complex numbers with
\begin{align\*}
0 < \mathrm{Im}(\tau) &< \varepsilon\\
|\mathrm{Re}(\tau) - x| &< \mathrm{Im}(\tau) (1 - C \mathrm{Im}(\tau))\\
|\mathrm{Im}(z)| &< \mathrm{Im}(\tau)^{1/2}\\
|\mathrm{Re}(z) - \alpha| &< \mathrm{Im}(\tau)^{1/2} (1 - C \mathrm{Im}(\tau)).
\end{align\*}
The $C \mathrm{Im}(\tau)$ terms are technical higher order corrections that we have inserted in order to absorb certain error terms in the iterative component of our analysis, but can be ignored at a first reading.
We then make the following claims:
1. The property $P(x,\alpha,\varepsilon)$ is invariant with respect to shifting $x$ or $\alpha$ by an integer. This is clear from the periodicity properties of $\vartheta$.
2. If (1) holds for some $x, \alpha$, then $P(x,\alpha,\varepsilon)$ holds for all sufficiently small $\varepsilon > 0$. This is a routine exercise in summation by parts, splitting
$ e^{\pi i \tau n^2 + 2\pi i z n}$ into $e^{\pi i x n^2 + 2\pi i \alpha n}$ times a factor which exhibits gaussian decay outside of the region $n = O( (\mathrm{Im} \tau)^{-1/2} )$ and has bounded total variation.
3. If $0 < |x| \leq 1/2$, $|\alpha| \leq 1/2$ and $P(x,\alpha,\varepsilon)$ holds for some $0 < \varepsilon < \varepsilon\_0$, then $P(-1/x, \alpha/x, 2\varepsilon)$ holds. This is a routine computation coming from the Jacobi identity
$$ \vartheta(z;\tau) = (-i\tau)^{-1/2} e^{-\pi i z^2/\tau} \vartheta(z/\tau;-1/\tau)$$
which can be viewed as a special case of the Poisson summation formula.
The main point is that the transformation $(z,\tau) \mapsto (z/\tau, -1/\tau)$ is approximately affine from the region $\tau = -1/x + O(\varepsilon)$, $z = \alpha/x + O(\varepsilon^{1/2})$ to the region $\tau = x + O( \varepsilon x^2 )$, $z = \alpha + O( \varepsilon^{1/2} x)$, up to multiplicative distortion factors of $1 + O( |\tau+1/x|)$ that simplify to $1 - O(\mathrm{Im}(\tau))$ in the region of interest, and which can be tolerated for $C$ large enough.
4. For $\delta$ sufficiently small depending on $\varepsilon\_0, C$, it is not possible for $P(x,\alpha,\varepsilon)$ to hold for any $x,\alpha \in [-1/2,1/2]$ if $\varepsilon$ is in the interval $[\varepsilon\_0, 2\varepsilon\_0]$. For, if this were possible for arbitrarily small $\delta$, then by sending $\delta$ to zero and using compactness we would see that $\vartheta(z,\tau)$ vanished on an open set, hence vanishes for all $z,\tau$ with $\mathrm{Im}(\tau)> 0$ by analytic continuation, which is absurd.
The four properties 1-4 imply that (1) cannot hold when $x$ is irrational. Indeed, if (1) was true for some irrational $x,\alpha$, then by property 2 we have $P(x,\alpha,\varepsilon)$ for some $0 < \varepsilon < \varepsilon\_0$. Using property 1 we can shift $x,\alpha$ to lie in $[-1/2,1/2]$, and then by property 3 we can then double $\varepsilon$ (altering $x,\alpha$ again, but keeping $x$ irrational). Iterating this, we eventually can place $\varepsilon$ in the range $[\varepsilon\_0, 2\varepsilon\_0]$, at which point we contradict property 4.
This argument actually gives the slightly stronger assertion that
$$ \limsup\_{N \to \infty}
N^{-1/2} |\sum\_{n \leq N} e^{\pi i x n^2 + 2 \pi i \alpha n}| \geq c$$
for all irrational $x$, real $\alpha$, and some absolute constant $c>0$.
| 9 | https://mathoverflow.net/users/766 | 452184 | 181,767 |
https://mathoverflow.net/questions/452185 | 10 |
>
> **Problem.** Is there a subset $X$ in the Euclidean plane such that $X$ is not contained in a line and for any points $a,b,c,d\in X$ with $a\ne b$ and $c\ne d$, the intersection $X\cap\overline{ab}$ is infinite and the intersection $X\cap\overline{ab}\cap\overline{cd}$ is not empty.
>
>
>
Here $\overline {uv}$ denotes the (unique) line containing distinct points $u,v$ in the Euclidean plane.
| https://mathoverflow.net/users/61536 | A projective plane in the Euclidean plane | Let $\ P^2(\mathbb Q)\ $ and $\ P^2(\mathbb R)\ $ be the projective planes over rationals and reals. Let $\,\ L\subseteq P^2(\mathbb R)\,\ $ be a straight line in the real plane such that
$$ L\cap P^2(\mathbb Q)\,\ =\,\ \emptyset, $$
and let projective map $\,\ F: P^2(\mathbb R)\to P^2(\mathbb R)\,\ $ map $\ L\ $ onto the ***infinite*** line. Then
$$ X\,\ :=\,\ F(P^2(\mathbb Q)) $$
is the required set (see the ***OP Question***).
| 16 | https://mathoverflow.net/users/110389 | 452190 | 181,768 |
https://mathoverflow.net/questions/442196 | 1 | Let $W$ be a standard one dimensional Brownian motion, and $X$ a continuous process adapted to $W$ such that $\int\_0^T X^2 \, ds < \infty$ almost surely for some $T > 0$.
Define for any sequence of partitions $\mathcal P\_n = \{t\_1^n, \dotsc t\_{k\_n} ^n\}$ of $[0, T]$, the elementary integral process $Y^n$ associated to $\mathcal P\_n$ by
$$Y\_t^n := \sum\_{i = 1}^{k\_n - 1} X\_{t^n\_i} (W\_{t^n\_{i+i} \wedge t} - W\_{t^n\_{i} \wedge t} \,).$$
**Question:** Is it true that for all partitions $\mathcal P\_n$ of $[0, T]$ with mesh going to $0$, we have
$$\sup\_{0 \leq t \leq T} \left\lvert\int\_0^t X\_s \, dW\_s - Y\_t^n\right\rvert \to 0$$
in probability?
| https://mathoverflow.net/users/173490 | Convergence in sup norm of elementary integrals to the Itô integral process | Let
\begin{equation\*}
X^{(n)}\_t=\sum\_{i=1}^{k\_n-1} X\_{t\_i^n}1\_{(t\_i^n,t\_{i+1}^n]}(t).
\end{equation\*}
Since $X=\{X\_t,0\le t\le T\}$ is a continuous adapted process, then
\begin{gather\*}
\lim\_{n\to\infty} \sup\_{0\le t\le T}|X^{(n)}\_t - X\_t |=0, \qquad \text{a.s.}, \\
\lim\_{n\to\infty} \int\_{0}^{T} (X^{(n)}\_t - X\_t )^2\,\mathrm{d}t=0, \qquad \text{a.s.},
\end{gather\*}
and
\begin{equation\*}
\lim\_{n\to\infty}\mathsf{P}\Big(\Big|\int\_{0}^{T} (X^{(n)}\_t - X\_t )^2\, \mathrm{d}t \Big| >d\Big)=0, \quad \forall d>0. \tag{1}
\end{equation\*}
Due to $(\int\_0^t (X^{n}\_s-X\_s)\mathrm{d}W\_s)^2$ is L-dominated by
\begin{equation\*}
\langle \int\_0^{\cdot} (X^{(n)}\_s-X\_s)\mathrm{d}W\_s \rangle\_t =
\int\_{0}^{t} (X^{(n)}\_s - X\_s )^2\,\mathrm{d}s.
\end{equation\*}
Using Lenglart's inequality(cf. S. W. He *et al., Semimartingale Theory and Stochastic Calculus*, Sci. Press and CRC(1992). **Th**9.26, pp240, or
J. Jacod, and A. N. Shiryayev, *Limit Theory for Stochastic Processes*, 2ed. Springer, 2003, **Lemma** I.3.30, pp35.), we get
\begin{align\*}
&\mathsf{P}\Big(\sup\_{0\le t\le T}\Big[\int\_0^t (X^{(n)}\_s-X\_s) \mathrm{d}W\_s \Big]^2 \ge\epsilon \Big)\\
&\quad\le \frac{d}{\epsilon}+\mathsf{P}\Big[\int\_{0}^{T} (X^{(n)}\_t - X\_t )^2\,\mathrm{d}t \ge d\Big]. \tag{2}
\end{align\*}
Now, by (1), letting $n\to\infty$ and $d\downarrow 0$ consecutively in (2),
we obtain immediately the following assertion,
\begin{equation\*}
\lim\_{n\to\infty}\mathsf{P}\Big(\sup\_{0\le t\le T}\Big|\int\_0^t (X^{(n)}\_s-X\_s) \mathrm{d}W\_s \Big|\ge\epsilon \Big)=0, \qquad \forall \epsilon>0.
\end{equation\*}
| 2 | https://mathoverflow.net/users/103256 | 452204 | 181,773 |
https://mathoverflow.net/questions/452211 | 7 | It’s well known that the heart of a t-structure is an abelian category. My question is that can we find some structure on a triangulated category which can “produce” an exact category in analogy with the t-structure? I would be appreciated if someone can answer this question or give me some related references.
| https://mathoverflow.net/users/510027 | structure in triangulated category similar to t-structure | In
*Jørgensen, Peter*, [**Abelian subcategories of triangulated categories induced by simple minded systems**](https://doi.org/10.1007/s00209-021-02913-5), Math. Z. 301, No. 1, 565-592 (2022). [ZBL1503.16015](https://zbmath.org/?q=an:1503.16015).
The following theorem due to Matthew Dyer is provided:
>
> Let C be a triangulated category and D be an extension closed subcategory of D such that $Hom^{-1}\_{D}(M, N)=0$ for all M, N in D. Then C has a natural
> structure of exact category (with short exact sequences obtained by suppressing the arrows of degree 1 in the distinguished triangles of C with vertices in D). Moreover, there are natural isomorphisms $Ext^{i}\_{C}(M, N)\cong Hom^{i}\_{D}(M, N)$ for $0 ≤ i ≤ 1$.
>
>
>
(Taken for convenience from Dyer's original paper [here](https://www3.nd.edu/%7Edyer/papers/extri.pdf)).
This is in direct analogy with the heart of a t-structure being an extension closed subcategory such that $Hom\_{D}^{-i}(M,N)=0$ for all $i>0$ and $M,N$ in $C$.
I think from here you could construct a structure analogous to a pair of aisles $(D^{\leq 0},D^{\geq 0})$ of a t-structure which gives you these sort of subcategories as 'hearts' in some sense. Of course these have to have weaker axioms, like you can't ask for the aisles to be closed under shifts but I'm not sure this will suffice or if something else breaks along the way that you need to avoid from the axioms.
I hope this is of help.
| 10 | https://mathoverflow.net/users/44499 | 452215 | 181,774 |
https://mathoverflow.net/questions/452213 | 0 | How to use the contraction mapping theorem to prove the following result: Let $X$ and $Y$ be Banach spaces, let $a>0$, and let $$B\_a=B\_a\left(z\_0\right)=\left\{z \in X:\left\|z-z\_0\right\| \leq a\right\}.$$ Suppose that $F$ is a $C^1$ map of $B\_a$ into $Y$, with $F^{\prime}\left(z\_0\right)$ invertible, and satisfying, for some $0<\theta<1$,
$$
\left\|F^{\prime}\left(z\_0\right)^{-1} F\left(z\_0\right)\right\| \leq(1-\theta) a,
$$
and
$$
\left\|F^{\prime}\left(z\_0\right)^{-1}\right\|\left\|F^{\prime}(z)-F^{\prime}\left(z\_0\right)\right\| \leq \theta \quad \text { for all } z \in B\_a .
$$
Then there is a unique solution in $B\_a$ of $F(z)=0$.
| https://mathoverflow.net/users/499114 | A contraction mapping theorem | With $G(z)=F'(z\_0)^{-1}(F(z))$ you reduce your problem to the following assertion:
>
> $\|G'(z)-id\|\le \theta$ for $\|z-z\_0\|\le a$ and $\|G(z\_0)\|\le (1-\theta)a$ imply that $G$ has a zero in $B\_a(z\_0)$.
>
>
>
Indeed, we are looking for a fixed point of $H(z)=z-G(z)$ in $B\_a(z\_0)$. From $\|H'(z)\|=\|id-G'(z)\|\le \theta$ in $B\_a(z\_0)$ and the mean value inequality we get $\|H(z)-H(y)\|\le \theta \|z-y\|$ for all $z,y\in B\_a(z\_0)$ and to conclude with the fixed point theorem we need $H(z)\in B\_a(z\_0)$ for $z\in B\_a(z\_0)$. For $z\in B\_a(z\_0)$ this follows from $$\|H(z)-z\_0\|\le\|H(z)-H(z\_0)\|+\|H(z\_0)-z\_0\| $$ $$\le \theta \|z-z\_0\|+\|G(z\_0)\| \le \theta a +(1-\theta)a=a.$$
| 3 | https://mathoverflow.net/users/21051 | 452216 | 181,775 |
https://mathoverflow.net/questions/452159 | 13 | I want to check if $$\left\lfloor \left( \sum\_{k=n}^{2n}{\frac{1}{F\_{2k}}} \right)^{-1} \right\rfloor =F\_{2n-1}~~(n\ge 3) \tag{$\*$}$$ where $\lfloor x \rfloor$ is th floor function.
The Fibonacci sequence is defined by $F\_1=1$, $F\_2=1$, $F\_{n+1}=F\_n+F\_{n-1}~(n\ge 2)$. Then we can get $$F\_n=\dfrac{\alpha^n-\beta^n}{\sqrt{5}}$$
where $\alpha=\dfrac{1+\sqrt{5}}{2}$ and $\beta=\dfrac{1-\sqrt{5}}{2}.$
The following are some of my attempts:
>
> For some example:
>
>
> $n=3$, the left hand is $5$, the right hand is $5.$
>
>
> $n=4$, the left hand is $13$, the right hand is $13.$
>
>
> $$\vdots$$
>
>
> $n=15$, the left hand is $514229$, the right hand is $514229.$
>
>
> It is all true. But as $n$ increases, the order of magnitude grows
> very rapidly.
>
>
> I ask one of my good friends to use a Python program to check $(\*).$
> He says it is true for $n\le 35$. When $n=36$, the Python says it is not true, But when $n= 37$, it is true again.
>
>
> Thus I change one way and I ask my fiend to use a Python program to
> check $$\left( \sum\_{k=n}^{2n}{\frac{1}{F\_{2k}}} \right)^{-1} =F\_{2n-1}~~(n\ge 3) \tag{$\*\*$}.$$
>
>
> Then the program shows it is true at least for $31\le n\le 51.$
>
>
> But as you see, the left hand of $(\*\*)$ is a decimal and the right hand of $(\*\*)$ is an integer.
>
>
>
So I do not know if it is because the order of magnitude on the left hand of $(\*)$ is growing very fast, $(\*)$ becomes not true due to some computer shortcomings.
Finally I wonder if $(\*)$ is true or false? Any help and references are greatly appreciated.
Thanks!
I have also posted it on MSE as [On the finite sum of reciprocal Fibonacci sequences](https://math.stackexchange.com/questions/4747937/on-the-finite-sum-of-reciprocal-fibonacci-sequences)
| https://mathoverflow.net/users/494116 | On the finite sum of reciprocal Fibonacci sequences | We need to prove, equivalently
$$\frac1{F\_{2n-1}+1}<\sum\_{k=n}^{2n}\frac1{F\_{2k}}\le \frac1{F\_{2n-1}}, $$
that is, by the above expression for $F\_{k}$, since $\beta=-\alpha^{-1}$, we need to check the double inequality
$$\frac1{\alpha^{2n-1}+\alpha^{-2n+1}+\sqrt5}<\sum\_{k=n}^{2n}\frac1{\alpha^{2k}-\alpha^{-2k}}\le \frac1{\alpha^{2n-1}+\alpha^{-2n+1}}. $$
To do so we bound below and above the middle sum the obvious way
$$\sum\_{k=n}^{2n} \frac1{\alpha^{2k}} <\sum\_{k=n}^{2n}\frac1{\alpha^{2k}-\alpha^{-2k}} \le \frac1{1-\alpha^{-4n}} \sum\_{k=n}^{2n}\frac1{\alpha^{2k}}.$$
By computation $\displaystyle\sum\_{k=n}^{2n} \frac1{\alpha^{2k}}= \alpha^{-2n+1}-\alpha^{-4n-1}$, so everything follows from
$$\frac1{\alpha^{2n-1}+\alpha^{-2n+1}+\sqrt5}\le \alpha^{-2n+1}-\alpha^{-4n-1} $$
and
$$\frac{\alpha^{-2n+1}-\alpha^{-4n-1}}{1-\alpha^{-4n}} \le \frac1{\alpha^{2n-1}+\alpha^{-2n+1}} $$
both easily checked (the former is true for any $n\ge0$, the latter for $n\ge3$).
| 9 | https://mathoverflow.net/users/6101 | 452222 | 181,776 |
https://mathoverflow.net/questions/452221 | 26 | This problem has been originally posted at [math.stackexchange](https://math.stackexchange.com/q/4746506/573047).
Since there are no answers and no comments there yet, I am crossposting it here to know if it is already known and tractable.
Here is the problem, slightly restricted with respect to the original:
starting from a multiset of $6$ integer numbers, we modify it repeatedly by replacing two elements $a,b$ at choice with $a+b,a+b$. Is it possible to make all elements equal for any choice of the original multiset?
Does the answer change if we replace "integer" with "rational"?
Note that the original poster at math.stackexchange seems to have a proof for $n=3,5$ elements.
As noted in comments below, it is impossible to equalize $a,a,b$ with $a \gt b \ge 0$, because you will have another sequence of the same form.
| https://mathoverflow.net/users/136218 | "Make all numbers equal" game | I claim that it is always possible to make all of the integers equal to each other. To see this, consider the allowable transformations
1. $(a,b,c,d,e,f)\mapsto(a+b,a+b,c,d,e,f)$,
2. $(a\_1,\dots,a\_6)\mapsto(a\_{\sigma(1)},\dots,a\_{\sigma(6)})$, and
3. $(a,b,c,d,e,f)\mapsto(a/n,b/n,c/n,d/n,e/n,f/n)$ whenever $a/n,b/n,c/n,d/n,e/n,f/n$ are all integers.
We shall write $(a,b,c,d,e,f)\mapsto^\*(g,h,i,j,k,l)$ if $(g,h,i,j,k,l)$ can be obtained from $(a,b,c,d,e,f)$ by using a sequence of allowable transformations. We observe that if $(a,b,c,d,e,f)\mapsto^\*(1,1,1,1,1,1)$, then can transform $(a,b,c,d,e,f)$ into $(n,n,n,n,n,n)$ for some $n$ using only transformations $1$ and $2$, but transformation 3 makes it easier to use induction to show that we can always get $(a,b,c,d,e,f)\mapsto(1,1,1,1,1,1)$.
We observe that $(a,b,c,d,e,f)\mapsto^\*(a+b+c+d,a+b+c+d,a+b+c+d,a+b+c+d,e+f,e+f)$.
Now set, $\alpha=a+b+c+d,\beta=e+f$. I claim that $(\alpha,\alpha,\alpha,\alpha,\beta,\beta)=(0,0,0,0,0,0)$ or $(\alpha,\alpha,\alpha,\alpha,\beta,\beta)\mapsto^\*(1,1,1,1,1,1)$ by induction on the square of the $\ell^2$ norm $4\alpha^2+2\beta^2$ of $(\alpha,\alpha,\alpha,\alpha,\beta,\beta)$.
Case 0: $\alpha=0,\beta=0$. This is trivial.
Case 1: $\alpha=0,\beta\neq 0$. In this case,
$(\alpha,\alpha,\alpha,\alpha,\beta,\beta)\mapsto^\*(\beta,\beta,\beta,\beta,\beta,\beta)\mapsto^\*(1,1,1,1,1,1)$.
Case 2: $\alpha\neq 0,\beta=0.$
$(\alpha,\alpha,\alpha,\alpha,\beta,\beta)\mapsto^\*(\alpha,\alpha,\alpha,\alpha,\alpha,\alpha)\mapsto^\*(1,1,1,1,1,1)$.
Case 3: $\alpha\neq 0,\beta\neq 0$, $\alpha$ is even.
If $\alpha$ is even, then
$(\alpha,\alpha,\alpha,\alpha,\beta,\beta)\mapsto^\*(\alpha,\alpha,\alpha,\alpha,2\beta,2\beta)\mapsto^\*(\alpha/2,\alpha/2,\alpha/2,\alpha/2,\beta,\beta)$ and
$(\alpha/2,\alpha/2,\alpha/2,\alpha/2,\beta,\beta)\mapsto^\*(1,1,1,1,1,1)$ by the induction hypothesis.
Case 4: $\alpha\neq 0,\beta\neq 0$, $\beta$ is even.
$(\alpha,\alpha,\alpha,\alpha,\beta,\beta)\mapsto^\*(2\alpha,2\alpha,2\alpha,2\alpha,\beta,\beta)\mapsto^\*(\alpha,\alpha,\alpha,\alpha,\beta/2,\beta/2)$, and
$(\alpha,\alpha,\alpha,\alpha,\beta/2,\beta/2)\mapsto^\*(1,1,1,1,1,1)$ by the induction hypothesis.
Case 5: $\alpha\neq 0,\beta\neq 0$, $\alpha,\beta$ are both odd.
Let $\alpha=2\gamma+1,\beta=2\delta+1.$ Then $(\alpha,\alpha,\alpha,\alpha,\beta,\beta)
\mapsto^\*(2(2\gamma+1),2(2\gamma+1),2(\gamma+\delta+1),2(\gamma+\delta+1),2(\gamma+\delta+1),2(\gamma+\delta+1))$
$\mapsto^\*(2\gamma+1,2\gamma+1,\gamma+\delta+1,\gamma+\delta+1,\gamma+\delta+1,\gamma+\delta+1)=(\alpha,\alpha,\frac{\alpha+\beta}{2},\frac{\alpha+\beta}{2},\frac{\alpha+\beta}{2},\frac{\alpha+\beta}{2}).$
If $\alpha=\beta$, then clearly $(\alpha,\alpha,\alpha,\alpha,\beta,\beta)
\mapsto^\*(1,1,1,1,1,1)$, and if $\alpha\neq\beta$, then we may use the induction hypothesis to conclude that $(\alpha,\alpha,\frac{\alpha+\beta}{2},\frac{\alpha+\beta}{2},\frac{\alpha+\beta}{2},\frac{\alpha+\beta}{2})\mapsto^\*(1,1,1,1,1,1).$
| 26 | https://mathoverflow.net/users/22277 | 452229 | 181,778 |
https://mathoverflow.net/questions/452189 | 16 | In FGA 3.V, there is a citation for
>
> Mumford D. and Tate J., Séminaire de géométrie algébrique, Harvard University, Spring term 1962 (à paraître).
>
>
>
This seems to be the same seminar mentioned by Grothendieck in a letter to M. Artin on the 23rd of February, 1963 (see footnote 2 of letter number 19, page 71, of [this digitisation](https://webusers.imj-prg.fr/%7Eleila.schneps/grothendieckcircle/Letters/AGMumford.pdf)), which says that
>
> Lichtenbaum’s notes on the lectures by Grothendieck, Mumford and Tate have not been published.
>
>
>
**Did these notes ever appear? Is there a better way to cite this seminar?**
| https://mathoverflow.net/users/73622 | Mumford–Tate 1962 "Algebraic geometry seminar" citation | I asked Steve Lichtenbaum, and he wrote "I started to write up the notes for the Grothendieck-Mumford seminar but then Grothendieck left and I never finished. The notes do not exist. They probably should not have been cited."
| 21 | https://mathoverflow.net/users/11926 | 452230 | 181,779 |
https://mathoverflow.net/questions/452228 | -2 | The context of this question is related to proving the first incompleteness by alternative ways related to Rosser's trick. So, for a proof by negation, we assume that $T$ is complete, and fulfills Gödel's criteria of effectively capturing all computable functions, etc..
Can $T$ decide on sentence $\sigma$ defined below?
$\sigma \iff \\\forall k (\operatorname {Proof}\_T(k, \ulcorner \sigma \urcorner) \to \\\forall s: \exists x [\operatorname {Proof}\_T(x,\ulcorner s \urcorner) \land x \leq k] \lor \exists y [\operatorname {Proof}\_T(y, \operatorname {neg} (\ulcorner s\urcorner)) \land y \leq k])$
In English this reads as: This sentence is only provable at lengths such that every sentence is either provable or disprovable below (non-strictly) it.
The idea is that, as defined, $\sigma$ can only be provable by proofs having non-standard Gödel codes, meaning that if $T$ is complete then $\sigma$ must be false, i.e. we have $T \vdash \neg \sigma$.
So, if we assume $T$ to be complete, then this criterion, I think, would separate standard from non-standards, so if a code of a proof doesn't fulfill that feature, i.e. not every sentence is decidable below it, then this code is standard!
However, I'm not sure of this later feature, that's why I'm not sure of the decidability of $\neg \sigma$ in $T$.
Of course this can be settled brutally by defining a minimal for such proofs, i.e. for example define:
${\frak K} = \min k : \forall s \, (\exists x [\operatorname {Proof}\_T(x,\ulcorner s \urcorner) \land x \leq k] \lor \exists y [\operatorname {Proof}\_T(y, \operatorname {neg} (\ulcorner s\urcorner)) \land y \leq k])$
Then define the diagonal $\pi$ as:
$\pi \iff \forall x (\operatorname {Proof}\_T(x,\ulcorner \pi \urcorner) \to x \geq {\frak K})$
And clearly $\pi$ would be undecidable in $T$.
>
> What is the status of $\sigma $ in $T$ should we assume that $T$ is complete and fulfill Gödel's criteria.
>
>
>
| https://mathoverflow.net/users/95347 | Can this Rosser-like trick also work as a proof of the first incompleteness theorem? | I see two major issues with the proposal.
First, if $\sigma$ is refutable, then in the standard model the biconditional will be vacuously true, since there will be no $k$ for which $\text{Proof}\_T(k,\ulcorner\sigma\urcorner)$. This seems to trivialize the idea.
But second, even if one were to fix that somehow, the more serious point is that PA proves that there can be no number $k$ that serves as a bound for the size of a code of a proof of every sentence $s$ or its negation. The reason is that any proof a sentence must mention that sentence, and from this it follows (using any of the standard proof predicates) that the code of any proof will be at least as great as the code of the sentence being proved. Because you are asking for $k$ to have the property that it bounds the proofs of every sentence or its negation, this is impossible — there can be no number $k$ with the property mentioned on the RHS of your biconditional.
From this point of view, any $\sigma$ that fulfills the biconditional will itself have to be refutable.
| 2 | https://mathoverflow.net/users/1946 | 452234 | 181,780 |
https://mathoverflow.net/questions/452027 | 5 | Let $\{x\_i\}\_{i\geq 1}$ be iid random elements of the sequence space $\ell^2(\mathbb{N})$;
assume that $\|x\_i\|\_2 \leq 1$ almost surely. Let $\Sigma = \mathbb{E}[x\_1 \otimes x\_1]$.
Define
$$
\Sigma\_n = \frac{1}{n} \sum\_{i=1}^n x\_i \otimes x\_i.
$$
I am interested in upper bounding the following lower tail
$$
\mathbb{P}\Big\{ \lambda\_{\rm max}(\Sigma - \Sigma\_n) > t \Big\}, \quad \mbox{for}~t > 0.
$$
What is a reasonably good bound for this quantity?
| https://mathoverflow.net/users/121486 | Lower tail of random rank one sums? | Warning: This is not a proper answer, just a dump of the thoughts I have had about this problem so far. Also, I'm not an expert in random matrix theory, so some bounds I'll be using may cry for improvement and if someone can do any of them better, I would appreciate both a helping hand and a criticism. We'll still get *something* in the end, but I'm in no way happy with that something, so, please, don't accept this post as an answer yet (or other people may get an impression that the problem is solved while it certainly isn't).
***The tools***
First of all, we'll need the Bernstein (a.k.a. Hoeffding, a.k.a....) inequality in the following form. Let $\xi$ be a real random variable on $[-1,1]$ with $E\xi=0$ and $E\xi^2=\lambda$. Then if $\xi\_1,\dots,\xi\_n$ are independent and identically distributed with $\xi$, we have for every $t>0$ and every $s\in(0,1)$
$$
P(\xi\_1+\dots+\xi\_n>nt)\le e^{ns(s\lambda-t)}\,.
$$
That comes straight from the consideration of the exponential moment $Ee^{s\xi}\le E(1+s\xi+s^2\xi^2)=1+s^2\lambda\le e^{s^2\lambda}$ and the usual chain of inequalities
$$
e^{snt}P(\xi\_1+\dots+\xi\_n>nt)\le Ee^{s(\xi\_1+\dots+\xi\_n)}=(Ee^{s\xi})^n\le e^{s^2\lambda n}\,.
$$
Next, we'll need the Shur bound for the operator norm of a matrix $B=(b\_{ij})$, which is
$$
\|B\|\le \sqrt{ \left(\sup\_i\sum\_j|b\_{ij}|\right)\left(\sup\_j\sum\_i|b\_{ij}|\right) }\,.
$$
Third, we'll need the fact that the operator norm of the operator $S=\sum\_{i} x\_i\otimes x\_i$ is at most the operator norm of the Gram matrix $G=(\langle x\_i,x\_j\rangle)\_{i,j}$.
That is because for every unit vector $v\in H$, we have
$$
|Sv|^2=\left|\sum\_i \langle x\_i,v\rangle x\_i\right|^2=
\sum\_{i,j} \langle x\_i,v\rangle G\_{ij} \langle x\_j,v\rangle
\\
\le \|G\| \sum\_i \langle x\_i,v\rangle^2 =\|G\|\langle Sv,v\rangle\le \|G\|\|S\|\,,
$$
so $\|S\|^2\le \|G\|\|S\|$, i.e., $\|S\|\le\|G\|$.
At last, we will need to recall that in the coupon collector problem with $d$ coupons, the typical (in any reasonable sense; say, median) time needed to collect all coupons is about $d\log d$.
The rest will be just some common sense that I'll introduce on the fly.
***Restatement***
We can restate the problem in the following way. What is the best universal (i.e., depending on $n$ and $t$ only) upper bound for the probability of the event that the operator
$$
\frac 1n\sum\_{i=1}^n x\_i\otimes x\_i-\frac 12 \Sigma+\frac t2 I = T-\frac 12 \Sigma+\frac t2 I
$$
is not positive definite?
We'll be interested in the regime of small $t$ (say, $t<1/2$).
***Example (presumably, the worst case scenario)***
We'll start with the following example. Let an integer $d$ be just below $1/t$. Consider the uniform distribution on the $d$ orthonormal basis vectors in $\mathbb R^d$. Then $\Sigma=d^{-1}I$, so our event definitely holds if at least one basis vector does not appear among $x\_i$. By the coupon collector mumbo-jumbo, this probability is uniformly separated from $0$ when $n<d\log d\approx \frac 1t \log\frac 1t$ and is about $d(1-\tfrac 1d)^n\approx \frac 1te^{-nt}$ afterwards, so $L(n,t)=\min(1,\tfrac 1t e^{-nt})$ is definitely a lower bound if you ignore a few absolute constant factors. I believe that it is actually the truth and we'll try to show that and prove enough to make this conjecture rather plausible, but I don't have the full argument yet.
***Reduction to a finite-dimensional space***
Our first reduction will be reducing the question to a $d$-dimensional subspace of our Hilbert space where $d=d(t)$ depends on $t$ only. We shall get $d(t)\approx t^{-2}$ at this first reduction though I suspect that one can get $d\approx t^{-1}$ immediately if one is just a bit more inventive.
Pick $\lambda>0$ and split our Hilbert space into the orthogonal sum of two $\Sigma$- invariant subspaces $H'\oplus H''$ such that $\Sigma\le\lambda I$ on $H''$ and $\Sigma\ge \lambda I$ on $H'$. Since the trace of $\Sigma$ is at most $1$, the dimension $d$ of $H'$ is at most $\lambda^{-1}$. Now take a unit vector $v=(v',v'')\in H$ and also split our random variable $x$ as $(x',x'')$. We have
$$
\langle (T-\tfrac 12\Sigma+\tfrac t2 I v,v\rangle=
\\
\frac 1n\sum\_i \langle v',x\_i'\rangle^2 + \frac 1n\sum\_i \langle v'',x\_i''\rangle^2+
\frac 2n\sum\_i \langle v',x\_i'\rangle\langle v'',x\_i''\rangle
\\
-\frac 12\langle \Sigma v',v'\rangle-\frac 12\langle \Sigma v'',v''\rangle+\frac t2(|v'|^2+|v''|^2)
\\
\ge \frac 34 \frac 1n\sum\_i \langle v',x\_i'\rangle^2 -\frac 12\langle \Sigma v',v'\rangle+\frac t2 |v'|^2
\\
-3\frac 1n\sum\_i \langle v'',x\_i''\rangle^2+(\tfrac t2-\lambda)|v''|^2
$$
where the inequality $2ab\ge -\frac 14 a^2-4b^2$ was used to get rid of the cross-term.
Out of the last two lines, the non-negativity of the first one corresponds pretty much to the original problem in the finite-dimensional space $H'$ except we have $\frac 34 T$ instead of $T$ (note that $P\_{H'}\Sigma P\_{H'}=E x'\otimes x'$ and we still have $|x'|\le 1$. Thus, if we show that with probability $\ge 1-ne^{-ct}$ the second of the last two lines is non-negative for every unit vector $v$, we will be able to concentrate on the finite-dimensional problem.
(note that $ne^{-ct}$ is essentially the same function as $t^{-1}e^{-ct}$ in my sense when truncated at $1$ from above).
The second line will certainly be non-negative if the norm of the operator $\frac 1n\sum\_i x\_i''\otimes x\_i''$ is not more than $\frac t6-\frac\lambda 3$. Switching to the Gram matrix, and using the Shur test, we see that this will certainly happen if
$$
\frac 1n\left[|x\_i''|^2+\sum\_{j:j\ne i}|\langle x''\_i,x''\_j\rangle|\right]\le \frac t6-\frac\lambda 3\,.
$$
Since $|x''\_i|^2\le 1$, we can reduce it to
$$
\frac 1{n-1}\sum\_{j:j\ne i}|\langle x''\_i,x''\_j\rangle|\le \frac t6-\frac\lambda 3-\frac 1n\,.
$$
The RHS for $n>\frac 1t\log \frac 1t$, $\lambda\le \frac t3$ and reasonably small $t$ is at least $0.1 t$. To estimate the LHS, we notice that for every fixed $x''\_i$, the i.i.d. random variables $\xi\_j=|\langle x''\_i,x''\_j\rangle|$, satisfy $E\xi\_j^2\le\lambda$ and, thereby, $E\xi\_j\le \sqrt{\lambda}$, so, by Bernstein,
$$
P\left(\frac 1{n-1}\sum\_{j:j\ne i}\xi\_j>\sqrt{\lambda}+\tau\right)\le e^{s(n-1)(s\lambda-\tau)}\,.
$$
Hence, choosing $\lambda=(0.05t)^2$, $\tau=0.05 t$ and $s=1$, we get our probability bound $e^{-cnt}$ for one row (or column: the Gram matrix is symmetric). The trivial union bound finishes the story then.
Thus, dropping the event of probability that is right in line with the conjecture, we can assume WLOG that we are working in the finite-dimensional space of dimension $\approx t^{-2}$.
This part will, probably, survive in the final version. At least, it definitely doesn't hurt to make this initial reduction step. The rest will, most likely, have to be modified a lot
***Crude net argument***
We thus want to estimate the probability that the operator $A=\frac 34 T-\frac 12\Sigma+\frac t2 I$ is not positive-definite on the $d$-dimensional space with $d\approx t^{-2}$. Note that the norm of $A$ is at most $1$, so it would be enough to check the inequality $\langle Av,v\rangle\ge \frac t4 |v|^2$ on, say, $\frac t8$-net on the unit $d$-dimensional sphere. For each unit vector $v$, the failure of the verification means that
$$
\frac 1n\sum\_{i}\xi\_i\ge (\tfrac 23 t+\tfrac 13 e(v))
$$
where $e(v)=E[\langle v,x\rangle^2]$ and $\xi\_i=e(v)-\langle v,x\_i\rangle^2$. Note that
$$
E\xi^2\le E \langle v,x\_i\rangle^4\le E\langle v,x\_i\rangle^2=e(v)
$$
(the subtraction of the mean only reduces the second moment), so, by Bernstein, the probability of the verification failure is at most
$
e^{ns(se(v)-[(\tfrac 23 t+\tfrac 13 e(v)])}=e^{-\frac 29 nt}
$
if we choose $s=\frac 13$.
That is just the decay we want. The issue is that the $t/8$-net in the $d\approx t^{-2}$-dimensional space is huge: it has about $e^{t^{-2}\log(1/t)}$ elements, so the trivial union bound gives us that extra factor, which means that we have proved our correct exponential decay $e^{-cnt}$ only for $n\ge t^{-3}\log\frac 1t$ while the conjecture is equivalent to saying that it holds for $n\ge t^{-1}\log\frac 1t$. The whole blood should be spilled now on reducing this 3 to 1 and I'd like to think a bit before posting the next morsel.
As I expected, Mark Rudelson had the right tools in his toolbox immediately. The key words are "matrix Bernstein inequality". You can get a ready reference in Vershinin's book but I, as usual, prefer to present a full exposition, so here is the argument.
***Matrix Bernstein inequality (the simplest version)*** Suppose that $X$ is a symmetric $d\times d$ random matrix and we have $n$ independent identically distributed with $X$ random symmetric matrices $X\_j$. Choose some positive number $\tau>0$ and write the chain of inequalities
$$
E \operatorname{Tr}[e^{\tau\sum\_{1\le i\le n}X\_i}]\le
E\operatorname{Tr}[e^{\tau\sum\_{1\le i\le n-1}X\_i}e^{\tau X\_n}]
\\
=
E\operatorname{Tr}[e^{\tau\sum\_{1\le i\le n-1}X\_i}Ee^{\tau X\_n}]\le
\|Ee^{\tau X}\| E\operatorname{Tr}[e^{\tau\sum\_{1\le i\le n-1}X\_i}]
$$
where in the first step the Golden-Thompson inequality $\operatorname{Tr}[e^{A+B}]\le\operatorname{Tr}[e^Ae^B] $ was used. A nice proof of it can be found on the Terence Tao blog: <https://terrytao.wordpress.com/2010/07/15/the-golden-thompson-inequality/>
Continuing all the way down and using the trivial inequality $\operatorname{Tr} A\le d\|A\|$ at the last step, we conclude that
$$
E \operatorname{Tr}[e^{\tau\sum\_{1\le i\le n}X\_i}]\le d\|Ee^{\tau X}\|^n\,.
$$
In particular, this implies that
$$
P(\lambda\_{\text{max}}(\sum\_i X\_i)\ge \omega n)\le d\|Ee^{\tau X}\|^ne^{-\omega\tau n}\,.
$$
We shall apply this first to $X\_i=x\_i\otimes x\_i$, $\tau=1$ in the situation when $Ex\otimes x\le \lambda I$ with $\lambda$ being a small multiple of $t\in(0,1)$ and $d\approx t^{-2}$. Then
$$
e^{X}=I+\frac{e^{|x|^2}-1}{|x|^2}x\otimes x\le I+(e-1)x\otimes x
$$
so $Ee^{\tau X}\le (1+(e-1)\lambda)I\le e^\lambda I$ and we conclude that
$$
P(\left\|\sum\_{i=1}^n x\_i\otimes x\_i\right\|>cnt)\le Ct^{-2}e^{(\lambda-ct)n}\,,
$$
which will allow us using the trickery from the first morsel to remove the invariant subspace of $\Sigma$ on which $\Sigma<\lambda I$ and reduce our considerations to the invariant subspace on which $\Sigma\ge\lambda I$ with $\lambda$ comparable with $t$ and $d$ comparable to $t^{-1}$ or less.
Now, in that subspace, we shall use $X\_i=I-\Sigma^{-1/2}x\_i\otimes x\_i\Sigma^{-1/2}=I-y\_i\otimes y\_i$ where $y\_i=\Sigma^{-1/2}x\_i$. Note that $Ey\otimes y=I$ and $|y\_i|^2\le Ct^{-1}$ now, so we can use $\tau=ct$ with small absolute $c>0$ and write
$$
e^{\tau X}=e^{ct}(I-\tfrac {1-e^{-ct|y|^2}}{|y|^2}y\_i\otimes y\_i)\le
e^{ct}(I-t\tfrac {1-e^{-cC}}{C}y\_i\otimes y\_i)\,,
$$
so, taking the expectation and then the norm, we get
$$
\|Ee^{\tau X}\|\le e^{ct}(1-t\tfrac{1-e^{-cC}}C)\le e^{2c^2Ct}
$$
provided that $cC<1, t<1$. Taking now $\omega=0.1$, say, we conclude that
$$
P(\lambda\_{\text{max}}(\sum\_i X\_i)>0.1 n)\le C't^{-1}e^{(2c^2C-0.1 c)tn}
$$
which, for small enough $c$, is what we wanted. It remains to note that the complement of the event in parantheses is
$
\sum\_i X\_i\le 0.1nI
$
or, equivalently, $\Sigma-\frac 1n\sum\_i x\_i\otimes x\_i\le 0.1\Sigma$.
That finishes the proof of the conjectured universal bound. Now we can think of what happens if the spectrum of $\Sigma$ decays faster.
| 7 | https://mathoverflow.net/users/1131 | 452247 | 181,781 |
https://mathoverflow.net/questions/452151 | 2 | If $X$ and $Y$ are random variables, then a **maximal coupling** of $X$ and $Y$ is a coupling $\left(X', Y'\right)$ such that $\mathbf{P}\left(X'=Y'\right)$ is maximal (that is, the probability that the coupled variables coincide is optimal). By $``$coupling$"$, I'm referring to a random vector $\left(X',Y'\right)$ such that $X'$ has the same distribution as $X$, and $Y'$ has the same distribution as $Y$.
>
> Are there any benefits of using a maximal coupling to describe distributional results regarding the marginals $X$ and $Y$? E.g., are the measures of dependence between $X$ and $Y$, such as covariance, mutual information, joint entropy, Kullback-Leibler distance, or $$\lvert\mathbf{P}\left(X'\in A\text{ and }Y'\in B\right)-\mathbf{P}\left(X'\in A\right)\mathbf{P}\left(Y'\in B\right)\rvert$$ going to be extreme in the maximal coupling?
>
>
>
| https://mathoverflow.net/users/95756 | Measures of dependence in a maximal coupling | $\newcommand{\R}{\mathbb R}\newcommand{\la}{\lambda}$Let $\mu$ and $\nu$ stand for the distributions of real-valued random variables (r.v.'s) $X$ and $Y$, respectively. Let $\Pi(\mu,\nu)$ denote the set of all probability distributions over $\R^2$ with marginals $\mu$ and $\nu$.
By Strassen's [Theorem 11](https://projecteuclid.org/journals/annals-of-mathematical-statistics/volume-36/issue-2/The-Existence-of-Probability-Measures-with-Given-Marginals/10.1214/aoms/1177700153.full), $\pi\in\Pi(\mu,\nu)$ is the distribution of a maximal coupling iff
\begin{equation}
\pi(D)=1-\|\mu-\nu\|,
\end{equation}
where $D:=\{(x,x)\colon x\in\R\}$ and $\|\mu-\nu\|$ is the total variation distance between $\mu$ and $\nu$.
It should be clear now that a maximal coupling is not unique. Indeed, let $(P,N)$ be the Hahn decomposition of the signed measure $\mu-\nu$ and then let $(\mu-\nu)\_+$ and $(\mu-\nu)\_-=(\nu-\mu)\_+$ be the Jordan positive and negative parts of $\mu-\nu$, so that $(\mu-\nu)\_+(A)=(\mu-\nu)(A\cap P)$ and $(\mu-\nu)\_-(A)=-(\mu-\nu)(A\cap N)$ for Borel $A\subseteq\R$. Let $\mu\wedge\nu:=\mu-(\mu-\nu)\_+=\nu-(\nu-\mu)\_+$. Let $\la$ be any measure on $\R^2$ with marginals $(\mu-\nu)\_+$ and $(\nu-\mu)\_+$. For all Borel $B\subseteq\R^2$, let
\begin{equation}
\pi(B)=(\mu\wedge\nu)(\text{pr}(B\cap D))+\la(B),
\end{equation}
where $\text{pr}(x,y):=x$ for $(x,y)\in\R^2$.
Then $\pi\in\Pi(\mu,\nu)$ and $\pi$ is the distribution of a maximal coupling. In particular, if $\mu$ and $\nu$ are mutually singular, then any $\pi\in\Pi(\mu,\nu)$ whatsoever is the distribution of a maximal coupling.
It should also be clear now that a maximal coupling is just a thing in itself, unlike the things you mentioned: the covariance, the mutual information, the joint entropy, the Kullback--Leibler distance.
The Kullback--Leibler distance actually depends only on the marginals $\mu$ and $\nu$; any choice of
$\pi\in\Pi(\mu,\nu)$ plays no role.
The covariance depends on Euclidean distances, which are of no consequence for a maximal coupling -- which latter depends on the Hamming distance $\rho(x,y):=1(x\ne y)$. The mutual information and the joint entropy also depend on the Hamming distance, and yet in general they attain their extremes not at a maximal coupling.
E.g., suppose that $\mu(\{x\})=P(X=x)=\dfrac x{m(2m+1)}\,1(x\in[2m])$ and $\nu(\{y\})=P(Y=y)=\dfrac {2m+1-y}{m(2m+1)}\,1(y\in[2m])$, where $m$ is a positive integer and $[n]:=\{1,\dots,n\}$ -- so that $(\mu\wedge\nu)(\{x\})=\dfrac {\min(x,2m+1-x)}{m(2m+1)}\,1(x\in[2m])$. So, for a maximal coupling $(X',Y')$ we have $$P(X'=Y')=(\mu\wedge\nu)(\R)=\dfrac{m+1}{2m+1}>\dfrac12.$$
On the hand, here a coupling $(X'',Y'')$ that maximizes the covariance and the mutual information (and minimizes the joint entropy) is given by the condition $Y''=2m+1-X''$ almost surely, so that $$P(X''=Y'')=0.\quad\Box$$
We see that the extremizers of the covariance, the mutual information, and the joint entropy are very different from maximal couplings.
| 2 | https://mathoverflow.net/users/36721 | 452250 | 181,783 |
https://mathoverflow.net/questions/452237 | 1 | This is a natural follow-up to my previous question, here: [A question regarding equational bases of the theory of the commutative and associative properties](https://mathoverflow.net/questions/409080/a-question-regarding-equational-bases-of-the-theory-of-the-commutative-and-assoc). As before, suppose we are working in the signature of a single binary operation $+$. Also as before, let $S$ be a set of equations which generate precisely the same equational theory generated by the set containing the commutative and associative equations: $\{x+y=y+x,x+(y+z)=(x+y)+z\}$. Now, suppose that the cardinality of $S$ is finite and greater than or equal to $3$. Must $S$ have at least $|S| - 2$ redundant axioms, where $|S|$ is the cardinality of $S$? I conjecture that it must, and moreover I make the stronger conjecture that there exists a subset $T$ of $S$ which has cardinality $|S|-2$, such that the set difference $S - T$ generates the same equational theory as the set of commutativity and associativity. In other words, not only are there $|S|-2$ redundant axioms, but the redundant axioms are "set-wise" redundant, not merely individually redundant. Is my stronger conjecture true? If not, is at least my weaker conjecture true?
| https://mathoverflow.net/users/43439 | Follow up to a question on equational bases of the theory of the commutative and associative properties | In more simple terms, you want to show that if $S$ is any set of equations equivalent to $\def\ac{\mathrm{AC}}\ac=\{x+y=y+x,x+(y+z)=(x+y)+z\}$, then there exists $T\subseteq S$ of size $2$ such that $T$ is already equivalent to $\ac$.
This follows from the answer to the linked question: since $\ac$ is finite, $S$ is equivalent to a finite subtheory; then if $T\subseteq S$ is a minimal finite subset equivalent to $\ac$, we cannot have $|T|\ge3$ by the linked answer, hence $|T|\le2$.
In fact, let me give a simpler argument that shows more: $S$ must “essentially” contain $\ac$ as a subset. More precisely, up to renaming of variables, it contains the commutative law $x+y=y+x$, and it contains an equation equivalent to the associative law $x+(y+z)=(x+y)+z$ modulo commutativity:
>
> **Proposition:** If $S$ is equivalent to $\ac$, then up to renaming of variables, $S$ contains the equation $x+y=y+x$, and an equation $t=s$ where $t\in\{x+(y+z),x+(z+y),(y+z)+x,(z+y)+x\}$ and $s\in\{(x+y)+z,(y+x)+z,z+(x+y),z+(y+x)\}$.
>
>
>
**Proof:**
Recall that an equation $t=s$ follows from $S$ iff it has an equational $S$-proof: a sequence of terms $t\_0,\dots,t\_n$ such that $t\_0$ is $t$, $t\_n$ is $s$, and each $t\_i=t\_{i+1}$ is an instance of an equation $e\in S$ (i.e., $t\_{i+1}$ results from $t\_i$ by replacing a subterm $u$ with a term $v$ where one of $u=v$ or $v=u$ is a substitution instance of $e$).
An equation $t=s$ is provable in $S\equiv\ac$ iff each variable has the same number of occurrences in $t$ and $s$. It follows that if $t\_0,\dots,t\_n$ is an $S$-proof, then all the terms $t\_i$ have the same number of occurrences of each variable.
Since $S\equiv\ac$, there is an $S$-proof of $x+y=y+x$. By the property above, the only terms that can occur in such a proof are $x+y$ and $y+x$, thus the only possibility is that $x+y=y+x$ is an instance of an equation $e\in S$. Since this equation is not an instance of any strictly shorter equation valid in $\ac$, it follows that $e$ is $x+y=y+x$ itself up to renaming of variables.
Likewise, there is an $S$-proof of $x+(y+z)=(x+y)+z$. The only terms that can occur in the proof are the $12$ terms with one occurrence of each of $x$, $y$, and $z$. We can partition the set of these terms as $T\_x\cup T\_y\cup T\_z$, where $T\_x=\{x+(y+z),x+(z+y),(y+z)+x,(z+y)+x\}$ consists of the terms whose “outermost variable” is $x$, and similarly for $T\_y$ and $T\_z$. The $4$ terms in $T\_x$ are equal modulo commutativity, and likewise for $T\_y$ and $T\_z$.
Now, since $x+(y+z)\in T\_x$ and $(x+y)+z\in T\_z$, an $S$-proof $t\_0,\dots,t\_n$ of $x+(y+z)=(x+y)+z$ must contain terms $t\_i,t\_{i+1}$ that belong to different $T\_{\dots}$ sets. By renaming variables if necessary, we may assume that $t\_i\in T\_x$ and $t\_{i+1}\in T\_z$. Again, the equation $t\_i=t\_{i+1}$ is an instance of an equation $e\in S$, but it is not an instance of any strictly shorter equation valid in $\ac$ (because $t\_i$ and $t\_{i+1}$ have no common subterms other than variables), hence $e$ is $t\_i=t\_{i+1}$ itself up to renaming of variables.
| 1 | https://mathoverflow.net/users/12705 | 452263 | 181,785 |
https://mathoverflow.net/questions/452264 | 2 | I would like to know some global geometry of the blowup of projective spaces.
Question: Let $Z\subseteq \mathbb{P}^n$ be a subvariety. How can I embed $BL\_Z\mathbb{P}^n$ in a projective space or product of projective spaces?
Examples:
1. Let $Z\subseteq \mathbb{P}^4$ be a union of two disjoint lines, then $BL\_Z\mathbb{P}^4\subseteq \mathbb{P}^2\times\mathbb{P}^2\times\mathbb{P}^1$, which is the blowup of $\mathbb{P}^2\times\mathbb{P}^2$ along a $\mathbb{P}^1\times\mathbb{P}^1$.
2. Let $Z\subseteq \mathbb{P}^3$ be a twisted cubic curve, then $BL\_Z\mathbb{P}^3\subseteq \mathbb{P}^3\_Y\times \mathbb{P}^2\_X$ is defined by
$$X\_0Y\_0+X\_1Y\_1+X\_2Y\_2=X\_0Y\_1+X\_1Y\_2+X\_2Y\_3=0. $$
I don't know whether there is a standard way to obtain these constructions or not. Thanks very much for your answer.
| https://mathoverflow.net/users/192152 | Blowing up the projective space along a subvariety | Let $I\_Z$ be the ideal sheaf of $Z$. If the sheaf $I\_Z(k)$ is globally generated (i.e., if hypersurfaces of degree $k$ cut out $Z$ as a scheme) and $m := h^0(I\_Z(k)) - 1$, then
$$
\mathrm{Bl}\_Z(\mathbb{P}^n) \subset \mathbb{P}^n \times \mathbb{P}^m.
$$
| 6 | https://mathoverflow.net/users/4428 | 452267 | 181,786 |
https://mathoverflow.net/questions/452055 | 4 | Let $X$ be a Polish space and $\mathcal{F}(X)$ the set of closed subsets of $X$ endowed with the Effros Borel structure, generated by sets of the form $\{F\in \mathcal{F}(X):F\cap U\neq \emptyset\}$, where $U$ ranges over the open subsets of $X$.
Question: Is the set $\mathcal{C}(X)$ of all *clopen* subsets of $X$ a Borel subset of $\mathcal{F}(X)$? I am particularly interested in the case when $X=\mathbb{N}^\mathbb{N}$ (which is, notably, not locally compact).
One attempt to show this is to note that a closed set $F$ is clopen if and only if there is another closed set $G$ such that $F\cap G=\emptyset$ and $F\cup G=X$, but the intersection operation is not Borel on $\mathcal{F}(X)$ unless $X$ is locally compact, and even then, this appears at best a $\mathbf{\Sigma}^1\_1$, or analytic, characterization. Another thought is to use that the set $\mathcal{O}(X)$ of open subsets of $X$ can be endowed with a Borel structure via complementation... but I don't see how that helps (the intersection of Borel spaces need not be Borel if their structures don't cohere in some way).
**Edit:** If $X$ is totally disconnected, we can fix a sequence $(U\_n)$ of basic clopen subsets, an element of $\mathcal{F}(X)^\mathbb{N}$, and say that $F\in\mathcal{F}(X)$ is clopen if and only if for every $x\in F$, there is an $n$ such that $x\in U\_n$ and $U\_n\subseteq F$. The subset relation is Borel on $\mathcal{F}(X)$ (see section 12.C of Kechris, Classical Descriptive Set Theory), so this is a $\mathbf{\Pi}^1\_1$ description of being clopen. Together with what was written above, this shows that "clopen" is a Borel property in $\mathcal{F}(X)$, provided $X$ is totally disconnected and locally compact, but still leaves out the case $X=\mathbb{N}^\mathbb{N}$, in which I am most interested.
| https://mathoverflow.net/users/16107 | Is the set of clopen subsets Borel in the Effros Borel space? | Here is a negative answer for $\mathbb{N}^\mathbb{N}$.
Given a countably-branching tree $T$, we built a new countably-branching tree $T'$ in two steps. First, for any $\sigma \in T$ we place $(\sigma + 2)$ into $T'$, where $\sigma + 2$ is just obtained by adding $2$ to any number appearing in $\sigma$. In the second step, for any $\sigma$ already in $T'$, we add all words of the form $\sigma0\tau$ for $\tau \in \mathbb{N}^{<\omega}$ to $T'$.
So basically, we shift every branch in $T$ two steps to the right, and then add the full subtree below the $0$-child and nothing below the $1$-child. We can compute an enumeration of $T'$ from an enumeration of $T$, and we can compute the characteristic function of $T'$ from the characteristic function of $T$. Moreover, $T'$ definitely has no dead ends.
An enumeration of $T'$ thus constitutes a code of $[T'] \in \mathcal{F}(\mathbb{N}^\mathbb{N})$ [where I am not thinking about just the Borel-structure on this space, but the concrete representation of the space of overt (closed) subsets of Baire space].
If $T$ is ill-founded, then from any point $p \in [T]$ we obtain a point $(p+2) \in \delta([T'])$ [here $p+2$ is the result of pointwise adding $2$ to each number in $p$, and $\delta([T'])$ denotes the boundary of $[T']$], as $((p+2)\_{\leq n}0^\mathbb{N})\_{n \in \mathbb{N}}$ converges to $(p+2)$ inside $[T']$, and $((p+2)\_{\leq n}1^\mathbb{N})\_{n \in \mathbb{N}}$ converges to $(p+2)$ outside of $[T']$. This means that $[T']$ is not clopen.
On the other hand, if $T$ is well-founded, then we find that for any $p \in \mathbb{N}^\mathbb{N}$ there is some $n \in \mathbb{N}$ such that $p \in [T']?$ only depends on $p\_{\leq n}$, ie $[T']$ is clopen.
It follows that being clopen is $\Pi^1\_1$-hard for elements of $\mathcal{F}(\mathbb{N}^\mathbb{N})$. The very same construction also shows that being clopen is $\Pi^1\_1$-hard for elements of $\mathcal{A}(\mathbb{N}^\mathbb{N})$, too. [In the latter space, we code sets as paths through trees given via their characteristic function, with dead ends being allowed.]
| 1 | https://mathoverflow.net/users/15002 | 452272 | 181,789 |
https://mathoverflow.net/questions/452142 | 6 | Let $X$ be a connected smooth complex algebraic variety and $Z=\bigcup\_{i=1}^r Z\_i$ be a union of smooth connected hypersurfaces, satisfying that each two intersect transversally. Assume for simplicity that $Z$ is connected and choose a point $x\in X\setminus Z$. As $Z$ has real codimension $2$, it is known that the map $\Phi:\pi\_1(X\setminus Z,x)\rightarrow \pi\_1(X,x)$ is a surjection. I am looking for a description of the kernel of this map.
I believe that this kernel should be the normalizer in $\pi\_1(X\setminus Z,x)$ of the subgroup generated by homotopy classes of loops $\tau\_i$ such that each $\tau\_i$ circles once around $Z\_i$. This is, for example, the case when $X=\mathbb{C}^n$ and $Z$ is the union of complex hyperplanes. See, for instance, [The fundamental group of the complement of a union of complex hyperplanes](https://link.springer.com/article/10.1007/BF01389187).
I have made some attempts at this, but haven't been successful thus far. My latest attempt goes as follows: Assume that $Z$ is a smooth connected hypersurface and take
a tubular neighborhood $Z\subset U\subset X$ (which we assume to contain $x$). Applying Seifert-van Kampen to the cover $X\setminus Z$ and $U$, we have an isomorphism:
$$(\pi\_1(X\setminus Z,x)\* \pi\_1(U,x))/N(\pi\_1(U\setminus Z,x)\rightarrow \pi\_1(X,x)$$.
As before, the map $\pi\_1(U\setminus Z,x)\rightarrow \pi\_1(U,x)$ is surjective. Hence, all elements coming from $\pi\_1(U,x))$ are trivial inside the amalgamated product. My feeling is that $U\setminus Z$ should deformation retract to a spherical bundle. For example, if $U$ is trivial then $U\setminus Z$ is homotopy equivalent to $Z\times \mathbb{S}^1$, and the loop corresponding to $(id, 1)\in \pi\_1(Z\times \mathbb{S}^1,x))\simeq \pi\_1(Z,x)\times \mathbb{Z}$ generates the kernel of $\Phi$. I would very much appreciate any help regarding this matter.
| https://mathoverflow.net/users/476832 | Presentation of the fundamental group of a complex variety | I found an affirmative answer to this question on [Complex reflection groups, braid groups, and Hecke algebras](https://www.degruyter.com/document/doi/10.1515/crll.1998.064/html?lang=en). The proof is in Appendix A1, specifically on page 181.
| 5 | https://mathoverflow.net/users/476832 | 452273 | 181,790 |
https://mathoverflow.net/questions/452269 | 7 | This is a slight variation of [this recommended blog post](https://karagila.org/2022/dual-df-maps/) by [Asaf Karagila](https://mathoverflow.net/users/7206/asaf-karagila). Let $A$ be a set. Then:
1. $A$ is said to be *Dedekind-finite* if every injective map $f:A\to A$ is also surjective.
2. $A$ is said to be *dually Dedekind-finite* if every surjective map $f:A\to A$ is also injective.
3. A map $g:A\to A$ is said to be *(dually) Dedekind-finite-to-one* if for all $b\in A$ the preimage $f^{-1}(\{b\})$ is (dually) Dedekind-finite.
**Question.** In ${\sf (ZF)}$, suppose that the set $A$ has the property, that every surjective map is (dually) Dedekind-finite-to-one. Does this imply that $A$ is (dually) Dedekind-finite?
| https://mathoverflow.net/users/8628 | Dedekind-finite-to-one vs Dedekind-finite | The answer is Yes.
For the first question, suppose that $A$ is Dedekind infinite and let $f$ be an injection from $\omega$ into $A$. Then the function that maps $f(2n)$ to $f(n)$ for each $n\in\omega$, maps $f(2n+1)$ to $f(0)$ for each $n\in\omega$, and leaves all other elements of $A$ fixed, is a surjection from $A$ onto $A$ which is not Dedekind-finite-to-one.
For the second question, suppose that $A$ is dually Dedekind infinite and let $f$ be a surjection from $A$ onto $A\cup\{b\}$, where $b\notin A$. For each $n\in\omega$, let $A\_n=(f^{n+1})^{-1}[\{b\}]$. Clearly, for all $m,n\in\omega$ with $m<n$, $f^{n-m}|A\_n$ is a surjection from $A\_n$ onto $A\_m$. Then the function $g$ that maps each $x\in A\_{2n}$ to $f^n(x)$ for all $n\in\omega$, maps each $x\in A\_{2n+1}$ to a fixed $a\in A$ for all $n\in\omega$, and leaves all other elements of $A$ fixed, is a surjection from $A$ onto $A$ which is not dually-Dedekind-finite-to-one. $g$ is not dually-Dedekind-finite-to-one, because the inverse image of $a$ under $g$ includes $B=\bigcup\_{n\in\omega}A\_{2n+1}$, and $B$ is dually Dedekind infinite since $f^2|B$ is a surjection from $B$ onto $B\cup\{b\}$.
| 7 | https://mathoverflow.net/users/101817 | 452287 | 181,792 |
https://mathoverflow.net/questions/452143 | 4 | Let $A\_n$ be the dual simplicial complex to the associahedron on $n$ letters. The complex $A\_n$ is thus a simplicial triangulation of an $(n-3)$-dimensional sphere. The vertices of $A\_n$ correspond to the ways of inserting one nontrivial parentheses into the expression $a\_1 a\_2 \cdots a\_n$, and the $(n-3)$-dimensional simplices correspond to ways of inserting the maximal number ($n-2$) of parentheses into this expression. For instance, the $1$-dimensional simplices of $A\_4$ are precisely
\begin{align\*}
&((a\_1 a\_2) a\_3) a\_4\\
&(a\_1 (a\_2 a\_3)) a\_4\\
&a\_1 ((a\_2 a\_3) a\_4)\\
&a\_1 (a\_2 (a\_3 a\_4))\\
&(a\_1 a\_2) (a\_3 a\_4)
\end{align\*}
Since $A\_n$ is a triangulation of an $(n-3)$-dimensional sphere, it can be oriented (in two ways, we pick one).
**Question**: This orientation should give an orientation on each $(n-3)$-dimensional simplex $\sigma$. Such an orientation corresponds to an ordering (defined up to the action of the alternating group) of the vertices of $\sigma$, i.e., to the pairs of parentheses in $\sigma$. How can we explicitly write down this orientation?
| https://mathoverflow.net/users/509958 | Orienting the dual of the associahedron | There are many combinatorial models for the associahedron -- parenthesizations of $n$ variables, triangulations of the $(n+1)$-gon, planar binary trees with $n$-leaves. I'll follow the OP's lead and use parenthesizations of $n$-variables.
First, let's remember how we label the vertices of the dual associahedron. We'll label the vertices as $v\_{ij}$ for $1 \leq i < j \leq n$, $(i,j) \neq (1,n)$. The vertex $v\_{ij}$ occurs in the facet corresponding to a multiplication if the product $a\_i a\_{i+1} \cdots a\_j$ is computed as a partial step in that multiplication. For example, say we have the parenthesization
$$((a\_1 a\_2) a\_3) (a\_4 a\_5).$$
When we compute this, we compute the partial products:
$$a\_1 a\_2,\ a\_1 a\_2 a\_3,\ a\_4 a\_5$$
and hence this parenthesization will correspond to vertices $v\_{12}$, $v\_{13}$ and $v\_{45}$.
You might notice that I didn't include a vertex $v\_{1n}$, which would correspond to the entire product $a\_1 a\_2 \cdots a\_n$. Let's make a new simpicial complex $A\_n$ which does use this vertex: So we have $\binom{n}{2}$ vertices $v\_{ij}$ for $1 \leq i < j \leq n$ and, for each parenthesization, we have a facet corresponding to the $n-1$ partial products occuring when we compute that parenthesization. So $A\_n$ is a closed $n-2$ dimensional ball, with the vertex $v\_{1n}$ at the center. And the ordinary dual associahedron is the boundary $\partial A\_n$.
With that preamble, here is how to orient $A\_n$, and its boundary $\partial A\_n$. Take each multiplication and insert the multiplication symbols, so our previous example turns into $((a\_1 \ast a\_2) \ast a\_3) \ast (a\_4 \ast a\_5)$. Go across the formula from left to right, and write down the partial products in the order that they are output by that multiplication symbol. In our example, we get $a\_1 a\_2$, $a\_1 a\_2 a\_3$, $a\_1 a\_2 a\_3 a\_4 a\_5$, $a\_4 a\_5$.
**Theorem** Ordering the vertices of each facet of $A\_n$ as listed here gives an orientation of the ball $A\_n$.
To orient the boundary $\partial A\_n$, we have to work a little harder: Take a parenthesization of $n$ variables. This corresponds to a facet $\sigma$ of $A\_n$ whose vertices are ordered as above. One of those vertices is $v\_{1n}$; let's say that $v\_{1n}$ is in the $k$-th position. Delete $v\_{1n}$ from the list of vertices of $\sigma$, giving a list of vertices of $\partial \sigma$. Assign that list of vertices, in that order, the sign $(-1)^{k-1}$. In the example above, we had the list of vertices $(v\_{12}, v\_{13}, v\_{15}, v\_{45})$. The vertex $v\_{15}$ is in the $3$-rd position, so we assign the sign $(-1)^{3-1}= +1$ to the ordering $(v\_{12}, v\_{13}, v\_{45})$ in $\partial A\_5$.
**Corollary** Ordering the vertices of each facet of $\partial A\_n$ as listed here gives an orientation of the sphere $\partial A\_n$.
Why do I call this a corollary? Because this is the standard recipe for how to orient the boundary of an oriented simplicial complex. So we just need to prove the Theorem, and the Corollary will follow.
**Proof of Theorem** We just need to take two adjacent facets and check that their orientation is compatible. Two adjacent facets differ by a single association -- say we change
$${\big(} (a\_{p+1} a\_{p+2} \cdots a\_q) \ast (a\_{q+1} a\_{q+2} \cdots a\_r) {\big)} \ast (a\_{r+1} a\_{r+2} \cdots a\_s) \qquad (1)$$
to
$$(a\_{p+1} a\_{p+2} \cdots a\_q) \ast {\big(} (a\_{q+1} a\_{q+2} \cdots a\_r) \ast (a\_{r+1} a\_{r+2} \cdots a\_s) {\big)} \qquad (2).$$
All the multiplication symbols which I haven't drawn correspond to the same vertices in both $(1)$ and $(2)$. Of the two symbols which I have drawn, the left symbol gives $v\_{(p+1) r}$ in $(1)$ and $v\_{(p+1) s}$ in $(2)$; the right symbol gives $v\_{(p+1) s}$ in $(1)$ and $v\_{(q+1) r}$ in $(2)$. So our ordering has changed from
$$(\cdots, v\_{(p+1) r}, \cdots, v\_{(p+1) s},\ \cdots)$$
to
$$(\cdots, v\_{(p+1) s}, \cdots, v\_{(q+1) r},\ \cdots).$$
If we switch the second ordering to $(\cdots, v\_{(q+1) r}, \cdots, v\_{(p+1) s},\ \cdots)$, we act by a single transposition, introducing a minus sign. Thus, if we rearrange the orders to coincide on the set of vertices we have in common, then the two orders get opposite signs. This is the condition to be an orientation. $\square$
I actually found this in a very different way, but this is answer is already too long, so I'll write more in comments.
| 2 | https://mathoverflow.net/users/297 | 452294 | 181,793 |
https://mathoverflow.net/questions/447472 | 11 | The following problem was considered in
Cohen and Kontorovich, "Local Glivenko-Cantelli",
<https://arxiv.org/abs/2209.04054>,
to appear in COLT 2023 (henceforth, CK'23).
Let $Y\_j$, $j\in\mathbb{N}$
be a sequence of independent
$\operatorname{Binom}(n,p\_j)$
random variables,
where $n\in\mathbb{N}$
and $1/2\ge p\_j\downarrow 0$
as $j\to\infty$.
Since $\mathbb{E} Y\_j=np\_j$,
let us consider
the centered, normalized process
$\bar Y\_j:=
n{^{-1}} Y\_j-p\_j
$.
Finally, we define
$\Delta\_n$ to be the
expected uniform absolute deviation:
$$
\Delta\_n :=
\mathbb{E}\sup\_{j\in\mathbb{N}}|\bar Y\_j|.
$$
CK'23
gave an exact
characterization
of the $p\in[0,{\frac12}]^{\mathbb{N}}\_{\downarrow0}$
(that is, $p\in[0,{\frac12}]^{\mathbb{N}}$
with $p\_j\downarrow 0$)
for which $\Delta\_n\to0$
as $n\to\infty$.
Namely, they defined the functional
$$
T(p) := \sup\_{j\in\mathbb{N}}\frac{\log (j+1)}{\log(1/p\_j)}
.
$$
and showed that
$\Delta\_n\to0$
iff
$T(p)<\infty$.
They also gave the finite-sample
bound
$$
\Delta\_n
\le
c\left(
\sqrt{\frac{S(p)}{n}}
+\frac{T(p)\log n}{n}
\right),
\qquad n\ge \mathrm{e}^3
\qquad(\*)
$$
where $c>0$ is an absolute constant
and
$$
S(p) := \sup\_{j\in\mathbb{N}}p\_j\log (j+1).
$$
We conjecture that the $\log n$
factor in $(\*)$ is superfluous. This conjecture is motivated by the
asymptotic
lower bounds
$$
\liminf\_{n\to\infty}
\sqrt n
\Delta\_n
\ge
c\sqrt{S(p)},
$$
$$
\liminf\_{n\to\infty}
n
\Delta\_n
\ge
cT(p)
,
$$
where $c>0$ is a universal constant.
This shows that beyond the conjecturally
removable $\log n$ factor,
$(\*)$ is tight up to constants.
**Open problem.** Is the
$\log n$ factor in
$(\*)$
necessary, or can it be removed?
**UPDATE** 21-Jul-2023.
This was presented as an open problem at COLT'23:
<https://proceedings.mlr.press/v195/cohen23b.html>
Within days of the announcement and independently of each other,
Moïse Blanchard, Václav Voráček, and Jaouad Mourtada
obtained essentially the same substantial improvement.
Formally, their results show that there is an absolute $c\_1>0$ such that for any constant $c>0$ there is an $n\_0$ such that for all $n>n\_0$ there is a sequence
$p\in[0,{\frac12}]^{\mathbb{N}}\_{\downarrow0}$
for which
$$
\Delta\_n(p) > c\sqrt{\frac{S(p)}{n}}
+
c\_1\frac{T(p)\log n}{n\log\log n}.
\qquad(\*\*)
$$
This certainly dashes any hope of removing the $\log n$ factor in $(\*)$ entirely; the best one could aim for is replacing it by
$\log(n)/\log\log(n)$.
This raises a number of fascinating questions. First: Is there a *fixed* sequence $p\_j$ (i.e., independent of $n$) for which
$$
\liminf\_{n\to\infty}
\frac{n\log\log n}{\log n}
\Delta\_n
\ge
cT(p)
,
$$
where $c>0$ is a universal constant?
Second: Is there a functional $\tilde T(p)$ and a monotonically increasing $\varphi:\mathbb{N}\to \mathbb{N}$ such that
$$
c\tilde T(p)
\le
\liminf\_{n\to\infty}
\varphi(n)
\Delta\_n
\le
\limsup\_{n\to\infty}
\varphi(n)
\Delta\_n
\le
c'\tilde T(p)
$$
for absolute constants $c,c'>0$?
**UPDATE** 25-Jul-2023. We congratulate
Moïse Blanchard on resolving the gap entirely, by
exhibiting a $p$ for which
$$
\Delta\_n(p) > c\sqrt{\frac{S(p)}{n}}
+
c\_1\frac{T(p)\log n}{n}
$$
(all constants and quantifiers as in $(\*\*)$).
Additionally, it has been pointed out to us by
Moïse, Václav, Jaouad
that the
"lower bound"
$
\liminf\_{n\to\infty}
n
\Delta\_n
\ge
cT(p)
$
is trivial, since for nontrivial $p$,
$\Delta\_n$ will be at least $1/\sqrt n$ for $n$ large enough, meaning that
$\liminf\_{n\to\infty}
n
\Delta\_n
=\infty$.
We have corrected this with the following non-trivial lower bound:
Theorem (Cohen, K., 2023):
There is an absolute constant $c>0$ such that
$$
\Delta\_n(p) \ge c\left(1\wedge\frac{T}{n}\right),
$$
for all $n\in\mathbb{N}$ and
$p\in[0,{\frac12}]^{\mathbb{N}}\_{\downarrow0}$.
Thus, currently, the two open questions are:
1. Is there a *fixed* sequence $p\_j$ (i.e., independent of $n$) for which
$$
\Delta\_n(p) \ge c\left(1\wedge\frac{T\log n}{n}\right),
$$
for some $c>0$?
2. Is there a functional $\tilde T(p)$ and a monotonically increasing $\varphi:\mathbb{N}\to \mathbb{N}$ such that
$$
\Delta\_n
\le
c\left(
\sqrt{\frac{S(p)}{n}}
+\frac{\tilde T(p)}{\varphi(n)}
\right)
$$
and
$$
\Delta\_n(p) \ge c'\left(1\wedge\frac{\tilde T(p)}{\varphi(n)}\right),
$$
for absolute constants $c,c'>0$
and all $n\in\mathbb{N}$,
$p\in[0,{\frac12}]^{\mathbb{N}}\_{\downarrow0}$?
| https://mathoverflow.net/users/12518 | Open problem: $\log n$ factor in Binomial empirical process | See the preprint by
Moïse Blanchard and Václav Voráček, titled "Tight Bounds for Local Glivenko-Cantelli", available here:
<https://arxiv.org/abs/2308.01896>
.
It clearly explains their construction.
| 1 | https://mathoverflow.net/users/12518 | 452296 | 181,794 |
https://mathoverflow.net/questions/451222 | 3 | We say that a [hypergraph](https://en.wikipedia.org/wiki/Hypergraph) $H=(V,E)$ has *property ${\bf B}$* if there is $S\subseteq V$ such that for all $e\in E$ with $|e|>1$ we have $S\cap e \neq \emptyset \neq e \setminus S$.
Let $(P,\leq)$ be a [partially-ordered set (poset)](https://en.wikipedia.org/wiki/Partially_ordered_set). For $p\in P$, let $d(p) = \{x\in P: x\leq p\}$. We assign to $(P,\leq)$ the *principal downset hypergraph* $H\_P = \big(P, D(P)\big)$ where $D(P) = \{d(p):p\in P\}$.
**Question.** Is it consistent with ${\sf (ZF)}$ that there is a poset $(P,\leq)$ such that $H\_P$ does not have property ${\bf B}$? And is the opposite also consistent?
(In other words: no matter how you color the elements of $P$ with "red" and "blue", you will get a principal downset $d(p)$ with more than $1$ element such that every element of $d(p)$ has the same color.)
| https://mathoverflow.net/users/8628 | Posets such that the collection of principal down-sets does not have property ${\bf B}$ | Let $M$ be the ordered Mostowski model (T. Jech, *The Axiom of Choice*, Section 4.5). Its set of atoms, $A$, has a linear order $\prec$ that makes it isomorphic to the rationals. Let $S\in M$ be a subset of $A$; then $S$ is a union of finitely many intervals. So if $S$ has a lower bound then $S$ is disjoint from (many) sets of the form $d(a)$, and if it does not have a lower bound then it contains (many) sets of the form $d(a)$. So in this model we have a linearly ordered set, $A$, such that $H\_A$ does not have property $B$.
The Jech-Sochor embedding theorem (Theorem 6.1 in Jech's book) can be used to transfer this to a model of $\mathsf{ZF}$.
| 4 | https://mathoverflow.net/users/5903 | 452297 | 181,795 |
https://mathoverflow.net/questions/452286 | 2 | Recall that a domain $D \subseteq \mathbb C$ is called *regular* if for each point $x \in \partial D$, we have $\mathbf P\_x\lbrack \tau\_D = 0\rbrack = 1$, where $\tau\_D = \inf\{t > 0 : B\_t \notin D\}$, and $(B\_t)\_{t \ge 0}$ is a Brownian motion.
1. Is the domain $\mathbb D \setminus [0, 1)$ regular?
2. Is every simply connected domain regular?
| https://mathoverflow.net/users/510096 | Is every simply connected domain regular? | Yes to both. Both were already answered here <https://math.stackexchange.com/questions/4389689/is-every-simply-connected-set-in-the-plane-regular-for-brownian-motion>
So for Q1 we can use the simple-arc criterion.
>
> The regularity of every boundary point of an open set in $\mathbb R^2$ is in fact strongly related to connectedness. However, that's more a property of the complement of the set: Problem 4.2.16 in [1](https://www.sciencedirect.com/science/article/pii/S0022247X18303433) which is:
>
>
>
>
> *Let $D\subset\mathbb R^2$ be open, and suppose that $a\in\partial D$ has the property that there exists a point $b\not=a$ in $\mathbb R^2\setminus D\,,$ and a simple arc in $\mathbb R^2\setminus D$ connecting $a$ to $b\,.$ Show that $a$ is regular.*
>
>
>
>
> The solution is provided in [1](https://www.sciencedirect.com/science/article/pii/S0022247X18303433) section 4.5. The unit disc minus the line segment $[0,1)\times\{0\}$ clearly satisfies the properties of $D$ in that problem.
>
>
>
>
> [1](https://www.sciencedirect.com/science/article/pii/S0022247X18303433) I. Karatzas, S. Shreve, *Brownian Motion and Stochastic Calculus.*
>
>
>
More generally for Q2, see Bass "probabilistic techniques in analysis" Prop. II.1.14
(or the article mentioned in the comments
["A remark on the probabilistic solution of the Dirichlet problem
for simply connected domains in the plane"](https://www.sciencedirect.com/science/article/pii/S0022247X18303433))
where they show that: *The Dirichlet problem is solvable for any simply connected domain in $\mathbb{C}$.*
| 2 | https://mathoverflow.net/users/99863 | 452303 | 181,797 |
https://mathoverflow.net/questions/452276 | 2 | Let $f$ be a :
1. $f\in\mathcal{C}^\infty(\mathbb{R},\mathbb{R})$,
2. for all $x> 0,~f(x)>0$,
3. for all $x< 0,~f(x)<0$,
I am struggling to find a bound for the distance between the root of $f$ and the root of its Gaussian convolution. Formaly,
>
> Let $f$ satisfies the previous properties and define $g$ as:
> $$g\_\sigma(x):=\int\_\mathbb{R}f(s)e^{-\frac{(x-s)^2}{2\sigma^2}}ds.$$
> From [this post](https://math.stackexchange.com/questions/4745372/zero-crossing-for-convolution/4745403?noredirect=1#comment10066982_4745403), we know that $g\_\sigma$ has a unique 0, denoted ${\lambda\_\sigma}$. How can we bound the convolution's zero $\lambda\_\sigma$ depending on $\sigma$ and the derivatives of $f$ ?
>
>
>
Numericaly, I think that $\lambda\_\sigma$ is linear in $\sigma$. But I have no clue to for the derivatives...
Any hint, ideas or solution will be highly appreciated ! Thank you very much !
Since $\lambda\_\sigma$ is a function of $f$ and $\sigma$, I have thought the implicit function theorem but I don't know how to apply it in this case.
| https://mathoverflow.net/users/510079 | Distance between root of $f$ and its Gaussian convolution | $\newcommand{\R}{\mathbb R}\newcommand{\la}{\lambda}\newcommand{\si}{\sigma}$After some simple rewriting, we see that $\la\_\si$ is the root $x=x\_\si$ of the equation $G(\si,x)=0$, where
\begin{equation\*}
G(\si,x):=Ef(x+\si Z)[=\si\sqrt{2\pi}\,g\_\si(x)\text{ if }\si>0]
\end{equation\*}
and $Z$ is a standard normal random variable.
To ensure that $G(\si,x)$ or, equivalently, $g\_\si(x)$ is finite for all real $x$ and all small enough $\si>0$, assume that
\begin{equation\*}
|f(x)|\le Ce^{Cx^2/2}
\end{equation\*}
for some real $C>0$ and all real $x$. We will also require that $f$ does not vanish at $\pm\infty$, so that
\begin{equation\*}
f\_a:=\inf\{|f(x)|\colon|x|\ge a\}>0
\end{equation\*}
for each real $a>0$ and
\begin{equation\*}
|f'(x)|+|f''(x)|\le Ce^{Cx^2/2}
\end{equation\*}
for the same real $C>0$ and all real $x$.
In view of your conditions on $f$, for each real $a>0$ and all real $x\ge a$ we have
\begin{equation\*}
\begin{aligned}
G(\si,x)&\ge Ef(x+\si Z)1(Z>0)+Ef(x+\si Z)1(Z<-x/\si) \\
&\ge f\_a/2-CEe^{Cx^2+C\si^2Z^2}1(Z<-x/\si)\to f\_a/2>0
\end{aligned}
\tag{10}\label{10}
\end{equation\*}
uniformly in real $x\ge a$ as $\si\downarrow0$. Similarly, for each real $a>0$ and all small enough $\si>0$ we have $G(\si,x)<0$ for all real $x\le-a$. Thus,
\begin{equation\*}
x\_\si\to0\quad\text{as } \si\downarrow0.
\end{equation\*}
Let $G\_1(\si,x)$ and $G\_2(\si,x)$ denote, respectively, the first and second partial derivatives of $G(\si,x)$ w.r.t. $\si$, so that
\begin{equation\*}
G\_1(0,x)=EZf'(x)=0
\end{equation\*}
and
\begin{equation\*}
\text{$G\_2(\si,x)=EZ^2f''(x+\si Z)\to f''(x)$ as $x\to0$ and $\si\downarrow0$.}
\end{equation\*}
So, by Taylor's expansion, for some $c\_\si$ between $0$ and $x\_\si$,
\begin{equation\*}
\begin{aligned}
0=G(\si,x\_\si)&=G(0,x\_\si)+G\_1(0,x\_\si)\si+G\_2(c\_\si,x\_\si)\si^2/2 \\
&=f(x\_\si)+0\,\si+(f''(0)+o(1))\si^2/2 \\
&=(f'(0)+o(1))x\_\si+0\,\si+(f''(0)+o(1))\si^2/2
\end{aligned}
\end{equation\*}
as $\si\downarrow0$.
Thus,
\begin{equation\*}
x\_\si=-\frac{f''(0)+o(1)}{2f'(0)}\,\si^2
\end{equation\*}
as $\si\downarrow0$, provided that $f'(0)\ne0$.
---
**Details on \eqref{10}:** For each real $a>0$, $\si\in(0,\min(a,(4C)^{-1/2}))$, and real $x\ge a$,
\begin{equation\*}
\begin{aligned}
G(\si,x)&=Ef(x+\si Z)1(x+\si Z>0)+Ef(x+\si Z)1(x+\si Z<0) \\
&\ge Ef(x+\si Z)1(Z>0)+Ef(x+\si Z)1(Z<-x/\si) \\
&\ge Ef(x)1(Z>0)-CEe^{C(x+\si Z)^2/2}1(Z<-x/\si) \\
&\ge f(x)P(Z>0)-CEe^{Cx^2+C\si^2Z^2}1(Z<-x/\si) \\
&\ge f\_a/2-Ce^{Cx^2}\int\_{-\infty}^{-x/\si}e^{(C\si^2-1/2)z^2}\,dz \\
&\ge f\_a/2-Ce^{Cx^2}\int\_{-\infty}^{-x/\si}e^{-z^2/4}\,dz \\
&\ge f\_a/2-C\sqrt2\,e^{Cx^2}e^{-(x/\si)^2/4} \\
&\ge f\_a/2-C\sqrt2\,e^{(C-1/(4\si^2))a^2} \underset{\si\downarrow0}\longrightarrow f\_a/2>0.
\end{aligned}
\end{equation\*}
| 3 | https://mathoverflow.net/users/36721 | 452305 | 181,798 |
https://mathoverflow.net/questions/451627 | 1 | On page 268 of Prof. John M Lee's book "Introduction to Smooth Manifolds" (second edition), it says if $E$, $M$ and $F$ are smooth manifolds with or without boundary, $\pi:E\to M$ is a smooth map and all the local trivializations $\Phi:\pi^{-1}(U)\to U\times F$ are diffeomorphisms, where $U$ is an open subset of $M$, then $\pi:E\to M$ is called a smooth fiber bundle.
My questions are
1. is there a problem if some or all of the smooth manifolds $E$, $M$ and $F$ have nonempty boundary? For example, if $E$ has empty boundary but $M$ and $F$ have nonempty boundary, and you happen to take $U$ in above to be an open submanifold with boundary, then $U\times F$ would be a smooth manifold with corner. But $\pi^{-1}(U)\subseteq E$ should not have a boundary?
2. [From this post](https://mathoverflow.net/questions/201670/ehresmann-fibration-theorem-for-manifolds-with-boundary) we know that Ehresmann fibration theorem, which states if "$E$ and $M$ are smooth manifolds without boundary and $f:E\to M$ is a smooth surjective submersion which is proper, then it is a smooth fiber bundle", should work when $E$ and $M$ are smooth manifolds with boundary by adding the extra assumption that the restriction $f|\_{\partial M}:\partial M\to\partial N$ is a (proper?) submersion. If it works, am I correct that in this case the fiber is still a smooth manifold without boundary?
3. I would like to know which case is impossible and what extra assumptions need to be imposed. For example, if all the smooth manifolds $E$, $M$ and $F$ have empty boundary, the definition of a smooth fiber bundle is fine; while if all the smooth manifolds $E$, $M$ and $F$ has nonempty boundary, it does not seem fine to me unless more assumption(s) is added.
4. Besides Prof. Lee's book and Prof. Liviu I Nicolaescu's book "Lectures on the Geometry of Manifolds" (third edition) (in Definition 3.4.50, for which $E$ is a smooth manifold with boundary and $M$ is a smooth manifold with empty boundary) I have not seen a definition of smooth fiber bundle where some or all of the spaces are smooth manifolds with boundary. Is there a reference (paper, monograph, or even textbooks) which has such a definition.
Thanks in advance.
| https://mathoverflow.net/users/41686 | On the definition of smooth fiber bundle and smooth manifolds with boundary | Allowing boundaries makes no big difference. Figure out, for example, the simple case of the first projection $\pi:D^2\times I\to D^2$ where $F=I$ is the compact interval and $M=D^2$ is the compact $2$-disk; the general case is not much more complicated.
First, there is nothing to change in the definition of a fibre bundle
in the case where $M$, $E$, $F$ have boundaries. You should still take $U$ to be an open subset of $M$ in the sense of general topology, so it can meet $\partial M$, but this makes no difference.
Second, one has
1. $\partial M=\partial F=\emptyset$ implies $\partial E=\emptyset$;
2. $\partial M\neq\emptyset$ and $\partial F=\emptyset$ implies that $E$ has a smooth boundary $\partial E=\pi^{-1}(\partial M)$, which is "vertical" i.e. the total space of a locally trivial fibre bundle of fibre $F$ over $\partial M$;
3. $\partial M=\emptyset$ and $\partial F\neq\emptyset$ implies that $E$ has a smooth boundary $\partial E$ which is "horizontal" i.e. the total space of a locally trivial fibre bundle of fibre $\partial F$ over $M$;
4. $\partial M\neq\emptyset$ and $\partial F\neq\emptyset$ implies that $E$ has a boundary $\partial E$ which is the union of the two precedent ones, and they intersect on the corners, which form the total space of a locally trivial fibre bundle of fibre $\partial F$ over $\partial M$.
Third, Eheresmann's theorem:
a) it goes without change in the case $\partial E=\pi^{-1}(\partial M)$.
b) If $M$ has no boundary, it takes the form "Assume $E$ is a smooth manifold with smooth boundary and $M$ is a smooth manifold without boundary and $\pi:→$ is a smooth surjective submersion which is proper, also assume that $\pi\vert\partial E:\partial E\to M$ also is a submersion, then $\pi$ is a smooth fiber bundle".
c) If $M$ has a boundary and $\partial E$ is more than $\pi^{-1}(\partial M)$,
it's a little more subtle, you have to control what happens close to the corners of $E$ by assuming that the corners $\partial^2E$ of $E$ split $\partial E$
into two components $\partial\_vE=\pi^{-1}(\partial M)$ and
$\partial\_hE$ such that $\pi\vert\partial\_hE$ is a submersion onto $M$,
and that $\pi\vert\partial^2E$ is a submersion onto $\partial M$.
| 1 | https://mathoverflow.net/users/105095 | 452306 | 181,799 |
https://mathoverflow.net/questions/452304 | 15 | I am interested in the amenability properties of infinite products of solvable groups. The following facts are well-known:
1. Any solvable group is amenable.
2. The class of solvable groups is closed under finite products and quotients (but not infinite products).
3. Infinite products of solvable groups of a fixed derived length are solvable.
4. Finite products and quotients of amenable groups are amenable (but not infinite products).
My specific question is whether an infinite product of solvable groups is still amenable?
The motivation arises from uniformity in multiple recurrence within ergodic theory. My coauthor and I wish to prove that the largeness of certain sets of multiple return times is independent of the underlying measure-preserving system *and* the acting group. Currently, we have achieved this for solvable groups of a fixed derived length, and more generally, for a class of uniformly amenable groups, where we use the device of ultraproducts as a key tool in our approach.
| https://mathoverflow.net/users/117822 | Is the infinite product of solvable groups amenable? | The free group $F\_2$ is residually nilpotent, meaning that the intersection of its lower central series is trivial, because the length of an element in the $k$th term of the lower central series is bounded below by $k$. It follows that $F\_2$ embeds in a direct product of nilpotent groups. So there is an infinite product of nilpotent groups that is not amenable.
| 21 | https://mathoverflow.net/users/460592 | 452307 | 181,800 |
https://mathoverflow.net/questions/452013 | 14 | This is probably old, a Chevalley level of old, but I'm not at all an expert in this field so I need help.
Let $G$ be a simply connected (almost) simple linear algebraic group defined over $K=\mathbb{F}\_{q}$. Is there a clean reference in the literature to the fact that there must be at least one regular semisimple element in $G(K)$?
I know from the literature that regular semisimple elements are dense in $G$ and in every maximal torus $T$ (for example §2.3 and §2.5 in Humphreys's *Conjugacy classes in semisimple algebraic groups*), but this ensures only that there are many elements over the algebraic closure $\overline{K}$, not specifically over $K$. I know that this is true for $\mathrm{char}(K)=0$, using unirationality and the fact that $K$ is infinite, as in [this question](https://mathoverflow.net/questions/307867/existence-of-regular-semisimple-elements-of-reductive-groups-in-characteristic-0). I even know how to presumably shoot my question with a cannon in a few different ways, for instance:
1. Simply connected (almost) simple groups are in 1-1 correspondence with Dynkin diagrams (see for instance §32 in Humphreys's *Linear algebraic groups*), so I could manually write down one regular semisimple element over $\mathbb{F}\_{q}$ for each such $G$.
2. Fleischmann and Janiszczak (1993) have formulas for the number of regular semisimple elements over $\mathbb{F}\_{q}$ at least for the classical $G$, i.e. of type $A,B,C,D$; they even announce on p. 484 that they obtained the number of semisimple conjugacy classes for $E$, so possibly they cover those cases as well. In any case, if there is a formula and its result is $>0$, then in particular there is one such element.
3. Lehrer (1992) showed among other things the "curious" result that, independently of characteristic, any semisimple simply connected $G$ has an odd number of regular semisimple conjugacy classes in $G(K)$ (see Corollary 3.5). Again, in particular if the number is odd then it is not $0$.
I would classify each of the routes above as "too strong for my question". I feel there must be some chapter-1-of-a-textbook kind of reference for "there exists a regular semisimple element over $\mathbb{F}\_{q}$", but I could not find one.
Do you know where to find it? Also, bonus content: is the fact true and referenced if I relax the hypotheses (connected, or semisimple, or reductive...)?
| https://mathoverflow.net/users/155467 | Existence of a regular semisimple element over $\mathbb{F}_{q}$ | See Proposition 7.1.4 in Dat, Orlik, and Rapoport, *[Period domains over finite and $p$-adic fields](https://doi.org/10.1017/CBO9780511762482)*, Cambridge tracts in Mathematics, vol. 183. While I don't have this book in front of me, I'm pretty sure that the result asserts the existence of elliptic semisimple elements in all finite reductive groups. Such elements are necessarily regular. Most groups are handled via a uniform argument, but several are treated as special cases. Technically, the proof contains a gap, as the case of $G\_2(\mathbb{F}\_2)$ was inadvertently omitted. Nonetheless, the result remains true in this case, where one can find a regular element by looking at the Coxeter torus $T$ of $\operatorname{SL}\_3$, which we can consider as a subgroup of $G\_2$. Every nontrivial element of $T(\mathbb{F}\_2)$ is regular in $G\_2$.
| 7 | https://mathoverflow.net/users/4494 | 452314 | 181,804 |
https://mathoverflow.net/questions/452331 | -4 | Let $F: S^2 \rightarrow \mathbb{R}^2$ be a continuous function. Does there exist a unit vector $v \in \mathbb{R}^2$ and a continuous function $f(x):S^2\rightarrow \mathbb{R}$ such that $f(x)>0$ on $S^2$ and
$$F(x)\neq f(x)v \ \ \ \ \forall x\in S^2?$$
| https://mathoverflow.net/users/115905 | Uncountable Cantor's diagonal argument on $S^2$ | Sure. Simply choose $f(x)=\|F(x)\|+1$, and pick an arbitrary unit vector $v\in\mathbb{R}^2$.
| 1 | https://mathoverflow.net/users/11919 | 452332 | 181,809 |
https://mathoverflow.net/questions/452337 | 1 | The following lemma is from the book *Discrete groups* by Ohshika.
>
> If a language $L$ is accepted by a non-deterministic automaton, then $L$ is regular, i.e., there exists a finite state automaton $M$ such that $L = L(M)$.
>
>
>
Proof. Let $M = (\Sigma,A,\mu,F,\Sigma\_0)$ be a non-deterministic automaton accepting $L$. For a subset $X$ of the state set $\Sigma$, we define its $\epsilon$-closure, denoted by $\overline{X}$, to be the set of states that can be reached from $X$ by tracking arrows labelled with $\epsilon$ in finite steps. (The states in $X$ are regarded as being reached in $0$ steps, hence contained in $\overline{X}$.) We say that a subset $X$ such that $\overline{X} = X$ is $\epsilon$-closed.
Let $Y$ be the set of $\epsilon$-closed subsets of $\Sigma$, where we regard the empty set also as being $\epsilon$-closed. We shall construct a finite state automaton whose state set is $Y$ as follows. For $a\in A$ and $X\in Y$, we define $Xa$ to be the $\epsilon$-closure of $\{s\in\Sigma\mid\text{ there exists }x\in X\text{ such that }(x,a,s)\in\mu\}$. We let the initial state be the $\epsilon$-closure of $\Sigma\_0$, and the final states the $\epsilon$-closed subsets containing at least one final state.
It's easy to see that such a finite state automaton accepts the same languages $L$ as the non-deterministic automaton $M$. $\square$
The problem with the proof is that the resulting automaton is not a finite state automaton. (The initial state is not a single string) I'm a bit worried because this fact is used quite frequently in the later part of the book. Is this statement true? If so, where can I find the reference?
$\bullet$ I'll add definitions if someone requires in the comment.
| https://mathoverflow.net/users/323920 | If a language $L$ is accepted by a non-deterministic automation, then $L$ is regular | [Wikipedia](https://en.wikipedia.org/wiki/Powerset_construction) and [Hopcroft and Ullman](https://en.wikipedia.org/wiki/Introduction_to_Automata_Theory,_Languages,_and_Computation) require a unique start state for their NFA's. You can transform an NFA with many start states into an NFA with just one by adding a new state $q\_0$ and connecting it to all $q$ in $\Sigma\_0$ by $\epsilon$-edges.
It looks like Ohshika is following [Word processing in groups](https://en.wikipedia.org/wiki/Word_Processing_in_Groups), which allows many start states (see Definition 1.2.4). He is also following their proof (see Theorem 1.2.7). Unfortunately, there appears to be a small typo/mistake in WPIGs on line -8 of page 14 - at that location in the proof they seem to think that NFAs have a single start state.
In any case, in their construction (and in Ohshika's) the start state for the new DFA "should be" the $\epsilon$-closure of the set of start states of the original NFA.
| 3 | https://mathoverflow.net/users/1650 | 452338 | 181,811 |
https://mathoverflow.net/questions/452295 | 5 | Let $U$ be an open simply connected subset of $\mathbb R^2$. Let $x$ be a boundary point of $U$.
Does then there always exist a continuous function $f\colon[0,1]\to\mathbb R^2\setminus U$ such that $x = f(0)\ne f(1)$?
That is, does then there always exist a nontrvial continuous curve starting at $x$ and staying outside $U$?
| https://mathoverflow.net/users/36721 | On the boundary of a simply connected set | Here's a variant of Fernando Muro's construction.
Let $X$ be the closure in $\mathbb R^2$of the graph of the function
$g(x):=x\sin\big(1/\sin(1/x)\big)$ defined on $[0,+\infty)\setminus\{1/n\pi:n\in \mathbb Z\_+\}$. Let $U:=\mathbb R^2\setminus X$, again a simply connected open set. The function $g$ has a set of essential discontinuities $\{1/n\pi:n\in \mathbb Z\_+\}$ that accumulates at $0$, but it is continuous at $0$, so $X$ has no points with $0$ first coordinate but $(0,0)$, and $(0,0)$ can be connected by continuous arcs to no other point of $X$ .
| 5 | https://mathoverflow.net/users/6101 | 452358 | 181,817 |
https://mathoverflow.net/questions/452355 | 0 | Let $K(\mathbb P^1)$ be the Grothendieck group of sheaves on $\mathbb P^1$. I want to show that the map $K^{{\rm PGL}(2)\times \{\pm 1\}}(\mathbb P^1) \to K(\mathbb P^1)$ is not onto. I read somewhere that the class of a vector bundle of odd degree on $\mathbb P^1$ is not equal to the class of a sheaf equivariant under ${\rm PGL}(2)\times \{\pm 1\}$. Could you explain me why this is the case?
| https://mathoverflow.net/users/12395 | Equivariant sheaves on $\mathbb P^1$ | Let me explain why the line bundle $\mathcal{O}(1)$ does not admit a $\mathrm{PGL}(2)$-equivariant structure. Indeed, if it does, then the vector space
$$
\mathrm{Hom}(\mathcal{O}, \mathcal{O}(1))
$$
would have a structure of a $\mathrm{PGL}(2)$-representation, compatible with the standard $\mathrm{SL}(2)$-representation structure. But this is not true, because the central element $-1 \in \mathrm{SL}(2)$ acts on this space non-trivially.
| 3 | https://mathoverflow.net/users/4428 | 452361 | 181,818 |
https://mathoverflow.net/questions/452369 | 2 | Is there a tabulation somewhere of the isomorphism classes of distributive lattices $L$ with $|L|=n$ for small $n$? Google has not found me one.
| https://mathoverflow.net/users/10366 | Classification of small finite distributive lattices | I have the same problem to collect posets with certain properties and for distributive lattices I have the following solution which was good enough for my purposes.
Here is a sage program to obtain all distributive lattices (in form of their leq matrices) on $n$ points which works for $n \leq 9$ in the sage online cell <https://sagecell.sagemath.org/> and works for about $n \leq 14$ with sage with a good computer (and alot of waiting).
```
n=5;
P=Posets(n-2)
P=[p.with_bounds() for p in P]
L=[p for p in P if p.is_lattice() and LatticePoset(p).is_distributive()]
display(len(L))
U=[p.lequal_matrix() for p in L]
display(U)
display(L[2])
```
the display(L[2]) command displays a distributive lattice of your choice and
the len(L) command gives the number of distributive lattices on $n$ elements.
| 2 | https://mathoverflow.net/users/61949 | 452371 | 181,822 |
https://mathoverflow.net/questions/452353 | 7 | Let $G$ be an infinite group. Let $N\_0$ be the set of all $x\in G$ for which the conjugacy class $\{y^{-1}xy: y\in G\}$ is a finite set. Clearly $N\_0$ is a normal subgroup. Iteratively, form an ascending transfinite sequence by
* for $n$ a non-limit ordinal, let $N\_n\subseteq G$ be the set of all $x\in G$ for which $xN\_{n-1}\in G/N\_{n-1}$ has finite conjugate class.
* for $\omega$ a limit ordinal, $N\_{\omega} = \bigcup\_{\alpha<\omega} N\_{\alpha}$.
There are two possibilities: either $(N\_\alpha)$ stops at $G$, or at a proper subgroup $G^{FC}$ of $G$. For convenience, let's call (only for this post) $G^{FC}$ the *[FC]-kernel* of $G$, and $G$ a *hyper-[FC] group* if $G=G^{FC}$. A simple observation is that $G/G^{FC}$ is an [[ICC]](https://en.wikipedia.org/wiki/Infinite_conjugacy_class_property) group. Let's call $G^{FC}$ a *minimal [FC]-kernel* if whenever $N\_0\subseteq H\subseteq G$ is another normal subgroup for which $G/H$ is [ICC], we have $H\supseteq G^{FC}$.
I apologize if the questions below are elementary for MO.
**Q0:** Could you please point me to a reference where I can find the proper terminology for the made-up names *hyper-[FC] group* & *[FC]-kernel*?
**Q1:** Is $G^{FC}$ a hyper-[FC] group in itself?
**Q2:** Is $G^{FC}$ always *minimal* in the sense above?
Thanks in advance.
---
**edit:** Following Sean Eberhard's suggestion, I've just came across the book "Finiteness Conditions and Generalized Soluble Groups" by Robinson. The terminology he used in his book is *FC-Hypercenter* for $G^{FC}$, and *FC-Hypercentral* if $G=G^{FC}$. It is a big deal(!) that the made-up names by yours truly were a close call after all.. The book explains various central series in detail for those who are curious about the similarities in the construction & the properties of X-hypercenters, and more..
| https://mathoverflow.net/users/164350 | Finite conjugacy classes | **Q0:** For references try starting with Robinson's *A course in the theory of groups*, starting around 14.5.5. There the FC center is defined and some characterizations are given for FC groups (groups with finite conjugacy classes). From there it is not a great leap to define the second FC center and so on, but I am not aware of any significant study of these concepts.
The answers to Q1 and Q2 are both positive and use a simple observation. Suppose $N\_1 \le N\_2$ are normal subgroups of $G$ and consider the natural map $G/N\_1 \to G/N\_2$. Then the FC-center of $G/N\_1$ maps into the FC-center of $G/N\_2$. Indeed, if an element has a finite conjugacy class modulo $N\_1$ then it certainly does so modulo the larger group $N\_2$.
**Q1:** Let me write $F(G)$ for the FC-center and $F^\alpha(G)$ for the iterated variants. Let $N = \bigcup\_\alpha F^\alpha(G)$ be the limit of the whole process. Then $N$ is a characterstic subgroup of $G$, as is $F^\alpha(N)$ for all $\alpha$. I claim inductively that $F^\alpha(G) \le F^\alpha(N)$ for all $\alpha$. The induction is obvious for zero and limit ordinals, and the case of successor ordinals follows from the observation above applied to $N/F^\alpha(G) \to N / F^\alpha(N)$.
**Q2:** Suppose $H$ is a normal subgroup of $G$ such that $G/H$ is ICC. I claim that $F^\alpha(G) \le H$ for all $\alpha$. Again it suffices to a consider a successor ordinal $\alpha+1$, and now we apply the observation above to $G / F^\alpha(G) \to G/H$, noting that $F(G/H)=1$.
In light of Q2, an alternative name for the "hyper FC center" might be the "ICC residual". It is the smallest normal subgroup $H$ such that $G/H$ is ICC.
| 8 | https://mathoverflow.net/users/20598 | 452372 | 181,823 |
https://mathoverflow.net/questions/452366 | 3 | The question concerns a very general setting and a very general inequality about KL divergence. I'm writing this thread to verify whether my intuition is correct.
Let $E\_1, E\_2$ be two measurable spaces, $f, g: E\_1 \rightarrow E\_2$ two measurable functions, and $A, B$ two random variables taking values in $E\_1$. Is the following inequality:
$D\_{KL}(f(A) || g(B)) \le D\_{KL}(A || B)$
always true?
In case it isn't, I'd like to ask whether the following inequality is always true:
$D\_{KL}(f(A) || f(B)) \le D\_{KL}(A || B)$
| https://mathoverflow.net/users/510181 | A general inequality for KL divergence of functions of variables | $\newcommand{\si}{\sigma}\newcommand{\F}{\mathcal F}\newcommand{\G}{\mathcal G}\newcommand{\pa}{\parallel}$The first inequality is obviously false in general, e.g. when $A=B$ but $f(A)$ differs from $g(B)=g(A)$ in distribution.
The second inequality is true. A bit more generally, let $\mu$ and $\nu$ be probability measures defined on a $\si$-algebra $\F$. Let $\mu\_\G$ and $\nu\_\G$ be the restrictions of $\mu$ and $\nu$ to a sub-$\si$-algebra $\G$ of $\F$.
>
> Then
> \begin{equation\*}
> D\_{KL}(\mu\_\G\pa\nu\_\G)\le D\_{KL}(\mu\pa\nu). \tag{1}\label{1}
> \end{equation\*}
>
>
>
Indeed,
\begin{equation\*}
D\_{KL}(\mu\pa\nu)=\int d\nu\,h\Big(\frac{d\mu}{d\nu}\Big),
\end{equation\*}
where $h(u):=u\ln u$ for $u\in(0,\infty)$, with $h(0):=0$ and $h(\infty)=\infty$.
Next, letting $E\_\nu(\cdot|\G)$ denote the conditional expectation given $\G$ w.r.t. $\nu$, for any nonnegative $\G$-measurable function $q$ we have
\begin{equation\*}
\int d\nu\_\G\,q\, E\_\nu\Big(\frac{d\mu}{d\nu}\Big|\G\Big)
= \int d\nu\,q\, E\_\nu\Big(\frac{d\mu}{d\nu}\Big|\G\Big)\,
=\int d\nu\,q\, \frac{d\mu}{d\nu}\,
= \int d\mu\,q= \int d\mu\_\G\,q;
\end{equation\*}
the second equality in the latter display holds by the definition of the conditional expectation. So,
\begin{equation\*}
\frac{d\mu\_\G}{d\nu\_\G}=E\_\nu\Big(\frac{d\mu}{d\nu}\Big|\G\Big).
\end{equation\*}
So, by Jensen's inequality applied to the convex function $h$,
\begin{multline\*}
D\_{KL}(\mu\_\G\pa\nu\_\G)
=\int d\nu\,h\Big(\frac{d\mu\_\G}{d\nu\_\G}\Big)
=\int d\nu\,h\Big(E\_\nu\Big(\frac{d\mu}{d\nu}\Big|\G\Big)\Big) \\
\le \int d\nu\,E\_\nu\Big( h\Big(\frac{d\mu}{d\nu}\Big)\Big|\G\Big)
=\int d\nu\,h\Big(\frac{d\mu}{d\nu}\Big)
=D\_{KL}(\mu\pa\nu),
\end{multline\*}
so that \eqref{1} follows.
Your second inequality now obtains by letting $\mu$ and $\nu$ be the distributions of $A$ and $B$, respectively, over the measurable space $(E\_1,\F)$ and letting $\G$ be the sub-$\si$-algebra of $\F$ generated by the measurable map $f$.
$\quad\Box$
| 2 | https://mathoverflow.net/users/36721 | 452373 | 181,824 |
https://mathoverflow.net/questions/452359 | 3 | $\newcommand\R{\mathbb R}$Let $U$ be an open subset of $\R^2$ such that the point $(0,0)$ is on the boundary of $U$. Let $f\colon[0,1]\to\R^2$ be the path that starts at $(0,0)$ and moves with a (say) constant speed, first horizontally right to $(2,0)$, then horizontally left to $(1,0)$, and finally goes counterclockwise twice over the unit circle. Let $g\colon[0,1]\to\R^2$ be another continuous path that also starts at $(0,0)$ and such that $g(t)\in U$ for all $t\in(0,1]$ and $|g(t)-f(t)|<1/16$ for all $t\in[0,1]$, where $|\cdot|$ is the Euclidean norm.
Can then the set $U$ be simply connected?
| https://mathoverflow.net/users/36721 | Can such a set be simply connected? | Let $I\_0$ be the interval where $f$ is a horizontal motion, and let $I\_1$ the interval where it is a circular motion. So up to reparametrisation (and adopting complex notation) $I\_0=[-3,0]$ and $I\_1=[0,4\pi]$, and $f(t)=\min(t+3,1-t)$ in $I\_0$ and $f(t)=e^{it}$ in $I\_1$.
Since $|g(3\pi/2)+i|<1/16$ and $|g(5\pi/2)-i|<1/16$ by connection there are points $t\_0\in I\_0$ and $ 3\pi/2< t\_1 < 5\pi/2$ such that $g(t\_1)=g(t\_0)$ **(\*)**.
This means that $g$ restricts to a closed arc on $[t\_0,t\_1]$, which is still $1/16$ close, up to reparametrisation, to the loop $f|[-2,2\pi]$, which is non contractible in $\mathbb C\setminus\{0\}$. Therefore so is the loop $g\_{|[t\_0,t\_1]}$, and a fortiori in $U\subset \mathbb C\setminus\{0\}$. Therefore $U$ is not simply connected.
**(\*)** This can be seen geometrically as a consequence of the Jordan curve theorem (note that $g(I\_0)$, as any image of a continuous curve, contains a simple curve with the same endpoints, and this can be enlarged to a simple closed curve bounding a domain containing $g(3\pi/2)$ and not $g(5\pi/2)$. Otherwise, without bothering about injectivity: $g\_{|I\_0}$ can be extended to a closed loop $\gamma$, juxtaposing to it an arc far from $g(I\_1)$, and such that $\text{ind}(\gamma,g(3\pi/2))=-1$ and $\text{ind}(\gamma,g(5\pi/2))=0$, which implies that $g\_{|[3\pi/2,5\pi/2]}$ must cross the curve $\gamma$ (necessarily in $g(I\_0)$) otherwise $\text{ind}(\gamma,g(t))$ would remain constant for $3\pi/2\le t \le 5\pi/2$
| 4 | https://mathoverflow.net/users/6101 | 452377 | 181,826 |
https://mathoverflow.net/questions/451805 | 6 | Suppose we have a compact three-manifold $M$, a codimension-one foliation $F$ of $M$, and a one-form on $\alpha$, chosen so that $F$ is tangent to $\alpha = 0$. We deduce that $\alpha \wedge d\alpha = 0$. Also, there is some one-form $\beta$ so that $d\alpha = \alpha \wedge \beta$.
The *Godbillon-Vey invariant* of $F$ is defined as the cohomology class of $\beta \wedge d\beta$. This class is independent of the choices made (at least up to sign).
Then we have the following question which is raised by Etienne Ghys:
**Conjecture:** The Godbillon-Vey invariant is a topological invariant.
See the paper É. Ghys, L’invariant de Godbillon–Vey, Astérisque (1989) 177–178, Exp. No. 706, Séminaire
Bourbaki, Vol. 1988/89, 155–181
That is, two topologically equivalent foliations have the same Godbillon-Vey class.
>
> What are some updates on this conjecture?
>
>
>
| https://mathoverflow.net/users/36688 | The current situation of the Godbillon-Vey invariant conjecture | Check out:
*Hilsum, Michel*, [**Functions with bounded variation and the class of Godbillon-Vey**](https://doi.org/10.1093/qmath/hav013), Q. J. Math. 66, No. 2, 547-562 (2015). [ZBL1401.57040](https://zbmath.org/?q=an:1401.57040).
| 4 | https://mathoverflow.net/users/1345 | 452384 | 181,829 |
https://mathoverflow.net/questions/452386 | 2 | Motivation:
-----------
I'm studying certain properties of conjugation in $SL(n,q)$. There's a nice number, a bit like a covering number, that one can associate with an arbitrary group. In writing a programme in GAP to compute this number, a shortcut comes to mind - I'm not going to share what exactly - that involves the [chief series](https://en.m.wikipedia.org/wiki/Chief_series) of the groups.
The Question:
-------------
>
> What, if anything, is the maximum length of the chief series of $SL(n,q)$?
>
>
>
Thoughts:
---------
I ran the following in GAP:
```
for n in [2..6] do
for q in Filtered([2..11],IsPrime) do
Print([n,q,Size(ChiefSeries(SL(n,q)))],"\n");
od;
od;
```
It produced . . .
```
[ 2, 2, 3 ]
[ 2, 3, 4 ]
[ 2, 5, 3 ]
[ 2, 7, 3 ]
[ 2, 11, 3 ]
[ 3, 2, 2 ]
[ 3, 3, 2 ]
[ 3, 5, 2 ]
[ 3, 7, 3 ]
[ 3, 11, 2 ]
[ 4, 2, 2 ]
[ 4, 3, 3 ]
[ 4, 5, 4 ]
[ 4, 7, 3 ]
[ 4, 11, 3 ]
[ 5, 2, 2 ]
[ 5, 3, 2 ]
[ 5, 5, 2 ]
[ 5, 7, 2 ]
```
. . . after which I terminated the calculation.
As you can see, the maximum length calculated is $4$ and is obtained by both $SL(2,3)$ and $SL(4,5)$. This is very little to go by, of course.
*What do I think it would be?*
I'd be pleasantly surprised if the maximum existed.
---
Please help :)
| https://mathoverflow.net/users/42153 | Is there a maximum length of the chief series of $SL(n,q)$? | Well, $PSL(n,q)$ is usually simple, so apart from $SL(2,3)$ you're looking at two plus the number of prime factors, counting multiplicity, of the greatest common divisor of $n$ and $q-1$. Of course, this is unbounded.
| 8 | https://mathoverflow.net/users/460592 | 452387 | 181,830 |
https://mathoverflow.net/questions/452385 | 1 | Let $H\_n$ be the $n$th probabilistic Hermite polynomial of degree n and $\eta = \exp(-x^2/2)/\sqrt(2 \pi)$ be the standard Gaussian density.
I would like to compute the integral $f\_n(x) = \int H\_n(x - z) \eta(z) dz$. Any hope to get a closed form expression?
Some ideas:
1. The $n$th Hermite polynomial $H\_n$ can be related to the n-th derivative of $\eta$
2. This integral is a convolution of $H\_n$ with the standard Gaussian kernel $\eta$.
| https://mathoverflow.net/users/510195 | Convolution of a Hermite polynomial with Gaussian kernel | Using the definition of the Hermite polynomials and then integrating by parts $n$ times, we get $f\_n(x)=x^n$.
Details:
\begin{equation}
\begin{aligned}
f\_n(x)&=\int(-1)^n\eta^{(n)}(x-z)\frac{\eta(z)}{\eta(x-z)}\,dz \\
&=(-1)^ne^{x^2/2}\int\eta^{(n)}(x-z)\ e^{-xz}\,dz \\
&=(-1)^{n-1}e^{x^2/2}x\int\eta^{(n-1)}(x-z)\ e^{-xz}\,dz \\
&\vdots \\
&=e^{x^2/2}x^n\int\eta(x-z)\ e^{-xz}\,dz \\
&=x^n.
\end{aligned}
\end{equation}
| 1 | https://mathoverflow.net/users/36721 | 452388 | 181,831 |
https://mathoverflow.net/questions/452365 | 4 | A factor $R$ is called stable if $M\_n(R)\cong R$ for all $n>0$. For the sake of this question, we call a factor *backwards stable* if $R\cong M\_n(S)$ implies $S\cong R$ where $S$ is allowed to be any other factor.
If $R$ is stable then it is backwards stable if $M\_n(R)\cong M\_n(S) \implies R\cong S$.
My question is which factors are backwards stable.
I'm particularly interested in hyperfinite type III factors. For hyperfinite type II factors I can show backwards stability using the uniqueness of the II$\_1$ factor.
As a side question, are there examples of factors that are not stable other than those of finite type I? For infinite factors, one can always find Cuntz isometries so that they should be stable and, by uniqueness, the hyperfinite II$\_1$ factor is stable. This leaves open the case of non-hyperfinite type II$\_1$ factors.
| https://mathoverflow.net/users/485160 | Backwards stable factors | For $\textrm{II}\_1$ factors, your definitions of backwards stable and stable are the same. The point here is that for a $\textrm{II}\_1$ factor $R$ you can talk about $M\_n(R)$ for all positive number $n$ instead of just positive integers. This is called amplification and is usually written as $R^n$. This is done by choosing an integer $m \geq n$ and a projection $p$ of trace $\frac{n}{m}$ in $M\_m(R)$, then defining $R^n = pM\_m(R)p$. The fact that two projections of the same trace are equivalent in a $\textrm{II}\_1$ factor ensures that this is well-defined and does not depend on the choice of $m$ or $p$. With this in mind, as $R = M\_n(R)^{\frac{1}{n}}$, $M\_n(R) \simeq M\_n(S) \Rightarrow R \simeq S$ always holds for $\textrm{II}\_1$ factors $R$ and $S$. As you have observed, this means stability implies backwards stability. The converse holds by observing that, if $R$ is backwards stable, as $R \simeq M\_n(R^{\frac{1}{n}})$ (this is a general fact about amplifications), $R \simeq R^{\frac{1}{n}}$. Then $R \simeq M\_n(R^{\frac{1}{n}}) \simeq M\_n(R)$, which shows $R$ is stable.
There are plenty of non-stable $\textrm{II}\_1$ factors. In fact, for a $\textrm{II}\_1$ factor $R$, the collection of positive real numbers $r$ for which $R \simeq R^r$ form a group under multiplication, which is known as the fundamental group of $R$. Plenty of $\textrm{II}\_1$ factors have fundamental groups that are not $\mathbb{R}\_+$. For example, all property (T) $\textrm{II}\_1$ factors have countable fundamental groups. They could even have trivial fundamental groups. (See, for example, chapter 18 in [https://www.math.ucla.edu/~popa/Books/IIunV15.pdf](https://www.math.ucla.edu/%7Epopa/Books/IIunV15.pdf) .) It is a famous open question as to what is the fundamental group of the free group factor $L(\mathbb{F}\_2)$, which is currently only known to be either trivial or $\mathbb{R}\_+$.
| 5 | https://mathoverflow.net/users/504602 | 452401 | 181,836 |
https://mathoverflow.net/questions/452421 | 1 | Let $S$ be the set of integers with largest prime factor bounded by a given positive integer $k$. Is there a formula for the asymptotic density of such a set $S$?
| https://mathoverflow.net/users/18659 | Prime factors bounded by $k$ | If $k$ is fixed, then the simple bound $|S\cap[1,x]|\leq(\log\_2 x)^{\pi(k)}$ shows that the aymptotic density of $S$ is zero. For stronger bounds, see Chapter III.5 in Tenenbaum: Introduction to analytic and probabilistic number theory.
| 8 | https://mathoverflow.net/users/11919 | 452422 | 181,840 |
https://mathoverflow.net/questions/452425 | 6 | I am not familiar with the definition of total positivity. I am not sure about the link between *log-concavity* and *total positivity*.
* In a paper [On Variation-Diminishing Integral Operators of the Convolution Type](https://www.jstor.org/stable/87970) of Schoenberg, he defines *Pólya frequency functions* as a *totally positive* one.
* In [Estimating a Polya Frequency Function](https://dept.stat.lsa.umich.edu/%7Emichaelw/PPRS/2007ims.pdf), Pal, Woodroofe, and Meyer define *Pólya frequency functions* as *log-concave* functions.
Then, my question is the following, what are the links between *total positivity* and *log-concavity* ? It seems that *total positivity* implies *log-concavity*, do we have the converse statement ?
| https://mathoverflow.net/users/510079 | Total positivity, log-concavity and Pólya frequency | $\newcommand{\R}{\mathbb R}$For a positive integer $r$, a measurable function $f\colon\R\to\R$ is called a Pólya frequency function of order $r$ (abbreviated as PF$\_r$) if the matrix $(f(x\_i-y\_j))\_{i,j,=1}^r$ is totally positive for all real $x\_i$ and $y\_j$ such that $x\_1\le\dots\le x\_r$ and $y\_1\le\dots\le y\_r$, that is, if $\det(f(x\_i-y\_j))\_{i,j,=1}^m\ge0$ for all integers $m=1,\dots,r$ and all real $x\_i$ and $y\_j$ such that $x\_1\le\dots\le x\_m$ and $y\_1\le\dots\le y\_m$. So, clearly PF$\_1\supseteq\,$PF$\_2\supseteq\cdots$.
Consider now the case $r=2$. Note that the set of all matrices of the form $(x\_i-y\_j)\_{i,j,=1}^2$ for real $x\_i$ and $y\_j$ such that $x\_1\le x\_2$ and $y\_1\le y\_2$ coincides with the set of all matrices of the form
$\begin{pmatrix}a&b\\c&d \end{pmatrix}$ for real $a,b,c,d$ such that $a+d=b+c$ and $\{a,d\}\subseteq[b,c]$.
So, if $f$ is log concave, then $f$ is PF$\_2$. Vice versa, if $f$ is PF$\_2$, then, taking $a=d=(b+c)/2$, we see that $f$ is midpoint log concave, and hence, being measurable, $f$ is log concave (see e.g. [Theorem II](https://www.ams.org/journals/tran/1919-020-01/S0002-9947-1919-1501114-0/)).
Thus, a function $f\colon\R\to\R$ is PF$\_2$ iff it is log concave.
However, it is easy to find examples of log-concave functions that are not in PF$\_3$, and hence not in PF$\_r$ for any natural $r\ge3$; for instance, take $f=1\_{(0,8)}$ and
\begin{equation}
(x\_1,x\_2,x\_3,y\_1,y\_2,y\_3)= (0, 1, 3, -5, -4, 0);
\end{equation}
then $\det(f(x\_i-y\_j))\_{i,j,=1}^3=-1\not\ge0$.
So, the definitions of Pólya frequency functions in the two papers referred to in your post are equivalent to each other for $r=2$ -- but not for $r\ge3$.
| 8 | https://mathoverflow.net/users/36721 | 452439 | 181,846 |
https://mathoverflow.net/questions/452442 | 1 | Let $d \in \mathbb N^\*,p \in [1, \infty]$ and $T>0$. Let
$$
F :[0, T] \to L^p (\mathbb R^d; \mathbb R\_{\ge 0}), t \mapsto F\_t
$$
be measurable. I would like to ask if there is a measurable function $G:[0, T] \times \mathbb R^d \to \mathbb R\_{\ge 0}$ such that
* $G(t, \cdot) \in L^p (\mathbb R^d; \mathbb R\_{\ge 0})$ for all $t \in [0, T]$.
* $\|G(t, \cdot) - F\_t\|\_{L^p} = 0$ for a.e. $t \in [0, T]$.
Thank you so much for your elaboration!
| https://mathoverflow.net/users/99469 | Does a measurable $F :[0, T] \to L^p (\mathbb R^d; \mathbb R_{\ge 0})$ have a "flattened" measurable version? | $\newcommand{\R}{\mathbb R}\newcommand{\ep}{\varepsilon}\newcommand{\LL}{\mathcal L}\newcommand{\si}{\sigma}$The answer is yes, at least for $p\in[1,\infty)$.
Indeed, $L^p(\R^d)$ is a separable metric space. So, for each real $\ep$ there is a countable measurable partition $(B\_{\ep,j})$ of $L^p(\R^d)$ such that for each $j$ we have $B\_{\ep,j}\ne\emptyset$ and the diameter of $B\_{\ep,j}$ is $\le\ep$. Pick any $y\_{\ep,j}$ in $B\_{\ep,j}$. For $(t,x)\in[0,T]\times\R^d$, let
\begin{equation}
G\_\ep(t,x):=\sum\_j y\_{\ep,j}(x)\,1(F(t)\in B\_{\ep,j}),
\end{equation}
where $F(t):=F\_t$. Then for any real $c$
\begin{equation}
\{(t,x)\in[0,T]\times\R^d\colon G\_\ep(t,x)>c\}
=\bigcup\_j F^{-1}(B\_{\ep,j})\times y\_{\ep,j}^{-1}((c,\infty))
\in\LL([0,T])\otimes\LL(\R^d),
\end{equation}
where $\LL(\cdot)$ denotes the Lebesgue $\si$-algebra. So, the function $G\_\ep$ is measurable, for each $\ep$.
Also, $\|G\_\ep(t,\cdot)-F(t)\|\_{L^p(\R^d)}\le\ep$ and hence $\|G\_\ep(t,\cdot)\|\_{L^p(\R^d)}\le\ep+\|F(t)\|\_{L^p(\R^d)}$ for each $t\in[0,T]$.
Since $F\colon[0,T]\to L^p(\R^d)$ is measurable and the norm on $L^p(\R^d)$ is continuous and hence measurable, we see that the function $[0,T]\ni t\mapsto w(t):=\dfrac1{1+\|F(t)\|^p\_{L^p(\R^d)}}\in[0,\infty)$ is measurable. So, for each real $\ep>0$ we have $G\_\ep\in L^p\_w([0,T]\times\R^d)$, where $L^p\_w([0,T]\times\R^d)$ is the space of all measurable functions $H\colon[0,T]\times\R^d\to\R$ with norm
$$\|H\|\_{L^p\_w([0,T]\times\R^d)}:=\Big(\int\_0^T dt\,w(t)\,\|H(t,\cdot)\|\_{L^p(\R^d)}^p\Big)^{1/p}<\infty.$$
For all integers $m,n$ such that $m\ge n\ge1$
\begin{equation}
\|G\_{1/m}-G\_{1/n}\|\_{L^p\_w([0,T]\times\R^d)}^p
=\int\_0^T dt\,w(t)\|G\_{1/m}(t,\cdot)-G\_{1/n}(t,\cdot)\|\_{L^p(\R^d)}^p\le(2/n)^pT\to0
\end{equation}
as $n\to\infty$. So, by the completeness of $L^p\_w([0,T]\times\R^d)$, for some sequence $(n\_k)$ of natural numbers going to $\infty$ there is a limit $G$ of $G\_{1/n\_k}$ in $L^p\_w([0,T]\times\R^d)$. Clearly, this limit $G$ satisfies your desired conditions. $\quad\Box$
| 2 | https://mathoverflow.net/users/36721 | 452445 | 181,848 |
https://mathoverflow.net/questions/452398 | 2 | Prove that:
$$ f(x) = \log\big(
{}\_2F\_1(a,\,b\,;\,c\,;\,x^{-1})\big),\;\;a,b,c>0
$$
is convex (and decreasing) on $(1,\infty)$.
It actually seems that the stronger result that $f\big((x+1)^{\beta}\big)$, $\beta>0$, is completely monotonic, is true. I saw a post on here proving a similar result using continued fractions to show that all the Taylor series coefficients are positive when $c\ge a+b$. I wonder if such an approach could used to show that the coefficients of a series expansion of $f\big((x+1)^{\beta}\big)$ have alternating signs for $x>0$ ( for arbitrary $a,b,c>0$ ). Or compute the inverse Laplace transform?
This result would imply that $f(\!\sqrt{x})$ is convex on $(1,\infty)$, which is equivalent to the function:
$$ g(x) = \frac{ \_2F\_1(a,\,b\,;\,c\,;\,\alpha x)}{ \_2F\_1(a,\,b\,;\,c\,;\,x)},\;\; 0<\alpha<1,\;\,a,b,c>0$$
being decreasing on $[0,1]$, which is what I originally wanted to show.
This result is important to show UMP properties of multiple determination coefficient tests.
| https://mathoverflow.net/users/510206 | Log convexity of hypergeometric function for $a,b,c>0$ | Using the series representation of the hypergeometric function, we see that $\_2F\_1(a,b\,;c\,;(1+x)^{-1})$, $x>0$, is the pointwise convergent limit of positive sums of completely monotonic functions for $a,b,c>0$, and is thus completely monotonic.
Since completely monotonic functions are log convex, it follows that $\log \big({}\_2F\_1(a,b\,;c\,;(x+1)^{-1}\big)$ is convex on $(0,\infty)$.
| 0 | https://mathoverflow.net/users/510206 | 452453 | 181,850 |
https://mathoverflow.net/questions/449715 | 6 | Given a Lie groupoid $G$, we can view it as representing a prestack on $\text{Mfld}$ by sending and manfold $M$ to the groupoid of smooth functors and smooth natural transformations
$$G(M) := \text{Fun}(M,G)$$
(viewing $M$ as a discrete groupoid). One of the multiple ways of improving this prestack to a stack is to instead use anafunctors and ananatural transformations. That is, we localize at coverings so that a manifold $M$ is sent to a groupoid containing objects of the form $M \leftarrow U \rightarrow G $ for coverings $U$ of $M$.
On the other hand, we can apply the plus construction to $G$ to get a stack. This is defined by
$$G^+(M) := \text{colim}\_{R \hookrightarrow M}\text{Psh}(R,G).$$
Here the colimit is taken over covering sieves of $M$. This seems to be doing the same thing as the anafunctors are. However, one would not expect to get a stack until the plus construction had been applied 3 times to $G$.
So, my question is: assuming these constructions really are doing the same thing, what about Lie groupoids lets us get away with only applying the plus construction once? Is this a general phenomenon with internal groupoids/geometric stacks, or is ther something special about the underlying site here?
| https://mathoverflow.net/users/139911 | Anafunctors vs the plus construction | The long-expected answer. $\DeclareMathOperator{\op}{op} \DeclareMathOperator{\Cat}{\mathbf{Cat}}\DeclareMathOperator{\Gpd}{\mathbf{Gpd}} \DeclareMathOperator{\disc}{disc}\DeclareMathOperator{\pr}{pr}$
Suppose for simplicity, we are in a site $(S,J)$ equipped with a subcanonical singleton pretopology $J$. This means every covering family consists of a single map that is an [effective epimorphism](https://ncatlab.org/nlab/show/effective+epimorphism) (the kernel pair exists because of the axioms of a pretopology, and the subcanonicity implies the map is a coequaliser of its kernel pair). For the purposes of your question, the category of smooth manifolds with surjective submersions as covers is such a site. Another way to get such a site is to consider an arbitrary subcanonical and *[superextensive](https://ncatlab.org/nlab/show/superextensive+site)* pretopology, then form the singleton pretopology consisting of the coproducts of the maps in the covering family. I'll come back to this later.
Recall that the definition of a prestack (for example Vistoli's *[Notes on Grothendieck topologies, fibered categories and descent theory](https://arxiv.org/abs/math/0412512)* Definition 4.6(i)) on an arbitrary site $(S,J)$ is a presheaf $F\colon S^{\op} \to \Cat$ such that for each object $M$ and covering family $\mathcal{U}$ of $M$ the comparison functor
$$
F(M) \to Desc\_F(\mathcal{U})
$$
to the category of [descent data](https://stacks.math.columbia.edu/tag/02ZC) is fully faithful. Let us spell out the definition of $Desc\_F(\mathcal{U})$ in the case that we have an internal groupoid $G$, and $F(M) = よ(G)(M) := \Gpd(S)(\disc(M),G)$, for $\disc$ denoting the internal groupoid with no non-identity arrows, and we have a singleton cover $j\colon U\to M$ in the pretopology $J$ as above.
An object of this category (following [Tag 026B](https://stacks.math.columbia.edu/tag/026B).(1)) is a internal functor $f\_0\colon \disc(U)\to G$ (equivalently, an arbitrary arrow $U\to G\_0$ of $S$, for $G\_0$ the objects of $G$), equipped with a natural isomorphism
$$
f\_1\colon f\_0\circ \pr\_1 \Rightarrow f\_0\circ \pr\_2,
$$
where $\pr\_i\colon U\times\_M U\to U$ projects on the $i^{th}$ coordinate. Such a natural iso is given by the data of an arrow $f\_1\colon U\times\_M U \to G\_1$ such that $(s\circ f\_1,t\circ f\_1) = (f\_0\circ \pr\_1,f\_0\circ \pr\_2)$. We also require that $f\_1$ satisfies as associativity constraint:
$$
\pr\_{23}^\*f\_1 \cdot \pr\_{12}^\*f\_1 = \pr\_{13}^\*f\_1\colon f\_0 \circ\pr\_1\Rightarrow f\_0\circ \pr\_3
$$
between these two functors $\disc(U\times\_M U\times\_M U) \to G$. Here $\cdot$ denotes the pointwise *internal* composition in $G$. This makes $f\_0$ and $f\_1$ the data of an internal functor from the Čech groupoid $\check{C}(U\to M)$ of the given cover, to $G$.
However, such a thing is precisely the data of an anafunctor $(U,f)\colon \disc(M)\leftarrow \check{C}(U\to M) \to G$! So we can identify the objects of the descent category with anafunctors, but this is not perhaps as helpful as it seems.
Next, let us calculate what the morphisms $(U,f)\to (U,g)$ are in the category (really, groupoid) of descent data (following [Tag 026B](https://stacks.math.columbia.edu/tag/026B).(2)). The data is a natural transformation $a\colon f\_0\Rightarrow g\_0\colon \disc(U)\to G$ (again, equivalently, a morphism $U\to G\_1$ in $S$ etc) satisfying the constraint that
$$
\pr\_2^\*a \cdot f\_1 = g\_1\cdot \pr\_1^\*a\colon f\_0\circ\pr\_1 \to g\_0\circ\pr\_2.
$$
This is precisely the definition of internal natural transformation between the two functors $f,g\colon \check{C}(U\to M)\to G$. Now this is *not* the same thing as a transformation between the anafunctors $(U,f)$ and $(U,g)$, the data of which would be a natural transformation $f\circ \pr\_1 \Rightarrow g\circ \pr\_2\colon \check{C}(U\times\_M U\to M) \to \check{C}(U\to M) \to G$.
Composition in this category of descent data agrees with the usual way of composing natural transformations, so that $Desc\_{よ(G)}(U\to M) = \Gpd(S)(\check{C}(U\to M),G)$. Moreover, the comparison functor $よ(G)(M) \to Desc\_{よ(G)}(U\to M)$ agrees with the precomposition functor $j^\*\colon \Gpd(S)(\disc(M),G) \to \Gpd(S)(\check{C}(U\to M),G)$.
So to show that $よ(G)$ is a prestack, we need to consider the action of the functor $j^\*$ on hom-sets. Fix then two morphisms $h,k\colon M \to G\_0$ of $S$ ($\Leftrightarrow$ functors $\disc(M)\to G$), and a natural transformation
$a\colon h\circ j \Rightarrow k\circ j$. First, note that the arrow component $(h\circ j)\_1\colon U\times\_M U \to G\_1$ is just $e\circ h \circ j'$ for $e\colon G\_0\to G\_1$ the unit map, and $j'\colon U\times\_M U\to M$ the projection. Similarly, $(k\circ j)\_1 = e\circ k \circ j'$.
The natural transformation is $a$ is given by the data of a function (abusing notation slightly) $a\colon U\to G\_1$ such that $\pr\_2^\*a \cdot (h\circ j)\_1 = (k\circ j)\_1\cdot \pr\_1^\*a$. Recall that here $\circ$ is composition of arrows in $S$, and $\cdot$ is pointwise composition of internal morphisms in $G$. Keeping this in mind, and using the calculation earlier, this gives
$$
\pr\_2^\*a \cdot (e\circ h \circ j') = (e\circ k \circ j')\cdot \pr\_1^\*a
$$
as natural transformations between functors $\check{C}(U\times\_M U)\to G$. But this amounts to saying two particular arrows $U\times\_M U \to G\_1$ are equal. What are these arrows?
Unwinding the above equation, each side is the internal composition of the component of a natural transformation with an identity arrow (because of the unit map $e$). But we need to calculate this in a slightly easier way.
We are now going to reason using generalised elements to check the two arrows $U\times\_M U \to G\_1$ are in fact equal.
Suppose that $(u\_1,u\_2)$ is a generalised element of $U\times\_M U$, satisfying $j(u\_1) = j(u\_2) =:m$. Then $\pr\_2^\*a(u\_1,u\_2) = a(u\_2)$ and $\pr\_1^\*a(u\_1,u\_2) = a(u\_1)$, generalised elements of $G\_1$. Now $a(u\_2) \colon h(j(u\_2)) \to k(j(u\_2))$, which is just $a(u\_2) = h(m) \to k(m)$ and likewise $a(u\_1) \colon h(m)=h(j(u\_1)) \to k(j(u\_1)) = k(m)$, as arrows in $G$. So $a(u\_1)$ and $a(u\_2)$ are parallel arrows. Then the condition they need to satisfy is nothing other than $a(u\_2)\cdot e\_{h(m)} = e\_{k(m)}\cdot a(u\_1)$, which is to say that $a(u\_2) = a(u\_1)$. This is true for arbitrary generalised elements, so by Yoneda we have $a\circ \pr\_2 = a\circ\pr\_1$, so that $a\colon U\to G\_1$ coequalises the parallel pair $\pr\_1,\pr\_2\colon U\times\_M U\rightrightarrows U$. But since $j$ is the coequaliser of this parallel pair, by subcanonicity of $J$, we have a *unique* $a'\colon M\to G\_1$ such that $a = j^\*a'$.
Thus $j^\*$ is fully faithful, and hence $よ(G)$ is a prestack.
Now, what if we don't want to use the singleton pretopology, but a general subcanonical superextensive pretopology? For instance, consider the case of the site of manifolds with open covers for covering families, or more generally, collections of submersions that are jointly-surjective. This is very close to, but not identical with, the site with only singleton covering families. In particular, one needs to consider the covering families given by coproduct injections, and of course everything generated from the combination of the two types of covering families. But, and this is the good point, an internal functor $\check{C}(\coprod\_{i\in I}U\_i\to M)\to G$ (given the extensivity of the ambient category $S$) is still equivalent to an object of the category of descent data! This is because a map $\coprod\_{i\in I}U\_i \to G\_0$ is equivalent to a collection of maps $\{U\_i \to G\_0\}\_{i\in I}$, and the definition of the descent category's objects in the non-singleton case amounts to defining a functor as above, and vice-versa. Similarly, the data of the natural transformation giving a morphism of descent data in the singleton case is equivalent to what it should be in the non-singleton case. The rest of the calculation goes through as before. One might also consider $\kappa$-extensivity for any given arity class $\kappa$ (eg finite extensivity, small extensivity, or even the degenerate case where we only consider "1-extensivity", which amounts to the original singleton case), and this all works fine.
So, to sum up: for an arbitrary subcanonical superextensive site $(S,J)$, the presheaf $よ(G)$ is a prestack for *any* internal groupoid $G$ in $S$ (so, for instance, one might wish to put some constraints of the source and target maps of $G$, but this is irrelevant to the above calculation). Note in particular that I haven't assumed $S$ has finite limits.
After this, one might want to consider the non-superextensive case. I believe that careful book-keeping should allow the above to go through, keeping track of the various components. In the end, one will use that the representable presheaf associated to the object $G\_1$ of $S$ is actually a sheaf, and the last calculation given checks that one has descent data for that sheaf. So I'm willing to claim one only needs $(S,J)$ subcanonical for this to work, and that the data of a topology is fine to use, rather than a pretopology. And, further, that really one only needs to know that the presheaf $よ(G\_1)$ (of sets) is a sheaf for the (pre)topology to get that the presheaf $よ(G)$ (of groupoids) is a prestack.
The only other generalisation I haven't touched on is when $G$ is instead an internal *category* that's not a groupoid. I believe this case should work too, since there was nothing in the above that actually *used* the fact $G$ was a groupoid. The category of descent data is then a category that's not a groupoid, but that doesn't affect anything, and I think the proof actually goes through unchanged. So, barring some tedious checking of book-keeping, I think I can claim:
**Theorem:** Given a site $(S,J)$ and an internal category $C$ in $S$ such that the representable presheaf on $S$ associated to the object of arrows $C\_1$ is a $J$-sheaf, the presheaf $よ(C)$ of categories is a prestack.
Now, to go back to the original question, this means that one only needs to apply the Grothendieck $+$-construction *once* to $よ(C)$ in order to get a stack on $(S,J)$.
Moreover, at least under the assumption that $(S,J)$ is subcanonical, and $J$ is singleton (or at the very least, superextensive), the [bicategory of internal anafunctors construction](http://www.tac.mta.ca/tac/volumes/26/29/26-29abs.html) gives a localisation of the 2-category $\Cat(S)$ (and analogously for the full sub-2-category $\Gpd(S)$) at the internal functors $\check{C}(U\to M) \to \disc(M)$ (using the more general version in *[The elementary construction of formal anafunctors](https://doi.org/10.52547/cgasa.15.1.183)*, using Example 2.11). ~~And we also have that the 2-category of stacks is *also* a localisation (from Pronk and Warren, *[Bicategorical fibration structures and stacks](http://www.tac.mta.ca/tac/volumes/29/29/29-29abs.html)*), and so the anafunctor bicategory and the 2-category of stacks are **equivalent** (here recalling we are in a subcanonical site, which Pronk and Warren assume). As a result objects of the the anafunctor bicategory essentially arise from performing the Grothendieck $+$-construction just once.~~
**EDIT** I need to correct that last statement. What is true is that the bicategory of internal categories, anafunctors and transformations is a localisation of the 2-category of internal categories, functors and transformations, and so is the 2-category of *presentable* stacks. Pronk and Warren's result is about localising the 2-category of prestacks (which the representable ones of the form $よ(G)$ form a full sub-2-category) at the local equivalences to get the 2-category of stacks. The local equivalences in the case of representable prestacks amount to (IIRC) the internal weak equivalences as in my work. The essential image of the 2-category of representable prestacks in the 2-category of all stacks under the localisation 2-functor of Pronk and Warren should be the 2-category of presentable stacks (I say should be, because I think I once checked this, and have promised to publish this result for a long time!). So you can combine all these ingredients to get that the 2-category of presentable stacks is equivalent to the anafunctor bicategory. It's a story I really need to write up properly one day, and it is to my shame that it hasn't happened yet...
| 4 | https://mathoverflow.net/users/4177 | 452456 | 181,852 |
https://mathoverflow.net/questions/452423 | 1 | The following equation may be meaningful, but how can we make it well-defined
$$\delta(x-a)\cdot\delta(x-b)=0$$
**Question**: How do we defined this equation? Or more broadly define product between generalized functions with certain restrictions. Whether this definition satisfies the product rule [$D(fg)=Df\cdot g+f\cdot Dg$].
---
**Added**: Maybe theory of hyperfunction can explain this, which I am not familiar with. Should its product satisfy the product rule?
| https://mathoverflow.net/users/510003 | Product of Dirac delta function | You can, of course, if you wish, consult the highly sophisticated treatise of Hörmander to verify that your result holds (for distinct values of $a$ and $b$—there is no serious text which claims this for the case $a=b$). Or you can refer to work which precedes this by decades and is completely elementary, in order to verify it as follows, without the use of functional analysis (to simplify the notation, we assume that $a=0$ and $b=1$). Then it is clear that the product exists and is equal to $0$ on each of the intervals $]-\infty,\frac 2 3[$ and $]\frac 1 3,\infty[$. We now piece these two distributions together (trivial case of “recollement des morceuses”) to get the zero distribution on the line.
We remark that the very elementary theory which justifies these simple manipulations is easily available online (reference below), where you will also find explicitly the following generalisations—just as easily proved:
1. if $f$ is a distribution on the line which equals a smooth function on a neighbourhood of $0$, then $f\delta\_0$ and indeed $f\delta\_0^{(n)}$ exist and are equal to zero if $f$ vanishes there.
2. If $a$ and $b$ are distinct, then $\delta^{(n)}(x-a)\delta^{(m)}(x-b)=0$.
The reference is to the site <https://jss100.campus.ciencias.ulisboa.pt> of the late portuguese mathematician J. Sebastião e Silva where you will find his text “Theory of Distributions” (the above results are in the fourth chapter “[Multiplication and Change of Variables](http://jss100.campus.ciencias.ulisboa.pt/Publicacoes/Textos-Didaticos-/Ens-JSS/TexDida_V3_III.1%20Theory%20of%20Distributions%20Chapter%204.pdf)”) under the "[Textos Didáticos](http://jss100.campus.ciencias.ulisboa.pt/Publicacoes/publicacoes.html)".
In response to the above request for an explicit formula for the product of a $C^n$ function and a distribution of order $n$, I can’t comment there but it is:
$$fg=\sum\_{k=0}^{n} \binom n k D^{n-k}(f^{(k)}G)
$$
where $f$ is $C^n$ and $g$ is a distribution of order $n$ of the form $D^n G $ with continuous $G$ ($D$ is the distributional derivative).
This is 4.1.2 in the reference given here. The most used version, where $g=\delta^{(n)}$, is 4.2.3.
| 4 | https://mathoverflow.net/users/510259 | 452459 | 181,854 |
https://mathoverflow.net/questions/452405 | 6 | Let $A$ be a C\*-algebra, and consider the infinite tensor power $A^{\otimes {J}}$, where $J$ is infinite (we consider the minimal or maximal tensor product). To any finite permutation, which is a bijection $\sigma\colon J\to J$ fixing all but finitely many elements of $J$, we can associate a \*-homomorphism $\hat{\sigma}\colon A^{\otimes {J}} \to A^{\otimes J}$ which sends the $i$-th copy of $A$ (in $A^{\otimes J}$) to the $\sigma(i)$-th copy of $A$.
Is there an element $x \in A^{\otimes J}$, other than $\lambda 1$ with $\lambda \in \mathbb{C}$, such that $\hat{\sigma}(x)=x$ for every finite permutation?
($\lambda 1$ is the trivial case: $\hat{\sigma}(\lambda 1) = \lambda \hat{\sigma}(1) = \lambda 1$).
---
From what I could do, it seems there is no such element, but I am unable to work out the details. Here are some inherent remarks.
1. If we consider $A^{\odot J}$, the pre-C\*-algebra given by the algebraic infinite tensor power, every element $x$ is represented in some $A^{\odot F}$ with $F$ finite (i.e., it's an element of the form $x \odot 1 \odot 1 \odot\dots$). Therefore, any $x \ne \lambda 1$ cannot satisfy $\hat{\sigma}(x)=x$.
2. Let $J= \mathbb{N}$. Consider $x \in A$, and imagine to construct $x^{\otimes \mathbb{N}}$. If $\lVert x \rVert<1$, we notice that $x^{\otimes n} \to x^{\otimes \mathbb{N}}$ as $n$ increases, so $\lVert x^{\otimes n}\rVert \to \lVert x^{\otimes \mathbb{N}}\rVert$. However, $\lVert x^{\otimes n} \rVert= \lVert x\rVert^n \to 0$: we conclude that $x^{\otimes N} =0$, which is a trivial case.
---
Some references:
Bruce E. Blackadar. Infinite tensor products of C\*-algebras. Pacific Journal of Mathematics,
72(2):313–334, 1977.
Tobias Fritz and Eigil Fjeldgren Rischel. Infinite products and zero-one laws in categorical probability. Compositionality 2. 2020.
| https://mathoverflow.net/users/510216 | Is there an element in an infinite tensor power of a C*-algebra that is invariant under finite permutations? | Thanks to the comments of @DiegoMartinez and @CalebEckhardt, I can answer my question.
Briefly, the answer is no: any $x$ satisfying the hypothesis is of the form $\lambda 1$.
Let us consider $x \in A^{\otimes J}$, where $J$ is infinite, such that $\hat{\sigma}(x) = x$ for every finite permutation $\sigma$. As suggested by @DiegoMartinez, we can proceed as follows. By definition of $A^{\otimes J}$, there is a sequence $a\_n \in A^{\otimes F\_n}\subset A^{\otimes J}$, where $F\_n$ is a finite set, such that $a\_n \to x$. For clarity, let us consider $\epsilon\_n>0$ such that $\lVert x-a\_n \rVert \le \epsilon\_n$ and $\epsilon\_n \to 0$.
We pick a finite permutation $\sigma$ such that $\sigma(F\_n)\cap F\_n=\emptyset$. Then
$$\lVert a\_n - \hat{\sigma}(a\_n)\rVert\le \lVert a\_n - x + \hat{\sigma} (x) - \hat{\sigma}(a\_n) \rVert \le \lVert a\_n - x \rVert + \lVert \hat{\sigma} \rVert \lVert x-a\_n \rVert \le 2 \epsilon\_n,$$
where we used $\hat{\sigma}(x) = x$ and $\lVert \hat{\sigma} \rVert=1$.
We now adopt a strategy similar to the one suggested by @CalebEckhardt. For any state $\Phi\colon A^{\otimes J} \to \mathbb{C}$, we notice that
$$\lVert \Phi(\hat{\sigma}(a\_n)) - \Phi(x)\rVert \le \epsilon\_n$$
for any finite permutation $\sigma$ (because $x= \hat{\sigma}(x)$).
Let us fix $\Phi$. Using the CPU map $\Psi := \Phi\_{\mid A^{\otimes \sigma(F\_n)}} \otimes \operatorname{Id}\_{A^{\otimes J\setminus \sigma(F\_n)}} \colon A^{\otimes J} \to A^{\otimes J\setminus \sigma(F\_n)}$, we obtain
$$\lVert a\_n - \Phi(\hat{\sigma}(a\_n))1 \rVert=\lVert \Psi (a\_n -\hat{\sigma}(a\_n)) \rVert\le\lVert a\_n -\hat{\sigma}(a\_n) \rVert\le 2\epsilon\_n.$$
Finally,
$$ \lVert x- \Phi(x)1 \rVert \le \lVert x-a\_n\rVert + \lVert a\_n - \Phi(\hat{\sigma}(a\_n))1 \rVert + \lVert \Phi(\hat{\sigma}(a\_n))1 - \Phi(x)1 \rVert \le 4 \epsilon\_n \to 0,$$
from which we conclude that $x=\Phi(x)1$.
| 3 | https://mathoverflow.net/users/510216 | 452472 | 181,858 |
https://mathoverflow.net/questions/452474 | 0 | Fix $p \in [1, \infty)$. Let $(L^p (\mathbb R^d), \|\cdot\|\_{L^p})$ be the Lesbesgue space of $p$-integrable real-valued functions on $\mathbb R^d$. Let ${\tilde L}^p (\mathbb R^d)$ be the space of Lebesgue measurable functions $f:\mathbb R^d \to \mathbb R$ such that
$$
\|f\|\_{\tilde L^p} := \sup\_{x \in \mathbb R^d} \|1\_{B(x, 1)} f\|\_{L^p} < \infty,
$$
where $B(x, 1)$ is the open unit ball centered at $x$.
>
> Is $(\tilde L^p (\mathbb R^d), \|\cdot\|\_{\tilde L^p})$ separable?
>
>
>
Thank you so much for your elaboration!
| https://mathoverflow.net/users/99469 | Is the space $L^p_{\text{loc}} (\mathbb R^d)$ separable w.r.t. the norm $\|f\|_{\tilde L^p} := \sup_{x \in \mathbb R^d} \|1_{B(x, 1)} f\|_{L^p}$? | $\newcommand\Z{\mathbb Z}\newcommand\R{\mathbb R}\newcommand\J{\mathcal J}$No. Consider first the case $d=1$. Let $\J$ denote the set of all subsets of $\Z$.
For $J\in\J$, let
$$f\_J:=\sum\_{j\in J}1\_{[j,j+1)},$$
so that $f\_J\in\tilde L^p(\R^d)$.
For any two distinct $J$ and $K$ in $\J$, we have $\|f\_J-f\_K\|\_{\tilde L^p(\R^d)}\ge1$. Since the set $\J$ is uncountable,
your space $\tilde L^p(\R^d)$ is not separable. (Indeed, if $S$ is a dense subset of $\tilde L^p(\R^d)$, then for each $J\in\J$ there is some $s\_J\in S$ such that $\|f\_J-s\_J\|\_{\tilde L^p(\R^d)}<1/2$. So, by the norm inequality, the $s\_J$'s must be pairwise distinct, so that the set $S$ is uncountable.)
The case $d\ge2$ is similar: then partition $\R^d$ into cubes rather than intervals.
| 3 | https://mathoverflow.net/users/36721 | 452475 | 181,859 |
https://mathoverflow.net/questions/452469 | 5 | In rational homotopy theory, one can study the rational homotopy and cohomology categories via an algebraic structure, respectively the rational Lie Algebra model and the Sullivan cdga model.
If I am not wrong, over a prime there is a similar theorem for $\mathbb{F}\_p$ cohomology, that is Mandell's Theorem: under adequate assumptions, $\mathbb{F}\_p$ cohomology together with its $E\_{\infty}$ structure does a job analogous to Sullivan cdga in rational homotopy theory.
I am definitely not confident in this field, but I don't understand if $p$-homotopy groups $\mathbb{F}\_p \otimes \pi\_n$ (or some $p$-adic friend) are determined by Mandell's model or not, and what is the analog of the Lie model.
What I want in the end is a generalization of the notion of coformality to the $p$-adic world, which I recall: $X$ is said to be coformal if there exists a (zig-zag of) quasi-isomorphism connecting the rational homotopy groups of $\Omega X$ with the free Lie Algebra on $C^\* X$.
| https://mathoverflow.net/users/140013 | Analogues of Sullivan Theory at a prime for coformality | Mandell shows that, under some hypotheses, the $\mathbb{F}\_p$-cochains detect the $\mathbb{F}\_p$-homotopy type in the sense that there is such an equivalence $X \simeq Y$, if and only if, there is an equivalence of $E\_\infty$-algebras $C\_\*(X; \mathbb{F}\_p) \simeq C\_\*(X;\mathbb{F}\_p)$. This is different than saying that $\mathbb{F}\_p$-homotopy theory is equivalent to the homotopy theory of $E\_\infty$-algebras in $\mathbb{F}\_p$-cochain complexes because it says nothing about essential surjectivity or equivalence of mapping spaces. Nonetheless, it is still reasonable to call a space $p$-formal if there is an equivalence of $E\_\infty$ algebras $C\_\*(X;\mathbb{F}\_p) \simeq H\_\*(X;\mathbb{F}\_p)$, and this is a useful, if rarely satisfied, condition. As an aside, Mandell showed this result actually be improved to an equivalence of homotopy theories if one instead takes coefficients in $\bar{\mathbb{F}}\_p$.
Mandell also shows that our only guess for the Lie model of $\mathbb{F}\_p$-homotopy theory, the Koszul dual of $C\_\*(X;\mathbb{F}\_p)$ is actually contractible. I expect this result implies there is not an easy way to detect homotopy groups from the $E\_\infty$-algebra model.
All is not lost! The correct way to model $p$-torsion information with Lie algebras was demonstrated by Heuts. Using chromatic homotopy theory, one can construct the $v\_n$-localization of spaces and spectra. These localizations can see $p$-torsion information when $n>0$ and when $n=0$ it coincides with rationalization. Heuts showed that $v\_n$-local spaces are modeled by Lie algebras in $v\_n$-local spectra. I don't think there is a definition of coformality in this context, but I would be very interested to see one.
| 7 | https://mathoverflow.net/users/134512 | 452480 | 181,860 |
https://mathoverflow.net/questions/452464 | 1 | Maple seems to suggest the following formula for $n>0$, $p \le q$:
\begin{align}
\frac{d^n}{d x^n} & {}\_p F\_q (a\_1,\ldots,a\_p;b\_1,\ldots,b\_q;1/x) \\[8pt]
= {} & (-1)^n \hspace{1pt} n!\hspace{2pt} \frac{\prod\_j a\_j}{\prod\_k b\_k}\hspace{2pt} x^{-n-1} {}\_{p+1} F\_{q+1} (n+1, a\_1+1, \ldots, a\_p+1; 2, b\_1+1, \ldots, b\_q+1; 1/x)
\end{align}
It also holds for $n=0$ if $1$ is added to the right hand side. As an example, for $n\ge 1$:
$$ \frac{d^n}{d x^n}\hspace{1pt} \cosh\big(2/\hspace{-1pt}\sqrt{x}\big) = (-1)^n \hspace{1pt} n!\hspace{2pt} 2\hspace{1pt} x^{-n-1} {}\_1 F\_2 (n+1\hspace{1pt}; 2,3/2\hspace{1pt}; 1/x)
$$
Does anyone know how this is derived? I haven't seen that formula in classic refs, wikis, or Wolfram. I think it must be based on Cauchy's integral formula, then a change of variable to turn it into a sort of Euler's integral formula for $\_{p+1} F\_{q+1}$ as a Beta scale mixture of $\_p F\_q$. I'm not sure how the details of the complex integration would work.
For $n=1$ it's the ordinary derivative formula with the chain rule. So I guess an induction argument would work.
However there is a very interesting more general formula the relates:
$$ \frac{d^n}{d x^n}\hspace{1pt}\normalsize \_p F\_q (\ldots;\ldots;x^{-m})$$
to a single term involving:
$$ \_{p+m} F\_{q+m} (\ldots;\ldots;x^{-m})$$
The additional $2 m$ parameters follow a simple pattern. Maybe this would follwo from an iteration for multiple poles in the Cauchy integral due to $m>1$? It doesn't seem to be easily amenable to induction.
There doesn't seem to be a similar formula for integral $m>1$, just the usual increase in terms with $n$. Though it's not clear--Maple returns a single term for the general $n$th derivative with positive integral $m$, but it evaluates to zero (and is incorrect).
For $x^{m/2}$ argument, with $m$ a positive integer, there seems to be an even more interesting single term formula as well, relating:
$$ \frac{d^n}{d x^n}\hspace{1pt}\normalsize \_p F\_q (\ldots; \ldots; x^{m/2})$$
to a single term involving:
$$ \_{2p+m-1}\hspace{-1pt} F\hspace{0.5pt}\_{2q+m} (\ldots;\ldots;4^{p-q-1} x^m)$$
where the *F* parameters are halved and incremented by 1 and 1/2, doubling their number, in addition to $2m-1$ patterned fraction parameters. It would be very interesting to know how that is derived.
Edit: the $x^{m/2}$ derivative actually has two terms, not 1. This seems to arise from doing termwise differentiation and then dividing the series into even and odd parts so that the $(mk/2+...)$ coefficients become $(mk+...)$ and $(mk+1/2+...)$ in the new series, which can then be expanded using the Gauss multiplication theorem. Similarly any rational power $m/l$.
| https://mathoverflow.net/users/510206 | $n$th Derivative of $_p F_q(a_1,...,a_p; b_1,...,b_q;x^{-m})$, $p \le q$ | The first displayed identity can be verified in a straightforward manner, by differentiating the power series for ${}\_{p} F\_{q} (a\_1,...,a\_p;b\_1,...,b\_q;1/x)$ (in the powers of $1/x$) term-wise $n$ times in $x$, and then comparing the coefficients of the resulting power series with the coefficients of the power series (in the powers of $1/x$) on the right-hand side of the identity. (Let me know if details on this are needed.)
Expression for the derivatives of any order of ${}\_{p} F\_{q} (a\_1,...,a\_p;b\_1,...,b\_q;x^c)$ can be obtained similarly, by term-wise differentiation.
| 2 | https://mathoverflow.net/users/36721 | 452490 | 181,862 |
https://mathoverflow.net/questions/452256 | 4 | $\DeclareMathOperator\cl{cl}\DeclareMathOperator\int{int}$A subset $A$ of a topological space $X$ is called regular closed if $A=\cl
\_{X}\int\_{X}A$.
The family of all regular closed sets of a topological space is denoted by $%
\mathcal{R}\left( X\right) $.
An ultrafilter $\mathcal{U}$ on $\mathcal{R}\left( X\right) $ is said to
converge to a point $p\in \beta X$ if $\left\{ p\right\} =\bigcap \left\{ \cl\_{\beta X}U:U\in \mathcal{U}\right\} $.
Lemma: Let $D$ be a dense subspace of a space $X$. Then the map $%
A\rightarrow \cl\_{X}A$ is a Boolean algebra isomorphism from $\mathcal{R}%
\left( D\right) $ onto $\mathcal{R}\left( X\right) $.
I think that the family $\mathcal{F}=\left\{ F\in \mathcal{R}\left( \beta
X\right) :p\in \int\_{\beta X}F\right\} $ is a filterbasis in $\mathcal{R}%
\left( \beta X\right) $. Therefore $\mathcal{F}$ can be imbedded in an
ultrafilter $\mathcal{U}$ in $\mathcal{R}\left( \beta X\right) $. Therefore $%
X\cap \mathcal{U}=\left\{ X\cap U:U\in \mathcal{U}\right\} $ is an
ultrafilter in $\mathcal{R}\left( X\right) $, and it converges to $p$.
My question is: for every $p\in \beta X$, does there exist a unique ultrafilter $%
\mathcal{U}$ in $\mathcal{R}\left( X\right) $ such that $\left\{ p\right\}
=\bigcap \left\{ cl\_{\beta X}U:U\in \mathcal{U}\right\} $, that is $%
\mathcal{U}$ converges to $p$?
Now, let $f:X\longrightarrow Y$ be a continuous map between Tychonoff spaces.
Then the Stone extension $\beta f:\beta X\longrightarrow \beta Y$ is defined as
follows: for $p\in \beta X$, there exists a unique $z$-ultrafilter $\mathcal{%
A}^{p}$ on $X$ with $p$, so is defined by $\left( \beta f\right) \left(
p\right) =\bigcap f^{\#}\mathcal{A}^{p}$, where $f^{\#}\mathcal{A}%
^{p}=\left\{ E\in Z\left( Y\right) :f^{-1}\left( E\right) \in \mathcal{A}%
^{p}\right\} $ (Gillman and Jerison, [Rings of continuous functions](https://doi.org/10.1007/978-1-4615-7819-2), p.85). In another article (K. Srivastava. [On the Stone–Čech compactification of an
orbit space](https://doi.org/10.1017/S0004972700003725)), it is defined by $\left( \beta f\right) \left( p\right)
=\bigcap\_{Z\in \mathcal{A}^{p}}\cl\_{\beta Y}f\left( Z\right) $. I guess
that's not quite right.
If, for every $p\in \beta X$, there exists a unique ultrafilter of regular
closed sets of $X$, then how can I define $\left( \beta f\right) \left(
p\right) $?
| https://mathoverflow.net/users/86099 | Stone–Čech compactification and an ultrafilter of regular closed sets | The family $\mathcal{F}=\{ F\in \mathcal{R}( \beta X) :p\in \operatorname{int}\_{\beta X}F\}$ is indeed a filterbase; it is a base for the neighbourhood filter at $p$. As [noted](https://mathoverflow.net/questions/452256/stone-%c4%8cech-compactification-and-an-ultrafilter-of-regular-closed-sets#comment1169517_452256) in the comments there need not be a unique ultrafilter that extends it; for example in $\beta\mathbb{R}$.
Take a point $p$ in $\mathbb{R}$ then $\mathcal{F}$ is generated by $\{[p-2^{-n},p+2^{-n}]:n\in\mathbb{N}\}$; you can add $(-\infty,p]$ or $[p,\infty)$ to $\mathcal{F}$ and still have a filter base in $\mathcal{R}$. So we have at least two $\mathcal{R}$-ultrafilters that extend $\mathcal{F}$.
Srivastava's definition is identical to the one in *Gillman and Jerison* and hence correct: note that if $Z\in\mathcal{A}^p$ and $E\in
f^\#\mathcal{A}^p$ then $Z\cap f^{-1}[E]\neq\emptyset$, hence $f[Z]\cap E\neq\emptyset$ as well. The latter implies that $\operatorname{cl}\_{\beta Y}f[Z]\cap f^\#\mathcal{A}^p\neq \emptyset$ too. So $\beta f(p)\in\bigcap\{\operatorname{cl}\_{\beta Y}f[Z]:z\in\mathcal{A}^p\}$. Next if $q\neq \beta f(p)$ then take $h:\beta Y\to[0,1]$ with $h(q)=1$ and $h(\beta f(p))=0$.
Then $h\circ f$ is continuous and $Z=\{x:h(f(x))\le\frac12\}$ is a member of $\mathcal{A}^p$, but $q$ is not in the closure of $f[Z]$, hence $\bigcap\{\operatorname{cl}\_{\beta Y}f[Z]:z\in\mathcal{A}^p\}=\{\beta f(p)\}$.
As an answer to the last question: there is already a definition, independent of the uniqueness of the $\mathcal{R}$-ultrafilters.
| 3 | https://mathoverflow.net/users/5903 | 452494 | 181,863 |
https://mathoverflow.net/questions/447782 | 10 | This is a cross-post of [this question from MSE](https://math.stackexchange.com/questions/4617947/serre-fibrations-between-spaces-of-embeddings-reference-request).
---
Given topological manifolds $M$ and $N$ of the same dimension, let $\operatorname{Emb}(M,N)$ denote the subspace of $\operatorname{Map}(M,N)$ consisting of the (topological) embeddings $M\to N$. Here $\operatorname{Map}(M,N)$ is equipped with the compact-open topology.
I am looking for a reference of the following assertion (if it is true):
>
> Let $M$ be a topological manifold of dimension $k$. The evaluation map $$\operatorname{Emb}(\mathbb{R}^k,M)\to M$$ at the origin $0\in\mathbb{R}^k$
> is a Serre fibration (or a Hurewicz fibration, preferably).
>
>
>
Thanks in advance.
| https://mathoverflow.net/users/144250 | Reference request - Fibrations between spaces of embeddings | It is a Serre fibration, and this is one of the rare cases where you can give an elementary argument and don't need something hard like [Edwards-Kirby](https://www.jstor.org/stable/1970753?origin=crossref), [Lees](https://www.projecteuclid.org/journals/bulletin-of-the-american-mathematical-society/volume-75/issue-3/Immersions-and-surgeries-of-topological-manifolds/bams/1183530549.full), or [Lashof](https://projecteuclid.org/journals/illinois-journal-of-mathematics/volume-20/issue-1/Embedding-spaces/10.1215/ijm/1256050168.full). I do not know a reference so let me outline the argument.
We construct a lift in a commutative square
$$\require{AMScd}
\begin{CD}
I^j \times \{0\} @>{f}>> \mathrm{Emb}(\mathbb{R}^k,M) \\
@VVV @V{\mathrm{ev}\_0}VV \\
I^j \times I @>{F}>> M
\end{CD}$$
by finding a continuous map $H \colon I^j \times I \to \mathrm{Homeo}\_c(M)$ such that $H(x,0) = \mathrm{id}\_M$ and $H(x,t)(F(x,0)) = F(x,t)$. Then the desired lift will be given by $(x,t) \mapsto H(x,t) \circ f(x)$.
Subdividing the domain $I^j \times I$ of $F$ into fine enough $(j+1)$-cubes such that $F$ sends each of them into a chart of $M$ homeomorphic to $\mathbb{R}^k$, we will construct $H$ inductively over these cubes and thus may assume that $M = \mathbb{R}^k$. We can then define $H(x,t)$ as given by translation by $F(x,t)-F(x,0) \in \mathbb{R}^k$, cut off by a suitable bump function.
| 4 | https://mathoverflow.net/users/798 | 452495 | 181,864 |
https://mathoverflow.net/questions/452341 | 3 | Suppose I have a functor of quasi-categories $f: \mathcal{C} \to \mathcal{D}$. I want to show a criterion like: "$f$ is an equivalence of $\infty$-categories if the homotopy fiber of $f$ satisfies something". In my case, I have something close to being the identity, yet I don't know the best approach to show it is an equivalence if I don't want to mess up too much with combinatorial stuff.
If we were in the Quillen model structure, it would be almost straightforward: up to substituting $f$ with a Kan fibration, we would have a fiber sequence $|\textrm{hofib}(f) |\to |\mathcal{C}| \to |\mathcal{D}|$ and the criterion would be "if $\textrm{hofib}(f)$ is weakly contractible, then $f$ is a weak equivalence". Does this generalize to the Joyal model structure? I have the impression I am missing some model-categorical proof since the statement is general.
PS Since I believe my $f$ is a Joyal fibration, we can ignore the eventual difference between homotopy fibers in the two contexts (I am starting to doubt the two model categories both underlie the $\infty$-category of simplicial sets).
| https://mathoverflow.net/users/140013 | A fiber-like method to show equivalence of infinity categories | An obvious necessary condition for $f$ to be a categorical equivalence is that $f$ is weakly equivalent to a (co)cartesian fibration of quasicategories, i.e., $f$ is an analogue of a [Street fibration](https://ncatlab.org/nlab/show/Street+fibration) of quasicategories.
Since $f$ is already a Joyal fibration, a plausible course of action is to check right away whether $f$ is a (co)cartesian fibration, i.e., has (co)cartesian lifts for morphisms.
If the above necessary condition holds, then $f$ is a categorical equivalence if and only if its fibers (which are automatically homotopy fibers because $f$ is a Joyal fibration) over every vertex of $\cal D$ are contractible Kan complexes, e.g., by Lurie's straightening-unstraightening theorem (Higher Topos Theory, Theorem 3.2.0.1, combined with Proposition 2.4.2.4).
| 3 | https://mathoverflow.net/users/402 | 452499 | 181,868 |
https://mathoverflow.net/questions/452509 | 20 | Say we have DC-λ where λ is some inaccessible cardinal. Is that enough to develop all of ordinary mathematics? If not, is there a strengthening that is but that nevertheless does not assume full choice high up in the universe of sets?
| https://mathoverflow.net/users/168572 | How much of the axiom of choice do you need in mathematics? | Your hypothesis is in a sense stronger than just assuming ZFC outright.
Namely, if we have $\lambda$-DC for some inaccessible cardinal $\lambda$, and ZF in the background, then in particular, we will have the full axiom of choice inside the universe $V\_\lambda$, consisting of all sets of rank less than $\lambda$, since $\lambda$-DC implies choice for all families of size less than $\lambda$. So $V\_\lambda$ will be a model of full ZFC.
Therefore, by simply jumping inside this universe $V\_\lambda$, we recover full ZFC for all the purposes of "ordinary mathematics." If all ordinary mathematics would take place inside such a universe, then the answer would be yes.
But let me note that it is somewhat subtle to define what one means by inaccessible cardinal in the absence of the axiom of choice, since the usual definition would be that $\lambda$ is an uncountable regular strong limit, but being a strong limit should mean that if $\kappa<\lambda$ then $P(\kappa)<\lambda$ as well, and so in particular, $P(\kappa)$ is well-orderable. But in this case it follows by definition that if $\lambda$ is inaccessible, then the axiom of choice holds in $V\_\lambda$ just as a consequence of inaccessibility.
In other words, we get ZFC in $V\_\lambda$ even without your $\lambda$-DC assumption. In this sense, the power of your hypothesis for ordinary mathematics consists of your having an inaccessible cardinal in the first place, rather than in your having $\lambda$-DC.
**Controversial counterpoint.** Meanwhile, let me also say that my view also is that we would make a fundamental disservice to mathematics and to ourselves by attempting to limit our mathematical conceptions to those ideas that have proved productive in the past, limiting ourselves to the ideas used in "ordinary" realms of mathematics. Set theory has discovered a vast new tranfinite realm of mathematical reality and fundamental principles that govern it, such as the axiom of choice but also large cardinals and many new strong principles with transformative global effects. In my view, the fact that those principles do not apply as much to the older "ordinary" questions do not show the impotence of the new ideas, as much as they show the impotence of the old ideas in capturing the vast new lands before us.
| 32 | https://mathoverflow.net/users/1946 | 452512 | 181,871 |
https://mathoverflow.net/questions/452360 | 2 | Let's take a $G\_2$ manifold $(M,\Phi)$, then we get a Spin$(7)$ manifold by taking $(M\times\mathbb{R},\Psi:=\Phi\wedge dt+\*\_M\Phi),$ where $t$ is the coordinate in the $\mathbb{R}$-direction. $\Phi\in\Omega^3(M),\Psi\in\Omega^4(M\times\mathbb{R})$ determining the corresponding $G\_2$ and Spin$(7)$ structures. Now these structures give us splitting of forms. E.g.,
\begin{align\*}
&\text{On } G\_2: \Lambda^4=\Lambda^4\_1\oplus \Lambda^4\_7\oplus \Lambda^4\_{27},\hspace{1 ex}\Lambda^3=\Lambda^3\_1\oplus \Lambda^3\_7\oplus \Lambda^3\_{27},\\
&\text{On Spin}(7) : \Lambda^4=\Lambda^4\_1\oplus \Lambda^4\_7\oplus \Lambda^4\_{27}\oplus \Lambda^4\_{35}
\end{align\*}
We start with a four form say, $\alpha\in\Omega^4(M\times\mathbb{R}).$ Now the component $\alpha\_7\in\Omega^4\_7(M\times\mathbb{R}).$ Moreover let's take $\alpha=\alpha\_1+\alpha\_2\wedge dt$, where $\alpha\_1\in\Omega^4(M)$ and $\alpha\_2\in\Omega^3(M)$.We can write $\alpha\_7=\gamma+\beta\wedge dt$ (at least pointwise), where $\gamma\in\Omega^4(M)$ and $\beta\in\Omega^3(M)$. My question is what are the components of $\gamma$ and $\beta$ in $\Omega^4(M)$ and $\Omega^3(M)$ in terms of the splitting coming from the $G\_2$ structure.
The question may seem a bit cryptic, I have done the part for a form in $\Omega^2\_7$. Hope this clears things up. The known descriptions for $\Omega^4\_7$ in the literature are not very calculation-friendly. That's where my difficulty arises.
On a $G\_2$ manifold $(M,\Phi)$, $\Omega^2=\Omega^2\_7\oplus \Omega^2\_{14},$ and for a $2$-form $\xi,\xi\_7=\frac{1}{3}\big(\xi+\*(\xi\wedge\Phi)\big).$
On a Spin-$(7)$ manifold $(N,\Psi)$, $\Omega^2=\Omega^2\_7\oplus \Omega^2\_{21},$ and for a $2$-form $\xi,\xi\_7=\frac{1}{4}\big(\xi+\*(\xi\wedge\Psi)\big).$
\begin{align\*}
&\xi\_{\Omega^2\_7(M\times\mathbb{R})}\\
&=\frac{1}{4}\big(\xi+\*(\xi\wedge(\Phi\wedge dt+\*\_M\Phi)\big)\\
&=\frac{1}{4}\big(\xi+\*\_M(\xi\wedge\Phi)+\*(\xi\wedge\*\_M\Phi)\big)\\
&=\frac{3}{4} \xi\_{\Omega^2\_7(M)}+\frac{1}{4}\*\_M(\xi\wedge\*\_M\Phi)\wedge dt
\end{align\*}
| https://mathoverflow.net/users/131004 | Decomposition of forms on a Spin$(7)$ manifold | Observe that, in a $\mathrm{Spin}(7)$-manifold, since $\mathrm{Spin}(7)\subset\mathrm{SO}(8)$, the $\mathrm{Spin}(7)$ decomposition $\Lambda^4 = \Lambda^4\_1\oplus \Lambda^4\_7\oplus\Lambda^4\_{27}\oplus\Lambda^4\_{35}$ must refine the splitting $\Lambda^4=\Lambda^4\_{+}\oplus\Lambda^4\_{-}$ into self-dual and anti-self-dual parts. Since $\Lambda^4\_1$ is self-dual, it follows that
$\Lambda^4\_{+} = \Lambda^4\_1\oplus \Lambda^4\_7\oplus\Lambda^4\_{27}$ and
$\Lambda^4\_{-} = \Lambda^4\_{35}$.
Thus, $\Lambda^4\_7$ consists of self-dual forms, and hence, under the $G\_2$-splitting $\alpha\_7 = \beta\wedge \mathrm{d}t + \gamma$, since $G\_2$ acts irreducibly on the $7$-dimensional representation, we must have $\beta\in\Lambda^3\_7(M)$ and $\gamma = -{\ast}\beta\in \Lambda^4\_7(M)$.
| 2 | https://mathoverflow.net/users/13972 | 452513 | 181,872 |
https://mathoverflow.net/questions/452412 | 3 | Lie's third theorem : Given any finite dimensional Lie algebra $\mathfrak{g}$, there exists a Lie group $G$ whose Lie algebra is equal to $\mathfrak{g}$.
In one of the talks, speaker mentions that this is easier to believe if one think of this in terms of differential graded manifolds.
Think of $\mathfrak{g}$ as a dg-manifold $\mathfrak{g}[1]$. Then, "the associated fundamental group" is the candidate for the Lie group.
Here, paths are to be seen as morphisms of dg-manifolds $T[1]I\rightarrow \mathfrak{g}[1]$, and homotopies are to be seen as morphisms of dg-manifolds $T[1](I\times I)\rightarrow \mathfrak{g}[1]$.
Can some one point me to a reference where this is discussed in detail.
| https://mathoverflow.net/users/118688 | Lie's third theorem via graded geometry | This is the Duistermaat–Kolk construction of a simply connected Lie group that integrates the given Lie algebra $\def\g{{\frak g}}\g$.
The starting observation is that for any simply connected Lie group $G$ the canonical morphism of group objects in diffeological spaces (or smooth sets) $$\def\Hom{\mathop{\rm Hom}}\Hom([0,1],G)\_0/\Hom([0,1]^2,G)\_0→G$$ that evaluates at 1 is an isomorphism.
Here $\Hom$ denotes the internal hom in sheaves of sets on smooth manifolds (or, equivalently, diffeological spaces), equipped with the group structure induced from $G$, and $\Hom([0,1],G)\_0$ denotes the subobject of $\Hom([0,1],G)$ consisting of paths that evaluate to $1∈G$ on 0, i.e., paths $[0,1]→G$ that start at $1∈G$.
Likewise, $\Hom([0,1]^2,G)\_0$ denotes the subobject of $\Hom([0,1]^2,G)$ comprising homotopies that do not move the endpoints, and the initial endpoint stays fixed at $1∈G$.
That is, paths that are homotopic relative endpoints are identified.
A key observation to make now is that both ingredients of the quotient can be recovered from their differentials (or derivatives).
That is, the above isomorphism can rewritten as
$$\Hom([0,1],\g)/\Hom([0,1]^2,\g^2)\_{0,\flat}→G.$$
To ensure that a smooth map $ω\colon[0,1]^2→\g^2$ is the derivative of a homotopy described above, we require that the endpoints stays fixed (encoded by the vanishing of the corresponding partial derivatives, depicted by the subscript $0$) and the curvature form $\def\d{{\rm d}}\d\,ω+[ω,ω]/2$ vanishes, depicted by the subscript $\flat$.
The advantage of this construction is that it is manifestly functorial, easy to describe, and readily generalizes to Lie ∞-algebras and Lie ∞-algebroids.
| 2 | https://mathoverflow.net/users/402 | 452516 | 181,874 |
https://mathoverflow.net/questions/452447 | 4 | Can one get cancellation in exponential sums such as, say,
$$
\sum\_{n\sim N} e(\lfloor n^\theta\rfloor^\beta),
$$
for fixed positive $\theta,\beta\not\in\mathbb Z$? When $\theta < 1$, it seems possible without issue by grouping together values taken on by $\lfloor n^\theta\rfloor$ and then using exponent pairs, but for $\theta > 1$, this obviously doesn't work.
| https://mathoverflow.net/users/40983 | Exponential sum involving floor function | Let me try to provide a partial answer in the case when $0<\beta<2$ much in line with what Terry suggested in the comments. Perhaps it is possible to extend this method to all $\beta>2$ but the computations get more complicated.
Using Taylor expansion and $\lfloor n^\theta\rfloor=n^\theta-\{n^\theta\}$ we obtain
$$
\lfloor n^\theta\rfloor^\beta = n^{\theta\beta}+\mathrm{o}\_{n\to\infty}(1)
$$
when $0<\beta<1$ and
$$
\lfloor n^\theta\rfloor^\beta = n^{\theta\beta}-\beta \{n^\theta\}n^{\theta(\beta-1)}+\mathrm{o}\_{n\to\infty}(1)
$$
when $1<\beta<2$. In the first case, one has $\sum\_{1\leq n\leq N}e(\lfloor n^\theta\rfloor^\beta)\approx\sum\_{1\leq n\leq N}e(n^{\theta\beta})$ and hence cancellation occurs exactly when $\theta\beta$ is not an integer (one can use the Kusmin-Landau inequality/ van der Corput method to obtain good bounds on the cancellation in this case, see for example the book by Graham and Kolesnik: <https://doi.org/10.1017/CBO9780511661976>).
So lets focus on the second case when $1<\beta<2$, which seems more complicated. Using the notation from the comments, let
$$
H(\mathbb{R})=\begin{pmatrix} 1 & \mathbb{R} & \mathbb{R} \\ 0 & 1 & \mathbb{R} \\ 0 & 0 & 1 \end{pmatrix}~~~\text{and}~~~H(\mathbb{Z})=\begin{pmatrix} 1 & \mathbb{Z} & \mathbb{Z} \\ 0 & 1 & \mathbb{Z} \\ 0 & 0 & 1 \end{pmatrix}.
$$
Consider $G=\mathbb{R}\times H(\mathbb{R})$ and $\Gamma=\mathbb{Z}\times H(\mathbb{Z})$ and take their quotient to obtain the nilmanifold $G/\Gamma=\mathbb{R}/\mathbb{Z}\times H(\mathbb{R})/H(\mathbb{Z})$. Consider the function
$$
F\left(u,\begin{pmatrix} 1 & x & z \\ 0 & 1 & y \\ 0 & 0 & 1 \end{pmatrix}\right)= e(u-z+x\lfloor y\rfloor),
$$
which is a Riemann integrable and well defined function on $X$. (This is because $(\{x\},\{y\}, \{z-x\lfloor y\rfloor\})$ is a fundamental domain for the Heisenberg nilmanifold, see here for details: <https://en.wikipedia.org/wiki/Nilmanifold>). Now consider the sequence
$$
g(n)=\left(n^{\theta\beta},\begin{pmatrix} 1 & \beta n^{\theta(\beta-1)} & \beta n^{\theta\beta} \\ 0 & 1 & n^\theta \\ 0 & 0 & 1 \end{pmatrix}\right)
$$
Note that with this choice of $X$, $F$,and $g$ we have
$$
\frac{1}{N}\sum\_{1\leq n\leq N}e(\lfloor n^\theta\rfloor^\beta) \approx \frac{1}{N}\sum\_{1\leq n\leq N}e(n^{\theta\beta}-\beta \{n^\theta\}n^{\theta(\beta-1)})=\frac{1}{N}\sum\_{1\leq n\leq N} F(g(n)).
$$
Due to the main result in <https://arxiv.org/abs/2006.02028> we know that the sequence $g(n)$ is uniformly distributed in the nilmanifold $X$ if and only if $(n^{\theta\beta},\beta n^{\theta(\beta-1)},n^\theta)$ is uniformly distributed in $\mathbb{T}^3=\mathbb{R}^3/\mathbb{Z}^3$. So if neither $\theta\beta$ nor $\theta(\beta-1)$ are integers then the sequence $g(n)$ is uniformly distributed in $X$ and hence
$$
\frac{1}{N}\sum\_{1\leq n\leq N} F(g(n))\rightarrow \int F\,d\mu\_X,
$$
where $\mu\_X$ is the natural (and normalized) Haar measure on $X$. But since $\int F\,d\mu\_X=0$, it follows that
$$
\frac{1}{N}\sum\_{1\leq n\leq N}e(\lfloor n^\theta\rfloor^\beta)\rightarrow 0.
$$
Note that if $\theta(\beta-1)$ is an integer then the sequence $g(n)$ is not uniformly distributed on the entire nilmanifold $X$ but rather distributes uniformly on a sub-nilmanifold $Y\subset X$. But due to the simple nature of $F$, it is not so hard to see that for any such sub-nilmanifold $Y$ one has $\int F\,d\mu\_Y=0$, which in turn still implies cancellation. So the only condition seems to be that $\theta\beta$ is not an integer.
Please let me know if this is helpful or if you have any questions.
| 4 | https://mathoverflow.net/users/510318 | 452535 | 181,880 |
https://mathoverflow.net/questions/452533 | 0 | Related to another question I asked, some questions came up, the most important is the following:
>
> Are there any 4-regular planar graphs without 2-cycles + 3-cycles?
>
>
>
Could someone draw an example if there is one? I couldn't find any paper that answers this question.
| https://mathoverflow.net/users/135618 | Even regular planar graphs without 2-cycles | In a connected planar 4-regular graph without cycles of length less than 4 we have $E=2V$ and $2E\geqslant 4F$ (since every face has at least 4 edges and every edge belongs to at most 2 faces). Thus $2E\geqslant 2V+2F$, contradiction to Euler formula $E+2=V+F$.
| 1 | https://mathoverflow.net/users/4312 | 452537 | 181,881 |
https://mathoverflow.net/questions/452476 | 6 | Let $X={\rm Gr}\_{n,k,{\Bbb R}}$ denote the Grassmannian of $k$-dimensional subspaces in ${\Bbb R}^n$.
We regard $X$ as an ${\Bbb R}$-variety with the set of complex points $X({\Bbb C})={\rm Gr}\_{n,k,{\Bbb C}}$.
We consider the twisted $\Bbb R$-forms $\_c X$ of $X$ where $c\in {\rm Aut\,} X\_{\Bbb C}$ is a $1$-cocycle.
They have the following property: $({}\_c X)\times\_{\Bbb R}{\Bbb C}\simeq X\times\_{\Bbb R}{\Bbb C}$.
>
> **Question 1.** What are the twisted $\Bbb R$-forms $\_c X$ of $X$ having ${\Bbb R}$-points?
>
>
>
>
> **Question 2.** What are references for the version of Question 1
> over an arbitrary field rather than over ${\Bbb R}$?
>
>
>
Concerning Question 1, I have found the following twisted forms $\_c X$,
for which I write the set of ${\Bbb R}$-points $({}\_c X)({\Bbb R})$.
1. ${\rm Gr}\_{n,k,{\Bbb R}}.$
2. For $n=2n',\ k=2k'$, the quaternionic Grassmannian ${\rm Gr}\_{n',k',{\Bbb H}}$
of $k'$-dimensional subspaces in ${\Bbb H}^{n'}$,
where ${\Bbb H}$ denotes the division algebra of Hamilton's quaternions.
3. For $n=2k$, the isotropic Hermitian Grassmannian whose ${\Bbb R}$-points are
the $k$-dimensional subspaces $W\subset{\Bbb C}^n={\Bbb C}^{2k}$ that are totally isotropic
with respect to the Hermitian form $\mathcal H$ with matrix
${\rm diag}(-1,\dots,-1,+1,\dots, +1)$ where $-1$ appears $k$ times
and also $+1$ appears $k$ times.
Here a subspace $W\subset {\Bbb C}^{2k}$ is called totally isotropic
if the restriction of $\mathcal H$ to $W$ is zero.
>
> **Question 3.** Do these exhaust all twisted real forms of $X$ having real points?
>
>
>
| https://mathoverflow.net/users/4149 | Twisted forms with real points of a real Grassmannian | **Updated**
Let $G:={\rm Aut}(X)^0$. Then it is well-known that
$$G\_{\mathbb C}= ({\rm Aut}(X)^0)\_{\mathbb C}= ({\rm Aut}(X)\_{\mathbb C})^0= {\rm Aut}(X\_{\mathbb C})^0\cong {\rm PGL}(n,\mathbb C).$$ So $G$ is a real form of ${\rm PGL}(n,\mathbb C)$ which means that $G$ is isomorphic to ${\rm PGL}(n,\mathbb R)$, ${\rm PGL}(n/2,\mathbb H)$, or ${\rm PU}(p,n-p)$. On the other hand, $X$ has a real point $x$ which means $X\cong G/P$ with $P=G\_x$. It follows from $G\_{\mathbb C}/P\_{\mathbb C}\cong X\_{\mathbb C}\cong {\rm Gr}\_{n,k,\mathbb C}$ that $P\_{\mathbb C}$ is a maximal parabolic which is defined over $\mathbb R$. Looking at the Satake diagram of $G$ it corresponds therefore to a non-compact self-conjugate simple root $\alpha\_k$. This gives the following possibilities
$$
G={\rm PGL}(n,\mathbb R), k=1,\ldots,n-1\quad\Rightarrow\quad\text{your first case}
$$
$$
G={\rm PGL}(n/2,\mathbb H), k=2,4\ldots,n-2\quad\Rightarrow\quad\text{your second case}
$$
$$
G={\rm PU}(p,n-p), k=p=n/2\quad\Rightarrow\quad\text{your third case}
$$
This argument should generalize to arbitrary fields. The first two cases combine to $X=$ space of $k$-dimensional subspaces of $D^n$ where $D$ is a central simple algebra.
| 4 | https://mathoverflow.net/users/89948 | 452543 | 181,883 |
https://mathoverflow.net/questions/452481 | 4 | One may see the modular interpretation of (points of) modular curves in the very first course on modular forms and modular curves. I am wondering if it is well-known that modular interpretation of the stalks of the structure sheaf at points of modular curves, i.e. if $Y$ is an open modular curve parameterising elliptic curves with some level structures and $x$ is a closed point, what can we say about $\mathcal{O}\_{Y,x}$?
| https://mathoverflow.net/users/44005 | Modular interpretation of the stalks of modular curves | I guess I'll make my comment into an answer so this question is no longer marked unanswered.
If the modular curve corresponds to a torsion-free subgroup, then the completion of the etale local ring at $x$ is the universal deformation ring of the object corresponding to $x$. In general, the completion of the etale local ring at a geometric point of a Deligne Mumford stack is the universal deformation ring at that point. The intuition for this is that the unique infinitesimal lifting property of etale morphisms exactly corresponds to the pro-representability property for universal deformation rings.
Thus as long as the etale local ring (i.e., strict henselization of the local ring) at $x$ is the same as the etale local ring of the moduli stack at $x$, then the same statement holds. For example, this is the case if $x$ does not lie above $0,1728$.
| 3 | https://mathoverflow.net/users/15242 | 452551 | 181,884 |
https://mathoverflow.net/questions/452539 | 1 | This is a repost from [MSE](https://math.stackexchange.com/questions/4744769/supremum-or-upper-bound-of-bivariate-function-involving-logarithms-and-combinato) because I got no answers there.
I have been trying to find the supremum of this bivariate function over a specific region. However, the expressions that I get are horrible. I tried Mathematica, but it did not provide a good answer as it gives you something that is clearly not the supremum, as it is smaller than some values I experimented. I hope someone here can help me. I am searching for an upper bound for the function $$\frac{\log\left(\frac{\binom{n}{d}}{2^{d}4}\right)}{d\log(\frac{n}{d})}$$ over the region $\{(n,d)\in\mathbb{N}\mid n>d>0\}.$ I do not need specifically the supremum (although, of course, that would be the best), any (constant) upper bound would also be helpful.
I tried feeding Mathematica with a real extension of this function developing the binomial coefficient using Euler's $\Gamma$ function, which is smooth over the region I am interested in, but Mathematica fails to provide a working upper bound: it gives something that I know it is not an upper bound because I know points in the region having higher value.
Could you help me bounding this function over the region of the integers provided? Does it in fact diverge? If so, how can I see that it diverges?
| https://mathoverflow.net/users/158098 | Supremum or upper bound of bivariate function involving logarithms and combinatorial coefficients or the gamma function over a region of the integers | Note that
\begin{equation\*}
\frac{\ln\frac{\binom{n}{d}}{2^{d}4}}{d\ln\frac{n}{d}}
\le r(n,d):=\frac{\ln\frac{\binom{n}{d}}{2^{d}}}{d\ln\frac{n}{d}};
\end{equation\*}
here and in what follows, $n$ and $d$ are integers such that $1\le d\le n-1$, as in the OP. So, it is enough to bound $r(n,d)$ from above by a universal constant.
[We have](https://en.wikipedia.org/wiki/Binomial_coefficient#Bounds_and_asymptotic_formulas)
\begin{equation\*}
\binom{n}{d}\le\sqrt{\frac{n}{2\pi d(n-d)}}\,2^{nH((d/n))},
\end{equation\*}
where $H(p):=-p\log\_2 p-(1-p)\log\_2(1-p)$.
So, letting
\begin{equation\*}
u:=\frac nd,
\end{equation\*}
we get
\begin{equation\*}
d\ge\max(1,\tfrac1{u-1}) \tag{1}\label{1}
\end{equation\*}
and
\begin{equation\*}
r(n,d)\le r\_1(d,u)+1+r\_3(u), \tag{2}\label{2}
\end{equation\*}
where
\begin{equation\*}
r\_1(d,u):=\frac{\ln\sqrt{\frac{n}{2\pi d(n-d)}}}{d\ln\frac{n}{d}}
=\frac12\frac{\ln\frac u{u-1}-\ln d}{d\ln u}
\end{equation\*}
and
\begin{equation\*}
r\_3(u):=\frac{g(u)-\ln2}{\ln u},\quad g(u):=(u-1)\ln\frac u{u-1}.
\end{equation\*}
Given \eqref{1}, we have the elementary inequalities $r\_1(d,u)\le1/2$ and $r\_3(u)\le1/2$; see details below.
So, by \eqref{2}, $r(n,d)\le2$. $\quad\Box$
**Details on the inequality $r\_1(d,u)\le1/2$:** Note that $r\_1(d,u)$ is decreasing in $d$ such that $r\_1(d,u)>0$. So, by \eqref{1}, (i) $r\_1(d,u)\le r\_1(1,u)=\frac12\frac{\ln u-\ln(u-1)}{\ln u}\le\frac12$ if $u-1\ge1$ and (ii) $r\_1(d,u)\le r\_1(\tfrac1{u-1},u)=\frac12\,(u-1)\le\frac12$ if $u-1\le1$.
**Details on the inequality $r\_3(u)\le1/2$:** Note that $g''(u)=\frac1{u^2(1-u)}<0$ (for $u>1$), so that $g$ is strictly concave. Also, $g(u)\to1$ as $u\to\infty$. So, $g$ is increasing and $g<1$ . Also, $g(2)=\ln2$ and hence for $u\in(1,2]$ we have $g(u)\le\ln2$, so that $r\_3(u)\le0\le1/2$. On the other hand, for $u\ge2$ we have $r\_3(u)\le\frac{1-\ln2}{\ln u}\le\frac{1-\ln2}{\ln2}=0.44\ldots<1/2$.
| 2 | https://mathoverflow.net/users/36721 | 452556 | 181,887 |
https://mathoverflow.net/questions/452547 | 5 | **Problem:**
Given a random isotropic unit vector in $\mathbb{R}^p$ for $p\ge2$, we are trying to compute (preferably exactly, otherwise to upper bound):
$$\mathbb{E}\_{\mathbf{w}\sim\mathcal{S}^{p-1}}\!\left[{w\_1}^{\!4}\,{w\_2}^{\!4}\right]\,.$$
Any help would be greatly appreciated!
---
**A related expectation we have already derived:**
Using an [answer](https://mathoverflow.net/a/449362/100796) to another question, we were able to compute
$\mathbb{E}\_{\mathbf{w}\sim\mathcal{S}^{p-1}}\!\left[{w\_1}^{\!2}\,{w\_2}^{\!2}\right]$ as follows:
\begin{align}1
&=\mathbb{E} \left(\sum\_{i=1}^p w\_i^2\right)^2 \\[10pt]
& =
p(p-1)\cdot
{\mathbb{E}\_{\mathbf{w}\sim\mathcal{S}^{p-1}}\!\left[{w\_{1}}^{\!2}\,{w\_{2}}^{\!2}\right]}+p\cdot \mathbb{E} w\_1^4
\\[10pt]
{\mathbb{E}\_{\mathbf{w}\sim\mathcal{S}^{p-1}}\!\left[{w\_{1}}^{\!2}\,{w\_{2}}^{\!2}\right]}
&
=\frac{1-p\cdot \mathbb{E} w\_1^4}{p(p-1)}\stackrel{(\*)}{=}
\frac{1}{p\left(p+2\right)}\,,\end{align}
where to show $(\*)$ we notice that
$w\_{1}^{2}=\frac{X}{X+Y}$ where $X\sim\chi^{2}\left(1\right)=\Gamma\left(\frac{1}{2},\frac{1}{2}\right)$ and $Y\sim\chi^{2}\left(p-1\right)=\Gamma\left(\frac{p-1}{2},\frac{1}{2}\right)$, and conclude that $w\_{1}^{2}=\frac{X}{X+Y}\sim B\left(\frac{1}{2},\frac{p-1}{2}\right)$ and
$\mathbb{E}\_{w\_1}\left[w\_1^4\right]
=\operatorname{Var}\left[w\_1^2 \right] + \left(\mathbb{E}\_{w\_1} \left[w\_1^2 \right] \right)^2= \dots =\frac{3}{p(p+2)}$.
| https://mathoverflow.net/users/100796 | Expectation of a function of two entries of an isotropic unit vector $\mathbb{E}_{\mathbf{w}\sim\mathcal{S}^{p-1}}\![{w_{1}}^{\!4}\,{w_{2}}^{\!4}]$ | As noted in the [previous answer](https://mathoverflow.net/a/449401/36721),
the joint distribution of $w\_1^2$ and $w\_2^2$ is the [Dirichlet distribution](https://en.wikipedia.org/wiki/Dirichlet_distribution#Definitions) with parameters $1/2,1/2,p/2-1$. [Therefore](https://en.wikipedia.org/wiki/Dirichlet_distribution#Moments), after some simple calculations we get
$$Ew\_1^4w\_2^4=\frac9{p(p+2)(p+4)(p+6)}. $$
| 3 | https://mathoverflow.net/users/36721 | 452559 | 181,889 |
https://mathoverflow.net/questions/452570 | 9 | Today, We call the Kronecker's Jugendtraum Hilbert's 12th problem. But, Hilbert's interpretation of the "Jugendtraum" was not that intended by Kronecker.
And Weber missed his chance to disprove Hilbert's claim in the third volume of his *Lehrbuch der Algebra*.
I read Schappacher's "[On the History of Hilbert's Twelfth Problem: A Comedy of Errors](https://www.emis.de/journals/SC/1998/3/pdf/smf_sem-cong_3_243-273.pdf)" But I can't understand well What Weber was overlooking.
| https://mathoverflow.net/users/167507 | What is Weber's mistake about Hilbert's 12th problem? | It would help to be clearer about where in Schappacher's paper this issue is brought up. I think you are asking about pages 256-258, yes?
The end of the 2nd paragraph on p. 257 says "Translating back to the characterization by ray class groups, Weber overlooked precisely the possibility of choosing different signs in $\pm 1$ modulo different prime factors of $\mathfrak m$." Is that the step you're asking about?
As an elementary analogue, if you want to solve $x^2 \equiv 1 \bmod m$ in integers where $m$ is an odd integer bigger than $1$, the answer is $\pm 1 \bmod m$ when $m$ is an odd prime power, but there are more answers than that mod $m$ when $m$ has more than one odd prime factor: if $m = p\_1^{e\_1}\cdots p\_ke^{e\_k}$, for distinct odd primes $p\_i$, solving $x^2 \equiv 1 \bmod m$ is equivalent to solving $x^2 \equiv 1 \bmod p\_i^{e\_i}$ for all $i$, and that is equivalent to solving $x \equiv \pm 1 \bmod p\_i^{e\_i}$ for all $i$. There are a total of $2^k$ different solutions mod $m$ to that system of congruences, which is more than the trivial solutions $x \equiv \pm 1 \bmod m$ if $k \geq 2$ ($m$ has more than one prime factor).
| 17 | https://mathoverflow.net/users/3272 | 452581 | 181,898 |
https://mathoverflow.net/questions/452584 | 4 | I’ve been looking for a quantitative notion of the Borel-Cantelli lemma along the following lines:
Let $(X,\Omega,p)$ be a probability space, and let $(A\_n)\_n\subseteq \Omega$ be a sequence of measurable sets s.t $p(A\_n)\geq 1-\delta$ for all $n$ (for some small $\delta>0$).
Then of course, the Borel-Cantelli lemma tells us that $p(\limsup A\_n)>0$, and in fact a trivial observation from the proof of the lemma shows $p(\limsup A\_n)\geq 1-\delta$.
However, I am interested in something further, relating to the lower density of sequences of indices to which elements belong: is there $\alpha>0$ s.t
$$p\left(\left\{x\in X: \exists n\_k\uparrow \infty\text{ with }\underline{d}((n\_k)\_k)\geq 1-\alpha\text{ s.t }x\in A\_{n\_k},\forall k\geq 0\right\}\right)\geq 1-\alpha?$$
where $\underline{d}(\cdot)$ denotes the lower-density of a sequence. In particular, can $\alpha>0$ be made small when $\delta>0$ is small?
Intuitively it seems to make sense, but so far no success in proving, and also searching online yielded nothing. Any interesting counter examples are also welcome.
Thanks.
| https://mathoverflow.net/users/70853 | Quantitative Borel-Cantelli | Imagine that for all $k=1,2,\ldots$ all events $A\_n$ for $n\in [k!,(k+1)!-1)$ are the same event $C\_k$. Then if $x$ belongs to all $A\_i$ along a subsequence of positive density yields that it belongs to all but finitely many $C\_k$. It may well appear that there is no such $x$.
| 4 | https://mathoverflow.net/users/4312 | 452587 | 181,900 |
https://mathoverflow.net/questions/452612 | 4 | Let $L$ be a restricted Lie algebra over a field of characteristic $p>0$. It is well known that the commutator subalgebra $[L,L]$ is not necessarily restricted (that is, closed under the $p$-map).
Does there exist any such counterexample where the restricted Lie algebra comes from an associative algebra?
I recall that an associative algebra over a field of characteristic $p>0$ can be regarded as a restricted Lie algebra via the Lie bracket $[x,y]=xy-yx$ and $p$-map given by ordinary $p$-exponentation.
| https://mathoverflow.net/users/17582 | Derived subalgebra of a restricted Lie algebra | There is no such a counterexample, as the derived subalgebra of the restricted Lie algebra associated to an associative algebra over a field of positive characteristic is always restricted.
For instance, this follows from Lemma 4.5 in Chapter 2 of the book "D. Passmann: The algebraic structure of group rings" ([MathSciNet](https://mathscinet.ams.org/mathscinet/article?mr=0798076)).
| 5 | https://mathoverflow.net/users/14653 | 452614 | 181,904 |
https://mathoverflow.net/questions/452600 | 3 | Let $(X,\mathcal B)$ be a matroid on the ground set $X=(x\_1,...,x\_n)$ and with set of bases $\mathcal B$, and let $P\subset\Bbb R^n$ be its [**matroid base polytope**](https://en.wikipedia.org/wiki/Matroid_polytope) (i.e. the convex hull of the characteristic vectors of the bases $B\in\mathcal B$).
The *circumcenter* of $P$ is the unique point $p\in\mathrm{aff}(P)$ that has the same distance to all vertices of $P$.
>
> **Question:** Does $P$ contain its circumcenter, perhaps even in its relative interior?
>
>
>
Here *relative interior* means the interior of $P$ considered as a subset of its affine hull.
**Example.** This is true for uniform matroids as their matrix base polytopes are hypersimplices.
| https://mathoverflow.net/users/108884 | Does a matroid base polytope contain its circumcenter? | I do not think so. Consider a uniform matroid on 3 elements $a, b, c$ of rank 2, and take 100 copies of $a$ (so, totally we have 102 elements). Then the matroid base polytope has full dimension 101, thus the circumcentre has all coordinates $1/51$. But each base has the sum of coordinates corresponding to $b$ and $c$ not less than 1, thus so does any their convex combination.
| 3 | https://mathoverflow.net/users/4312 | 452616 | 181,905 |
https://mathoverflow.net/questions/452623 | 0 | Start by introducing the finite sums
$$A\_n:=\sum\_{m=1}^nq^m\prod\_{j=1}^{m-1}(1-q^j) \qquad \text{and} \qquad
B\_n:=\sum\_{m=1}^nq^m\prod\_{j=m+1}^n(1-q^j).$$
An algebraic proof is *facile*: Clearly, $A\_1=B\_1=q$. Note
$A\_{n+1}=A\_n+q^{n+1}\prod\_{j=1}^n(1-q^j)$ while $B\_{n+1}=B\_n+q^{n+1}-q^{n+1}B\_n$. By induction $A\_n=B\_n$, so it remains to verify $\prod\_{j=1}^n(1-q^j)=1-B\_n$ or $\sum\_{\pmb{{\color{blue}{m=0}}}}^nq^m\prod\_{j=m+1}^n(1-q^j)=1$ or
$\sum\_{m=0}^n\frac{q^m}{(q)\_m}=\frac1{(q)\_n}$. The latter is valid for $n=0$ and if true for $n$ then $\sum\_{m=0}^{n+1}\frac{q^m}{(q)\_m}=\frac{q^{n+1}}{(q)\_{n+1}}+\sum\_{m=0}^n\frac{q^m}{(q)\_m}=\frac{q^{n+1}}{(q)\_{n+1}}+\frac1{(q)\_n}=\frac{q^{n+1}}{(1-q^{n+1})(q\_n)}+\frac1{(q)\_n}=\frac1{(q)\_{n+1}}$. $\square$
It's worth remarking that $A\_{\infty}=B\_{\infty}=1+\sum\_{k\in\mathbb{Z}}(-1)^{k-1}q^{\omega\_k}$ where $\omega\_k=\frac{k(3k+1)}2$ are the *pentagonal numbers*.
>
> **QUESTION.** Can you furnish a direct combinatorial proof for $A\_n=B\_n$?
>
>
>
| https://mathoverflow.net/users/66131 | A combinatorial proof: where art thou? | Consider all partitions $\lambda$ with distinct parts not exceeding $n$ and sum up $(-1)^{r(\lambda)+1}q^{|\lambda|}$, where $r$ goes for the number of parts. You may count the sum by fixing the smallest part, or by fixing the largest part, getting representations $B\_n, A\_n$ respectively.
| 7 | https://mathoverflow.net/users/4312 | 452625 | 181,909 |
https://mathoverflow.net/questions/452621 | 2 | Assume $X=\operatorname{Spec}(A)$ is connected and normal (especially integral), and let $g:\eta \hookrightarrow X$ be the inclusion of the generic point of $X$. In Milne's LEC script on Etale Cohomology (Example 12.4, page 81) is claimed that the stalk of derived direct image sheaf of sheaf $\mathcal{F}$ on etale site $(\eta)\_{et}=\operatorname{Spec}k(X))\_{et} $ in a geometric point $\overline{x}= \operatorname{Spec}k(x)^{\text{sep}} \to X$ is given by
$$ (R^rg\_\* \mathcal{F})\_{\overline{x}} =H^r(\operatorname{Spec}K\_{\overline{x}}, \widetilde{\mathcal{F}}) $$
where where $K\_{\overline{x}}$ is the field of fractions of $\mathcal{O}\_{X,\overline{x}}^{\text{et}}$, where the latter is the stalk of $\overline{x}$ in etale sheaf, ie the inductive limit of the system of connected etale affine neighborhoods $(U,u) \to X$ of $\overline{x}$.
$\widetilde{\mathcal{F}}$ is the pullback of $\mathcal{F}$ with respect the natural composition $\operatorname{Spec} \text{Frac}(\mathcal{O}\_{X,\overline{x}}^{\text{et}})
\to \operatorname{Spec}\mathcal{O}\_{X,\overline{x}}^{\text{et}} \to X$
**Question:** Why this identity holds? Can it be elaborated out in detail?
My considerations: In general we have for more general $g:Y \to X$ the stalk formula
$$ (R^rg\_\* \mathcal{F})\_{\overline{x}} = \varinjlim\_{(U,u) \text{ etale nbhd of } \overline{x}} H^r(U\_Y, U\_Y^\* \mathcal{F}) $$
where as before inductive limit is taken over the system of connected etale affine neighborhoods $(U,u) \to X$ of $\overline{x}$ and $U\_Y:= U \times\_X Y$.
Let's come back to $Y=\eta$ the generic point of $X$. By definition $\operatorname{Spec}\mathcal{O}\_{X,\overline{x}}^{\text{et}}=\varprojlim \_{\text{ etale nbhd of } \overline{x}}(U,u)= \operatorname{Spec} \varinjlim A\_U$
where $U=\operatorname{Spec}A\_U$. Moreover we have $(\varprojlim U) \times\_X \eta=\varprojlim (U\times\_X \eta)$
So once we know that $A\_U \otimes\_A k(X)= \text{Frac}(A\_U)$ we win, and it would be the exactly then the case if the generic fiber $\operatorname{Spec}A\_U \otimes\_A k(X)$ - which is etale over $\eta= \operatorname{Spec} k(X) $ by base change stability of etaleness- would be also *connected*. But why this should be the case here?
Does somebody see why the generic fiber $\operatorname{Spec}A\_U \otimes\_A k(X)$ at $\eta$ of an connected etale affine neighborhoods $(U,u)=\operatorname{Spec}A\_U \to X$ of $\overline{x}$ is connected, ie equals to a Spec of a field? Or at least that evenry such etale $A$-algebra is *dominated* by one whose generic fiber is connected.
If that's not the case, how else one should deduce that $ (R^rg\_\* \mathcal{F})\_{\overline{x}} =H^r(\operatorname{Spec}K\_{\overline{x}},\widetilde{\mathcal{F}} ) $ as claimed above in Milne's script?
| https://mathoverflow.net/users/108274 | Calculate stalk of etale derived pushforward sheaf (Milne's LEC) | This is because $U$ is again normal [Tag [033C](https://stacks.math.columbia.edu/tag/033C)], so it is irreducible since it is connected [Tag [033M](https://stacks.math.columbia.edu/tag/033M)]. Clearly the generic point of $U$ maps to the generic point of $X$, and conversely any point mapping to the generic point of $X$ is generic in $U$ since étale maps are (locally) quasi-finite [Tag [03WS](https://stacks.math.columbia.edu/tag/03WS)]. By irreducibility, there is only one generic point, so there is also only one point in the generic fibre.
| 2 | https://mathoverflow.net/users/82179 | 452629 | 181,911 |
https://mathoverflow.net/questions/451824 | 6 | Fix a complex semisimple Lie algebra $\mathfrak{g}$. Denote by $W$ the corresponding Weyl group, with length function $\ell$ and Bruhat order $\leq$. Let $\lambda$ be an integral anti-dominant weight. In the paper TOWARDS THE KAZHDAN-LUSZTIG CONJECTURE by Gabber and Joseph, they claim that there is a formula, due to D Vogan,
$$P\_{w,w'}(q)=\sum\_{k=0}^{\infty}q^{\frac{\ell(w')-\ell(w)-k}{2}}\dim\operatorname{Ext}\_{\mathcal{O}}^k(M(w\cdot \lambda),L(w'\cdot \lambda))$$
that is equivalent to the Kazhdan-Lusztig conjecture. Here $w'\leq w$ are elements in the Weyl group,$P\_{w,w'}$ is the Kazhdan-Lusztig polynomial, $M(w\lambda)$ is the Verma module of highest weight $w\lambda$ and $L(w'\lambda)$ is the simple module of highest weight $w'\lambda$.
I want to know how to derive this formula from Kazhdan-Lusztig conjecture. Moreover, I want to ask if there is a clear reference (say, by Vogan) on this.
| https://mathoverflow.net/users/466793 | An alternative form of the Kazhdan-Lusztig conjecture | Let $G$ be the simply-connected Lie group with Lie algebra $\mathfrak g$, let $B\subset G$ be the Borel subgroup, and let $X=G/B$ be the flag variety. There are equivalences $$\mathrm{Mod}\_f(\mathfrak g,B,\chi\_\lambda)\xrightarrow{-\otimes\_{U(\mathfrak g)}D\_X}\mathrm{Mod}\_c(D\_X,B)\xrightarrow{DR\_X}\mathrm{Perv}(\mathbb C\_X,B),$$
where the first equivalence is the Beilinson-Bernstein correspondence, and the second equivalence is the Riemann-Hilbert correspondence.
Note that $X$ has a stratification by $B$-orbits, indexed by the Weyl group: for $w\in W$, let $X\_w:=BwB/B$, of dimension $\ell(w)$. Then under the equivalence $M(w\lambda)$ corresponds to $\mathbb C\_{X\_w}[\ell(w)]$ and $L(w\lambda)$ corresponds to $IC(\overline X\_w,\mathbb C\_{X\_w})$.
Moreover, Kazhdan-Lusztig "Schubert varieties and Poincaré duality" shows
$$P\_{y,w}(q)=\sum\_{i}(\dim H^{i-\ell(w)}(IC(\overline X\_w,\mathbb C\_{X\_w}))\_{yB})q^{i/2}.$$
Here,
$$
\begin{align\*}
H^{i-\ell(w)}(IC(\overline X\_w,\mathbb C\_{X\_w})\_{yB})&=\mathrm{RHom}\_{\mathrm{Perv}(\mathbb C\_X,B)}(\mathbb C\_{X\_y},IC(\overline X\_w,\mathbb C\_{X\_w})[i-\ell(w)])\\
&=\mathrm{RHom}\_{\mathcal O}(M(y\lambda),L(w\lambda)[i+\ell(y)-\ell(w)])\\
&=\mathrm{Ext}\_{\mathcal O}^{-i-\ell(y)+\ell(w)}(M(y\lambda),L(w\lambda)).
\end{align\*}$$
(most of this is in Hotta-Takeuchi-Tanisaki *D-modules, Perverse Sheaves, and Representation Theory*.)
| 3 | https://mathoverflow.net/users/123673 | 452633 | 181,913 |
https://mathoverflow.net/questions/452632 | 4 | Show that $f(x)\ge 0$ for $0\le x \le 1$, where:
$$f(x) = \arccos(x)^2 -8x(5x^2-2) \sqrt{1-x^2}\arccos(x)+36 x^8-112 x^6+93 x^4-17 x^2$$
The endpoints are $f(0)=\pi^2/4$ and $f(1)=0$. Plotting verifies that the function is positive, but it is not monotonic, or concave. The maximum is the root of a complicated function, so it is not clear how to separate into sub-intervals. There seems to be no way to separate the function into the sum of two simpler parts that are positive.
The problem is equivalent to showing that $\log\hspace{-1pt}\big( h(r)\big)$ is concave on $(0,1]$, where:
$$h(r) = r^{-1}\Big(\pi -2 (1-2r ) \sqrt{r}\sqrt{1-r}-2 \arccos\big(\hspace{-2pt}\sqrt{r}\big)\Big),\quad 0< r \le 1 $$
Update: based on a comment, the problem can be reduced to showing that $g(x) \le 0$ for $1/2 \le x \le 1$, where:
$$g(x) = x\big(x^6-\mbox{$\frac{7}{3}$} x^4 + \mbox{$\frac{103}{72}$}x^2-\mbox{$\frac{25}{144}$}\big)\sqrt{1-x^2}+\mbox{$\frac{5}{9}$}\big(x^4-\mbox{$\frac{19}{20}$}x^2+\mbox{$\frac{7}{80}$}\big)\arccos(x)$$
Since $g$ is monotonic, this is equivalent to showing that $w(x)\ge 0$ on $[1/2,1]$, where:
$$w(x) = x(40 x^2-19)\sqrt{1-x^2}\arccos(x)-144 x^8+378 x^6-323 x^4+93 x^2-4$$
This may seem somewhat obscure, but it turns out to be important for proving a significant theorem on uniqueness of solutions to unmixing linear combinations of independent random variables. One part of the solution I posted in another question. I believe I have a solution for that, though it involves working with a piecewise function and verifying monotonicity of 60 essentially elementary functions. The other part, involving the integral over the radial direction comes down to proving that this function is positive (derived and extracted from a more complicated expression).
Does anyone know how to show that a function like this is positive, or have an idea that I could pursue? I'm asking here because I have no idea how to proceed.
| https://mathoverflow.net/users/510206 | Non-negativity of a complicated function | We have to show that
\begin{equation\*}
g\overset{\text{(?)}}\le0 \text{ on }[1/2,1].
\end{equation\*}
Note that $p(x):=x^4-\frac{19}{20}x^2+\frac{7}{80}$ is of the same sign as $x-x\_\*$ for $x\in[1/2,1]$, where
$x\_\*:=\frac{1}{2} \sqrt{\frac{1}{10} \left(19+\sqrt{221}\right)}=0.920\dots$.
For $x\in[1/2,1]\setminus\{x\_\*\}$, let
\begin{equation\*}
h(x):=\frac{g(x)}{p(x)}.
\end{equation\*}
It suffices to show that
\begin{equation\*}
h\overset{\text{(?)}}\ge0 \text{ on }[1/2,x\_\*) \tag{10}\label{10}
\end{equation\*}
and
\begin{equation\*}
h\overset{\text{(?)}}\le0 \text{ on }(x\_\*,1]. \tag{20}\label{20}
\end{equation\*}
For $x\in[1/2,1)\setminus\{x\_\*\}$,
\begin{equation\*}
h''(x)=\frac{x h\_2(x)}{28800 p(x)^3\,\sqrt{1-x^2}},
\end{equation\*}
where
\begin{equation\*}
h\_2(x):=-345600 x^{14}+1266240 x^{12}-1823120 x^{10}+1301740 x^8-478566
x^6+85096 x^4-5777 x^2-497<0
\end{equation\*}
for $x\in[1/2,1]$.
So, $h$ is convex on $[1/2,x\_\*)$ and concave on $(x\_\*,1]$.
We have $h(1)=0=h'(1)$. So, \eqref{20} follows.
Also, $h'(3/4)=0.119\ldots>0$ and hence for all $x\in[1/2,x\_\*)$ we have $h(x)\ge h(3/4)+h'(3/4)(x-3/4)
\ge h(3/4)+h'(3/4)(1/2-3/4)=0.102\ldots>0$. So, \eqref{10} follows as well. $\quad\Box$
| 3 | https://mathoverflow.net/users/36721 | 452636 | 181,915 |
https://mathoverflow.net/questions/452074 | 6 | Let $A$ be an analytic subset of a complex manifold $M$ and $O\_{M}$ be the sheaf of complex analytic functions on $M$. The sheaf of ideals $\mathcal{J}\_{A}$ is defined as the subsheaf of $O\_{M}$ whoes stalk at each point $x\in M$ consists of germs of analytic functions vanishing on $A$. If $x\notin A$, then $\mathcal{J}\_{A,x}=O\_{M,x}$.
On page 99 of Demailly's book "[Complex analytic and differential geometry](https://www-fourier.ujf-grenoble.fr/%7Edemailly/manuscripts/agbook.pdf)", he proved following theorem concerning the coherence of the ideal sheaf $\mathcal{J}\_{A}$:
**Theorem([Cartan-Oka 1950]).** $\mathcal{J}\_{A}$ is coherent.
Since coherece is a local property, we can assume that $A\_{x}$(the germ of $A$ at $x$) is irreducible, i.e. $\mathcal{J}\_{A,x}$ is a prime ideal of $O\_{M,x}$. Then the proof of this theorem relies heavily on the fact that we can choose an appropriate coordinate system centered at $x$ such that $O\_{M,x}/\mathcal{J}\_{A,x}$ is a finite integral extension of $O\_{d,x}:=\mathbb{C}\{z\_{1},\dots,z\_{d}\}$ for some integer $d\leq Dim M$. Indeed, it can be shown that for any prime ideal $\mathcal{J}$, we can find some interger $d$ such that $\mathcal{J}\cap O\_{d,x}=\{0\}$ whence $O\_{d,x}$ naturally embedded into $O\_{M,x}/\mathcal{J}\_{A,x}$. Moreover it is finitely generated as an $O\_{d,x}$-module by the family of monomials $z\_{d+1}^{\alpha\_{d+1}}\dots z\_{n}^{\alpha\_{n}}$.
As $\mathcal{J}\_{A,x}$ is prime, $O\_{M,x}/\mathcal{J}\_{A,x}$ is an entire ring. We denote by $\tilde{f}$ the class of any germ $f\in O\_{M,x}$ in $\mathcal{J}\_{A,x}$, by $M\_{A}$ and $M\_{d}$ the quotient fields of $\mathcal{J}\_{A,x}$ and $O\_{d,x}$ respectively. Then $M\_{A}=M\_{d}[\tilde{z}\_{d+1}\dots\tilde{z}\_{n}]$ is a finite algebraic extension of $M\_{d}$. Let $q=[M\_{A}:M\_{d}]$ be its degree of extension and let $\sigma\_{1},\dots,\sigma\_{q}$ be the $q$ embeddings of $M\_{A}$ over $M\_{d}$ in an algebraic closure $\overline{M\_{A}}$. Since a factorial ring is integrally closed in its quotient field, every element of $M\_{d}$ which is integral over $O\_{d,x}$ lies in fact in $O\_{d,x}$. By the primative element theorem, there exists a linear form $u(z'')=c\_{d+1}z\_{d+1}+\dots+c\_{n}z\_{n}$, such that $M\_{A}=M\_{d}[\tilde{u}]$. As $\tilde{u}$ is integral over the integrally closed ring $O\_{d,x}$, the unitary irreducible polynomial $W\_{u}$ of $\tilde{u}$ over $M\_{d}$ has coefficents in $O\_{d,x}$, i.e. $W\_{u}=W\_{u}(z',T)$ where $z'$ represents the first $d$ complex variables.
Moreover, let $\delta(z')\in O\_{d,x}$ be the discriminant of $W\_{u}(z',T)$. Then it is easy to show that for every element $g$ of $M\_{A}$ which is integral over $O\_{d}$ we have $\delta g\in O\_{d}[\tilde{u}]$.(Demailly proved this claim as lemma 4.15 on page 94). In particular, for $k\geq d+1$, we have $\delta z\_{k}\in O\_{d}[\tilde{u}]$. Or more precisely, there exists an unique polynomial $B\_{k}\in O\_{d}[T]$ with degree less than $q$ such that $\delta z\_{k}=B\_{k}(z',\tilde{u}(z''))$.
**Then he claimed that $\delta$ and the coefficents of the polynomials $W\_{u}(z',T)$ and $B\_{k}(z',T)$ all can be expressed as polynomials in the $c\_{j}$'s with coefficients in $O\_{d,x}$, because all of them can be expressed in terms of the elementary symmetric functions of the $\delta\_{k}\tilde{u}$'s.**
Since $W\_{u}(z',T)$ is the irreducible polynomial of $\tilde{u}$ over the quotient field $M\_{d}$ of $O\_{d}$, his claim is obvious. But for the other polynomial $B\_{k}(z',T)$, I really cannot fill in the gap. The only thing I know is that the coefficients $b\_{j}$'s of the polynomial $B\_{k}$ can be calculated by the cramer rule according to his proof of lemma 4.15. But I really don't know how to show that they are elementary symmetric polynomials of the $\delta\_{k}\tilde{u}$'s with coefficients in $O\_{d,x}$.
| https://mathoverflow.net/users/480953 | A question on Demailly's proof of coherence of ideal sheaf | We need the concept of multi-symmetric polynomial which is a straight forward generalization of symmetric polynomial. The definitions and relevant results concerning it can be found in [this paper](https://%20arxiv.org/pdf/math/0205233.pdf).
Indeed, the coefficients $\delta b\_{j}$ of $B\_{k}$ can be expressed as a multi-symmetric polynomial of $\sigma\_{i}\tilde{u}$ and $\sigma\_{i}\tilde{z}\_{k}$. So it is quite misleading to say that they are just symmetric polynomials of $\sigma\_{i}\tilde{u}$. And it was proved in the paper that the ring of multi-symmetric polynomials $\mathbb{C}[x\_{1},\dots,x\_{n},y\_{1},\dots,y\_{n}]^{S\_{n}}$ can be generated by
$$
\bigg\{ e\_{\alpha\beta}=\sum\_{i=1}^{n}x\_{i}^{\alpha}y\_{i}^{\beta}\Big|\alpha,\beta \in \mathbb{N},\alpha+\beta\leq n
\bigg\}.
$$
And it is easy to see that $e\_{\alpha\beta}\in O\_{d,x}[c\_{k+1},\dots,c\_{n}]$.
| 1 | https://mathoverflow.net/users/480953 | 452641 | 181,916 |
https://mathoverflow.net/questions/452607 | 2 | A smooth $m$-dimensional submanifold of $\mathbb{R}^{d}$ is said to be $k$-*ruled* if it is foliated by $k$-dimensional planes, called *rulings*.
Let $M$ be a $k$-ruled submanifold. Then $M$ can be parametrized (locally) by a smooth map $\sigma \colon (U\subset \mathbb{R}^{m-k}) \times \mathbb{R}^{k} \to \mathbb{R}^{d}$ given by
$$\sigma(u\_{1}, \dotsc, u\_{m-k}, v\_{1}, \dotsc, v\_{k}) = \xi(u\_{1}, \dotsc, u\_{m-k}) +\sum\_{j=1}^{k} v\_{j}X\_{j}(u\_{1}, \dotsc, u\_{m-k}),$$
where $\xi$ is a "base" submanifold and $(X\_{1}, \dotsc, X\_{k})$ an orthonormal frame along $\xi$.
In particular, if $k=m-1$, then we can always choose the curve $\xi$ to be orthogonal to all the $(m-1)$-planes, i.e., such that
$$\langle \xi', X\_{j} \rangle =0 \quad \text{for all $j=1,\dotsc, m-1$}.$$
I wonder if the same holds in general.
**Question.** Does there exist a parametrization of $M$ such that
$$\bigl\langle \frac{\partial \xi}{\partial u\_{1}}, X\_{j}\bigr \rangle = \dotsb = \bigl\langle \frac{\partial \xi}{\partial u\_{m-k}}, X\_{j}\bigr \rangle =0 \quad \text{for all $j=1,\dotsc, k$}?$$
| https://mathoverflow.net/users/74033 | Parametrization of $k$-ruled submanifold: can we choose the base to be orthogonal to the rulings? | The answer is already 'no' in the first nontrivial case: A 3-manifold in $\mathbb{R}^d$ (where $d>3$) that is ruled by lines (i.e., $k=1$).
One can see this as follows: As the OP notes, one can write $\sigma$ as above in the form
$$
\sigma(u\_1,u\_2,v\_1) = \xi(u\_1,u\_2) + v\_1 X(u\_1,u\_2)
$$
where, without loss of generality, one can assume that $X\cdot X = 1$
and that the three $\mathbb{R}^d$-valued mappings $X,\ \partial\_{u\_1}\xi,\ \partial\_{u\_2}\xi$ are everywhere linearly independent. It's easy to see that a surface $S$ in the domain of $\sigma$ will map via $\sigma$ to a surface in $\mathbb{R}^d$ that is orthogonal to the $X$-ruling if and only if the surface is an integral surface of the $1$-form
$$
\theta = X\cdot \mathrm{d}\sigma = \mathrm{d} v\_1 + (X\cdot \partial\_{u\_1}\xi)\,\mathrm{d}u\_1
+ (X\cdot \partial\_{u\_2}\xi)\,\mathrm{d}u\_2\,.
$$
(Note the use of $X\cdot X = 1$, which implies $X\cdot \mathrm{d}X = 0$.)
In particular, the $2$-form
$$
\mathrm{d}\theta = \bigl((\partial\_{u\_1}X\cdot \partial\_{u\_2}\xi)-(\partial\_{u\_2}X\cdot \partial\_{u\_1}\xi)\bigr)\,\mathrm{d}u\_1\wedge \mathrm{d}u\_2
$$
must vanish on such a surface.
However, it's easy to write down specific examples for which the function $\bigl((\partial\_{u\_1}X\cdot \partial\_{u\_2}\xi)-(\partial\_{u\_2}X\cdot \partial\_{u\_1}\xi)\bigr)$ is nowhere vanishing. In such examples, no such surface $S$ can exist because the vanishing of $\theta$ and $\mathrm{d}u\_1\wedge \mathrm{d}u\_2$ on $S$ implies that the three differentials $\mathrm{d}u\_1$, $\mathrm{d}u\_2$, and $\mathrm{d}v\_1$ satisfy two independent linear relations on $S$, which is impossible.
| 1 | https://mathoverflow.net/users/13972 | 452647 | 181,918 |
https://mathoverflow.net/questions/452644 | 8 | In a given domain $\Omega$, we have: $\Delta u=-\lambda u$ with $u>0$. Does this mean that $u$ is a principal eigenfunction for $\Delta$ in $\Omega$?
Also, more generally, does this also apply for $Lu=a^{ij}u\_{ij}+b^iu\_i+cu$?
I tried to prove it by using variational way through considering $\frac{\int|Du|^2}{\int u^2}$, but I didn’t figure out if it’s right.
| https://mathoverflow.net/users/348579 | Are all positive eigenfunctions principal eigenfunctions? | **Answer.** Yes, for appropriate boundary conditions (e.g., Dirichlet or Neumann) the Laplace operator on bounded domains with sufficiently smooth boundary has no positive eigenfuntions, except for those that belong to the leading eigenvalue.
Here are the details:
**Part 0.** Notation.
Consider an $L^p$-space over a $\sigma$-finite measure space. For an element of $f$ of an $L^p$-space over of $\sigma$-finite measure space, where $p \in [1,\infty)$, I'll use the notation $f \ge 0$ to mean $f(\omega) \ge 0$ for almost all $\omega$. I'll use the notation $f \gg 0$ to mean $f(\omega) > 0$ for almost all $\omega$ (some people use the notation $f > 0$ for that, but from an order theoretic point of view this notation is a bit unfortunate).
**Part 1.** A Perron-Frobenius (or Krein-Rutman) type result for operators which "improve" the positivity of functions:
**Theorem.** Let $(\Omega,\mu)$ be a $\sigma$-finite measure space and let $p \in [1,\infty)$ and assume that $L^p(\Omega,\mu)$ is non-zero. Let $T: L^p(\Omega,\mu) \to L^p(\Omega,\mu)$ be a compact linear operator with the following property: for non-zero $0 \le f \in L^p(\Omega,\mu)$ one has $Tf \gg 0$.
Then the spectral radius os $T$ satisfies $r(T) > 0$, it is an eigenvalue with an eigenvector $u \gg 0$, and it is the only eigenvalue of $T$ which has an eigenvector $v \ge 0$.
*Proof.* That fact that $r(T) > 0$ is a special case of a theorem of Ben de Pagter; see the main result of de Pagter's 1986 paper "[Irreducible compact operators](https://zbmath.org/0607.47033)".
The fact that $r(T)$ is an eigenvalue with an eigenvector $u \ge 0$ is the classical Krein-Rutman theorem. The fact that $u$ actually satisfies $u \gg 0$ is an immediate consequence of the positivity improving assumption on $T$. The fact that $r(T)$ is the only eigenvalue of $T$ with an eigenvector $v \ge 0$ can, for instance, be found in Theorem V.5.2(iv) and its corollary (both on page 329) of Helmut H. Schaefer's 1974 book "[Banach lattices and positive operators](https://zbmath.org/0296.47023)". $\square$
**Part 2.** Application to the Laplace operator with local boundary conditions.
Consider a bounded domain $\Omega$ in $\mathbb{R}^d$ and assume that the boundary of $\Omega$ is sufficiently smooth to ensure that the Laplace operator with the boundary conditions that we consider in the following has compact resolvent.
Let $\Delta: L^2(\Omega) \supseteq \operatorname{dom}(\Delta) \to L^2(\Omega)$ denote the Laplace operator, where the domain $\operatorname{dom}(\Delta)$ encodes the boundary conditions.
We assume the boundary conditions to be one of the classical choices, for instance Dirichlet or Neumann or mixed boundary conditions. Then we have the following result:
**Corollary.** If $\lambda$ is an eigenvalue of $\Delta$ with an eigenfunction $u \ge 0$, then $\lambda$ is the leading eigenvalue of $\Delta$.
*Proof.*
Let $\mu \in \mathbb{R}$ be a number that is strictly larger than the largest eigenvalue of $\Delta$.
Then the resolvent operator $(\mu I - \Delta)^{-1}: L^2(\Omega) \to L^2(\Omega)$ satisfies the assumptions of the operator $T$ in the theorem in part 1. Moreover, the spectral mapping theorem for the resolvent tells us that the mapping $z \mapsto \frac{1}{\mu - z}$ maps the eigenvalues of $\Delta$ $1$-to-$1$ to the eigenvalues of $(\mu I - \Delta)^{-1}$ and that the eigenvectors are thereby preserved. In particular, $\frac{1}{\mu -\lambda\_0}$ is the spectral radius of $(\mu I - \Delta)^{-1}$.
So by the theorem in part 1, the claim follows. $\square$
**Part 3.** Further remarks.
* The same arguments apply to elliptic operators with coefficients and with lower order terms.
* For self-adjoint operators with compact resolvent (like the Laplace operator with Dirchlet or Neumann or mixed boundary conditions), there is a simpler argument that was pointed out in [Christian Remling' answer](https://mathoverflow.net/a/452653/102946).
An advantage of the argument sketched above is that it also works for non-self adjoint operators which one gets, for instance, by considering second-order operators with lower order terms.
* The same result is no longer true for the Laplace operator with non-local boundary conditions. For instance, consider the one-dimenional domain $\Omega = (0,1)$ and the Laplace operator with the boundary conditions $u'(1) = \mathrm{e} u'(0)$. Then the constant function $1$ is an eigenvector for the eigenvalue $0$. But $1$ is also an eigenvalue (with $\exp$ as an eigenfunction).
* The compactness assumption in the theorem in part 1 can, in the case $p=2$, be replaced with a self-adjointness assumption; see Theorem 2.2 in the 2015 paper "[Note on basic features of large time behaviour of heat kernels](https://zbmath.org/1369.58017)" by Keller, Lenz, Vogt, and Wojciechowski.
(One can apply this reference to $-T$ to obtain the above theorem for self-adjoint but non-compact $T$. However, that's actually a bit of a detour, since the result in this reference is itself designed to deal with unbounded operators.)
| 12 | https://mathoverflow.net/users/102946 | 452650 | 181,920 |
https://mathoverflow.net/questions/452519 | 1 | Define the family of densities:
$$p(\phi;\theta) = \Big(f\big(\hspace{-1pt}\cos(\phi-\theta)\big) - f\big(\hspace{-1pt}\cos(\phi+\theta)\big)\Big)\hspace{0.5pt} \frac{\sin(2\phi)}{\sin(2\theta)}, \quad 0 \le \phi,\theta\le \pi/2$$
where $f(x)=g(x^2)$ with $g$ non-negative, increasing, and convex or concave, on $[0,\infty)$. (Interestingly these densities indeed have the same measure for all $\theta$ whenever $g$ is concave or convex, which can be shown by an integral representation.)
Edit: As shown in the answer, these conditions are not sufficient. I believe it is true though if additionally $g^{(3)}(x) \ge 0$, for $x>0$.
Show that $p(\phi;\theta)$ has a monotone likelihood ratio (decreasing for concave $g$, increasing for convex $g$). I.e., for $0\le\theta\_1 < \theta\_2\le\pi/2$:
$$ h(\phi) = \frac{f\big(\hspace{-1pt}\cos(\phi-\theta\_2)\big) - f\big(\hspace{-1pt}\cos(\phi+\theta\_2)\big)}{f\big(\hspace{-1pt}\cos(\phi-\theta\_1)\big) - f\big(\hspace{-1pt}\cos(\phi+\theta\_1)\big)}$$
is monotonic on $[0,\pi/2]$.
Examples of functions are $f(x) = |x|^p$, $1\le p<2$, or for $p>2$. (For $p=2$, $p(\phi;\theta) = \sin^2(2\phi)$.) And the function $f(x) = \log( \cosh(x))$, which is concave in $x^2$ (i.e. $\log(\cosh(\sqrt{x}))$ is concave on $[0,\infty)$), and twice differentiable (unlike $|x|^p$, $p<2$).
This result is important to prove uniqueness of stable optima in blind source separation and deconvolution with strongly sub- and super-gaussian input densities, using the Karlin-Rubin theorem. The result can be proved for $f(x)=x^4$ by simplifying the derivative expression. This corresponds to using kurtosis as the the cost function, and the uniqueness result is already known in this case. For $f(x) = |x|$, the likelihood ratio is non-increasing, constant around $\phi=0$ and $\phi=\pi/2$.
| https://mathoverflow.net/users/510206 | Monotone likelihood ratio of a family of densities with compact support | $\newcommand{\ep}{\varepsilon}$This conjecture is not true in general.
Indeed, suppose the "convex" part of your conjecture is true. Then (letting $x:=\phi$, $t:=\theta\_1$, and $\theta\_2\downarrow\theta\_1=t$) we see that for any strictly increasing convex smooth function $g$ and all $x$ and $t$ in $(0,\pi/2)$ we would have $h\_2(g;x,t):=\partial\_x\partial\_t\,\ln(g(\cos^2(x-t))-g(\cos^2(x+t))\ge0$. (Note that for all $x$ and $t$ in $(0,\pi/2)$ we have $\cos^2(x-t)-\cos^2(x+t)=\sin2x\,\sin2t>0$, so that $h\_2(g;x,t)$ is well defined.)
For real $\ep>0$ and real $u$, let $u\_{+;\ep}:=\frac12(u+\sqrt{\ep^2+u^2})$, an "$\ep$-smoothed" version of $u\_+:=\max(0,u)$. For $c$ and $c\_\*$ in $[0,\infty)$, let $g\_{c\_\*,\ep}(c):=(c-c\_\*)\_{+;\ep}$.
Then the function $g\_{c\_\*,\ep}$ is strictly increasing, convex, and smooth on $[0,\infty)$. However, $h\_2(g;x,t)=-44051.358\ldots\not\ge0$ if $g=g\_{c\_\*,\ep}$, $c\_\*=\frac12$, $\ep=\frac1{1000}$, $x=\frac{39}{100}$, and $t=\frac{118}{100}$. So, the "convex" part of your conjecture is not true in general.
---
Suppose now the "concave" part of your conjecture is true. Then for any strictly increasing concave smooth function $g$ and all $x$ and $t$ in $(0,\pi/2)$ we would have $h\_2(g;x,t)\le0$.
For $c$ and $c\_\*$ in $[0,\infty)$, let $G\_{c\_\*,\ep}(c):=c-\sqrt{\ep^2+(c-c\_\*)^2}$.
Then the function $G\_{c\_\*,\ep}$ is strictly increasing, concave, and smooth on $[0,\infty)$. However, $h\_2(G;x,t)=32614.565\ldots\not\le0$ if $G=G\_{c\_\*,\ep}$, $c\_\*=\frac12$, $\ep=\frac1{1000}$, and $x=\frac{39}{100}=t$. So, the "concave" part of your conjecture is not true in general either.
$\quad\Box$
| 1 | https://mathoverflow.net/users/36721 | 452659 | 181,922 |
https://mathoverflow.net/questions/452657 | -3 | I was wondering whether it would be possible to do propositional logic without any rules of inference and assumptions (except modus ponens).
I have the following axioms:
1. $ p \to (q \to p) $
2. $ (p \to (q \to r)) \to ((p \to q) \to (p \to r)) $
3. $ (\lnot p \to \lnot q) \to (q \to p) $
4. MP: From $ p $ and $ p \to q $, infer $ q $.
The other connectives ($ \lor $ and $ \land $) are defined like this:
$ p \lor q := \lnot p \to q $
$ p \land q := \lnot (p \to \lnot q) $
For example, [double negation](https://en.wikipedia.org/wiki/Double_negation#Proofs) uses [hypothetical syllogism](https://en.wikipedia.org/wiki/Hypothetical_syllogism): from $ p \to q $ and $ q \to r $, infer $ p \to r $.
Moreover, I can't find proofs for (or prove myself) common rules of inference in natural deduction, such as:
* $ (p \to q) \to (\lnot q \to \lnot p) $
* $ p \to \lnot \lnot p $
* $ p \to (p \lor q) $
* $ (p \land q) \to p $
* $ ((p \to q) \land (q \to r)) \to (p \to r) $
My question is whether it is possible to not use rules of inference and assumptions (except MP) in propositional logic, or that you can't prove these statements without them.
I found this question, but it does not have any answers yet: <https://math.stackexchange.com/questions/2468217/prove-double-negation-introduction-without-elimination-in-a-propositional-logic>
Example proof
-------------
I hope this clarifies some things, here's something that I managed to prove:
1. $ p \to ( p \to p ) $ (Axiom 1)
2. $ p \to ( ( p \to p ) \to p ) $ (Axiom 1)
3. $ ( p \to ( ( p \to p ) \to p ) ) \to ( ( p \to ( p \to p ) ) \to ( p \to p ) ) $ (Axiom 2)
4. $ ( p \to ( p \to p ) ) \to ( p \to p ) $ (MP 2, 3)
5. $ p \to p $ (MP 1, 4)
| https://mathoverflow.net/users/510414 | Propositional logic without rules of inference and assumptions (except MP) | I'll do one of them, $p \to (p \vee q)$.
**Define** $a \vee b$ as $\lnot b \to a$.
We want to prove
$$
p \to (p \vee q) .
$$
This is now merely alternate notation for
$$
p \to (\lnot q \to p) .
$$
But this is an instance of Axiom 1.
| 1 | https://mathoverflow.net/users/454 | 452661 | 181,923 |
https://mathoverflow.net/questions/452656 | 4 | Let $f = (f\_0,f\_1,\ldots,f\_n,\ldots) \in \mathcal{P}(\mathbb N)$ be a probability distribution on $\mathbb N$ and denote by $$\hat{f}(z) = \sum\_{n\geq 0} z^n f\_n$$ for its probability generating function. It has been shown [in this post](https://math.stackexchange.com/questions/220705/how-to-obtain-probability-function-from-its-probability-generating-function) that we have the following Bromwich's inversion formula (using complex integration over the positively oriented unit circle $\mathbb{S}^1$): $$\sum\limits\_{i > n} f\_i = \frac{1}{2 \pi i} \oint\_{\mathbb{S}^1} \frac{1-\hat{f}(\xi)}{1-\xi}\cdot \frac{\mathrm{d} \xi}{\xi^{n+1}} \label{1}\tag{1}$$ for each $n\geq 0$. Note that the left hand side of equation \eqref{1} represents the probability $\mathbb{P}(X > n)$ for a $f$-distributed random variable $X$. Now my question is, since we have the obvious asymptotic limit $$\lim\limits\_{n \to \infty} \mathbb{P}(X > n) = 0.$$ How can we show that the right hand side of equation \eqref{1} also tends to zero as $n \to \infty$ (**without using the probabilistic interpretation of the contour integration**)?
| https://mathoverflow.net/users/163454 | Show that $\frac{1}{2 \pi i} \oint_{\mathbb{S}^1} \frac{1-\hat{f}(\xi)}{1-\xi}\cdot \frac{\mathrm{d} \xi}{\xi^{n+1}} \to 0$ as $n \to \infty$ | $\newcommand\R{\mathbb R}$Assume that the integral
$$I\_n:=\oint\_{\mathbb{S}^1} \frac{1-\hat{f}(\xi)}{1-\xi}\,\frac{d\xi}{\xi^{n+1}}
=\int\_0^{2\pi} \frac{1-\hat{f}(e^{it})}{1-e^{it}}\,\frac{e^{it}\,i\,dt}{e^{i(n+1))t}} $$
exists in the Lebesgue sense -- that is,
$$\int\_0^{2\pi} \Big|\frac{1-\hat{f}(e^{it})}{1-e^{it}}\Big|\,dt<\infty. $$
Then
$$I\_n=\int\_0^{2\pi} \frac{1-\hat{f}(e^{it})}{1-e^{it}}\,\frac{\,i\,dt}{e^{int}}\to0$$
(as $n\to\infty$) by the [Riemann–Lebesgue lemma](https://en.wikipedia.org/wiki/Riemann%E2%80%93Lebesgue_lemma).
**Detail:** The Riemann–Lebesgue lemma states that $\int\_\R g(t)e^{-itx}\,dt\to0$ as $|x|\to\infty$ if $\int\_\R |g(t)|\,dt<\infty$. Here the Riemann–Lebesgue lemma is used with $g(t):=\frac{1-\hat{f}(e^{it})}{1-e^{it}}\,i\,1(0<t<2\pi)$.
| 5 | https://mathoverflow.net/users/36721 | 452662 | 181,924 |
https://mathoverflow.net/questions/452322 | 14 | 1. $\DeclareMathOperator\pcf{pcf}$For simplicity say $\aleph\_\omega$ is a strong limit. Let $A=\pcf\{\aleph\_n:n\in\omega\}$. Then it follows from basic properties of pcf operation that $X\subseteq A\rightarrow\pcf(X)\subseteq A$, and that pcf is a Kuratowski closure operator on $A$, which gives $A$ a topology. Is $A$ zero-dimensional? Is it compact Hausdorff? I thought I was able to prove that it has finite intersection property for closed sets and hence compact, but [PCF structures of height less than $\omega\_3$](https://www.jstor.org/stable/20799313) by Er-rhaimini and Velickovic says the pcf space is "locally compact", which confused me a bit. **Edit:** I believe $A$ is compact Hausdorff, and combined with the fact that it's scattered (because if $X\subseteq A$ then $\min X$ is isolated in $X$) we know it's totally disconnected, which for compact Hausdorff space is equivalent to zero-dimensional.
2. The above paper also mentions the pcf space is scattered, and hence one can define a Cantor–Bendixson rank. Is this related to the proof that $\max A<\aleph\_{\omega\_4}$, say the proof in Jech? If not can the proof be interpreted topologically in any way?
3. This question is probably not so topological, but I hesitate to ask it separately. If $2^{\aleph\_0}<\aleph\_\omega$, then by pcf theory $\max\pcf\{\aleph\_n:n\in\omega\}=\aleph\_\omega^{\aleph\_0}\leq\aleph\_{\omega+\omega}^{\aleph\_0}=\max\pcf\{\aleph\_{\omega+n}:n\in\omega\}$. Is it possible to prove $\max\pcf\{\aleph\_n:n\in\omega\}\leq\max\pcf\{\aleph\_{\omega+n}:n\in\omega\}$ directly?
| https://mathoverflow.net/users/499983 | Topology and pcf theory | For (1) and (2), you may want to look at Burke and Magidor's [exposition](https://core.ac.uk/download/pdf/82500424.pdf) of pcf theory, as they adopt a topological stance at a couple of places. For my money, the part of the ``$\aleph\_{\omega\_4}$ Theorem'' where the topological flavor is strongest is when you prove there is no left-separated subspace of ${\rm pcf}(A)$ of cardinality $|A|^+$. There are some pieces of the proof that can be phrased in terms of CB rank, but I am not sure how much insight that gives overall. (But that's a matter of taste, I'm sure.)
Regarding (3), you want to be a little careful with your computations, as $2^{\aleph\_0}<\aleph\_\omega$ does not necessarily imply that $\aleph\_{\omega+\omega}^{\aleph\_0}$ is equal to $\max{\rm pcf}(\{\aleph\_{\omega+n}:n<\omega\})$.
To characterize the cofinality of $([\aleph\_{\omega+\omega}]^{\aleph\_0},\subseteq)$ in terms of pcf theory, you need to compute $\max{\rm pcf}(\{\aleph\_{\alpha+1}:\alpha<\omega+\omega\})$ (using all the infinite regular cardinals less than $\aleph\_{\omega+\omega}$) and not just the tail of order-type $\omega$. If we just look at
$\max{\rm pcf}(\{\aleph\_{\omega+n}:n<\omega\})$ instead, then we get the cofinality of the structure $([\aleph\_{\omega+\omega}]^{\aleph\_\omega},\subseteq)$.
From the point of view of general pcf theory, there is a sort of ``inverse monotonicity'' going on in this situation.
If $A$ is an interval of regular cardinals with $|A|<\min(A)$, then ${\rm{pcf}}(A)$ is also an interval of regular cardinals with cardinality less than $|A|^{+4}$. In particular, $\rm{pcf}(A)$ does not have a weakly inaccessible accumulation point and so ${\rm{pcf}}({\rm{pcf}}(A))={\rm{pcf}}(A)$ whenever $A$ is a progressive interval of regular cardinals.
Now let $A = \{\aleph\_n:n<\omega\}$. If $\max\rm{pcf}(A)>\aleph\_{\omega+\omega}$ then $\{\aleph\_{\omega+n}:n<\omega\}$ is contained in $\rm{pcf}(A)$ hence ${\rm{pcf}}(\{\aleph\_{\omega+n}:n<\omega\})$ is a subset of $\rm{pcf}(A)$. Thus, in the "interesting" case, pcf theory tells you that
$$\max{\rm{pcf}}(\{\aleph\_{\omega+n}:n<\omega\})\leq \max{\rm pcf}(\{\aleph\_n:n<\omega\}).$$
and therefore if
$$ \aleph\_{\omega+\omega}<\max{\rm pcf}\{\aleph\_n:n<\omega\}$$
then
$$\max{\rm pcf}(\{\aleph\_{\omega+n}:n<\omega\})\leq\max{\rm pcf}(\{\aleph\_{\alpha+1}:\alpha<\omega+\omega\}=\max{\rm pcf}(\{\aleph\_n:n<\omega\}).$$
But in the particular setting you asked about, the above paragraphs are just a fancy way of saying that
$${\rm cf}([\aleph\_{\omega+\omega}]^{\aleph\_0},\subseteq)={\rm cf}([\aleph\_\omega]^{\aleph\_0},\subseteq)\cdot {\rm cf}([\aleph\_{\omega+\omega}]^{\aleph\_\omega},\subseteq),$$
and if $\aleph\_{\omega+\omega}<\theta = {\rm cf}([\aleph\_\omega]^{\aleph\_0})$, then
$$\theta\leq{\rm{cf}}([\aleph\_{\omega+\omega}]^{\aleph\_0},\subseteq) \leq {\rm cf}([\theta]^{\aleph\_0},\subseteq)= {\rm cf}([\aleph\_\omega]^{\aleph\_0},\subseteq)=\theta.$$
I don't know anything about the value of ${\rm{pp}}(\aleph\_{\omega+\omega})$ in various models where ${\rm{pp}}(\aleph\_\omega)>\aleph\_{\omega+\omega+1}$ (just ignorance on my part), but my assumption is that someone (possibly Gitik or one of his students?) will have analyzed this. I conjecture that it's going to be easier to maintain ${\rm pp}(\aleph\_{\omega+\omega})=\aleph\_{\omega+\omega+1}<{\rm pp}(\aleph\_\omega)$ than it is to blow it up to match ${\rm pp}(\aleph\_\omega)$, as this second alternative would involve adding a long scale in $\prod\_{n<\omega}\aleph\_{\omega+n}$ in addition to the one you need to add in $\prod\_{n<\omega}\aleph\_n$.
| 7 | https://mathoverflow.net/users/18128 | 452682 | 181,931 |
https://mathoverflow.net/questions/452691 | 2 | Does the Riemann hypothesis predict an upper bound for
$$\left|f(x)-\left(\operatorname{li}(x)-\frac{x}{\log x} \right)\right|,\quad x\ge 2\tag{1}$$
where
$$f(x)=\sum\limits\_{n=2}^x \frac{\Lambda(n)}{\log^2(n)}\tag{2}$$
and $\Lambda(n)$ is the [von Mangoldt function](https://en.wikipedia.org/wiki/Von_Mangoldt_function)?
| https://mathoverflow.net/users/110710 | Does the Riemann hypothesis predict a bound for this prime-counting function? | The Riemann hypothesis is *equivalent* to the following statement:
$$f(x)=\mathrm{li(x)}-\frac{x}{\log x}+O(\sqrt{x}),\qquad x\geq 2.$$
Note that
$$\mathrm{li(x)}=\mathrm{li(2)}+\frac{x}{\log x}-\frac{2}{\log 2}+\int\_2^x\frac{dt}{\log^2 t},$$
hence the claim is that the Riemann hypothesis is equivalent to
$$f(x)=\int\_2^x\frac{dt}{\log^2 t}+O(\sqrt{x}),\qquad x\geq 2.\tag{$\ast$}$$
**1.** The Riemann hypothesis implies $(\ast)$ as follows:
\begin{align\*}
f(x)&=\int\_{2-}^x\frac{d\psi(t)}{\log^2 t}\\
&=\int\_2^x\frac{dt}{\log^2 t}+\int\_{2-}^x\frac{d(\psi(t)-t)}{\log^2 t}\\
&=\int\_2^x\frac{dt}{\log^2 t}+\left[\frac{\psi(t)-t}{\log^2 t}\right]\_{2-}^x+2\int\_2^x\frac{\psi(t)-t}{t\log^3 t}\,dt.
\end{align\*}
Here $\psi(t)-t=O(\sqrt{t}\log^2 t)$ by the Riemann hypothesis, hence we conclude $(\ast)$.
**2.** Let us denote by $g(x)$ the integral in $(\ast)$. Then $(\ast)$ implies RH as follows:
\begin{align\*}
\psi(x)&=\int\_{2-}^x(\log^2 t)\,df(t)\\
&=\int\_2^x(\log^2 t)\,dg(t)+\int\_{2-}^x(\log^2 t)\,d(f(t)-g(t))\\
&=x-2+\left[(\log^2 t)\,(f(t)-g(t))\right]\_{2-}^x-2\int\_2^x\frac{\log t}{t}(f(t)-g(t))\,dt.
\end{align\*}
Here $f(t)-g(t)=O(\sqrt{t})$ by $(\ast)$, hence we infer that $\psi(x)=x+O(\sqrt{x}\log^2 x)$, which is a form of the Riemann hypothesis.
| 9 | https://mathoverflow.net/users/11919 | 452694 | 181,936 |
https://mathoverflow.net/questions/446460 | 2 | **EDIT 1:** All topological groups in this question are assumed to be second countable. In particular, this forces every group to be metrizable and every Lie group to have at most countably many components. In particular, all discrete groups are countable. The general case (where a Lie group is just assumed to be locally Euclidean and nothing else) could also be interesting, but for this question the second countable case is enough.
First some terminology:
A closed normal subgroup $N$ of a topological group $P$ is *co-Lie* in $P$, if $P/N$ is a (finite dimensional real) Lie group.
A *locally compact pro-Lie group* $P$ is a locally compact group such that every identity neighborhood $U\subseteq P$ contains a subgroup $N$ which is co-Lie in $P$. (This is equivalent to the fact that $P$ is the projective limit of some family of Lie groups.)
Now, there is one version of the famous Gleason-Yamabe theorem which states that
"Every locally compact group $G$ contains an open subgroup $P$ which is a pro-Lie group."
Easy examples show that such a subgroup $P$ is not unique. So, the question arises if there is a maximal one. The obvious thing to try would be to use Zorn's Lemma, but it is not clear to me if the union of a chain of open pro-Lie groups is again a pro-Lie group (it is certainly open, that is not the problem).
Any ideas on this?
**EDIT 2:** Note that in the case where $G$ itself is pro-Lie, it is obvious that there is a maximal open pro-Lie subgroup; just take $G$ itself. This solves the problem for the cases where
* $G$ is abelian
* $G$ is compact
* $G$ is connected (or has a finite number of components),
because in all of these cases it is well-known that $G$ is pro-Lie.
So, if there is a counter-example, it has to be a locally compact group $G$ which is non-abelian, non-compact and has infinitely many components.
| https://mathoverflow.net/users/153400 | Does every locally compact group G contain a maximal open subgroup P which is a pro-Lie group? | After thinking about the problem for some time, I came up with a counter-example, so as YCor wrote in his comment, it is indeed true that there is not always a maximal open pro-Lie subgroup.
Let $A$ be a nontrivial compact group, for example $\mathbb Z/2\mathbb Z$ or the circle group $\mathbb R /\mathbb Z$. Then the countable power $A^{\mathbb N}$ is also compact. Now let $\Gamma$ be the discrete group of all permutations on $\mathbb N$ with finite support. It is clear that $\Gamma$ acts on $A^{\mathbb N}$ by permuting coordinates, so we may form the semi-direct product:
$$ G := A^{\mathbb N} \rtimes \Gamma $$
which is a locally compact group as a product of a compact and a discrete group (and tdlc if $A$ is so). It is relatively easy to see that $A^{\mathbb N}$ has no co-Lie subgroups which are $\Gamma$ invariant, so $G$ is not a pro-Lie group.
However, since $\Gamma$ is a countable union of finite permutation groups, $G$ is a countable union of compact groups which are all open in $G$. Compact groups are pro-Lie groups, so we have an increasing sequence of open pro-Lie subgroups which does not have an upper bound. So, there is no maximal open pro-Lie subgroup of $G$.
| 3 | https://mathoverflow.net/users/153400 | 452706 | 181,938 |
https://mathoverflow.net/questions/452681 | 13 | Let $A$ and $G$ be two groups. Let $\alpha:G\rightarrow\operatorname{Bij}(A)$ be a group homomorphism and $\beta: A\rightarrow\operatorname{Bij}(G)$ an anti-homomorphism satisfying some conditions given in [Wikipedia](https://en.wikipedia.org/wiki/Zappa%E2%80%93Sz%C3%A9p_product#External_Zappa%E2%80%93Sz%C3%A9p_products). The actions $\alpha$ and $\beta$ induce a group $A\mathbin{\_{\alpha}{\bowtie}\_{\beta}} G$ called the [*Zappa-Szép product*](https://en.wikipedia.org/wiki/Zappa-Sz%C3%A9p_product). This group is a natural generalization of the semidirect product of two groups in which neither of the factors is required to be normal. I know some examples of semidirect products induced by nontrivial actions and isomorphic to direct products. However, I couldn't find a group $H=A\mathbin{\_{\alpha}{\bowtie}\_{\beta}} G$ isomorphic to the direct product $A\times G$ where the actions $\alpha$ and $\beta$ are both nontrivial. So I will be thankful if someone provides me an example. Notice that I have asked a similar question in [*math.stackexchange*](https://math.stackexchange.com/questions/4751563/zappa-sz%c3%a9p-product-vs-direct-product).
| https://mathoverflow.net/users/95592 | Is it possible for a direct product to be isomorphic to the Zappa–Szép product? | Let $S$ be a nonabelian group with a fixed-point-free automorphism $\alpha$. (Such groups for which $\alpha$ has prime order are necessarily nilpotent. I think the smallest example is the nonabelian group of order $7^3$ and exponent $7$, which has a fixed-point-free automorphism of order $3$.)
Now let $G = S \times S$ and define the (diagonal) subgroups $H$ and $K$ of $G$ by
$$ H = \{ (s,s) : s \in S \},\quad K = \{(s,\alpha(s)) : s \in S \}.$$
Then $H \cong K \cong S$, $H \cap K = \{1\}$ and $G = HK$, so $G$ is the internal Zappa-Szép product of $H$ and $K$, but neither $H$ nor $K$ is normal in $G$, so this is not equal to a semidirect product.
| 18 | https://mathoverflow.net/users/35840 | 452712 | 181,942 |
https://mathoverflow.net/questions/452654 | 2 | Given an odd natural integer $2a-1$ with $a\geq 1$, associate to it recursively the composition $\psi(1)=\emptyset$ and $\psi(2^{-n}a)+(n+\delta\_{>1}(m))$ if $a=2^n m$ with $m$ odd where $\delta\_{>1}(1)=0$ and $\delta\_{>1}(m)=1$ otherwise.
For $2a-1=37$ we get for example
\begin{align\*}
\psi(2\cdot 19-1)&=\psi(2\cdot 2\cdot 5-1)+(0+\delta\_{>1}(19))\\
&=\psi(2\cdot 3-1)+(1+\delta\_{>1}(5))+1\\
&=\psi(2\cdot 2\cdot 1-1)+2+1\\
&=\psi(2\cdot 1-1)+(1+\delta\_{>1}(1))+2+1\\
&=1+1+2+1\\
\end{align\*}
First values for $\psi$ are given by
$$\begin{array}{r|ccccccccc}
n&1&3&5&7&9&11&13&15&17\\
\hline
\psi(n)&\emptyset&1&1+1&2&1+1+1&1+2&2+1&3&1+1+1+1\\
\end{array}$$
$\psi$ is one-to-one: the composition $n\_1+n\_2+\cdots+n\_l$
is the image of $(\cdots(((2^{1+n\_1}-1)2^{n\_2}-1)2^{n\_3}-1)\cdots)2^{n\_l}-1$.
The partition $1+1+2+1$ corresponds to
$$(((2^{1+1}-1)2^1-1)2^2-1)2^1-1=37.$$
For $n>0$, it sends the $2^{n-1}$ odd integers $2^n+1,2^n+3,\dots,2^{n+1}-1$ to the $2^{n-1}$ compositions of sum $n$ in an order-preserving way for the lexicographic order on compositions.
Restricting $\psi$ to integers $\geq 5$ which are congruent to $1$ modulo $4$ (or to $3$ modulo $4$) and considering summands of compositions as coefficients of continued fraction expansions, we get of course a bijection onto all rationals in $(0,1)$.
*Is there a good reference for the bijection $\psi$ (or some close relative)?*
| https://mathoverflow.net/users/4556 | A bijection between odd natural integers and compositions | More or less a comment. I'd say the closest known bijection is just given by the binary representation, namely, if I understand it correctly, for e.g. $37$ we do $3$ times the last power of $2$ less than $37$, here $3\cdot32=96$, and the binary representation of the difference to $37$, here $96-37=59=2^5+2^4+2^3+2^1+2^0$, and finally the increments in the (decreasing) sequence of exponents, here $(5-4={\bf1},4-3={\bf 1},3-1={\bf 2},1-0={\bf 1})$.
| 3 | https://mathoverflow.net/users/6101 | 452714 | 181,943 |
https://mathoverflow.net/questions/452707 | 2 | Consider over $\mathbb{C}$. Let $(X,\mathcal{O}(1))$ be a smooth projective scheme with an ample polarisation. Let $P(t):=\chi(X,\mathcal{O}(t))$ denote the Hilbert polynomial of $\mathcal{O}\_X$. Choose a decomposition $P(t)=I(t)+Q(t)$ such that $I(t),Q(t)$ are also Hilbert polinomials. If a sheaf of ideals $\mathcal{I}\subseteq\mathcal{O}\_X$ has Hilbert polynomial $I(t)$, then the sheaf $\mathcal{O}\_X/\mathcal{I}$ has Hilbert polynomial $Q(t)$.
By smoothness, we have that $\mathcal{O}\_X$ is Gieseker stable. Then any ideal $\mathcal{I}\subseteq\mathcal{O}\_X$ is stable, because any subsheaf of a rank one pure sheaf is Gieseker stable.
Consider the following moduli scheme of stable sheaves and Hilbert scheme
$$\mathrm{M}^{I(t),\mathrm{s}}:=\{\textrm{stable sheaves on }X\textrm{ of Hilbert polynomial }I(t)\}\\
\mathrm{Hilb}\_X^{Q(t)}:=\{\textrm{subschemes }Z\hookrightarrow X\textrm{ s.t. }\mathcal{O}\_Z\textrm{ has Hilbert polynomial }Q(t)\}.$$
For a closed subschem $Z\hookrightarrow X$, the associated ideal $\mathcal{I}\_Z:=\ker(\mathcal{O}\_X\to \mathcal{O}\_Z)$ is stable of Hilbert polynomail $I(t)$, defining a point of $\mathrm{M}^{I(t),\mathrm{s}}$. This defines a morphism
$$\mathrm{Hilb}\_X^{Q(t)}\to \mathrm{M}^{I(t),\mathrm{s}},\quad (Z\hookrightarrow X)\mapsto \mathcal{I}\_Z. $$
The question is when is the morphism an isomorphism? It seems that it is an isomorphism when $X=\mathbb{P}^3$ and the Hilbert scheme parametrises curves according to discussions in this [MO](https://mathoverflow.net/questions/223618/hilbert-schemes-and-moduli-of-ideal-sheaves). Does it suffices to assume that $X$ is smooth, and $Z\hookrightarrow X$ has higher codimension?
| https://mathoverflow.net/users/105537 | When is the morphism from the Hilbert scheme to the moduli scheme of stable sheaves an isomorphism? | This works for any smooth projective variety $X$ under the assumption
$$
\mathrm{Pic}^0(X) = 0
$$
and any $Z$ of codimension at least 2. For the proof see Lemma B.5.6 in Kuznetsov, Alexander G.; Prokhorov, Yuri G.; Shramov, Constantin A. Hilbert schemes of lines and conics and automorphism groups of Fano threefolds. Jpn. J. Math. 13 (2018), no. 1, 109--185.
| 1 | https://mathoverflow.net/users/4428 | 452720 | 181,945 |
https://mathoverflow.net/questions/451365 | 1 | This is a cross-post from [this other question](https://math.stackexchange.com/q/4726158/610053) that I asked ~1 month ago in the mathematics forum, with no reaction. I am still stuck on this, looking for references or approaches to proofs. I hope I have a bit more luck in this forum! let me know if any clarifications would be needed as I am not a professional mathematician.
---
### Definitions:
Consider the set of orthogonal matrices in the compact Stiefel manifold:
$\mathcal{M}\_{N, k} := \{o \in \mathbb{R}^{N \times k} : \quad o^\top o = I\_k \}$
We also define the projector $P\_i$ of a matrix $o\_i \in \mathcal{M}\_{N, k}$ as:
$P\_i = o\_i o\_i^\top$
And the *principal angles* $\theta\_{i\rightarrow j}$ between 2 matrices $o\_i, o\_j$ (see e.g. [Absil, Mahony, Sepulchre, 2003](https://www.cis.upenn.edu/%7Ecis6100/Diffgeom-Grassmann.Absil.pdf) or [Qiu, Zhang, Li, 2005, s.2](https://eeqiu.people.ust.hk/wp-content/uploads/2021/04/Unitarily-Invariant-Metrics-on-the-Grassmann-Space.pdf)) as:
$SVD[o\_i^\top o\_j] = U \quad cos(\theta\_{i\rightarrow j}) \quad V^T$
Finally, consider the following (squared and normalized) Frobenius projective metric between any 2 matrices $(o\_i, o\_j)$ in $\mathcal{M}\_{N, k}$ (see e.g. [Edelman, Arias, Smith, 1998, section 4.3](https://arxiv.org/abs/physics/9806030)):
$d\_{pF}^2(o\_i, o\_j) = \frac{1}{k} \lVert P\_i - P\_j \rVert\_F^2 = \frac{1}{k} \lVert sin^2(\theta\_{i\rightarrow j}) \rVert\_F^2 = 1 - \frac{1}{k} \lVert cos ^2(\theta\_{i\rightarrow j}) \rVert\_F^2 = 1 - \frac{1}{k} \lVert o\_i^\top o\_j \rVert\_F^2 \in [0, 1]$
---
### Setup:
1. Draw a random pair $(o\_1, o\_2) \sim (\mathcal{M}\_{N, k} \times \mathcal{M}\_{N, k})$ uniformly from the Haar measure (e.g. using the [QR method](https://finmath.stanford.edu/%7Ecgates/PERSI/papers/subgroup-rand-var.pdf))
2. Compute $d\_{pF}^2(o\_1, o\_2)$
3. Repeat steps 1 and 2 for different random pairs, and average the results. We observe that $\frac{1}{k} \lVert o\_i^\top o\_j \rVert\_F^2$ converges to $\frac{k}{N}$. Or equivalently, $\mathbb{E}[d\_{pF}^2] = 1 - \frac{k}{N}$
---
### Question:
I imagine this is a well known result in the literature. Where could I find it?
Alternatively, how could I prove it?
The approach I thought of so far seems rather cumbersome, and I am not sure if it would be correct:
1. Take the joint distribution $\mathcal{X}$ of principal angles for a Haar-uniform random orthogonal matrix in $\mathbb{R}^{N \times N}$ (presented e.g. in [Rummler, 2002](https://core.ac.uk/download/pdf/159153606.pdf)).
2. Truncate $\mathcal{X}$ to $k$ entries, and calculate the normalized squared cosine, i.e. $\mathcal{Y} = \frac{1}{k} cos^2(\mathcal{X}\_k)$
3. Then, $\mathbb{E}[\mathcal{Y}]$ should equal $\frac{k}{N}$ (?)
Again, this is probably not necessary since I assume this result is known and I'm just missing the reference. Thanks in advance!
| https://mathoverflow.net/users/172514 | Expected value of the projective metric between random orthogonal Stiefel matrices in $\mathbb{R}^{N \times k}$ equals $1 - \frac{k}{N}$ | Fortunately, there's nothing deep going on here.
We'll use slightly different notation. Let $\mathbf{Q} \in \mathbb{R}^{N \times k}$ be a random matrix drawn uniformly from the Stiefel manifold of $N \times k$ orthonormal frames. In particular,
* The columns of $\mathbf{Q}$ are orthonormal.
* The marginal distribution of each column $\mathbf{q}\_i$ is the uniform distribution over the Euclidean unit sphere, hence is isotropic: $\mathbb{E} [\mathbf{q}\_i \mathbf{q}\_i^\* ] = N^{-1} \mathbf{I}\_N$, where $\mathbf{I}\_N$ is the $N \times N$ identity matrix.
The orthogonal projector onto the range of $\mathbf{Q}$ takes the form $\mathbf{P} = \mathbf{QQ}^\*$, where $\*$ is the transpose. The key observation here is that
$$
\mathbb{E}[ \mathbf{P} ] = \sum\_{i=1}^k \mathbf{E}[ \mathbf{q}\_i \mathbf{q}\_i^\* ] = \sum\_{i=1}^k \frac{1}{N} \mathbf{I}\_N = \frac{k}{N} \mathbf{I}\_N.
$$
Given two independent realizations $\mathbf{Q}\_1, \mathbf{Q}\_2$, we form the associated orthogonal projectors $\mathbf{P}\_1, \mathbf{P}\_2$. Write $\mathbb{E}\_1, \mathbb{E}\_2$ for the expectations with respect to the randomness in $\mathbf{Q}\_1, \mathbf{Q}\_2$ respectively. Using the definition of the Frobenius norm in terms of the trace, linearity, and independence, we find that
\begin{multline\*}
\mathbb{E} \Vert \mathbf{P}\_1 \mathbf{P}\_2 \Vert\_{\mathrm{F}}^2
= \mathbb{E} \operatorname{trace}[ \mathbf{P}\_2 \mathbf{P}\_1^2 \mathbf{P}\_2 ]
= \mathbb{E} \operatorname{trace}[ \mathbf{P}\_2 \mathbf{P}\_1 \mathbf{P}\_2 ] \\
= \mathbb{E}\_2 \operatorname{trace}[ \mathbf{P}\_2 \mathbb{E}\_1[ \mathbf{P}\_1] \mathbf{P}\_2 ]
= \frac{k}{N} \mathbb{E}\_2 \operatorname{trace}[ \mathbf{P}\_2^2 ]
= \frac{k^2}{N^2} \operatorname{trace}[ \mathbf{I}\_N ]
= \frac{k^2}{N}.
\end{multline\*}
This is the core part of the computation.
Finally, we compute the expected Frobenius projective distance between the pair:
\begin{multline\*}
\mathbb{E} \mathrm{dist}\_{\mathrm{F}}^2(\mathbf{Q}\_1, \mathbf{Q}\_2)
= \mathbb{E}\big[ 1 - k^{-1} \Vert \mathbf{Q}\_1^\* \mathbf{Q}\_2 \Vert\_{\mathrm{F}}^2 \big] \\
= \mathbb{E}\big[ 1 - k^{-1} \Vert \mathbf{P}\_1 \mathbf{P}\_2 \Vert\_{\mathrm{F}}^2 \big]
= 1 - k^{-1} (k^2 / N)
= 1 - k / N.
\end{multline\*}
That's it.
| 1 | https://mathoverflow.net/users/510467 | 452727 | 181,947 |
https://mathoverflow.net/questions/452723 | 3 | For any finite, simple, undirected graphs $G, H$ we denote by $G\times H$ their [categorical product](https://en.wikipedia.org/wiki/Tensor_product_of_graphs). For any graph $G$ we let $G^1 = G$ and for $n\geq 1$ we let $G^{n+1} = G \times G^n$.
It is easy to see that the sequence of the power chromatic numbers $(\chi(G^n))\_{n\in\mathbb{N}}$ is decreasing and so it is eventually constant with a unique value we denote by $\chi^\infty(G)$.
**Question.** Is $\chi^\infty(G)$ [computable](https://en.wikipedia.org/wiki/Computable_function) for finite graphs?
| https://mathoverflow.net/users/8628 | The sequence of the power chromatic numbers $(\chi(G^n))_{n\in\mathbb{N}}$ | Yes.
First note that $\chi(G^k) \leq \chi(G)$. We can colour $(x\_1, \ldots, x\_k)$ with $c(x\_1)$ where $c$ is a colouring of $G$.
Vice versa $\chi(G^k) \geq \chi(H) = \chi(G)$, where $H$ is the induced subgraph of $G^k$ on the vertices $\{(v, v, \ldots, v) \mid v \in V(G)\}$.
Thus $\chi(G^k) = \chi(G)$, which is computable.
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