parent_url
stringlengths 37
41
| parent_score
stringlengths 1
3
| parent_body
stringlengths 19
30.2k
| parent_user
stringlengths 32
37
| parent_title
stringlengths 15
248
| body
stringlengths 8
29.9k
| score
stringlengths 1
3
| user
stringlengths 32
37
| answer_id
stringlengths 2
6
| __index_level_0__
int64 1
182k
|
|---|---|---|---|---|---|---|---|---|---|
https://mathoverflow.net/questions/427958 | 2 | Let $W$ be a standard one dimensional Brownian motion, and let $X$ be the solution to the SDE
$$dX\_t = \sigma(X\_t) \, dW\_t \;, \quad X\_0 = x\_0\;.$$
where $\sigma:\mathbb R \to \mathbb R$ is a Lipschitz continuous function, and $x\_0 \in \mathbb R$ is a fixed constant.
For every $\varepsilon > 0$, let $A\_\varepsilon$ denote the event
$$\{\underset{0 \leq t \leq \varepsilon}{\text{max}} W\_t \geq 1\} \;, $$
and let $\mathbb P^\varepsilon$ be the probability measure given by
$$\mathbb P^\varepsilon (E) = \frac{\mathbb P(E \cap A\_\varepsilon)}{\mathbb P(A\_\varepsilon)} \;, $$
for all events $E$.
We denote by $\mathbb E\_{\mathbb P^\varepsilon}$ the expectation under $\mathbb P^\varepsilon$.
Consider the solution $Y$ to the deterministic ODE
$$dY\_t = \sigma(Y\_t) \, dt \; , \quad Y\_0 = x\_0.$$
**Question:** Is it true that
$$\lim\_{\varepsilon \to 0} \, \mathbb E\_{\mathbb P^\varepsilon} \big [\underset{0 \leq t \leq \varepsilon}{\sup} |X\_t - Y\_{t/\varepsilon} | \, \big] = 0?$$
| https://mathoverflow.net/users/173490 | Short time limits for SDE | Let $\tau = \inf\{ t>0 : W\_t = 1 \}$. The conjecture is **true** and the essence of the proof outlined below appears to be the following peculiar property of the hitting time $\tau$: $$
\lim\_{\epsilon \searrow 0} P\_{\epsilon}[\tau > \epsilon-\epsilon^{3/2}] = 1 \;. $$
(This can be computed directly using the fact that distribution of $\tau$ is [inverse gamma](https://en.wikipedia.org/wiki/Inverse-gamma_distribution) with parameters $1/2$ and $1/2$.) In order to leverage this property, one must carefully split up $e\_t:= X\_t - Y\_{t/\epsilon}$ as outlined below.
>
> **Theorem**. It holds that: $$\lim\_{\epsilon \searrow 0} E[ \sup\_{0 \le t \le \epsilon} |e\_t|^2 ] = 0 \;.$$
>
>
>
**Proof**.
The proof shows that for any $T \in [0, \epsilon]$ $$
E\_{\epsilon} [ \sup\_{0 \le t \le T} |e\_t|^2 ] \le C\_1(\epsilon) + C\_2 (\epsilon) \int\_0^T E\_{\epsilon} [ \sup\_{0 \le r \le s} |e\_r|^2 ] ds
$$ where $C\_1(\epsilon)$ and $C\_2 (\epsilon) $ are non-negative and $\lim\_{\epsilon \searrow 0} C\_1(\epsilon) =0$ and $\lim\_{\epsilon \searrow 0} C\_2(\epsilon) \epsilon = O(1)$. By Grönwall's inequality, $$
E\_{\epsilon} [ \sup\_{0 \le t \le \epsilon} |e\_t|^2 ] \le C\_1(\epsilon) \exp(C\_2 (\epsilon) \epsilon) \;,
$$ and then passing to the limit gives the required result. The remaining details follow.
By Itô's formula, \begin{align\*}
& |e\_t|^2 = \mathrm{I} + \mathrm{II} + \mathrm{III} \quad \text{where} \\
& \mathrm{I}:= \frac{2}{\epsilon} \int\_0^t e\_s (\sigma(X\_s) - \sigma(Y\_{s/\epsilon})) ds \;, \\
& \mathrm{II}:= \frac{2}{\epsilon} \int\_0^t e\_s \sigma(X\_s) (\epsilon dW\_s - ds) \;, \\
& \mathrm{III}:= \int\_0^t \sigma(X\_s)^2 ds \;.
\end{align\*}
---
*Estimate for $\mathrm{I}$.*
This term exclusively contributes to $C\_2(\epsilon)$.
Since $\sigma$ is $L$-Lipschitz for some $L>0$ $$
I \le \frac{2 L}{\epsilon} \int\_0^t |e\_s|^2 ds
$$ and thus $$
\sup\_{0 \le t \le \epsilon} I \le \frac{2 L}{\epsilon} \int\_0^{\epsilon} |e\_s|^2 ds \le \frac{2 L}{\epsilon} \int\_0^{\epsilon} \sup\_{0 \le r \le s} |e\_r|^2 ds \;.
$$
Thus, $C\_2(\epsilon) = 2 L / \epsilon$.
---
*Estimate for $\mathrm{II}$.*
This term contributes to $C\_1(\epsilon)$, and here is where we leverage the aforementioned peculiar property of $\tau$.
\begin{align\*}
& \lim\_{\epsilon \searrow 0} E\_{\epsilon} [ \sup\_{0 \le t \le \epsilon} \left| \mathrm{II} \right| ] =
\lim\_{\epsilon \searrow 0} E\_{\epsilon} [ \sup\_{0 \le t \le \epsilon} \left| \mathrm{II} \right| \mathbf{1}\_{ \{ \tau > \epsilon - \epsilon^{3/2} \} } ] \\
& \quad =
\lim\_{\epsilon \searrow 0} E [ \sup\_{0 \le t \le \epsilon} \left| 2 \int\_0^t e\_s \sigma(X\_s) dW\_s \right| ] = 0
\end{align\*}
Here we took 3 steps that are explained in detail below.
*In the first step*, we used Cauchy-Schwarz to show that $$
\left( E\_{\epsilon} \sup\_{0 \le t \le \epsilon} |\mathrm{II}| \mathbf{1}\_{\{\tau < \epsilon - \epsilon^{3/2} \} } \right)^2 \le \underbrace{E[\sup\_{0 \le t \le \epsilon} |\mathrm{II}|^2 ]}\_{\to O(1)} \, \underbrace{P\_{\epsilon}[ \tau < \epsilon - \epsilon^{3/2}]}\_{\to 0}
$$
*In the second step*, we used a natural splitting and the triangle inequality to write, \begin{align\*}
& E\_{\epsilon} \sup\_{0 \le t \le \epsilon} |\mathrm{II}| \mathbf{1}\_{\{\tau > \epsilon - \epsilon^{3/2} \} } \le \mathrm{II}\_a + \mathrm{II}\_b \quad \text{where} \\
& \mathrm{II}\_a := E\_{\epsilon} \sup\_{0 \le t \le \epsilon} \frac{2}{\epsilon} \left| \int\_0^{t \wedge \tau} e\_s \sigma(X\_s) (\epsilon dW\_s - ds) \right| \mathbf{1}\_{\{\tau > \epsilon - \epsilon^{3/2} \} }\\
& \mathrm{II}\_b := E\_{\epsilon} \sup\_{0 \le t \le \epsilon} \frac{2}{\epsilon} \left| \int\_{t \wedge \tau}^t e\_s \sigma(X\_s) (\epsilon dW\_s - ds) \right| \mathbf{1}\_{\{\tau > \epsilon - \epsilon^{3/2} \} }
\;.
\end{align\*}
To estimate these terms, there are two cases to consider:
* *Case 1: $t \le \tau$*. Then $\mathrm{II}\_b=0$ and $\mathrm{II}\_a$ can be written in terms of the piece of the Brownian bridge up to time $t$; and,
* *Case 2: $t > \tau$*. Then $t>\epsilon - \epsilon^{3/2}$ and hence $\mathrm{II}\_b=O(\epsilon^{1/2})$ and again $\mathrm{II}\_a$ can be written in terms of the piece of the Brownian bridge up to time $\tau$.
In other words, conditioned on the event $(\tau < \epsilon)$, the law of $\epsilon W\_s - s$ is equal to the law of a standard Brownian bridge.
*In the third and last step*, we used [Doob's martingale inequality](https://en.wikipedia.org/wiki/Doob%27s_martingale_inequality#Further_inequalities), [Itô isometry](https://en.wikipedia.org/wiki/It%C3%B4_isometry), and (standard) a priori bounds on $X\_t$ and $Y\_{t/\epsilon}$ over $(0,\epsilon)$. Since the estimate of this term is almost identical to the estimate of $\mathrm{III}$ given below, the details are suppressed.
---
*Estimate for $\mathrm{III}$.*
This term also contributes to $C\_1(\epsilon)$. Noting that $\sigma$ is $L$-Lipschitz, \begin{align\*}
E\_{\epsilon} [ \sup\_{0 \le t \le \epsilon} \mathrm{III} ] &= E\_{\epsilon}[ \int\_0^{\epsilon} [ \sigma(X\_s)^2 ds ] \\
&\le 2 \epsilon \sigma(0)^2 + 2 L^2 E\_{\epsilon}[ \int\_0^{\epsilon} |X\_s|^2 ds ] \\
&\le 2 \epsilon ( \sigma(0)^2 + L^2 |x\_0|^2 ) + 2 L^2 \epsilon \int\_0^{\epsilon} \sigma(X\_s)^2 ds \\
& \quad + 2 L^2 \epsilon \int\_0^{\epsilon} X\_s \sigma(X\_s) dW\_s
\end{align\*}
and as long as $2 L^2 \epsilon \le 1/2$, it follows that
$$
E\_{\epsilon} [ \sup\_{0 \le t \le \epsilon} \mathrm{III} ] \le 4 \epsilon ( \sigma(0)^2 + L^2 |x\_0|^2 ) + 4 L^2 \epsilon \int\_0^{\epsilon} X\_s \sigma(X\_s) dW\_s
$$ The last term in this expression can be treated in a similar way as the last step in the estimate for $\mathrm{II}$, namely [Doob's martingale inequality](https://en.wikipedia.org/wiki/Doob%27s_martingale_inequality#Further_inequalities), Cauchy-Schwarz, [Itô isometry](https://en.wikipedia.org/wiki/It%C3%B4_isometry), and (standard) a priori bounds on $X\_t$ over $(0,\epsilon)$. In particular, \begin{align\*}
\left( E\_{\epsilon} [ \sup\_{0 \le t \le \epsilon} \left| \int\_0^t X\_s \sigma(X\_s) d W\_s \right| ] \right)^2 &\le
E \sup\_{0 \le t \le \epsilon} \left| \int\_0^t X\_s \sigma(X\_s) dW\_s \right|^2\\
&\le 4 E \left| \int\_0^{\epsilon} X\_s \sigma(X\_s) dW\_s \right|^2 \\
&\le 4 E \int\_0^{\epsilon} X\_s^2 \sigma(X\_s)^2 ds \\
&\le 4 \tilde{C}\_2 (1+ x\_0^4) e^{\tilde{C}\_1 \epsilon} \epsilon
\end{align\*}
where in turn we used Cauchy-Schwarz, Doob's martingale inequality with $p=2$, Itô's isometry, and then an a priori bound on the second/fourth moment of $X\_t$ over $(0, \epsilon)$.
$\Box$
| 2 | https://mathoverflow.net/users/64449 | 428708 | 173,788 |
https://mathoverflow.net/questions/428467 | 3 | Let $D\ne 0$ be a linear differential operator with constant coefficients acting on either real or complex valued functions on $\mathbb{R}^n$.
**Is it true that the equation $$Du=f$$
is solvable in any open ball, when $f,u\in C^\infty$?**
| https://mathoverflow.net/users/16183 | Solvability of general linear PDE with constant coefficients | The answer is yes. More generally, a (non-zero) constant coefficient differential operator $P$ induces a surjective map $P:C^\infty(X)\to C^\infty(X)$ for an open set $X\subset \mathbb{R}^n$ if and only if $X$ is $P$-convex for supports. A reference for this result is Theorem 10.6.6 and Corollary 10.6.8 in Hörmander's "The Analysis of Linear PDOs", volume 2.
An open set $X$ is $P$-convex for supports if for every compact set $K\subset X$ there is another compact set $K'\subset X$ such that, if $\phi \in C^\infty\_c(X)$ is such that $\text{supp}\,P(-D)\phi \subset K$, then $\text{supp}\,\phi \subset K'$. The basic result (Theorem 10.6.2 in the aformentioned book) is that every convex open set $X\subset \mathbb{R}^n$ is $P$-convex for supports for every operator $P\not=0$. In particular, $P:C^\infty(X)\to C^\infty(X)$ is surjective for every convex open set $X\subset \mathbb{R}^n$.
| 4 | https://mathoverflow.net/users/78745 | 428709 | 173,789 |
https://mathoverflow.net/questions/428657 | 0 | Let $\mathcal{P}$ be the set of real-valued and strictly stationary processes with expectation zero and finite variance, i.e.:
\begin{equation}
\mathcal{P}:=\left\{ X = (X\_t)\_{t \in \mathbb{Z}} \, : \, X \hbox{ is strictly stationary, } \mathbb{E} X\_t = 0 \hbox{ and } \mathbb{E}[X\_t^2]< \infty, \, \forall\, t \in \mathbb{Z} \right\}
\end{equation}
*Remark:* for any stochastic process $X$, we consider $Q$ the Law of a stochastic process according [this](https://en.wikipedia.org/wiki/Law_(stochastic_processes)). We denote $X \sim Q$.
I'm trying to show whether or not $\mathcal{P}$ is closed according to the Mallows metric:
Let $X = (X\_t)\_{t \in \mathbb{Z}} \sim P$ and $Y = (Y\_t)\_{t \in \mathbb{Z}}\sim Q$ be two stochastic processes. In order to define the Mallows metric, for all $m\in \mathbb{N}$, let $\mathcal{M}\_m$ be the random vectors $(\tilde{X},\tilde{Y})$ having marginals $P\circ\pi\_{1,...,m}^{-1}$ and $Q\circ\pi\_{1,...,m}^{-1}$, where $\pi\_{1,...,m}( (X\_t)\_{t \in \mathbb{Z}} )= (X\_{1},..., X\_{m})$ . So:
$$d( (X\_t)\_{t \in \mathbb{Z}},(Y\_t)\_{t \in \mathbb{Z}})= \sum\_{m=1}^\infty d^{(m)}(P\circ\pi\_{1,...,m}^{-1}, Q\circ\pi\_{1,...,m}^{-1})2^{-m}$$
where
$$d^{(m)}(P\circ\pi\_{1,...,m}^{-1}, Q\circ\pi\_{1,...,m}^{-1}) = \inf\_{(\tilde{X},\tilde{Y})\in \mathcal{M}\_m}{(E||\tilde{X}-\tilde{Y}||^2)^{\tfrac{1}{2}}}.$$
Some hint?
| https://mathoverflow.net/users/478920 | Show that the set of strictly stationary, mean zero and finite variance stochastic processes is closed (or not) | $\newcommand{\Z}{\mathbb Z}\newcommand{\PP}{\mathcal D}\newcommand{\R}{\mathbb R}$Your function $d$ is not a metric, for two reasons: (i) there may be many processes $(X\_t)\_{t\in\Z}$ with the same distribution $P$ and (ii) your function $d$ does not take into account the values of $X\_t$ for negative $t\in\Z$. So, your $d$ is, not a metric, but a [pseudometric](https://en.wikipedia.org/wiki/Pseudometric_space), which does not allow one to identify limits uniquely.
We can fix these deficiencies as follows: Let $\PP$ denote the set of the distributions of the processes in $\mathcal P$.
Given $P$ and $Q$ in $\PP$, for any natural $m$ let
\begin{equation}
P\_m:=P\circ\pi\_{-m,\dots,m}^{-1},\quad Q\_m:=Q\circ\pi\_{-m,\dots,m}^{-1},
\end{equation}
where $\pi\_{r,\dots,s}((x\_t)\_{t\in\Z}):=(x\_r,\dots,x\_s)$ for any given integers $r,s$ such that $r\le s$.
Let
\begin{equation}
d(P,Q):=\sum\_{m=1}^\infty d^{(m)}(P\_m,Q\_m)2^{-m},
\end{equation}
where $d^{(m)}$ is the [Wasserstein metric](https://en.wikipedia.org/wiki/Pseudometric_space) of order $2$.
We want then to show that $\PP$ is closed with respect to the metric $d$.
Suppose now that we have a sequence $(P^{(n)})$ in $\PP$ such that $d(P^{(n)},Q)\to0$ (as $n\to\infty$) for some probability measure $Q$ (on the [cylindrical $\sigma$-algebra](https://en.wikipedia.org/wiki/Cylindrical_%CF%83-algebra)) over $\R^\Z$. Then for each natural $m$ we have $d^{(m)}(P^{(n)}\_m,Q\_m)\to0$. So, by the well-known [characterization of the convergence in the Wasserstein metric](https://math.stackexchange.com/questions/3042947/characterization-of-wasserstein-convergence), $P^{(n)}\_m\to Q\_m$ weakly,
$\int\_{\R^{\Z\_m}} x\_t^2\,Q\_m(dx)=\lim\_n\int\_{\R^{\Z\_m}} x\_t^2\,P^{(n)}\_m(dx)<\infty$, and
$\int\_{\R^{\Z\_m}} x\_t\,Q\_m(dx)=\lim\_n\int\_{\R^{\Z\_m}} x\_t\,P^{(n)}\_m(dx)=\lim\_n0=0$ for $t\in\Z\_m:=\{-m,\dots,m\}$.
So, $\int\_{\R^\Z} x\_t\,Q(dx)=0$ and $\int\_{\R^\Z} x\_t^2\,Q(dx)<\infty$ for all $t\in\Z$, and
$P^{(n)}\_{r,s}\to Q\_{r,s}$ weakly for any given integers $r,s$ such that $r\le s$, where $P^{(n)}\_{r,s}:=P^{(n)}\circ\pi\_{r,\dots,s}^{-1}$ and $Q\_{r,s}:=Q\circ\pi\_{r,\dots,s}^{-1}$.
By the stationarity, $P^{(n)}\_{r+1,s+1}=P^{(n)}\_{r,s}$ for all suitable $r,s,n$. Letting now $n\to\infty$, we conclude that $Q\_{r+1,s+1}=Q\_{r,s}$, so that $Q$ is the distribution of a stationary process. Also, as we saw, $\int\_{\R^\Z} x\_t\,Q(dx)=0$ and $\int\_{\R^\Z} x\_t^2\,Q(dx)<\infty$ for all $t\in\Z$. So, $Q\in\PP$.
We conclude that $\PP$ is closed, as desired.
| 1 | https://mathoverflow.net/users/36721 | 428716 | 173,793 |
https://mathoverflow.net/questions/428714 | 3 | The Rademacher functions are an explicit iid sequence with Bernoulli law. Does it exist an explicit construction of an iid sequence with uniform law?
| https://mathoverflow.net/users/7294 | Uniform iid sequence | $\newcommand\N{\mathbb N}$Let $(r\_k\colon k\in\N)$ be the sequence of the [Rademacher functions](https://en.wikipedia.org/wiki/Rademacher_system). Re-enumerate this sequence into a two-way array $(r\_{i,j}\colon(i,j)\in\N^2)$. So, the $r\_{i,j}$'s are iid random variables (r.v.'s, defined on the standard probability space over the interval $[0,1]$) each uniformly distributed on the two-point set $\{-1,1\}$.
For each $i\in\N$, define the r.v. $U\_i$ on the standard probability space over the interval $[0,1]$ by the formula
$$U\_i:=\sum\_{j=1}^\infty\frac{r\_{i,j}}{2^j}.$$
Then the $U\_i$'s are iid r.v.'s each uniformly distributed on $[-1,1]$.
---
Moreover, if $F$ is any cumulative distribution function (cdf) and $X\_i:=F^{-1}(\frac{1+U\_i}2)$ for $i\in\N$, then the $X\_i$'s are iid r.v.'s each with cdf $F$. Here, as usual,
$$F^{-1}(u):=\inf\{x\in\mathbb R\colon F(x)\ge u\}$$
for $u\in(0,1)$.
So, one can explicitly get the distribution of any sequence of iid real-valued r.v.'s from the single sequence of the Rademacher functions.
| 6 | https://mathoverflow.net/users/36721 | 428718 | 173,795 |
https://mathoverflow.net/questions/428633 | 8 | Are there complete finitely axiomatizable first order theories (with equality) with arbitrarily high computational complexity?
Here, arbitrarily high (computational) complexity means that for every computable predicate $P$ one can choose a theory $T$ (as above) such that $P$ is polynomial time computable from $T$.
*Variations:*
- A variation is to drop equality.
- A natural weakening is to replace 'complete' with 'decidable'.
- Existence of finitely axiomatizable theories with arbitrary c.e. Turing degrees was proved by William Hanf in "Model-theoretic methods in the study of elementary logic". The proof might be adaptable to give complete (or at least decidable) finitely axiomatizable theories with arbitrarily high complexity.
Without finite axiomatization, there are examples:
- If $P⊆ℕ$ is decidable and every finite pattern occurs in $P$, i.e. $∀k \, ∀S⊆k \, ∃n \, ∀k'<k \, (P(n+k') ⇔ S(k'))$, then the monadic second order theory of $(ℕ, <, P)$ is decidable. Proof sketch: Reduce a sentence to acceptance of $P$ by a deterministic infinite automaton; the structure of $P$ ensures that we eventually reach a terminal strongly connected component of states, and reach all of its states infinitely often.
- We can also encode (with linear time encoding and decoding) a decidable predicate on strings for decidability together with S2S (the monadic second order theory of binary strings with functions $s→s0$ and $s→s1$ but no other operations). I suspect S2S with a decidable predicate $P$ is decidable if $P$ is sufficiently generic, specifically $∀S \, ∀k∈ℕ \, ∀(\text{infinite branch } T)$ $∃t∈T \, ∀v (|v|<k) (P(tv) ⇔ S(v))$ where $|v|$ is the length of $v$.
Without first order logic, there are finitely presented equational theories of arbitrarily high complexity, including I believe the word problem for a finitely presented group.
With first order logic, for appropriate computations I think the theory of the computation graph with a sequential write-only output tape is decidable, but I am not sure how to formalize and prove it. Note that first order logic can only express local properties, and for bounded graph degree, formulas can be converted to Hanf normal form. Perhaps, there are also groups with finitely axiomatizable theories of arbitrarily high complexity.
As an aside, for "natural" (not necessarily finitely axiomatizable) decidable theories, the highest complexity that I know is iterated exponential, which occurs for S2S, S1S, WS2S, WS1S, and related theories. For natural decidable problems more generally, reachability in vector addition systems has Ackermann complexity.
| https://mathoverflow.net/users/113213 | Decidable theories with arbitrary complexity | There are complete decidable finitely axiomatizable theories of arbitrary high computational complexity.
As Dmytro already sketched in his question there are complete decidable theories of arbitrary high computational complexity. Now we want to switch to finitely axiomatizable theories. This could be achieved using a theorem of Peretyatkin (see his book [1] and also a review of his book [2]; I haven't read the book myself and taking the formulation of the theorem from [2]):
**Theorem**(Peretyat'kin) There is a method of constructing, effectively in the c.e. index m of T, a finitely axiomatizable model-complete theory F(m) and an effective interpretation I of the theory T in F(m) such that I induces one-one correspondece between the completions of T and F(m).
Since a theory is complete and consistent iff it has a unique completion, if we apply this to a complete decidable theory $T$, then we would get a complete finitely axiomatizable theory $T'$ interpreting $T$. And, of course, the presence of the interpretation guarantees that $T$ is polynomial time reducible to $T'$. As James Hanson pointed in the comments, any complete finitely axiomatizable theory is decidable, hence we get complete decidable finitely axiomatizable theories of arbitrary high computational complexity.
*Comment:* Dmytro's example of complete decidable theories of arbitrary high computational complexity could be slightly simplified if instead of the monadic theory of $(\mathbb{N},<,P)$ the elementary theory of $(\mathbb{N},S,P)$ would be considered; then decidability could be proved by a more elementary quantifier elimination argument instead of Büchi automatons, see <https://mathoverflow.net/a/426389/36385>. There also an option to use the technique from my paper with Juvenal Murwanashyaka and Albert Visser [3]. There we define theory $\mathsf{J}$ (based on a theory earlier considered by Janiczak) and a polynomial time computable sequence of sentences $\mathsf{A}\_i$ indexed by naturals $i$ such that $\mathsf{J}$ is consistent with any propositionally consistent Boolean combination of $\mathsf{A}\_i$ and over $\mathsf{J}$ any sentence is equivalent to a Boolean combination of $\mathsf{A}\_i$. Now given a decidable set $P$ we could consider theory $$\mathsf{J}+\{\mathsf{A}\_i\mid i\in P\}\cup\{\lnot\mathsf{A}\_i\mid i\not\in P\},$$
that is a complete consistent decidable theory such that $P$ is polynomial time reducible to it.
P.S. As I already indicated in the comments, I have learned about the book of Peretyat'kin from Albert Visser.
[1]Mikhail, G. Peretyat'kin. "Finitely Axiomatizable Theories." Siberian School of Algebra and Logic. Consultants Bureau, New York (1997).
[2]Vivienne Morley. Review of M.G. Peretyat'kin "Finitely Axiomatizable Theories.". <https://doi.org/10.2307/2586818>
[3]Juvenal Murwanashyaka, Fedor Pakhomov, and Albert Visser. "There are no minimal essentially undecidable Theories." arXiv preprint arXiv:2207.08174 (2022). <https://arxiv.org/abs/2207.08174v2>
| 7 | https://mathoverflow.net/users/36385 | 428749 | 173,803 |
https://mathoverflow.net/questions/428642 | 3 | Let $K$ be a finite extension field of $\mathbb{Q}\_p$. Let us consider a semiabelian variety $G$ defined over $K$, i.e there exists an extension of an abelian variety $B$ and a torus $T$ defined over $K$
$$0 \rightarrow T \rightarrow G \rightarrow A \rightarrow 0$$
It is know [that](https://mathoverflow.net/questions/112710/is-the-n-torsion-of-an-extension-of-an-abelian-variety-by-a-torus-finite-and-fl) $G[n]$ is finite flat. My question is whether we can have an exact sequence
$$0 \rightarrow T[p^\infty](L) \rightarrow G[p^\infty](L) \rightarrow A[p^\infty](L) \to 0 \;?$$
where $X[p^\infty](L)$ is $p$-power torsion points of $X$ over any extension $L$ of $K$.
In particular, do we have that if $G[p^\infty](L)$ is finite then is $A[p^\infty](L)$?
| https://mathoverflow.net/users/146212 | $p$-power torsion of semiabelian variety | $\newcommand{\Spec}{\mathrm{Spec}}\newcommand{\oL}{\overline{L}}\newcommand{\bG}{\mathbb{G}}\newcommand{\bZ}{\mathbb{Z}}\newcommand{\cL}{\mathcal{L}}$Not in general. The sequence of $p$-divisible groups $0\to T[p^{\infty}]\to G[p^{\infty}]\to A[p^{\infty}]\to 0$ is exact as a sequence of sheaves of abelian groups on the
étale site of $\Spec\, K$ hence there is an exact sequence
$$0\to T[p^{\infty}](L)\to G[p^{\infty}](L)\to A[p^{\infty}](L)\xrightarrow{\delta} H^1(G\_L,T[p^{\infty}](\oL))$$
I will describe the connecting homomorphism $\delta$ in terms of the Kummer map in the case $T=\bG\_m$ and then will say how to witness that it does not vanish in general.
Recall the Kummer map: the short exact sequence $0\to A[p^n]\to A\to A\to 0$ of fppf sheaves on $\Spec K$ gives rise to the exact sequence $0\to A[p^n](L)\to A(L)\xrightarrow{p^n}A(L)\to H^1(G\_L,A[p^n](\oL))$ and hence an embedding $\kappa\_{A,p^n}: A(L)/p^n\to H^1(G\_L,A[p^n](\oL))$.
The data of an extension $0\to \bG\_m\to G\to A\to 0$ is equivalent to the choice of a degree zero line bundle $\cL$ on $A$ (to a line bundle $\cL$ corresponds the semi-abelian variety $G\_{\cL}$ which is the total space of $\cL$ equipped with the unique group structure compatible with the projection down to $A$). Consider the corresponding point $[\cL]\in (\mathrm{Pic}^0A)(L)=A^{\vee}(L)$ of the dual abelian variety, and send it along the Kummer map $\kappa\_{A^{\vee},p^n}$ to get a Galois cohomology class $\kappa\_{A^{\vee},p^n}([\cL])\in H^1(G\_L,A^{\vee}[p^n](\oL))$
**Lemma.** The homomorphism $\delta|\_{A[p^n](L)}$ is the cuprpoduct with the class $\kappa\_{A^{\vee},p^n}([\cL])$ followed by the Weil pairing:
$$A[p^n](L)\xrightarrow{\kappa\_{A^{\vee},p^n}([\cL])\,\cup\,-}H^1(G\_L,A^{\vee}[p^n](\oL)\otimes A[p^n](\oL))\to H^1(G\_L,\mu\_{p^n}(\oL))\to H^1(G\_L,\mu\_{p^{\infty}}(\oL))$$
Before sketching a proof, let me remark that the important information that we will get from this lemma is that $\delta$ varies non-trivially when variying the line bundle $L$, thanks to the Weil pairing being perfect; this is the point of rewriting $\delta$ like this, even if the formula appears more complicated than the definition of $\delta$.
Idea of the proof of the lemma: The Weil pairing identifies $A^{\vee}[p^n]$ with the internal group scheme of homomorphisms $Hom(A[p^n],\mu\_{p^n})$ and it follows from the construction of the pairing in Theorem 8.1.3 [here](http://virtualmath1.stanford.edu/%7Econrad/249CS15Page/handouts/abvarnotes.pdf) that for a degree zero line bundle $\cL$ the class of the extension $0\to\mu\_{p^n}\to G\_{\cL}[p^n]\to A[p^n]\to 0$ of etale sheaves of $\bZ/p^n$-modules in $\mathrm{Ext}^1\_{L\_{et}}(A[p^n],\mu\_{p^n})=H^1(G\_L,Hom(A[p^n],\mu\_{p^n})(\oL))=H^1(G\_L,A^{\vee}[p^n](\oL))$ coincides with the image of $[\cL]$ under the Kummer map. This implies the statement because, by definition, $\delta$ is the cup-product with this class in $H^1(G\_L,Hom(A[p^n],\mu\_{p^n})(\oL))$
We can now give an example when $\delta$ is non-zero and therefore the map $G[p^{\infty}](L)\to A[p^{\infty}](L)$ is not surjective. Fix an abelian variety $A$ and an integer $n$ and take $L$ to be a large enough finite extension of $K$ such that $A[p^n](L)=A[p^n](\overline{K})$ and $A^{\vee}[p^n](L)=A^{\vee}[p^n](\overline{K})$. Note that this condition says that the Galois actions of $G\_L$ on $A[p^n](\oL)$ and $A^{\vee}[p^n](\oL)$ are trivial (and therefore so is the action on $\mu\_{p^n}(\oL)$).
It remains to choose the line bundle $\cL$ appropriately: if our goal was to prove only that the composition $A[p^n](L)\xrightarrow{\kappa\_{A^{\vee},p^n}([\cL])\,\cup\,-}H^1(G\_L,A^{\vee}[p^n](\oL)\otimes A[p^n](\oL))\to H^1(G\_L,\mu\_{p^n}(\oL))$ is non-zero, we could have taken any $\cL\in A^{\vee}(L)$ whose image under the Kummer map is non-zero. Indeed, by the triviality of the Galois action on $A[p^n](\oL)$ and $A^{\vee}[p^n](\oL)$ we factor them out of Galois cohomology: $H^1(G\_L,A^{\vee}[p^n](\oL)\otimes A[p^n](\oL))=H^1(G\_L,\bZ/p^n)\otimes A^{\vee}[p^n](L)\otimes A[p^n](L)$ and the claim follows from the Weil pairing being non-degenerate.
To make sure that the image of $A[p^n](L)$ in $H^1(G\_L,\mu\_{p^{\infty}}(\oL))$ is still non-zero we need to choose $\cL$ with a bit more care as the map $H^1(G\_L,\mu\_{p^n}(\oL))\to H^1(G\_L,\mu\_{p^{\infty}}(\oL))$ has a non-zero kernel given by $\mu\_{p^{\infty}}(L)/(p^n\cdot \mu\_{p^{\infty}}(L))$. But $\mathrm{rk}(A(L)\otimes\bZ\_p)=\dim A\cdot [L:\mathbb{Q}\_p]$ while that kernel is a cyclic group, so we can always choose $\cL$ that will do the job.
| 2 | https://mathoverflow.net/users/39304 | 428753 | 173,805 |
https://mathoverflow.net/questions/428750 | 8 | In [[KM63]](https://www.maths.ed.ac.uk/%7Ev1ranick/papers/kervmiln.pdf), Kervaire and Milnor introduced the *group of homotopy spheres*. Its elements are h-cobordism classes of smooth homotopy $n$-spheres under the summation induced by connected sum. Further, the trivial element is $S^n$ and this group is denoted by $\Theta^n$.
They proved that $\Theta^n$ is finite unless $n=3$, in particular $\Theta^4$ is trivial.
This should be an ambiguous question but I wonder this provides a positive clue for the smooth Poincaré conjecture in dimension 4.
| https://mathoverflow.net/users/475366 | Kervaire-Milnor group of homotopy spheres and smooth Poincaré conjecture | That a homotopy 4-sphere is h-cobordant to $S^4$ is in principle a step towards proving the 4-dimensional Poincaré conjecture. But it's known from Donaldson's work that the h-cobordism theorem is false for simply connected closed $4$-manifolds. Indeed the step that fails is in cancelling handles that homologically cancel, and that issue would come up in trying to prove the Poincaré conjecture starting from an h-cobordism.
So unless there's something special about trivializing an h-cobordism between homotopy spheres that doesn't hold for slightly more complicated $4$-manifolds, I'd say that this fact isn't much of a clue.
Just an opinion, of course!
| 12 | https://mathoverflow.net/users/3460 | 428758 | 173,806 |
https://mathoverflow.net/questions/428755 | 10 | I was recently compiling some notes for an undergrad-level course on number theory, and I went over the proof of the fact that $(\mathbb{Z}/p\mathbb{Z})^\times$ is cyclic for any prime $p$: it's a finite abelian group and thus the direct sum of cyclic groups, and the fact that $X^r - 1\in (\mathbb{Z}/p\mathbb{Z})[X]$ has at most $r$ zeros forces $(\mathbb{Z}/p\mathbb{Z})^\times$ to be cyclic itself. It's a completely nonconstructive proof, and there aren't many tools available at that level for digging any deeper into the problem.
So, what is the current state of the art for primitive roots mod $p$? I'm not a number theorist, and I don't know much about the problem aside from Artin's conjecture (which, as far as I know, is still a conjecture). I think it's known that it holds for infinitely many primes unconditionally and all but a small, finite number modulo a particular form of the generalized Riemann hypothesis, but those results are both several decades old. (Even the most relevant questions on this site I could find are more than a decade old.) Are there any newer results or any new methods that look promising, or is this just an untractable problem for now? Or, for that matter, is the last result I mentioned considered a satisfying answer? (To clarify, I'm asking for my own benefit and am not looking for an undergrad-level answer.)
| https://mathoverflow.net/users/61829 | State of the art for primitive roots | Artin's conjecture on primitive roots has a qualitative version and quantitative version. The qualitative version says if $a \in \mathbf Z$ is not $-1$ or a perfect square then $a \bmod p$ is a primitive root mod $p$ for infinitely many $p$. The quantitative version says for such $a$ that the number of $p \leq x$ such that $a \bmod p$ is a primitive root mod $p$ is asymptotic to $c\_ax/\log x$, where $c\_a$ is an explicit positive constant depending on $a$. The quantitative version was completely proved by Hooley assuming GRH for zeta-functions of a certain infinite set of number fields (no "all, but a small finite number" as you wrote).
In what I write below, I will be referring to the qualitative version of the conjecture ("infinitely many").
You wrote "I think it's known that it holds for infinitely many primes unconditionally". The correctness of that statement depends on the kind of result you are after.
1. For *no* specific integer $a$ in $\mathbf Z$ is it known (without using GRH) that $a \bmod p$ is a primitive root for infinitely many $p$.
2. In 1984, Gupta and Murty showed there are infinitely many $a$ such that Artin's conjecture is true for $a$, but they could not make that infinite list of $a$ completely explicit. See Theorem 4 in Murty's Math. Intelligencer article on Artin's conjecture for a precise statement, or the original paper of Gupta and Murty [here](https://eudml.org/doc/143169), where you'll see the main result relies on quantitative estimates, but not involving the constant $c\_a$ from the quantitative form of Artin's conjecture.
3. In 1985, Heath-Brown proved that Artin's primitive root conjecture is true for all but *at most two* prime values of $a$ and all but *at most three* squarefree $a > 1$. For example, at least one of $2$, $3$, or $5$ is a primitive root mod $p$ infinitely often. Certainly we expect the conjecture is true for every prime value of $a$ and every squarefree value of $a > 1$ with no exceptions at all, but that can't be shown without using GRH as in Hooley's work.
There are a *lot* of generalizations of the original Artin primitive root conjecture: replace powers of a single integer mod $p$ by powers of a finite set of integers mod $p$ (or a finite set of rational numbers mod $p$, ignoring the finitely many $p$ dividing one of the denominators), by powers of an algebraic integer modulo prime ideals, by multiples of the mod $p$ reductions of a rational point on an elliptic curve over $\mathbf Q$, and so on. Many of these generalizations can be proved using some version of GRH, or weaker versions can be proved without GRH using sieve theory.
| 28 | https://mathoverflow.net/users/3272 | 428764 | 173,809 |
https://mathoverflow.net/questions/428785 | 6 | I'm trying to produce a toy version of the RH Weil conjecture. Solving this could help me to get a good start at understanding where the $1/2$'s come in here, ideally without having to prove the Hard Lefschetz theorem.
I'm restricting attention to the scheme $X=\mathbb{F}\_p \mathbb{P}^1$. Write $x = \text{Spec}(\mathbb{F}\_p)$, $y = \text{Spec}(\mathbb{F}\_p^{closure})$, and $Y = X \times\_{x} y$. My goal is to show the RH Weil conjecture for $Y$, but I suspect that the fact that $Y$ is $1$ dimensional projective space simplifies things a great deal. Meanwhile, I've worked through how to show that
1. $l$-adic cohomology has a poincare duality for $k \mathbb{P}^1$.
2. $l$-adic cohomology has a Kunneth formula and proper base change theorem.
3. There is a Lefschetz fixed point theorem for $X$ and $Y$.
My hope here is that there is to get a proof that
>
> Let $F$ be the Frobenius endomorphism of $Y$. Then the eigenvalues of $F$ acting on $H^r(Y, \mathbb{Q}\_l)$ have absolute value $q^{r/2}$.
>
>
>
I have also shown that the inner product $H^\*(Y, \mathbb{Q}\_l) \otimes H^\*(Y, \mathbb{Q}\_l) \rightarrow \mathbb{Q}\_l(2)$ is fixed by the Frobenius map.
I'm hoping that since $X$ and $Y$ here are pretty simple spaces, that I don't need the Hard Lefschetz theorem. Could this be possible?
[TEST\_1](https://www.google.com/?client=safari)
| https://mathoverflow.net/users/30211 | Zeta function of $X = \mathbb{F}_p \mathbb{P}^1$ | Deligne proved the hard Lefschetz theorem as a consequence of the Weil conjectures, not vice versa.
The RH of the Weil conjectures is a statement about the Weil zeta function. For the projective line, this is easy to prove without étale cohomology at all, only by counting points.
But of course you could ask for a proof that the eigenvalues of Frobenius on $H^i$ have size $q^{i/2}$ (Deligne's RH) in this special case. This is indeed much easier than the general case!
Since $Y$ is connected, it's straightforward to check from the definitions that $H^0 (Y, \mathbb Q\_\ell) =\mathbb Q\_\ell$, with trivial Frobenius action. So this verifies the $i=1$ case since $q^{0/2}=1$.
Next one checks that $H^1(Y,\mathbb Q\_\ell)=0$, which vacuously verifies the $i=1$ case. Probably the easiest way to do this is to use the relation between the first étale cohomology and the étale fundamental group, and show the étale fundamental group is trivial by Riemann-Hurwitz.
Finally, using the Poincaré duality $H^0(Y,\mathbb Q\_\ell) \times H^2(Y, \mathbb Q\_\ell) \to \mathbb Q\_\ell(-1)$ (not $\mathbb Q\_\ell(2)$), one sees that $H^2(Y,\mathbb Q\_\ell) \cong \mathbb Q\_\ell(-1)$ and thus the unique Frobenius eigenvalue on it is $q$, verifying the $i=2$ cases since $q^{2/2}=q$.
Whether that explains where the $1/2$ comes in sufficiently well is up to you! If not, I would look at the case of abelian varieties, where $H^1$ is nontrivial, and the RH can be proved by the theory of endomorphisms of abelian varieties.
| 10 | https://mathoverflow.net/users/18060 | 428786 | 173,816 |
https://mathoverflow.net/questions/428781 | 18 | Is there a $C^\infty$-smooth embedding $\gamma : I \to \mathbb{R}^3$ so that there is no real analytic $2$-dimensional submanifold $M \subset \mathbb{R}^3$ with $\gamma(I)\subset M$?
| https://mathoverflow.net/users/1540 | Smooth curve in $\mathbb{R}^3$ not contained in real analytic surface? | Here is an example: Let $\gamma:\mathbb{R}\to\mathbb{R}^3$ be defined by $\gamma(t) = \bigl(t,\exp(-1/t^2),0\bigr)$ for $t<0$, $\gamma(0) = (0,0,0)$ and $\gamma(t) = \bigl(t,0,\exp(-1/t^2)\bigr)$ for $t>0$. Then I claim that there is no nonsingular real-analytic surface $M\subset\mathbb{R}^3$ that contains the image of $\gamma$.
To see this, suppose that such an $M$ does exist and let $u\in S^2$ be a unit normal to $M$ at the origin. If $u\not=(0,0,\pm1)$, then $M$ and the plane $z=0$ intersect transversely at $(0,0,0)$ along a real-analytic curve that contains $\gamma\bigl((-\epsilon,0]\bigr)$ for some $\epsilon>0$, which is clearly impossible. Similarly, we reach a contradiction if $u\not=(0,\pm1,0)$. Thus, $M$ does not exist.
| 20 | https://mathoverflow.net/users/13972 | 428800 | 173,821 |
https://mathoverflow.net/questions/428797 | 4 | Let $A$ be a $C^\*$ algebra. A $C^\*$ subalgebra $C\subset A$ is said to be $C^\*$ algebraic complemented of $A$ if there exist a $C^\*$ subalgebra $D\subset A$ with $A=C\oplus D$ and the obvios mapping $A\mapsto C\oplus D$ preserves all algebraic structure.
In this question we consider the particular case $A=\ell^\infty$. Inspired by the result of Joram Lindenstrauss On complemented subspace of m, Israel Journal of Mathematics, volume 5, pages153{156, 1967. which says that the only infinit dimensional complemented subspace of $\ell^\infty$ are those Banach subspace which are isomorphism to $\ell^\infty$ we ask the following $C^\*$ algebraic question:
>
> Is it true to say that an infinite dimensional $C^\*$ subalgebra of $\ell^\infty$ is $C^\*$ algebraic complemented if and only if it is $C^\*$ isomorphic to $\ell^\infty$?
>
>
>
**Note:** This would suggest characterization of all $C^\*$ algebras whose only infinite dimensional $C^\*$ algebraic complemented subalgebras are isomorphic to itself.
| https://mathoverflow.net/users/36688 | A $C^*$ algebraic analogy of the concept of complemented subspace in the particular case of $\ell^\infty$ | Suppose $l^\infty \cong C \oplus D$ (a C${}^\*$-direct sum). Then both $C$ and $D$ must have units, which will be projections in $l^\infty$, and hence must be of the form $1\_X$ and $1\_{\mathbb{N}\setminus X}$ for some $X \subseteq \mathbb{N}$. Then every element of $C$ must be supported on $X$ and every element of $D$ on $\mathbb{N}\setminus X$, so $C = l^\infty(X) \subseteq l^\infty$. If $C$ is infinite-dimensional it clearly must be C${}^\*$-isomorphic to $l^\infty$.
The converse is false: let $C \subset l^\infty$ consist of all those sequences satisfying $a\_{2n} = a\_{2n+1}$ for all $n$. This is clearly C${}^\*$-isomorphic to $l^\infty$ but it does not have the above form and therefore is not C${}^\*$-complemented.
| 7 | https://mathoverflow.net/users/23141 | 428804 | 173,822 |
https://mathoverflow.net/questions/428619 | 2 | Let $R$ be a regular local ring and $M$ a finitely generated reflexive $R$-module. When $R$ has dimension 2, then $M$ is a free $R$-module. This is discussed in [Reflexive modules over a 2-dimensional regular local ring](https://mathoverflow.net/questions/22943/reflexive-modules-over-a-2-dimensional-regular-local-ring).
Suppose $R$ has dimension 3. Are there any results about the structure of $M$ in this case?
| https://mathoverflow.net/users/61621 | Structure of reflexive modules over regular local rings | If $R$ is a Gorenstein (or even Cohen-Macaulay and Gorenstein in codimension $1$) local ring, then $M$ is reflexive if and only if it is a second syzygy (see e.g. [this answer](https://math.stackexchange.com/q/3487112)). As regular rings are Gorenstein, this directly extends the result you quote for regular local rings of dimension $2$. Indeed, a regular local ring of dimension $2$ has global dimension $2$, so second syzygies are free.
| 2 | https://mathoverflow.net/users/155965 | 428813 | 173,825 |
https://mathoverflow.net/questions/428807 | 4 | Given an integral positive-definite rank $n$ quadratic form $f$, one can use the algorithm in Conway and Sloane (Chapter 15, [SPLaG](https://link.springer.com/book/10.1007/978-1-4757-6568-7)) to efficiently determine if the genus of $f$ contains more than one spinor genus. My question is: given two (integral positive-definite) forms $f,g$ in the same genus, such that the genus contains more than one spinor genus, can one efficiently determine if the forms lie in distinct spinor genera?
I am interested in $n\geq5$.
| https://mathoverflow.net/users/164802 | Computing spinor equivalence for positive definite forms | Yes: one can write down the explicit local transformations at all primes where they are not both unimodular and evaluate the spinor norms and the local automorphism groups. Magma in fact claims to implement such an algorithm, but I don't have personal experience with it.
| 7 | https://mathoverflow.net/users/6084 | 428826 | 173,832 |
https://mathoverflow.net/questions/428824 | 1 | If $G = (V, E)$ is a simple, undirected graph and $T \subseteq V$, let $$N(T) = \{v \in V: \{v, t\}\in E \text{ for some }t\in T\}.$$
Given $v\in V$ we let $N\_0(v) = \{v\}$ and $N\_{k+1}(v) = N\_k(v) \cup N(N\_k(v))$ for all $k\geq 1$. The *iterated degree sequence* of $v$, denoted by $(\text{deg}\_k(v))\_{k\in\omega}$, is defined by $$\text{deg}\_k(v) = |N\_k(v)|\text{ for every }k\in \omega.$$
To every finite graph $G = (V,E)$ we associate the *iterated degree matrix* $\mathbb{D}(G) \in \mathbb{N}^{n\times n}$ (where $n=|V|$) in the following way: for every $v\in V$, take the first $n$ elements of its iterated degree sequence; order these $n$-element integer vectors [lexicographically](https://en.wikipedia.org/wiki/Lexicographic_order), and put these lexicographically ordered vectors in the matrix.
**Question.** Are there finite $G\_i = (V\_i, E\_i)$ for $i = 1,2$ with $|V\_1| = |V\_2|$, $\mathbb{D}(G\_1) = \mathbb{D}(G\_2)$, but $\chi(G\_1) \neq \chi(G\_2)$?
| https://mathoverflow.net/users/8628 | Do graphs with identical degree matrix have the same chromatic number? | There are such examples. Take two graphs with the same degrees but different chromatic number (for example, a cycle of length 6 and two triangles). Add a vertex of full degree to both.
| 5 | https://mathoverflow.net/users/4312 | 428827 | 173,833 |
https://mathoverflow.net/questions/428738 | 1 | For $n\geq 2$, $P\mathbb{R}^n$ is a simple example of finite polyhedron with finitely generated simple fundamental group. I was wondering if someone could give me an example of a finite polyhedron with infinite finitely generated simple fundamental group. Thanks in advance.
Here by a simple group I mean in the group theoretical sense.
| https://mathoverflow.net/users/114476 | Examples of finite polyhedra with finitely generated simple fundamental group | As suggested in the comments, what you are asking for is essentially the presentation complex of a finitely presented, infinite, simple group. Thus it suffices to exhibit a presentation for such a group. Some are known, but not many.
Probably the easiest examples are Thompson's groups $T$ and $V$. Google gives me a link to an explicit finite presentation for $T$ in §11 of some [notes](http://www.macs.hw.ac.uk/%7Eaal7/docs/Introduction%20to%20Thompson%27s%20Groups%20F,%20T%20and%20V.pdf) of Levine, based on the classic notes of Cannon, Floyd and Parry.
Even more remarkable examples were constructed by Burger and Mozes. Their examples are CAT(0) amalgams of free groups, and in particular their presentation complex is aspherical. [This survey](https://arxiv.org/abs/1709.05949) of Caprace is a good place to start learning about these. It looks like the smallest known example is an amalgam of free groups of the form $F\_7\*\_{F\_{49}}F\_7$ (where the subscripts indicate the ranks of the free groups). [UPDATE: Carl-Fredrik Nyberg Brodda points out in comments that there is now an example of the form $F\_3\*\_{F\_{11}}F\_3$.]
Finally, if you would be satisfied with a group without non-trivial *finite* quotients, then *Higman's group*
$\langle a,b,c,d\mid bab^{-1}a^{-2}, cbc^{-2}b^{-2}, dcd^{-1}c^{-2},ada^{-1}d^{-2} \rangle$
provides a fairly digestible example.
| 4 | https://mathoverflow.net/users/1463 | 428843 | 173,838 |
https://mathoverflow.net/questions/428840 | 6 | This question is basically a lift to MO of a part of [an old MSE question](https://math.stackexchange.com/questions/4221319/if-a-structure-has-a-definable-ordered-pairing-function-must-it-have-a-definabl). That question asked, roughly, for the model-theoretic relationship between ordered and unordered pairing functions. To keep things precise, given a structure $\mathcal{A}$ say that
* $\mathcal{A}$ **has a pairing function** iff there is a formula $\varphi(x,y,z)$ with parameters from $\mathcal{A}$ such that $$\mathcal{A}\models\forall x,y\exists!z\varphi(x,y,z)\wedge \forall x,y,x',y',z(\varphi(x,y,z)\wedge\varphi(x',y',z)\leftrightarrow x=x'\wedge y=y'),$$ and
* $\mathcal{A}$ **has an unordered pairing function** iff there is a formula $\varphi(x,y,z)$ with parameters from $\mathcal{A}$ satisfying the same condition but with the clause $x=x'\wedge y=y'$ replaced by $(x=x'\wedge y=y')\vee (x=y'\wedge y=x')$.
Any of the usual definitions of set-theoretic ordered pairs show that every structure that has an unordered pairing function also has a pairing function. The converse, however, is open:
>
> Is there a structure $\mathcal{A}$ which has a pairing function but does not have an unordered pairing function?
>
>
>
There is a natural guess at such a structure: namely, $$\mathcal{G}=(\mathbb{N};g)$$ where $g$ is a "sufficiently generic" bijection $\mathbb{N}^2\rightarrow\mathbb{N}$ with respect to the poset of finite partial injections $\mathbb{N}^2\rightarrow\mathbb{N}$ ordered by reverse inclusion as usual. I suspect there's actually an easy argument here, but I'm not seeing it:
>
> Is $\mathcal{G}$ as defined above indeed an example of the type of structure desired?
>
>
>
| https://mathoverflow.net/users/8133 | Are ordered and unordered pairing functions "definably equivalent?" | The answer is no. Consider absolutely free algebra $\mathfrak{A}=(A;0,\langle \cdot,\cdot\rangle)$ with one binary function $\langle \cdot,\cdot\rangle$ and one generator $0$. Clearly, $\langle \cdot,\cdot\rangle$ is a pairing function on $\mathfrak{A}$. As I will show below there are no definable unordered pairing functions on this structure.
The elementary theory of $\mathfrak{A}$. admits quatifier elimination in the language extended by functions $\pi\_1(x),\pi\_2(x)$, where $\pi\_1(0)=\pi\_2(0)=0$, $\pi\_1(\langle x,y\rangle)=x$, and $\pi\_2(\langle x,y\rangle)=y$ (this follows from a more general result of Mal'tsev[1]).
Suppose for a contradiction that we have an unordered pairing function $\varphi(x,y,z)$. We put in quantifier-free form. Since over $\mathfrak{A}$ we could transform equality $\langle t,u\rangle=v$ to the conjunction $t=\pi\_1(v)\land u=\pi\_2(v)$, we equivalently transform $\varphi(x,y,z)$ to a quantifier-free formula without function $\langle \cdot,\cdot\rangle$. Hence each term in $\varphi(x,y,z)$ is of the form $\pi\_{i\_1}\ldots\pi\_{i\_m}(w)$, where $w\in \{x,y,z\}$, or of the form $\pi\_{i\_1}\ldots\pi\_{i\_m}(0)$; further we transform all the terms of the latter form to just $0$.
Let terms $t\_n(x)$ be $$t\_0(x)=x\text{ and }t\_{n+1}(x)=\langle t\_n(x),0\rangle.$$ By choosing $n$ to be equal to the maximal length of $\pi$-function chains in $\varphi$-terms we observe that $\varphi(t\_n(x),t\_n(y),z)$ is equivalent to a quantifier-free formula $\psi(x,y,z)$, where all terms are either $0,x,y$, or of the form $\pi\_{i\_1}\ldots\pi\_{i\_m}(z)$. Of course, $$\mathfrak{A}\models \forall x,y\exists!z \;\psi(x,y,z).$$ Now it is easy to see that there should be a term $u(x,y)$ using just $0$ and $\langle\cdot,\cdot\rangle$ such that $$\mathfrak{A}\models \forall x,y,z( \psi(x,y,z)\land x\ne y\land x\ne 0\land y\ne 0 \to z=u(x,y)).$$
However, for any $a\ne b$, $a\ne 0$, $b\ne 0$ from $\mathfrak{A}$, clearly $\mathfrak{A}\models u(a,b)\ne u(b,a)$. We should have had
$$\mathfrak{A}\models \forall z\; (\psi(a,b,z)\mathrel{\leftrightarrow} \psi(b,a,z)).$$
However, $\mathfrak{A}\models \psi(a,b,u(a,b))$ and $\mathfrak{A}\not\models \psi(b,a,u(a,b))$, contradiction.
[1]Anatolii I. Mal’tsev, On the elementary theories of locally free universal algebras, Doklady Akademii Nauk SSSR 138 (1961), no. 5, pp. 1009–1012 (in Russian), English translation in: Soviet Mathematics – Doklady 2 (1961), no. 3, pp. 768–771.
| 4 | https://mathoverflow.net/users/36385 | 428845 | 173,840 |
https://mathoverflow.net/questions/428838 | 8 | What are some examples of Tate rings $R$ (i.e. Huber rings with with topologically nilpotent units) which are Noetherian but *not* strongly Noetherian ($R$ is strongly Noetherian iff for all $n \in \mathbb{N}$, the corresponding $R$-algebra of convergent power series $R\{x\_1, ..., x\_n\}$ is Noetherian) ?
| https://mathoverflow.net/users/143390 | Noetherian but not strongly Noetherian | The only example I know of occurs in [*On Hausdorff completions of commutative rings in rigid geometry*](https://www.sciencedirect.com/science/article/pii/S0021869311000524) by Fujiwara, Gabber, Kato (and according to the intro the example is due to Gabber). The example is the subject of $\S$8.3, in their language they give an example of a $t$-adically complete ring $A\_0$ which is Noetherian outside $(t)$ but $A\_0\langle x\rangle$ is not noetherian outside $(t)$. In this setting $A:=A\_0[1/t]$ has a unique structure of a Tate-Huber ring making $A\_0$ a ring of definition, and then $A$ being Noetherian is the same as $A\_0$ being Noetherian outside $(t)$, so one gets a Noetherian Tate ring $A$ for which $A\langle x\rangle$ is not Noetherian. In addition they prove in Lemma 8.3.3 that $A$ is in fact a field (but of course not a nonarchimedean field).
| 8 | https://mathoverflow.net/users/82848 | 428846 | 173,841 |
https://mathoverflow.net/questions/428170 | 0 | I would like to know if, under Ramanujan conjecture, the following three distributions are known or conjectured to match:
1. the distribution of spacings between Satake parameters of an L-function $F$ at all unramified primes on the unit circle
2. the distribution of the spacings between non trivial zeros of $F$ under the analogue of RH for $F$ on a circle of the Riemann sphere whose stereographic projection is the critical line
3. the spacings of eigenvalues of some random matrix along the lines of Katz-Sarnak philosophy.
| https://mathoverflow.net/users/13625 | Spacings of Satake parameters under Ramanujan conjecture | If the first item refers to the set of Satake parameters for $X \le p <2 X$ for large $X$, then that distribution will not match the others. This is because the Satake parameters for different primes do not "see" each other: they will not have (quadratic) repulsion as is expected or known in the other two cases.
| 3 | https://mathoverflow.net/users/10220 | 428861 | 173,845 |
https://mathoverflow.net/questions/428860 | 3 | What is the number of $3$-CNF (conjunctive normal form) formulas with $n$ sentential variables and what is the fraction of satisfiable ones? I consider two formulas the same if they are syntactically the same modulo repeated clauses. Thus $$(x\_1\lor x\_1\lor \lnot x\_1)\text{ and } (x\_1\lor x\_1\lor \lnot x\_1)\land (x\_1\lor x\_1\lor \lnot x\_1)$$ are the same while $$(x\_1\lor x\_1\lor \lnot x\_1)\text{ and } (x\_1\lor x\_1\lor \lnot x\_1)\land (x\_1\lor \lnot x\_1\lor x\_1)$$ are **not**. Now, what is the number of these equivalence classes of formulas in $n$ variables and what is the fraction of satisfiable ones? If the exact numbers cannot be given, just some recurrence formulas will do. Does the $\lim$ of the fraction of satisfiable ones go to $0$ as $n\to \infty$?
| https://mathoverflow.net/users/489947 | The number of $3$-CNF formulas in $n$-variables and the fraction of satisfiable ones | Regarding the fraction of satisfiable 3-CNF formulas in $n$ variables, it is widely believed that there is a phase transition that occurs depending on how many clauses there are compared to the number of variables. To be precise, it is conjectured (but not yet proven) that if there are more than $\alpha n$ clauses, then the formula is almost surely unsatisfiable, and if there are less than $\alpha n$ clauses then the formula is almost surely satisfiable (where $\alpha$ is around 4.2667). This phase transition has been established for $k$-SAT when $k$ is large. See [Proof of the satisfiability conjecture for large $k$](https://arxiv.org/abs/1411.0650) by Ding, Sly, and Sun.
| 5 | https://mathoverflow.net/users/2233 | 428877 | 173,850 |
https://mathoverflow.net/questions/428855 | 0 | Let $S\_{d-1}$ denote the unit sphere in $\mathbb{R}^d$ and let $(Z\_x)\_{x \in S\_{d-1}}$ be a gaussian process with mean zero and covariance structure given by the square of the scalar product, i.e.
$$
\operatorname{Cov}[Z\_x, Z\_y] = \langle x,y \rangle^2 = (x \cdot y)^2 \ .
$$
I am interested in the distribution of $\max\limits\_{x \in S\_{d-1}} Z\_x$ and most of the literature I can find gives only bounds. Can someone point me to some relevant literature or give me some tipps on how to describe the distribution of $\max\limits\_{x \in S\_{d-1}} Z\_x$?
Any help is much apprechiated.
| https://mathoverflow.net/users/409412 | Maximum of a certain Gaussian field | Your field is the $2$-spin, that is can be represented as $Z\_x=\sum\_{i,j} J\_{ij} x\_i x\_j$, where $J\_{i,j} $ are iid Gaussian (up to symmetry, ie $J\_{i,j}=J\_{j,i}$ and the diagonal has twice the variance of off-diagonal. In short, $J$ is a GOE matrix. In particular, the maximum you seek is the top eigenvalue of $J$, whose distribution is well known for $d$ large.
| 1 | https://mathoverflow.net/users/35520 | 428879 | 173,851 |
https://mathoverflow.net/questions/428852 | 10 | Let $G$ be a countable group. A Følner sequence is a sequence of finite subsets $(F\_n)\_n$ such that
$$\lim\_{n\to\infty} \frac{|KF\_n \mathbin\triangle F\_n|}{|F\_n|} = 0$$
for each fixed finite subset $K \subset G$. Such a sequence exists if and only if the group is [amenable](https://en.wikipedia.org/wiki/Amenable_group). The sequence is said to satisfy Tempelman's condition if there is a constant $C \geq 1$ such that $|F\_n^{-1}F\_n| \leq C|F\_n|$ for every $n$. Let us say $(F\_n)\_n$ is a *Tempelman* sequence if it is a Følner sequence satisfying Tempelman's condition. Not every countable amenable group admits a Tempelman sequence; see [Hochman - Averaging sequences and abelian rank in amenable groups](https://arxiv.org/abs/math/0601432). Every countable nilpotent group has a Tempelman sequence.
My question is:
>
> If $(F\_n)\_n$ is a Tempelman sequence, is the sequence $(F\_n^{-1}F\_n)\_n$ necessarily a Følner sequence? Or: can a Tempelman sequence $(F\_n)\_n$ always be found such that $(F\_n^{-1}F\_n)\_n$ is Følner?
>
>
>
This is obviously true in $\mathbb{Z}^d$, where if $F\_n = [0,n)^d \subset \mathbb{Z}^d$ then $F\_n^{-1}F\_n = (-n,n)^d$. It seems like it should be true in general; if $K$ is fixed and $n$ is large then intuitively $|KF^{-1}\_nF\_n \mathbin\triangle F\_n^{-1}F\_n| \lesssim C|KF\_n \mathbin\triangle F\_n|$ should follow from $|F\_n^{-1}F\_n| \leq C|F\_n|$. Yet, I cannot find a proof.
I can see how it is done under a modified condition: suppose for every $n$ there is a subset $C\_n \subset G$ such that $|C\_n| \leq C$ and $F\_n^{-1}F\_n \subset F^{-1}\_nC\_n$. If it makes things easier, we can assume that $F\_n^{-1} = F\_n$ (in which case $(F\_n)\_n$ is both left Følner and right Følner).
Proof: assume that $e\in K$ and $F = F\_n = F\_n^{-1}$ is large. Any element $g \in KFF \setminus FF$ decomposed as $g = kf\_1f\_2$ in any way necessarily has $kf\_1 \notin F$. Then, $FF \subset FC$ implies each $g \in KFF\setminus FF$ may be decomposed as $g = kfc$. There is therefore a map $g \mapsto (kf,c)$ which embeds $KFF \setminus FF$ in $(KF \setminus F) \times C$; the result follows.
| https://mathoverflow.net/users/489544 | If $(F_n)_n$ is a Følner sequence satisfying Tempelman's condition, is $(F_n^{-1}F_n)_n$ also Følner? | The answer to the first question is negative, even for the integers $G = \mathbb{Z}$. The point is that Følner sequences $F\_n$ are stable under modification by small sets, but the difference set $F\_n^{-1} F\_n$ can be significantly affected by such a modification. For a concrete example, take
$$ F\_n = \{0,\dots,n^2\} \cup \{ 2n^2 + 2nj: 0 \leq j < n \} \cup \{ -2n^2 - 2i: 0 \leq i < n\}.$$
This is a modification of $\{0,\dots,n^2\}$ by $O(n)$ elements and is thus a Følner sequence; it is also a dense subset of $\{-4n^2,\dots,4n^2\}$ and thus obeys the Tempelman condition. On the other hand, $F\_n^{-1} F\_n$ has cardinality $\asymp n^2$ and contains the even elements of $\{4n^2,\dots,6n^2-2\}$ but not the odd elements, so will not be Følner. (This counterexample is not symmetric, but it is not difficult to modify the construction to give a symmetric counterexample; we leave this as an exercise to the interested reader.)
On the other hand, given a Følner sequence $F\_n$ obeying the Tempelman condition, one can use a standard pigeonholing argument to find a sequence of subsets $F'\_n$ of $F\_n$ with $|F\_n \backslash F'\_n| = o(|F\_n|)$ (so that $F'\_n$ is also a Følner sequence and will also obey the Tempelman condition) such that ${F'}\_n^{-1} {F'}\_n$ is Følner; in the above example, this would amount to "trimming" the unwanted portions $\{ 2n^2 + 2nj: 0 \leq j < n \} \cup \{ -2n^2 - 2i: 0 \leq i < n\}$ of $F\_n$ and only keeping the "core" portion $\{0,\dots,n^2\}$. A sketch of the construction is as follows. Let $K \subset G$ be a finite symmetric neighborhood of the identity and let $k \geq 2$ be a natural number. For $n$ large enough one can use the Følner condition to find $F''\_n \subset F\_n$ with $|F\_n \backslash F''\_n| \leq \frac{1}{k} |F\_n|$ and $F''\_n \cdot K^k \subset F\_n$. The sets $K^i \cdot (F''\_n)^{-1} \cdot F''\_n \cdot K^{i}$ for $i=0,\dots,k$ are then increasing with cardinality at most $|F\_n^{-1} \cdot F\_n| \leq C |F\_n|$, so by the pigeonhole principle one can find $0 \leq i < k$ such that
$$ |K^{i+1} \cdot (F''\_n)^{-1} \cdot F''\_n \cdot K^{i+1}| \leq
|K^{i} \cdot (F''\_n)^{-1} \cdot F''\_n \cdot K^{i}| + \frac{C}{k} |F\_n|.$$
If we then set $F'\_n := F''\_n \cdot K^i$ then $F'\_n$ is a subset of $F\_n$ with $|F\_n \backslash F'\_n| \leq \frac{1}{k} |F\_n|$ and
$$ |(K \cdot (F'\_n)^{-1} \cdot F'\_n) \backslash ((F'\_n)^{-1} \cdot F'\_n)|
\leq \frac{C}{k} |F\_n|.$$
From this it is an easy matter to verify that $(F'\_n)^{-1} \cdot F'\_n$ will obey the Følner condition (letting $K, k$ grow sufficiently slowly with $n$).
| 11 | https://mathoverflow.net/users/766 | 428884 | 173,852 |
https://mathoverflow.net/questions/428878 | 15 | What are the feasible $2^n$-tuples of entropies $h(S) := H(X\_{i\_1},\dots,X\_{i\_{|S|}})$ where $X\_1,\dots,X\_n$ are discrete random variables with some (unknown) joint probability distribution as $S=\{i\_1,\dots,i\_{|S|}\}$ ranges over the subsets of $\{1,\dots,n\}$? Here $X\_{i\_1},\dots,X\_{i\_{|S|}}$ denotes the compound random variable that in classic information theory texts might be written as $XY$, $XYZ$, etc. and $H(\cdot)$ is the standard entropy $\sum p \log 1/p$.
Of course $h(\{\,\}) = 0$ and $S\_1 \subseteq S\_2$ implies $h(S\_1) \leq h(S\_2)$; furthermore, nonnegativity of conditional mutual information gives $$h(S\_1 \cup S\_2 \cup S\_3) + h(S\_3) \leq h(S\_1 \cup S\_3) + h(S\_2 \cup S\_3).$$ Are there other constraints, linear or otherwise?
Assuming the answer is “no”, I’d be interested in knowing a combinatorial characterization of the extreme rays of the convex polyhedral cone of achievable $2^n$-tuples.
| https://mathoverflow.net/users/3621 | Information inequalities | Yes. The set of $2^n$ (or $2^n-1$ excluding the empty set) dimensional vectors formed by entropies is called the *entropic region* [1]. Inequalities on the entropic region not implied by the nonnegativity of conditional mutual information are called *non-Shannon-type inequalities*. The first such inequality for $n=4$ was given in:
[1] Zhen Zhang and Raymond W Yeung, "On characterization of entropy function via information inequalities", IEEE Trans. Inf. Theory 44, 4 (1998), pp. 1440-1452.
Since then, many more non-Shannon-type inequalities were discovered. Remarkably, there are infinitely many such inequalities even for $n=4$, as shown in:
[2] Frantisek Matúš, "Infinitely many information inequalities", in 2007 IEEE ISIT (2007), pp. 41-44.
Characterizing the entropic region is still a major open problem in information theory (even for $n=4$). The problem might even be undecidable depending on how you formulate it.
| 21 | https://mathoverflow.net/users/489962 | 428887 | 173,854 |
https://mathoverflow.net/questions/428875 | 4 | Suppose $K$ is a centrally symmetric convex body in $\mathbb{R}^n$ and $E$ is the John's ellipsoid, the ellipsoid of maximal volume inside $K$.
If $E$ and $K$ have exactly $2n$ contact points, say $(\pm x\_i)\_{i=1}^{n}$, do $(x\_i)\_{i=1}^n$ form an orthonormal basis for the Euclidean norm indiuced by $E$?
Naively, this statement seems true in two dimensions, but I don't know how to prove it. Or my intuition could be wrong.
Edit: removed another question (whether all points on $\partial K$ extreme points implies exactly $2n$ contact points) as it has an easy negative answer (in comments). Hopefully the remaining question is not as trivial.
| https://mathoverflow.net/users/69275 | Contact points for John's ellipsoid | Looks true. A necessary and sufficient condition for these points (let $E$ be a standard ball) is that the identity operator $I$ is a non-negative linear combination of projectors $P\_i$ on lines through $x\_i$: $$I=\sum c\_i P\_i.\quad\quad\quad\quad\quad(\heartsuit)$$
If $x\_i$'s are linearly dependent, multiply $(\heartsuit)$ by a vector $y$ orthogonal to all $x\_i$'s to get a contradiction. If not, denote by $(z\_i)\_{i=1}^n$ a biorthogonal system to $x\_i$'s and multiply $(\heartsuit)$ by $z\_j$ to get $z\_j=c\_jP\_jz\_j$. This means that $z\_j$ is an eigenvector of $P\_j$ with non-zero eigenvalue, thus $z\_j$ is parallel to $x\_j$. This is what you need.
| 8 | https://mathoverflow.net/users/4312 | 428890 | 173,856 |
https://mathoverflow.net/questions/428891 | 6 | **Background**: Let $G$ be a reductive $\mathbb F\_q$-group and let $X$ be the variety of Borel subgroups of $G$. By the Bruhat decomposition, the $G$-orbits in the space $X\times X$ (with diagonal action) are indexed by elements $w$ of the universal Weyl group $W$ of $G$. Let $\cal O(w)$ be the orbit corresponding to $w$. The *Deligne-Lusztig variety* $X(w)$ is defined to be the intersection of $\cal O(w)$ with the graph of the Frobenius endomorphism of $X$.
Deligne and Lusztig argue that $X(w)$ is of pure dimension $\ell(w)$, the length of $w$, by a transversality argument. My question is about this argument. I assume the argument to be standard, since Deligne and Lusztig devote only a sentence to it, but there's one point I'm having trouble with.
Two subvarieties of an ambient variety are transverse at a point $x$ if the sum of their tangent spaces at $x$ generates the ambient tangent space. When $x$ is a point of transversality, there is a formula for the codimension of the intersection of the varieties (near $x$): it is the sum of the codimensions of the subvarieties.
Since Frobenius has derivative zero, any intersection point of $\cal O(w)$ and the graph of Frobenius is a point of transversality. It follows (by the previous paragraph and the formula for the dimension of $\cal O(w)$) that if $\cal O(w)$ and the graph of Frobenius intersect, then their intersection, the Deligne-Lusztig variety $X(w)$, has pure dimension $\ell(w)$.
**Question**: (Why) does $\cal O(w)$ intersect the graph of Frobenius at all? In other words, (why) is the Deligne-Lusztig variety $X(w)$ non-empty?
**Additional thoughts**: It's easy to see that $X(1) = X(\mathbb F\_q)\neq\varnothing$. Using the dimension estimate, you can also show that $X(w\_0)\neq\varnothing$, where $w\_0$ is the longest element of $W$: indeed, $X(w\_0)$ is a dense subvariety of $X$. However, I don't see how to conclude from formal properties of the Bruhat stratification and the dimension estimate that the intersection must be nonempty.
| https://mathoverflow.net/users/174855 | (Why) are Deligne-Lusztig varieties nonempty? | $\newcommand\FF{\overline{\mathbb F}}\DeclareMathOperator\Fr{Fr}$The Lang (or Lang–Steinberg) map $G(\FF) \to G(\FF)$ given by $g \mapsto g^{-1}\Fr(g)$ is surjective. In particular, there is some $g \in G(\FF)$ such that $g^{-1}\Fr(g)$ represents $w$. If $B$ is a rational Borel containing the reference torus, then $g B\_{\FF} g^{-1}$ and $\Fr(g B\_{\FF} g^{-1})$ are in relative position $w$.
| 13 | https://mathoverflow.net/users/2383 | 428892 | 173,857 |
https://mathoverflow.net/questions/428882 | 4 | There is a notion of 'oidification' in category theory which characterises many object versions of mathematical objects. For example:
* magmas $\rightarrow$ magmoids
* loops $\rightarrow$ loopoids
* groups $\rightarrow$ groupoids
* rings $\rightarrow$ ringoids
And in reverse, one object magmoids are magmas and so on. This is also referred to as horizontal categorification in contrast to vertical categorification.
NLab mentions that heaps have a many object oidification: heapoids. But do not give an explicit characterisation. It's not immediately obvious to me what this is, unlike the mathematical objects mentioned above, heaps are characterised by a ternary operation.
Hence I'm just looking for pointers to a definition, preferably accessible online as I don't have access to a math library.
| https://mathoverflow.net/users/35706 | What is the definition of a heapoid? | The claim that heapoids exist was added to the nLab page in [revision 3](https://ncatlab.org/nlab/revision/heap/3) by [Toby Bartels](http://tobybartels.name/), so you could ask him what he had in mind.
I can speculate that a heapoid would have
* a set of objects.
* families of morphims $f:x\to y$.
* for any $f:x\to y$, $g:z\to y$, and $h:z\to w$, a ternary composite $t(h,g,f) : x\to w$ (note the reversal of direction in $g$).
* for any $f:x\to y$ and $g:x\to z$, we have $t(g,f,f) = g$.
* for any $f:x\to y$ and $g:z\to y$, we have $t(g,g,f) = f$.
* for any $f:x\to y$, $g:z\to y$, $h:z\to w$, $k:u\to w$, and $\ell : u\to v$, we have $t(\ell,k,t(h,g,f)) = t(t(\ell,k,h),g,f)$.
Then any groupoid would become a heapoid with $t(h,g,f) = h\circ g^{-1}\circ f$, just as any group becomes a heap in a similar way.
| 7 | https://mathoverflow.net/users/49 | 428898 | 173,859 |
https://mathoverflow.net/questions/428902 | 0 | Let $u$ be a subharmonic function in a domain $\Omega$ pf $\mathbb{C}$. The functions $u\_{j} := \max(u, -j)$ still subharmonic. Let $\mu := \Delta u$ and $\mu\_{j} := \Delta u\_{j}$ be the associated Riesz measures (which are positive). Let $B$ be a borelian of $\Omega$.
**Question :** is it true that
$$
\int\_{B} 1\_{\{\phi > -j\}}\mu\_{j} \to \int\_{B} 1\_{\{\phi > -\infty\}}\mu\quad\text{ as }\quad j \to +\infty\;?
$$
---
I try to prove it in the following way : let $B$ be a borelian. We then have
$$
\begin{split}
1\_{\{\phi > -\infty\}}( \mu )(B) - 1\_{\{\phi > -j\}}(\mu\_{j})(B) & = 1\_{\{\phi > -\infty\}}(\mu )(B) - 1\_{\{\phi > -j\}}(\mu )(B) \\
&\quad + 1\_{\{\phi > -j\}}(\mu)(B) - 1\_{\{\phi > -j\}}(\mu\_{j})(B)
\end{split}
$$
The quantity $1\_{\{\phi > -\infty\}}(\mu)(B) - 1\_{\{\phi > -j\} }(\mu)(B)$ converges to $0$ by monotone convergence. It then remains to estimate : $1\_{\{\phi > -j\}}(\mu)(B) - 1\_{\{\phi > -j\}}(\mu\_{j})(B) = 1\_{\{\phi > -j\}}(\mu - \mu\_{j})(B)$. But I don't have enough information on the measures $\mu\_{j}$ and $\mu$ to estimate this.
I wish you a good day.
| https://mathoverflow.net/users/339613 | Convergence of Riesz measure of SH function | This is not true. Take $u(z)=(2\pi)^{-1}\log|z|$, so that the Riesz measure $\mu=\delta\_0$. Then the Riesz measure $\mu\_j$ of
$u\_j\max\{ u,-j\}$ is the uniform measure on the circle
$\{ z:|z|=e^{-j}\}$. Now let $B$ be the open right half-plane,
so $\mu\_j(B)=1/2$ while $\mu(B)=0$. If you the choose the closed right half-plane for $B$, then $\mu\_j(B)=1/2$ while $\mu(B)=1$.
The correct statement is that
$$\int\phi\, d\mu\_j\to\int\phi\, d\mu,$$
for every continuous function with bounded support, that is the weak convergence of measures. This follows from three facts: 1. Laplace operator is continuous in the space $D'$ of Schwartz distributions, 2. For subharmonic functions, convergence in $D'$ is equivalent to convergence in
$L^1\_{\rm{loc}}$, and 3. For (positive) measures, convergence in $D'$ is equivalent to the weak convergence. The reference for these facts is
L. Hormander, The analysis of linear partial differential operators, I.
Your statement with Borel sets can be true only
if $\mu(\partial B)=0$.
| 2 | https://mathoverflow.net/users/25510 | 428907 | 173,860 |
https://mathoverflow.net/questions/428912 | 2 | **Real-life motivation.** My eldest son attended a football (soccer) class with $17$ other students, so $18$ students in total. There were $6$ students with year of birth (yob) $n$, then $6$ more with yob $n+1$ and the remaining $6$ had yob $n-1$. In the evening, my son told me that at some point, all $18$ students sat in a circle such that for every student $x$, the years of birth of $x$ and his/her two neighbors covered all $3$ possible years of birth. I found this amazing until I found a trivial solution.
**Formalization.** Let $G= (V, E)$ be a simple, undirected graph. For $v\in V$ let $N(v) = \{w\in V: \{v,w\} \in E\}$ be the set of neighbors of $v$. For a positive integer $k>1$, we call a graph $k$-*regular* if $|N(v)| = k$ for all $v\in V$. We call a $k$-regular graph *satisfying* if there is a map $f:V\to \{1,\ldots,k+1\}$ such that $$f\big(N(v)\cup\{v\}\big) = \{1,\ldots,k+1\} \text{ for all } v\in V.$$
Obviously, for any integer $k>1$ the complete graph $K\_{k+1}$ is satisfying.
**Question.** (Only the first needs to be answered for acceptance.)
1. If $k>1$ is an integer, is there a non-complete satisfying connected $k$-regular graph $G=(V,E)$?
2. If $k>1$ is an integer and $G=(V,E)$ is $k$-regular and satisfying, is $|V|$ necessarily a multiple of $k+1$?
| https://mathoverflow.net/users/8628 | Assigning numbers $\{1,\ldots,k+1\}$ in a balanced way in a $k$-regular graph | The satisfying graphs are precisely the [covering graphs](https://en.wikipedia.org/wiki/Covering_graph) of $K\_{k+1}$. General theory of covering spaces implies that, since $K\_{k+1}$ is connected, each point in the graph has the same number of preimages under a covering map, so every satisfying graph indeed has the number of vertices divisible by $k+1$.
The theory of covering spaces also answers the question of existence - connected covering spaces are in bijection with subgroups of the fundamental group of $K\_{k+1}$, which is free on $|E(K\_{k+1})|-|V(K\_{k+1})|+1=\frac{k(k-1)}{2}$ generators. This group has many subgroups, including many of finite index, so we get many finite satisfying $k$-regular graphs.
| 4 | https://mathoverflow.net/users/30186 | 428916 | 173,862 |
https://mathoverflow.net/questions/428924 | 11 | Let $G$ be a topological group (we can assume that $G$ is countable and discrete) and let $\beta(G)$ be the Stone–Čech compactification of $G$. It is known that $\beta(G)$ can be turned into a left topological semigroup.
1. What are the invertible (left/right/both) elements of $\beta(G)$ as a semigroup?
2. Is right invertiblility the same as left invertibility? (That is, if $xy=1$ does this mean that $yx=1$?)
| https://mathoverflow.net/users/113200 | Stone–Čech compactification as a semigroup | Corollary 4.33 of Hindman and Strauss's book on Algebra in the Stone Cech Compactification says that if $S$ is an infinite cancellative (discrete) semigroup, then the nonprincipal ultrafilters in $\beta S$ form a two-sided ideal. In particular every left invertible element is invertible and the units of $\beta S$ and $S$ coincide. If $S$ is a group this means there are no left or right invertible elements outside of $S$.
| 8 | https://mathoverflow.net/users/15934 | 428930 | 173,864 |
https://mathoverflow.net/questions/428927 | 2 | Let $W$ be the Weyl group of a semisimple algebraic group $G$. $I$ be the simple roots. $J\subset I$ generate a parabolic subgroup of $W$ denote by $W\_J$. $w^J$
is the shortest representative of $w$ in $W/W\_J$. Suppose we have $u^J\le v^J$, here $\le$ is the Bruhat order. Let $M=\{ x\in W\_J| u^{J}x\le v^{J}\}$. My question is whether $M=\{id \}$ always hold?
| https://mathoverflow.net/users/147080 | Parabolic subgroup of Weyl group | No. In $A\_2$ with $I=\{s\_1,s\_2\}$, take $J=\{s\_1\}$ and $u^J=id$ and $v^J=s\_1s\_2$. Then $M=W\_J$ because $s\_1\le s\_1s\_2$.
| 6 | https://mathoverflow.net/users/5519 | 428932 | 173,865 |
https://mathoverflow.net/questions/427328 | 4 | Let $\zeta\_n = e^{i2\pi/n}$. What is the group of all units in the integral cyclotomic ring $\mathbb{Z}[\zeta\_n]$?
Here I like to know all the group elements for small $n$'s. For $n=1$ and $n=2$, the group is given by $\{1,-1\}$. For $n=3$, the group appears to be $\{\pm 1,\pm \zeta\_3, \pm \zeta\_3^2\}$. For $n=4$, the group is $\{\pm 1,\pm \zeta\_4\}$. What are the groups for larger $n$?
| https://mathoverflow.net/users/17787 | The group of all units of integral cyclotomic ring | As you suspect, identifying all the units in cyclotomic rings of order greater than $4$ is a non-trivial problem (I am ignoring order $6$ because, of course $\mathbb Z[\zeta\_6]$ is the same as $\mathbb Z[\zeta\_3]$, and similarly for all other orders twice an odd number.)
We can get just a taste of what is involved by considering order $5$, that is $\mathbb Z[\zeta\_5]$ = $\mathbb Z[\zeta\_{10}]$. Taking into account arithmetic inverses, we can identify a set of units given by $\zeta\_{10}^k$ where $\zeta\_{10}$ is any primitive tenth root of unity and $k$ is any integer. This gives a finite set of ten units.
But wait, there's more. If we add any complex conjugate pair, that is $\zeta\_{10}^k+\zeta\_{10}^{-k}$ for $k$ not a multiple of $5$, we will get a real sum having one of the forms
$\pm\tau,\pm(1/\tau)$
where $\tau=(1+\sqrt5)/2$ is the golden ratio (I am using an older notation here to avoid confusion with $\phi$ for the Euler totient function, which soon enters the discussion). And of course the presence of these reciprocal pairs implies that they are units, and so are their powers and products with the units derived from the tenth roots of unity. We are suddenly confronted with an infinite set all of whose elements must be units in $\mathbb Z[\zeta\_5]$:
$\color{blue}{\zeta\_{10}^k\tau^m}$
where $\zeta\_{10}, k, \tau$ are defined above and $m$ is another integer independent of $k$.
If we plot these points on a two-dimensional complex plane, we see that they are actually a subset of the tenfold symmetric quasilattice (the set of *all* elements in this ring would appear as the entire quasilattice). As such, they cannot all appear at unit distance from the origin when rendered onto the paper. We need the parent lattice having $\phi(5)=4$ dimensions to see geometrically that these elements all have equal norms.
Might there be more points than even these? Since $\phi(5)=4$, or in geometric terms four dimensions are sufficient to render the true unit norms, the set described in blue above (products formed from the primitive roots and one additional generator) turns out to be all the units in $\mathbb Z[\zeta\_5]$. Similarly we may construct four-dimensionally based sets using the roots of unity and one additional generator for $\mathbb Z[\zeta\_8]$ and $\mathbb Z[\zeta\_{12}]$ as $\phi(8)$ and $\phi(12)$ are also $4$ (again, $10$ is twice an odd number and thus $\mathbb Z[\zeta\_{10}]$ is not a distinct ring). For $\mathbb Z[\zeta\_{12}]$ the additional generator would actually be a complex number, as the real quasiperiodic ratio we might expect ($2+\sqrt3$) splits into two nonprimitive factors in the full cyclotomic field. We end with these:
$\mathbb Z[\zeta\_5]: \zeta\_{10}^k[(1+\sqrt5)/2]^m$
$\mathbb Z[\zeta\_8]: \zeta\_8^k(1+\sqrt2)^m$
$\mathbb Z[\zeta\_{12}]: \zeta\_{12}^k(1+\zeta\_{12})^m$
But these are the only cases which are that simple. With other orders having higher Euler totient functions, we need more complex (and geometrically higher-dimensional) structures to capture all the units.
| 3 | https://mathoverflow.net/users/86625 | 428934 | 173,866 |
https://mathoverflow.net/questions/428936 | 2 | For $\Omega$ a bounded open set of $\mathbf{R}^d$, denote $\mathrm{d}\_\Omega:x\mapsto \mathrm{d}(x,\partial\Omega)$ the distance-to-boundary function.
If $\Omega$ is convex, a short argument [recalled here](https://mathoverflow.net/questions/291308/convexity-of-distance-to-boundary-function) by Anton Petrunin proves that $\mathrm{d}\_\Omega$ is a concave function inside $\Omega$. In particular, if $\Omega$ is smooth then in a neighborhood of $\partial\Omega$ and inside $\Omega$, the hessian matrix of $\mathrm{d}\_\Omega$ is non-positive.
I would like to know if outside $\Omega$ (but still in a neighborhood of $\partial\Omega$), this hessian matrix still has a sign or equivalently if $\mathrm{d}\_\Omega$ is convex or concave in $\mathbf{R}^d\setminus \Omega$.
I am a bit disturbed because on the one hand the short argument above does not seem to apply outside $\Omega$ (in a way or another) but at the same time I would expect that the convexity of the "enclosed" volume defined by a closed surface (here $\partial\Omega$) could be guessed directly from it (or using its distance function).
| https://mathoverflow.net/users/27767 | Concavity/convexity of distance-to-boundary function | The signed distance function $f=\pm\textrm{dist}\_{\partial \Omega}$ is concave everywhere;
here $f=\textrm{dist}\_{\partial \Omega}$ inside and $f=-\textrm{dist}\_{\partial \Omega}$ outside.
The proof is straightforward, for inside you know it, for outside it is very similar:
Assume $B(x,r\_x)\cap \Omega\ni p$ and $B(y,r\_y)\cap \Omega\ni q$.
It is sufficient to show that
$$B(\tfrac{x+y}2,\tfrac{r\_x+r\_y}2)\cap \Omega\ne \varnothing.$$
The latter follows since $\tfrac{p+q}2\in B(\tfrac{x+y}2,\tfrac{r\_x+r\_y}2)$.
| 2 | https://mathoverflow.net/users/1441 | 428946 | 173,868 |
https://mathoverflow.net/questions/428926 | 5 | To motivate things, let me start with a special case of the question I'm interested in. Let $\mathsf{In}(x)\equiv$ "$x$ is an inaccessible cardinal."
>
> **Question 1**: Is it consistent with the theory $$\mbox{$\mathsf{ZFC}$ + "There is a transitive model of $\mathsf{ZFC}$ + 'There are $2$ inaccessibles'"}$$ that there is a transitive set $A\models\mathsf{ZFC}$ and an $\alpha\in\mathsf{In}^A$ such that no transitive set $B\supseteq A$ has $B\models\mathsf{ZFC}+\mathsf{In(\alpha)}+ \exists \beta>\alpha[\mathsf{In}(\beta)]$?
>
>
>
(Of course to make this answerable in principle I'm assuming the consistency of $\mathsf{ZFC}$ + "There is a transitive model of $\mathsf{ZFC}$ + 'There are $2$ inaccessibles'" in the first place.)
Intuitively, I'm imagining building a model of $\mathsf{ZFC}$ stage-by-stage and changing the large cardinal requirements as we go ("actually, the client wants *two* inaccessibles"); this question is my attempt to make precise the naive question of whether we can be forced to 'undo previous progress' for non-consistency-strength-flavored reasons.
---
More generally, suppose we have two formulas $P(x),Q(y)$ in the language of set theory. Say that $(P,Q)$ is a **fragile pair** iff the following hold:
* $\mathsf{ZFC}$ proves that both $P$ and $Q$ only ever (if at all) hold of ordinals.
* The conjunction of the following axioms is consistent:
+ each axiom of $\mathsf{ZFC}$,
+ there is a transitive model of $\mathsf{ZFC+\exists \alpha<\beta[P(\alpha)\wedge Q(\beta)]}$, and
+ there is a transitive $A$ and an $\alpha\in A$ such that $$A\models\mathsf{ZFC+P(\alpha)}$$ but there is no transitive $B\supseteq A$ such that $$B\models \mathsf{ZFC} + P(\alpha) + \exists \beta>\alpha[ Q(\beta)].$$
In this language, and letting $\mathsf{In}(x)\equiv$ "$x$ is an inaccessible cardinal," Question 1 above is asking whether $(\mathsf{In,In})$ is a fragile pair.
>
> **Question 2**: Heuristically speaking, when should we expect a pair $(P,Q)$ of naturally-occurring large cardinal properties to be (non-)fragile?
>
>
>
This question is so vague that it deserves some unpacking. In my experience, questions like Q1 above tend to have very "coarse" answers in the sense that they are applicable to a broad class of properties in place of the specific ones considered. So, in fact, I expect that any answer to Q1 will already make progress on Q2; I'm asking Q2 separately to open the door to partial results which don't directly bear on Q1 or its more direct analogues.
I'm especially interested, re: Q2, in the case where $P$ and $Q$ are compatible with $\mathsf{V=L}$. To see why this is (potentially) special, consider the following simple proof that $(\mathsf{In},\mathsf{Meas})$ is fragile where $\mathsf{Meas}(x)\equiv$ "$x$ is a measurable cardinal:" if $A\models$ "$\mathsf{V=L}$ and $\alpha$ is the least inaccessible" then there is no end extension of $A$ with a measurable in which $\alpha$ stays inaccessible. More generally, $(P,\mathsf{Meas})$ is fragile whenever $P$ is compatible with $\mathsf{V=L}$, via the same argument. But nothing like this argument seems to work for weaker large cardinal properties.
| https://mathoverflow.net/users/8133 | Fragility of large cardinals with respect to transitive end extensions | Question 1: Yes, in fact if $M\models$ZFC+"There is a transitive model of $T$" where $T$ is the theory ZFC + "There are two distinct inaccessibles", then there is such an $(A,\alpha)$ - just let $A$ be the least model of ZFC + "There is an inaccessible". Then $A=L\_\gamma$ for some ordinal $\gamma$, and there is a unique $\alpha\in A$ such that $A\models$"$\alpha$ is inaccessible". Let $\alpha$ be this. Then $(A,\alpha)$ has the property, since $A$ is pointwise definable, and note that if $A\subseteq B$ and $B$ is transisitve and $B\models$ZFC+"$\alpha$ is inaccessible", then $\mathrm{Ord}^B=\gamma=\mathrm{Ord}^A$, and $B\models$"$\alpha$ is the unique inaccessible", since otherwise letting $\beta\in B$ be such that $\alpha\neq\beta$ and $B\models$"$\beta$ is inaccessible", then actually $\alpha<\beta$ and $L\_\beta\models$ZFC+"$\alpha$ is inaccessible", contradicting the minimality of $A$.
Here is a more general argument that covers the previous case and various other ones. Assume all the hypotheses of fragility for $(P,Q)$ excluding the last clause, i.e. regarding the transitive set $A$. Assume also that
ZFC + "$P(x)\wedge Q(y)\wedge x<y$" proves (i) "$x$ is uncountable" and (ii) "$V\_y\models$ ZFC + $P(x)$". Then $(P,Q)$ is fragile.
For just let $(\alpha,\beta)$ be the lexicographically least pair of ordinals such that $\alpha<\beta$ and there is a transitive set $M$ with $\alpha,\beta\in M$ and such that $M\models$ ZFC + "$P(\alpha)\wedge Q(\beta)$".
Then by assumption (ii), $V\_\beta^M\models$ ZFC + $P(\alpha)$.
But then working in $M$, we can find a countable elementary $X\preccurlyeq V\_\beta^M$ with $\alpha\in X$. Let $A$ be the transitive collapse of $X$
and $\pi:A\to V\_\beta^M$ the uncollapse. Let $\bar{\alpha}=\pi^{-1}(\alpha)$. By assumption (i), $\alpha$ is uncountable in $M$, so $\bar{\alpha}<\alpha$.
Now $A$ is transitive and $A\models$ ZFC + $P(\bar{\alpha})$. But by the minimality of $(\alpha,\beta)$, there is no $\bar{\beta}$ and transitive $B$ such that $B\models$ ZFC + $P(\bar{\alpha})\wedge Q(\bar{\beta})$
(and in particular no such $B$ with $A\subseteq B$).
So e.g. if $P(x)$ says "$x$ is Woodin" and $Q(y)$ says "$y$ is measurable" then $(P,Q)$ is fragile.
| 5 | https://mathoverflow.net/users/160347 | 428948 | 173,869 |
https://mathoverflow.net/questions/428931 | 6 | Can there be applications of graph theory, combinatorics etc. in PDEs mainly hydrodynamics?
Tried my luck with Google's search engine, didn't show much info.
I guess you can try to use these features of GT and combinatorics in QFT (there's a section in Kaku's book on QFT on forests and skeletons from GT).
And what sort of applications are there?
I am asking, since I am thinking of taking courses in infinite combinatorics, axiomatic set theory and forcing this coming semester, and would like to know if it's not redundant. (As someone who likes to learn a lot it won't be redundant, but I guess I am asking would it have applications to PDE in Hydrodynamics).
I once in my search found a book on categories in continuum mechanics by Lawvere, so perhaps there are such attempts to use algebra and combinatorics in analysis.
<https://link.springer.com/book/10.1007/BFb0076928>
| https://mathoverflow.net/users/13904 | Can there be an application of discrete mathematics in PDEs, mainly the ones used in hydrodynamics? | 1. The inviscid Hopf-Burgers equation is at the heart of combinatorics of convex polytopes surrounding compositional inversion. See, for example, the MO-Q "[Why is there a connection between enumerative geometry and nonlinear waves?](https://mathoverflow.net/questions/145555/why-is-there-a-connection-between-enumerative-geometry-and-nonlinear-waves/181534#181534)". For more on the combinatorics of associahedra, the inviscid HB eqn., compositional inversion (and quantum field theory, operads, ...), see links in OEIS [A133437](https://oeis.org/A133437).
2. The KdV equation and its soliton solution is rife with combinatorics associated with the Eulerian numbers [A008292](https://oeis.org/A008292). See, for example, the two pdfs presented in my post "[The Elliptic Lie Triad: KdV and Riccati Equations, Infinigens, and Elliptic Genera](https://tcjpn.wordpress.com/2015/10/12/the-elliptic-lie-triad-kdv-and-ricattt-equations-infinigens-and-elliptic-genera/)".
3. See also papers by Yuji Kodama and Lauren Williams relating the combinatorics of the Eulerian numbers and Grassmannians to hydrodynamics (see refs in [this MO-A](https://mathoverflow.net/questions/339214/applications-of-number-theory-in-dynamical-systems/339226#339226)).
4. See OEIS [A036039](https://oeis.org/A036039) for links to connections between the KP hierarchy, which generalizes the KdV hierarchy, and combinatorics.
5. The entry for the refined Eulerian numbers [A145271](https://oeis.org/A145271) illustrates the relations among flow equations and reps for compositional inversion involving several classic number arrays in combinatorics. See the Cf. section of the entry for links to partition polynomials for compositional inversion that have combinatorial models such as noncrossing partitions (central to many expositions on free probability theory and random matrices) and phylogenetic trees (and balls in bins), as well as the associahedra.
Edit (Sept. 2, 2022):
6. For vector fields acting on an analytic function,
$$e^{tg(x)\partial\_x} W(x) = W[f^{(-1)}(f(x)+t)],$$
with $g(x) = 1/f'(x) = 1/(\partial\_x f(x))$. Then
$$\partial\_t \; e^{tg(x)\partial} W(x) = g(x)\partial\_x \; e^{tg(x)\partial} W(x) = \partial\_t W[f^{(-1)}(f(x)+t)],$$
published by Charles Grave around 1850, predating Lie (Abel published related formulas even earlier). This flow equation is at the basis of the the analysis underlying [OEIS A145271](https://oeis.org/A145271) and [A139605](https://oeis.org/A139605), and the iterated Lie infinitesimal generator $g(x)\partial\_x$ has an interpretation in terms of Cayley's analytic trees, among other algebraic-combinatorial-topological constructs for special reps of $g(x)$, such as the associahedra as noted in the ref for item 1 above.
7. The exponential generating function of a Sheffer polynomial sequence (see, e.g., "[How are Sheffer polynomials related to Lie theory?](https://mathoverflow.net/questions/415441/how-are-sheffer-polynomials-related-to-lie-theory/415451#415451)"), umbrally presented with $(p.(x))^n = p\_n(x)$, is
$$e^{tR} p\_{0}(x) = e^{tR} \; 1 = e^{tp.(x)} = S(x,t),$$
where $ R$ is a differential raising op defined by $R \; p\_n(x) = p\_{n+1}(x)$.
Then the associated p.d.e. is
$$\partial\_t S(x,t) = \partial\_t e^{tR} \; 1 = R e^{tR} \; 1 = R \; S(x,t) .$$
The family of binomial Sheffer sequences include the Stirling polynomials of the first and second kinds, of much importance in operational calculus and combinatorics in particular in normal ordering of products of $x$ and $D\_x$. Examples are given in the MSE-Q "[Is there a formula similar to $f(x+a) = e^{a\frac{d}{dx}}f(x)$ to express $f(\alpha\cdot x)$?](https://math.stackexchange.com/questions/116633/is-there-a-formula-similar-to-fxa-ea-fracddxfx-to-express-f-al/126744#126744)".
For the normal sub-group of Sheffer polynomials, the Appell polynomials, of which the families of Hermite polynomials are examples important in analysis, combinatorics, and physics (see, e.g., "[Why is the Gaussian so pervasive in mathematics?](https://mathoverflow.net/questions/40268/why-is-the-gaussian-so-pervasive-in-mathematics/383402#383402)", $R$ is a differential raising / creation op of the form $R =x + \sum\_{n \geq 0} b\_n \frac{D\_x^n}{n!}$. The lowering / destruction / annihilation op $L$ is $D\_x$; i.e., $D\_x p\_n(x) = n p\_{n-1}(x)$. The action of the exponential map on the identity gives the e.g.f. of the Appell sequence; that is, $e^{tR}1 = f(t)e^{xt} =e^{a.t}e^{xt} = e^{tp.(x)}$, umbrally, with $f(0) = p\_0(t) =1$. Another common example of an Appell sequence is the set of Bernoulli polynomials, see "[Intuitive explanation why "shadow operator" $\frac D{e^D-1}$ connects logarithms with trigonometric functions?](https://mathoverflow.net/questions/380142/intuitive-explanation-why-shadow-operator-frac-ded-1-connects-logarithms/380189#380189)".
For the relation of an Appell Sheffer sequence and the Riemann zeta function to the fractional calculus of Heaviside, see "[Lie group heuristics for a raising operator for $(-1)^n \frac{d^n}{d\beta^n}\frac{x^\beta}{\beta!}|\_{\beta=0}$](https://math.stackexchange.com/users/27786/tom-copeland)" and "[What's the matrix of logarithm of derivative operator ($\ln D$)? What is the role of this operator in various math fields?](https://mathoverflow.net/questions/382735/whats-the-matrix-of-logarithm-of-derivative-operator-ln-d-what-is-the-rol/383563#383563)" (and links therein).
The Laguerre and associated Laguerre polynomials are general Sheffer sequences and also examples of confluent hypergeometric functions that are central in many discussions in combinatorics and physics--Laguerre histoires (see. e.g., "[An involution on restricted Laguerre histories and its applications](https://arxiv.org/abs/2208.11627)" by Chen and Fu), [rook polynomials](https://en.wikipedia.org/wiki/Rook_polynomial), and the radial part of wave functions for the hydrogen atom. (The [Lah polynomials](https://en.wikipedia.org/wiki/Lah_number) are normalized Laguerre polynomials of order -1, a binomial Sheffer sequence, and a family of Hermite polynomials is related to the Laguerre polynomials of orders 1/2 and -1/2.)
| 5 | https://mathoverflow.net/users/12178 | 428957 | 173,870 |
https://mathoverflow.net/questions/428862 | 1 | The permanent of an $n$-by- $n$ matrix $A=\left(a\_{i j}\right)$ is defined as
$$
\operatorname{perm}(A)=\sum\_{\sigma \in S\_{n}} \prod\_{i=1}^{n} a\_{i, \sigma(i)}
$$
The sum here extends over all elements $\sigma$ of the symmetric group $S\_{n}$ i.e. over all permutations of the numbers $1,2, \ldots, n$.
$$
\operatorname{perm}\left(\begin{array}{ll}
a & b \\
c & d
\end{array}\right)=a d+b c
$$
Given number $N$ and matrix $A$, is it possible to check for $N$ being permanent of the $A$ in polynomial time? Or are fast verification algorithms not guaranteed for *#P-complete* problems?
| https://mathoverflow.net/users/489949 | Deciding if given number is a permanent of matrix | First, there is a subtlety: the permanent of *nonnegative* integer matrices is computable in [**#P**](https://complexityzoo.net/Complexity_Zoo:Symbols#sharpp), and it is **#P**-complete even for $\{0,1\}$-matrices. However, the permanent of general integer matrices is only in [**GapP**](https://complexityzoo.net/Complexity_Zoo:G#gapp) (i.e., it is the difference of two **#P**-functions); I believe it is **GapP**-complete, though I cannot find a good reference at the moment.
The graph of the permanent (or permanent verification, as the question puts it)
$$\mathrm{PermGraph}=\{(A,N):A\in\mathbb Z^{n\times n},N=\operatorname{perm}(A)\}$$
belongs to the class [$\mathbf{C\_=P}$](https://complexityzoo.net/Complexity_Zoo:C#cequalsp), which consists of decision problems $L$ such that
$$\tag1x\in L\iff f(x)=h(x)$$
for some $f\in\mathbf{\#P}$ and $h\in\mathbf{FP}$, or equivalently, such that
$$x\in L\iff g(x)=0$$
for some $g\in\mathbf{GapP}$.
In fact, PermGraph is $\mathbf{C\_=P}$-complete. This follows easily from Valiant’s reduction: #3SAT is #P-complete under parsimonious reductions, and given a 3CNF $\phi$ with $s(\phi)$ satisfying assignments, Valiant constructs a $\{-1,0,1,2,3\}$-matrix $A\_\phi$ such that
$$\operatorname{perm}(A\_\phi)=4^{t(\phi)}s(\phi)$$
for a certain polynomial-time polynomially bounded function $t$. Thus, for $L\in\mathbf{C\_=P}$ expressed as (1), there is a polytime function $x\mapsto\phi\_x$ such that $f(x)=s(\phi\_x)$, and we have
$$x\in L\iff(A\_{\phi\_x},4^{t(\phi\_x)}h(x))\in\mathrm{PermGraph}.$$
Now, what does this tell us about the complexity of PermGraph? The classes **#P**, **GapP**, and their decision version [**PP**](https://complexityzoo.net/Complexity_Zoo:P#pp) have more-or-less the same complexity (they are polynomial-time *Turing*-reducible to each other). The class $\mathbf{C\_=P}$ seems to be weaker than that, nevertheless it is “half-way through” towards **#P**: first, observe that $\mathbf{C\_=P}$ includes the class [**UP**](https://complexityzoo.net/Complexity_Zoo:U#up) of NP-problems that have at most one witness (and this relativizes: $\mathbf{UP}^X\subseteq\mathbf{C\_=P}^X$ for any oracle $X$); then we have
$$\mathbf{PP\subseteq UP^{C\_=P}\subseteq C\_=P^{C\_=P}}.$$
Indeed, if $L\in\mathbf{PP}$, there are $f\in\mathbf{\#P}$ and $h\in\mathbf{FP}$ such that
$$\tag2x\in L\iff f(x)\ge h(x)\iff\exists y\:(f(x)=y\land y\ge h(x)),$$
where $f(x)=y\land y\ge h(x)$ is a $\mathbf{C\_=P}$ predicate, and the witness $y$ is unique if it exists.
In particular, if $\mathrm{PermGraph}\in\mathbf P$, then $\mathbf{C\_=P=P}$, thus $\mathbf{PP=P}$, thus the whole counting hierarchy [**CH**](https://complexityzoo.net/Complexity_Zoo:C#ch) collapses to **P** (and $\mathbf{FCH=\#P=FP}$).
More generally, let $F$ be any **#P**-hard function, and $G\_F$ its graph. Then an argument similar to (2) shows that
$$\mathbf{PP\subseteq UP}^{G\_F},$$
therefore (using $\mathbf{PP^{UP}=PP}$)
$$G\_F\in\mathbf P\implies\mathbf{CH=UP\cap coUP}.$$
I’m not sure whether one can bring this down to $\mathbf{CH=P}$ in this case.
| 2 | https://mathoverflow.net/users/12705 | 428959 | 173,872 |
https://mathoverflow.net/questions/428928 | 4 | If we have $Z\subset X$ a closed irreducible subscheme of an integral scheme $X$ (which you can take to have various further niceness properties if you want), one can take its generic point $\eta\_Z$ and localize at it, to get a local ring with quotient field the quotient field of $Z$.
However, intuitively, I'd like to be able to "take a small neighborhood of $Z$" and not just of $\eta\_Z$, i.e. I'd like a canonically constructed subscheme of $X$ whose "special fiber" is $Z$ and doesn't contain any points of $X$ besides the ones which specialize to points of $Z$. Certainly in general, I wouldn't expect this to be possible, so I don't expect a general functorial construction like usual localizations. But are there known cases when something like this exists?
I'm particularly interested in when $\pi:X\to S$ is a flat family or something similarly nice, and $Z$ is a section of $\pi$, so that what I'm asking for is basically like the "relative local ring of the point $Z$ over the base $S$". Then I'd also like to ask that the pullback of my "localization at $Z$" to any fiber of $\pi$ over a closed point of $S$ yields the localization of that fiber at the point where it intersects $Z$.
In this case, if $X$ is a trivial bundle over $S$, like if $X=S\times\_k Y$ for some $Y$, then obviously this is possible; I'm wondering if it's possible more generally.
(I'd also appreciate an explicit counterexample to it being possible in general flat families, since I don't expect it to be possible in that generality, but find it difficult to prove it's impossible in any particular case.)
| https://mathoverflow.net/users/120548 | When is it possible to localize a scheme along a closed subscheme? | Let $W \subset X$ denote the set of points specializing to $Z$ with the induced topology. Denote $\mathcal{O}\_W$ the pullback of $\mathcal{O}\_X$. Then $W = (W, \mathcal{O}\_W)$ is a locally ringed space and we can ask:
Question: is $W$ a scheme?
Usually not: for example if $X = \mathbf{P}^2\_k$ and $Z$ is a line, then this is false. (Hint: show that there aren't any nonempty affine opens.)
Sometimes yes: if $Z = \{x\}$ consists of a closed point, then $W$ is the spectrum of the local ring $\mathcal{O}\_{X, x}$.
Here is a criterion: suppose that we can write $W = \bigcap\_{i \in I} U\_i$ as the intersection of quasi-compact opens with the following properties: (a) $\forall i, j \in I, \exists k \in I$ such that $U\_k \subset U\_i \cap U\_j$ and (b) $\forall i, j \in I$ if $U\_i \subset U\_j$ then $U\_i \to U\_j$ is an affine morphism. Then $W = \lim U\_i$ is a scheme.
If $X$ is a smooth projective surface then (a) and (b) hold if $Z$ is the fibre of a nonconstant morphism from $X$ to a curve or if $Z$ can be blown down (to a point on another algebraic surface -- does not work if the contraction gives an algebraic space).
If $X$ is a smooth projective surface over the algebraic closure of a finite field and $f : X \to S$ is a nonconstant morphism to a smooth projective curve and $Z$ is the image of a section of $f$, then it is often the case that the normal bundle of $Z$ in $X$ is negative (think about elliptic surfaces for example). Then by Artin you can contract $Z$ and hence $W$ is a scheme.
Anyway, it is fun to play around with these ideas, but I do not know a good way to work with $W$ like this. Many people would, I think, instead consider the formal completion of $X$ along $Z$ and work with that instead.
| 5 | https://mathoverflow.net/users/152991 | 428961 | 173,873 |
https://mathoverflow.net/questions/428720 | 4 | I am wondering if there is a reference for the following:
Let $G$ be a finite group, and suppose that $f\colon M\rightarrow N$ is a continuous and $G$-equivariant map. Here $M$ and $N$ are finite dimensional $G$-manifolds, where $M$ possibly has boundary (I only care when $M$ is compact if that helps). Suppose also that $A\subset M$ is a closed $G$-invariant subset of $M$ on which $f$ is smooth, in that there exists an open neighborhood $U\subset M$ containing $A$ and a smooth equivariant function $h\colon U\rightarrow N$ such that $h\rvert\_A=f\rvert\_A$. Is it true that for any $\epsilon>0$, there exists a smooth equivariant map $f\_\epsilon\colon M\rightarrow N$ such that
1. $\lvert f\_\epsilon(x)-f(x)\rvert<\epsilon$ for all $x\in M$ and
2. $f\_\epsilon\rvert\_A=f\rvert\_A$?
| https://mathoverflow.net/users/489804 | Equivariant Whitney approximation | This is Corollary 1.12 in
*Wasserman, A. G.*, [**Equivariant differential topology**](http://dx.doi.org/10.1016/0040-9383(69)90005-6), Topology 8, 127-150 (1969). [ZBL0215.24702](https://zbmath.org/?q=an:0215.24702).
The proof is essentially the same as the one given by Peter Michor in [his answer](https://mathoverflow.net/a/136765/8103) to [the question linked in the comments by Nick L](https://mathoverflow.net/q/136749/8103).
**Added later:** In fact the above reference does not treat the relative case. A textbook reference for the result in the generality asked for is Theorem 4.2 in Chapter VI of
*Bredon, Glen E.*, Introduction to compact transformation groups, Pure and Applied Mathematics, 46. New York-London: Academic Press. XIII,459 p. $ 21.00 (1972). [ZBL0246.57017](https://zbmath.org/?q=an:0246.57017).
| 3 | https://mathoverflow.net/users/8103 | 428971 | 173,875 |
https://mathoverflow.net/questions/428974 | 1 | Broadly speaking, I have a radial distribution on $\mathbb R^n$, i.e., the pdf only depends on the $\ell\_2$-norm of the argument. I would like to obtain an expression for the pdf in the form $\int\_{w=0}^\infty h(w) \cdot 1\_{w \mathbb B^n} \ dw$ with $h(w) \geq 0$, where $w \mathbb B^n = \{\mathbf x \in \mathbb R^n:|\mathbf x|\_2 \leq w\}$ is the radius $w$ ball and $1\_{w \mathbb B^n}$ is the indicator function for this set. I would like to know if there is a theory that justifies the sort of argument I outline below: in particular, is there a sense in which one can integrate over parametrized families of distributions, and further apply some variant of integration by parts? Below, I will describe my problem more specifically, as well as my "heuristic guess" for the right answer.
The pdf of a $n$-dimensional Gaussian vector $N(\mathbf{0},I\_n)$ (i.e., each coordinate is an independent $N(0,1)$) is $ f(\mathbf x) = \frac{1}{(2\pi)^{n/2}} \exp\left(-|\mathbf x|\_2^2\right)$. This can be written as
$$f(\mathbf x) = \frac{1}{(2\pi)^{n/2}}\int\_{w=0}^\infty \exp\left(-w^2\right) \cdot 1\_{w\mathbb S^{n-1}}(\mathbf x) \ dw,$$
where $w\mathbb S^{n-1} = \{\mathbf x \in \mathbb R^n:|\mathbf x|\_2 = w\}$ and $1\_{w \mathbb S^{n-1}}$ denotes the indicator function for this set.
Intuitively, this can be viewed as a convex combination for $f$ in terms of indicator functions for spheres. Instead, I would like to obtain an expression for $f$ as a convex combination of indicator functions for balls, i.e., an expression like
$$f(x) = \int\_{w=0}^{\infty} h(w) \cdot 1\_{w \mathbb B^n}(\mathbf x) \ dw $$
where $h(w)$ is a nonnegative function.
My approach is to try to use a sort of "integration by parts". Heuristically, if $u\_S$ denotes the uniform distribution over a set $S \subseteq \mathbb R^n$, it seems reasonable to me that an expression like
$$ u\_{w\mathbb B^n} = \frac{n}{w^n} \int\_{v=0}^w v^{n-1} \cdot u\_{v \mathbb S^{n-1}} \ dv$$
should hold. As justification, consider the random variable $X$ defined by sampling a uniformly random point from the ball of radius $w$ and then outputting its norm: we have $\Pr[X \leq v] = \frac{v^n}{w^n}$ so the pdf is $\frac{n}{w^n} \cdot v^{n-1}$. Thus, if we define a function $b(w) = \frac{w^n} n u\_{w \mathbb B^n}$ (whose output is a function on $\mathbb R^n$) something like the Fundamental Theorem of Calculus should imply $b'(w) = v^{n-1} u\_{v \mathbb S^{n-1}}$. However, as the set $\mathbb S^{n-1}$ has measure $0$, this seems sketchy to me.
In any case, if one is willing to accept this reasoning, we could also try to write
$$ f(\mathbf x) = \int\_{v=0}^{\infty} \phi(v) u\_{v \mathbb S^{n-1}} \ dv,$$
where $\phi(v)$ denotes the pdf of a $\chi\_2$-distribution with $n$ degrees of freedom. (Again, the idea is to think of sampling a Gaussian by first sampling its norm and then outputting a uniformly random point from the sphere of that radius.) Then by defining $a(x) := \phi(v)/v^{n-1}$ we could differentiate $a(v)$ and then by some version of integration by parts arrive at an expression like
$$ f(\mathbf x) = \int\_{v=0}^{\infty} a'(v) \frac{v^n}{n} u\_{v \mathbb B^n} \ dv.$$
I am wondering if it is possible to make this sort of approach rigourous. Ideally, I would like to know if there is any general theory explaining how to apply integration by parts to parametrized families of distributions, where the integration is over the parameter of the families.
| https://mathoverflow.net/users/490036 | Integration by parts for indicator of a sphere to indicator of a ball | $\newcommand\R{\mathbb R}$Let $f$ be a radial pdf on $\R^n$, so that
$$f(x)=g(|x|)\tag{1}\label{1}$$
for some function $g\colon[0,\infty)\to\R$ and all $x\in\R^n$, where $|x|:=|x|\_2$. Then your desired respresentation
$$f(x)=\int\_0^\infty h(w) \cdot 1\_{w \mathbb B^n} \ dw\tag{2}\label{2}$$ with $h\ge0$ can be rewritten as
$$g(|x|)=\int\_0^\infty dw\,h(w)\,1(|x|\le w)$$
for all $x\in\R^n$ and then further as
$$g(u)=\int\_u^\infty dw\,h(w)$$
for all real $u\ge0$, with $h\ge0$.
So, your desired representation \eqref{2} for $f$ as in \eqref{1} holds if and only if $g$ is a nonincreasing absolutely continuous function such that $g(u)\to0$ as $u\to\infty$. Moreover, then $h=-g'$ almost everywhere.
---
The formula
$$f(\mathbf x) = \frac{1}{(2\pi)^{n/2}}\int\_{w=0}^\infty \exp\left(-w^2\right) \cdot 1\_{w\mathbb S^{n-1}}(\mathbf x) \ dw\tag{3}\label{3}$$
for $ f(\mathbf x) = \frac{1}{(2\pi)^{n/2}} \exp\left(-|\mathbf x|\_2^2\right)$ in the very beginning of your argument is incorrect. Indeed, the integral in \eqref{3} is in fact
$$\int\_0^\infty \exp\left(-w^2\right) \,1(w=|\mathbf x|\_2) \, dw=0\ne f(\mathbf x).$$
| 2 | https://mathoverflow.net/users/36721 | 428979 | 173,877 |
https://mathoverflow.net/questions/428960 | 5 | Let $f: \mathbb R^n \to \mathbb R$ be a bounded measurable function, and $g: \mathbb R \to \mathbb R$ an absolutely continuous function.
**Question:** Is it true that if $x \in \mathbb R^n$ is a Lebesgue point of $f$, then $x$ is a Lebesgue point of $g \circ f$?
*Notes:*
1. Here $g \circ f$ is the composite function ”f then g”.
2. We use the “strong” definition of Lebesgue points, as given [here](https://en.m.wikipedia.org/wiki/Lebesgue_point).
| https://mathoverflow.net/users/173490 | Does postcomposition with an absolutely continuous function preserve Lebesgue points? | We only need $g$ to be continuous. For $\varepsilon>0$ we can find $u\in C^1(\mathbb{R})$ such that $$\sup\_{s\in\mathbb{R}}|g(s)-u(s)|<\varepsilon .$$
For all $h>0$
$$\frac{1}{2h}\int\_{x-h}^{x+h}|g(f(t))-g(f(x))|d{t}\le 2\varepsilon+\frac{K}{2h}\int\_{x-h}^{x+h}|f(t)-f(x)|d{t}$$
where $K=\sup\_{s\in H}|u'(s)|<\infty$ and $H$ is any bounded convex set containing $f(\mathbb{R})$. This implies if $x$ is a lebesgue point of $f$ then $$\limsup\_{h\to 0 }\frac{1}{2h}\int\_{x-h}^{x+h}|g(f(t))-g(f(x))|d{t}\le 2\varepsilon$$ since $\varepsilon$ was arbitrary we are done.
| 3 | https://mathoverflow.net/users/481665 | 428986 | 173,879 |
https://mathoverflow.net/questions/428983 | 1 | I am studying exceptional isomorphisms recently, which arise due to the coincidence in Dynkin diagram.
I saw two forms of expressing the exceptional isomorphisms, one is isomorphisms between the spin group and the corresponding group, established by spin or half-spin representations; the other is isomorphism from symmetric squares, exterior squares or tensor squares of the relevant group to the special orthogonal group. But in this case, I don't know how to construct the induced symmetric bilinear form on the vector space. And in the case ${\bigwedge}^2\operatorname{Sp}(4,\mathbb{C})\cong \operatorname{SO}(5,\mathbb{C})$, I think the dimension of exterior square of a 4-dimensional vector space is 6, and I am not clear how it gets mapped into $\operatorname{SO}(5)$.
| https://mathoverflow.net/users/350297 | Exterior square of $\operatorname{Sp}(4,\mathbb{C})$ is isomorphic to $\operatorname{SO}(5,\mathbb{C})$ | $\DeclareMathOperator\SL{SL}\DeclareMathOperator\SO{SO}\DeclareMathOperator\Sp{Sp}\DeclareMathOperator\Ker{Ker}$Let's start with the double cover $\SL(4) \to \SO(6)$. Let $V$ be the standard $4$-dimensional representation of $\text{SL}(4)$, and let $\bigwedge^2 V$ be its tensor square. We have a non-degenerate symmetric bilinear form $\bigwedge^2 V \times \bigwedge^2 V \to \bigwedge^4 V \cong \mathbb{C}$ by $\langle \alpha, \beta \rangle = \alpha \wedge \beta$. (This is symmetric because even degree wedges commute with each other.) So the action of $\SL(4)$ preserves this symmetric bilinear form, and we get a map $\SL(4) \to \SO(6)$. This can be checked to be a double cover (the kernel is $\pm \mathrm{Id}$ and a computation with Lie algebras checks that the map is surjective).
Now, let $\omega : V \times V \to \mathbb{C}$ be a symplectic form and let $\Sp(4)$ be the subgroup of $\SL(4)$ preserving this form. Then $\omega$ induces a linear map $L\_{\omega} : \bigwedge^2 V \to \mathbb{C}$, and $\Sp(4)$ preserves the $5$-dimensional space $\Ker(L\_{\omega})$ inside $\bigwedge^2 V$. So $\Sp(4)$ maps to $\SO(5)$, sitting inside $\SO(6)$, and this is again a double cover.
| 7 | https://mathoverflow.net/users/297 | 428990 | 173,883 |
https://mathoverflow.net/questions/428944 | 3 | First let me recall Stone duality in terms of propositional logic.
Let $L$ and $K$ be propositional signatures (i.e., sets of propositional variables). Let $T$ be a propositional theory over $L$ and $S$ a propositional theory over $K$. An *interpretation* of $T$ in $S$ is a map $I\colon L\to \{\text{$K$-sentences}\}$ (which extends to a map $\{\text{$L$-sentences}\}\to \{\text{$K$-sentences}\}$, which we also denote by $I$) such that $S\models I(\phi)$ for all $\phi\in T$. Two interpretations $I$ and $I'$ of $T$ in $S$ are *homotopic* if $S\models I(p)\leftrightarrow I'(p)$ for all $p\in L$. Let me denote the category of all propositional theories (over all propositional signatures) and homotopy classes of interpretations by $\mathrm{PropTh}$.
There is a contravariant functor $\mathrm{Mod}\colon \mathrm{PropTh}\to \mathrm{Top}$, which maps a propositional theory $T$ over a propositional signature $L$ to the set $\mathrm{Mod}(T)$ of all models $M\colon L\to \{0,1\}$ of $T$ equipped with the topology generated by the sets of the form
$$\{M\colon L\to \{0,1\}\mid M\models T\cup \{\phi\}\}$$
for each $L$-sentence $\phi$. Stone's duality theorem states two things:
* This functor is fully faithful, i.e., for all propositional theories $T$ and $S$, the canonical map from the set of all homotopy classes of interpretations of $T$ in $S$ to the set of all continuous maps $\mathrm{Mod}(S)\to \mathrm{Mod}(T)$ is bijective.
* A topological space lies in the essential image of this functor if and only if it is compact, totally disconnected, and Hausdorff.
I wonder what happens if we replace the category $\mathrm{PropTh}$ by the larger category $\mathrm{PropClass}$ of propositional classes:
A *propositional class* is a pair $(L, W)$, where $L$ is a propositional signature and $W\subseteq \{0,1\}^L$ is a set of $L$-structures. An *interpretation* between propositional classes $(L, W)$ and $(K, V)$ is a map $I\colon L\to \{\text{$K$-sentences}\}$ such that for each $K$-structure $M\colon K\to \{0,1\}$ in $V$, the induced $L$-structure $IM:=M \circ I\colon L\to \{0,1\}$ is in $W$. Two interpretations $I$ and $I'$ between $(L, W)$ and $(K, V)$ are *homotopic* if for any $K$-structure $M$ in $V$ and all $p\in L$, $IM\models p$ if and only if $I'M\models p$ (i.e., $IM=I'M$). Denote the category of all propositional classes and homotopy classes of interpretations by $\mathrm{PropClass}$.
Note that there is a fully faithful functor from $\mathrm{PropTh}$ to $\mathrm{PropClass}$, sending a propositional theory $T$ over $L$ to the propositional class $(L,\{\text{models of $T$}\})$. In this sense, propositional classes generalize propositional theories.
There is a canonical contravariant functor from $\mathrm{PropClass}$ to $\mathrm{Top}$, sending a propositional class $(L, W)$ to the set $W$ equipped with the topology generated by the sets of the form
$$\{M\in W\mid M\models \phi\}$$
for each $L$-sentence $\phi$.
**Questions:**
1. Is that functor fully faithful? If not, is there a way to make it fully faithful, maybe by replacing the category $\mathrm{Top}$?
2. Can one describe the essential image of that functor?
| https://mathoverflow.net/users/476804 | An extension of Stone duality | Let $(L,W)$ be a propositional class, so $W\subseteq 2^L$. The topology you assign to $W$ is exactly the subspace topology inherited from $2^L$, where $2$ gets the discrete topology and $2^L$ gets the product topology. So the essential image of your functor is just the subcategory of $\mathsf{Top}$ consisting of all spaces which are subspaces of Cantor cubes. These are exactly the Hausdorff [zero-dimensional](https://en.wikipedia.org/wiki/Zero-dimensional_space) spaces (where zero-dimensional means that there is a basis of clopen sets).
I believe the functor $(L,W)\mapsto W$ is faithful, but it is not full. For a counterexample, let $L = \{p\}$, and let $K = \{q\_n\mid n\in \omega\}$. Let $W$ be the set containing both $L$-structures, so as a topological space it is a discrete space with $2$ points. Let $V$ be the set of $K$-structures in which exactly one variable $q\_n$ is true. As a topological space it is a countably infinite discrete space. Then there are continuum-many continuous maps $V\to W$, but only countably many interpretations of $W$ in $V$ (one for each $K$-sentence).
I think the natural extension of Stone duality here should be a contravariant equivalence between $\mathsf{PropClass}$ and the category $\mathsf{Stone}\_D$ of Stone spaces (compact, Hausdorff, zero-dimensional) with a distinguished dense set. An arrow $(X,D)\to (Y,E)$ is a continuous map $f\colon X\to Y$ such that $f(D)\subseteq E$. The functor maps $(L,W)$ to $(\mathrm{Mod}(\mathrm{Th}(W)),W)$, where $\mathrm{Th}(W)$ is the set of all propositional sentences true in every structure in $W$.
The idea is simple: An interpretation of $(L,W)$ in $(K,V)$ is exactly the same data as an interpretation of $\mathrm{Th}(W)$ in $\mathrm{Th}(V)$, with the additional requirement that $I(V)\subseteq W$. By ordinary Stone duality, it corresponds to a continuous map $\mathrm{Mod}(\mathrm{Th}(V))\to \mathrm{Mod}(\mathrm{Th}(W))$. And $V$ is dense in $\mathrm{Mod}(\mathrm{Th}(V))$ (indeed, the latter is the closure of $V$ in $2^L$) and similarly for $W$. To see that the essential image is exactly $\mathsf{Stone}\_D$, note that every Stone space $X$ is homeomorphic to $\mathrm{Mod}(T)$ for some $L$-theory $T$. The homeomorphism maps a dense set $D$ in $X$ to a set $W$ of models of $T$, and $T = \mathrm{Th}(W)$ by density.
| 2 | https://mathoverflow.net/users/2126 | 428993 | 173,885 |
https://mathoverflow.net/questions/428991 | 18 | I've been trying to read [Kapustin–Witten - Electric–Magnetic Duality And The Geometric Langlands Program](https://arxiv.org/abs/hep-th/0604151) recently, as someone whose mathematical interests are in the Langlands program. I have some physics background, but not including string theory. I'm looking to understand "branes", which play a big part in the paper, and so far, in this context, here is my understanding (please feel free to correct any misconceptions):
Kapustin–Witten formulate a TQFT with gauge group $G$ on a spacetime of the form $\Sigma\times C$, where $C$ is going to play the role of the curve of the geometric Langlands correspondence. $\Sigma$ is a Riemann surface with boundary $\partial\Sigma$, and we are going to look at an effective field theory on $\Sigma$ after compactification on $C$. The resulting fields on $\Sigma$ are going to have to satisfy the Hitchin equations, whose solutions are parametrized by the Hitchin moduli space $\mathcal{M}\_{H}(G,C)$. The assignment of the value of the field on $\Sigma$ is going to described by maps from $\Sigma$ into $\mathcal{M}\_{H}(G,C)$. I believe this is what physicists refer to as a "sigma model".
As a side remark, $\mathcal{M}\_{H}(G,C)$ has many fascinating properties, for example it is hyperkahler, and related to both $\operatorname{Bun}\_{G}(C)$ and $\operatorname{Loc}\_{G}(C)$, and its mirror pair concerns the Langlands dual ${}^{L}G$ in place of $G$. This makes it of interest in the Langlands program.
As far as I understand, a *brane* is a condition on where $\partial \Sigma$ gets sent to in $\mathcal{M}\_{H}(G,C)$, i.e. a submanifold of $\mathcal{M}\_{H}(G,C)$. There are actually two TQFT's in consideration here, an "A-model" and a "B-model", and for the A-model the submanifold $\partial \Sigma$ can get sent to is a Lagrangian (or more generally coisotropic) submanifold, while for the B-model it is a complex submanifold.
But for some reason it is more than just what I described — in fact, a brane in the B-model (or a B-brane) really consists of a coherent sheaf. Hence the appearance of the derived category of coherent sheaves in the formulation of homological mirror symmetry.
My first question is: **Why are B-branes coherent sheaves?** Apparently this is related to some sort of "boundary modification" (see 5.2 of Frenkel's survey [Gauge Theory and Langlands Duality](https://arxiv.org/abs/0906.2747)), but I don't know what these are, and I would be happy to see an explanation or elaboration.
Furthermore, branes (both A-branes and B-branes) form a category. There is supposed to be some sort of physical intuition for the morphisms between branes, but it is not clear to me what that is. For A-branes, the corresponding category is supposed to be some enlargement of the Fukaya category, whose morphisms are Floer chain groups.
This is my second question: **What is the "physical intuition" behind morphisms of branes, why are they the Floer chain groups for A-branes, and why do branes form a category?**
Morphisms of branes are actually pretty important in another way, as this is how Kapustin–Witten get from A-branes to D-modules on $\operatorname{Bun}\_{G}(C)$ (in terms of which the geometric Langlands correspondence is stated). This is constructed by means of a "canonical coisotropic A-brane" (which is $\mathcal{M}\_{H}(G,C)$ itself considered as an A-brane) whose endomorphisms are supposed to give a sheaf differential operators after some sort of localization or sheafification process. Kapustin's notes [Lectures on Electric–Magnetic Duality and the Geometric Langlands Program](http://www.ctqm.au.dk/research/MCS/LectureNotesKapustin.pdf) say this is related to "insertion of vertex operators" (I don't really understand what this means). This also somehow looks reminiscent of Beilinson–Drinfeld's work producing D-modules (see 9.2 of [Frenkel - Lectures on the Langlands Program and Conformal Field Theory](https://arxiv.org/abs/hep-th/0512172)) although as far as I know that makes no reference to branes.
Finally, my third question: **Why do endomorphisms of the canonical coisotropic A-brane give us differential operators?**
| https://mathoverflow.net/users/85392 | What are "branes", and why do they form a category? | Let me start by putting your questions into a bit more context. Kapustin and Witten's story occurs within string theory, a theory of 1-dimensional extended objects. Strings may be "closed," forming loops, or "open," with two boundary points. As a string evolves in time, it traces out a 2-dimensional "worldsheet," which is your $\Sigma$. Moreover, string theory is a *quantum* theory, and so the state of a string is given by a vector in a vector space, and the dynamics of the string are determined by summing over all possible trajectories (worldsheets) the string can take.
Now, what are (Dirichlet) branes in string theory? For closed strings, all you need to specify their dynamics is a target space for them to move around in, which for you is $\mathcal{M}\_H(G, C)$. However, for open strings with two endpoints, one must also specify boundary conditions. These boundary conditions are given by branes, and constrain the dynamics so that the string endpoints may only move about inside the brane.
One is free to choose different boundary conditions (branes) for the left and right ends of the string! Thus, given two branes $X$ and $Y$, we may consider the dynamics of strings with one endpoint on $X$ and the other on $Y$. The states of such a string form a vector space. Moreover, given three branes $X, Y$, and $Z$, we can consider the physical process in which strings stretching from $X$ to $Y$ and from $Y$ to $Z$ join along $Y$ to produce a string stretching from $X$ to $Z$. Thus, we have a composition map, and we may define a linear category whose objects are branes, and whose morphism spaces are vector spaces of open string states.
Now, there are a few caveats needed to make this quite correct. First of all, string states are subject to gauge redundancies, which by a standard procedure (BV-BRST) means that that "vector spaces of string states" are really chain complexes, whose cohomology encode the "physically meaningful" states. Moreover, composition is only associative up to the gauge redundancy, and rather than a category, we obtain an $A\_\infty$ category, where the $n$-ary composition is defined by the physical process in which $n$ strings join into one.
All of this makes sense in an arbitrary string theory, but your questions are more specifically in the context of *topological* string theory. Topological string theory is a topological shadow of more physical string theories, defined by severly restricting the dynamics of the strings, and focusing only on the space of physical ground states (i.e., states of zero energy).
In the A-model, strings are restricted to move along holomorphic curves, with their boundaries on Lagrangian branes. Thus, the objects are Lagrangian branes, and the morphisms are the space of string ground states: there is a potential ground state at each point of intersection (since the distance between the branes is proportional to the energy of a string stretched between), but these ground states may quantum tunnel into each other via dynamical processes (holomorphic curves). To answer your second question: **The quantum-corrected space of ground states of A-model strings stretched between Lagrangian branes is the Floer homology.**
The B-model is more restrictive than the A-model: the only allowed dynamics of the string are trivial (i.e., the only allowed worldsheets are constant maps). However, the boundary conditions in the B-model are more interesting: they are given by complex subvarieties of any dimension. There is actually a bit more data: the endpoint of the string is a charged particle, and couples to a dynamical gauge field living on the brane. This gauge field is given by a holomorphic vector bundle over the complex subvariety, whose dimension is the "number" of branes sitting along that subvariety. The space of ground states stretching between the branes is the space of sheaf morphisms; for example, the morphisms for a stack of $N$ branes intersecting a stack of $M$ branes transversely at a point is given by $Hom(\mathbb{C}^N, \mathbb{C}^M) \cong \mathbb{C}^{NM}$, since one must choose which of the $N$ or $M$ branes the string starts or ends on.
A general coherent sheaf is basically just a complex vector bundle over a subvariety, but if you want a more precise picture, recall that a coherent sheaf is locally given by the cokernel of a morphism $E \to F$ of complex vector bundles. The physical picture is that $F$ is a stack of branes, while $E$ is a stack of *anti*-branes, and the morphism denotes a possible value for the "tachyon field" valued in the vector bundle $Hom(E, F)$. The tachyon denotes a physical process by which the branes and anti-branes can partially annihilate, leaving behind the cokernel: a coherent sheaf. Thus, **A coherent sheaf is the most general result of brane/anti-brane annihilations in the B-model, taking into account their gauge bundles.** For the record, this process is also why D-brane charges are given by K-theory classes: virtual differences between vector bundles.
For your third question, I don't think I can give a complete answer, but one thing to say is this. While before I said that A-branes were given by Lagrangian submanifolds, this is actually too restrictive. In fact, the real condition is that the submanifold be coisotropic. Moreover, just as with B-branes, A-branes also should have gauge fields (bundles) living on them. For Lagrangian branes, these bundles must be flat, and in particular we may turn them off (take them to be trivial). However, for coisotropic branes of higher dimension, the A-model topological constraint actually prevents us from fully turning off the gauge field.
| 16 | https://mathoverflow.net/users/490054 | 429006 | 173,887 |
https://mathoverflow.net/questions/428938 | 16 | In every treatment of Grothendieck sites I can find, flasque sheaves are not defined in the way one would naïvely expect from ordinary sheaf cohomology; namely instead of saying that "restriction maps surject," one says something about Čech cohomologies vanishing. The problem is then to set up the theory of flasque sheaves, one needs to already know some rather heavy facts about Čech cohomology that you don't need in the point-set case.
**Question.** Suppose that $\mathcal{F}'$ is an abelian sheaf on a Grothendieck site all of whose restriction maps are surjective. Does there exist a quick proof that if
$$0 \to \mathcal{F}'\to \mathcal{F}\to\mathcal{F}''\to 0$$
is exact, for $\mathcal{F}$, $\mathcal{F}''$ arbitrary abelian sheaves, then $\mathcal{F}(U)\to\mathcal{F}''(U)$ surjects for any $U$?
In the point-set world, I would prove this using Zorn's lemma in a way similar to the Hahn–Banach theorem, and the characterization that "a sheaf morphism is surjective if and only if for each $U$ and each $s \in \mathcal{F}''(U)$, there is an open cover of $U$ by objects $U\_i$ and $t\_i \in \mathcal{F}(U\_i)$ so that $\phi(U\_i)(t\_i) = s\rvert\_{U\_i}$." Unless I have made some mistake, I think this characterization of surjectivity is still true for sites, but I can't make my Zorn's lemma/Hahn–Banach argument work (namely, to construct an upper bound for a chain, I want to ‘take a union’ but I can't figure out any way to make that work here).
| https://mathoverflow.net/users/490010 | Zorn's lemma for Grothendieck sites | The statement you formulate is not true in this generality.
The idea of the following counterexample is to exploit the fact that the assumption "all restriction maps along morphisms of the site are surjective", applied to a thin site (no parallel morphisms), does not ensure that the sheaf is flabby (flasque) when regarded as a sheaf on the locale presented by the thin site.
Consider the topological space / locale $X = \mathbb{C}\setminus \{0\}$.
Let $\mathcal{F} = \mathcal{F}''$ be the abelian sheaf of continuous functions with values in the (multiplicative) group $\mathbb{C}^\times$.
Let $f : \mathcal{F} \to \mathcal{F}''$ be the sheaf homomorphism that takes the pointwise square of any section, that is, $f(s)(x) = (s(x))^2$ for $s \in \mathcal{F}(U)$, $x \in U$.
Then $f\_U : \mathcal{F}(U) \to \mathcal{F}''(U)$ is surjective as long as $U$ is simply connected -- in particular, $f$ is an epimorphism -- but $f\_X : \mathcal{F}(X) \to \mathcal{F}''(X)$ is not surjective. (Roots of nonzero functions exist locally but not globally.) The kernel $\mathcal{F}'$ of $f$ is the abelian sheaf of continuous (= locally constant) functions with values in $\{-1, 1\}$.
Now, $\mathcal{F'}$ is not a flabby (flasque) sheaf. But as a site of definition for $\mathrm{Sh}(X)$ we can take any basis of the topology of $X$, for example the *connected* open subsets of $X$. For connected $U, V$ with $U \subseteq V$, the restriction map $\mathcal{F}'(V) \to \mathcal{F}'(U)$ is surjective (even bijective). At the same time, there is a connected open subset $U$ such that $f\_U$ is not surjective, namely $U = X$. Thus, we have found a Grothendieck site and an exact sequence $0 \to \mathcal{F}' \to \mathcal{F} \to \mathcal{F}'' \to 0$ of abelian sheaves on it that violates the statement in the question.
| 14 | https://mathoverflow.net/users/166281 | 429010 | 173,888 |
https://mathoverflow.net/questions/428661 | 7 | In this article:
>
> Canfell, M. J. "Completion of Diagrams by Automorphisms and Bass′ First Stable Range Condition." Journal of algebra 176.2 (1995): 480-503.
>
>
>
the author defines a ring $R$ to have the unique generator property on right ideals if for all pairs $a,b\in R$, $aR=bR$ implies $a=bu$ for some unit $u\in R$. (In other articles it is just abbreviated to UGP. In fact the definition seems to go back all the way to Kaplansky.)
The author comments that they do not know if the condition is left-right symmetric. In my brief search, I was unable to unearth any resources declaring that this had been decided one way or the other since then.
Is anyone aware if it is now known if UGP is or isn't symmetric?
---
*Update* :
Also in
>
> Khurana, Dinesh, and T. Y. Lam. "Rings with internal cancellation." Journal of Algebra 284.1 (2005): 203-235.
>
>
>
the authors there remark that it is unknown at that time.
| https://mathoverflow.net/users/19965 | Symmetry of unique generator property | Lam informed me that, as far as he knew, this problem was still open. However, the example below shows that the condition is **not** left-right symmetric.
Let
$$
R=\mathbb{F}\_2\langle a,b,c\, :\, a^2=ab=ac=ba=b^2=bc=0,\ ca=b,\ cb=a\rangle.
$$
The ideal generated by $a$ and $b$ is nilpotent of index $2$; call that ideal $J=\mathbb{F}\_2a+\mathbb{F}\_2b$. We see that $R/J\cong \mathbb{F}\_2[c]$ is an integral domain and Jacobson semisimple, so $J$ is the Jacobson radical of $R$. As the only unit in $\mathbb{F}\_2[c]$ is $1$, and an element of $R$ is a unit if and only if its image is a unit modulo the Jacobson radical, we see that $U(R)=1+J$.
Note that $Ra=Rb$ (since $ca=b$ and $cb=a$). However, $U(R)a=(1+J)a=\{a\}\not\ni b$, so $R$ is not a left UGP ring.
Now, suppose that $rR=sR$ for some elements $r,s\in R$. (We can quickly reduce to the case that they are nonzero.) The relations allow us to write $r$ uniquely in the form $r=f(c)+\alpha a+\beta b$ for unique polynomial $f(x)\in \mathbb{F}\_2[x]$ and unique $\alpha,\beta\in \mathbb{F}\_2$. Similarly, $s=g(c)+\gamma a+\delta b$.
Now, looking at $rR=sR$ modulo $J$, we get $f(c)\mathbb{F}\_2[c]=g(c)\mathbb{F}\_2[c]$. As $\mathbb{F}\_2[c]$ is an integral domain, it has the UGP property. Its only unit is $1$, and so $f(c)=g(c)$.
Write $r=st$ for some $t\in R$. If $f(c)\neq 0$, then $t\in 1+J$, and hence $t$ is a unit. If $f(c)=0$, then $r,s\in J$, and so
$$
r\mathbb{F}\_2 =rR=sR=s\mathbb{F}\_2.
$$
Thus $r=s$ in this case.
| 7 | https://mathoverflow.net/users/3199 | 429013 | 173,889 |
https://mathoverflow.net/questions/429009 | 7 | So rank $1$ local Langlands is special in as that it is given by the Artin map
$$\text{GL}\_1(K)\to G\_K^{ab},$$
whereas in the higher rank (to the best of my knowledge) there doesn't exist a map
$$\text{GL}\_n(K)\to G\_K$$
which realizes the rank $n$ local Langlands. In fact, I think I've read that there exists a good reason that no such map can exist. I can't find a source claiming this though, so here I am.
Is there a reason that no "higher" Artin map can exist?
| https://mathoverflow.net/users/152554 | Non-existence of "higher" Artin map | There is no way of reformulating local Langlands for $n > 1$ in terms of such a map.
Local Langlands is a bijection between irreducible smooth representations of $\operatorname{GL}\_n(K)$, and $n$-dimensional Frobenius-semisimple Weil–Deligne representations. However, if $n > 1$ then an irreducible smooth $\operatorname{GL}\_n(K)$-rep will **almost always be infinite-dimensional**, so there's no way of getting it by composing a Galois representation with a map from $\operatorname{GL}\_n(K) \to G\_K$. (Moreover, the "irreducibility" constraint also trips you up: there are lots of $n$-dimensional Weil–Deligne reps which are not irreducible, but they still correspond to something irreducible on the $\mathrm{GL}\_n$ side, so they couldn't be given by composing a reducible representation with a map.)
| 10 | https://mathoverflow.net/users/2481 | 429018 | 173,892 |
https://mathoverflow.net/questions/428935 | 8 | $\DeclareMathOperator\ncl{ncl}$When I attended a geometric group theory summer school, a question asked by the speaker reminded me of an old [question](https://mathoverflow.net/questions/171776/two-relator-products-of-cyclic-groups) but shaped in a different manner:
>
> Given two nontrivial groups $A,B$ and $w \in A \ast B$. For any $k\geq 2$, is the quotient
> $(A\ast B)/\ncl(w^k)$ nontrivial, where $\ncl(w^k)$ is the normal closure of $w^k$?
>
>
>
This is true if $B = \mathbb{Z}$ and the projection of $w$ to $B =\mathbb{Z}$ is $\pm 1$. For $k\geq 4$, this was handled by a theorem of Howie.
It is not true in general when $k =1$. For example, let $A = \mathbb{Z}\_2$ and $B = \mathbb{Z}\_3$
and $w = ab^{-1}$. When $w$ becomes a relation, $a,b$ must be conjugate, but $a,b$ originally have orders $2$ and $3$, resp. It follows that $a = b$ are trivial elements and the group is trivial. I'm particularly interested in the following question
>
> When $k=1$, are there counterexamples if $A,B$ are torsion-free?
>
>
>
| https://mathoverflow.net/users/114032 | Nontriviality of one-relator products | Here is a partial answer: The Kervaire-Laudenbach Conjecture states that, for any group $A$, $(A\ast\mathbb{Z})/\operatorname{ncl}(w)$ is non-trivial. This was proven by Klyachko for torsion free groups $A$; see Fenn and Rourke, *Klyachko's methods and the solution of equations over torsion-free groups*, Enseign. Math. (2) 42 (1996), no. 1-2, 49–74 ([doi](http://dx.doi.org/10.5169/seals-87871)). Hence:
>
> There are no counter-examples for $A$ torsion-free and $B\cong Z$.
>
>
>
This can be easily generalised: A group is *indicable* if it surjects onto $\mathbb{Z}$. If $B$ is indicable, so $\phi:B\twoheadrightarrow\mathbb{Z}$, then the map $\phi$ induces a map $\overline{\phi}:(A\ast B)/\operatorname{ncl}(w)\twoheadrightarrow(A\ast\mathbb{Z})/\operatorname{ncl}(u)$ for some word $u\in A\ast\mathbb{Z}$. By Klyachko's theorem, we know the image is non-trivial so we immediately have:
>
> There are no counter-examples for $A$ torsion-free and $B$ indicable.
>
>
>
---
Klyachko's proof is really pretty, based on a "funny property" of spheres which "is so simple and funny" that, he claims, it could be included in a collection of puzzles or suggested as a problem for a school mathematical tournament. He phrased it in terms of car crashes, and you can find a MathOverflow discussion on it [here](https://mathoverflow.net/q/2372/6503). The paper of Fenn and Rourke I cited above is the best place to read about it though.
---
Added later: I asked Jim Howie if he knew what the status of this problem is, and he said he thought it open and very difficult (so there are no *known* counter-examples).
He also pointed to two papers of his - one with Brodskii from 1993, and the other with Edjvet from 2021. These prove that when $A,B$ are torsion-free, then they embed into $(A\*B)/\operatorname{ncl}(w)$ if $w$ has free product length $\le 6$ (first paper) and $\le 8$ (second paper). These are innocuous-sounding results, but the one extra step took ~28 years, and both took quite a bit of work to prove.
Embedding results like these are typically referred to as *Freiheitssatz* and seem to be the main approach to problems of this form. For example the main result of Klyachko's paper is a Freiheitssatz for when $w$ has a specific form; the other cases follow from an easy observation. A Freiheitssatz is a priori stronger than a non-triviality result would be, however Jim points out that it is not clear how much stronger.
The papers are:
S. D. Brodskii and J. Howie, *One-relator products of torsion-free groups*, Glasgow Math. J. 35 (1993) 99-104 ([doi](https://doi.org/10.1017/S0017089500009617)).
M. Edjvet and J. Howie, *On singular equations over torsion-free groups* Internat. J. Algebra Comput. 31 (2021), 551-580 ([doi](https://doi.org/10.1142/S0218196721500272), [arXiv](https://arxiv.org/abs/2001.07634.pdf).
| 5 | https://mathoverflow.net/users/6503 | 429023 | 173,894 |
https://mathoverflow.net/questions/428962 | 3 | Let $X\subset\mathbb{P}^N$ be an irreducible complex variety. Fix an integer $a\geq 2$ and call $P\_a$ the following property: given $x\_1,\dots,x\_a\in X$ general points there exists an irreducible curve $C\subset X$ such that $x\_1,\dots,x\_a\in C$, $C$ spans a projective space of dimension $2a-1$ and $\deg(C) > 2a-1$.
For instance, $P\_2$ means that two general points of $X$ can be connected by a curve of degree at least $4$ lying in $\mathbb{P}^3$.
I would like to ask whether there is some natural condition one could put on $X$ (especially when $X$ is rational) for $P\_a$ to hold?
Thank you.
| https://mathoverflow.net/users/14514 | Varieties connected by curves in projective spaces of small dimension | I am writing an answer to the question in the comments. Let $a$ be a positive integer. Let $X\subset \mathbb{P}^N$ be a linearly nondegenerate, smooth, projective variety. Let $\alpha \in \text{Hom}(\text{Pic}(X)/\text{Pic}^0(X),\mathbb{Z})$ be the class of a rational curve in $X$. Denote by $a$ the $\mathcal{O}\_{\mathbb{P}^N}(1)|\_X$-degree of $\alpha$. Assume that for $a$ general points of $X$ there exists an irreducible curve of degree $\alpha$ in $X$ that contains those points. This implies the inequality, $$\langle c\_1(T\_X),\alpha \rangle \geq a(\text{dim}(X)-1) - (\text{dim}(X)-3).$$ For smooth complete intersections in $\mathbb{P}^N$ of type $(d\_1,\dots,d\_c)$, this is equivalent to the inequality, $$\frac{N-3-c}{a} + c+2 \geq d\_1 + \dots + d\_c,$$ and this inequality is also sufficient if the characteristic is $0$ and the complete intersection is general in moduli.
Inside the space of irreducible, genus-$0$ stable maps of class $2\alpha$ containing $a$ general points, consider the locus of those whose linear span has dimension at most $2a-1$. By the usual theory of the Thom-Porteous formula, the codimension of every irreducible component of this locus is at most $N+1-2a$. On the other hand, the locus of double covers of curves of class $\alpha$ is in this locus and has codimension $\langle c\_1(T\_X),\alpha \rangle - 2$. Thus, whenever the following inequality holds, $$\langle c\_1(T\_X),\alpha \rangle \geq N+3-2a,$$ then the double covers deform / generize to irreducible, genus-$0$ stable maps that are not double covers (hence are birational to their image) and whose linear span has dimension at most $2a-1$.
In particular, if $X$ is a smooth, complete intersection in $\mathbb{P}^N$ of type $(d\_1,\dots,d\_c)$, the inequality above holds if $$a(N+3-(d\_1+\dots+d\_c)) \geq N+3.$$ For $a=1$, of course this is correct: linearly degenerate "conics" must be double covers of lines. For $a=2$, this recovers the inequality in the comments: $d\_1+\dots+d\_c$ is no greater than $(N+3)/2$.
| 1 | https://mathoverflow.net/users/13265 | 429028 | 173,895 |
https://mathoverflow.net/questions/428870 | 5 | Let G be a graph with $n \ge 3$ nodes and with average degree $3\le k\le n$. What are the largest and smallest girth G can have?
| https://mathoverflow.net/users/36886 | Largest girth of a graph of average degree k | You can find your answer in the paper:
**Generalized Girth Problems in Graphs and Hypergraphs**
By Uriel Feige and Tal Wagner.
Let $A\_2(n,k)$ denotes the optimal upper bound for girth of graphs with $n$ nodes and average degree $k$. So, it is proved that we have
$$A\_2(n,k)=\Theta(\log\_{k-1}(n))$$ and it is best possible. Also, finding the best possible constant $c$ of the $log^{n}\_{k-1}$ is an open problem and the best upper bound is $c\leq 2$ and the best lower bound is $c\geq 4/3$.
By Erdos work we have:
$$A\_2(n,k)\leq 4n/k$$.
For some good results you can see the girth of Tanner graphs of Low-density parity check codes and the conditions for having special girth in bipartite graphs.
| 3 | https://mathoverflow.net/users/19885 | 429032 | 173,897 |
https://mathoverflow.net/questions/425477 | 6 | Let $(M, g)$ be a compact Riemannian manifold and $f: M \rightarrow \mathbb{R}$ be a Morse-Bott function, i.e. the set a critical points of $f$, $Crit(f)$, has connected components which are smooth manifolds and such that $T\_x Crit(f) = \ker \nabla^2\_x f$,
(where $\nabla^2\_x f: T\_x M \rightarrow T\_xM$ is the linear operator obtained via $g$ from the hessian $f\_{\*\*,x} : T\_x M \times T\_x M \rightarrow \mathbb{R}$ defined as $f\_{\*\*,x}(v, w) = v(W(f))$ for $W \in \Gamma(TM)$ any extension of $w$ (this is well defined and symmetric at critical points) )
Let $\nabla f \in \Gamma(TM)$ be defined by $g(\nabla f, w) = w(f)$. Consider the flow of $-\nabla f$ denoted $\phi\_t$.
How to prove that $\phi\_t(y)$ converges when $t \rightarrow + \infty $ and it converges in the critical set of $f$?
Ps: This question appeared under bounty for 500 points here <https://math.stackexchange.com/questions/4265927/the-flux-of-a-the-negative-gradient-flow-of-a-morse-bott-function-on-a-compact-m?noredirect=1#comment9393403_4265927> but didn't receive any answer so far.
| https://mathoverflow.net/users/172459 | The negative gradient flow of a Morse-Bott function on a compact manifold converges to a critical point? | I will assume that "converges in the critical set of $f$" is asking that if $f$ is MB then $\phi\_t(y)$ (the flowlines) converge as $t\to\infty$ to $y\_\infty$ a critical point (of course, depending on $y$).
---
For any function $f$ on $(M,g)$ a closed Riemannian manifold, note that
$$
\tag{\*} \frac{d}{dt} f(\phi\_t(y)) = - |\nabla f|^2(\phi\_t(y).
$$
Since $f$ is bounded, we can integrate this from $0$ to $\infty$ to find that
$$
\int\_0^\infty |\nabla f|^2(\phi\_s(y)) ds < \infty.
$$
>
> Claim: $|\nabla f|^2(\phi\_s(y)) \to 0$ as $s\to\infty$.
>
>
>
One way to see this is to note that
$$
\frac{d}{ds} |\nabla f|^2(\phi\_s(y)) = 2 D^2 f(\nabla f,\nabla f)|\_{\phi\_s(y)}
$$
is bounded and to prove that if $u \in L^2([0,\infty))$ has $|u'|\leq C$ then $u(s)\to 0$ as $s\to\infty$ (if $u$ is $>\epsilon$ far out at infinity, it will be $>\epsilon/2$ on a definite interval by the derivative estimate, this will contribute too much to the $L^2$ norm). This proves the claim.
In particular, for any sequence $t\_i\to\infty$, we can pass to a subsequence (compactness of $M$) so that $\phi\_{t\_i}(y) \to y\_\infty$. By the above claim, $|\nabla f|(y\_\infty) = 0$.
Of course, well-known examples (the goat tracks) show that $y\_\infty$ need not be the UNIQUE limit (without some further assumption on $f$).
The usual assumptions that would suffice are either Morse, Morse Bott, or real analytic.
---
Now, let us assume that $f$ is MB. There are several ways to conclude that $y\_\infty$ is the limit of $\phi\_t(y)$. A good tool is the MB lemma, which says that you can find coordinates where $f$ is exactly a quadratic form, with $+1,-1,0$ eigenvalues. (Note that the metric need not be Euclidean in these coordinates, so you can't conclude that the gradient of $f$ is as simple as one might expect).
There are various ways to argue, I will describe one below. My proof is based on the "Lojasiewicz method" (but is much simpler due to the Morse Bott assumption + the MB lemma). I am sure that it is possible to argue in a more direct manner, but it's a bit annoying to handle the various cases (e.g., $x\_1\gg x\_2$, etc) and this way is a bit smoother (although somewhat indirect).
For simplicity, let me assume that $n=3$ and in some neighborhood $U$ of $y\_\infty$, we can choose coordinates $x\_1,x\_2,x\_3$ so that $y\_\infty$ corresponds to $(0,0,0)$
$$
f(x) = -x\_1^2 + x\_2^2.
$$
Then, locally the critical set of $f$ is $x\_1=x\_2=0$. (The general case is essentially identical.)
In the coordinates with respect to the MB neighborhood, let me choose some constant $C$ so that the Riemannian metric satisfies
$$
C^{-1}\delta \leq g \leq C \delta
$$
for $\delta$ the Euclidean metric in these coordinates.
Then, we find that $|\nabla f|\_g \geq C^{-1} |\nabla f|\_\delta$, so indeed
$$
|\nabla f|\_g^2 \geq c(x\_1^2+x\_2^2).
$$
Thus, we find that
$$
\tag{\*\*} |\nabla f|\_g^2 \geq c|f|.
$$
I claim that if $f$ satisfies (\*\*) then the limit $y\_\infty$ is unique. (Morse Bott will not be used again).
We compute
$$
\frac{d}{dt} f(\phi\_t(y)) = - |\nabla f|^2(\phi\_t(y)) \leq - c f(\phi\_t(y))
$$
(This is valid as long as $\phi\_t(y)$ remains in $U$, the neighborhood where (\*\*) holds.) We can assume that $\phi\_0(y) \in U$ (just shift time). Below, we will assume that $\phi\_0(y)$ is very close to $y\_\infty$ (by a final shift in time).
Integrating the above ODE we find
$$
f(\phi\_t(y)) \leq C e^{-ct}
$$
as long as $\phi\_t(y)$ remains in $U$.
We can also compute differently:
$$
\frac{d}{dt} f(\phi\_t(y))^{1/2} = - \frac 12 f^{-1/2}|\nabla f|^2(\phi\_t(y)) \leq -c|\nabla f|(\phi\_t(y)).
$$
In particular,
$$
\textrm{length}(\phi\_t(y)|\_{[0,T]}) = \int\_0^T |(\phi\_t(y)'|\_g = \int\_0^T |\nabla f|\_g(\phi\_t) \leq - c \int\_0^T \frac{d}{dt} f(\phi\_t(y))^{1/2} \leq C (f(\phi\_0(y))^{1/2} - f(\phi\_T(y))^{1/2}) \leq C (f(\phi\_0(y))^{1/2}.
$$
This shows that as long as $f(\phi\_0(y)$ is sufficiently close to $y\_\infty=0$ (which we can arrange), $\phi\_t(y)$ cannot travel far enough to leave the neighborhood $U$. Thus, the above analysis is valid for all times.
Repeating the same calculation, we find
$$
\textrm{length}(\phi\_t(y)|\_{[t,\infty]}) \leq C f(\phi\_t(y))^{1/2} \leq C e^{-ct/2}.
$$
This proves that $y\_\infty$ is the unique limit point. Indeed, assume that $y\_\infty'$ is also a limit point. Then, the curve $\phi\_t(y)|\_{[t,\infty]}$ will have to pass infinitely many times between points close to $y\_\infty$ and points close to $y'\_\infty$. This will force the curve to have infinite length (but we proved it has finite length).
| 6 | https://mathoverflow.net/users/1540 | 429039 | 173,899 |
https://mathoverflow.net/questions/429011 | 5 | *Although self-contained, this question is a follow-up to [this earlier one](https://mathoverflow.net/questions/428926/fragility-of-large-cardinals-with-respect-to-transitive-end-extensions/428948#428948). Also, thanks to Fedor Pakhomov for fixing a trivial early version of this question!*
Work in $\mathsf{ZFC}$ + "There is a measurable cardinal," and let $\Omega$ be the least measurable. (Measurability is massive overkill here, I just want to make sure I have plenty of "room," so to speak.) Say that a sentence $\sigma$ is an **inaccessibility safety net** iff $V\_\Omega$ satisfies each of the following statements:
* $\mathsf{ZFC+\sigma}$ has arbitrarily large transitive models, and
* whenever $M$ is a transitive model of $\mathsf{ZFC+\sigma}$ and $\alpha$ is $M$-inaccessible there is a transitive end extension $N\supseteq M$ such that $N\models\mathsf{ZFC}$, $\alpha$ is $N$-inaccessible, and $M\cap\mathsf{Ord}<N\cap\mathsf{Ord}$.
As far as I can tell, Farmer S's argument at the above-linked question - which relied on [pointwise-definable models](https://arxiv.org/abs/1105.4597) - does not prove that inaccessibility safety nets cannot exist (consider e.g. $\sigma\equiv\forall x[\mathsf{V\not=HOD}(x)]$).
>
> **Question**: Is there an inaccessibility safety net?
>
>
>
In some sense an inaccessibility safety net has to axiomatize (a superset of) the consequences which the existence of a "large" inaccessible would have for a "small" inaccessible. Of course this is rather vague, but hopefully it helps motivate this question.
| https://mathoverflow.net/users/8133 | Upwards-fragility of inaccessibles (again) | There are no sentences $\sigma$ with this property.
For any $\sigma$ having transitive models we could
consider the $<\_L$-least well-founded model $E\_\sigma$ of
$\mathsf{ZFC}+\sigma$. Let $M\_\sigma$ be the transitive collapse of $E\_\sigma$. To answer the question it is sufficient to prove the following claim.
**Claim.** Suppose $\mathsf{N}\models \mathsf{ZFC}$ and $\mathsf{Ord}\cap N>\mathsf{Ord}\cap M\_\sigma$. Then all $M\_\sigma$-ordinals are countable from the perspective of $N$.
Note that in particular the claim implies that no $M\_\sigma$-ordinal could be inaccessible from the point of view of $N$.
First observe that provably in $\mathsf{ZFC}$, if $E\_\sigma$ exists, then it is countable. Indeed, we reason in $\mathsf{ZFC}$ and assume for a contradiction that $E\_\sigma$ is uncountable. Consider $E'$ that is an isomorphic copy of a countable elementary submodel of $E\_\sigma$ such that the domain of $E'$ is $\omega$. By condensation lemma $E'\in L\_{\omega\_1}$ and hence $E'$ is $<\_L$ below any uncountable set, contradicting the minimality of $E\_\sigma$.
Next observe that the property $\varphi(\gamma)$ asserting "there exists a well-founded models of $\mathsf{ZFC}+\sigma$, whose height is $\le \gamma$" is absolute for transitive models of $\mathsf{ZFC}$. For this we use the machinery of $\beta$-proofs, see [1] for definitions. Namely we consider the canonical $\beta$ pre-proof $P$ for the following sequent in two sorted language containing the special sort for ordinals $o$, the second sort which we treat as the sort of sets, predicate $<$ for $o$, predicate $\in$ for sets, and a unary function symbol $f$ from $o$ to sets: the sequent contains the negation of all axioms of $\mathsf{ZFC}$, the negation of $\sigma$ (both formulated in the second sort), and the negation of the axiom asserting that $f$ is an order-preserving embedding of $o$ into $\mathsf{Ord}$. Now if for a given ordinal $\gamma$, the proof-tree $P\_\gamma$ is well-founded, then there are no well-founded models of $\mathsf{ZFC}+\sigma$ of the height $\le \gamma$. And from any infinite path through $P\_\gamma$ we could construct a well-founded model of $\mathsf{ZFC}+\sigma$ of the height $\le \gamma$. The trees $P\_\gamma$ have very simple definitions, in particular uniformly, for limit $\gamma$, they are $\Delta\_1$-definable in $L\_\gamma$. Thus we could reduce this model existence question to the question about ill-foundedness of the mentioned trees and the latter property is clearly absolute.
Now we are ready to prove the claim. If $\mathsf{Ord}\cap N>\mathsf{Ord}\cap M\_\sigma$, then, by absoluteness proved above, in $N$ there is some transitive model of $\mathsf{ZFC}+\sigma$. Hence $N$ thinks that there is a countable transitive model of $\mathsf{ZFC}+\sigma$ and thus by Shoenfield's absoluteness $L^N$ thinks that there is a countable transitive model of $\mathsf{ZFC}+\sigma$. Hence $E\_\sigma\in N$. Inside $N$ we define transitive collapse of $E\_\sigma$ and observe that it gives us $M\_\sigma$. Thus $M\_\sigma\in N$ and $M\_\sigma$ is a countable transitive model from the perspective of $N$. This immediatelly implies the claim.
[1] J.-Y. Girard. $\Pi^1\_2$-logic, part 1: Dilators. Ann. Math. Logic, 21(2):75 – 219, 1981.
| 7 | https://mathoverflow.net/users/36385 | 429054 | 173,902 |
https://mathoverflow.net/questions/429003 | 20 | The category of smooth manifolds (SmoothMfld) can be thought of the Cauchy completion of the category $U$ of open subsets of Euclidean spaces (with smooth maps) [1]. This fact is shocking to me as it provides an **intrinsic** definition of smooth manifolds.
If this fact can be generalized to manifolds equipped with other types of structures, we'd have a whole new perspective of what a manifold ought to be.
Alas, the proof given in [1] only works for smooth manifolds: Modulo some categoricaly nonsense, the crux of the proof (given in [1]) lies in the fact that the fixed point set of any idempotent in $U$ again has a smooth manifold structure. This essentially requires the use of tangent space and, more importantly, the inverse function theorem
Still, it doesn't prove that it fails for other cases. Thus this question: **Is the category (X-Mfld) the Cauchy completion of the corresponding $U$**, for X being Topological, PiecewiseLinear, Complex, Analytic.. etc?
**Reference**
[1] nLab authors, "[chapter 4: The category of smooth manifolds](https://ncatlab.org/nlab/revision/Karoubi%20envelope/35#the_category_of_smooth_manifolds)", [*Karoubi envelope* (Revision 35)](http://ncatlab.org/nlab/revision/Karoubi%20envelope/35), August 2022.
| https://mathoverflow.net/users/124549 | Manifolds as Cauchy completed objects | (Expanding on Phil Tosteson’s [comment](https://mathoverflow.net/questions/429003/manifolds-as-cauchy-completed-objects#comment1103552_429003).) **No: the Cauchy-completion characterisation doesn’t hold for the PL, topological, or complex-analytic cases.**
The key technical point is that split idempotents are always *absolute*, i.e. preserved by all functors, in particular the forgetful functor to $\mathrm{Top}$. This says that any splitting of an idempotent must be precisely (up to iso) the subspace of fixpoints, as you’d expect.
With this in hand, it’s easy to check that **in the PL and topological categories, the subspace of fixpoints of an idempotent isn’t generally a manifold, so the category of manifolds isn’t Cauchy-complete**. Take for instance the PL idempotent retracting $\newcommand{\R}{\mathbb{R}}\R^2$ onto the lines $x = \pm y$, sending $(x,y)$ to $\min(\left|x\right|,\left|y\right|)(\newcommand{\sg}{\operatorname{sg}}\sg(x),\sg(y))$ (where $\sg(x)$ denotes the sign of $x$, in $\{1,-1\}$).
Contrariwise, **in the complex-analytic category, not every manifold arises as the splitting of an idempotent**, since not every complex manifold embeds into some $\mathbb{C}^n$. For example, connected compact complex manifolds have no nonconstant holomorphic functions to $\mathbb{C}$; this follows easily from the maximum modulus principle.
| 19 | https://mathoverflow.net/users/2273 | 429055 | 173,903 |
https://mathoverflow.net/questions/346571 | 5 | The boundary of an $r$-neighborhood of the convex core of hyperbolic $n$-manifold is smooth, by page 73 of Hyperbolic Manifolds and Kleinian Groups. The authors does not provide a proof for this fact. This does not look very obvious as the $r$-neighborhood of a convex set in a Riemannian manifold may not be smooth. One example is that:
the convex hull of two intersecting geodesic segment in hyperbolic 2-space, one going from (0,1) to $\infty$, one is a quarter circle from $(0,1)$ to $(1,0)$. Its $r$-neighborhood has 3 parts. By observing the curvature of each part we can see that its boundary is not smooth.
If it is not smooth, what is the maximal possible regularity of the boundary.
This paper <https://link.springer.com/article/10.1007/BF02921327> talks about how to perturb the boundary of a cusp and tube in a controlled way so that it becomes smooth. Would similar argument work in this case?
If it is true by some special property of the boundary of the convex core for hyperbolic manifolds, I would also wonder whether this is true for negatively curved manifold with pinched sectional curvature $\kappa \in [−b,−1]$, where $b > 1$. The classical paper Geometrical finiteness with variable negative curvature by Bowditch does not seem to discuss the regularity issues of the boundary of the $r$-neighborhood of the convex core. Any reference for this issue, preferably with a proof would help. In general the $r$-neighborhood of a convex set with piecewise smooth boundary is not smooth.
| https://mathoverflow.net/users/136031 | Smoothness of boundary of $r$-neighborhood of convex core | Just coming across this 3 years too late, but thought I'd mention for posterity that I don't think the boundary of the r-nbhd is smooth. It's not so hard to prove it's $C^{1,1}$. Intuitively, it's $C^1$ because every point on the boundary of the r-nbhd is on the boundary of a r-ball centered in the convex core. That entire r-ball is contained in the r-nbhd of the convex core, and the r-nbhd is also convex, so you have a support plane, and hence you have a well defined tangent plane at that point. It's also not hard to show that these tangent planes vary lipschitzly. Walter, "Some analytical properties of geodesically convex sets", is a good reference in a more general Riemannian setting.
You're probably not going to do much better than $C^{1,1}$, though. For instance, in $3$-dimensions, if the bending lamination of a convex core boundary component has a closed leaf, locally the r-nbhd of the convex core is just going to look like the r-nbhd of the intersection of a pair of half-spaces, and its boundary won't be $C^2$, sort of like in the example you suggested. Probably the boundary of the r-nbhd is $C^2$ exactly when the convex core has totally geodesic boundary, which is unusual.
I think almost everything above (except the last sentence, which I'm less confident about) works fine in variable negative curvature, too.
| 3 | https://mathoverflow.net/users/74169 | 429056 | 173,904 |
https://mathoverflow.net/questions/429052 | 13 | Let $G$ be a finite group, $A$ some coefficients (e.g. $A = \mathbb{F}\_2$ or $\mathbb{Z}$), and write $\mathrm{H}^\bullet\_{\mathrm{gp}}(G; A)$ for the (ordinary) group cohomology of $G$ with coefficients in $A$. Recall that a class $\alpha \in \mathrm{H}^\bullet\_{\mathrm{gp}}(G; A)$ is *essential* if it is nonzero, but the restriction $\alpha|\_S \in \mathrm{H}^\bullet\_{\mathrm{gp}}(S; A)$ is zero for *all* proper subgroups $S \subsetneq G$. For example, a standard lemma shows that $G$ has no essential cohomology if $G$ is not a $p$-group for some prime $p$ (as the collection of Sylow $p$-subgroups together detect all cohomology). On the other hand, for a cyclic group of prime order, all classes are essential, just because the group has no subgroups.
>
> **General question:** Which groups have essential cohomology *of low degree*?
>
>
>
The question is basically trivial in degree $1$. Indeed, $\mathrm{H}^1\_{\mathrm{gp}}(G; A) = \hom(G, A)$, and a nonzero homomorphism takes a nonzero value on some element, so the only groups with essential cohomology in degree $1$ are cyclic (and these do have essential cohomology). Moreover, it's approximately true that an abelian group has essential cohomology of degree $d$ when it has rank $d$; compare [Which groups have undetectable third U(1)-cohomology?](https://mathoverflow.net/questions/378205/which-groups-have-undetectable-third-u1-cohomology).
>
> **Specific question:** Which groups have essential classes in $\mathrm{H}^2\_{\mathrm{gp}}(-; \mathbb{F}\_2)$?
>
>
>
| https://mathoverflow.net/users/78 | Which finite groups have low-degree essential cohomology? | Call a finite $p$ group $G$ *$p$--central* if every element of order $p$ is central.
In a 1997 Comm. Math. Helv. paper, A. Adem and D. Karagueuzian proved that every $p$--central group has essential mod $p$ cohomology. I got at this theorem in a different way in [N.J.Kuhn, Adv. Math. 216 (2007), 387--442], with a result that defines a number $e(G)$ that gives the degree of a particular essential element with other nice properties (e.g. it is acted on trivially by all Steenrod operations).
So what is this number $e(G)$? Let $C$ be the (unique) maximal elementary abelian subgroup of $G$: the nontrivial elements of $C$ are all the elements of $G$ of order $p$. Then $e(G)$ is determined by the image of the restriction
$ H^\*(G;\mathbb Z/p) \rightarrow H^\*(C;\mathbb Z/p)$, which is always a sub-Hopf algebra of $H^\*(C;\mathbb Z/p)$. More precisely, $e(G)$ is the top degree of the finite dimensional Hopf algebra $H^\*(C;\mathbb Z/p) \otimes\_{H^\*(G;\mathbb Z/p)} \mathbb Z/p$, and is often quite easy to compute.
$e(G)$ is often quite easy to compute, and, in tables in an appendix of the paper, I list the values of $e(G)$ for all $2$--central $2$--groups of order up through 64. In particular, the following indecomposable 2-groups have $e(G)=2$, and thus essential classes in degree 2: groups of order 32 with Hall-Senior numbers 19, 21, 29, and 30, and groups of order 64 with Hall-Senior numbers 38, 39, 41, 64, 65, 140, and 141.
A general result would go: **Proposition** Let $G$ be a $2$--central group with $C$ of rank 2. If the image of the restriction map (as above) is polynomial on two classes in dimension $2$, then $e(G)=2$ and so there is essential cohomology in degree 2.
None of this is very trivial. Read my paper (and a later follow-up of mine on nilpotence in cohomology) if your are interested. The theory is worked out at odd primes too; I was just obsessed with $2$-groups as I looked for examples. I should also say that I was not looking for the essential cohomology class of lowest degree; indeed, my class is likely the essential class of greatest degree.
| 7 | https://mathoverflow.net/users/102519 | 429062 | 173,906 |
https://mathoverflow.net/questions/428970 | 0 | For any integer $k\geq 2$, a $k$-regular linear set system is a set ${\cal E}\subseteq {\cal P}(\omega)$ such that $|e| = k$ for all $e\in {\cal E}$, and moreover, for all $a\neq b\in\omega$ there is exactly one $e\in {\cal E}$ such that $\{a,b\}\subseteq e$. ([There is indeed](https://mathoverflow.net/a/428639/8628) a $k$-regular linear set system on $\omega$ for any given integer $k\geq 2$.)
We say that a set system ${\cal A}\subseteq {\cal P}(\omega)$ is *vertex-transitive* if for all $a, b\in \omega$ there is a bijection $\varphi:\omega\to\omega$ such that
1. $\varphi(a) = b$, and
2. for all $X\in{\cal A}$ we have $\varphi(X) \in {\cal A}$ as well as $\varphi^{-1}(X)\in {\cal A}$.
**Question.** If $k\geq 2$ is an integer and ${\cal E}\subseteq{\cal P}(\omega)$ is a $k$-regular linear system, is ${\cal E}$ necessarily vertex-transitive?
| https://mathoverflow.net/users/8628 | Are $k$-regular linear set systems vertex-transitive? | **Introduction**
I'll provide plenty of non-transitive examples (*just be patient, please*).
Let me introduce a useful and more general notion of *selector system* instead of *linear set system*; then, I will specialize it to the *linear* ones. even in the special case, we can equivalently consider an arbitrary infinite countable set in place of $\omega$.
**Remark** We will see that my complete (2 n)-systems are the same as the OP's linear set systems.
---
---
**Definition of the systems and extensions**
A selector system is an ordered triple $\ (X\ \cal B\,\ \sigma)\ $ (*please, do not edit my comma-free notation*), where $\ X\ $ is an arbitrary set, $\ \cal B\subseteq 2^X,\ $ and $\ \sigma:B\to2^X\ $ satisfies two axioms:
$$ \forall\_{B\in\cal B}\quad B\cap\sigma(B)\ =\ \emptyset $$
and
$$\forall\_{A\ B\in\cal B}\quad(A\subseteq B\cup\sigma(B)\,\ \Rightarrow
\,\ A\cup\sigma(A)\ =\ B\cup\sigma(B)) $$
Selector system $\ (Y\ \cal D\,\ \phi)\ $ extends selector system
$\ (X\ \cal B\,\ \sigma)\ \Leftarrow:\Rightarrow $ $ X\subseteq Y,
\ \cal B\subseteq\cal D,\ \sigma=\phi|\cal B,\ $ and
$$ \forall\_{B\in \cal B}\,\forall\_{D\in\ \cal D}
\quad (D\subseteq B\cup\sigma(B)\,\ \Rightarrow\,\ B\cup\sigma(B)
\ =\ D\cup\phi(D)) $$
---
---
(k n)-**systems and their *free style* purposeful extensions**
Let $\ 1<k<n\ $ be natural numbers. In this section, let selector system $\ (X\ \cal B\,\ \sigma)\ $ be such that
$\ \emptyset\ne\cal B\subseteq$ $\binom Xk,\ $
$\ |\sigma(B)|=n-k,\ $ and
$$ \binom{B\cup \sigma(B)}k\subseteq \cal B $$
for every $\ B\in\cal B.\ $ Then we can construct an extension
$\ (Y\ \cal D\,\ \phi)\ $ that is similar (as we will see soon), and that has also the following ***purposeful property***:
$$ \binom Xk\ \subseteq\ \cal D $$
**Let's do it:**
Let set $\ T\ $ has cardinality $\ |T|=n-k,\ $ and let $\ Y\ $ be the
following disjoint "$\cup$":
$$ Y\,\ :=\,\ X\ \cup\ \left(\binom Xk\setminus\cal B \right)\times T, $$
and
$$ \cal D\ :=\ \cal B\,\cup
\bigcup\_{b\in\binom Xk\setminus\cal B}\binom{b\cup\phi(b)}k,$$
*(sometimes the MO software forces calligraphic $\ \cal X\ $ in place of simple $\ X$)*
where $\ \phi\ $ extends $\ \sigma\ $ as follows:
$$ \forall\_{b\in\ \binom X2\setminus\cal B}\quad \phi(b)\ :=
\ \{b\}\times T $$
and
$$ \forall\_{b\in\ \binom Xk\setminus\cal B}\,
\forall\_{c\in\binom{b\cup\phi(b)}k}\quad\phi(c):=(b\cup\phi(b))\setminus c
$$
Otherwise, $\ \phi(B)=\sigma(B)\ $ for $\ b\in\cal B.$
Thus, if $\ X\ $ is countable then so is $\ Y.$
The above extension was done in the ***free style***.
---
---
**Complete selector** (k n)-**systems**
Now, we get an entire class of infinite countable ***complete*** selector systems
$\ (X\_\infty\ \cal B\_\infty\ \sigma\_\infty) $ such that
$\ B\_\infty=\binom {X\_\infty}k,\ $ and
$$ \forall\_{b\in\binom {X\_\infty}k} |\sigma\_\infty(b)|=n-k $$
***Remark**: In general, the complete systems don't have to have the triple index $\infty$ in their names.*
We obtain a system as this as an extension of arbitrary countable system
$$ (X\_0\ \cal B\_0\ \sigma\_0)\ :=\ (X\ \cal B\,\ \sigma), $$
where
$$ \binom Xk\setminus\cal B\ \ne\ \emptyset $$
Then we obtain $\ X\_{s+1}\ \cal B\_{s+1}\ \sigma\_{s+1}\ $ from
$\ (X\_s\ \cal B\_s\ \sigma\_s)\ $ the way we have obtained
$\ (Y\ \cal D\,\ \phi)\ $ from $\ (X\ \cal B\,\ \sigma)\ $ for every $\ s\in\mathbb N.$ Now we define:
$$ X\_\infty\ :=\ \bigcup\_{s=0}^\infty\, X\_s\qquad\text{and}
\qquad\cal B\_\infty\ :=\ \bigcup\_{s=0}^\infty\,\cal B\_s, $$
and let $\ \sigma\_\infty\ $ be the common extension of all $\ \sigma\_s,\ $
onto $\ \cal B\_\infty.$
THe above complete extension was done in the ***complete free style***.
---
---
**No certain intermediate subsystems**
Let $\ (X\_0\ \cal B\_0\ \sigma\_0)\ :=\ $ $(X\ \cal B\,\ \sigma)\ $ be an
(k n)-system, and let $\ (X\_\infty\ \cal B\_\infty\ \sigma\_\infty)\ $ be
the complete extension of $\ (X\ \cal B\,\ \sigma)\ $ done in the complete
free style.
We also assume the non-completeness condition:
$$ \binom Xk\setminus\cal B\ne\emptyset $$
---
**Theorem** There does not exists an intermediate **complete**
subsystems $\ (Y\ \cal D\,\ \phi\ ) $ such that
$$ \qquad 0\ <\ |Y\setminus X|\ <\ \infty\qquad\qquad\qquad\qquad (\#) $$
---
**Proof**
By (#), there exists $\ s\in\mathbb N\ $ such that
$$ Y\setminus X\_s\ =\ \emptyset\ \ne\ Y\setminus X\_{s-1} $$
Let $\ x\in Y\setminus X\_{s-1}.\ $ Then, for arbitrary
$\ L\subseteq\binom{X\_{s-1}}{k-1}\ne\emptyset\ $ and $\ p:=\{x\}\cup L\ $
we have
$$ \emptyset\ \ne\ \sigma\_{s+1}(p)\ \subseteq\ X\_{s+1}\setminus X\_n $$
Thus, $\ p\notin\binom Yk.\quad$ **End of Proof**
---
---
**Non-transitive examples**
Let (k n)-system $\ (X\_1\ \cal B\_{-1}\ \sigma\_1)\ $ be complete, i.e.
$$ B\_{-1}\ =\ \binom{X\_{-1}}k $$
The examples can be taken from finite geometries or trivial (finite or nor)
when $\ |X\_{-1}|=n\ $ hence necessarily $\ \cal B\_{-1}:=\binom{X\_{-1}}k$.
Let $\ e\ $ and $\ X\_0\ :=\ X\_0\cup\{e\}\ $ be such that
$\ e\in X\_0\setminus X\_{-1}.$
Then we get an induce non-complete (k n)-system
$\ (X\_0\ \cal B\_0\ \sigma\_0),\ $ where
$$ X\_0:=X\_{-1}\cup{e}\qquad\text{and}\qquad\cal B\_0=\cal B\_{-1}\qquad
\text{and}\qquad\forall\_{B\in\cal B\_0}\ \sigma\_0(B):=\sigma\_{-1}(B) $$
---
**Theorem** In every such situation as above, if $\ X\_0|<\infty\ $
then
$\ (X\_\infty\ \cal B\_\infty\ \sigma\_\infty)\ $ is an infinite complete
(k n)-system that is not vertex transitive.
---
**Proof** Indeed, there is no authomorphism $\ X\_\infty\to X\_\infty $
can map any element of $\ X\_{-1}\ $ to any element of
$\ X\_\infty\setminus X\_0.$ **End of Proof**
| 1 | https://mathoverflow.net/users/110389 | 429067 | 173,908 |
https://mathoverflow.net/questions/429068 | 6 | Is there a compact metric space $X$ of covering dimension $2$ without a Lipschitz surjection on $[0,1]^2$?
For a space $X$ with Hausdorff dimension greater than $2$, we have a negative answer (see Theorem 1.1 in article <https://arxiv.org/pdf/1203.0686.pdf>).
| https://mathoverflow.net/users/490101 | Lipschitz mappings, covering dimension | Let $\ n\in\mathbb N.\ $ It's convenient to consider the injective metric
in $\ \mathbb R^n,\ $ it's Lipschitz equivalent to the Euclidean metrics.
By dimension, let's mean the *topological dimension* dim (say, *covering* -- for metric compact spaces, *topological* has only one standard meaning).
---
**Theorem** For every metric compact space $\ X,\ $ with a topologically agreeing metrics $\ d,\ $ such that $\ \dim(X)\ge n,\ $
there exists a Lipschitz surjection
$\ f:X\to[-1;1]^n.$
---
**Proof** Due to the dimension $\ \ge n\ $ there are
$\ (F\_k:\ k=\pm1 \ldots \pm n)\,\ $ closed subsets of $\ X\ $ such that
$$ \forall\_{k=1}^n\quad F\_{-k}\cap F\_k\ =\ \emptyset $$
and
$$ \bigcap\_{k=1}^n\,S\_k\ \ne\ \emptyset $$
for arbitrary closed separators S\_k between $\ F\_{-k}\ $ and $\ F\_k,\ $
where $\ 1\le k\le n.$
Let $\ g\_k:F\_{-k}\cup F\_k\to[-1;1]\ $ be defined by $\ g\_k(x)=-1\ $
for every $\ x\in F\_{-k},\ $ and $\ g\_k(x)=1\ $ for every $\ x\in F\_k,\ $
whenever $\ 1\le k\le n.\ $ These maps are obviously Lipschitz. Since $\ [-1;1]\ $ is metrically injective (is a **metric** AR), there are
Lipschitz maps $\ f\_k:X\to [-1;1]\ $ that extend the respective
$\ g\_k,\ $, hence
$$ f\ :=\ \triangle\_{k=1}^n f\_k:X\to[-1;1]^n $$
is a Lipschitz map.
Such continuous maps that start with the non-separable collections
$\ F\_{\pm k}\ $ are well-known and easily seen to be surjective
(or even *universal*).
**End of Proof**
| 7 | https://mathoverflow.net/users/110389 | 429069 | 173,909 |
https://mathoverflow.net/questions/428532 | 2 | **Some motivation:**
Let $W$ be a standard Brownian motion, and $X$ an integrable process with respect to $W$, i.e. progressively measurable with respect to the natural filtration of $W$ and square integrable on compacts almost surely.
It is known that if $X$ is continuous, then for any sequence of partitions $\mathcal P\_n$ of $[0, T]$ with mesh going to $0$, the Riemann sums
$$\sum\_{i = 0}^{K\_n - 1} X\_{t^n\_i} (W\_{t^n\_{i+1}} - W\_{t^n\_i})$$
converge in probability to the Itô integral
$$\int\_0^T X\_t \, dW\_t.$$
Here we have written $\mathcal P\_n := \{t^n\_0, \dots, t^n\_{K\_n}\}$.
Now it follows that if we take a suitable subsequence of partitions, the Riemann sums converge almost surely to the Itô integral. But I am wondering how almost sure convergence fails to begin with! It is unclear to me how a certain choice of partitions could lead to the limit fluctuating instead of existing.
**Questions:**
1. Can anyone provide an explicit example of an almost surely continuous process $X$, integrable with respect to $W$, and a sequence of partitions $\mathcal P\_n$ with mesh going to $0$ such that
$$\lim\_{n \to \infty} \sum\_{i = 0}^{K\_n - 1} X\_{t^n\_i} (W\_{t^n\_{i+1}} - W\_{t^n\_i})$$
fails to exist with some positive probability, or even full probability?
2. For a given $X$, are there some growth conditions on $\mathcal P\_n$ that ensure the integral would converges a.s.?
| https://mathoverflow.net/users/173490 | Divergence of Riemann sums in the Itô integral | Here is an example of an integrand $X$ and a deterministic sequence of partitions such that almost surely, the limit fails to exist.
Indeed, consider $X = W$. The associated Riemann sums are
$$\lim\_{n \to \infty} \sum\_{[u, v] \in \mathcal P\_n} W\_u (W\_{v} - W\_{u}).$$
We may rewrite this as
$$\frac{1}{2} W^2\_T - \frac{1}{2} \sum\_{[u, v] \in \mathcal P\_n} (W\_u - W\_v)^2.$$
Now, Exercise 1.15 in Peres and Mortens' *Brownian Motion* yields a sequence of deterministic partitions $\mathcal P\_n$ such that $\limsup\_{n \to \infty} \frac{1}{2} \sum\_{[u, v] \in \mathcal P\_n} (W\_u - W\_v)^2 = \infty$, almost surely, whence the result follows.
| 3 | https://mathoverflow.net/users/479223 | 429074 | 173,911 |
https://mathoverflow.net/questions/429073 | 2 | I’ve recently realised there is a subtlety in Girsanov’s theorem that I don’t really understand.
Consider a standard one dimensional Brownian motion $W$, and consider the SDE
$$dZ\_t = \mu(t, Z\_t) \, dt + \sigma(t, Z\_t) \, dW\_t \, , \,Z\_0 = x\_0 \, \, \, \text{(Equation 1)}$$
for some $x\_0 \in \mathbb R$, where $\mu, \sigma: [0, \infty) \times \mathbb R \to \mathbb R$ are Lipschitz continuous.
Denote by $\mathbb P$ the probability measure under which $W$ is a standard Brownian motion. Suppose we have an equivalent probability measure $\mathbb Q$ under which $W$ is no longer a standard Brownian motion, but a semimartingale.
We may still consider Equation 1 under $\mathbb Q$ as a semimartingale SDE.
Suppose $X$ solves Equation 1 under $\mathbb P$, and $Y$ solves Equation 1 under $\mathbb Q$.
**Question:** Is it true that we still have $X = Y$ up to indistinguishability? That is, do we have $X\_t = Y\_t$ for all $t \in [0, \infty)$, ($\mathbb P$, and hence $\mathbb Q$) almost surely?
It seems that this result is used implicitly in transforming SDE via Girsanov’s theorem, but it is not obvious to me at all.
| https://mathoverflow.net/users/173490 | A question related to Girsanov’s theorem | I managed to find an answer to this problem - indeed the answer is yes. The key ingredient is a theorem in Protter’s *Stochastic Integration and Differential Equations* (Chapter 2, Theorem 14) which states the following:
>
> For every fixed process $X$, integrable with respect to a semimartingale $S$, the stochastic integrals $\int X\, dS$ under $\mathbb P$ and $\mathbb Q$ are indistinguishable as processes.
>
>
>
The indistinguishability as processes part is key, and allows us to conclude that solutions to SDE too are invariant, as follows:
Since $X$ solves Equation 1 under $\mathbb P$, by definition we have
$$X = X\_0 + \int\_0^\cdot \mu(s, X\_s) \, ds + \int\_0^\cdot \sigma(s, X\_s) \, dW\_s$$
under $\mathbb P$, almost surely.
By indistinguishability of the stochastic integral on the RHS, we have also
$$X = X\_0 + \int\_0^\cdot \mu(s, X\_s) \, ds + \int\_0^\cdot \sigma(s, X\_s) \, dW\_s$$
under $\mathbb Q$, almost surely.
So $X$ solves Equation 1 under $\mathbb Q$. By uniqueness to solutions of SDE, it is *the* solution, i.e. $X = Y$ up to indistinguishability, and we are done.
| 3 | https://mathoverflow.net/users/173490 | 429077 | 173,912 |
https://mathoverflow.net/questions/429076 | 4 | Let $(M, g)$ be a compact Riemannian manifold. Suppose $M$ admits a smooth free circle action (Denote the circle group by $G$. The action $G$ on $M$ is not necessarily isometric) and the orbit space $B$ is another closed manifold, i.e., we have a smooth principal circle bundle $G\hookrightarrow M\rightarrow B$.
My question is:
(1) With the Riemannian metric $g$ being fixed for $M$, can we find another free $G$-action on $M$ so that the new action is isometric?
(2) If so, can we even find a new free circle action on $M$ so that the map $M\rightarrow M/G$ is harmonic?
| https://mathoverflow.net/users/169860 | Smooth circle actions on Riemannian manifolds and harmonicity of quotient map | I don't think so. I suspect that this action does not exists for generic metric $g$ on manifolds which admit circle actions.
Here is the thing that I have in mind.
Consider the two torus. This clearly has a free $G=S^1$ action. Now equip the torus with a metric $g$, and consider the scalar curvature $f$ of the metric. Suppose that there is an isometric action on the torus for the metric. Then the level sets of $f$ must be mapped to themselves by this action. So construct a metric which has a unique maximal curvature somewhere. This must be a fixed point of the action.
| 4 | https://mathoverflow.net/users/12156 | 429080 | 173,913 |
https://mathoverflow.net/questions/429043 | 3 | Prove: There exist real numbers a,b, such that 1<a<b and for any integer $n\ge 2$, there exist positive integers $x\_1\ge x\_2\ge ...\ge x\_{\lfloor{\frac{n}{2}}\rfloor}$such that $n!=\Pi\_{k=1}^{\lfloor\frac{n}{2}\rfloor}x\_k$ and $x\_k\in (a^\frac{n}{k},b^{\frac{n}{k}})$.
I've tried to list all the prime factors of n! but that's too complicated. Notice that the function $x^\frac{n}{k}$ is too strange. It means there are many small factors and a few large factors.
| https://mathoverflow.net/users/472630 | Factorising n! fitting $x^\frac{n}{k}$ | Your initial guess about prime factors can produce a solution. More precisely, it is known that
$$
n!=\prod\_{p\leq n} p^{v\_p(n!)},
$$
where the product is taken over prime numbers and
$$
v\_p(n!)=\left[\frac{n}{p}\right]+\left[\frac{n}{p^2}\right]+\ldots
$$
Let now for any $1\leq k\leq n/2$
$$
x\_k=\prod\_{\alpha, p: p^\alpha\leq n/k}p.
$$
The product is over prime values of $p$ and positive integers $\alpha$. The inequality $x\_{k}\geq x\_{k+1}$ is trivial. Product of all $x\_k$ is equal to $n!$ by the factorisation above: the factor $p$ that corresponds to a given $d=p^\alpha$ appears in $x\_k$ for $d\leq n/k$, there are $\left[\frac{n}{p^\alpha}\right]$ such $k$. This means that
$$
\prod\_k x\_k=\prod\_{\alpha,p} p^{\left[\frac{n}{p^\alpha}\right]}=\prod\_{p\leq n} p^{v\_p(n!)}=n!,
$$
as needed. Finally, let $\Lambda(d)$ denote the von Mangoldt function, i.e. $\Lambda(d)=\ln p$ if $d=p^\alpha$ for some $p$ and $\alpha$ and $\Lambda(d)=0$ otherwise. Then we clearly have
$$
\ln x\_k=\sum\_{d\leq n/k} \Lambda(d)=\psi(n/k).
$$
Here $\psi(x)$ is Chebyshev's psi function, the summatory function of $\Lambda$. It is well-known that there are $c,C>0$ such that for $x\geq 2$ we have
$$
cx\leq \psi(x)\leq Cx.
$$
In particular, one can take $1<a<e^c$, $b>e^C$ and get the desired bound:
$$
a^{n/k}\leq \exp(\psi(n/k))=x\_k\leq b^{n/k}.
$$
| 7 | https://mathoverflow.net/users/101078 | 429083 | 173,915 |
https://mathoverflow.net/questions/428863 | 0 | I am trying to characterize a set of distributions that satisfy two conditions. It is easy to characterize distributions fitting each of those conditions separately, but I am unable to make progress on characterizing the intersection. Specifically:
Let's have $X\_0,X\_1$ independent random variables over the interval $[0, 1]$ and let $F\_0, F\_1 : [0,1] \to [0,1]$ be their cumulative distribution functions. We can assume that both $F\_0$ and $F\_1$ are invertible.
Condition 1 entails:
$$
\forall x \in [0,1]: F\_1(x) = 2x - F\_0(x)
$$
This has implications for $F\_0$ as not all valid $F\_0$ will lead to valid $F\_1$. Specifically $F\_1$ is a valid invertible CDF as long as:
$$
\forall x \in [0,1]: 0 \leq 2x - F\_0(x) \leq 1 \\
F\_0'(x) < 2
$$
The problem is that Condition 2 works with the inverse CDFs:
$$
\forall x \in [0,1]: F^{-1}\_1(x) = \sqrt{2x - 2 F^{-1}\_0(x) + (F^{-1}\_0(x))^2}
$$
Similarly, this has additional implications - $F^{-1}\_1$ will be a valid inverse CDF as long as $F^{-1}\_0$ is valid and:
$$
\begin{align}
\forall x, 0 \leq x \leq \frac{1}{2} : F^{-1}\_0(x) &\leq 1 - \sqrt{1 - 2x} \\
\forall x, \frac{1}{2} \leq x \leq 1 : F^{-1}\_0(x) &\geq 1 - \sqrt{2 - 2x} \\
(F^{-1}\_0)^\prime(x) (F^{-1}\_0(x) - 1) &\geq -1
\end{align}
$$
I am unable to either re-express condition 1 in terms of the inverses - the closest I get is $x = F^{-1}\_0(2x - F\_1(x))$. I am similarly unable to transform condition 2 into a condition on the non-inverted CDFs.
This leaves me with equations that involve both a function and its inverse. Which I am unable to solve. I tried plugging that into Mathematica's `RSolve` but it couldn't solve it either.
Incorporating the implied conditions on $F\_0$ into the solution is not necessary, they can stay separate, I am just sharing them in case they are helpful for making progress at combining condition 1 and condition 2.
From the way the criteria are constructed, I know for sure that there are at least two solutions:
$$
F\_0(x) = F\_1(x) = x
$$
and
$$
F\_0(x) = 2x - x^2, F^{-1}\_0(x) = 1 - \sqrt{1-x} \\
F\_1(x) = x^2, F^{-1}\_1(x) = \sqrt{x}
$$
It would surprise me if those were the only solutions.
Does this look hopeless or is there some strategy to attack this? Even finding additional solutions without claiming that all possibilities are exhausted would help.
| https://mathoverflow.net/users/105908 | A set of equations for two (cumulative distribution) functions and their inverses | So I figured it out.
first substituting $x = F\_1(s)$ into condition 2:
$$
s = \sqrt{2 F\_1(s) - 2 F\_0^{-1}(F\_1(s)) + (F\_0^{-1}(F\_1(s)))^2} \\
s^2 = 2 F\_1(s) - 2 F\_0^{-1}(F\_1(s)) + (F\_0^{-1}(F\_1(s)))^2
$$
This is a quadratic equation for $F\_0^{-1}(F\_1(s))$ we solve it (only one branch of the solutions falls into $[0,1]$) and apply $F\_0$ to both sides of the solution:
$$
F\_0^{-1}(F\_1(s)) = 1 - \sqrt{1 - 2 F\_1(s) + s^2} \\
F\_1(s) = F\_0(1 - \sqrt{1 - 2 F\_1(s) + s^2}| 0)
$$
We now substitute condition 1 $F\_1(s) = 2s - F\_0(s)$ and thus obtain the equation:
$$
F\_0(x) = 2x - F\_0(1 - \sqrt{1 - 4x + 2 F\_0(x) + x^2}) \tag{1}
$$
Choosing $x\_0$ and $v\_0 = F\_0(x\_0)$, equation 1 implies a sequence of argument-value pairs that are fixed by that choice, i.e.:
$$
x\_{i + 1} = 1 - \sqrt{1 - 4x\_i + 2 v\_i + x\_i^2} \\
F\_0(x\_{i + 1}) = v\_{i + 1} = 2x\_i - v\_i
$$
We obtain that $x\_i = x\_{i+1} \iff x\_i = v\_i$, i.e. in this case the constraint on functional behaviour is local and does not induce any restrictions elsewhere. What about the other cases?
I plotted a bunch of those sequences and they looked quite quadratic. In fact it turns out that
$$
v\_i = -x\_i^2 + 2 x\_i +c \implies v\_{i + 1} = -x\_{i + 1}^2 + 2 x\_{i + 1} +c
$$
So if the starting point lies on such a quad curve, the whole subsequence does. And any point on the plane lies on exactly one such quad curve.
Now any $F\_0$ has to be a union of such sequences. If we assume that $F\_0$ is continuous (which we do), it means that it can only "switch" between the $F\_0(x) = x$ line and the quadratic curves at points where they intersect. Two quadratic curves differing only by constant never intersect each other, so we only need to care about intersections of the line and the quadratics which happens at:
$$
x = \frac{1}{2}\left( 1 \pm \sqrt{1 + 4c} \right)
$$
We also need to have the intersection within the $[0,1]$ interval, so we get $-\frac{1}{4} \leq c \leq 0$. Finally, we need $F\_0(0) = 0, F\_0(1) = 1$. This gives us a family of solutions:
$$
F\_0(s) = \begin{cases}
-x^2 + 2x + c & \frac{1}{2}\left( 1 \pm \sqrt{1 - 4c} \right) < x < \frac{1}{2}\left( 1 + \sqrt{1 + 4c} \right) \\
x & \text{otherwise}
\end{cases}
$$
where at $c = -\frac{1}{4}$ we get one of the solutions I knew previously and at $c = 0$ we have the other solution. All the intermediate solutions do in fact satisfy all the constraints, so they are valid for the original question and (assuming continuity) those are the only solutions.
| 0 | https://mathoverflow.net/users/105908 | 429089 | 173,916 |
https://mathoverflow.net/questions/429087 | 1 | How to calculate this sum $$\sum^\infty\_{n=0}\frac{1}{n^4+n^2+1}.$$
Thank you in advance
| https://mathoverflow.net/users/172078 | Closed formula for this sum $\sum^\infty_{n=0}\frac{1}{n^4+n^2+1}.$ | $\newcommand{\Ga}{\Gamma}$Martin Sleziak gave references to several answers to this question. All of those answers seem to involve complex analysis.
Here is how this sum can be evaluated (almost) without complex analysis: Using partial fraction decomposition, as e.g. in [this answer](https://math.stackexchange.com/a/966977/96609), we reduce the problem to that of finding two sums of the form
\begin{equation}
s:=\sum\_{n=0}^\infty\Big(\frac1{n+a}-\frac1{n+b}\Big)
\end{equation}
for certain complex $a$ and $b$ not in the set $\{0,-1,-2,\dots\}$.
We have
\begin{equation}
\begin{aligned}
s&=\sum\_{n=0}^\infty\int\_0^1 dx\,(x^{n+a-1}-x^{n+b-1}) \\
&=\int\_0^1 dx\,\sum\_{n=0}^\infty(x^{n+a-1}-x^{n+b-1}) \\
&=\int\_0^1 dx\,\frac{x^{a-1}-x^{b-1}}{1-x}=\lim\_{h\downarrow0}I(h),
\end{aligned}
\end{equation}
where
\begin{equation}
\begin{aligned}
I(h)&:=\int\_0^1 dx\,\frac{x^{a-1}-x^{b-1}}{(1-x)^{1-h}} \\
&=\frac{\Ga(a)\Ga(h)}{\Ga(a+h)}-\frac{\Ga(b)\Ga(h)}{\Ga(b+h)} \\
&=\frac{\Ga(a)\Ga(b+h)-\Ga(b)\Ga(a+h)}{h}\,\frac{\Ga(1+h)}{\Ga(a+h)\Ga(b+h)} \\
&\to\frac{\Ga(a)\Ga'(b)-\Ga(b)\Ga'(a)}{\Ga(a)\Ga(b)}
\end{aligned}
\end{equation}
as $h\downarrow0$, by l'Hospital's rule.
So,
\begin{equation}
\sum\_{n=0}^\infty\Big(\frac1{n+a}-\frac1{n+b}\Big)
=\frac{\Ga(a)\Ga'(b)-\Ga(b)\Ga'(a)}{\Ga(a)\Ga(b)}
=\psi(b)-\psi(a),
\end{equation}
where $\psi:=\Ga'/\Ga$ is the digamma function.
---
Another elementary way to find the sum is as follows: For natural $N$,
\begin{equation}
\begin{aligned}
\sum\_{n=0}^{N-1}\frac1{n+a}&=\frac d{da}\sum\_{n=0}^{N-1}\ln(n+a) \\
&=\frac d{da}\ln\prod\_{n=0}^{N-1}(n+a) \\
&=\frac d{da}\ln\frac{\Ga(N+a)}{\Ga(a)}=\psi(N+a)-\psi(a).
\end{aligned}
\end{equation}
So,
\begin{equation}
\begin{aligned}
&\sum\_{n=0}^\infty\Big(\frac1{n+a}-\frac1{n+b}\Big) \\
&=\lim\_{N\to\infty}\Big(\sum\_{n=0}^{N-1}\frac1{n+a}-\sum\_{n=0}^{N-1}\frac1{n+b}\Big) \\
&=\psi(b)-\psi(a)+\lim\_{N\to\infty}(\psi(N+a)-\psi(N+b)) \\
&=\psi(b)-\psi(a).
\end{aligned}
\end{equation}
| 3 | https://mathoverflow.net/users/36721 | 429090 | 173,917 |
https://mathoverflow.net/questions/429098 | 3 | Let $\Omega \subset \mathbb R^2$ be a *bounded* $C^2$ domain. Is $\Omega$ then finitely connected? As I learned [recently](https://mathoverflow.net/questions/421658/what-is-a-finitely-connected-domain#comment1083842_421658bounded) a domain in $\mathbb R^2$ is finitely connected iff “[its] complement has finitely many components (holes)“.
The first (mostly smooth) bounded domain with infinitely many holes that comes to mind is
$$\Omega = B\_1(0) \setminus \overline{\bigcup\_{n=2}^\infty B\_{1/n^2}(1/n)},$$
however this domain fails to be $C^2$ at $0$. (Here $B\_{r}(x)$ is the ball with radius $r$ centered at $x$.)
Similar constructions should run into the same issue. In particular, suppose that $\Omega$ is not finitely connected and enumerate the holes by $A\_1, A\_2, \dots$. For $n \in \mathbb N$, take $x\_n \in \partial A\_n$. As $\partial \Omega$ is compact, there exists a converging subsequence. I doubt that $\Omega$ can be $C^2$ at this limit point. However, I fail to make this precise.
My main interest is to apply a result holding for finitely connected $C^2$ domains to a bounded domain and thus I would welcome a reference stating that bounded $C^2$ domains are always finitely connected. This is also the reason I require the domains to be $C^2$ but the same question can of course also be asked for Lipschitz domains, for instance.
**Edit:** A domain is $C^2$ if it can locally be written as the set of points above the graph of a $C^2$ function. A domain is bounded if it is contained in $B\_R(0)$ for some $R > 0$. A bounded $C^2$ domain is a domain which is both bounded and $C^2$.
| https://mathoverflow.net/users/44919 | Is every planar bounded $C^2$ domain finitely connected? | Yes. It is finitely connected. The boundary of your domain is a compact one-dimensional manifold and therefore it consists of a finite number of curves diffeomorphic to circles. You can find a proof of classification of compact one dimensional manifolds in:
**J.W. Milnor,** *Topology from the differentiable viewpoint.* Based on notes by David W. Weaver University Press of Virginia, Charlottesville, Va. 1965.
| 6 | https://mathoverflow.net/users/121665 | 429103 | 173,924 |
https://mathoverflow.net/questions/429116 | 1 | A functional Hilbert space $\mathscr H=\mathscr H(\Omega)$ is a Hilbert space of complex valued functions on a (nonempty) set $\Omega$, which has the property that point evaluations are continuous i.e. for each $\lambda\in \Omega$ the map $f\mapsto f(\lambda)$ is a continuous linear functional on $\mathscr H$. The Riesz representation theorem ensure that for each $\lambda\in \Omega$ there is a unique element $k\_{\lambda}\in \mathscr H$ such that $f(\lambda)=\langle f,k\_{\lambda}\rangle$ for all $f\in \mathscr H$. The collection $\{k\_{\lambda} : \lambda\in \Omega\}$ is called the reproducing kernel of $\mathscr H$. For $\lambda\in \Omega$, let $\hat{k\_{\lambda}}=\frac{k\_{\lambda}}{\|k\_{\lambda}\|}$ be the normalized reproducing kernel of $\mathscr H$.
For a bounded linear operator $A$ on $\mathscr H$, we define the following norm:
\begin{align\*}
N(A):=\sup\{\big|\langle T\widehat{k}\_{\lambda},\widehat{k}\_{\lambda}\rangle\big|: \lambda\in\Omega\}.
\end{align\*}
>
> It is well-known that $w(A^n)=w^n(A)$ holds for any selfadjoint operator $A$, $w(.)$ denotes the numerical radius. Now, assume that $A$ is a selfadjoint operator. Is
> $$N(A^n)=N^n(A)\;?$$
>
>
>
| https://mathoverflow.net/users/113054 | Is $N(A^n)=N^n(A)\;$ for a self-adjoint operator $A$? | No. If $A=P$ is a projection, then $P^2=P$, but in general you won't have $N(P)=0$ or $1$. In fact, you could just take $\Omega=\{ 1,2\}$, so $\mathcal H\cong\mathbb C^2$, and $P$ as the projection onto $(1,1)$.
| 3 | https://mathoverflow.net/users/48839 | 429117 | 173,925 |
https://mathoverflow.net/questions/429139 | 1 | Let $\mathbb{H}$ be the upper half plane model of hyperbolic geometry. Let $\Gamma$ be the Fuchsian group such that $\mathbb{H}/\Gamma$ is the compact orientable surface of genus $2$.
Suppose $\Gamma = \langle g\_1,g\_2,g\_3,g\_4 \mid g\_1g\_2g\_1^{-1}g\_2^{-1}g\_3g\_4g\_3^{-1}g\_4^{-1}=1 \rangle$ where $g\_1,g\_2,g\_3,g\_4$ are hyperbolic translations.
In B. Maskit's book on Kleinian Groups, it is given that the normal form of a hyperbolic translation is given by the matrix
$$\frac{1}{x-y}\begin{pmatrix}xk^{-1}-yk&xy(k-k^{-1})\\ k^{-1}-k & xk-yk^{-1} \end{pmatrix}$$
where $x$ and $y$ are fixed points of the translation and $k^2$ is the multiplier.
Then we have four 3 tuples $(x\_i,y\_i,k\_i)$ for $i=1,2,3,4$ for our $g\_i$'s. Now the fundamental domain of $\Gamma$ will be a hyperbolic octagon.
The area of the fundamental domain will be $4\pi$.
Can we determine some identity or equation in terms of the parameters $x\_i,y\_i, k\_i$ and $4\pi$?
Is it possible to find the index of $\Gamma$ in the $(2,3,7)$ triangle group in terms of the parameters $x\_i,y\_i, k\_i$?
| https://mathoverflow.net/users/490039 | Area of fundamental domain of Fuchsian group and index of a Fuchsian group in the triangle group | EDIT: My answer (below) is for the original question. The current question has been modified to include my answer.
---
The area of a fundamental domain for $\Gamma$ is $4\pi = -2\pi \chi(\mathbb{H} / \Gamma)$. There are various proofs; most people think of this as a consequence of the Gauss-Bonnet theorem.
---
To answer your next question:
>
> Is it possible to find the index of $\Gamma$ in the $(2,3,7)$ triangle group in terms of the parameters?
>
>
>
When $\Gamma$ is a subgroup at all (which happens only for [very very very special](https://johncarlosbaez.wordpress.com/2013/05/25/42/#comment-29515) values of the parameters) it has index $84$. This is because the area of the $(2,3,7)$ orbifold is $\frac{\pi}{21}$.
| 4 | https://mathoverflow.net/users/1650 | 429140 | 173,930 |
https://mathoverflow.net/questions/429152 | 3 | Let $X \subset \mathbb{C}^n$ be a smooth complex affine variety of dimension $m$. Let
$$I\_X \subset \mathbb{C}[z\_1, \ldots, z\_n]$$
be its ideal. Note that it is not always possible to find a set of generators
$$I\_X = \langle g\_1, \ldots, g\_r \rangle$$
such that $r = n - m$ is the codimension. In general, we only have $r \ge n - m$. But can we do this locally? That is:
**Question.** *Can we cover $X$ by distinguished open sets $X\_f$ such that by viewing $X\_f$ as an affine variety in $\mathbb{C}^{n + 1}$, its ideal is generated by $n + 1 - m$ elements?*
If not, is there a result of that flavour that I'm missing?
| https://mathoverflow.net/users/479013 | Locally, the minimal number of ideal generators is the codimension | I am just writing an answer to correct my mistake!
This is true for smooth subvarieties, roughly by the Jacobian criterion. For each closed point, the rank of the Jacobian matrix at that point equals the codimension $m$. So if you choose $m$ defining relations whose partial derivative vectors give $m$ linearly independent rows in the Jacobian matrix (or columns, depending how you write things), then those $m$ relations locally define $X\_f$.
My comment above was answering a different question: this is not true if you allow $X$ to be singular. There are "topological" consequences of being a set-theoretic local complete intersection, e.g., the connectedness theorem of Hartshorne. This prevents certain singular varieties from being set-theoretic complete intersections (notice, this connectedness theorem does not prevent curves in $3$-space from being set-theoretic complete intersections, and it is open whether all curves in $3$-space are set-theoretic complete intersections).
| 6 | https://mathoverflow.net/users/13265 | 429165 | 173,938 |
https://mathoverflow.net/questions/429161 | 7 | I am reading Colombeau's book "New Generalized Functions and Multiplication of Distributions" and he uses the notation $C^\infty({C^\infty}'(\Omega))$ out of nowhere.
Here $\Omega$ is any open subset of some $\mathbb{R}^n$ and $C^\infty(\Omega)$ is the space of complex-valued $C^\infty$ functions on $\Omega$. ${C^\infty}'(\Omega)$ is the dual space of $C^\infty(\Omega)$.
So, my questions are:
1. What topology is given on $C^\infty(\Omega)$ in general? Here $\Omega$ doesn't have to be bounded at all.. We need to deal with this issue in order to define ${C^\infty}'(\Omega)$.
2. What topology is given on the dual space ${C^\infty}'(\Omega)$ then? This is necessary to define the notion of "$C^\infty$" for the functions whose domain is ${C^\infty}'(\Omega)$.
Could anyone clarify?
| https://mathoverflow.net/users/56524 | How should I understand the "$C^\infty$ functions" whose domain is the dual of $C^\infty(\mathbb{R}^n)$? | The standard topology on $C^\infty(\Omega)$ is that of uniform convergence on compact subsets of $\Omega$ of all derivatives which is given by the seminorms $\|f\|\_{K,n}=\sup\{|\partial^\alpha f(x)|: x\in K, |\alpha|\le n\}$ with $K\subseteq \Omega$ compact and $n\in \mathbb N$. This topology makes $C^\infty(\Omega)$ a nuclear Fréchet space. The standard topology on the dual is the topology of uniform convergence on all bounded (in this case, this is the same as on all compact) sets. This makes $C^\infty(\Omega)'=\mathscr E'(\Omega)$ (in Schwartz' notation, it is the space of distributions with compact support in $\Omega$) a very nice locally convex space (complete, ultrabornological, nuclear,...).
| 8 | https://mathoverflow.net/users/21051 | 429173 | 173,939 |
https://mathoverflow.net/questions/429176 | 4 | Let $V$ be a vector space and $\|\cdot \|\_1$ and $\|\cdot\|\_2$ two norms on $V$.
Let $\|\cdot\|\_+$ be given by
$$ \|v\|\_+ := \inf\_{v = v\_1 + v\_2} \|v\_1\|\_1 + \|v\_2\|\_2 $$
It is well-known that $\|\cdot\|\_+$ is a norm on $V$. ${}{}{}{}$
Is it true that the (closed) unit ball of $\|\cdot\|\_+$ is the closed convex hull of the union of those of $\|\cdot\|\_1$ and $\|\cdot\|\_2$?
* If so: where can I find a proof?
* If not: is there an easy characterization of the unit ball of $\|\cdot\|\_+$ in terms of that of $\|\cdot\|\_1$ and $\|\cdot\|\_2$?
| https://mathoverflow.net/users/3948 | Unit ball of the sum space | Let $B\_+,B\_1,B\_2$ denote the closed unit balls w.r. to $\|\cdot\|\_+,\|\cdot\|\_1,\|\cdot\|\_2$, respectively. Let $C$ be the convex hull of $B\_1\cup B\_2$. Let $\bar C$ be the closure of $C$ (w.r. to the "norm" $\|\cdot\|\_+$ -- which is actually only a semi-norm in general, as pointed out by [Jochen Wengenroth](https://mathoverflow.net/a/429306/36721)).
Take any $v\_1\in B\_1$. Then $v\_1=v\_1+0$ and $\|v\_1\|\_+\le\|v\_1\|\_1 + \|0\|\_2=\|v\_1\|\_1\le1$. So, $B\_1\subseteq B\_+$. Similarly, $B\_2\subseteq B\_+$. So, $C\subseteq B\_+$ and $\bar C\subseteq B\_+$.
Vice versa, take any $v\in B\_+^\circ$, the interior of $B\_+$, so that $\|v\|\_+<1$. Then there are some $v\_1$ and $v\_2$ in $V$ such that $v=v\_1+v\_2$ and $\|v\_1\|\_1 + \|v\_2\|\_2<1$. Let $t:=\|v\_1\|\_1$, so that $t\in[0,1)$ and $\|v\_2\|\_2<1-t\in(0,1]$.
If $t=0$, then $v=v\_2$, $\|v\|\_2=\|v\_2\|\_2<1$, and hence $v\in B\_2\subseteq\bar C$.
It remains to consider the case when $t\in(0,1)$. Then $v=tu\_1+(1-t)u\_2$, where $u\_1:=\frac{v\_1}t\in B\_1$ and $u\_2:=\frac{v\_2}{1-t}\in B\_2$.
So, in either case, $v\in C\subseteq\bar C$, for each $v\in B\_+^\circ$.
So, $B\_+^\circ\subseteq\bar C$ and hence $B\_+\subseteq\bar C$.
Thus, $B\_+=\bar C$, as desired.
$\quad\Box$
---
One may want to note here that the semi-norm $\|\cdot\|\_+$ is the [infimal convolution](https://en.wikipedia.org/wiki/Convex_conjugate#Infimal_convolution) of the norms $\|\cdot\|\_1$ and $\|\cdot\|\_2$.
| 6 | https://mathoverflow.net/users/36721 | 429180 | 173,940 |
https://mathoverflow.net/questions/428726 | 2 | Suppose a group $G$ acts on an infinite set $X$ and $X$ has no non-empty $G$-paradoxical subsets. Is it possible for $X$ to have non-trivial $G$-paradoxical subsets modulo finite sets? I.e., can there be infinite subsets $E\_1,...,E\_n$ such that $E\_i \cap E\_j$ is finite when $i\ne j$, and $\tau\_1,\dots,\tau\_n \in G$ such that $E\_1\cup \dots \cup E\_n$, $\tau\_1 E\_1 \cup \dots \cup \tau\_m E\_m$ and $\tau\_{m+1} E\_{m+1} \cup \dots \cup \tau\_n E\_n$ differ only by finite sets for some $m$?
**Note:** When $G$ acts transitively (for any $x$ and $y$ there is a $\tau\in G$ such that $y=\tau x)$, the answer is negative. For let $E=\bigcup E\_i$. Because $G$ has no paradoxical sets, there is a $G$-invariant finitely additive measure $\mu:2^X \to [0,\infty]$ with $0<\mu(E)<\infty$. Because $G$ acts transitively, all singletons have equal measure, and since $E$ is infinite and has finite measure, all finite sets must have measure zero. Paradoxicality of $E$ modulo finite sets would then imply that $\mu(E)=\mu(E)+\mu(E)$, a contradiction.
| https://mathoverflow.net/users/26809 | Paradoxical decomposition modulo finite sets | No it's not possible.
Recall the notation. For $G$ acting on a set $X$, subsets $Y,Y'$ of $X$ are called $G$-congruent if $\exists g\in G$ such that $Y'=gY$, and piecewise $G$-congruent if there are partitions $(Y\_i)\_{1\le i\le n}$ of $Y$, $(Y'\_i)\_{1\le i\le n}$ of $Y$ such that $Y\_i$ and $Y'\_i$ are $G$-congruent for all $i$. The subset $Y$ of $X$ is $G$-paradoxical if there is a partition $Y=Y\_1\sqcup Y\_2$ such that $Y$ is piecewise $G$-congruent to both $Y\_1$ and $Y\_2$.
Then Tarski's theorem says that $Y$ is not $G$-paradoxical iff there exists a finitely additive $G$-invariant measure $\mathcal{P}(X)\to [0,\infty]$ mapping $Y$ to $1$.
Next one can define near (piecewise) $G$-congruent, near $G$-paradoxical by allowing "modulo finite subsets".
**Proposition** $Y\subset X$ is near $G$-paradoxical iff it can be written as union of a $G$-paradoxical subset and a finite set.
Proof: $\Leftarrow$ is trivial. Let us prove $\Rightarrow$ by contraposition. That is, assume that $Y\smallsetminus F$ is not $G$-paradoxical for every finite subset $F$.
Let $Y\_0$ be the set of $y\in Y$ such that $Gy\cap Y$ is finite. Say that a finite subset $F$ of $Y$ is $G$-saturated if $GF\cap Y\subset F$ (this forces $F\subset Y\_0$).
For each $G$-saturated $F$, by Tarski's theorem, let $m\_F:\mathcal{P}(X\smallsetminus GF)\to [0,\infty]$ be a finitely additive $G$-invariant measure mapping $Y\smallsetminus F$ to 1; extend it to $\mathcal{P}(X)$ as being zero on $GF$.
Let $m$ be a limit point of $m\_F$ when $F$ tends to $Y\_0$. So $m$ is a finitely additive $G$-invariant measure mapping $Y\smallsetminus F$ to 1; moreover $m$ vanishes on every singleton of $Y\_0$. Then $m$ vanishes on every singleton (if $y\in Y\smallsetminus Y\_0$ then $Gy\cap Y$ is infinite).
So $Y$ is not near $G$-paradoxical. The proof is complete.
**Corollary** If every nonempty $Y\subset X$ is non-paradoxical, then every infinite $Y\subset X$ is non-near-paradoxical.
This answers the question.
| 2 | https://mathoverflow.net/users/14094 | 429188 | 173,942 |
https://mathoverflow.net/questions/429168 | 7 | $\newcommand\CAR{\mathit{CAR}}\newcommand\Cl{\mathbb C\mathit l}$This question will be rather long and it will be my attempt to finally clarify many issues concerning CCR, CAR and Clifford algebras together with the Fock spaces and the classification of the corresponding von Neumann algebras appearing in this way. I must say that I'm very confused by all this stuff. Let me summarize what I have learned so far. Let $V$ be a Hilbert space (over $\mathbb{R}$ I suppose—at least for the Clifford algebra):
0. First of all, let me emphasize the slight difference between the so called CAR (canonical anticommutation relation) algebra over a Hilbert space and the Clifford algebra (in the $C^\*$-algebraic version, to be denoted by $\Cl^\*(V)$: these two are very closely related. However, note that the Clifford algebra is generated by all $j(v)$'s where $j:V \to \Cl^\*(V)$ is the natural embedding (in other words this algebra is the universal $C^\*$-algebra satisfying Clifford relations) while the CAR algebra is generated by all creation and anihilation operators (abstractly) so is generated by a priori larger set of relations.
>
> **Question 1** Is it true that abstractly $\CAR(V)$ and $\Cl^\*(V)$ are isomorphic as $C^\*$-algebras?
>
>
>
1. Now it is shown in Plymen's book „Spinors in Hilbert space‟ that given the von Neumann version of the Clifford algebra (over infinite dimensional $V$) acting via the left regular representation (meaning the GNS representation coming from the trace) one obtains the hyperfinite $II\_1$ factor. This $II\_1$ hyperfinite factor is unique (up to isomorphism) and admits various incarnations: one of them is an infinite tensor product of $M\_2(\mathbb{C})$. This means that $\operatorname{v.N.}\Cl(V)$ acts naturally on the infinite tensor product.
2. On the other hand, there are plenty of nonequivalent representations of this Clifford algebra, constructed with the help of the Fock space. The construction of this Fock space requires the choice of the so called unitary (or complex) structure $J$ on our *real* Hilbert space $V$. Different choices of $J$ *can* lead to inequivalent representations and the condition for equivalence is the content of Shale–Stinespring theorem and is of the form that $J-K$ should be a Hilbert–Schmidt operator.
3. One shows that Fock representations are irreducible: therefore (the v.N. version of) our Clifford algebra will be type $I\_{\infty}$. However, we also stated that in the left regular representation our Clifford algebra is just the infinite tensor product $\bigotimes M\_2(\mathbb{C})$ which is of type $II\_1$. Therefore a natural question arises:
>
> **Question 2** a) Is there a way to find a representation $\pi$ of $\Cl^\*(V)$ such that $(\pi(\Cl^\*(V)))''$ is of type $III$? Is the Hilbert space of this representation some kind of infinite tensor product (or wedge product)?
>
> b) Is there a some sort of *canonical* isomorphism between Fock space and the infinite tensor product (possibly depending whether we take full, or symmetric or antisymmetric Fock space) allowing us to see Fock representation as a representation on this infinite symmetric tensor product?
>
>
>
4. Recall that the Fock representation $\pi$ is defined as $\pi(v)=a^+(v)+a^-(v)$ and turns out to be irreducible: thus the von Neumann algebra generated by $\{\pi(v), v \in V\}$ will be the same as the von Neumann algebra generated by $\{a^+(v),a^-(v): v \in V \}$ both being type $I\_{\infty}$. Note that the family $\{\pi(v): v \in V\}$ consists from self adjoint operators but the operators $\pi(v)$ do not commute with each other—unlike in the bosonic case for which I would like to turn now.
5. So in the bosonic case the analogue of the *algebraic* version of Clifford algebra is the so called Weyl algebra—however one can show that it is impossible to represent a Weyl algebra as bounded operators in some Hilbert space (unlike for Clifford algebra)—therefore one considers the associated canonical commutation relation in the *integrated* form. One then considers the *universal* $C^\*$-algebra generated by these relations. Recall that for the Clifford algebra we had a particular way to represent it as a type $II\_1$ hyperfinite factor (using the trace).
>
> **Question 3** a) Is there a canonical „left regular representation‟ picture for CCR? If the answer is yes, what is the corresponding von Neumann algebra (its type? Is it a factor)?
>
> b) Is there a natural and canonical way to represent CCR algebra as an infinite tensor product?
>
>
>
6. For CCR one can again speak about Fock representations and this representation in the bosonic case turns out to be again irreducible: therefore the von Neumann algebra in this representation would be again of type $I\_{\infty}$. But note the subtle difference here: the family of field operators $\pi(v):=a^+(v)+a^{-}(v)$ while unbounded, still consists from self adjoint operators but this time this family is *commutative*! Therefore it should produce some measure: I suspect that this family should produce some variant of Gaussian measure but still I would like to know the details:
>
> **Question 4** a) If we consider a von Neumann algebra generated by the family $\{\pi(v): v \in V\}$ it should be naturally isomorphic to $L^{\infty}(X,\mu)$. Can one identify the space $(X,\mu)$ concretely?
>
> b) is there a way to arrive at some canonically defined measure starting from CAR instead of CCR? If yes, what is the corresponding measure space? (Some relation with the Ising model?)
>
>
>
7. Finally there is also a *free* (sometimes called *full*) version of the Fock space where one does not use any symmetrization and antisymmetrization. Here the creation and anihilation operators satisfy the following commutation relation: $a^-(u)a^+(v)=\langle u,v \rangle$. Quite surprisingly (at least for me) Voiculescu in his paper „Symmetries of some reduced free product of $C^\*$-algebras‟ has proved that the von Neumann algebra generated by $\{a^+(v)+a^-(v): v \in V\}$ in the Fock representation is isomorphic with the von Neumann algebra generated by the free group where the numbers of generators is equal to $\dim V$. So let me ask a final question:
>
> **Question 5** What von Neumann algebra we get by taking the von Neumann algebra $\{a^+(v),a^-(v): v \in V\}''$ generated by all creation and anihilation operators in the *free* Fock space?
>
>
>
| https://mathoverflow.net/users/24078 | CCR vs. CAR vs. Clifford algebras, infinite tensor products and type of the corresponding von Neumann algebra | Wow, that's a lot of questions. For a more in-depth discussion you might look at my book *Mathematical Quantization*, particularly Section 2.5 and Chapters 4 and 7. But anyway, let me start with the most interesting questions, 2b and 3a.
Question 2b. Are Fock spaces infinite tensor products? Yes. First, remember that full infinite tensor products of Hilbert spaces are nonseparable and horrible. The version of ``tensor product'' you want to look at is, for any family of Hilbert spaces $H\_i$ with distinguished unit vectors $u\_i$, the closed subspace of their full tensor product generated by the elements $\bigotimes v\_i$ with each $v\_i \in H\_i$ and $v\_i = u\_i$ for all but finitely many $i$.
Given that definition of tensor product, let $X$ be a countable set and let $\{e\_0, e\_1\}$ be the standard basis of $\mathbb{C}^2$. Then $\bigotimes\_X \mathbb{C}^2$, using the above definition with $u\_i = e\_0$ for all $i \in X$, has a natural orthonormal basis indexed by all finite subsets of $X$. To each such subset $S$ we associate the vector $\bigotimes v\_i$ where $v\_i = e\_1$ if $i \in S$ and $v\_i = e\_0$ if $i \not\in S$. These vectors constitute an orthonormal basis of $\bigotimes\_X \mathbb{C}^2$.
This answer is already getting a bit long, so I'll leave it to you to check that the span of the basis vectors coming from singleton subsets of $X$ can be identified with $l^2(X)$; and for arbitrary $n$, the span of the basis vectors coming form subsets of $X$ of size $n$ can be identified with the antisymmetric $n$-fold tensor power of $l^2(X)$. That is, $$\bigotimes\_X \mathbb{C}^2 \cong H\_0 \oplus H\_1 \oplus \cdots$$
where $H\_n$ is the $n$-fold antisymmetric tensor power of $l^2(X)$. I.e., it is the fermionic Fock space over $l^2(X)$.
This relates to quantum field theory in the following way. In QFT we have a Hilbert space $H\_x$ at each point $x$ of space, or relativistically at each point of some maximal spacelike surface. And morally, a state of the field is an element of the tensor product of all these Hilbert spaces. If you take space to be a discrete set $X$, and let the field have two possible states at each point (one ground state and one excited state), then we can take the field Hilbert space to literally be $\bigotimes\_X \mathbb{C}^2 \cong \mathcal{F}\_a(l^2(X))$. This is a fermionic field. If space is not discrete, $X = \mathbb{R}^3$ for example, then the natural Hilbert space for the field is $\mathcal{F}\_a(L^2(\mathbb{R}^3))$, which is morally $\bigotimes\_{\mathbb{R}^3}\mathbb{C}^2$ for some notion of ``measurable tensor product''.
Going back to the discrete case, if we give the field infinitely many states $e\_n$ ($n \in \mathbb{N}$), spanning a copy of $l^2$, at each point of $X$, then the natural orthonormal basis now corresponds to finite *multisubsets* of $X$ and we get an isomorphism $\bigotimes\_X l^2 \cong \mathcal{F}\_s(l^2(X))$. This is the bosonic case.
So yes, Fock spaces of the two types can be identified literally with infinite tensor powers of either $\mathbb{C}^2$ or $l^2$ over a discrete set, or morally with infinite tensor products over a measure space.
Question 3a. Do CCR algebras have canonical ``left regular'' representations? Yes. First, let $H$ be a finite-dimensional real Hilbert space. For each $v \in H$ define unbounded self-adjoint operators $Q\_v$ and $P\_v$ on $L^2(H)$ by
$$Q\_vf(w) = \langle w,v\rangle f(w)\qquad{\rm and}\qquad P\_vf(w) = -i\hbar\frac{\partial f}{\partial v}(w).$$
Then $CCR(H)$ is the C\*-algebra generated by the unitary operators $e^{iQ\_v}$ and $e^{iP\_v}$ for $v \in H$.
When $H$ is infinite you can't understand $L^2(H)$ as a genuine $L^2$ space, but you can take it to be a direct limit of genuine $L^2$ spaces for all finite subspaces of $H$. I'm just going to refer you to Section 7.2 of my book for details.
I guess I can answer some of the other questions too.
Question 1. Yes, Theorem 3.19 in [these notes](https://www.mathematik.uni-muenchen.de/%7Eschotten/qftcs/qed%20cliff%20car.pdf). This is basically just a calculation.
Question 2a. It's a theorem of Glimm (``Type I C${}^\*$-algebras'', *Ann. of Math.* **73** (1961), 572-612) that for any C${}^\*$ algebra $A$, either $\pi(A)''$ is type I for every representation or $\pi(A)''$ can be any type.
Question 3b. Literally yes for $CCR(H)$ when $H$ is finite-dimensional; it's naturally the tensor product of the algebras $CCR(E)$ where $E$ ranges over all the one-dimensional subspaces of $H$ generated by the elements of some fixed orthonormal basis. Morally this is still true when $H$ is infinite-dimensional, but not literally because (in the $L^2(H)$ model mentioned above) the operators $e^{iQ\_v}$ and $e^{iP\_v}$, for any $v$ not in the span of finitely many basis vectors, cannot be approximated in norm by tensor products.
Question 4a. Yeah, in the $L^2(H)$ model again you'd now be looking at the C${}^\*$-algebra generated by just the operators $e^{iQ\_v}$. These all commute, and the C${}^\*$-algebra they generate is just the algebra of almost periodic functions on $H$ acting as multiplication operators. For finite-dimensional $H$ the von Neumann algebra they generate would just be $L^\infty(H)$, acting by multiplication. For infinite-dimensional $H$ this doesn't really work and I think the von Neumann algebra there is pretty horrible.
Question 4b. I guess the obvious thing to do here is to go back to the $\mathcal{F}(l^2(X))$ model and identify the finite subsets of $X$ with elements of the product space $\{0,1\}^X$, giving $\{0,1\}$ counting measure.
I haven't thought about Question 5.
| 6 | https://mathoverflow.net/users/23141 | 429222 | 173,953 |
https://mathoverflow.net/questions/429214 | 2 | **Background**
Say we have an optimization problem $$\min\_x f(x) = g(x) + h(x)$$
where $g$ is differentiable and convex, and $h$ are convex but not necessarily differentiable. If $g$ is the mean squared error function, we can use proximal algorithms to solve optimizations of this type, using the proximal operator, given by $$\text{prox}(x) = \text{argmin}\_z \frac{1}{2}\left\| x-z\right\|^2\_2+h(z)$$
**Problem**
Suppose we know the proximal operator (so the solution to the optimization problem) for two problems of the same form such that $h\_1$ and $h\_2$ are convex but not necessarily differentiable:
1. $\min\_x \frac{1}{2}\left\| y-x\right\|^2 + h\_1(x)$, so that we know $\text{prox}\_{h\_1}(x)$ for this problem.
2. $\min\_x \frac{1}{2}\left\| y-x\right\|^2 + h\_2(x)$, so that we know $\text{prox}\_{h\_2}(x)$ for this problem.
If we now combine the two to give us a new optimization problem
$$\min\_x \frac{1}{2}\left\| y-x\right\|^2 +h\_1(x)+ h\_2(x)$$
**My questions**
1. Can we work out a proximal operator $\text{prox}\_{h\_1, h\_2}$ to solve this new problem using the two operators we already have?
2. Do we require any assumptions on separability? Do we need the new penalty $h' = h\_1 + h\_2$ to be a separable function (i.e, are $h\_1$ and $h\_2$ separable to each other)? What happens if this is not the case?
**Example**
To give a more concrete example, we can look at the sparse-group lasso, given by
$$\min\_{\beta}\left( \frac{1}{2n}\left\|y-X\beta \right\| + (1-\alpha)\lambda\sum\_{l=1}^m \sqrt{p\_l}\left\|\beta^{(l)} \right\|\_2 + \alpha \lambda \left\| \beta\right\|\_1 \right),$$
where $h\_1(\beta)=(1-\alpha)\lambda\sum\_{l=1}^m \sqrt{p\_l}\left\|\beta^{(l)} \right\|\_2$ is the group lasso penalty and $h\_2(\beta)=\alpha \lambda \left\| \beta\right\|\_1$ is the lasso penalty. Therefore, could we use the proximal operators for the lasso and group lasso to derive a proximal operator for SGL? I understand that we can also derive the proximal operator directly for SGL, but I'm adding this as an example to make the question more concrete.
In this case, do we have separability between the lasso and group lasso penalties? Do we need it to help?
| https://mathoverflow.net/users/479197 | Can we use the solution to two optimisation problems to solve a third, bigger, one? | I think what you want is the alternating direction method of multipliers (ADMM), which is a type of proximal method.
The formulas you've written down are really close to ADMM.
Rather than try to minimize the difficult objective $f(x) = g(x) + h(x)$ directly, we instead introduce an auxiliary variable $y$ and minimize
$$f(x, y) = g(x) + h(y)$$
subject to the constraint that $x = y$.
Now, there are several ways we might enforce this constraint; to do so, we'll use the augmented Lagrangian
$$\mathscr L\_\rho(x, y, \lambda) = g(x) + h(y) - \langle \lambda, x - y\rangle + \frac{\rho}{2}\|x - y\|^2.$$
The vector $\lambda$ is a Lagrange multiplier and $\rho$ is a scalar penalty factor.
We can complete the square to write this in a slightly more revealing form:
$$\ldots = g(x) + h(y) + \frac{\rho}{2}\left\|x - y - \rho^{-1}\lambda\right\|^2 - \frac{1}{2\rho}\|\lambda\|^2.$$
At this point we can see that finding a minimizer of the augmented Lagrangian just with respect to $x$ or $y$ separately amounts to evaluating a proximal operator of $g$ or $h$ with a shifted argument.
You've already assumed that both of these proximal operators are computable.
Finally, there's the question of what to do about the Lagrange multipliers $\lambda$.
For that you can use a gradient ascent step at every iteration.
I found the fact that this works without any line search to be deeply mysterious and very few references dig into this detail at all.
The book [Constrained Optimization and Lagrange Multiplier Methods](https://www.mit.edu/%7Edimitrib/Constrained-Opt.pdf) by Dmitri Bertsekas does a good job explaining why with some nice theory behind it and the author makes the book available for free on his website.
This is all described in further detail in section 4.4.1 of [Proximal Algorithms](https://web.stanford.edu/%7Eboyd/papers/pdf/prox_algs.pdf) by Parikh and Boyd.
I also wrote a bit about using ADMM to solve inverse coefficient problems for PDE with total variation regularization [here](https://shapero.xyz/posts/admm/) which, although distinct from the lasso problem you've described, still has some of the same $\ell^1$ kind of character to it.
| 2 | https://mathoverflow.net/users/49417 | 429226 | 173,956 |
https://mathoverflow.net/questions/427802 | 5 | Assume that BSD holds for number fields. Let $E/\mathbf{Q}$ be an elliptic curve. For simplicity, let's assume it has Mordell-Weil rank zero. Let $F\_1/\mathbf{Q}$ and $F\_2/\mathbf{Q}$ be finite, abelian, disjoint extensions, and let $F=F\_1.F\_2$ denote their compositum.
>
> **Question:** Under what conditions can we conclude that
> $$ \mathrm{rank}\_{\mathbf{Z}}(E/F)=\mathrm{rank}\_{\mathbf{Z}}(E/F\_1) + \mathrm{rank}\_{\mathbf{Z}}(E/F\_2) ?$$
>
>
> In other words, are there any sufficient conditions to ensure that any Mordell-Weil rank growth in $F$ actually in arises either $F\_1$ or $F\_2$?
>
>
>
On the one hand, this seems too good to be true. On the other hand, *if we assume BSD for number fields*, it seems that we should be able to make an argument using the factorization of $L$-functions as follows:
Let $\mathrm{Gal}(F\_1/\mathbf{Q})=G\_1$ and $\mathrm{Gal}(F\_2/\mathbf{Q})=G\_2$. Since $F\_1 \cap F\_2 = \mathbf{Q}$, we have $\mathrm{Gal}(F/\mathbf{Q})=G\_1 \times G\_2$. Using $^\hat{}$ to denote character groups, it seems we should have
\begin{align\*}
L(E, F, s) &= \prod\_{\chi\in \hat{G}} L(E, \mathbf{Q}, s, \chi)\\ &\simeq \prod\_{\chi\_1\in \hat{G\_1}}\prod\_{\chi\_2\in \hat{G\_2}} L(E, \mathbf{Q}, s, \chi\_1 \chi\_2)\\ &=L(E, F\_1, s)\times L(E, F\_2, s) \times\prod\_{\substack{\chi\_1\in \hat{G\_1} \\ \chi\_1\neq 1}}\prod\_{\substack{\chi\_2\in \hat{G\_2} \\ \chi\_2\neq 1}} L(E, \mathbf{Q}, s, \chi\_1 \chi\_2)
\end{align\*}
where the products are taken over the irreducible characters. Thus, we have the desired additivity if we can control the terms of the form $L(E, \mathbf{Q}, s, \chi\_1 \chi\_2)$.
Are there any hypotheses which push this across the finish line?
---
Edit: Ariel Weiss has posted an answer which highlights the following situation:
$$E=X\_0(11), F\_1=\mathbf{Q}(\sqrt{-7}), F\_1=\mathbf{Q}(\sqrt{-8}), F=\mathbf{Q}(\sqrt{-7},\sqrt{-8})$$ Then we have
$$\mathrm{rank}\_{\mathbf{Z}}(E/\mathbf{Q})=0, \mathrm{rank}\_{\mathbf{Z}}(E/F\_1)=1, \mathrm{rank}\_{\mathbf{Z}}(E/F\_2)=1, \mathrm{rank}\_{\mathbf{Z}}(E/F)=2$$ This is precisely the sort of situation I'm looking to capture in general.
| https://mathoverflow.net/users/10547 | Additivity of Elliptic Curve Rank over Compositum of Fields | If you're willing to expand your question a bit, there are interesting relations to be found. More precisely, let $E/K$ be an elliptic curve defined over a number field, and let $L/K$ be a Galois extension such that $G(L/K)$ has a non-trivial idempotent relation (also sometimes called a Brauer relation). For example, this will be true if if $G(L/K)$ is a non-cyclic abelian group. Each idempotent relation yields a corresponding relation among the ranks of $E(F)$ as $F$ ranges over the fields lying between $K$ and $L$. This, and much more, may be found in the paper
*Kani, Ernst; Rosen, Michael*, [**Idempotent relations among arithmetic invariants attached to number fields and algebraic varieties**](http://dx.doi.org/10.1006/jnth.1994.1014), J. Number Theory 46, No. 2, 230-254 (1994). [ZBL0853.14011](https://zbmath.org/?q=an:0853.14011).
| 3 | https://mathoverflow.net/users/11926 | 429239 | 173,958 |
https://mathoverflow.net/questions/429242 | 0 | Let $\Omega$ be a smooth bounded domain, $H^1(\Omega) :=\{u: u, Du\in L^2(\Omega)\},$
and $H^1\_0(\Omega)$ is the closure of $C^{\infty}\_{c}(\Omega)$ in $H^1(\Omega)$. Define:
* $\sup\_{\partial\Omega } u:=\inf\{a :(u-a)^+\in H^1\_0(\Omega)\}$
* $ess \sup\_{\Omega} u:= \inf\{a :(u-a)^+=0, a.e.~ in~ \Omega\}$
The weak maximum principle tells me that if $-\Delta u \leq 0$ and $u\in H^1(\Omega)$ then \begin{equation}ess \sup\_{\Omega} u \leq \sup\_{\partial\Omega } u.\end{equation}
From the trace theorem, we know that for any $u\in H^1(\Omega),$ the trace of $u$ called $Tr(u)$ is a measurable function on $\partial \Omega$, hence we can define $ess \sup\_{\partial\Omega} Tr(u).$
* I'm confused about the difference of $ess \sup\_{\partial\Omega} Tr(u)$ and $\sup\_{\partial\Omega } u$, could the same be true in the weak maximum principle by replacing $\sup\_{\partial\Omega } u$ with $ess \sup\_{\partial\Omega} Tr(u)$.
| https://mathoverflow.net/users/166368 | Sobolev space and weak maximum principle | I may have overlooked something, but I think $\operatorname{esssup}\_{\partial \Omega} \operatorname{Tr} u = \sup\_{\partial \Omega} u$ for all $u \in H^1(\Omega)$.
This follows at once from the well-known characterization $H^1\_0(\Omega) = \{u \in H^1(\Omega) \mid \operatorname{Tr} u = 0\}$ and the fact that $(\operatorname{Tr}u - a)^+ = \operatorname{Tr} ((u-a)^+)$ for all $u \in H^1(\Omega)$. The last equation is easily verified for $u \in C^{\infty}(\Omega)$, then extended to $u \in H^1(\Omega)$ by continuity of $H^1(\Omega) \rightarrow L^2(\partial \Omega) : u \mapsto (\operatorname{Tr}u - a)^+$ and $H^1(\Omega) \rightarrow L^2(\partial \Omega) : u \mapsto \operatorname{Tr} (u-a)^+$.
| 1 | https://mathoverflow.net/users/490262 | 429245 | 173,959 |
https://mathoverflow.net/questions/429240 | 4 | Let $H$ be a Hilbert space, which we interpret as a space of quantum states.
* If $U(t):H\to H$ is a unitary **norm-continuous** one-parameter group with $U(0)=I$, (essentially) *Cauchy's functional equation* says it has the form $$U(t)=e^{itA}$$ for some **bounded** self-adjoint Hamiltonian $A$.
* If $U(t):H\to H$ is a unitary **strongly-continuous** one-parameter group with $U(0)=I$, *Stone's theorem* says it has the form $$U(t)=e^{itA}$$ for some **unbounded** self-adjoint Hamiltonian $A$.
Now let $\mathcal{S}\_1$ be the set of trace-class operators of trace $1$ on $H$, which we interpret as states in an open quantum system.
* If $\mathcal{T}(t):\mathcal{S}\_1\to\mathcal{S}\_1$ is a trace-preserving completely positive **norm-continuous** one-parameter group with $\mathcal{T}(0)=I$, *Lindblad's theorem* says it has the form $$\mathcal{T}(t)=e^{t\mathcal{L}}$$ where $$\mathcal{L}(\rho)=-i[A,\rho]+\sum\_{k=0}^{\infty}L\_k\rho L\_k^\*+L\_k^\*L\_k\rho$$ is a Lindbladian with **bounded** $A$, $L\_k$ where $A$ is self-adjoint and $\sum\_{k=0}^{\infty}L\_k^\*L\_k$ is strongly summable.
My question is:
**What can we say in this case when $\mathcal{T}(t)$ is only strongly continuous?**
| https://mathoverflow.net/users/142740 | "Open systems" version of Stone's Theorem for one-parameter groups of quantum operations | This has been generalized by Brian Davies to the general case, the article is
Davies, E.B.: Quantum dynamical semigroups and the neutron diffusion equation. Rep. Math.
Phys. 11(2), 169–188 (1977)
A more modern introduction are chapters 5-7 in this book
<https://www.cambridge.org/core/books/quantum-stochastics/0E5897239F5430E41858ACB936DC4386>
| 4 | https://mathoverflow.net/users/490271 | 429246 | 173,960 |
https://mathoverflow.net/questions/429241 | 2 | Let $f: X \rightarrow S$ be a finitely presented morphism of schemes and let $$E = \{s \in S \mid \text{ $X\_s$ is a point with residue field $\kappa(s)$ } \}$$
Is $E$ a constructible set?
The basic motivation for my asking this is in to check, for dominant $f$, whether there exists an affine monomorphism $S' \rightarrow X \rightarrow S$ such that $S'$ is $0$-dimensional and dominates $X$. If $E$ is constructible, then it suffices to look at the generic points of $X$ (or any other subset whose closure contains the generic points).
If we remove the condition on the residue field, then the resulting set certainly is constructible, since e.g. both "$0$-dimensional" and "geometrically irreducible" are constructible properties of fibers.
My reference for this sort of thing is usually appendix $E$ in Görtz and Wedhorn's Algebraic Geometry I. I don't know of any constructible properties that control residue fields in this way, but the property in question does seem to me like it should be a decent "descent" property.
Linking [this relevant MO question](https://mathoverflow.net/questions/57802/what-properties-define-open-loci-in-families?rq=1) for convenience.
| https://mathoverflow.net/users/97635 | Constructibility of the locus of points where the fiber is an isomorphism modulo nilpotents | Let $k$ be a field of characteristic $p>0$, $X=S=\operatorname{Spec}k[t]$ the affine line, and $f:X\to X$ the $p$-th power map. Then $E$ is dense but does not contain the generic point, so it is not constructible.
**[Added August 29]** On the other hand, if $S$ is a noetherian $\mathbb{Q}$-scheme, then $E$ is constructible. Indeed, it suffices to see that if $S$ is integral then either $E$ or $S\smallsetminus E$contains a nonempty open set. We may assume $X$ reduced (passing to $X\_\mathrm{red}$ does not change $E$, and $X\_\mathrm{red}\to S$ is still finitely presented). If $E$ contains the generic point $\eta$, then $f$ is birational and we are done. If not, then the fiber $X\_\eta$ is either empty, or positive-dimensional, or finite (reduced) over $\kappa(\eta)$ of degree $>1$. Each of these properties extends to a neighborhood of $\eta$, whence the claim. (The characteristic zero assumption is used via the fact that "reduced"="geometrically reduced".)
Whether one can extend this to arbitrary $\mathbb{Q}$-schemes is unclear to me.
| 6 | https://mathoverflow.net/users/7666 | 429249 | 173,961 |
https://mathoverflow.net/questions/428895 | 6 | Suppose that $H = H(X)$ is some quantity growing with $X$. Are there any bounds on $$F(X, H) = \min\_{X < n\le X + H} \omega(n)?$$
It isn't hard to obtain a lower bound $\max\_{x\sim X} F(X, H)\gg \frac{\log X}{H\log\log X}$. Also, for $H\gg X^\delta$, I believe it follows from a standard lower bound sieve the bound $F(X, H)\ll \delta^{-1}$. Is anything better known in either direction?
| https://mathoverflow.net/users/40983 | Upper bound on minimum number of prime factors in short intervals | There is some useful information in the paper [P. Erdôs and I. Kátai. On the growth of some additive functions of small intervals. Acta Math. Hungar. 33 (1979), 345-359.](https://P.%20Erd%C3%B4s%20and%20I.%20K%C3%A1tai.%20On%20the%20growth%20of%20some%20additive%20functions%20of%20small%20intervals.%20Acta%20Math.%20Hungar.%2033%20(1979),%20345-359.)
Let $$O\_{k}(n)=\max \_{j=1, \ldots, k} \omega(n+j), \quad o\_{k}(n)=\min \_{j=1, \ldots, k} \omega(n+j).$$
Then (see *I. Kátai. Local growth of the number of the divisors of consecutive integers. Publ. Math. Debrecen 18 (1971), 171-175.*) for every $\varepsilon>0$, apart from a set of $n$'s having zero density, the inequalities
$$
O\_{k}(n) \leqq(1+\varepsilon) \varrho\left(\frac{\log k}{\log \log n}\right) \log \log n, \quad o\_{k}(n) \geqq(1-\varepsilon) \bar{\varrho}\left(\frac{\log k}{\log \log n}\right) \log \log n
$$
hold for every $k=1,2, \ldots$ Here $\varrho(u)$ $(u \ge 0)$ is defined as the inverse function of $\psi(z)=z \log \frac{z}{e}+1$ defined in $z \ge 1$, and $\bar{\varrho}(n)$ $(n \ge 0)$ is the inverse function of the same $\psi(z)$ defined in $0<z \le 1$. In the same paper it was conjectured that
$$
O\_{k}(n) \geqq(1-\varepsilon) \varrho\left(\frac{\log k}{\log \log n}\right) \log \log n
$$
and
$$
o\_{k}(n) \leqq(1+\varepsilon) \bar{\varrho}\left(\frac{\log k}{\log \log n}\right) \log \log n
$$
They prove that for every $\varepsilon>0$ these irequalities hold for every $k \ge 1$, apart from a set of $n$'s having zero density.
| 1 | https://mathoverflow.net/users/5712 | 429255 | 173,962 |
https://mathoverflow.net/questions/428557 | 4 | $\newcommand\R{\mathbb R}$Let $0<d\_1<\cdots<d\_k<\infty$ and let $m\_1,\dots,m\_k$ be any integers $\ge1$. Let $n:=m\_1+\dots+m\_k-1$.
Let $d$ denote the Euclidean distance in $\R^n$.
>
> Do then there always exist pairwise disjoint subsets $S\_1,\dots,S\_k$ of $\R^n$ of respective cardinalities $m\_1,\dots,m\_k$ such that
> $$d(x,y)=d\_j$$
> for all $j\in\{1,\dots,k\}$ and all $(x,y)\in T\_j\times S\_j$ with $x\ne y$, where $T\_j:=S\_1\cup\dots\cup S\_j$?
>
>
>
This conjecture is obviously true for $k=1$ and seems not hard to prove for $k=2$.
---
Reviewing the edits suggested by MattF., one can restate the conjecture in the following way:
>
> Let $n$ be any natural number. Suppose that $0<d\_1\le\cdots\le d\_{n+1}<\infty$.
>
>
>
>
> Do then there always exist pairwise distinct points $x\_1,\dots,x\_{n+1}$ in $\R^n$ such that
> $$d(x\_i,x\_j)=d\_j$$
> for all integers $i,j$ such that $1\le i<j\le n+1$?
>
>
>
| https://mathoverflow.net/users/36721 | About Euclidean distances | Finite [ultrametric spaces](https://en.wikipedia.org/wiki/Ultrametric_space) [allow isometric embeddings into $L\_2$](https://core.ac.uk/download/pdf/82490074.pdf).
| 3 | https://mathoverflow.net/users/32454 | 429256 | 173,963 |
https://mathoverflow.net/questions/429257 | 3 | Can anyone suggest to me some references for studying triangle groups? Especially the existence of finite index subgroups, subgroups isomorphic to fundamental groups of compact surfaces etc.
| https://mathoverflow.net/users/490039 | Reference for triangle groups | It seems that you are interested in “Fenchel’s conjecture” stating that (essentially) all two-orbifolds are finitely (orbifold) covered by surfaces. The case of triangle orbifolds is the hardest, and was solved by Ralph Fox (with, it seems, a few errors). See the 1983 paper by Chau titled [*A note concerning Fox’s paper on Fenchel’s conjecture*](https://doi.org/10.2307/2045442).
| 4 | https://mathoverflow.net/users/1650 | 429267 | 173,965 |
https://mathoverflow.net/questions/429273 | 6 | Let $X$ be a set and $f: X \to X$ a function. A point $x \in X$ is, of course, said to be **periodic** for $f$ if $x \in \{f(x), f^2(x), \ldots\}$.
Now suppose that $X$ is a topological space and $f$ is continuous. In private, I've been saying that a point $x \in X$ is **topologically periodic** for $f$ if $x \in \overline{\{f(x), f^2(x), \ldots\}}$.
But "topologically periodic" is just a name I made up.
>
> What is the standard name for "topologically periodic" point?
>
>
>
Note that I'm looking for an adjective that applies to *points*, not *functions*. I'm aware of terms such as quasiperiodic function and almost periodic function, but they don't answer my question.
In case it helps, I'm working in less generality than arbitrary topological spaces and continuous maps. In my situation, $X$ is a compact metric space and $f$ is distance-decreasing ($d(f(x), f(y)) \leq d(x, y)$). Under those hypotheses, my definition of "topologically periodic" may have other equivalent formulations that are more familiar.
| https://mathoverflow.net/users/586 | What is the name for a point that is periodic to within $\varepsilon$? | Such a point is called **recurrent**.
---
It was stated by Alp Uzman in a comment that the usual definition of recurrent is that $x$ is in its own $\omega$-limit set. This is the definition I had in mind, and under a suitable separation axiom it's obvious that this is equivalent to what was asked, which is why I just wrote the statement. Actually you don't need any separation axiom. I'll just write a full proof.
The **$\omega$-limit set** $\omega(x, f)$ of $x$ is the set of accumulation points of the sequence $I\_x = (x, f(x), f^2(x), \ldots)$. (In the reference given by Uzman, read literally the first definition of the $\omega$-limit set given is that it's the accumulation points of the set $\{f^i(x) \;|\; i \in \mathbb{N}\}$. This would not be equivalent, and is "wrong", but it's clear from the discussion that the author means accumulation points of the sequence.)
We need an auxiliary notion. For two points $x, y$, write $y \leq x$ if every open set containing $x$ also contains $y$. We need two lemmas, which are left as exercise.
>
> Lemma A. Continuous functions preserve $\leq$ in the sense that $y \leq x \implies f(y) \leq f(x)$.
>
>
>
>
> Lemma B. The $\leq$-relation is transitive.
>
>
>
>
> Theorem. Let $X$ be a topological space and $f : X \to X$ a continuous function. Then $x$ is in $\omega(x, f)$ if and only if $x \in \overline{\{f^i(x) \;|\; i \geq 1\}}$.
>
>
>
Proof. Clearly $\omega(x, f) \subset \overline{\{f^i(x) \;|\; i \geq 1\}}$, so it suffices to show $x \in \overline{\{f^i(x) \;|\; i \geq 1\}} \implies x \in \omega(x, f)$. Suppose thus $x \in \overline{\{f^i(x) \;|\; i \geq 1\}}$.
First, suppose that $f^n(x) \leq x$ for some $n \geq 1$. Then by Lemma A, $f^{(k+1)n}(x) \leq f^{kn}(x)$ for all $k$. By Lemma B, $f^{kn}(x) \leq x$ for all $k$. Then clearly $x$ is even an accumulation point of the subsequence $(x, f^k(x), f^{2k}(x), \ldots)$ -- all points of this set are in every neighborhood of $x$.
Next, suppose $f^n(x) \not\leq x$ for all $n \geq 1$. For all $n \geq 1$ let $V\_n$ be a neighborhood of $x$ not containing $f^n(x)$, and $W\_n = \bigcap\_{i = 1}^n V\_i$. Let $U \ni x$ be any neighborhood. By the assumption on $x$, for any $n$ we have $f^m(x) \in U \cap W\_n$ for some $m \geq 1$. Necessarily $m > n$, so in particular we find infinitely many $m$ such that $f^m(x) \in U$. Square.
| 13 | https://mathoverflow.net/users/123634 | 429276 | 173,971 |
https://mathoverflow.net/questions/429270 | 4 | Let $f: X\rightarrow S$ be a flat quasi-projective morphism, where $X$ is a smooth variety, and $S$ is a discrete valuation ring. Then we know that $f$ is proper morphism if and only if it satisfies the well-known valuative criterion of properness. The evaluative criterion of properness says the following.
Given a DVR $T$ (with the closed point $0$) and maps $T\rightarrow S$ and $T\setminus 0\rightarrow X$ such that the obvious square commutes, there exists a unique lift $T\rightarrow X$ such that all the triangles commute.
My questions are the following.
1. If we know that the map $f$ is flat, then is it enough to check the criterion for DVR T (over $S$) such that the composite map $T\setminus 0\rightarrow X\rightarrow S$ is flat?
2. Suppose $f: X\rightarrow S$ is any flat morphism (may not be quasi-projective) then is it enough to check the criterion for DVR T (over $S$) such that the composite map $T\setminus 0\rightarrow X\rightarrow S$ is flat?
| https://mathoverflow.net/users/490296 | Relative valuative criteria of properness for flat morphisms | Let $S = \operatorname{Spec} R$ where $R$ is a DVR with generic point $\eta$ and closed point $0$. A homomorphism $R \to T$ is flat if and only if it is injective and more generally, $f : X \to S$ is flat if and only if $X$ is equal to the scheme theoretic closure of the generic fiber $X\_{\eta}$.
In particular, $T$ is a DVR with fraction field $K$, then $R \to T$ is flat if and only if $R \to K$ is injective, so that $K$ is an extension of the fraction field of $R$. Thus, a test diagram
$\require{AMScd}$
\begin{CD}
\operatorname{Spec}K @>>> X\\
@VVV @VV f V\\
\operatorname{Spec} T @>>> S
\end{CD}
in the valuative criterion satisfies the flatness condition if and only if $\operatorname{Spec} K$ maps to the generic fiber of $f$. It follows that the generic fiber of $f$ is proper over $\eta$ so the only thing that can go wrong is that is the central fiber of $f$ is "missing points".
Since $f$ is quasi-projective we can take the closure of $X$ inside $\mathbb{P}^n\_S$ for some $n$ and obtain a flat and proper map $g : \bar{X} \to S$. It suffices to show that $X = \bar{X}$. Let $x \in \bar{X}\_0$ be a point of the special fiber. By flatness of $g$, there exists a point $y \in \bar{X}\_\eta = X\_\eta$ specializing to $x$. Since everything is Noetherian, we can witness this specialization via a DVR $T$ and a map $\operatorname{Spec} T \to \bar{X}$ with generic point $\operatorname{Spec} K$ mapping to $y$ and closed point mapping to $x$. Then $T$ is flat over $R$ so by assumption we have an extension $\operatorname{Spec} T \to X \subset \bar{X}$ with $\operatorname{Spec}K \mapsto y$. Since $\bar{X}$ is separated, these two maps $\operatorname{Spec}T \to \bar{X}$ must agree so $x \in X$ and $X = \bar{X}$.
**Edit:** If we drop quasi-projectivity but assume that $f$ is separated, then by Nagata's Compactification Theorem, there exists an open immersion $X \subset \bar{X}$ and a proper morphism $g : \bar{X} \to S$ extending $f$. Then we can run the argument as before. Since any map between separated schemes ($X$ is a variety and $S$ is affine) is separated, then we are done.
If we ask the same question for general schemes where $f$ is not separated or $X$ is not quasi-compact, then I imagine things can go wrong but I'm not sure.
| 5 | https://mathoverflow.net/users/12402 | 429278 | 173,972 |
https://mathoverflow.net/questions/429285 | 3 | Let $y\in \mathbb{R}$ and $\mathbf{x},\mathbf{z}\in\mathbb{R}^p$ be *random variable* and *random vectors*. Assume $y=f(\mathbf{x}+\mathbf{z},\mathbf{z})$ for some function $f$.
Is the following statement correct?
If $\mathbf{z}\perp \!\!\! \perp y$ and $\mathbf{z}\perp \!\!\! \perp\mathbf{x}$, where $\perp \!\!\! \perp$ means *statistical independence*, then there exists a function $g$ such that $y=f(\mathbf{x}+\mathbf{z},\mathbf{z})=g(\mathbf{x})$.
| https://mathoverflow.net/users/99157 | Independent input feature z can be removed: if y=f(x+z,z), then y=g(x)? | A simple counter-example: take $p=1$, $x=+1$ or $x=-1$ with probability 1/2, $z=+1$ or $z=-1$ with probability 1/2, independently of $x$.
Define $y=xz$ [or, if you wish, $f(x+z,z)=(x+z)z-z^2$]; one has $y=+1$ or $-1$ with probability 1/2, independently of either $x$ or $z$; hence, there is no function $g(x)$ such that $y=g(x)$, although $z$ is independent of $x$ and $y$.
| 4 | https://mathoverflow.net/users/11260 | 429290 | 173,975 |
https://mathoverflow.net/questions/429287 | -9 | **Notation:**
$$ \{x\}\ :=\ x-\lfloor x\rfloor $$
---
APF-functions $\ \tau(n)\ $ for $\ 2<n\in\mathbb N,\ $ and $\ \xi(n)\ $ for $\ 3<n\in\mathbb N,\ $ are defined as follows:
$$ \tau(n)\ :=\ \sum\_{k=2}^{n-1}\,\left\{\frac nk\right\}\qquad\qquad\text{and}\qquad
\qquad\xi(n)\ :=\ \sum\_{k=2}^{\lfloor\sqrt n\rfloor}\ \left\{\frac nk\right\} $$
These functions are sensitive to their arguments being or not a prime.
Locally, primes seem (how true is it?) dominate over their neighborhoods. In this spirit,
**Question:** Do you already know or can you prove non-obvious results about
the APF-functions $\ \tau\ $ and $\ \xi\,?$
| https://mathoverflow.net/users/110389 | Summatory functions for fractional parts | Your $\tau(n)$ and $\xi(n)$ are essentially the same as the [divisor summatory function](https://en.wikipedia.org/wiki/Divisor_summatory_function), often denoted by $\sigma(n)$. Indeed, we have
$$\sigma(n)=\sum\_{m=1}^n\sum\_{k\mid m}1=\sum\_{k=1}^n\left[\frac{n}{k}\right]=2\sum\_{k=1}^{\left[\sqrt{n}\right]}\left[\frac{n}{k}\right]-\left[\sqrt{n}\right]^2.$$
That is,
$$\sigma(n)=\tau(n)+n+1=2\xi(n)+2n-\left[\sqrt{n}\right]^2.$$
Note that we have rather precise estimates for $\sigma(n)$, this is what the Dirichlet divisor problem is about. See the above Wikipedia page for more details. For example, Huxley (2003) proved for any $\varepsilon>0$ that
$$\sigma(n)=n\log n+(2\gamma-1)n+O\_\varepsilon(n^{131/416+\varepsilon}).$$
In particular, these functions $\sigma$, $\tau$, $\xi$ are **not sensitive** to their arguments being prime or not prime.
**Added.** In my response, $[x]$ denotes the integral part of $x$, not the fractional part as in the OP's post. Sorry about that. At any rate, it is straightforward to relate the sum of the integral parts of $n/k$ and the sum of the fractional parts of $n/k$, because the sum of $n/k$ has a well-known asymptotic expansion.
| 7 | https://mathoverflow.net/users/11919 | 429292 | 173,976 |
https://mathoverflow.net/questions/315913 | 2 | Let $M\_1$ and $M\_2$ be two real positive-semidefinite matrices. Is there any algorithm to compute a permutation matrix $P$ to minimize $\| M\_1-PM\_2P^T \|\_F^2$ or equivalently to maximize $trace(M\_1PM\_2P^T)$?
To be simple, for $i=1,2$, further assume $M\_i=Q\_iQ\_i^T$, where $Q\_i$ has orthonormal columns.
| https://mathoverflow.net/users/99157 | matching two positive-semidefinite matrices | It is a graph-matching problem, which is NP-hard. But some solutions available if relaxing the perturbation matrix assumption. See example at <https://doi.org/10.1016/j.patcog.2016.07.015>
| 4 | https://mathoverflow.net/users/84817 | 429300 | 173,981 |
https://mathoverflow.net/questions/429308 | 1 | I asked the same on [math.Stackexchange](https://math.stackexchange.com/q/4520055/721298).
I have $n$ (say $n = 300$) vectors $v\_1,\dots,v\_n$. Each of them has $K$ coordinates (say $K = 30$). For vector $v\_j$ I denote it's coordinates as $v\_{j1},\dots,v\_{jK}$. I want select $M$ (say $M = 5$) vectors ${v\_{\hat s\_1}}, \dots, v\_{\hat s\_M} \in \{v\_1,\dots,v\_n\}$ such that
$$\sum\_{k=1}^{K}\max\_{1 \le m \le M} v\_{\hat s\_mk} = \max\_{1 \le s\_1 \le n,\dots, \le s\_m \le n}\sum\_{k=1}^{K}\max\_{1 \le m \le M} v\_{s\_mk}.$$
What's the name for this type of questions? How do I (approximately) solve those fast? (I need to solve like 30 per minute)
| https://mathoverflow.net/users/32454 | Given a set of vectors how to pick $M$ such that sum of maximums of coordinates is maximized? | At very least the problem can be approached via (M)ILP by introducing indicator variables $q\_{jk}$ telling whether the corresponding coordinate $v\_{jk}$ is present in the sum of max, and $r\_j$ telling whether the vector $v\_j$ is selected. This results in the following formulation:
$$\begin{cases}
\sum\_{j=1}^n \sum\_{k=1}^K v\_{jk} q\_{jk} \longrightarrow \max\\
\sum\_{j=1}^n q\_{jk} = 1\qquad (k\in[K]), \\
\sum\_{k=1}^K q\_{jk} \leq Kr\_j\qquad (j\in[n]), \\
\sum\_{j=1}^n r\_j = M, \\
q\_{jk}, r\_j\in\{0,1\}\qquad (j\in[n],\ k\in[K]).
\end{cases}$$
| 1 | https://mathoverflow.net/users/7076 | 429315 | 173,987 |
https://mathoverflow.net/questions/429322 | 3 | Let $\mathbf{y},\mathbf{x},\mathbf{z}$ are real-valued *random vectors* with possibly different dimensions. Assume $\mathbf{y}=f(\mathbf{x},\mathbf{z})$ for some function $f$.
If $\mathbf{z} \perp\!\!\!\perp \{\mathbf{y},\mathbf{x}\}$ (i.e., $\mathbf{z}$ is *statistically* independent of $\mathbf{y}$ and $\mathbf{x}$ **jointly**), is it true that there exists a function $g$ such that $\mathbf{y}=f(\mathbf{x},\mathbf{z})=g(\mathbf{x})$?
If it is not true, how about the simpler case: $y=f(\mathbf{x}+\mathbf{z},\mathbf{z})=g(\mathbf{x})$ for $y\in \mathbb{R}$ and $\mathbf{x},\mathbf{z}\in \mathbb{R}^p$?
| https://mathoverflow.net/users/99157 | $\mathbf{y}=f(\mathbf{x},\mathbf{z})=g(\mathbf{x})$ if $\mathbf{z}\perp \!\!\! \perp \{\mathbf{y},\mathbf{x}\}$ jointly? | $\newcommand\R{\mathbb R}$The answer is yes. According to standard convention, write $X,Y,Z$ in place of $\mathbf x,\mathbf y,\mathbf z$, respectively, reserving the corresponding lower-case letters $x,y,z$ for the corresponding values of $X,Y,Z$.
So, $Z$ is independent of $(X,Y)$. We also have $Y=f(X,Z)$ for some Borel function $f$. So, for any Borel sets $A,B,C$ in the respective spaces,
$$P(X\in A,f(X,Z)\in B,Z\in C) \\
=P(X\in A,f(X,Z)\in B)\,P(Z\in C). \tag{1}\label{1}$$
Since the independence of $Z$ from $(X,Y)$ implies the independence of $Z$ from $X$, we rewrite \eqref{1} as
$$\int\_A P(X\in dx)P(f(x,Z)\in B,Z\in C) \\
=\int\_A P(X\in dx)P(f(x,Z)\in B)\,P(Z\in C); \tag{2}\label{2}$$
this rewriting holds by the Tonelli theorem, which implies that
$$\begin{aligned}Eh(X,Z)&=\iint P(X\in dx,Z\in dz) h(x,z) \\
&=\iint P(X\in dx)P(Z\in dz) h(x,z) \\
&=\int P(X\in dx)\int P(Z\in dz) h(x,z)\\
&=\int P(X\in dx)E h(x,Z)
\end{aligned}$$
for any nonnegative Borel function $h$ (provided that $X$ and $Z$ are independent). In our case, we apply this to $h(x,z)=1(x\in A,f(x,z)\in B,z\in C)$ on the left and to $h(x,z)=1(x\in A,f(x,z)\in B)$ on the right.
Since the Borel set $A$ in \eqref{2} is arbitrary, we see that
$$P(f\_x(Z)\in B,Z\in C)
=P(f\_x(Z)\in B)\,P(Z\in C) \tag{3}\label{3}$$
for almost all (a.a.) $x$ (with respect to the probability measure that is the distribution of $X$), where $f\_x(z):=f(x,z)$. Letting now $C=f\_x^{-1}(B)$, we get
$$P(f\_x(Z)\in B)
=P(f\_x(Z)\in B)^2\tag{4}\label{4}$$
and hence $P(f\_x(Z)\in B)\in\{0,1\}$ for any (say) open ball $B$ and a.a. $x$. So, $P(f\_x(Z)\in B)\in\{0,1\}$ for a.a. $x$ and all open balls $B$ of rational radii and rational coordinates of the centers -- at once (since there are only countable many such balls). So, for a.a. $x$, the support of the distribution of the random variable $f\_x(Z)$ is a singleton set (by Lemma 1 below), so that for some real $g\_x$
$$P(f(x,Z)\ne g\_x)=P(f\_x(Z)\ne g\_x)=0.$$
So,
$$P(Y\ne g\_X)=P(f(X,Z)\ne g\_X)=\int P(X\in dx)P(f\_x(Z)\ne g\_x)=0.$$
Thus indeed, $Y=g\_X$ almost surely.
---
**Lemma 1:** Let $V$ be a random vector in $\R^d$ such that $P(V\in B)\in\{0,1\}$ for all open balls $B$ in $\R^d$ of rational radius and rational coordinates of the center. Then the support $S\_V$ of the distribution of $V$ is a singleton set.
*Proof:* Suppose that contrary: that there are two distinct points $u,v$ in $S\_V$, so that the distance between $u$ and $v$ is $d>0$. Let $B\_u$ and $B\_v$ be the open balls centered at $u$ and $v$ of radii $d/4$, so that the shortest distance between $B\_u$ and $B\_v$ is $d/2>0$. Since $u,v$ are in $S\_V$, we have $P(V\in B\_u)>0$ and $P(V\in B\_v)>0$. By the continuity of probability (or the Fatou lemma), there exist **disjoint** open balls $C\_u$ and $C\_v$ of rational radii and rational coordinates of the centers that are close enough to $B\_u$ and $B\_v$ so that $P(V\in C\_u)>0$ and $P(V\in C\_v)>0$.
Then
$$0<P(V\in C\_u)\le1-P(V\in C\_v)<1,$$
which contradicts the condition that $P(V\in C\_u)\in\{0,1\}$. $\quad\Box$.
| 4 | https://mathoverflow.net/users/36721 | 429327 | 173,990 |
https://mathoverflow.net/questions/429330 | 5 | A very important theorem in mathematical physics is Poincaré’s recurrence theorem.
As you recall, this theorem states that given a dynamical system $(M , \phi , \mu)$ with $ \mu M < +\infty$, for every $A$ measurable the set of points of $A$ for which it exists some $N>0$ such that $\phi^Na \notin A$ for $n \geq N$ has measure zero.
This theorem applies, for example, to Hamiltonian dynamics, hence it has non trivial physical implications. For example a gas expanding in an empty room and then going back to the initial position.
What I would like to ask is: what are the physical implications of Poincaré theorem especially in statistical physics?
Is it possibile to use Poincaré’s recurrence in order to argue for a sort of circularity of time? In particular, can the evolution of the universe from its initial state be seen as dynamical system satisfying the hypotheses of Poincaré’s recurrence theorem?
I am writing it here on mathoverflow, because i) I see no difference between mathematics and physics ii) I am under the (possibly wrong) impression that not many physicists care much about Poincaré recurrence, while mathematicians do.
**edit** I have found this physics stackexchange quesiton of 8 years ago which asks one part of my question <https://physics.stackexchange.com/questions/94122/is-poincare-recurrence-relevant-to-our-universe>
| https://mathoverflow.net/users/139854 | Poincaré recurrence and its implications for statistical physics and the arrow of time | Since the question is about physical implications of Poincaré recurrence one should take both quantum effects and gravitational effects into consideration. Quantum mechanics does not spoil the recurrences, any finite quantum mechanical system evolves quasi-periodically ([Wikipedia](https://en.wikipedia.org/wiki/Poincar%C3%A9_recurrence_theorem#Quantum_mechanical_version) has a simple proof).
Gravity provides more complications, which are not fully resolved because we lack a quantum theory of gravity. Black holes, once formed, grow if they are colder than the surrounding space, and thereby preempt the recurrence. See [Gravity can significantly modify classical and quantum Poincare recurrence theorems](https://arxiv.org/abs/1611.00792?context=quant-ph).
---
As an aside, the impression of the OP "that not many physicists care much about Poincaré recurrence" is not quite true. It is difficult to observe this effect in the laboratory, but it is an active topic of research. A recent publication, [Recurrences in an isolated quantum many-body system,](https://arxiv.org/abs/1705.08231) has received much attention (see this [news item](https://www.sciencedaily.com/releases/2018/02/180222145053.htm)).
| 11 | https://mathoverflow.net/users/11260 | 429334 | 173,992 |
https://mathoverflow.net/questions/429319 | 5 | Let $A$ be a separable $C^\*$-algebra and let $\omega$ be a state on $A$.
Then there is an "orthogonal" probability measure $\mu$ on the pure state space $P(A)$ of $A$ such that $\omega(x) = \int\_{P(A)} \psi(x) \, d\mu(\psi)$ [Takesaki 1, IV.6.28]).
If I understand correctly the orthogonality of $\mu$ means that the GNS representation of $\omega$ is (unitarily equivalent to) a direct integral:
$$
(H\_\omega,\pi\_\omega) = \int\_{P(A)}^\oplus (H\_\psi,\pi\_\psi) \, d\mu(\psi)
$$
and $\Omega\_\omega = \int\_{P(A)}^\oplus \Omega\_\psi\,d\mu(\psi)$ (see for example [Takesaki 1, IV.8.31].
But doesn't this imply that the von Neumann algebra $\pi\_\omega(A)''$ takes the form $$\pi\_\omega(A)'' = \int\_{P(A)}^\oplus\pi\_\psi(A)'' \,d\mu(\psi) = \int\_{P(A)}^\oplus B(H\_\psi) \,d\mu(\psi)$$
(because $\psi$ is pure one has $\pi\_\psi(A)''=B(H\_\psi)$).
This would imply that $\pi\_\omega(A)''$ is a type I von Neumann algebra since it can be written as a direct integral of type I factors, right?
The argument must be wrong since not every state on a separable $C^\*$-algebra is a type I state (see the answer to this question [Factor states on C\*-algebras](https://mathoverflow.net/q/391566/485160)).
Any help would be much appreciated!
| https://mathoverflow.net/users/485160 | Separable C* algebras and type I states | I don't think $\pi\_\omega(A)''$ has that form. For example, take $A = M\_2$ and let $\omega$ be the normalized trace. Then $\omega = \frac{1}{2}(\psi\_1 + \psi\_2)$ where $\psi\_i(x) = \langle xe\_i, e\_i\rangle$, for $x \in M\_2$ and $\{e\_1,e\_2\}$ the standard basis of $\mathbb{C}^2$. That is, $\omega$ is the integral $\int \psi\, d\mu(\psi)$ where $\mu = \frac{1}{2}(\delta\_{\psi\_1} + \delta\_{\psi\_2})$. Then $\pi\_\omega(M\_2)'' \cong \pi\_{\psi\_1}(M\_2)'' \cong \pi\_{\psi\_2}(M\_2)'' \cong M\_2$, so $\pi\_\omega(M\_2)'' \not\cong \pi\_{\psi\_1}(M\_2)'' \oplus \pi\_{\psi\_2}(M\_2)''$.
We do have an isomorphism between $H\_\omega$ and $H\_{\psi\_1} \oplus H\_{\psi\_2}$ which takes $\pi\_\omega(x)$ to $\pi\_{\psi\_1}(x) \oplus \pi\_{\psi\_2}(x)$, but of course that doesn't make $\pi\_\omega(A) \cong \pi\_{\psi\_1}(A) \oplus \pi\_{\psi\_2}(A)$.
| 7 | https://mathoverflow.net/users/23141 | 429337 | 173,993 |
https://mathoverflow.net/questions/429341 | 7 | I'm reading Michael Shulman's articles on cohomology in HoTT [here](https://homotopytypetheory.org/2013/07/24/cohomology/) and [here](https://ncatlab.org/nlab/show/spectral+sequences+in+homotopy+type+theory), as well as Floris van Doorn's thesis [here](https://florisvandoorn.com/papers/dissertation.pdf).
Given $E: Z \to \mathsf{Spectrum}$ a family of spectra over a homotopy type $Z$, they define the parametrized (or twisted) cohomology of $Z$ with coefficients in $E$ by
$$H^n(Z; E) :\equiv \pi\_0\left( \prod\_{x \in X} E\_n(x) \right)$$
where $\prod\_{x \in X} E\_n(x)$ denotes the appropriate type of sections. In the special case that we have a family of abelian groups $A: X \to \mathsf{AbGroup}$ then the composite with the Eilenberg-Mac Lane construction $H: \mathsf{AbGroup} \to \mathsf{Spectra}$ gives us a parametrized family of Eilenberg-Mac Lane spectra $HA: X \to \mathsf{Spectra}$. The corresponding twisted cohomology is cohomology with local coefficients.
Though they don't discuss it in their articles, I am wondering if we can get long exact sequences in parametrized cohomology starting from a cofibre sequence of types? I can see how you do it for unparametrized spectra, but the pi-types in parametrized cohomology make it a bit more confusing.
Specifically, suppose that I have a homotopy cofibre sequence $X \xrightarrow{f} Y \xrightarrow{q} C\_f$ equipped with a parametrized spectrum $E: C\_f \to \mathsf{Spectrum}$.
Clearly I can pull back along $q$ and $q \circ f$ to get parametrized spectra $q^\*E:=E \circ q: Y \to \mathsf{Spectrum}$ and $f^\*q^\*E:=E \circ q \circ f: X \to \mathsf{Spectrum}$. Does the above cofibre sequence give me a long exact sequence in parametrized cohomology?
| https://mathoverflow.net/users/56938 | Long exact sequences for parametrized cohomology | $\require{AMScd}$Yes. It is better to consider the more general situation of a pushout square of types, since pushout squares (unlike cofiber sequences) are stable under base change. Then the difference between parametrized and unparametrized disappears, since one can work in the $\infty$-topos over the pushout.
More generally, consider a pushout square
\begin{CD}
A @>>> B\\
@VVV @VVV\\
C @>>> D
\end{CD}
in an $\infty$-topos $T$ and a family of spectra $E$ over $D$. Then the induced square of global sections (where the pullbacks of $E$ are implicit)
\begin{CD}
\Gamma(D,E) @>>> \Gamma(C,E)\\
@VVV @VVV\\
\Gamma(B,E) @>>> \Gamma(A,E)
\end{CD}
is cartesian, hence gives rise to a long exact sequence of parametrized cohomology groups
$$
\dots\to \pi\_{n+1}\Gamma(A,E) \to \pi\_n\Gamma(D,E) \to \pi\_n\Gamma(B,E)\times \pi\_n\Gamma(C,E)\to\dots.
$$
The statement for a family of spectra $E$ is equivalent to the statement for each family of types $\Omega^{\infty-i}E$ (which gives the above long exact sequence up to the product of $\pi\_{-i}$'s), so it suffices to consider a family of types. By the descent property of $\infty$-topoi, we have cartesian square of $\infty$-categories
\begin{CD}
T\_{/D} @>>> T\_{/C}\\
@VVV @VVV\\
T\_{/B} @>>> T\_{/A}
\end{CD}
The cartesian square of global sections is then a special case of the assertion that mapping spaces in a limit of $\infty$-categories are the limits of the mapping spaces.
Alternatively, we can pass to spectrum objects to obtain a cartesian square of stable $\infty$-categories, and there is an analogous assertion about mapping spectra in a limit of stable $\infty$-categories.
| 7 | https://mathoverflow.net/users/20233 | 429350 | 173,997 |
https://mathoverflow.net/questions/429351 | 3 | Let $\widetilde{M}\_k^{\leq \ell}$ be the space of weight $k$ depth $\leq \ell$ quasimodular forms, and $\widetilde{M}\_{k,\mathbb R}^{\leq \ell}$ be a subspace of $\widetilde{M}\_k^{\leq \ell}$ whose elements have real Fourier coefficients.
Let $D=\frac{1}{2\pi i}\frac{d}{d\tau}=q\frac{q}{dq} : \widetilde{M}\_k^{\leq \ell} \to \widetilde{M}\_{k+2}^{\leq \ell+1}$ be the differential operator. If $k>2$, then we have:
**Theorem**. $\widetilde{M}\_k^{\leq 1}=M\_k \oplus DM\_{k-2}$.
It is obvious that $M\_{k,\mathbb R} \oplus DM\_{k-2,\mathbb R} \subseteq \widetilde{M}\_{k,\mathbb R}^{\leq 1}$, and I believe that $M\_{k,\mathbb R} \oplus DM\_{k-2,\mathbb R} = \widetilde{M}\_{k,\mathbb R}^{\leq 1}$. More generally, we have
**Theorem**. $\widetilde{M}\_k^{\leq \frac{k}{2}}=\bigoplus\_{j=0}^{\frac{k}{2}-2} D^j M\_{k-2j} \oplus \mathbb C D^{\frac{k}{2}-1}E\_2$.
and I hope to prove $\widetilde{M}\_{k,\mathbb R}^{\leq \frac{k}{2}}=\bigoplus\_{j=0}^{\frac{k}{2}-2} D^j M\_{k-2j,\mathbb R} \oplus \mathbb R D^{\frac{k}{2}-1}E\_2$ but I'm having trouble. Does anyone have an idea?
| https://mathoverflow.net/users/123157 | Decomposition of real quasimodular forms of depth 1 | There is an involution of $\widetilde M\_k^{\le l}$, $$i : \; f(\tau) \mapsto \overline{f(-\overline{\tau})}, \quad \sum\_{n=0}^{\infty} a\_n q^n \mapsto \sum\_{n=0}^{\infty} \overline{a\_n} q^n$$ which commutes with $D$, whose fixed points are exactly the quasimodular forms with real Fourier coefficients. So if $f \in \widetilde M\_{k, \mathbb{R}}^{\ell}$ is decomposed $$f = \sum\_{j=0}^{\ell - 2} D^j f\_j + c \cdot D^{\ell - 1} E\_2, \quad f\_j \in M\_{k - 2j}, \; c \in \mathbb{C}$$ then $$f = i(f) = \sum\_{j=0}^{\ell - 2} D^j i(f\_j) + \overline{c} \cdot D^{\ell - 1} E\_2.$$
Since the sum $\widetilde M\_k^{\le \ell} = \bigoplus\_{j = 0}^{\ell - 2} D^j M\_{k - 2j} \oplus \mathbb{C} D^{\ell - 1} E\_2$ is direct, we have $$f\_j = i(f\_j) \; \text{for all} \; j \; \text{and} \; c = \overline{c},$$ so $f\_j \in M\_{k - 2j, \mathbb{R}}$ and $c \in \mathbb{R}$.
| 3 | https://mathoverflow.net/users/490352 | 429355 | 173,998 |
https://mathoverflow.net/questions/429346 | 3 | Let $f:[0,1]\to [0,1]$ be continuous. Let—
$$B\_n(f)(x)=\sum\_{k=0}^n f(k/n) {n \choose k} x^k (1-x)^{n-k},$$
be the *Bernstein polynomial* of $f$ of degree $n$.
This question relates to the difference between two Bernstein polynomials of the same function $f(x)$, namely the question of finding a simple and explicit function $\phi(f, n)$ such that— $$|B\_{2n}(f)(x) - B\_{n}(f)(x)| \le \phi(f, n) \text{ whenever } 0 \le x \le 1. \tag{1}$$
I have not been able to find results answering this question. However:
* Butzer (1953) showed a very similar result on a family of operators, one of which is $2 B\_{2n}(f) - B\_n(f)$, but without explicit error bounds.
* ~~For the polynomials $2x(1-x)$ and $2x^2(1-x)$, the left-hand side of $(1)$ appears to converge to 0 at the rate $O(1/n^2)$. I suspect this is the case whenever $f(x)$ has a continuous second derivative.~~
Motivation
----------
Answers to this question may help solve another question of mine, on finding [bounds on the expected value of a hypergeometric random variable](https://mathoverflow.net/questions/429037/bounds-on-the-expectation-of-a-function-of-a-hypergeometric-random-variable).
Questions
---------
Let $M(f, r)$ equal the maximum absolute value of $f$ and its derivatives up to the $r$-th derivative.
1. Given that $f$ has a continuous second derivative, can $\phi(f, n)$ equal $CM(f, 2)/n^2$ and/or $C/n^2$? In either case, find an upper bound for $C$.
2. Given that $f$ is continuous, what is an explicit upper bound for $\phi(f, n)$, perhaps given the modulus of continuity of $f$?
3. Does the same bound for $C$ as in question 1 hold if, instead—
* $f$ has a Lipschitz-continuous derivative, and
* $M(f, r)$ equals the maximum absolute value of $f$ and the Lipschitz constants of $f$ and its derivative?
4. Does the same bound for $C$ as in question 1 hold if, instead—
* $f'$ is in the *Zygmund class*, that is, for some constant $D\ge 0$ it has the property $|f'(x) + f'(y) - 2f'((x+y)/2)| \le D\epsilon$ for every $\epsilon>0$, whenever $x$ and $y$ are in $[0, 1]$ and $|x-y|\le\epsilon$, and
* $M(f, r) = \max(D, \max |f|, \max |f'|)$?
It would be nice if the answers to questions 1 and 4 can easily carry over to the problem of finding an upper bound for $|B\_{n+1}(f)(x) - B\_{n}(f)(x)|$.
References
----------
* Butzer, Paul (1953) "[Linear Combinations of Bernstein Polynomials](https://doi.org/10.4153/CJM-1953-063-7)". Canadian Journal of Mathematics, 5, 559-567. doi:10.4153/CJM-1953-063-7, [MR0058023](https://mathscinet.ams.org/mathscinet-getitem?mr=MR0058023), [Zbl 0051.05002](https://zbmath.org/?q=an%3A0051.05002).
| https://mathoverflow.net/users/171320 | Explicit bounds on the difference between Bernstein polynomials | Some comments.
The estimate $\|B\_{2n} f-B\_n f\|\_\infty \leq c \omega (f, n^{-1/2})$, ($\omega$ is the modulus of continuity of $f$) follows from $\|f-B\_n f\|\_\infty \leq c \omega (f, n^{-1/2})$. Both estimates can be improved to $\frac C n$ if $f \in C^1([0,1]$. However, since
$$ n(f-B\_nf) \to -\frac{x(1-x)}{2}f''$$
an estimate $\|f-B\_n f\|\_\infty =o(1/n)$ gives $f(x)=ax+b$ (see Chapter 1 of the book of Lorentz, Bernstein polynomials).
If $\|B\_{2n} f-B\_n f\|\_\infty \leq Cn^{-\alpha}$, then $\|B\_{2^{k+1}n} f-B\_{2^k n} f\|\_\infty \leq C2^{-\alpha k}n^{-\alpha}$ and summinng over $k$, $\|f-B\_nf \|\_\infty \leq C n^{-\alpha}$ and $\alpha \leq 1$ unless $f$ is linear.
| 0 | https://mathoverflow.net/users/150653 | 429359 | 173,999 |
https://mathoverflow.net/questions/429356 | 2 | Let $I \subset \mathbb{R}$ be an interval, and $f\_n, f: I \rightarrow \mathbb{R}$ a convex function; then its Legendre transform is the function $f^{\ast}: I^{\ast} \rightarrow \mathbb{R}$ defined by
$$
f^{\ast}\left(x^{\ast}\right)=\sup \_{x \in I}\left(x^{\ast} x-f(x)\right), \quad x^{\ast} \in I^{}
$$
where sup denotes the supremum, and the domain $I^{\ast}$ is
$$
I^{\ast}=\{ x^{\ast} \in R: \sup \_{x \in I}\left(x^{\ast} x-f(x)\right)<\infty \}.$$
The transform is always well-defined when $f(x)$ is convex.
$$Question$$
Assume that $\frac{1}{n}f\_{n}(x) \to f(x)$. Is it true that $\sup\_{x \in I}\left(x^{\ast} x-\frac{1}{n}f\_{n}(x)\right) \to f^{\ast}(x^{\ast})?$
| https://mathoverflow.net/users/127839 | Continuity of Legendre transform | Conditions in terms of the function $f$ that are simultaneously necessary and sufficient for such convergence are given in [this paper](https://www.heldermann.de/JCA/JCA26/JCA261/jca26005.htm) and in its [arXiv version](https://arxiv.org/abs/1307.3806). The form of those conditions depends on whether either or both endpoints of the interval $I$ are finite or not.
In particular, by Corollary 2a of that paper, the desired convergence will always hold in your setting if both endpoints of the interval $I$ are finite.
| 1 | https://mathoverflow.net/users/36721 | 429369 | 174,002 |
https://mathoverflow.net/questions/429332 | 6 | Given a $d$-ball $\mathcal{S}^{d}$, let $P\_n$ a set of $n$ points selected uniformly at random on the boundary $\mathcal{S}^{d-1}$ of $\mathcal{S}^{d}$. Let $\mathcal{C}\_n$ the convex hull of $P\_n$. We denote by $V(\mathcal{S}^{d})$ and $V(\mathcal{C}\_n)$ the volume of $\mathcal{S}^d$ and $\mathcal{C}\_n$ respectively.
---
**Question:** Can we prove that, for all constant values $a\in (0,1)$, $n$ has to grow superpolynomially in $d$ (for $d\to\infty$) to satisfy $\mathbb{E}[V(\mathcal{C}\_n)]\ge aV(\mathcal{S}^{d})$ ?
---
---
***Note:*** *Do you please have any reference for the convex hull $\mathcal{C}\_n^\*$ of the $n$ points of $\mathcal{S}^{d-1}$ maximizing $V(\mathcal{C}\_n)$ (so that we can hopefully avoid to calculate the expectation $\mathbb{E}[V(\mathcal{C}\_n)]$ if $n=\omega(\mathrm{poly}(d))$ to satisfy $V(\mathcal{C}\_n^\*)\ge aV(\mathcal{S}^{d})$ for any $a\in(0,1)$ when $d\to\infty$) ?*
| https://mathoverflow.net/users/115803 | $d$-ball approximation for $d\gg 1$ with a convex hull of random points on its boundary | It's true deterministically in $P\_n$. In fact, here's a proof that there is some $c$ so that if $n \leq e^{cd}$ then we have that $V(C\_n)/V(S^d) \to 0$.
We'll work with the ball of radius $1$. For a given configuration $P\_n$, let $p$ denote the probability that a point chosen uniformly from the sphere lies in the $C\_n$; then $p = V(C\_n)/V(S^d)$. Let $X$ be a point chosen uniformly at random form the ball $S^d$.
If $X \in C\_n$, then there must be a point $y \in P\_n$ so that $\langle y , \frac{X}{\|X\|} \rangle \geq \|X\|$; otherwise, then hyperplane with normal vector $X/\|X\|$ separates $X$ from the set $P\_n$. Union bounding over all points we have $$ \mathbb{P}(X \in C\_n) \leq \sum\_{y \in P\_n} \mathbb{P}\left( \left\langle y, \frac{X}{\|X\|} \right\rangle \geq \| X \|\right) = n \mathbb{P}\left( \left\langle Y, \frac{X}{\|X\|} \right\rangle \geq \| X \|\right)$$ where in the last line we used spherically symmetry of $X$, and we let $Y$ be chosen uniformly at random from the sphere.
It is thus sufficient to show that this probability decays exponentially. Note that $\mathbb{P}(\|X \| < 1/2) = 2^{-d}$ and so $$\mathbb{P}\left( \left\langle Y, \frac{X}{\|X\|} \right\rangle \geq \| X \|\right) < 2^{-d} + \mathbb{P}\left( \left\langle Y, \frac{X}{\|X\|} \right\rangle \geq 1/2\right).$$ By rotational symmetry of $Y$, we may assume WLOG that $X/\|X\|$ is the first coordinate vector $e\_1$. We may sample $Y$ by considering $d$ i.i.d. standard Gaussians $Z = (Z\_1,\ldots,Z\_d)$ and taking $Y = Z/\|Z\|$. Thus we have that $\langle Y, e\_1 \rangle$ has the same distribution as $\frac{Z\_1}{\|Z\|}$. We have that $\mathbb{P}(\|Z\| < \sqrt{d}/2) \leq e^{-cd}$, and so $$ \mathbb{P}\left(\frac{Z\_1}{\|Z\|} \geq 1/2\right) \leq e^{-cd} + \mathbb{P}(Z\_1 \geq \sqrt{d}/4) \leq e^{-c'd}.$$
| 4 | https://mathoverflow.net/users/69870 | 429382 | 174,005 |
https://mathoverflow.net/questions/429376 | 4 | In a comment to this [question](https://mathoverflow.net/questions/429009/non-existence-of-higher-artin-map), David Loeffler asked if one can show that the (local) Artin map
$$K^\times \to G\_K^{ab}$$
does not have a lift to $G\_K$. Probably this wouldn't be canonical, but I can't show that such a lift doesn't exist.
So does there exist such a lift?
| https://mathoverflow.net/users/152554 | Existence of lift of (local) Artin map | No, there does not, except possibly a few small corner cases. The problem is finding somewhere for the torsion in $K^\times$ to go.
If $K$ is a finite extension of $\mathbf{Q}\_p$, then there exists a number field $\mathscr{K}$ and a prime $v$ of $\mathscr{K}$ above $p$ such that $\mathscr{K}\_v = K$; and hence we obtain an embedding $G\_K \hookrightarrow G\_{\mathscr{K}} \hookrightarrow G\_{\mathbb{Q}}$. But $G\_{\mathbb{Q}}$ is known to have no nontrivial elements of finite order except the conjugacy class of complex conjugation of order 2 (see [this question](https://mathoverflow.net/questions/6802/element-in-the-absolute-galois-group-of-the-rationals)).
So $K^\times$ cannot embed into $G\_K$ unless $(K^\times)\_{\mathrm{tors}}$ has order exactly 2, which can only happen if $p = 2$ or $p = 3$ (and I am too tired this evening to think about those cases).
| 5 | https://mathoverflow.net/users/2481 | 429386 | 174,007 |
https://mathoverflow.net/questions/429126 | 6 | Let $G$ be a compact Lie group with Lie algebra $\mathfrak{g}$. Let $<.,>$ denote a $G$-invariant inner product on $\mathfrak{g}$.
Let $(M,\omega)$ be a symplectic compact manifold endowed with a hamiltonian action of $G$, and let $\mu : M \longrightarrow \mathfrak{g}^\*,$ be a moment map associated to this action. We fix a Riemannian metric $g$ on $M$.
I have read somewhere that every gradient flow line $\phi\_x(t)$ of the norm-squared of the moment map $\mu$ begins and ends at a critical point, i.e $\lim\_{t \rightarrow + \infty} \phi\_t(x) $ and $\lim\_{t \rightarrow - \infty} \phi\_t(x) $ exist, and they are both critical points of the norm-squared of $\mu$. but I couldn't find a proof of this fact anywhere, Hopefully someone can help.
(This is a follow-up question to This one: [The negative gradient flow of a Morse-Bott function on a compact manifold converges to a critical point?](https://mathoverflow.net/questions/425477/the-negative-gradient-flow-of-a-morse-bott-function-on-a-compact-manifold-conver/429039#429039))
| https://mathoverflow.net/users/172459 | The norm-squared of a moment map behaves like a Morse-Bott function | As requested I am submitting my comment as an answer.
The desired statement is proved in [Gradient flow of the norm squared of a moment map](https://www.e-periodica.ch/cntmng?pid=ens-001%3A2005%3A51%3A%3A14) by Eugene Lerman who attributes the proof to Duistermaat.
| 4 | https://mathoverflow.net/users/33141 | 429390 | 174,009 |
https://mathoverflow.net/questions/428728 | 1 | Let $T'$ be the Tsirelson space dual in the Figiel-Johson construction, and $B\_{T'}$ the unit closed ball. Then, $B\_{T'}$ satisfies the next properties:
(i) $B\_{T'}\subset B\_{l^\infty}.$ Each vector basis $e\_n$ belongs to $B\_{T'}.$
(ii) $\forall \mathbf{x} \in B\_{T'},\ \mathbf{y} \in \mathbb{R}^{\mathbb{N}},\ $ if $|\mathbf{y}|\leq |\mathbf{x}|$ (pointwise) then $\mathbf{y} \in B\_{T'}$
(iii) If $\{\mathbf{y}\_i\}\_{i=1}^n$ is a block basic sequence of $B\_{T'},$ then $\dfrac{1}{2}Q\_n(\sum\_{i=1}^n\mathbf{y}\_i)\in B\_{T'},$ where $Q\_n$ denote the natural projection of $T'$ onto $[e\_k]\_{k=n}^\infty.$
The original contruction (by Tsirelson) took $A$ to be the smallest set in $l^\infty$ enyoing properties (i) through (iii), and let $K$ denote the clousure of $A$ respect to the topology of pointwise convergence. Since it is easily seen $K\subset c\_0,$ it follows that $K$ is weakly compact. Finally, the closed convex hull $V$ of $K$ defines a weakly compact convex set in $c\_0$, which inherits properties (i) through (iii). The minimality of $A$ then forces
$$V=B\_{T'}.$$
I can easily see that $V\subset B\_{T'}$ because of the minimailty argument, but I don't get to see the other one.
| https://mathoverflow.net/users/489810 | About the constructions of Tsirelson space | IIRC Figiel and I reasoned this way:
The dual norm to Tsrilson's original norm has the property that for any vector $x$ in $c\_{00}$,
$$
\|x\| \ge \|x\|\_{c\_0} \vee (1/2) \sup \sum\_{k=1}^n \|E\_k x\|,
$$
where the sup is over all admissible $(E\_k)$.
But, by construction, the FJ norm is the smallest unconditional norm satisfying this inequality. The inclusion you see yields that the dual to Tsirelson's original norm is smaller than the FJ norm, so the two norms are the same.
BTW: You should have said that the bulk of your post (absent the question) is taken word for word from page 17 in the Casazza-Shura book.
| 4 | https://mathoverflow.net/users/2554 | 429401 | 174,012 |
https://mathoverflow.net/questions/429399 | 1 | "Rule 150" is a fascinating one-dimensional and simple [cellular automaton](https://en.wikipedia.org/wiki/Cellular_automaton) giving raise to some quite chaotic behaviour. This is the starting point of this question.
Let $\{0,1\}^\mathbb{Z}$ denote the set of all functions $x:\mathbb{Z}\to \{0,1\}$. Let $+$ denote addition modulo $2$ on $\{0,1\}$ (corresponding to [${\sf XOR}$](https://en.wikipedia.org/wiki/XOR_gate) in computer science).
Define a function $f:\{0,1\}^\mathbb{Z} \to \{0,1\}^\mathbb{Z}$ by $x \mapsto f(x)$ where $f(x):\mathbb{Z}\to\{0,1\}$ is defined by $$f(x)(i) = x(i) + x(i-1) + x(i+1) \text{ for all }i\in \mathbb{Z}.$$ Inductively define $f^{(0)}(x) = x$, and $f^{(n+1)}(x) = f(f^{(n)}(x))$ for all integers $n\geq 0$.
**Question.** For what integers $n>1$ is there $x\in \{0,1\}^\mathbb{Z}$ such that $f^{(n)}(x) = x$ but $f^{(k)}(x) \neq x$ for all integers $k\geq 1$ with $k<n$?
| https://mathoverflow.net/users/8628 | Possible finite periodicities of "Rule 150" in the infinite setting | As already mentioned, this cellular automaton is Rule 150.
Rule 150 is an example of a bipermutive cellular automaton, meaning that it is both left permutive (the value of $f(x)(i)$ can be permuted by permuting the value of $x(i-1)$) and right permutive (the value of $f(x)(i)$ can be permuted by permuting the value of $x(i+1)$). Due to bipermutivity, by Theorem 6.3 in <https://core.ac.uk/download/pdf/236376428.pdf> Rule 150 is conjugate to the shift map $\sigma$ on $\{0,1,2,3\}^\mathbb{N}$ defined by $\sigma(x)(i)=x(i+1)$ for $x\in\{0,1,2,3\}^\mathbb{N}$, $i\in\mathbb{N}$ (conjugacy means that there is a bijection $\phi:\{0,1\}^\mathbb{Z}\to \{0,1,2,3\}^\mathbb{N}$ such that $\phi\circ f = \sigma\circ \phi$). For all integers there is a sequence $x\in\{0,1,2,3\}^\mathbb{N}$ such that $\sigma^{(n)}(x)=x$ but $\sigma^{(k)}(x)\neq x$ for $k<n$, so due to conjugacy the same holds for the function $f$.
| 9 | https://mathoverflow.net/users/178593 | 429414 | 174,017 |
https://mathoverflow.net/questions/429312 | 7 | Let $A$ be a C\*-algebra that has no one-dimensional irreducible representations, that is, there is no (closed) two-sided ideal $I\subseteq A$ such that $A/I\cong\mathbb{C}$.
Let $J$ denote the (not necessarily closed) two-sided ideal generated by additive commutators in $A$:
$$
J:=\{ \sum\_{k=1}^n a\_k[b\_k,c\_k]d\_k : a\_k,b\_k,c\_k,c\_k\in A\}.
$$
>
> Question: Is $A=J$?
>
>
>
The answer is `Yes', if $A$ is unital and in some other cases, but does it hold in general? Note that $J$ is a dense, two-sided ideal (thus contains the Pedersen ideal of $A$), and that $A/J$ is a commutative algebra.
| https://mathoverflow.net/users/24916 | Commutator ideal in nonunital C*-algebra | The answer is NO. Rordam and Robert [MR3072284](https://mathscinet.ams.org/mathscinet-getitem?mr=3072284) have found
a sequence $(A\_n)\_n$ of simple unital infinite dimensional C\*-algebras
such that $\prod A\_n$ has a nonzero character.
(Thanks are due to Yasuhiko Sato for informing me of this.)
Thus the following is true: For every $m$ and $C>1$, there is $n=n(m,C)$
such that $1=\sum\_{k=1}^m a\_k[b\_k,c\_k]d\_k$ in $A\_n$ implies
$\sum\_{k=1}^m\|a\_k\| \|b\_k\| \|c\_k\| \|d\_k\| > C$.
Now consider the $c\_0$-sum $A:=\bigoplus\_m A\_{n(m,m^2)}$.
Then $(m^{-1})\_m \in A$ cannot be expressed as
$\sum\_{k=1}^l a\_k[b\_k,c\_k]d\_k$ in $A$, because it would imply
$\sum\_{k=1}^l \|a\_k(m)\| \|b\_k(m)\| \|c\_k(m)\| \|d\_k(m)\| \geq m$ for every $m\geq l$.
On the the hand, it is easy to show that there are $m$ and $C>1$
that satisfies the following:
For every von Neumann algebra without nonzero abelian direct summand,
one has $1=\sum\_{k=1}^m a\_k[b\_k,c\_k]d\_k$ for some
$a\_k,b\_k,c\_k,d\_k$ with $\sum\_{k=1}^m\|a\_k\| \|b\_k\| \|c\_k\| \|d\_k\| < C$.
By the Hahn--Banach separation theorem, this implies the following:
For every $A$ without nonzero characters and every $x\in A$,
there are infinite sequences $a\_k,b\_k,c\_k,d\_k$ such that
$\sum\_{k=1}^\infty\|a\_k\| \|b\_k\| \|c\_k\| \|d\_k\| \le C\|x\|$
and $x = \sum\_{k=1}^\infty a\_k[b\_k,c\_k]d\_k$.
| 9 | https://mathoverflow.net/users/7591 | 429423 | 174,018 |
https://mathoverflow.net/questions/429405 | 2 | I have a question about a proof from Jacob Lurie's [Higher Topos Theory](https://people.math.harvard.edu/%7Elurie/papers/highertopoi.pdf) (p 45):
**Proposition 1.2.12.5.** Let $\mathcal{C}$ be a simplicial set. Every strongly final object (see below what means) of
$\mathcal{C}$ is a final object of $\mathcal{C}$. The converse holds if $\mathcal{C}$ is an $\infty$-category.
>
> *Proof:* Let $[0]$ denote the category with a single object and a single morphism.
> Suppose that $Y$ is a strongly final vertex of $\mathcal{C}$. Then there exists a retraction of $\mathcal{C}^{ \triangleright }$ onto $\mathcal{C}$ carrying the cone point to $Y$ . Consequently, we obtain
> a retraction of (H-enriched) homotopy categories from $h\mathcal{C} \* [0]$ to $h\mathcal{C}$ carrying the unique object of $[0]$ to $Y$ . This implies that $Y$ is final in $h\mathcal{C}$, so that $Y$ is a final object of $\mathcal{C}$.
>
>
>
(The converse part in the proof I understand)
On used notations:
-$\mathcal{C}^{ \triangleright }$ is the *right cone* defined to be the join $\mathcal{C} \* \Delta^0$.
-*Definition 1.2.12.3.:* A vertex $X$ of simplicial set $\mathcal{C}$ is called *strongly final* if the projection $ \mathcal{C}\_{/ X} \to \mathcal{C}$ is a trivial fibration of simplicial sets. In other words if any map $f\_0 : \partial \Delta^n \to \mathcal{C}$ such that $f\_0(n) = X$ can be extended to a map $f : \Delta^n \to \mathcal{C}$.
*Question:* Why does the assumption that $Y$ is a strongly final vertex of $\mathcal{C}$ imply that there exists a
retraction of $\mathcal{C}^{ \triangleright }$ onto $\mathcal{C}$ carrying the cone point to $Y$? Can its construction be given explicitly or (if not) which abstract result is involved to deduce the existence of this retraction?
| https://mathoverflow.net/users/108274 | Strongly final vertex $Y$ in a simplicial set gives a retraction of $\mathcal{C}^{ \triangleright }$ onto $\mathcal{C}$ | I think this can be done pretty concretely. According to Definition 1.2.8.1, there is an obvious inclusion $i:C \to C\ast \Delta^0 = \mathcal{C}^{ \triangleright }$. Now you need the retraction.
The key is Proposition 1.2.9.2, which implies that $Hom\_{sSet}(C,C/Y) = Hom\_p(C\ast \Delta^0,C)$ where $p: \Delta^0 \to C$ has $im(p) = Y$ (here $Hom\_p$ means we only want $f: C\ast \Delta^0 \to C$ such that $f|\_{\Delta^0} = p$).
Now we have to use what it means to be strongly final. We know that the projection $C/Y \to C$ is a trivial fibration. This means it induces the following equivalence of hom sets $Hom\_{sSet}(C,C)\simeq Hom\_{sSet}(C,C/Y)$, since every $C \in sSet$ is cofibrant. The identity $id\_C$ is taken by the composition $Hom\_{sSet}(C,C)\simeq Hom\_{sSet}(C,C/Y) = Hom\_p(C\ast \Delta^0,C)$ to the desired retraction, $r$. It is an easy and valuable exercise to check that $r\circ i = id\_C$. Note the universal property in 1.2.9.2.
| 1 | https://mathoverflow.net/users/11540 | 429425 | 174,019 |
https://mathoverflow.net/questions/429363 | 2 | Let $\mathscr{H}^m$ be the $m$ dimensional Hausdorff measure in $\mathbb{R}^n$, $m\leq n$. Is it true that for $\mathscr{H}^m$-almost every point $p$ on a Lipschitz manifold $M$ of dimension $m$ embedded in $\mathbb{R}^n$ there is a neighborhood $B\_\epsilon(p)\subset \mathbb{R}^n$ such that $M\cap B\_\epsilon(p)=graph(u)\cap B\_\epsilon(p)$ for some
$$
u:p+T\_pM\to p+(T\_pM)^\perp
$$
defined on the (approximate) tangent space of $p$ (which exists for a.e. $p$) with $Lip(u)<\epsilon$?
Please note that the question is trivial if we ask $u$ only to be a Lipschitz function without the requirement that it satisfies $Lip(u)<\epsilon$.
Edit: Since a Lipschitz manifold can be defined in several ways not all equivalent to each other, here we can just consider the simplest case of $M=graph(f)$ for some Lipschitz function $f:\mathbb{R}^m\to\mathbb{R}^{n-m}$. If it helps, we can also set $m=1$ and $n=2$, so that $f:\mathbb{R}\to\mathbb{R}$.
| https://mathoverflow.net/users/351083 | $(1+\epsilon)$-bilipschitz parametrization of Lipschitz manifold | The answer to my question is no, and a counter-example is provided by the staircase function of the fat-Cantor set, which is a Lipschitz function $f:[0,1]\to\mathbb{R}$ with the property that $\{f'=0\}$ is dense and $\{f'\geq1\}$ has positive measure. Therefore, for $\epsilon>0$ small enough, a function $u$ as in the body of the question cannot exist around all points $(x,f(x))$ with $x\in\{f'\geq1\}$. This set has positive $\mathscr{H}^1$-measure (in fact, at least $\sqrt{2}\mathscr{L}^1(\{f'\geq1\})$, since for all measurable $A\subset [0,1]$ it holds $\mathscr{H}^1(G(A))=\int\_{[0,1]}1\_A \sqrt{1+|f'(x)|^2}dx$, where $G:[0,1]\to[0,1]\times\mathbb{R}$ is given by $G(x)=(x,f(x))$.
See also this question <https://math.stackexchange.com/questions/3380703/where-is-a-fat-cantor-staircase-differentiable>.
| 1 | https://mathoverflow.net/users/351083 | 429438 | 174,022 |
https://mathoverflow.net/questions/429302 | 2 | Let $K$ be an imaginary quadratic field and let $F$ be a finite extension of $K$. Let $E$ be an elliptic curve over $F$ with CM by $K$. Suppose that $p$ is a prime that splits as $p=\pi\pi^\*$ in $K$. Then for each place $v$ of $F$, we have the local Kummer map
$$ \kappa\_v: E(F\_v) \, \otimes \mathbf{Q}\_p/\mathbf{Z}\_p \hookrightarrow H^1(F\_v, E[\pi^{\infty}]). $$
*My question is*: what is the image of $\kappa\_v$ for various places $v$ of $F$? I know that if $v$ does not lie over $\pi$ or $\pi^\*$, then the image of $\kappa\_v$ is zero. But what about the cases $v = \pi$ or $v = \pi^\*$? Is the image of $\kappa\_v$ different in these two cases?
| https://mathoverflow.net/users/394740 | Image of Kummer map for CM Elliptic curves | There are three cases:
1. If $v$ is a finite place of $F$ not lying over $\pi$ or $\overline{\pi}$, then the image of the Kummer map $\kappa\_v$ is zero.
2. If $v$ is a finite place of $F$ lying over $\overline{\pi}$, it turns out that the image of the Kummer map $\kappa\_v$ is *still* zero.
3. The tricky case is when $v$ is a finite place of $F$ lying over $\pi$. Here, the image of the Kummer map $\kappa\_v$ *may be non-zero*. The precise description of the image is more complicated and it seems to involve formal groups in some way.
These results are in pages 7-9 of this paper by Murty and Ouyang:
[http://staff.ustc.edu.cn/~yiouyang/ouyang-murty.pdf](http://staff.ustc.edu.cn/%7Eyiouyang/ouyang-murty.pdf)
| 0 | https://mathoverflow.net/users/394740 | 429445 | 174,023 |
https://mathoverflow.net/questions/429440 | 2 | Is there a standard or "simple" contact structure on the $3$-dimensional torus $T^3$, like there are for example for the Eucliden space and the $3$-sphere?
My first thought was to consider a structure descending to $T^3$ from the standard structure defined by the form
$$ \omega\_{std} = dz + x\wedge dy$$
on $\mathbb{R^3}$, but this doesn't work because $\omega\_{std}$ is not invariant under the action of $\mathbb{Z}^3$. Also, if such a structure were to descend to $T^3$, the Reeb flowlines would consist of periodic curves parallel to the $\partial\_z$ direction, and in particular it would be transversal to a $2$-dimensional torus $T^2$. That, on the other hand, can not happen, because the contact condition would imply the existence of an exact volume form on $T^2$, contradicting Stokes theorem.
I know there are contact structures on $T^3$ because of a theorem of Thurston's that states that every closed smooth $3$ manifold supports a contact structure, but I couldn't find any explicit examples for the torus. I am particularly interested in explicit descriptions of the associated Reeb field and its flow.
| https://mathoverflow.net/users/142154 | Standard contact forms on the torus | Consider the $1$-form on $\mathbb{R}^3$ given by
$$ \alpha = \cos(2\pi z) dx + \sin(2\pi z) dy, $$
where $(x,y,z)$ are the standard coordinates on $\mathbb{R}^3$. This is a contact form on $\mathbb{R}^3$ with Reeb vector field given by $$R\_{\alpha} = \cos(2\pi z) \partial\_x + \sin(2\pi z) \partial\_y.$$
This vector field can be integrated explicitly and its integral curve passing through $(x\_0,y\_0,z\_0)$ at time $t=0$ is given by $$\phi(t) = (x\_0 + \cos(2\pi z\_0)t, y\_0 + \sin(2\pi z\_0)t, z\_0). $$
Consider the standard action of $\mathbb{Z}^3$ on $\mathbb{R}^3$ by translations. The $1$-form $\alpha$ is invariant under this action and, therefore, descends to a $1$-form on the quotient $\mathbb{R}^3/\mathbb{Z}^3 \simeq T^3$. This gives an example of a contact structure on $T^3$.
More generally, given any $k \in \mathbb{Z} \smallsetminus \{0\}$, one can define a contact form on $\mathbb{R}^3$ by
$$ \cos(2\pi k z) dx + \sin(2\pi k z) dy. $$
This contact form is invariant under the above action of $\mathbb{Z}^3$ and, therefore, descends to a contact form on $T^3$.
| 3 | https://mathoverflow.net/users/13022 | 429453 | 174,025 |
https://mathoverflow.net/questions/429198 | 1 | Let $X$, $Y$, $Z$ be discrete random variables, with $Y$ and $Z$ independent. Does the following equality hold **if $Z$ is independent also of $X$**?
$$
\max\_{f\_{Y,Z}} \big\{ \ I(X; f\_{Y,Z}(Y,Z)) \ \big\} = \max\_{f\_X, f\_Y} \big \{ \ I(X; f\_Y(Y), f\_Z(Z)) \ \big \}
$$
where the maximization is taken over all **non-injective**, deterministic functions.
P.S.: See [this](https://mathoverflow.net/questions/426330/maximization-of-information-over-set-of-non-injective-functions) for the inequality version of the question.
| https://mathoverflow.net/users/101100 | Maximization of information over set of non-injective functions (Equality) | No. Let $Y,Z$ be iid Bernoulli(1/2), and let $X=Y+Z$ mod 2, which induces pairwise independence. The only non-injective deterministic functions $f\_Y,f\_Z$ are constants, rendering the RHS zero. For the LHS, we can take $f\_{Y,Z}(Y,Z)$ equal to the the binary AND of $Y$ and $Z$, which is not injective, and not independent of $X$. Hence, the LHS is something positive, showing there can be strict inequality.
| 1 | https://mathoverflow.net/users/99418 | 429463 | 174,030 |
Subsets and Splits