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https://mathoverflow.net/questions/414240 | 0 | Can someone please provide a reference (starting point) for analysing recurrence/transience of random walks on graphs with general edge weights? Looking into random walks that are known to be NOT reversible.
| https://mathoverflow.net/users/473751 | Recurrence criterion for non-reversible random walks on general infinite (locally finite) graph with unequal edge weights | The most powerful and flexible method to prove recurrence or transience for nonreversible walks is the method of Lyapunov functions.
One of the origins of the method is Foster's criteria for recurrence and positive recurrence. [1], [2]. See Theorem 1 page 2 in [3] (or page 40 of [9]) for a concise statement of the recurrence criterion.
The method is expounded with many examples and references in the books [4], [5], [6]. Some specific applications to delicate problems are in [3], [4].
[1] F. G. Foster: On the stochastic matrices associated with certain queuing processes. Ann. Math. Statistics 24 (1953) 355–360.
[2] <https://en.wikipedia.org/wiki/Foster%27s_theorem>
[3] [https://mathweb.ucsd.edu/~pfitz/downloads/courses/spring05/math280c/foster.pdf](https://mathweb.ucsd.edu/%7Epfitz/downloads/courses/spring05/math280c/foster.pdf)
[4] Fayolle, Guy, Vadim Aleksandrovich Malyshev, and Mikhail Vasilʹevich Menʹshikov. Topics in the constructive theory of countable Markov chains. Cambridge university press, 1995.
[5] Menshikov, Mikhail, Serguei Popov, and Andrew Wade. Non-homogeneous random walks: Lyapunov function methods for near-critical stochastic systems. Vol. 209. Cambridge University Press, 2016.
[6] S.P. Meyn and R.L. Tweedie: Markov Chains and Stochastic Stability. SpringerVerlag, London, 1993.
[7] Asymont, Inna M., Guy Fayolle, and Μ. V. Menshikov. "Random walks in a quarter plane with zero drifts: transience and recurrence." Journal of applied probability 32, no. 4 (1995): 941-955.
[8] Peres, Yuval, Serguei Popov, and Perla Sousi. "On recurrence and transience of self-interacting random walks." Bulletin of the Brazilian Mathematical Society, New Series 44, no. 4 (2013): 841-867.
[9] [https://www.math.stonybrook.edu/~rdhough/mat639-spring17/lectures/lecture12.pdf](https://www.math.stonybrook.edu/%7Erdhough/mat639-spring17/lectures/lecture12.pdf)
| 0 | https://mathoverflow.net/users/7691 | 414956 | 169,211 |
https://mathoverflow.net/questions/414082 | 1 | Specifically my question is the following: Let $P$ be a Peano space. If $(P,\sigma,\mu)$ and $(P,\sigma,\nu)$ are both nonatomic probability measures, does there exist a continuous function $f:P\to P$ such that $f\_\# \mu(A)=\nu(A)$ for all $A\in \sigma$?
Keep in mind that if $(X,B)$ and $(Y,C)$ are measurable spaces, $f: X \to Y$ is a measurable map, and $\mu$ is a measure on $B$, then we write $f\_\#\mu$ for the push forward measure on $C$, which is defined by $f\_\#\mu(A) = \mu(f^{-1}(A))$ for all $A \in C$.
Any insights or sources would help.
| https://mathoverflow.net/users/473399 | A question about pushforward measures and Peano spaces | In general the answer to this problem is negative: if the measure $\mu$ has connected support and the measure $\nu$ has disconnected support, then for any continuous map $f:P\to P$ the measure $f\_\# \mu$ will be supported by a connected subset of $P$ and hence will be not equal to the measure $\nu$.
Let us recall that the *support* $\mathrm{supp}(\mu)$ of a measure $\mu$ on $P$ is the closed set of all points $x\in P$ such that every neighborhood $O\_x$ of $x$ has measure $\mu(O\_x)>0$.
On the other hand, if $\mu,\nu$ are nonatomic probability measures on Peano continua $X,Y$ and $\mathrm{supp}(\nu)=Y$, then there exists a continuous map $f:X\to Y$ such that $\nu=f\_\# \mu$. The construction of the map $f$ consists of five steps:
1. Construct a continuous map $\varphi:[0,1]\to Y$ such that for every nonempty open set $U\subseteq [0,1]$ the image $\varphi[U]$ has nonepty interior and hence has measure $\nu(\varphi[U])>0$. The construction of $\varphi$ uses a piece of graph theory.
2. Choose any probability measure $\lambda$ on $[0,1]$ such that $\varphi\_\#\lambda=\nu$ and observe that $\mathrm{supp}(\lambda)=[0,1]$.
3. Choose a compact zero-dimensional set $C\subseteq X$ such that every nonempty relatively open subset $U\subseteq C$ has measure $\mu(U)>0$. Then $C$ is homeomorphic to the Cantor set and hence admits a continuous surjective function $\psi\_0:C\_0\to[0,1]$ such that $|\psi\_0^{-1}(y)|\le 2$ for every $y\in[0,1]$. Using the Baire Category Theorem, choose a continuous function $\psi:X\to [0,1]$ such that $\psi{\restriction}\_C=\psi\_0$ and for every $y\in [0,1]$ the preimage $\psi^{-1}(y)$ has measure $\mu(\psi^{-1}(y))=0$. Then $\psi\_\#\mu$ is a nonatomic measure on $[0,1]=\mathrm{supp}(\psi\_\#\mu)$.
4. Choose a monotone homeomorphism $h:[0,1]\to[0,1]$ such that $h\_\#(\psi\_\#\mu)=\lambda$.
5. The function $f=\varphi\circ h\circ\psi:X\to Y$ will have the desired property: $f\_\#\mu=\nu$.
| 1 | https://mathoverflow.net/users/61536 | 414959 | 169,212 |
https://mathoverflow.net/questions/414770 | 21 | Consider the following partial order. The objects are unordered tuples $\{V\_1,\ldots,V\_m\}$, where each $V\_i \subseteq \mathbf{R}^n$ is a nontrivial linear subspace and $V\_1 \oplus \cdots \oplus V\_m = \mathbf{R}^n$. We say that $\{V\_1,\ldots,V\_m\} \geq \{W\_1,\ldots,W\_i\}$ if for each $j=1,\ldots,m$ there exists $1 \leq k \leq i$ such that $V\_j \subseteq W\_k$. (I.e. if $\{V\_1,\ldots,V\_m\}$ is a finer decomposition than $\{W\_1,\ldots,W\_i\}$.) This partial order has a minimal element, $\{\mathbf{R}^n\}$, which is the trivial decomposition. If we remove it, what is the homotopy type of this partial order?
Some things that are known about similar problems.
1. If we're working over $\mathbb{F}\_1$, i.e. with subsets of $\{1,\ldots,n\}$, then there is both a minimal and a maximal object. If we remove both of these then the homotopy type of the partial order is a wedge of $S^{n-3}$'s.
2. If instead we work with ordered decompositions, then this is a barycentric subdivision of Ruth Charney's split building, so it is a wedge of $S^{n-2}$'s.
Edit: clarity and notation.
| https://mathoverflow.net/users/1378 | What is the homotopy type of the poset of nontrivial decompositions of $\mathbf{R}^n$? | Let me write $V$ for a finite-dimensional vector space over some field (the field will not play a role), and $\mathsf{P}(V)$ for the poset described in the question, which I consider as a category. Let me rephrase the order relation: $\{V\_1, \ldots, V\_n\} \geq \{W\_1, \ldots, W\_m\}$ if and only if each $W\_j$ is a direct sum of $V\_i$'s. The nerve of this poset is $(\mathrm{dim}(V)-2)$-dimensional, so to see that it is a wedge of spheres one must show that it is $(\mathrm{dim}(V)-3)$-connected. I think this is true.
Let $\mathsf{S}(V)$ denote the following category. It has objects given by pairs $([n], f : [n] \to Sub(V))$ consisting of a standard set $[n] := \{1,2,\ldots, n\}$ and a function from $[n]$ to the set of proper vector subspaces of $V$, such that $\bigoplus\_{i \in [n]} f(i) = V$. A morphism in $\mathsf{S}(V)$ from $([n], f)$ to $([n'], f')$ is given by a surjection $e : [n] \to [n']$ such that $f'(i) = \bigoplus\_{j \in e^{-1}(i)} f(j)$.
There is a functor $F : \mathsf{S}(V) \to \mathsf{P}(V)$ given by sending $([n], f)$ to the unordered collection $\{f(i)\}\_{i \in [n]}$.
Now let me do something a bit odd: choose a total order $\prec$ on the set of vector subspaces of $V$. Then we can consider objects of $\mathsf{P}(V)$ as given by lists of subspaces $(V\_1, \ldots, V\_n)$ with $V\_1 \prec V\_2 \prec \cdots \prec V\_n$, and such that $\bigoplus\_{i=1}^n V\_i = V$. Attempt to define a functor $G : \mathsf{P}(V) \to \mathsf{S}(V)$ by
$$G(V\_1, \ldots, V\_n) := ([n], i \mapsto V\_i)$$
on objects. If $(V\_1, \ldots, V\_n) \geq (W\_1, \ldots, W\_m)$ in $\mathsf{P}(V)$ then each $W\_j$ is a direct sum of $V\_i$'s. Define a function $e : [n] \to [m]$ by $e(i) = j$ if $V\_i \subset W\_j$; this gives a morphism in $\mathsf{S}(V)$ from $([n], i \mapsto V\_i)$ to $([n], j \mapsto W\_j)$. To check that this is indeed a functor suppose that $(W\_1, \ldots, W\_m) \geq (U\_1, \ldots, U\_\ell)$, with $e'(j) = k$ when $W\_j \subset U\_k$. Then we indeed have $e' e(i) = k$ when $V\_i \subset W\_{e(i)} \subset U\_k$, so $G$ is indeeed a functor.
The conclusion of this discussion is that $\mathsf{P}(V)$ is a retract of $\mathsf{S}(V)$, so it suffices to show that $\mathsf{S}(V)$ is $(\mathrm{dim}(V)-3)$-connected.
The full name of $\mathsf{S}(V)$ is $\mathsf{S}^{E\_\infty}(V)$, the $E\_\infty$-splitting category defined by Galatius, Kupers, and I in Definition 17.17 of [Cellular $E\_k$-algebras](https://arxiv.org/abs/1805.07184) (applied to the symmetric monoidal groupoid of finite-dimensional vector spaces; the way I have described it here is not identical to that definition, but is an equivalent category). It has a cousin $\mathsf{S}^{E\_1}(V)$ which turns out to be precisely (the category of simplices of) Charney's split building. In particular $\mathsf{S}^{E\_1}(V)$ is $(\mathrm{dim}(V)-3)$-connected by Charney's theorem.
It would take too long to explain all the details here, but the theory developed in that paper shows that the collection of all $\Sigma^2 \mathsf{S}^{E\_\infty}(V)$'s can be obtained from the collection of all $\Sigma^2 \mathsf{S}^{E\_1}(V)$'s by an iterated bar construction, and in particular given Charney's connectivity result (for **all** $V$) it follows that the $\Sigma^2 \mathsf{S}^{E\_\infty}(V)$ are also $(\mathrm{dim}(V)-1)$-connected, so the $\mathsf{S}^{E\_\infty}(V)$ are homologically $(\mathrm{dim}(V)-3)$-connected. (It should not be hard to show it is simply-connected, by hand.) I'm happy to discuss the details by e-mail, if you like.
(Also, now that I believe it is true I expect there must be a more elementary way to deduce it from Charney's theorem.)
Edit:
1. As @inna says in the comments, the functor $F : \mathsf{S}(V) \to \mathsf{P}(V)$ is actually an equivalence, because there is a natural isomorphism $\mathrm{Id} \Rightarrow F \circ G$ given by applying the unique permutation necessary to put things in the order given by $\prec$.
2. Let me try to give some references for what I said in the final paragraph. All references are to [Cellular $E\_k$-algebras](https://arxiv.org/abs/1805.07184):
Writing $\mathsf{G}$ for the symmetric monoidal category of finite-dimensional vector spaces and linear isomorphisms, we work in the category $\mathsf{sSet}\_\*^\mathsf{G}$ of functors from $\mathsf{G}$ to pointed simplicial sets. This is again symmetric monoidal by Day convolution, and the functor
$$\mathbb{t}(V) = \begin{cases}
S^0 & V \neq 0\\
\* & V=0
\end{cases}$$
is a nonnital commutative monoid, and hence also a nonunital $E\_\infty$-algebra. (In the paper this object is called $\underline{\*}\_{>0}$.)
There is another character involved: the derived $E\_k$-indecomposables $Q^{E\_k}\_\mathbb{L}(\mathbb{t})$ for $1 \leq k \leq \infty$, which are again objects of $\mathsf{sSet}\_\*^\mathsf{G}$. For $V$ an object of $\mathsf{G}$ we write $H\_{V,d}^{E\_k}(\mathbb{t}) := H\_d(Q^{E\_k}\_\mathbb{L}(\mathbb{t})(V))$.
The ingredients I have in mind are now:
a. Combining Proposition 17.4 and Lemma 17.10 shows that
$$\Sigma Q^{E\_1}\_\mathbb{L}(\mathbb{t})(V) \simeq \Sigma^2 \mathsf{S}^{E\_1}(V)$$
(in fact by Section 17.5 this can be desuspended once) and hence Charney's theorem shows that $H\_{V,d}^{E\_1}(\mathbb{t})=0$ for $d < \dim(V)-1$.
b. Theorem 14.4 (this is the application of the bar construction result, but is packaged so one doesn't explicitly have to think about that) applied with $\rho(V) := \mathrm{dim}(V)$ shows that $H\_{V,d}^{E\_\infty}(\mathbb{t})=0$ for $d < \dim(V)-1$ too.
c. Corollary 17.23 shows that $Q^{E\_\infty}\_\mathbb{L}(\mathbb{t})(V) \simeq \Sigma \mathsf{S}^{E\_\infty}(V)$ so the above translates to $\tilde{H}\_\*(\mathsf{S}^{E\_\infty}(V))=0$ for $\* \leq \mathrm{dim}(V)-3$.
| 11 | https://mathoverflow.net/users/318 | 414964 | 169,214 |
https://mathoverflow.net/questions/414962 | 3 | Let $\mathbb{N}$ denote the set of non-negative integers. We say that a sequence $f:\mathbb{N}\to \{0,1\}$ is *normal* if every finite $\{0,1\}$-sequence appears in $f$.
Let the *swapping operation* $\sigma:\mathbb{N}\to \mathbb{N}$ be defined by swapping each even integer with its successor - that is, $\sigma(2n) = 2n+1$ and $\sigma(2n+1) = 2n$ for all $n\in \mathbb{N}$.
**Question.** Is there a normal binary sequence $f:\mathbb{N} \to \{0,1\}$ such that $f \circ \sigma$ is not normal?
| https://mathoverflow.net/users/8628 | Is normalcy preserved under the swapping operation? | If every string appears in $f$ consecutively, then every string appears consecutively in $f \sigma$.
(So *yes* in answer to the title question, *no* in answer to the question as phrased in the question body :)
(Thanks to Alessandro Della Corte for corrections in the comments below!)
To see this, suppose that every string appears consecutively in $f$. Let $S$ be a string of even length. Since $S 0 S$ appears consecutively in $f$, it follows that $S$ appears consecutively in $f$ starting from an even position. (We could equally have used $S1S$.)
Since every string $S$ of even length appears consecutively in $f$ starting from even position, it's also the case that $S \circ \sigma$ appears consecutively in $f$ starting from an even position. So every string $S$ of even length appears consecutively in $f \circ \sigma$ (starting from even position). As every string is a consecutive substring of a string of even length, it follows that every string appears consecutively in $f \circ \sigma$.
| 9 | https://mathoverflow.net/users/2362 | 414966 | 169,215 |
https://mathoverflow.net/questions/414707 | 1 | Let $W$ be a standard Brownian motion and $\mathcal F\_t$ its natural filtration.
Suppose $\theta, A$ are positive $L^1$ random variables independent of $\mathcal F\_t$.
Let $Y\_t$ be the process
$$Y\_t := A \sin \, (\theta t) + W\_t $$
and denote by $\mathcal Y\_t$ its natural filtration.
**Question:** Is it true that $\mathbb E[A| \mathcal Y\_t] \to A$, and $\mathbb E[\theta| \mathcal Y\_t] \to \theta$ almost surely?
| https://mathoverflow.net/users/173490 | Filtering the period and amplitude of a sine wave corrupted by noise | By the Levy zero-one law, the question is equivalent to deciding whether $A$ and $\theta$ are measurable with respect to $\mathcal F\_\infty$, the common refinement of all the $\mathcal F\_t$. The answer is positive.
For $\alpha>0,\beta\ge 0$, consider
$$Z\_n=Z\_n(\alpha,\beta):=Y\_{\beta+(2n+1)\alpha }-Y\_{\beta+(2n-1)\alpha }=U\_n+X\_n$$
where, for $n=0,1,2,\ldots$, the sequence
$$U\_n:=A\sin\Bigl(\theta (\beta+(2n+1)\alpha)\Bigr)-A\sin\Bigl(\theta (\beta+(2n-1)\alpha)\Bigr)=2A\sin(\alpha \theta) \cdot \cos(\theta\beta+2n\theta\alpha)$$
is an almost periodic sequence obtained as a function of the (zero entropy) rotation by angle $2\alpha\theta$, and
$$X\_n:=W\_{\beta+(2n+1)\alpha }-W\_{\beta+(2n-1)\alpha }$$
is a sequence of i.i.d. Gaussian variables with mean 0 and variance $2\alpha$.
In [1], H. Furstenberg defined disjointness of dynamical systems and proved in Theorem I.2 that i.i.d. processes and zero-entropy processes are disjoint.
In Theorem 1.5 of the same paper, he showed that if $(U\_n)\_{n \ge 1}$ and $(X\_n)\_{n \ge 1}$ are sequences
of integrable real random variables which define two disjoint stationary
processes, then the sum sequence $(U\_n+X\_n)\_{n \ge 1}$ determines $(U\_n)\_{n \ge 1}$.
A generalization is in Theorem 7 of [2].
In particular, from the sequence $\{Z\_n(\alpha,\beta)\}\_{n \ge 1}$ we can obtain
$(U\_n)\_{n \ge 1}$ =$(U\_n(\alpha,\beta))\_{n \ge 1}$ for every $\alpha$ and a.e. $\beta$, and using continuity, for all $\beta$. Then $\theta=\pi/(4\alpha\_\*)$, where
$\alpha\_\*=\min\{\alpha>0: U\_1(\alpha,0)=0\}$ and then $A$ is easy to determine.
Theorem 1.6 in [3] gives another proof.
For an elementary direct argument,
define the deterministic function $f\_0(\alpha)=\lim\_n \frac{1}{n}\sum\_{k=1}^n Z\_n(\alpha,0)^2$
and observe that the smallest positive $\alpha$ where $f\_0(\alpha)\alpha^{-2}$ is minimized
is $\alpha\_{min}=\pi/\theta$. Moreover, $$f\_0(\alpha\_{min}/2)=4A^2+\alpha\_{min}$$
which yields $A$.
[1] H. Furstenberg (1967), Disjointness in ergodic theory, minimal sets, and a problem in
Diophantine approximation, Math. systems theory 1, pp. 1-49.
[https://mathweb.ucsd.edu/~asalehig/F\_Disjointness.pdf](https://mathweb.ucsd.edu/%7Easalehig/F_Disjointness.pdf)
[2] Furstenberg, Hillel, Yuval Peres, and Benjamin Weiss. "Perfect filtering and double disjointness." In Annales de l'IHP Probabilités et statistiques, vol. 31, no. 3, pp. 453-465. 1995.http://www.numdam.org/article/AIHPB\_1995\_\_31\_3\_453\_0.pdf
[3] Lev, Nir, Ron Peled, and Yuval Peres. "Separating signal from noise." Proceedings of the London Mathematical Society 110, no. 4 (2015): 883-931.
<http://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.749.2820&rep=rep1&type=pdf>
| 2 | https://mathoverflow.net/users/7691 | 414967 | 169,216 |
https://mathoverflow.net/questions/414867 | 4 | I am trying to find a standard reference for the natural analogue of root subgroups (and their properties) in twisted Chevalley groups.
Let me first recall the classical set-up. According to Steinberg's lecture notes, every simple untwisted Chevalley group $G$ can be obtained as follows. Consdider a finite-dimensional, complex simple Lie algebra $L$ together with a Chevalley basis and a finite field $\mathbb{F}$. Associate to each root $\alpha$ a one-parameter subgroup $x\_\alpha : \mathbb{F} \longrightarrow \operatorname{GL}(L) : t \mapsto x\_\alpha(t)$, called the root subgroup of $\alpha$, in the usual way (using the exponential of the ad-nilpotent matrices, interpreted over the field $\mathbb{F}$). Then each such $x\_\alpha$ is an injective homomorphism from the additive group $(\mathbb{F},+)$ of the field $\mathbb{F}$ to the group $\operatorname{GL}(L)$ and $G$ is (isomorphic to) the group that is generated by the elements $x\_\alpha(t)$ (where $\alpha$ ranges over the roots and where $t$ ranges over the elements of the field). These root subgroups are moreover compatible with field automorphisms in the obvious way.
The case of twisted Chevalley groups is more complicated to describe. We assume that the Dynkin diagram and the field allow us to define a distinguished automorphism $\sigma = \gamma \cdot \phi$ of the group $G$ in the usual way (with $\gamma$ a symmetry of the Dynkin diagram and with $\phi$ a suitable field automorphism), so that we may define the twisted Chevalley group associated to $G$ and $\sigma$ as $G^\sigma$, the subgroup of $G$ that is fixed elementwise by $\sigma$.
I would now like to have a collection of injective homomorphisms $y\_B : (\mathbb{F},+) \longrightarrow G^\sigma : t \mapsto y\_B(t)$ that are compatible with field automorphisms and that generate $G^\sigma$, as in the classical case. Here, the field $\mathbb{F}$ is the defining field, (or possibly the subfield $\mathbb{F}\_0$ of $\mathbb{F}$ that is fixed by the distinguished field automorphism $\phi$).
The obvious choice of $x\_\alpha$ does not work, since the $x\_\alpha$ need not be contained in $G^\sigma$. A natural, but slightly naive, approach would be to consider a $\langle \gamma \rangle$-orbit B and to then define
$y\_B : \mathbb{F}\_0 \longrightarrow G^\sigma : t \mapsto x\_\alpha(t)$ (for an orbit $B = \{ \alpha \}$ of length $1$),
$y\_B : \mathbb{F} \longrightarrow G^\sigma : t \mapsto x\_{\alpha}(t) \cdot x\_{\gamma^1(\alpha)}(\phi^1(t))$ (for an orbit $B = \{ \alpha , \gamma(\alpha) \}$ of length $2$),
$y\_B : \mathbb{F} \longrightarrow G^\sigma : t \mapsto x\_{\alpha}(t) \cdot x\_{\gamma^1(\alpha)}(\phi^1(t)) \cdot x\_{\gamma^2(\alpha)}(\phi^2(t))$ (for an orbit $B = \{ \alpha , \gamma(\alpha) , \gamma^2(\alpha) \}$ of length 3).
>
> **Question:** *Is there a standard text that constructs a collection of one-parameter
> subgroups $t \mapsto y\_B(t)$ of $G^\sigma$ satisfying the following properties?*
>
>
>
* **Property 1:** Each $y\_B$ is an injective homomorphism.
* **Property 2:** The group $G^\sigma$ is generated by the elements $y\_B(t)$.
* **Property 3:** The one-parameter groups $y\_B$ are compatible with field automorphisms, in the sense that if the field automorphism $A$ is induced by the automorphism $a$ of the field $\mathbb{F}$, then $A(y\_B(t)) = y\_B(a(t))$ for all $t \in \mathbb{F}$.
Assuming that the $y\_B$ are indeed defined using the above $\langle \sigma \rangle$-orbits, Property 1 would *seem* to follow from the "uniqueness of expression" in the product formulae for the classical case, while Property 3 would follow immediately.
Gorenstein and Lyons make a vague reference to such a construction in [The local structure of finite groups of characteristic 2 type: Part I, Chapter I, Paragraph 4]. It also looks like E. Abe has indeed tried "something along those lines" in [Coverings of twisted Chevalley groups over commutative rings. Sci. Rep. Tokyo Kyoiku Daigaku Sect. A 13 (1977), no. 366-382, 194–218.] But the text is difficult to understand and I am still unsure about the conventions / assumptions / generality of it.
Any help would be appreciated.
| https://mathoverflow.net/users/203598 | Twisted root subgroups in twisted Chevalley groups (reference request) | Tom's answer is the complete story with two kinds of exceptions: (a) $G^\sigma$ is a Suzuki-Ree-type group -- ${^2}B\_2(2^a)$, ${^2}F\_4(2^a)$, or ${^2}G\_2(3^a)$; or (b) there exists a root $a\_0$ such that $a\_0+\gamma(a\_0)$ is again a root.
In case (a), the symmetry of order $2$ of the Dynkin diagram leads to an automorphism of $G$ which is not in general of order $2$, but can be composed with a suitable field automorphism to give the desired automorphism $\sigma$ of order $2$. This process is described in Steinberg's lecture notes.
In case (b), $\sigma$ arises as you describe, but Property 1 fails: $\gamma\_B$ is not a homomorphism for the orbit $B$ containing $a\_0$. This is an issue for ${^2}A\_{2n}(q)$.
Actually, case (a) is a subcase of case (b), but I've separated out the Suzuki-Ree-type groups since their construction has extra complications.
As a result there are actually 7 types of "twisted root groups," of which you have described three. Three of the other four are not abelian. They are all $p$-groups, where $p$ is the characteristic of the underlying fields.
A fourth type occurs in the odd-dimensional unitary groups, when $a\_0$ and $\gamma(a\_0)$ are fundamental roots in an $A\_2$-subsystem of the root system of $G$. (You can see from the Dynkin diagram that there is at least one such $a\_0$.) This type has nilpotence class $2$. A fifth type gives the Sylow $2$-subgroups of the Suzuki groups ${^2}B\_2(q)$, and also appears in the groups ${^2}F\_4(q)$. It too has class $2$, with one small exception. A sixth type is abelian and also occurs in ${^2}F\_4(q)$, and the seventh type, of class $3$ with one small exception, gives the Sylow $3$-subgroups of the Ree groups ${^2}G\_2(q)$.
This is discussed in some detail in Chapter 2, particularly in Section 2.4, of: D. Gorenstein, R. Lyons, and R. Solomon, The Classification of the Finite Simple Groups, Number 3. Part I, Chapter A: Almost Simple $K$-Groups. Mathematical Surveys and Monographs vol. 40, American Mathematical Society, 1998.
And Properties 2 and 3 remain true in all cases.
| 2 | https://mathoverflow.net/users/99221 | 414968 | 169,217 |
https://mathoverflow.net/questions/414965 | 0 | Let $u,v,x \in \mathbb R^d$ be three unit vectors. I found a very complicated proof that $(v^Tx)^2-(u^Tx)^2 \leq 1-(u^Tv)^2$.
That is $\lVert uu^T-vv^T\rVert^2\_2 = 1-(u^Tv)^2$, or that $f(v,x)\leq f(v,u)+f(u,x)$ where $f(v,u)=1-(v^Tu)^2$ is the squared sine of the angle between $u$ and $v$.
Is there a one-liner proof?
Say, using the (spherical?) law of cosines or the Haversine formula? Induced norm for positive semi-definite matrices?
Edit: Thank you all for the quick answers. I am confused by the counter examples. I tried to cite Lemma 27 in a paper (I think from STOC'15):
<https://arxiv.org/abs/1606.05225>
The eigenvalue of $||uu^T-vv^T||\_2$ seems to be correct by
[How to find the eigenvalues of $xx^T-yy^T$](https://math.stackexchange.com/questions/3913336/how-to-find-the-eigenvalues-of-x-xt-y-yt)
Edit 2: I assumed that all the conjectures in the question are equivalent but maybe I was wrong. I took the Schur Decomposition $USU^T$ of $uu^T-vv^T$ to get $\max\_{||x||=1} ||(uu^T-vv^T)x||^2=||USU^Tx||^2=S\_{1,1}^2$ and assumed it is the same $\max\_{||x||=1} |x^T(uu^T-vv^T)x|^2=|x^TUSU^Tx|=S\_{1,1}^2$. Then I noticed that $|x^T(uu^T-vv^T)x|^2=|(x^Tu)^2-(v^Tx)|$. Not sure what went wrong.
Summary: As GH from Mo noted below, I forgot a squared root in the right hand side and the statement is wrong. Hope to get your help also in the fixed version [here](https://mathoverflow.net/questions/417022/prove-that-vtx2%E2%88%92utx2-leq-sqrt1%E2%88%92utv2-for-any-unit-vectors-u-v)
| https://mathoverflow.net/users/476110 | Prove that $(v^Tx)^2-(u^Tx)^2 < 1-(u^Tv)^2$ for any unit vectors $u$, $v$, $x$ | The inequality as stated is false, but it is true that
$$\langle v,x\rangle^2+\langle v,u\rangle^2\leq 1+|\langle x,u\rangle|.$$
Moreover, the right-hand side is optimal in the sense that it is the maximum of the left-hand side over the unit vectors $v$. More generally, if $u\_1,\dots,u\_R$ are any vectors in a Hilbert space, then
$$\max\_{\langle v,v\rangle=1}\sum\_{r=1}^R|\langle v,u\_r\rangle|^2$$
equals the largest eigenvalue of the Gram matrix $(\langle u\_s,u\_t\rangle)\_{1\leq s,t\leq R}$. For details, see [these notes](https://users.renyi.hu/%7Egharcos/bombieri-halasz-montgomery.pdf) on the Bombieri-Halász-Montgomery inequality.
| 3 | https://mathoverflow.net/users/11919 | 414969 | 169,218 |
https://mathoverflow.net/questions/414961 | 1 | The generator an OU process is given by
$$A = \operatorname{tr}(QD^2)+\langle Bx,D\rangle.$$
This one possesses an invariant measure given by
$$d\mu(x) = b(x) \ dx \text{ with } b(x) = \frac{1}{(4\pi)^{N/2} \vert Q\_{\infty} \vert^{1/2}} e^{-\langle Q\_{\infty}^{-1}x,x \rangle /4},$$
where $Q\_{\infty}= \int\_0^{\infty} e^{sB} Q e^{sB^\*} \ ds.$
I read that $A$ is self-adjoint on $L^2(\mathbb R^n, d\mu)$ if and only if $QA^\*=AQ,$ but I have never seen an argument for this.
In particular, what I am mainly interested in, is there anywhere an explicit representation of the adjoint operator of $A$?
| https://mathoverflow.net/users/150549 | Adjoint operator of OU generator | The argument is a bit tricky. Assume for simplicity that $Q=I$, then $BQ\_\infty+Q\_\infty B^\*=-I$. Next decompose $A$ as a self-adjoint part plus a remainder, namely introduce the form $a(u,v)=\int\_{\mathbb R^n} \nabla u \nabla v \, d\mu$ which corresponds to $A\_1=\Delta-\frac 12 Q\_\infty ^{-1} x \cdot \nabla$ and write $A=A\_1+C$ with $C=(Bx+\frac 12 Q\_\infty^{-1} x) \cdot \nabla$.
A computation shows that $C$ is skew-adjoint in $L^2\_\mu$, so that $A$ is self-adjoint iff $C=0$ or $B=-\frac 12 Q\_\infty^{-1}$, in particular $B=B^\*$. On the other hand, if $B=B^\*$, we can diagonalize $B$ and check that $B=-\frac12 Q\_\infty^{-1}$ so that $A=A\_1$ is self-adjoint. This answers to the first question. Concerning the second, $A^\*=A\_1-C$, but I dooubt that one can compute it explicitely without knowing $Q\_\infty$.
| 1 | https://mathoverflow.net/users/150653 | 414972 | 169,219 |
https://mathoverflow.net/questions/414973 | 17 | [Van der Waerden's theorem](https://en.wikipedia.org/wiki/Van_der_Waerden%27s_theorem) states that any colouring of the integers in a finite number of colours has monochromatic arithmetic progressions of arbitrary length. [Szemerédi's Theorem](https://en.wikipedia.org/wiki/Szemer%C3%A9di%27s_theorem) is a dramatic strengthening of this result and says that any set of positive integers with positive natural density contains arbitrary length arithmetic progressions. Clearly the latter theorem implies the former but not the other way round.
The proof of Van der Waerden's theorem although elementary is not simple and perhaps quite challenging to discover. However it would follow from the following theorem:
>
> **Theorem:**
> Let $A$ be subset of the integers such that $|A\cap [n]|\geq n/2$ for all $n>0$. Then $A$ contains an arithmetic progression of arbitrary length.
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>
>
This is weaker than Szemeredi's theorem but seems only slightly stronger than Van der Waerden. Hence my question:
>
> **Is there a proof of the above theorem that is substantially simpler than the proof of Szemeredi's Theorem?**
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>
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| https://mathoverflow.net/users/7113 | A proof of Van der Waerden's theorem using a weakened form of Szemeredi's theorem | As (implicitly) observed already in Szemerédi's celebrated paper
*Szemerédi, Endre*, [**On sets of integers containing no (k) elements in arithmetic progression**](http://dx.doi.org/10.4064/aa-27-1-199-245), Acta Arith. 27, 199-245 (1975). [ZBL0303.10056](https://zbmath.org/?q=an:0303.10056).
(and perhaps previously), Szemerédi's theorem for a fixed density $0 < \delta\_0 < 1$ (such as $\delta\_0 = 1/2$), when combined with van der Waerden's theorem, implies Szemerédi's theorem for arbitrary density $\delta > 0$. This is because, once one is given a subset $A$ of integers in $\{1,\dots,N\}$ of density $\delta$, it is not difficult to use the probabilistic method to find $O\_{\delta,\delta\_0}(1)$ translates of $A$ (by shifts randomly selected between $-N$ and $N$) which cover this interval to density at least $\delta\_0$. If one can find a sufficiently long arithmetic progression inside this union of translates, then by van der Waerden's theorem, at least one of these translates also contains a long progression, which gives Szemerédi's theorem for that density $\delta$.
As a consequence of this argument, the gap in difficulty between Szemerédi's theorem for a fixed density $0 < \delta\_0 < 1$ (but arbitrary lengths $k$) and for arbitrary densities (and arbitrary lengths) is basically no greater than the difficulty required to prove van der Waerden's theorem (which can be proved in a page or two).
EDIT: the situation is very different if instead one fixes the length $k$ of the progression. As pointed out in comments, Szemerédi's theorem is now easy for very large densities such as $\delta > 1-1/k$, and the difficulty increases as the density lowers (although several proofs of Szemerédi's theorem proceed by a downward induction on density now commonly known as the **density increment argument**). However, in most proofs, the increase in difficulty as $\delta$ decreases is negligible compared to the increase in difficulty as $k$ increases; for instance the $k=3$ case of the theorem, first established by Roth, is substantially easier than the $k>3$ cases. So the van der Waerden reduction given above, which trades the small-$\delta$ difficulty for the large-$k$ difficulty, is generally not useful in practice (in particular, it is largely incompatible with any attempt to induct on $k$, which tends to be a key component of most approaches to this theorem).
| 36 | https://mathoverflow.net/users/766 | 414978 | 169,220 |
https://mathoverflow.net/questions/414981 | 1 | Let $X$ be a one dimensional Ito diffusion given by
$$X\_t = b \,W\_t$$
where $b$ is a constant, and $W$ is a standard Brownian motion.
Let $B$ be another Brownian motion independent of $W$, and define the *observation process* $Y$ by
$$Y\_t = X\_t + B\_t.$$
Fix $T > 0$. A choice of *sampling times* is simply a choice of real numbers $0 \leq t\_1 \leq \dots \leq t\_n \leq T$.
**Question:** For fixed $n > 1$, what choice of sampling times $t\_1, \dots, t\_n$ minimises the expression
$$\mathbb E\left [ |\mathbb E[X\_T| \sigma(Y\_{t\_1}, \dots, Y\_{t\_n})] - X\_T | \right ]?$$
Where $\sigma(Y\_{t\_1}, \dots, Y\_{t\_n})$ denotes the sigma algebra generated by the $Y\_{t\_i}$.
**Remark:** It is not certain that there exists a minimiser - to prove existence first, it would suffice to show that the given function is continuous in the $t\_i$ and apply compactness.
| https://mathoverflow.net/users/173490 | What are the optimal times to sample a process? | Write $$Z\_t = W\_t - b B\_t,$$ so that $Y\_t$ and $Z\_t$ are independent Brownian motions, $$X\_t = b W\_t = b \cdot \frac{b Y\_t + Z\_t}{1 + b^2} \, ,$$ and the question asks for the distance between $X\_T$ and
$$ \mathbb E[X\_T | \sigma(Y\_{t\_1},\ldots,Y\_{t\_n})] = b \cdot \frac{b Y\_{t\_n} + 0}{1 + b^2} \, . $$
This distance is of course
$$ \frac{b}{1 + b^2} \mathbb E[|b (Y\_{t\_n} - Y\_T) - Z\_T|] , $$
which is minimised when $t\_n = T$.
| 3 | https://mathoverflow.net/users/108637 | 414985 | 169,222 |
https://mathoverflow.net/questions/410224 | 2 | **Short version.** If $X$ is a diffusion with generator $L$ and the Lebesgue measure is invariant for $X$, then $L^\*$ has no term of order zero and it corresponds to another diffusion $X^\*$. Denoting by $p$ and $p^\*$ their kernels,
>
> do we have $p^\*\_t(x,y)=p\_t(y,x)$?
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>
>
I am interested in full answers, but I would assume there is a reference for this.
**In details.** Let $X$ be a diffusion, solution to
$$\mathrm dX\_t = b(X\_t)\mathrm dt + \sum\_i\sigma\_i(X\_t)\mathrm dW^i\_t.$$
I am in a somewhat regular case: $b$ and all the $\sigma\_i$ are smooth, the diffusion is Feller (the semigroup acts on continuous functions vanishing at infinity) and hypoelliptic, i.e. the vector fields satisfy the [parabolic Hörmander condition](https://en.wikipedia.org/wiki/H%C3%B6rmander%27s_condition) when written in Stratonovich form (so the kernel is well-defined and smooth). I am also interested in cases with stronger hypotheses.
This diffusion has generator $L:f\mapsto \sum\_kb^k\partial\_kf+\frac12\sum\_i\sum\_{k\ell}\sigma\_i^k\sigma\_i^\ell\partial\_{k\ell}f$, with (formal) adjoint
$$ L^\*:f\mapsto
- \sum\_k\partial\_k\big(b^kf\big)+\frac12\sum\_i\sum\_{k\ell}\partial\_{k\ell}\big(\sigma\_i^k\sigma\_i^\ell f\big)
= \frac12\sum\_i\sum\_{k\ell}\sigma\_i^k\sigma\_i^\ell\partial\_{k\ell}f + (\text{lower order}).$$
If $L^\*1=0$, i.e. there are no terms of order zero (the Lebesgue measure is preserved), this $L^\*$ is the generator of another diffusion
$$\mathrm dX^\*\_t = b^\*(X\_t)\mathrm dt + \sum\_i\sigma\_i(X\_t)\mathrm dW^i\_t$$
with the same noise components. (I am abusing notation a bit, in the sense that $b^\*$ is not the adjoint of $b$, since the former depends on the $\sigma\_i$.)
Since $L$ is hypoelliptic, $X$ started at $x$ admits a smooth kernel $(t,y)\mapsto p\_t(x,y)$, i.e.
$$\mathbb P\_x(X\_t\in A)=\int\_Ap\_t(x,y)\mathrm dy.$$
We can check (easier in Stratonovich form) that hypoellipticity of $L$ equivalent to that of $L^\*$, so $X^\*$ also has a smooth kernel $p^\*$.
>
> **Question.**
> Is it true that for any $f,g$ smooth with compact support (say), we have
> $$ \int f(x)p^\*\_t(x,y)g(y)\mathrm dx\mathrm dy = \int f(x)p\_t(y,x)g(y)\mathrm dx\mathrm dy? $$
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>
>
I call this a question about time reversals because it should be true that $t\in[0,T]\mapsto X\_{T-t}$ has the same distribution as $t\in[0,T]\mapsto X^\*\_t$ when the distribution of the initial point is the Lebesgue measure (whatever that means), so $p\_t(y,x)$ is the time reversal of the kernel $p\_t(x,y)$ of $X$, and $p^\*$ is the kernel of the time reversal $X^\*$ of $X$.
[Haussmann and Pardoux](https://projecteuclid.org/journals/annals-of-probability/volume-14/issue-4/Time-Reversal-of-Diffusions/10.1214/aop/1176992362.full) (open access) prove results about time reversals that can be used to answer my question in the positive if we have instead a finite equilibrium measure $\mu$ (i.e. $L^\*\mu=0$ when we consider the adjoint with respect to $\mu$).
**Formal proof.** At the formal level, I see the following reasoning as evidence that the result should be true.
Let $f,g$ be nice functions, and consider the derivative of the integral
$$I:s\in(0,t)\mapsto \int p^\*\_s(x,y)p\_{t-s}(z,y)g(z)\mathrm dy.$$
The Fokker-Planck equation asserts that $\partial\_tp = L^\*\_yp$ and $\partial\_tp^\* = L^{\*\*}\_yp^\* = L\_yp^\*$ ($y$ is the second variable), so
$$I'(s) = \int\Big((L\_yp^\*)\_s(x,y)p\_{t-s}(z,y) - p^\*\_s(x,y)(L^\*\_yp)\_{t-s}(z,y)\Big)\mathrm dy = 0 $$
since $L^\*$ is the adjoint of $L$ (this is more or less the same as saying that $s\mapsto p\_{t-s}(z,X\_s)$ is a local martingale, for any initial condition $X^\*\_0=x$ and point $z$, although this latter fact is actually true and easy to prove).
Hence we must have $I(0)=I(t)$ with
$$ I(0) = \int \delta\_x(y)p\_t(z,y)\mathrm dy
= p\_t(z,x) $$
and
$$ I(t) = \int p^\*\_t(x,y)\delta\_z(y)\mathrm dy
= p^\*\_t(x,z). $$
Maybe there are standard arguments that can prove the derivative and limits taken above are valid, but I have not been able to find them either myself or in the literature.
**Edit: The *dual process* approach.**
If I am not mistaken, one approach suggested by Mateusz Kwaśnicki in his answer is to go in the direction $X\mapsto\hat X\mapsto \hat L$ rather than $X\mapsto L^\*\mapsto X^\*$. It would be sufficient to carry through the following steps.
1. Prove that there exists a dual process $\hat X$ (more or less the resolvent of the processes are in duality), in the sense of section 13.2 of [CW].
2. Prove that the semigroup of $\hat X$ is given by $P\_tf(x)=\int p\_t(y,x)f(y)\mathrm dy$.
3. Prove that $\hat X$ satisfies the dual SDE, so in fact $\hat X=X^\*$.
These all come with some level of difficulty, but there are a few results in [CW] that may help. For instance, for 1. we can use Theorem 15.5, and all that remains to show is that the Lebesgue measure is invariant (in the strong sense $\int\mathbb P\_x(X\_t\in A)\mathrm dx=\mathrm{leb}(A)$) and that there exists a “cofine topology” in the sense of Definition 15.4. Point 2. is very close to the duality of the semigroups, as in Proposition 13.6, and point 3., once we have point 2., reduces to having enough functions in the domain of $\hat L$.
Although the route seems sound, altogether the results in [CW] look very measure-theoretic to me (unsurprisingly) and I am having trouble relating them to the smooth quantities I am considering.
[CW] Chung, Walsh, [Markov Processes, Brownian Motion, and Time Symmetry](https://doi.org/10.1007/0-387-28696-9).
| https://mathoverflow.net/users/129074 | Is $p_t(y,x)$ the kernel of the time reversal of the diffusion $X$, for $p_t(x,y)$ the kernel of $X$? | This is an immediate corollary of Theorem 3.50 in
>
> D. W. Stroock, [An introduction to the analysis of paths on a Riemannian manifold](https://bookstore.ams.org/surv-74-s) (2000).
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> Mathematical Surveys and Monographs. 74. Providence, RI: American Mathematical Society (AMS). xvii, 269 p.
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>
See also Theorem 4.17 for a manifold version of it.
| 1 | https://mathoverflow.net/users/129074 | 415001 | 169,225 |
https://mathoverflow.net/questions/414940 | 1 | I've posted this question [Tangent space of $G \times\_H M$](https://math.stackexchange.com/questions/4365925/tangent-space-of-g-times-h-m) in MSE, but didn't get any answer. My question is the following:
Let $G$ be a Lie group and let $H$ be a Lie subgroup of $H$. Let $M$ be a smooth manifold on which $H$ acts from the left.
Let's consider the action of $H$ on $G \times M$: $$h((g,m)):= (gh,h^{-1}m), \quad h \in H , g \in G , m \in M, $$ and define the manifold $Z$ to be the quotient $G \times\_H M$.
If we fix $(g,m) \in G \times M$, what is the induced equivalence relation on $T\_gG \times T\_mM$?
(My background is not so good in differential geometry, so please be patient with me.)
| https://mathoverflow.net/users/172459 | The tangent space of $G \times _H M$ | When moving to the tangent space level, the general rule is to take the derivative (of everything you can). Just from this I'd expect something like the equivalence on $T\_gG\oplus T\_mM$ should be such that $gX - Xm = 0$ for every $X\in \mathfrak h$.
(As a side note, it may be better to write $(g,m).h$ because it's a right action. Or at least change it to $h.(g,m):=(gh^{-1},hm)$. I will stick with the way you have it for now.)
In detail, we want to explicitly describe the kernel of $\mathrm d \pi$, since this is exactly the $H$ relation you seek.
Let $h(t) = \exp(tX)$ be a path in $H$ with tangent vector $X\in \mathfrak h$, and consider the corresponding path in $G\times M$, $h(t).(g,m) = (gh(t),h(t)^{-1}m)$. The tangent vector is $gX - Xm\in T\_gG\oplus T\_mM$.
This whole path is projected to the same point $[(g,m)]\in Z$, so its tangent vector is in the kernel of $\mathrm d\pi$. Also since you are just quotienting by $H$ to get to $Z$, it is exactly these types of paths which make up the whole kernel. You can also convince yourself by counting dimensions. You should have exactly $\dim(H)$ conditions for the equivalence.
| 3 | https://mathoverflow.net/users/145436 | 415020 | 169,231 |
https://mathoverflow.net/questions/415021 | 4 | Let $a,b \in \mathbb R$ and consider the functional $J$ on $X$:
$$J[u] = \int\_0^1 \left( (u'(x))^2 -a)^2 + b \ln (1+ u^2(x))\right) dx$$
Providing reasons specify if the $\inf J$ over $X$ is attained of not
1. $a>0, b \geq 0, X =\{ u \in C^1([0,1]); u(0)=0, u(1)=0 \}$
2. $a>0, b > 0, X =\{ u \in AC([0,1]); u(0)=0, u(1)=0 \}$
3. $a<0, b \geq 0, X =\{ u \in AC([0,1]); u(0)=0, u(1)=1 \}$
**My proof**
I am trying to apply the following theorem to find the answer.
**Special version of Tonelli’s theorem**
Assume that the function $f(x,u): [a,b] \times \mathbb{R} \to \mathbb{R}$, $g(x, \xi): [a,b] \times \mathbb{R} \to \mathbb{R}$ are continuous, $f$ is bounded below, $g$ is convex in $\xi$ and satisfies
$$\exists r>1,\, \exists C>0\,\, \text{such that}\,\, g(x,\xi) \ge C| \xi|^r,\,\, \forall (x, \xi) \in [a,b] \times \mathbb{R}.$$
Then there exists a minimizer of the functional
$$J[u] = \displaystyle\int\_a^b (f(x,u(x)) + g(x,u'(x))) dx$$ in the space $X= \{ u \in AC([a,b]); u(a)=\alpha, u(b)= \beta \}.$
Here I considered $f(x,u(x)) = b \ln (1+u^2(x))$ and $g(x,u'(x))= (u'(x)^2 -a)^2$. Am I in the right path? I am not sure on what should I do next.
| https://mathoverflow.net/users/471464 | Which set of functions admits the existence of the minimizer? | The answer to (3) is yes. Indeed, then all the conditions of what you call "special version of Tonelli’s theorem" (proved in [this answer](https://mathoverflow.net/a/413645/36721)) are satisfied.
The answer to (2) is no. Indeed, for natural $n$ and $x\in[0,1]$, let
$$u\_n(x):=\sqrt a\,d(x,E\_n),$$
where $E\_n$ denotes the set $\{0,\frac1n,\frac2n,\dots,\frac nn\}$ and $d(x,E\_n)$ denotes the shortest distance from $x$ to the set $E\_n$. Then $u\_n\in AC[0,1]$ and
$$J[u\_n]=b\int\_0^1\ln(1+u\_n^2(x))\, dx\le b\ln(1+a/(4n^2))\to0$$
as $n\to\infty$. Since $J[\cdot]\ge0$, it follows that $\inf J[\cdot]=0$. However, this zero infimum is not attained at any $u$ -- if $J[u]=0$, then $u=0$ and hence $J[u]=a^2>0$.
The answer to (1) is no as well. This follows because the $u\_n$'s as above can be appropriately approximated by $C\_1$ functions.
| 3 | https://mathoverflow.net/users/36721 | 415026 | 169,232 |
https://mathoverflow.net/questions/415012 | 5 | In the previous post [What is the smallest set of real continuous functions generating all rational numbers by iteration?](https://mathoverflow.net/q/412923/7113) I asked for the smallest set of continuous real functions that could generate $\mathbb Q$ by iteration starting from $0$. Surprisingly one continuous function suffices and this can also be Lipschitz and possibly analytic.
Given that we are applying the function only to rational numbers to generate our sequence the question arises whether it is possible that the function restricted to the rationals is also computable:
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> Is there a continuous function $f:\mathbb{R}\rightarrow \mathbb{R}$ that generates $\mathbb{Q}$ by iteration where $f$ restricted to $\mathbb{Q}$ is computable?
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| https://mathoverflow.net/users/7113 | Is $\mathbb{Q}$ the orbit of a continuous function that is computable when restricted to $\mathbb{Q}$? | Yes. This answer is based on the answers to your previous question.
Start with a computable ergodic map $T$ (D. Thomine constructs an example [here](https://mathoverflow.net/a/404564)). For every basic open neighborhood $B\_i$, the set $D\_i = \bigcup\_n T^{-n}(B\_i)$ is a dense effectively open set, uniformly in $i$. So some computable real $p$ meets all the $D\_i$, meaning $X = \{ T^n(p) : n \in \omega\}$ is dense.
There is a computable, order-preserving bijection $g: X \to \mathbb{Q}$ (back-and-forth argument), and we may assume that $g(p) = 0$. $g$ induces a (bi-computable) homeomorphism $G: \mathbb{R} \to \mathbb{R}$.
Define $f = G\circ T\circ G^{-1}$.
---
Constructing the bijection:
We'll build a computable sequence of rationals $(q\_n)\_{n \in \omega}$ and define $g(T^n(p)) = q\_n$.
Begin with $q\_0 = 0$. Then compute enough of $p$ and $T(p)$ to determine how they are ordered ($p < T(p)$ or $T(p) < p$). If $p < T(p)$, choose $q\_1$ to be a rational greater than 0; otherwise, choose $q\_1$ to be a rational less than 0. In either case, choose $q\_1$ to be the appropriate rational with the smallest Gödel number.
Then compute enough of $p$, $T(p)$ and $T^2(p)$ to determine how they are ordered, and choose $q\_2$ to be a rational in the same relative position to $q\_0$ and $q\_1$. Again, choose the appropriate rational of smallest Gödel number.
Etc.
This is all a computable process, so $g$ is computable. It's a total order-preserving injection by construction.
Surjectivity is by induction on Gödel number. For a rational $r$, by the inductive hypothesis all rationals of smaller Gödel number are in the range of $g$. So fix an $n$ such that all rationals of smaller Gödel number occur in $q\_0, \dots, q\_{n-1}$, and assume $r$ does not occur in this list, as otherwise we are done. Define $C = \{i < n : q\_i < r\}$. By the density of $X$, there is an $m \ge n$ with $T^i(p) < T^m(p)$ for all $i \in C$, and $T^m(p) < T^i(p)$ for all $i < n$ with $i \not \in C$. Fix the least such $m$. Then by construction, $q\_m = r$.
| 6 | https://mathoverflow.net/users/32178 | 415027 | 169,233 |
https://mathoverflow.net/questions/415007 | 23 | Did anyone ever consider a "function" or "distribution" $F(x)$ with the following property:
its integral $\int\_a^b F(x)\,dx=0$ for any finite interval $(a,b)$ but $\int\_{-\infty}^\infty F(x)\,dx=1$? It definitely can be seen as a limit of smooth functions, and its Fourier transform will be a finction, equal to $1$ at $0$ and otherwise $0$.
I think such object for sure should be considered and named by someone.
| https://mathoverflow.net/users/10059 | Anti-delta function? | There's quite a bit on this in Harold Jeffreys's unusual book *Theory of Probability*, in which such a function is one of the more prominent examples of “improper priors.” In [Robert, Chopin, and Rousseau – Harold Jeffreys's Theory of Probability Revisited](https://arxiv.org/abs/0804.3173) we find this:
>
> For a 21st century reader, Jeffreys’s *Theory of Probability* is nonetheless puzzling for its lack of formalism, including its difficulties in handling improper priors, its reliance on intuition, its long debate about the nature of probability, and its repeated attempts at philosophical justifications. The title itself is misleading in that there is absolutely no exposition of the mathematical bases of probability theory in the sense of Billingsley (1986) or Feller (1970): “Theory of Inverse Probability” would have been more accurate. In other words, the style of the book appears to be both verbose and often vague in its mathematical foundations for a modern reader.
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Consider the upper half-plane $\{(m,s): -\infty<m<+\infty,\, 0<s<+\infty\}$ with the measure $dm\,ds/s.$ Pretend that when multiplied by some infinitely small constant (“constant” = not depending on $m$ or $s$) this becomes a probability measure on the upper half-plane. Now suppose that the conditional probability distribution of $X\_1,\dotsc,X\_n$ given $m$ and $s$ is that they are independent and normally distributed with expected value $m$ and variance $s$ (variance, not standard deviation). What then is the conditional distribution of $(m,s)$ given $X\_1,\dotsc,X\_n$? It is a perfectly ordinary probability distribution. Jeffreys proposes that that is the conditional probability distribution of the expectation and variance of a population from which $X\_1,\dotsc,X\_n$ is a random sample, when one has no other information about their values than what one gets from that sample.
Another example is the measure $c\,dp/\bigl(p(1-p)\bigr)$ whose integral over $(0,1)$ is $1$. Suppose $X\_1,\dotsc,X\_n$ are conditionally independent given $p$ and $\Pr(X\_k=1\mid p) = p$ and $\Pr(X\_k=0\mid p)=1-p$ for $k=1,\dotsc,n$. Then the conditional distribution of $p$ given $X\_1,\dotsc,X\_n$ is $\text{constant}\times p^{X\_1+\dotsb+X\_n-1}(1-p)^{n-(X\_1+\dotsb+X\_n)-1} \, dp,$ which is a perfectly ordinary probability distribution except when $X\_1+\dotsb+X\_n\in\{0,n\}$.
I think Jeffreys's doctorate was in mathematics, and he was a professor of astronomy for some time, and spent a lot of time on seismology and earth sciences, and in the '50s claimed to be able to tell the difference between seismic waves resulting from nuclear tests and those from earthquakes when the U.S. government was denying that that could be done.
The physicist Edwin Jaynes defended improper priors in a number of his writings.
| 22 | https://mathoverflow.net/users/6316 | 415029 | 169,234 |
https://mathoverflow.net/questions/415022 | 17 | In Cohen's article, [The Discovery of Forcing](https://projecteuclid.org/journals/rocky-mountain-journal-of-mathematics/volume-32/issue-4/The-Discovery-of-Forcing/10.1216/rmjm/1181070010.full), he says that "one cannot prove the
existence of any uncountable standard model in which AC holds, and
CH is false,"
and offers the following proof.
>
> If $M$ is an uncountable standard
> model in which AC holds, it is easy to see that $M$ contains all countable ordinals. If the axiom of constructibility is assumed, this means that
> all the real numbers are in $M$ and constructible in $M$. Hence CH holds.
>
>
>
But this argument, on the surface of it, invokes $V = L$. Can we eliminate the use of $V = L$? The discussion in a [related MO question](https://mathoverflow.net/questions/151234) seems to come close to answering this question, but doesn't directly address it.
| https://mathoverflow.net/users/3106 | Must uncountable standard models of ZFC satisfy CH? |
>
> Remarks (2) and (3) are added in this edit.
>
>
>
What Cohen's quoted proof outline is leaving implicit is the following statement in which $\mathrm{Con}(T)$ means "$T$ is consistent".
$(\*)$ Assuming $\mathrm{Con(ZF + SM)}$, $\mathrm{V} \neq \mathrm {L}$ is not provable from $\mathrm{ZF + SM}$, where $\mathrm{SM}$ stands for the statement "there is standard (i.e., well-founded) model of ZF".
$(\*)$ is an immediate consequence of the the well-known fact that $\mathrm{Con(ZF + SM)}$ implies $\mathrm{Con(ZF + SM + V = L)}$. This well-known fact, in turn, follows from absoluteness considerations: if $\mathcal{M}\models \mathrm{ZF + SM}$, then $\mathrm{L}^{\mathcal{M}} \models \mathrm{ZF + SM+V=L}$, where $\mathrm{L}^{\mathcal{M}}$ is the constructible universe as computed in $\mathcal{M}$.
By the way: The quoted statement of Cohen in his article is phrased as the theorem below on pages 108-109 of his book "Set Theory and the Continuum Hypothesis". In Cohen's terminology SM stands for the statement "there is standard (i.e., well-founded) model of $\mathrm{ZF}$".
**Theorem.** *From* $\mathrm{ZF + SM}$ *or indeed from any axiom system containing* $\mathrm{ZF}$ *which is consistent with* $\mathrm{V = L}$, *one cannot prove the existence of an uncountable standard model in which* $\mathrm{AC}$ *is true and* $\mathrm{CH}$ *is false, nor even one in which* AC *holds and which contains nonconstructible real numbers*.
Three remarks are in order:
**Remark (1)** In unpublished work, Cohen and Solovay noted that one can use forcing over a countable standard model of ZF to build *uncountable* standard models of $\mathrm{ZF}$ (in which AC fails by Cohen's aforementioned result). Later, Harvey Friedman extended their result by showing that every countable standard model of $\mathrm{ZF}$ of (ordinal) height $\alpha$ can be generically extended to a model with the same height but whose cardinality is $\beth\_{\alpha}$ (*Friedman, Harvey*, [**Large models of countable height**](http://dx.doi.org/10.2307/1997333), Trans. Am. Math. Soc. 201, 227-239 (1975). [ZBL0296.02036](https://zbmath.org/?q=an:0296.02036)).
**Remark (2)** It is easy to see (using the reflection theorem and relativizing to the constructible universe) that, assuming the consistency of $\mathrm{ZF + SM}$, the theory $\mathrm{ZF + SM}$ + "there is no uncountable standard model of $\mathrm{ZFC}$" is also consistent.
**Remark (3)** Within $\mathrm{ZF}$ + "there is an uncountable standard model $\mathcal{M} \models \mathrm{ZFC+V=L}$ such that $\omega\_3^{\mathcal{M}}$ is countable", one can use forcing to build a generic extension $\mathcal{N}$ of $\mathcal{M}$ that violates $\mathrm{CH}$; thus $\mathcal{N}$ *is an uncountable standard model of* $\mathrm{ZFC + \lnot CH}$. More specifically, the assumption of countability of $\omega\_3^{\mathcal{M}}$, and the fact that GCH holds in $\mathcal{M}$, assures us that there exists a $\mathbb{P}$-generic filter over $\mathcal{M}$, where $\mathbb{P}$ is the usual notion of forcing in $\mathcal{M}$ for adding $\omega\_2$ Cohen reals. Thus, in the presence of the [principle](https://en.wikipedia.org/wiki/Zero_sharp) "$0^{\sharp}$ exists" (which is implied by sufficiently large cardinals, and implies that every definable object in the constructible universe is countable) there are lots of uncountable standard models of $\mathrm{ZFC + \lnot CH}$ .
| 23 | https://mathoverflow.net/users/9269 | 415033 | 169,235 |
https://mathoverflow.net/questions/415050 | 1 | Do there exist continuous functions $c\_n\colon\mathbb R^+\to\mathbb R$ and $\gamma\colon\mathbb R^+\to\mathbb R$ such that $\lim\_{x\to\infty}\gamma\left(x\right)=0$ and the following equation is true for all $\alpha,x>0$
$$
e^{-\alpha x}=\sum\_{n=1}^{\infty}c\_n\left(\alpha\right)\gamma\left(x\right)^n
$$
If yes, how regular can we make $\gamma$, can it be smooth? Analytic? What about the $c\_n$?
**Clarifying edit:** all the $x$-dependence is in $\gamma$ and all the $\alpha$ dependence is in the $c\_n$
| https://mathoverflow.net/users/104719 | Can we write $e^{-\alpha x}$ as $\sum_{n=0}^\infty c_n\left(\alpha\right)\gamma\left(x\right)^n$ such that $\lim_{x\to\infty}\gamma\left(x\right)=0$ | The answer is **no**.
Suppose that $c\_n(\alpha)$ and $\gamma(x)$ with the desired properties exist. Necessarily $\gamma$ is one-to-one, and hence $$e^{-\alpha \gamma^{-1}(x)} = \sum\_{n = 0}^\infty c\_n(\alpha) x^n.$$ Since the right-hand side converges for some $x > 0$, it defines a function $$f\_\alpha(x) = \sum\_{n = 0}^\infty c\_n(\alpha) x^n$$ which is real-analytic near $0$ and not identically equal to zero. In particular, $f\_\alpha(x) \sim c\_{k(\alpha)}(\alpha) x^{k(\alpha)}$ as $x \to 0$, where $k(\alpha)$ is the index of the first non-zero coefficient $c\_n(\alpha)$. This means that $$e^{-\gamma^{-1}(x)} \sim c\_{k(\alpha)}(\alpha) x^{k(\alpha)/\alpha},$$ and therefore $k(\alpha) / \alpha$ is constant. Since $k(\alpha)$ is integer-valued, this is not possible, unless $k(\alpha) = 0$. But this means that $e^{-\gamma^{-1}(x)}$ has a positive right limit at $0$, contrary to the assumption that $\gamma^{-1}(x) \to \infty$ as $x \to 0^+$.
| 3 | https://mathoverflow.net/users/108637 | 415053 | 169,239 |
https://mathoverflow.net/questions/414943 | 2 | Let $k$ be a field and $A$ be a central simple algebra over $k$. It's known that $A$ has a splitting field (i.e. a field $K/k$ such that $A\_K\cong M\_n(K)$ for some $n$) which is finite and Galois.
This allows us to define a *reduced norm* $N:A\to k$ which is given by the determinant $M\_n(K)\to K$ and then descended (via Galois descent) to $k$.
I wonder if we can do the same thing using fpqc descent but avoiding the need for the existence of a finite Galois splitting field, and using only that $\overline{k}$ splits $A$, which is way simpler. (Of course that's not for didactic reasons, since we're exchanging a hard theorem for a harder one. That's just for my curiosity.)
| https://mathoverflow.net/users/131975 | Existence of reduced norms for CSAs using fpqc descent | You can define the reduced norm using fpqc descent. Have a look at chapter III, section 1.2 in Knus' "[Quadratic and Hermitian Forms over Rings](https://doi.org/10.1007/978-3-642-75401-2)", for instance.
| 2 | https://mathoverflow.net/users/86006 | 415065 | 169,242 |
https://mathoverflow.net/questions/415071 | 2 | Let $T$ be a given (finite) tree.
**Question 1:** Is it always possible to add edges to $T$ to obtain a $2$-connected outerplanar supergraph $G$?
**Question 2:** If the answer to Question #1 is negative, can the trees for which it is possible be characterized?
**Question 3( Defect form of Question 1):** Let $T$ be a *rooted* tree with root vertex $v\_{0} \in V(T)$. Is it always possible to add edges to $T$ to obtain a $2$-connected planar graph $G$ with a plane embedding in which $v\_{0}$ is the only internal vertex?
| https://mathoverflow.net/users/22051 | Completing a tree to a 2-connected outerplanar graph | Yes. Pick an arbitrary vertex to be the root. Consider the sequence of vertices $v\_1, v\_2, \ldots$ produced by a pre-order traversal of the rooted tree, adding edges $v\_i - v\_{i+1}$ where they don't already exist. Finally, add an edge back from the last vertex to the root, if it doesn't already exist. This cycle defines the outer face and gives 2-connectivity.
| 2 | https://mathoverflow.net/users/46140 | 415075 | 169,246 |
https://mathoverflow.net/questions/414315 | 6 | A follow-up question to [Alternating subgroups of $\mathrm{SU}\_n $](https://mathoverflow.net/questions/414265/alternating-subgroups-of-mathrmsu-n).
$\DeclareMathOperator\PU{PU}\DeclareMathOperator\PSU{PSU}\DeclareMathOperator\PSL{PSL}$Let $ \PU\_m $ be the projective unitary group, a compact simple adjoint Lie group corresponding to the root system $ A\_{m-1} $.
Let $ \PSL\_n(q) $ be the finite simple group of Lie type $ A\_{n-1}(q) $ given by taking the special linear group with entries from the field with $ q $ elements and modding out by the center.
Let $ \PSU\_n(q^2) $ be the finite simple group of Lie type $ ^2 A\_{n-1}(q^2) $ given by taking the special unitary group with entries from the field with $ q^2 $ elements and modding out by the center.
$ \PU\_2 $ contains a $ 60 $ element subgroup isomorphic to $ A\_5 \cong \PSL\_2(4) \cong \PSL\_2(5) $. It is a maximal closed subgroup of $ \PU\_2 $ (the only closed subgroup containing it is the whole group).
$ \PU\_3 $ contains a subgroup of order $ 360 $ isomorphic to $ A\_6 \cong \PSL\_2(9) $, it is maximal. Also $ \PU\_3 $ contains a group of order $ 168 $ isomorphic to $ \PSL\_2(7)\cong \PSL\_3(2) $, it is maximal.
There is also a 60 element $ A\_5 \cong \PSL\_2(4) \cong \PSL\_2(5) $ subgroup of $ \PU\_3 $ but that is already in $ SO\_3(\mathbb{R}) $ so it is not maximal. For more details see
<https://math.stackexchange.com/questions/497853/closed-lie-subgroups-of-su3>
The reference [Hanany and He - A Monograph on the Classification of the Discrete Subgroups of SU(4)](https://arxiv.org/abs/hep-th/9905212) from ([The finite subgroups of SU(n)](https://mathoverflow.net/questions/17072/the-finite-subgroups-of-sun)) shows that $ \PU\_4 $ contains a subgroup of order $ 25{,}920 $ isomorphic to $ \PSU\_4(2)\cong PSp\_4(3) $, it is maximal.
Also $ \PU\_4 $ contains a maximal $ A\_7 $ which in turn contains a group of order $ 360 $ isomorphic to $ \PSL\_2(9) \cong A\_6 $, and a group of order $ 168 $ isomorphic to $ \PSL\_2(7)\cong \PSL\_3(2) $,as well as a group of order 60 isomorphic to $ A\_5 \cong \PSL\_2(4) \cong \PSL\_2(5) $. For more details see
<https://math.stackexchange.com/questions/4535647/maximal-closed-subgroups-of-su-4>
That leads me to ask: Does $ \PU\_m $ always have a maximal $ \PSL\_n(q) $ or $ \PSU\_n(q) $?
| https://mathoverflow.net/users/387190 | Finite simple groups and $ \operatorname{SU}_n $ | One can consult the tables of Bray-Holt-Roney Dougal to work out the subgroups of $\mathrm{PSU}\_m$.
1. For $m=5$, we have copies of $PSL\_2(11)$ and $PSU\_4(4)$.
2. For $m=6$ we have $PSL\_2(11)$, $PSL\_3(4)$ and $PSU\_4(9)$.
3. For $m=7$ we have $PSU\_3(9)$.
4. For $m=8$ we have $PSL\_3(4)$.
5. For $m=9$ we have $PSL\_2(19)$.
6. For $m=10$ we have $PSL\_2(19)$ and $PSL\_3(4)$.
7. For $m=11$ we have $PSL\_2(23)$ and $PSU\_5(4)$.
8. For $m=12$ we have $PSL\_2(23)$ and $PSL\_2(9)$.
So far, so good. Bray-Holt-Roney Dougal stops at dimension 12. For dimensions 13 to 15 one consults tables in Anna Schroeder's PhD thesis (St Andrews, 2015) and there are examples in those dimensions. For dimensions 16 and 17 one consults the thesis of Daniel Rogers (Warwick, 2017). For 16 there is an example, but there are no interesting subgroups (i.e., finite subgroups not contained in a closed positive-dimensional subgroup) of $\mathrm{PSU}\_{17}$ at all, never mind linear or unitary ones.
So it seems $m=17$ is the first case where things go wrong.
| 8 | https://mathoverflow.net/users/152674 | 415082 | 169,248 |
https://mathoverflow.net/questions/415062 | 4 | Let finite group $G$ act on a finite set $X$ and hence on colorings $Y^X$, where $Y=\{1,2,\ldots,k\}$ is a set of colors. The Burnside-Pólya-Redfield-etc. counting theorem says that the number of orbits of $G$ acting on $Y^X$ is
$$ \frac{1}{\#G} \sum\_{g \in G} k^{c(g)},$$
where for $g\in G$, we use $c(g)$ to denote the number of cycles of the permutation $g\colon X \to X$.
Note that this number $P(k)$ is a polynomial in $k$, and it is an integer, so by the basic theory of integer-valued polynomials there are integral constants $a\_j \in \mathbb{Z}$ such that $P(k) = \sum\_j a\_j\binom{k}{j}$. (Recall $\binom{x}{j} := x(x-1)\ldots(x-(j-1))/j!$.)
**Question**: is there a simple formula for these $a\_j$?
This must've been considered before, but a cursory Googling did not lead me to anything.
| https://mathoverflow.net/users/25028 | Integer-valued polynomials from Pólya counting | Okay, I understand Nate's comment now and am posting this elucidation of it as a community wiki answer.
Recall that an *ordered set partition* of $X$ into $j$ blocks is an ordered tuple $(T\_1,T\_2,\ldots,T\_j)$ of non-empty subsets $\varnothing \neq T\_i \subseteq X$ that are pairwise disjoint and whose union is all of $X$. Since $G$ acts on $X$ it acts naturally on the ordered set partitions of $X$. Moreover, to any ordered set partition $(T\_1,T\_2,\ldots,T\_j)$ and choice of subset $\{y\_1 < y\_2 < \cdots < y\_j\} \subseteq Y$ of $j$ colors, we can associate the coloring $f\colon X \to Y$ determined by $f(x) = y\_i$ iff $x \in T\_i$. And two such colorings are equivalent under the action of $G$ exactly when the ordered set partitions are equivalent.
This shows
$$ P(k) = \sum\_{j\geq 1} (\textrm{$\#$ of orbits of $G$ acting on ordered set partitions of $X$ into $j$ blocks}) \cdot \binom{k}{j},$$
i.e., that
$$ a\_j = \textrm{$\#$ of orbits of $G$ acting on ordered set partitions of $X$ into $j$ blocks}.$$
This is a perfectly fine formula that shows the $a\_j$ are nonnegative integers, but it's a bit different from the Burnside's Lemma formula for $P(k)$ in that the Burnside's Lemma formula is "local" in the sense of looking at each $g\in G$ individually. I guess it would be hard to give a similar "local" formula for the $a\_j$.
| 8 | https://mathoverflow.net/users/25028 | 415089 | 169,250 |
https://mathoverflow.net/questions/415092 | 17 | Suppose that $a\_1,\dots,a\_n,b\_1,\dots,b\_n$ are iid random variables each with a symmetric non-atomic distribution.
Let $p$ denote the probability that there is some real $t$ such that $t a\_i \ge b\_i$ for all $i$.
It [was shown](https://mathoverflow.net/a/415074/36721) that
$$p=\frac{n+1}{2^n}.$$
Can this be proved by a combinatorial/symmetry argument?
| https://mathoverflow.net/users/36721 | Can this probability be obtained by a combinatorial/symmetry argument? | If I understand correctly, $c\_i := b\_i/a\_i$ should also be symmetric and non-atomic.
Then the result holds if there exists $t$ so that for all $i$
* $t \geq c\_i$ if $a\_i > 0$;
* $c\_i \geq t$ if $a\_i < 0$.
Reorder the indices so that $|c\_1| < |c\_2| < \dots < |c\_n|$.
There are $2^n$ ways to assign signs to $a\_i$ for each $i \in [n]$.
The only assignments for which an appropriate $t$ exists are those where the signs of $a\_1,\dots,a\_k$ are positive and the signs of $a\_{k+1},\dots,a\_n$ are negative with $k \in \{0,1,2,\dots,n\}$.
There are $n+1$ such assignments, so the desired probability is $(n+1)/2^n$.
| 25 | https://mathoverflow.net/users/7717 | 415102 | 169,253 |
https://mathoverflow.net/questions/415043 | 1 | Let $\Omega(n)$ be the total number of prime factors of $n$ (e.g $\Omega(12)=\Omega(2\*2\*3)=3$).
Consider these probabilities (probably they have a name):
* $P\_{>}(n)$ the probability that $\Omega(n+1) > \Omega(n)$
* $P\_{<}(n)$ the probability that $\Omega(n+1) < \Omega(n)$
* $P\_{=}(n)$ the probability that $\Omega(n+1) = \Omega(n)$
And the same probabilities, with an increment of 2; e.g.:
* $P'\_{=}(n)$ the probability that $\Omega(n+2) = \Omega(n)$
What are the values of these limits (if known):
$\lim\_{n \to \infty} P\_{=}(n)$ i.e. the probability that a successor of a number $n$ have the same number of factors of $n$
$\lim\_{n \to \infty} P'\_{=}(2n)$ i.e. the the probability that the even successor of an even number $n$ have the same number of factors of $n$
$\lim\_{n \to \infty} P'\_{=}(2n-1)$ i.e. the the probability that the odd successor of an odd number $n$ have the same number of factors of $n$
Are they all zero?
| https://mathoverflow.net/users/35419 | Probability of an equal number of factors of a successor | Loosely speaking, $\Omega(n)$ is approximately normal with mean and variance $\log\log n$. (See <https://en.wikipedia.org/wiki/Erd%C5%91s%E2%80%93Kac_theorem>) We expect $\Omega(n)$ and $\Omega(n+1)$ to behave like independent random variables. So we expect that all your probabilities containing an equal sign have a limit of zero, while the probabilities containing $>$ or $<$ are expected to approach $1/2$.
| 3 | https://mathoverflow.net/users/12947 | 415110 | 169,254 |
https://mathoverflow.net/questions/415039 | 5 | Consider a (countable) group $G$, a subgroup $H\leq G$ of finite index, and a unitary representation $\pi:G\to \mathcal{U}(\mathcal{H})$.
>
> If the center of the von Neumann algebra $\pi(H)''$ is finite dimensional, does this imply that the center of $\pi(G)''$ is finite dimensional as well?
> Is it at least true assuming that the representation $\pi$ is induced from a representation of $H$?
>
>
>
Remarks:
* As far as I'm concerned, $H$ can be assumed to be normal if it makes things simpler.
* If $\pi$ is of type I, then the question is whether $\pi$ decomposes to finitely many irreducible representations provided that the restriction $\mathrm{Res}\_H^G\pi$ does. This is obviously true.
| https://mathoverflow.net/users/104498 | The center of a representation von Neumann algebra, and finite index subgroups | The answer is YES and it follows from the general theory of finite-index inclusions of von Neumann subalgebras (a la Jones, Pimsner, Popa...), which says that finite-dimensionality of the center is preserved under taking a finite-index subalgebra/extension. This is perhaps an overkill and so I will try a layman's proof.
**Definition.** An inclusion $M\subset N$ of von Neumann algebras is of finite-index if there is a (necessarily faithful normal) conditional expectation $E$ from $N$ onto $M$ such that $E\geq\lambda\,\mathrm{id}\_N$ for some $\lambda>0$.
Note that $E$ maps $Z(N)$ into $Z(M)$ and that a von Neumann algebra $A$ is finite-dimensional iff $\mathbb{C}1\subset A$ is of finite-index. Thus if $M\subset N$ is of finite-index, then
$$\dim Z(M)<\infty\Rightarrow\dim Z(N)<\infty.$$
**Example.** If $H\le G$ is of finite-index, then the inclusion $\pi(G)'\subset\pi(H)'$ is of finite-index with the conditional expectation given by
$$E(x)=\frac{1}{[G:H]}\sum\_{g\in G/H}\pi(g)x\pi(g)^\*,$$
which satisfies $E\geq\frac{1}{[G:H]}$. Hence $\dim Z(\pi(G)'') <\infty\Rightarrow\dim Z(\pi(H)'') <\infty$.
We want to prove the converse. As I said in the beginning, one way is to use von Neumann algebra machinery (the standard form and the basic construction) to "flip" the inclusion. Here is an alternative way. By passing to a finite-index subgroup, we may assume that $H$ is normal. Take a natural realization (which I'm too lazy to write) of the induced representation $\mathrm{Ind}\_H^G\,\pi$ on $\ell\_2(G/H)\otimes\mathcal{H}$
that satisfies
$$(\mathrm{Ind}\_H^G\,\pi)(H)''\subset(\mathrm{Ind}\_H^G\,\pi)(G)''\subset B(\ell\_2(G/H)) \otimes\pi(H)''.$$
Then one can show the commutant inclusion ${\mathbb C}1\otimes\pi(H)'\subset (\mathrm{Ind}\_H^G\,\pi)(H)'$ is of finite-index (via the conditional expectation coming from the trace on $B(\ell\_2(G/H))$) and so the intermediate inclusion ${\mathbb C}1\otimes\pi(H)'\subset (\mathrm{Ind}\_H^G\,\pi)(G)'$ is also of finite-index. From this and the fact that $\pi\subset\mathrm{Ind}\_H^G\,\pi$,
one obtains
$$\dim Z(\pi(H)'') <\infty\Rightarrow\dim Z(\pi(G)'') <\infty.$$
| 7 | https://mathoverflow.net/users/7591 | 415115 | 169,255 |
https://mathoverflow.net/questions/415120 | 7 | It is well known that the codomain functor $$cod:\mathcal{C}^\to\to\mathcal{C}$$ from the [arrow category](http://nlab-pages.s3.us-east-2.amazonaws.com/nlab/show/arrow+category) of a category $\mathcal{C}$ to itself is a [fibration](http://nlab-pages.s3.us-east-2.amazonaws.com/nlab/show/Grothendieck%20fibration) iff $\mathcal{C}$ has binary pullbacks.
>
> Is there another functor that always exists for which promotion to a fibration is equivalent to $\mathcal{C}$ having a terminal object?
>
>
>
I naively tried $!:\mathcal{C}\to{\bf 1}$ and $\{X\}:{\bf 1}\to\mathcal{C}$ without any luck.
This would be of interest for the obvious reason -- having binary pullbacks and a terminal object is equivalent to being finitely complete, so a functor answering the above question positively would allow for a 'fully fibrational' characterization of finite completeness. Further, the construction below yields a functor that always exists such that promotion to a fibration is equivalent to having all pullbacks -- this together with Alexanders answer below offers a fully fibrational characterization of completeness, as originally desired.
---
Thanks to Alexander and Andrej for pointing out in the comments that $cod$ as above being a fibration only yields binary pullbacks, not arbitrary ones.
**EDIT**: I was able to fix the following construction to actually yield a functor that always exists such that promoting this functor to a fibration is equivalent to $\mathcal{C}$ having all pullbacks. (The first correction worked; we want them all to be the same commutative square universally, so that all paths from the pullback object into the object we're pulling back over are equal.)
>
> For a category $\mathcal{C}$, consider the category $Sink(\mathcal{C})$ whose objects are [sinks](http://nlab-pages.s3.us-east-2.amazonaws.com/nlab/show/sink) in $\mathcal{C}$ indexed over ordinals $\alpha\in{\bf O\_n}$ $$\{f\_i:X\_i\to X\}\_{i<\alpha}$$ and whose arrows are only defined from sinks indexed over $1$ to arbitrary sinks $$\hat h:\{f:A\to B\}\to\{g\_i:C\_i\to D\}\_{i<\alpha}$$ given by ordered pairs $$\hat h=\big(\{h\_i:A\to C\_i\}\_{i<\alpha},h:B\to D\big)$$ such that all the obvious squares commute, so $$h\circ f=g\_i\circ h\_i$$ for all $i<\alpha$. We have an obvious functor $$cod:Sink(\mathcal{C})\to\mathcal{C}$$ $$\{f\_i:X\_i\to X\}\_{i<\alpha}\mapsto X,$$ $$\hat h\mapsto h$$ and this functor is a fibration iff $\mathcal{C}$ has all pullbacks.
>
>
>
We can identify $\mathcal{C}^\to$ with the subcategory of $Sink(\mathcal{C})$ indexed over $1$.
| https://mathoverflow.net/users/92164 | A fibration equivalent to having a terminal object | A category $C$ has a terminal object if and only if the canonical functor $C \star \mathbf{1} \to \mathbf{1} \star \mathbf{1} = \mathbf{2}$ is a fibration.
| 10 | https://mathoverflow.net/users/57405 | 415123 | 169,257 |
https://mathoverflow.net/questions/414631 | 0 | Let $E\to X$ be a holomorphic vector bundle. Denote by $\mathbb{P}(E)\to X$ its projectivisation and $\mathcal{O}\_E(1)\to \mathbb{P}(E)$ the associated tautological line bundle.
I would like to know whether we can characterize the fact that $E$ is Hermitian-Eintein using the bundle $\mathcal{O}\_E(1)$.
* On one hand we have a correspondance between Finsler metric on $E$ and Hermitian metrics on $\mathcal{O}\_E(1)$.
* On the other hand there exists a theorem which states "$E$ admits a Finsler-Einstein metric iff it admits a Hermitian-Einstein metric" (see Geodesic-Einstein metrics and nonlinear stabilities by Feng–Liu–Wan, Trans. AMS 2019, [link at AMS site](https://www.ams.org/journals/tran/2019-371-11/S0002-9947-2018-07658-9/home.html))
From this it seems that we should be able to characterize the existence of Hermitian-Einstein metrics on $E$ through the geometry of $\mathcal{O}\_E(1)$. But there is a priori no notion of Hermitian-Einstein metric on $\mathcal{O}\_E(1)$ as $\mathbb{P}(E)$ is no canonically Kahler.
| https://mathoverflow.net/users/102114 | Characterize Hermitian-Einstein metric on $E$ using the tautological bundle $\mathcal{O}_E(1)$ | This is exactly the proposition 3.5 of the mentioned paper:
*If $\mathcal{O}\_{P(E^\*)}(1)$ admits a geodesic-Einstein metric, then the induced $L^2$ metric on $E$ is a Hermitian-Einstein metric.*
| 0 | https://mathoverflow.net/users/102114 | 415136 | 169,261 |
https://mathoverflow.net/questions/415139 | 6 | *Motivation: define a concrete Abelian category as a category with a univalent and injective functor in $\mathrm{Ab}^I$ (such that all homological concepts in it coincide with simple set-theoretic concepts $\mathrm{Ab}^I$)*
Is it true that any abelian category is isomorphic to a subcategory $\mathrm{Ab}^I$ for some small category $I$? (I know Mitchell's embedding, but I don't see if it's useful here). The subcategory is closed with respect to taking kernels and cokernels. The isomorphism must be additive (as far as I understand, it follows from this that it preserves kernels, cokernels, exact sequences).
If yes, then is it true that any Abelian category is isomorphic to a full subcategory $(R-\mathrm{Mod})^I$ for some ring $R$ and some small category $I$ (but this seems much more doubtful to me).
| https://mathoverflow.net/users/148161 | Is any abelian category a subcategory of $\mathrm{Ab}^I$? | Observe that $\bigoplus\_I : \textbf{Ab}^I \to \textbf{Ab}$ is a conservative exact functor: it is right exact by general nonsense, it preserves monomorphisms (because e.g. $\bigoplus\_{i \in I} A\_i$ is naturally a subgroup of $\prod\_{i \in I} A\_i$), and it is conservative because $\bigoplus\_{i \in I} A\_i \cong 0$ implies each $A\_i \cong 0$.
It seems to me that you are looking for a conservative exact functor $\mathcal{A} \to \textbf{Ab}$, so this observation implies that allowing $I$ with more than one element does not add any generality.
For small abelian categories $\mathcal{A}$, the Freyd–Mitchell embedding theorem gives a fully faithful exact functor $\mathcal{A} \to R\textbf{-Mod}$ (where $R$ is a not necessarily commutative ring), so every small abelian category admits a conservative exact functor $\mathcal{A} \to \textbf{Ab}$.
In particular, $\mathcal{A}$ will be isomorphic to a (not necessarily full) abelian subcategory of $\textbf{Ab}$, by which I mean a subcategory that is closed under finite direct sums/products, kernels, and cokernels.
If $\mathcal{A}$ is not required to be small then one has to do more work.
The existence of a conservative exact functor $\mathcal{A} \to \textbf{Ab}$ is itself a size restriction on $\mathcal{A}$.
For example:
**Proposition.**
If there is a conservative exact functor $\mathcal{A} \to \textbf{Ab}$, then $\mathcal{A}$ is locally small and wellpowered.
*Proof.*
A conservative exact functor is automatically faithful, and $\textbf{Ab}$ is locally small, so $\mathcal{A}$ must also be locally small.
Similarly, a conservative exact functor induces embeddings of subobject lattices, and $\textbf{Ab}$ is wellpowered, so $\mathcal{A}$ must also be wellpowered. ◼
Maybe you don't believe in categories that are not locally small and wellpowered.
Even so, I am not aware of any embedding theorems that work for arbitrary locally small and wellpowered abelian categories.
(I am also not aware of counterexamples.
Perhaps there is some large cardinal axiom that implies it can be done.)
In practice, it is not necessary to embed the entire category to prove the theorems you want – you can usually find a small abelian subcategory containing all the objects and morphisms you need for your theorem and then you can embed that subcategory.
Asking for an embedding of the whole category at once is being greedy – like asking for a global holomorphic chart of a complex manifold when local charts suffice.
Here are some embedding theorems I know for non-small abelian categories.
**Theorem.**
If $\mathcal{A}$ is a Grothendieck abelian category then there is a conservative exact functor $\mathcal{A}^\textrm{op} \to \textbf{Ab}$ that has a left adjoint.
*Proof.*
It is a well-known theorem that Grothendieck abelian categories have injective cogenerators.
But an injective cogenerator of $\mathcal{A}$ is precisely an object $I$ such that $\textrm{Hom}\_\mathcal{A} (-, I) : \mathcal{A}^\textrm{op} \to \textbf{Ab}$ is a conservative exact functor.
Furthermore, representable functors in cocomplete categories automatically have a left adjoint, and $\mathcal{A}^\textrm{op}$ is indeed cocomplete. ◼
Maybe contravariance is jarring.
But $\textbf{Ab}$ is itself a Grothendieck abelian category, so the theorem (or Pontryagin duality) gives us a conservative exact functor $\textbf{Ab}^\textrm{op} \to \textbf{Ab}$, and composing them yields a conservative exact functor $\mathcal{A} \to \textbf{Ab}$.
The following is a small generalisation.
**Theorem.**
If $\mathcal{A}$ is a locally small abelian category and has a small generating set, then there is a conservative exact functor $\mathcal{A}^\textrm{op} \to \textbf{Ab}$.
*Proof.*
The hypothesis implies there is a small full abelian subcategory $\mathcal{B}$ containing the given small generating set of $\mathcal{A}$.
We get a fully faithful functor $\mathcal{A} \to [\mathcal{B}^\textrm{op}, \textbf{Ab}]$, but in any case it is not automatically exact.
Let $\textbf{Lex} (\mathcal{B}, \textbf{Ab})$ be the full subcategory of left exact functors $\mathcal{B}^\textrm{op} \to \textbf{Ab}$.
Then, the earlier functor factors as a fully faithful exact functor $\mathcal{A} \to \textbf{Lex} (\mathcal{B}, \textbf{Ab})$.
But $\textbf{Lex} (\mathcal{B}, \textbf{Ab})$ is a Grothendieck abelian category, so we may apply the earlier theorem to conclude. ◼
| 5 | https://mathoverflow.net/users/11640 | 415150 | 169,265 |
https://mathoverflow.net/questions/414747 | 2 | I was reading the following [paper](https://eudml.org/doc/286980) which claims to generalize Borel spectral sequence for non-compact Stein fibers. However, I don't understand how the following bundle fits into the picture:
$$
(\mathbb{C}^\times)^2\xrightarrow{\mathbb{C}} \mathbb{T}
$$
$\mathbb{T}$ is a compact torus.
Here $\mathbb{C}$ acts on $(\mathbb{C}^\times)^2$ as a multiplicative subgroup of the form:
$$
\mathbb{C} \times (\mathbb{C}^\times)^2 \ni (z,t\_1,t\_2)\mapsto (e^{iz}\cdot t\_1,e^{z}\cdot t\_2).
$$
If I am not mistaken the fiber is Stein and acyclic. However, Dolbeault cohomology of $\mathbb{T}$ differs drastically from that of $(\mathbb{C}^\times)^2$. Am I missing something?
| https://mathoverflow.net/users/143549 | Borel spectral sequence with non-compact fibers | The second page of the Borel spectral sequence in your example is
$$
E\_2^s=\bigoplus\_{p,q} \mathrm H^{s-q,q}(\mathbf T,\Omega^{p+q-s}(\mathbf C))=\bigoplus\_{p,q} \mathrm H^{s-q,q}(\mathbf T)\otimes\_{\mathbf C}\Omega^{p+q-s}(\mathbf C)
$$
and comprises Dolbeault cohomology of the compact complex torus $\mathbf T$ with values in the infinite-dimensional complex vector spaces $\Omega^{p+q-s}(\mathbf C)$ of all holomorphic $(p+q-s)$-forms on the Stein manifold $\mathbf C$ (infinite-dimensional when $p+q-s=0,1$). That makes it a lot more plausible to converge to the ordinary Dolbeault cohomology of $(\mathbf C^\star)^2$ with values in $\mathbf C$!
| 3 | https://mathoverflow.net/users/85592 | 415152 | 169,266 |
https://mathoverflow.net/questions/415151 | 4 | Let $f=(u,v)\in \mathscr{D}'(U,\mathbb{C})$ be a distribution, where $U\subset\mathbb{C}=\mathbb{R}^2$ is an open set and $u$ and $v$ are the projection of $f$ onto the real and imaginary axis (ie $\langle f,\phi\rangle=\langle u,\phi\rangle+i\langle v,\phi\rangle$). Suppose that
$$
\frac{\partial}{\partial \overline{z}}f=0\qquad\text{in U,}
$$
where $\frac{\partial}{\partial \overline{z}}=\frac{1}{2}\bigg(\frac{\partial}{\partial x}+i\frac{\partial}{\partial y}\bigg)$ and the derivatives are in distributional sense.
Does it follow that $f$ is holomorphic in the classical sense, ie $f\in C^\infty(U,\mathbb{C})$ and the Cauchy-Riemann equations are satisfied?
The obvious idea would be to mollify, get holomorphic functions and then take the limit. But how can we conclude that the limit is still holomorphic?
| https://mathoverflow.net/users/351083 | $\frac{\partial f}{\partial \overline{z}}=0$ in distributional sense implies $f$ is holomorphic | I've just realized that, if $f$ is in $L^1\_{loc}$ and not just in $\mathscr{D}'$, my question can be answered using Weyl's lemma for harmonic functions. Indeed, from $\frac{\partial f}{\partial \overline{z}}=0$ it easily follows that $\Delta u=\Delta v=0$, and then Weyl's lemma implies that $u$ and $v$ are smooth. But then $f$ is smooth and satisfies the Cauchy-Riemann equations, thus is holomorphic.
EDIT: As pointed out by Ben McKay in the comments, the hypothesis that $f\in L^1\_{loc}$ is not necessary.
| 4 | https://mathoverflow.net/users/351083 | 415153 | 169,267 |
https://mathoverflow.net/questions/415159 | 15 | $\DeclareMathOperator\Hom{Hom}$Fix a commutative ring $R$. For $R$-modules $M$ and $N$, there is an inclusion of abelian groups $\Hom\_R(M,N) \to \Hom\_{\mathbb{Z}}(M,N).$ Are there conditions on $R$ that will ensure this is an inclusion of a direct summand (preferably with a section natural in $M$ and $N$)?
(I have tried googling this question, but all the words are too common. Pointers to the right terminology would also be appreciated!)
| https://mathoverflow.net/users/131360 | R-module hom a direct summand of Z-module hom? | There exists a natural section if and only if $R$ is *separable* over $\mathbb Z$ (more generally over $k$, if you have a $k$-algebra $R$ and are asking about the morphism $\hom\_R(M,N)\to \hom\_k(M,N)$). If you're only asking for a section, no naturality, then I don't know any reasonable condition, but maybe someone can chime in and say something about it.
Here's a proof of my claim about separability (I'll do it over a base commutative ring $k$, because $\mathbb Z$ adds nothing specific here): $\hom\_k(M,N)\cong \hom\_R(R\otimes\_k M, N)$ and the forgetful map $\hom\_R(M,N)\to \hom\_k(M,N)$ corresponds to precomposition by the action map $R\otimes\_k M\to M$ under this equivalence.
So by the Yoneda lemma, a natural section $\hom\_R(R\otimes\_k M, {-}) \to \hom\_R(M,{-})$ is the same thing as a natural section $M\to R\otimes\_k M$ to the action map $R\otimes\_k M\to M$ (note that we use naturality in $N$ to get a section $M\to R\otimes\_k M$, and naturality in $M$ to get that this section is natural).
We now apply this to $M = R$, we get a section $R\to R\otimes\_k R$ to the multiplication. Now, this is very weak as is, but this section is natural in endomorphisms of $R$ as a left $R$-module, i.e. in multiplications on the right by elements of $R$. If you unravel what this naturality means, it will imply that $R\to R\otimes\_k R$ is a section of $R\otimes\_k R^\text{op}$-modules ($R$-bimodules over $k$), which is what is called a "separability idempotent" for $R$. Recall:
>
> **Definition** : A $k$-algebra $R$ is called separable (over $k$) if the multiplication map $R\otimes\_k R^\text{op}\to R$ admits a $R\otimes\_k R^\text{op}$-module section.
>
>
>
The converse is not hard to show: if $R$ is separable, then the action map has a natural section $M = R\otimes\_R M \to (R\otimes\_k R)\otimes\_R M = R\otimes\_k M$, and so $\hom\_R(M,N)\to \hom\_R(R\otimes\_k M, N) = \hom\_k(M,N)$ has a natural section too.
**Examples of separable algebras**:
* Finite étale extensions of $k$ are separable.
* Matrix rings over commutative rings are separable over their base (they all "look like this" étale-locally).
* A group ring $k[G]$ is separable if and only if $\lvert G\rvert\in k^\times$.
* Finally, separable algebras are closed under many familiar operations: taking the center, finite products, taking tensor products (over the same base or over a different base).
| 21 | https://mathoverflow.net/users/102343 | 415162 | 169,269 |
https://mathoverflow.net/questions/415165 | 2 | Consider an arbitrary finite set of orthogonal-projection matrices (symmetric, idempotent, etc.) in $\mathbb{R}^{n\times n}$.
We draw two matrices $Q,P$ uniformly and i.i.d. from this set.
**Question:** Is the following expected matrix $\mathbb{E}\_{P,Q} \left[I-2P+QPQ\right]$ necessarily positive semi-definite?
Clearly, one such matrix is not necessarily PSD. For instance,
$P=\begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix},Q=\begin{bmatrix}0 & 0 \\ 0 & 1\end{bmatrix}$ do not yield a PSD matrix together, but I'm asking about the expectation rather than about an arbitrary instantiation.
| https://mathoverflow.net/users/100796 | Expected matrix created from two random orthogonal-projection matrices | Found a counter example.
If we take the set to be $\left\{
\begin{bmatrix}1 & 0 \\ 0 & 0\end{bmatrix},
\frac{1}{2}\begin{bmatrix}1 & 1 \\ 1 & 1\end{bmatrix}
\right\}$, then we have
$\mathbb{E}\_{P,Q} \left[I-2P+QPQ\right]=
\frac{1}{16}\begin{bmatrix}1 & -5 \\ -5 & 11\end{bmatrix}$, whose eigenvalues are $-0.06694174, 0.81694174$.
| 0 | https://mathoverflow.net/users/100796 | 415166 | 169,270 |
https://mathoverflow.net/questions/415145 | 1 | Suppose we have an equivalence of triangulated categories $\Phi : \mathcal{A} \to \mathcal{B}$. Let $G$ be a finite group. Are there any methods/conditions for specifying when one has an induced equivalence $\Phi^G : \mathcal{A}^G \to \mathcal{B}^G$ of the associated $G$-equivariant categories?
The particular case I'm interested in is when $\mathcal{A}$ is a semiorthogonal component of $D^b(X)$ and $\mathcal{B}$ is a semiorthogonal component of $D^b(X')$, where $X$ and $X'$ are smooth projective complex varieties. The group $G$ in my case can be something simple like $\mathbb{Z}/2$.
| https://mathoverflow.net/users/142073 | Inducing an equivalence of $G$-equivariant categories | You need some assumptions. But in the case of admissible subcategories, if their equivalence is given by a Fourier-Mukai functor $\Phi\_K$ and the object $K \in D^b(X \times X')$ is $G$-equivariant, then $\Phi^G$ can be defined.
See arXiv:1403.7027, Theorem 6.9 for a general statement of this sort.
| 0 | https://mathoverflow.net/users/4428 | 415168 | 169,271 |
https://mathoverflow.net/questions/415182 | 1 | This seemingly simple problem is doing my head in. I have tried doing a proof by induction over the edges of the graph but I just cannot seem to get it to work.
I am trying to prove that in a finite directed graph $G = (V, E)$ that obeys the following rules, any mode number of outgoing edges cannot exceed 2.
The rules:
* The graph has $|V| = n$ vertices - I've called them $0, 1, ..., n-2, n-1$.
* The graph has $|E| ≤ n$.
* Every vertex has 0 or 1 incoming edges.
* No vertex has an edge from itself to itself.
* The graphs are constructed by a process of adding edges between pairs of vertices according to the following rules:
+ An edge can be added from any vertex $u$, but only to $v ≠ u$ when $v$ has no incoming edges already.
+ Adding an edge from $u$ to $v$ when $u$ has no incoming edges means an edge must also be added from $v$ to $u$
An example of a graph that follows the rules, with $n = 2$:
* $V = \{0, 1\}$
* $E = \{ (0, 1), (1, 0) \}$
* Mode = 1
+ there are 2 vertices with 1 outgoing edge
+ therefore the modal number of outgoing edges is 1.
Another example of a graph that follows the rules, with $n = 4$:
* $V = \{0, 1, 2, 3\}$
* $E = \{ (0, 1), (1, 0), (0, 2), (1, 3) \}$
* Set of all modes = $\{0, 2\}$ (multimodal - see below)
+ there are 2 vertices with 0 outgoing edges
+ there are 2 vertices with 2 outgoing edges
+ therefore there are two modal numbers of outgoing edges: 0, and 2.
The way I tried to formulate the inductive step was to let $T\_r$ denote the set of nodes that have $r$ outbound edges, and then to say:
For some $i$ in $\{0, 1, 2\}$, for all $j$ in $\{3, 4, ..., n-2, n-1\}$, $|T\_i| > |T\_j|$.
But I tie myself in knots trying to cover all the cases in making the inductive step.
Any help appreciated.
I feel like I'm going down a blind alley on this one. I haven't been able to come up with a counterexample either...
| https://mathoverflow.net/users/82417 | How to prove that the modal number(s) of outgoing edges cannot exceed 2? | The sum of outdegrees is at most $n$. Thus if certain value $d\geqslant 3$ appears, say, $k>0$ times, the value 0 must appear at least $(d-1)k>k$ times, and do $d$ is not modal.
| 4 | https://mathoverflow.net/users/4312 | 415185 | 169,275 |
https://mathoverflow.net/questions/415187 | 2 | Let $A$ and $B$ denote $C^{\ast}$-algebras. Let $\lVert\cdot\rVert\_h$ and $\lVert\cdot\rVert\_{\text {max}}$ denote the Haagerup norm and max $C^\*$-norms on $ A \otimes B$, respectively. I am looking for a reference/proof for the following result:
>
> $\lVert\cdot\rVert\_{\text{max}} \leq \lVert\cdot\rVert\_h$.
>
>
>
| https://mathoverflow.net/users/129638 | Need reference for: $\lVert\cdot\rVert_{\text{max}} \leq \lVert\cdot\rVert_h$ | Let $v \in A \otimes B$, where $A\otimes B$ is the algebraic tensor product.
The Haagerup norm of $v$ is the infimum of the expressions $\sqrt{\|\sum\_{i=1}^{r} x\_{i} x\_{i}^{\ast}\|}\cdot \sqrt{\|\sum\_{i=1}^{r} y\_{i}^{\ast} y\_{i} \|}$, taken over all decompositions $v=\sum\_{i=1}^{r} x\_{i} \otimes y\_{i}$ ($r$ can be arbitrarily large).
On the other hand, the maximal norm is the supremum of the norm $\|\sum\_{i=1}^{r} \pi(x\_i) \sigma(y\_{i})\|\_{B(H)}$ over all pairs of representations $\pi: A \to B(H)$, $\sigma: B \to B(H)$ with commuting ranges. Note that $\sum\_{i=1}^{r} \pi(x\_i) \sigma(y\_{i})$ can be interpreted as the product of the row $(\pi(x\_1),\dots, \pi(x\_r))$ and the column $(\sigma(y\_1),\dots, \sigma(y\_r))^{T}$, so we get can estimate $\|\sum\_{i=1}^{r} \pi(x\_i) \sigma(y\_{i})\|\_{B(H)}$ by
$
\sqrt{\|\sum\_{i=1}^{r} \pi(x\_{i} x\_{i}^{\ast})\|\_{B(H)}} \cdot \sqrt{\|\sum\_{i=1}^{r} \sigma(y\_{i}^{\ast} y\_{i})\|\_{B(H)}}.
$
As representations of $C^{\ast}$-algebras are contractive, this is clearly not greater than the quantity defining the Haagerup norm.
It might be useful to note that we did not even have to use the fact the representations $\pi$ and $\sigma$ had commuting ranges.
| 7 | https://mathoverflow.net/users/24953 | 415195 | 169,277 |
https://mathoverflow.net/questions/415096 | 5 | Classification of simple (or simple-restricted) Lie algebras over algebraically closed fields in positive characteristic is studied for a long time. Today, we know all finite-dimensional simple (or simple-restricted) Lie algebras over algebraically closed fields of characteristic $p \ge 5$. But, how about over **finite fields**? I don't know either the results of classifications over finite fields or some examples. Does anyone know some references?
| https://mathoverflow.net/users/167704 | Classification of simple Lie algebras over finite fields | The classification of simple finite-dimensional Lie algebras over finite fields is a very hard task and only few results are known. However, I suggest to have a look at the following paper by Bettina Eick and references therein: [Some new simple Lie algebras in characteristic 2](https://doi.org/10.1016/j.jsc.2010.05.003).
| 4 | https://mathoverflow.net/users/14653 | 415206 | 169,282 |
https://mathoverflow.net/questions/415203 | 4 | I came across this equation in my research (related to reaction diffusion system):
$$\frac{d^2y}{dr^2}+B\operatorname{sech}^2(r) \frac{dy}{dr} + Cy = 0$$
where $B$ and $C$ are constants. Can it be solved analytically?
| https://mathoverflow.net/users/476359 | Non-constant-coefficient second-order linear ODE | Making the change of the independent variable $x=e^{it}$, we obtain
$$x^2w''+xw'+\frac{4iBx^3}{(x^2+1)^2}w'-Cw=0.$$
This equation has $2$ regular singularities $0,\infty$, and irregular
at $\pm i$. Therefore
for generic $B$ and $C$ its solutions cannot be expressed in terms of functions
of hypergeometric type, or any other usual special functions.
| 8 | https://mathoverflow.net/users/25510 | 415216 | 169,285 |
https://mathoverflow.net/questions/415225 | 5 | Say we're working in a symmetric monoidal $\infty$-category $\mathcal{S}$, and $A$ is an associative algebra in it. For instance,
$$\mathcal{S}\ =\ \text{dg vector spaces},\ \ \ A\ =\ \text{a dg algebra}.$$
Then:
1. $\text{Mod}(A)=\text{Mod}\_\mathcal{S}(A)$ is just a plain category, no extra structure (i.e. an $E\_0$ category).
2. $A$ is a *bi*algebra iff $\text{Mod}\_\mathcal{S}(A)$ is an $E\_1$ category, i.e. is monoidal, and if the forgetful functor is monoidal.
3. $A$ is a *quasitriangular* *bialgebra* (e.g. quantum group) iff $\text{Mod}\_\mathcal{S}(A)$ is an $E\_2$ category, i.e. is braided monoidal, and if the forgetful map to $\mathcal{S}$ is a braided monoidal functor.
4. (This next step is what my question is about)
**Question:** What about for slightly higher $n$, like $E\_3,E\_4,E\_5,...$? Is there a simple definition of what $A$ has to be in that case? Are there examples of such $A$ coming up in real life? Quantum groups are extremely interesting and deep objects, so one imagines that the next ones along will be even more interesting.
| https://mathoverflow.net/users/119012 | $\text{Mod}(A)$ is an $E_n$ category $\Leftrightarrow$ $A$ is an ??? algebra | $E\_0$ is not quite "no extra structure" : you know who $A$ is inside $Mod(A)$, so it's a *pointed* category (more generally, an $E\_0$ object in $\mathcal S$ is an object with a "unit" $\mathbb 1\to X$).
For a different version of your question, the answer is "$A$ is $E\_{n+1}$ iff $Mod(A)$ is $E\_n$", where "iff" can be made more precise : the space of $E\_n$-structures on the $E\_0$-category $Mod(A)$ (i.e. the space of $E\_n$-structures where $A$ is the unit) is equivalent to the space of $E\_{n+1}$-structures on $A$ extending its $E\_1$-structure.
But that does not seem to be the question you're asking, you seem to be asking about monoidal structures on $Mod(A)$ such that the forgetful functor $Mod(A)\to \mathcal S$ is *strict* monoidal - at least that's how I understand your answer for $n=1$ (you say "bialgebra", where the previous paragraph hints at $E\_2$-algebra, which shows we aren't thinking of the same thing).
So what's the answer for your version of the question ? To answer this, I'll make a few observations. But before that, I'll set the stage, where probably the assumptions are not necessary but they'll make things easier: I'll assume $\mathcal S$ is presentably symmetric monoidal, so that I can work in $\mathcal S$-modules in $Pr^L$ and have a nicely behaved Lurie tensor product $\otimes\_\mathcal S$.
1- The functors appearing in your structure are of the form $Mod(A)\otimes\_\mathcal S Mod(A)\to Mod(A)$, over $\mathcal S \simeq \mathcal S\otimes\_\mathcal S\mathcal S\to \mathcal S$, or variations thereon with more tensor products. One of the nice properties of $\otimes\_\mathcal S$ is that $Mod(A)\otimes\_\mathcal S Mod(B) \simeq Mod(A\otimes B)$. In particular your functors are of the form $Mod(A\otimes A)\to Mod(A)$, over $\mathcal S$. Because limits are preserved and reflected by the forgetful functors, your functors preserve both limits and colimits.
2- In particular, the full subcategory of $Mod\_\mathcal S(Pr^L)\_{/\mathcal S}$ on objects of the form $Mod(A)\overset{forgetful}\to \mathcal S$ is closed under tensor products where the monoidal structure on $Mod\_\mathcal S(Pr^L)\_{/\mathcal S}$ is the usual one on $C\_{/A}$ where $A$ is an algebra: the tensor product is given by $(X\to A)\otimes (Y\to A) := (X\otimes Y\to A\otimes A\to A)$. You can make this more precise and actually make this into a monoidal category (as monoidal as $A$ is in $C$). It's closed under tensor products, but also every functor there is both in $Pr^R$ and in $Pr^L$.
3- The structure you're interested in is the structure of an $E\_n$-algebra on $Mod(A)\to \mathcal S$, viewed in $Mod\_\mathcal S(Pr^L)\_{/\mathcal S}$. By the point just above, this is the same as the structure of an $E\_n$-coalgebra in $Mod\_\mathcal S(Pr^L)\_{\mathcal S/}$ on the object $\mathcal S\overset{-\otimes A}\to Mod(A)$
4- Lurie proves in *Higher Algebra* that $Alg(\mathcal S)\to Mod\_\mathcal S(Pr^L)\_{\mathcal S/}$ is an equivalence of symmetric monoidal $\infty$-categories. So the structure you're after is exactly that of an $E\_n$-coalgebra in algebras.
For $n=1$ you recover the notion of bialgebra. For $n=2$, I'm not sure what a quasitriangular bialgebra is, but if your condition is an "iff", then it should be the same data as an algebra $A$ together with a comultiplication $A\to A\otimes A$ (which is an algebra map) which is $E\_2$-cocommutative as an algebra map, in a highly coherent sense. What comes later is just more and more cocommutativity for this map.
| 8 | https://mathoverflow.net/users/102343 | 415233 | 169,290 |
https://mathoverflow.net/questions/415232 | 1 | Let $k$ be a field and $A$ a finite dimensional $k$-algebra. Given a sequence of inclusions $M\_1 \subseteq M\_2 \subseteq \dots$ of $A$-modules consider the direct limit $M:= \bigcup\_{i=1}^\infty M\_i$. For a finite dimensional module $X$ suppose we have $X\subseteq M$. Then $X$ is generated by finitely many elements $x\_1, x\_2, \dots, x\_n$. Hence we find $m>0$ such that $x\_i\in M\_m$ for all $m$. It follows that $X\subseteq M\_m$. Now consider the dual setting:
Let $M\_1 \twoheadleftarrow M\_2 \twoheadleftarrow \dots $ be a sequence of $A$-modules, let $M$ be its inverse limit and suppose we find an epimorphism $M\twoheadrightarrow X$, where $X$ is a finite dimensional $A$-module. Is it true in general, that we obtain an epimorphism $M\_m \twoheadrightarrow X$ for some $m>0$?
| https://mathoverflow.net/users/145920 | Epimorphism going out of an inverse limit into a finite dimensional module | The answer is no. First, let me point out that the $A$ here plays no role: because each $M\_{i+1}\to M\_i$ is an epimorphism, $M\to M\_i$ is one too, and so if there is a $k$-linear factorization $M\to M\_i\to X$, then it is automatically $A$-linear.
So we can simply focus on $k$-vector spaces and worry about the analogous statement. Let $f: M\to X$ denote our morphism. Let $K\_i\subset M$ be the kernel of $M\to M\_i$. $K\_i$ is a nonincreasing sequence of subspaces of $M$, therefore the same is true of $f(K\_i)$ inside $X$.
Because $X$ is finite dimensional, this nonincreasing sequence of subspaces stabilizes. $f$ factors through $M\_i$ if and only if $f(K\_i) = 0$, so the point is to find an example where no such $i$ exists.
For instance, take $k$ to be a finite field, pick a nonprincipal ultrafilter $\mathcal U$ on $\mathbb N$ and define $k^\mathbb N\to k$ by $u\mapsto \lim\_\mathcal U u$ (we take the limit along the ultrafilter in a topological sense using the discrete topology on $k$, which is compact). More concretely, this partitions $\mathbb N$ as $\coprod\_{x\in k} u^{-1}(x)$ and decides which of these is in $\mathcal U$. Write $k^\mathbb N$ as $\lim\_n k^n$. The kernel $K\_n$ is $0^n\times \prod\_{k> n}k$ and the image of $f$ on that is always $k$: take the sequence $u$ which is $0$ until $n$, and $1$ afterwards.
| 2 | https://mathoverflow.net/users/102343 | 415238 | 169,292 |
https://mathoverflow.net/questions/415237 | 1 | Let $M$ be a compact riemannian manifold equipped with a geodesic distance and let $\mathcal{B}(M)$ be the borel sigma algebra generated by the geodesic distance. Let $(\Omega,\mathcal{F},\mathbb{P})$ be a probability space that arises from a compact metric space $(\Omega,d)$ and $\mathcal{F}$ is the borel sigma algebra. Let $\tilde{\Omega}$ be a countable dense subset of $\Omega$.
We would like to answer to the following question:
Let:
$$f \colon (\tilde{\Omega},\mathcal{F}\_{\rvert\tilde{\Omega}}) \to (M,\mathcal{B}(M))$$
be a measurable function.
Is it possible to construct a function
$$g \colon (\Omega,\mathcal{F}) \to (M,\mathcal{B}(M)) $$
with the property that $g$ is measurable, $g\_{\rvert\tilde{\Omega}}= f$ and,
if $\omega \in \Omega\setminus \tilde{\Omega}$, then
$$g(\omega)= \lim\_{n \to \infty}f(\omega\_n) $$
for a sequence $ \{\omega\_n\}\_{n \in \mathbb{N}} \subset \tilde{\Omega}$ such that $\omega\_n \to \omega$ as $n \to \infty$?
| https://mathoverflow.net/users/297294 | Extension of measurable function from dense subset | $\newcommand\R{\mathbb R}\newcommand\om\omega\newcommand\Om\Omega\newcommand\tom{{\tilde\omega}}\newcommand\tOm{\tilde\Omega}\newcommand{\ep}{\varepsilon}\newcommand{\de}{\delta}\newcommand{\vpi}{\varphi}\newcommand\ol\overline\newcommand{\N}{\mathbb N}\newcommand{\Q}{\mathbb Q}$Take any $m\in\text{Im}(f)$ and then let
$$g(\om):=
\begin{cases}
f(\om)&\text{ if }\om\in\tOm, \\
m&\text{ if }\om\in\Om\setminus\tOm.
\end{cases}$$
The $g$ will have the desired properties.
---
The OP has changed the question, thus invalidating the answer above.
The answer to the changed question is still yes, but the construction is much more complicated.
Since the manifold $M$ is compact, it has a finite [atlas](https://en.wikipedia.org/wiki/Differentiable_manifold#Atlases) $A:=\{(U\_i,\vpi\_i)\colon i\in[k]\}$, where $[k]:=\{i\in\N\colon i\le k\}$, with bounded images $\vpi\_i(U\_i)$ of the $U\_i$'s. Without loss of generality (wlog), $A$ is a refinement of another atlas $\{(V\_i,\psi\_i)\colon i\in[k]\}$ of $M$ in the sense that $\ol{U\_i}\subseteq V\_i$ and $\vpi\_i=\psi\_i|\_{U\_i}$ for all $i$, where $\ol S$ denotes the closure of a set $S$ and $|\_S$ denotes the restriction to $S$.
For each $i\in[k]$, let
\begin{equation}
\hat\Om\_i:=f^{-1}(U\_i), \quad
\Om\_i:=\ol{\hat\Om\_i}\setminus\bigcup\_{j\in[k-1]}\ol{\hat\Om\_j}, \quad
\tOm\_i:=\hat\Om\_i\cap\Om\_i.
\end{equation}
Then
\begin{equation}
\Om=\bigcup\_{i\in[k]}\Om\_i
\end{equation}
(because $M=\bigcup\_{i\in[k]}U\_i$, whence $\tOm=\bigcup\_{i\in[k]}\hat\Om\_i$, whence $\Om=\ol\tOm=\ol{\bigcup\_{i\in[k]}\hat\Om\_i}=\bigcup\_{i\in[k]}\ol{\hat\Om\_i}=\bigcup\_{i\in[k]}\Om\_i$);
* the $\Om\_i$'s are pairwise disjoint and measurable;
* $\tOm\_i$ is countable and dense in $\Om\_i$ for each $i$ (because the $\ol{\hat\Om\_i}$'s are closed sets);
* $f(\tOm\_i)\subseteq U\_i$ for each $i$.
So, it suffices to construct the function $g$ locally -- that is, separately on each $\Om\_i$. So, to simplify notations in what follows, fix any $i\in[k]$ and let
* $U:=U\_i$, $\vpi:=\vpi\_i$, and $\psi:=\psi\_i$.
* Also, re-define $\Om$ and $\tOm$ by letting them, respectively, denote $\Om\_i$ and $\tOm\_i$, from now on.
In other words, we are going to omit the subscript ${}\_i$ everywhere in what follows.
Let $n$ denote the dimension of the manifold $M$, so that
\begin{equation}
\vpi(m)=(\vpi^1(m),\dots,\vpi^n(m))\in\R^n
\end{equation}
for some real-valued functions $\vpi^1,\dots,\vpi^n$ and all $m\in U$.
For $j\in[n]$, let
\begin{equation}
h\_j:=\vpi^j\circ f,\quad\text{so that }h\_j\colon\tOm\to\R.
\end{equation}
Take any real $\ep>0$. For each $\om\in\Om$, let
\begin{equation}
H\_\ep(\om):=\vpi(f(\tOm\cap B\_\om(\ep)))
=\{h(\tom)\colon\tom\in\tOm\cap B\_\om(\ep)\},
\end{equation}
where
\begin{equation}
h(\tom):=\vpi(f(\tom))=(h\_1(\tom),\dots,h\_n(\tom))
\end{equation}
and
\begin{equation}
B\_\om(\ep):=\{\tau\in\Om\colon d(\tau,\om)<\ep\}.
\end{equation}
Since the map $\vpi$ is continuous and $f(\tOm\cap B\_\om(\ep))\subseteq\ol U\subseteq M$, the set $\ol{H\_\ep(\om)}$ is a compact subset of $\R^n$. So, there is a unique maximum, say
\begin{equation}
\ol h\_\ep(\om)=(\ol h\_{1,\ep}(\om),\dots,\ol h\_{n,\ep}(\om))\in\R^n,
\end{equation}
of the set $\ol{H\_\ep(\om)}$ with respect to the ([total/linear](https://en.wikipedia.org/wiki/Total_order)) [lexicographic order](https://en.wikipedia.org/wiki/Lexicographic_order).
Note that for any real $c\_1$
\begin{equation}
\begin{aligned}
&\ol h\_{1,\ep}(\om)\le c\_1 \\
&\iff \\
&\forall\tom\in\tOm\quad \tom\in B\_\om(\ep)\implies h\_1(\tom)\le c\_1,
\end{aligned}
\end{equation}
which can be rewritten as
\begin{equation}
\begin{aligned}
&\ol h\_{1,\ep}(\om)\le c\_1 \\
&\iff \\
&\forall\tom\in\tOm\quad \om\in B\_\tom(\ep)\implies h\_1(\tom)\le c\_1,
\end{aligned}
\end{equation}
Letting $\Q\_{++}:=\Q\cap(0,\infty)$,
by induction on $n$ one can show that for $n\ge2$ and any real $c\_1,\dots,c\_n$
\begin{align\*}
&\ol h\_{1,\ep}(\om)\le c\_1\ \&\ \cdots\ \&\ \ol h\_{n,\ep}(\om)\le c\_n \\
&\iff \\
& \begin{aligned}
& \ol h\_{1,\ep}(\om)\le c\_1\ \&\ \cdots\ \&\ \ol h\_{n-1,\ep}(\om)\le c\_{n-1} \\
&\&\ \forall\de\_n\in\Q\_{++}\ \exists \de\_1,\dots,\de\_{n-1}\in\Q\_{++}\
\forall\tom\in\tOm\quad \\
&\quad\ \om\in B\_\tom(\ep) \\
&\quad\ \implies
(h\_1(\tom)\ge c\_1-\de\_1\ \&\ \cdots\ \&\ h\_{n-1}(\tom)\ge c\_{n-1}-\de\_{n-1} \\
&\qquad\qquad \implies
h\_n(\tom)\le c\_n+\de\_n).
\end{aligned}
\end{align\*}
Therefore and because the balls $B\_\tom(\ep)$ are measurable whereas the sets $\Q\_{++}$ and $\tOm$ are countable, we conclude that the $\R^n$-valued function $\ol h\_\ep$ is measurable.
Moreover, since the set $H\_\ep(\om)$ is nondecreasing in $\ep$, so is the maximum $\ol h\_\ep(\om)$ of the set $\ol{H\_\ep(\om)}$. So, there is a pointwise limit $\ol h$ of $\ol h\_\ep$: for each $\om\in\Om$,
\begin{equation}
\ol h(\om):=\lim\_{\ep\downarrow0}\ol h\_\ep(\om)
\end{equation}
and, hence, by the continuity of the map $\psi^{-1}$,
\begin{equation}
g(\om):=\psi^{-1}(\ol h(\om))=\lim\_{\ep\downarrow0}\psi^{-1}(\ol h\_\ep(\om)).
\end{equation}
The function $\ol h$ is measurable, as the pointwise limit of the measurable functions $\ol h\_\ep$. Also, the continuous map $\psi^{-1}$ is measurable. So, $g=\psi^{-1}\circ \ol h$ is measurable.
The maximum $\ol h\_\ep(\om)$ of the set $\ol{H\_\ep(\om)}$ is of course in $\ol{H\_\ep(\om)}$ and hence $\ol h\_\ep(\om)=\lim\_{r\to\infty}h(\tom\_{\ep,\om,r})$ -- for each real $\ep>0$, each $\om\in\Om$, and some sequence $(\tom\_{\ep,\om,r})\_{r\in\N}$ in $\tOm\cap B\_\om(\ep)$.
Since $\psi^{-1}$ is continuous and $\psi^{-1}(h(\tom))=f(\tom)$ for all $\tom\in\tOm$, we have
\begin{equation}
g(\om)=\lim\_{\ep\downarrow0}\psi^{-1}(\lim\_{r\to\infty}h(\tom\_{\ep,\om,r}))
=\lim\_{\ep\downarrow0}\lim\_{r\to\infty}\psi^{-1}(h(\tom\_{\ep,\om,r}))
=\lim\_{\ep\downarrow0}\lim\_{r\to\infty}f(\tom\_{\ep,\om,r}).
\end{equation}
So, for each $\om\in\Om$ there is a sequence $(r\_t(\om))\_{t\in\N}$ in $\N$ such that
\begin{equation}
g(\om)
=\lim\_{t\to\infty}f(\tom\_{1/t,\om,r\_t(\om)}).
\end{equation}
Moreover, $\tom\_{1/t,\om,r\_t(\om)}\in B\_\om(1/t)$ and hence $\tom\_{1/t,\om,r\_t(\om)}\to\om$ as $t\to\infty$. $\quad\Box$
*Remark:* Some of the conditions here can be relaxed. In particular, we do not need $M$ to be a Riemannian manifold or any manifold at all. In particular, it suffices that $M$ be any topological space admitting a finite open cover $\{U\_i\colon i\in[k]\}$ with homeomorphisms $\psi\_i$ of $\ol{U\_i}$ onto compact domains $D\_i\subset\R^{n\_i}$, for each $i\in[k]$.
| 1 | https://mathoverflow.net/users/36721 | 415242 | 169,294 |
https://mathoverflow.net/questions/415244 | 2 | By trying to extend certain limit properties of sequences from compact subsets to the entire set, I cam up with something that can be formed in the following question.
Let $a\_{mn}$ be a double sequence of nonnegative real numbers. I want to be able to switch order of iterated limits in the following form
$$\limsup\_{m \to \infty}\limsup\_{n \to \infty} a\_{mn}=\limsup\_{n \to \infty}\limsup\_{m \to \infty} a\_{mn}$$
under the following conditions
* $a\_{mn}$ is increasing in $m$ for every $n$
* $\limsup\_{m \to \infty} a\_{mn}=\lim\_{m \to \infty} a\_{mn}=a\_{n}, \forall n$
* $\limsup\_{n \to \infty} a\_{mn} \leq a\_m, \forall m$ where $a\_m$ is increasing in $m$
I tried all of the ideas from [here](https://math.stackexchange.com/questions/15240/when-can-you-switch-the-order-of-limits), but without any success. Any helpful ideas or references?
| https://mathoverflow.net/users/121671 | Switching order of limits in double sequences | $a\_{mn}=1(m\ge n)$ is a counterexample. (Here, $1(A)$ is the indicator of an assertion $A$. That is, $1(A):=1$ if $A$ is true and $1(A):=0$ if $A$ is false.)
If you insist on understanding "increasing" in the strict sense, then $a\_{mn}=1(m\ge n)-1/m-1/n$ is a counterexample.
| 7 | https://mathoverflow.net/users/36721 | 415245 | 169,295 |
https://mathoverflow.net/questions/415246 | 2 | I am looking for a reference to the Epstein zeta function. For the Riemann zeta function, there is Titchmarsh's treatment. However, I do not know of any references regarding the Epstein zeta function and I'll be thankful to anyone who points me to a seminal paper/book that talks about properties of the Epstein zeta function and why it is an interesting object of study. This question is probably both broad and imprecise, but I hope it will attract an interesting set of answers.
In addition, I was wondering if there are some "interesting" conjectures that drive research regarding the Epstein zeta function (or where the Epstein zeta-function naturally occurs).
| https://mathoverflow.net/users/392272 | Resources and outstanding conjectures about the Epstein zeta function | I would recommend Harold Stark's papers on the Epstein Zeta function, such as "[L-functions and character sums for quadratic forms, I](https://doi.org/10.4064/aa-15-3-307-317)", Acta Arith. vol 14 (1967/68), PP. 35–50, and "[On the zeros of Epstein's zeta function](https://doi.org/10.1112/S0025579300008007)", Mathematika vol 14 (1967), pp. 47–55.
These contributed to the understanding of the Deuring–Heilbronn phenomenon, and the solution of class number problems for class number 1, 2, and 3.
An earlier reference (in German) is Deuring's "[Zetafunktionen quadratischer Formen](https://doi.org/10.1515/crll.1935.172.226)", J. Reine Angew. Math., vol 172,
(1934), pp. 226-252.
---
As to open problems, here's a suggestion, with some number theory background for context. There is a very elegant theorem about discriminants of binary quadratic forms with one class per genus, which motivates the desire to classify them. The congruence class of a prime number $p$ modulo $d$ determines which form of discriminant $-d<0$ represents $p$ if and only if there is one class per genus \cite{Cox}.
Such discriminants which are congruent to $0$ modulo $4$ are of course Euler's *numeri idonei* or idoneal numbers. Euler expected there would be infinitely many such discriminants, and was surprised to be unable to find more. It was Gauss who conjectured that the only such discriminants are the 65 examples (not necessarily fundamental) known to Euler. There are also 65 known fundamental discriminants (not necessarily even) with one class per genus. The existence of a 66th is still an open problem. By genus theory we know that for discriminants with one class per genus, the class group satisfies
$$
\mathcal C(-d)\cong \left(\mathbb Z/2\right)^{g-1},
$$
where $g$ is the number of prime divisors of $d$. Obviously $d$ is bigger than the absolute value of the smallest fundamental discriminant with $g$ prime divisors,
$$
d\_g\overset{\text{def.}}=3\cdot4\cdot5\cdot7\dots \cdot p\_g.
$$
From lower bounds on the size of $p\_g$, the $g$th prime and on $\theta(x)=\sum\_{p<x}\log(p)$, one can show that
$$
d\_g>g^g.
$$
Since $
2^{g-1} \ll \sqrt{g^g},
$
lower bounds for the class number which we expect to be true rule out the possibility for one class per genus for large $g$. Chowla used this idea to show that the number of classes per genus tends to infinity with $d$.
In 1973, Peter Weinberger showed that on GRH, no fundamental discriminant $-d<-5460$ has one class per genus, and unconditionally there is at most one more such $d$. Weinberger used Tatuzawa's version of Siegel's Theorem to deduce there is at most one such $d$ bigger than $d\_{11}=401120980260\approx 4\times 10^{11}$, and sieving to eliminate the $d<d\_{11}$.
In contrast, Oesterle explicitly observed that the lower bound due to Goldfeld-Gross-Zagier is not strong enough to finish the classification of discriminants with one class per genus:
$
\log(g^g) $ is $\ll 2^{g-1}
$.
Iwaniec and Kowalski observed that even the full strength of the Birch Swinnerton-Dyer conjecture, "the best effective lower bounds which current technology allows us to hope for" would not suffice, as $ \log(g^g)^r$ is $\ll 2^{g-1}$ for any $r$. In fact, the outlook is still more bleak: Watkins observed that if the discriminant $-d$ is divisible by all the primes up to $(\log\log d)^3$ (as $d\_g$ certainly is), the product over primes dividing $d$ in the Goldfeld-Gross-Zagier lower bound is so small the resulting bound is worse than the trivial bound.
The Deuring-Heilbronn phenomenon says that the existence of a Landau-Siegel zero for a Dirichlet $L$-function $L(s,\chi\_{-d})$ would affect the position of zeros of other $L$-functions. In this context, the low lying complex zeros of $\zeta(s)L(s,\chi\_{-d})$ for class number 1 are shown to be on the critical line in the papers of Deuring and Stark above. This is the local version of 'modified GRH'.
Moreover the location of the zeros on the critical line is constrained; they are forced to be nearly periodic.
An interesting application of this phenomenon can be seen in Stark's PhD thesis. Stark used explicit values for the zeros of $\zeta(s)$ to show no discriminant $-d$ had class number $1$ in the range $163<d<\exp(5.6\cdot 10^9)$.
It would be interesting to adapt the methods of Stark, (and later improvements of Montgomery-Weinberger, and of Watkins) to the study of discriminants with one class per genus. Traditionally looks at binary quadratic forms $Q(x,y)=ax^2+bxy+cy^2$ which are reduced; in this case $a$ is the minimum nonzero value represented by $Q$. However if there is one class per genus, it is more convenient to look at forms with $a|d$, $a<\sqrt{d}$; the classes are in one to one correspondence with such divisors. These are the so-called 'ambiguous' forms. Most such forms are reduced, and those which are not are still close enough to being reduced that we get good estimates.
So here's a suggestion for an open problem on the Epstein zeta function: adapt the methods of Stark's thesis (for the Epstein zeta function attached to the. principal form, i.e. $h=1$) to Epstein zeta functions attached to ambiguous forms, to be able to use zeros of $\zeta(s)$ to eliminate the possibility of one class per genus, up to some very large value of $d$.
I am not suggesting this is straightforward - I spent some time thinking about this and got nowhere.
| 4 | https://mathoverflow.net/users/6756 | 415248 | 169,296 |
https://mathoverflow.net/questions/415250 | 2 | The starting point of my question is the following fact: suppose $G$ is a finite group and let $H,K \leq G$ be arbitrary subgroups, then there exists an isomorphism of $G$-sets as follows
\begin{equation}
G/H \times G/K \cong \coprod\_{H \backslash G/K}G/(H^g\cap K).
\end{equation}
The proof is not difficult: given an element $(xH,yK)$ we see it has stabilizer $xHx^{-1}\cap yKy^{-1}$ which is isomorphic to $H^{x^{-1}y}\cap K$. Then we map the orbit of $(xH,yK)$ to the coset $G/(H^{x^{-1}y}\cap K)$ indexed at the element $Hx^{-1}yK$ of the double coset. It is basic group calculus to verify that the function provided in this way is bijective and $G$-equivariant.
This fact has the following consequence in equivariant stable homotopy theory: it allows us to show that, for a fixed family $\mathcal{F}$ of subgroups of $G$, then $\text{Thick}(G/H\_+ : H \in \mathcal{F})$, the thick subcategory of compact $G$-spectra generated by the cosets $G/H\_+$ for all $H \in \mathcal{F}$, is a tensor ideal. In fact, knowing that the thick subcategory of compact $G$-spectra is generated by $G/K\_+$ for the various $K \leq G$, using the above formula we see the smash product $G/H\_+\wedge G/K\_+$ decomposes as a wedge of cosets $G/L\_+$ where $L \in \mathcal{F}$.
I wanted to prove the same claim in the case $G$ is a compact Lie group and the subgroups are assumed to be closed. In this situation we could define a map as above, going to the coproduct of the orbits, but we have no guarantee it is continuous. Indeed, I believe that the topology of the double coset $H\backslash G/K$ should be taken into account to produce a homeomorphic $G$-space.
It is know that the double coset admits a decomposition as follows: first we consider $G/K$ with the obvious left $H$-action, the stabilizer of $gK$ is given by $H\_{gK}=H\cap gKg^{-1}$. We say that two orbits have the same type if the stabilizers of their points are conjugate in $H$. Also, we say that an orbit $H\backslash L\_1$ has type lesser than or equal to $H \backslash L\_2$ if $L\_2$ is conjugate to a subgroup of $L\_1$. Given that, it has been proved that the double coset $H\backslash G/K$ can be decomposed as a disjoint union of manifolds of possibly various dimensions. Each of these manifolds is given by double cosets in the same orbit type and they could be not connected. Moreover, the closure of an orbit type manifold is given by the manifolds with lesser or equal orbit type. Nevertheless, the number of types and the connected components of each manifold are finite.
This is described in the paper "[The Transfer and Compact Lie Groups](https://doi.org/10.2307/1998687)" of Feshbach, where he gives a reference to Bredon's "Introduction to compact transformation groups" for a proof of the finiteness of the connected components and the orbit types.
I tried to to use the manifold decomposition of the double coset to produce a suitable $G$-space and concoct an homeomorphism with the product to cosets, but without results. I am not able to present an explicit formula as in the finite case.
Then I fell back into just showing that the product $G/H\times G/K$ admits a structure of finite $G$-CW complex: since the stabilizers are again in the form $G/L\_+$ with $L$ subconjugated to $H$ and $K$ the induced suspension spectrum must belong to the thick subcategory generated by the various $G/L\_+$. However, I did not manage to find an appropriate result in the literature stating that the product of two $G$-CW complexes is still a $G$-CW complex in the compact case.
Indeed, when we deal with the product of cells and try to show that a generic product $(G/H \times D^{n\_1}) \times (G/K \times D^{n\_2})$ admits a decomposition in equivariant cells in the end we have to treat the product of cosets $G/H\times G/K$, so we go back to the previous problem.
To conclude, I have two questions:
1. Can you provide a complete reference for a result showing that the product of two $G$-CW complexes is still a $G$-CW complex in the general case where $G$ is a compact Lie group?
2. Does there exist some sort of known analogue of the double coset decomposition formula in the finite case?
| https://mathoverflow.net/users/131453 | Double coset decomposition for compact Lie groups | Your question (1) may not be true as stated. One approach that I like is to think of the product of two $G$-CW complexes first as a $(G\times G)$-CW complex, which works because $G/H\times G/K \approx (G\times G)/(H\times K)$. Then you can consider it as a $G$-space by restricting along the diagonal $G\to G\times G$. However, in "[Restricting the transformation group in equivariant CW complexes](http://projecteuclid.org/euclid.ojm/1200782103)," Sören Illman showed that the restriction of a $G$-CW complex to an $H$-space does not generally give an $H$-CW complex. He then gave a construction of an $H$-homotopy equivalent $H$-CW complex with nice properties.
Related, and related to your question (2), is Illman's paper "[Existence and uniqueness of equivariant triangulations of smooth proper $G$-manifolds with some applications to equivariant Whitehead torsion](https://doi.org/10.1515/crll.2000.054)." One consequence of this paper is that $G/H\times G/K$ has a $G$-triangulation, so a $G$-CW structure. The problem that arises when you look at a product of two $G$-CW complexes is that the triangulations of the products of orbits don't have to interact well with the original CW structures.
| 4 | https://mathoverflow.net/users/58888 | 415260 | 169,300 |
https://mathoverflow.net/questions/415236 | 3 | Theorem 1 of Milnor's paper "A procedure for killing homotopy groups of differentiable manifolds" states that two manifolds are in the same cobordism class if and only if they can be obtained from each other by a sequence of surgeries (meaning they are $\chi$-equivalent, following Milnor's terminology).
In his proof that cobordism implies $\chi$-equivalence, he utilizes a Morse function on the cobordism manifold, although not explicitly. Is there a reference containing a more modern reformulation of this proof via Morse functions? I would also like to avoid introducing the language of attaching handles if possible.
| https://mathoverflow.net/users/475761 | A question on the proof of Theorem 1 in Milnor's "Killing Homotopy Groups" | If you want to avoid the language of attaching handles there is the treatment in Milnor's Lectures on the h-cobordism theorem. He proves that a cobordism with a Morse function with only one critical point (an `elementary cobordism') gives rise to a surgery, which is the statement you're looking for. To a modern reader, the handles are there even if that terminology isn't used.
I remember hearing as a grad student that Milnor wrote the lectures in that way because Morse didn't like handles (which was the way that Smale described his argument) or was worried about doing handle attachments smoothly. So (according to this possibly urban legend) Milnor phrased things in a way that Morse (who again allegedly attended the lectures) the statements were all about eliminating critical points of a Morse function rather than cancelling handles and so on. I would be interested if anyone can provide any evidence for this story.
| 3 | https://mathoverflow.net/users/3460 | 415267 | 169,303 |
https://mathoverflow.net/questions/415221 | 2 | Let $G$ be a $d$-regular graph. Let $d= \lambda\_1 \ge \lambda\_2 \ge \dots \ge \lambda\_n \ge -d $ be the eigenvalues of the adjacency matrix of $G$, and set $\lambda = \max (|\lambda\_2| , |\lambda\_n|) $ We say that $G$ is a *Ramanujan graph* if $\lambda \le 2 \sqrt{d-1}$.
[According to Wikipedia](https://en.wikipedia.org/wiki/Ramanujan_graph), if $\mathbb{F}\_q$ is a finite field of order $q$, and $S= \{ f(x) : x\in \mathbb{F}\_q \}$ is the image of a degree 2 or degree 3 polynomial $f(x)$ over $\mathbb{F}\_q$ such that $S = -S$, then the Cayley graph for $\mathbb{F}\_q$ with generators $S$ is a Ramanujan graph. Is there a standard reference for this fact?
I am wondering about more general formulations, especially to multivariate polynomials, and zero sets of polynomials rather than images.
The specific example I have in mind is the following. Let $p$ be a prime and set $S= \{ (x,y) \in \mathbb{F}\_p^2 : x^2 + y^2 = 1 \}$. It seems that the Cayley graph on $\mathbb{F}\_p^2$ with generators $S$ is Ramanujan. A number theorist friend sketched a proof of this for me recently. The proof involved classical Weil estimates for Kloosterman sums in the case $p \cong 1 \pmod{4}$ and base change to reduce to the classical case when $p \cong 3 \pmod{4}$. Comment: the degree of this graph is $d=p-1$ when $p \cong 1 \pmod{4}$, and $d=p+1$ when $p \cong 3 \pmod{4}$.
I am wondering if this follows directly from more general constructions for Ramanujan graphs that are well known to experts (i.e. Cayley graphs made from varieties over finite fields), or if this example is new.
Update: it turns out this specific example is already known. I added an answer below. I am still wondering about a reference that explains how to get Ramanujan graphs from Cayley graphs over finite fields (and suitable polynomials), as in the statement from Wikipedia.
| https://mathoverflow.net/users/4558 | Ramanujan graphs from varieties over finite fields | The assumptions mean that the eigenvalues of the Cayley graph are of the form $\sum\_{x\in S} \psi (x)$ for $\psi \colon \mathbb F\_q\to \mathbb C^\times$ an additive character, which is $\sum\_{x\in \mathbb F\_q}\psi(f(x))$ since $S$ is taken with multiplicity.
For sums of this type, we have the bound
$$\left|\sum\_{x\in \mathbb F\_q}\psi(f(x))\right| \leq (\deg f-1) \sqrt{q}$$ over any finite field $\mathbb F\_q$ of characteristic greater than $\deg f$, for any nontrivial character $\psi$.
This bound makes the Cayley graph Ramanujan when $\deg f = 2$ or $3$.
This bound is due to Gauss in the degree $2$ case, and is due to Weil in the higher degree case (in [the same paper that the bound for Kloosterman sums is from](https://www.jstor.org/stable/88420), equation (5), although the translation from that reference is annoying).
| 3 | https://mathoverflow.net/users/18060 | 415271 | 169,305 |
https://mathoverflow.net/questions/415266 | 1 | So, I've been reading about the lonely runner conjecture, but I have to admit that my knowledge of diophantine approximation is so limited that I wouldn't be able to properly explain the term.
So let's consider the simplest case, which is $n$ "real" "linear" runners in a circle ($=\mathbb R/\mathbb Z$).
Let me try and view this problem from the angle of equidistribution on tori; I don't know a lot about this subject either, but I think I remember having read about this in a book by Tao when the library was still open (ie. before Corona). So suppose we got $n$ runners with speeds $v\_1, \ldots, v\_n$. Choose a $t > 0$ such that $α := t (v\_1, \ldots, v\_n)$ has only irrational components. Then the sequence $x\_n := nα$ should be equidistributed on the $n$-torus, and therefore we should be able to approximate any configuration of runners by choosing a point within the $n$-torus and using equidistribution.
I'm only vaguely familiar with the case of polynomials, but this should in principle generalise to polynomial functions under certain assumptions on the leading coefficient, right?
Is this approximately how one would approach the problem from that angle? Is there any reference?
I'd be enormously grateful for any comments on this matter.
Thank you very much in advance.
| https://mathoverflow.net/users/58976 | The lonely runner conjecture and equidistribution on tori | A corrected version of this argument is contained in Section 4 of [Six Lonely Runners](https://doi.org/10.37236/1602) by Bohman, Holzman, and Kleitman.
Their argument shows, using equidistribution, that the case of the lonely runner conjecture with rational speeds implies the general case.
| 5 | https://mathoverflow.net/users/18060 | 415276 | 169,307 |
https://mathoverflow.net/questions/415284 | 3 | As in [Deformations of Hirzebruch surfaces and toric action](https://mathoverflow.net/questions/69339/deformations-of-hirzebruch-surfaces-and-toric-action),
the Hirzebruch surface $F\_n$ can be deformed into $F\_{n-2m}$ ($0<2m\leq n$) under the fibration given by
$$
M=\{([x\_0:x\_1],[y\_0:y\_1:y\_2],t)\in \mathbb{P}^1\times \mathbb{P}^2 \times \mathbb{C} \mid x^n\_0y\_1−x^n\_1y\_0+tx^{n−m}\_0x^m\_1y\_2=0\}
$$
over $\mathbb{C}$. What I would like to know is the explicit equations that give the negative section of each Hirzebruch surface in this situation. Of course that of the central fiber $M\_0$, which is a ($-n$)-section, is given by $y\_0=y\_1=0$ in $M\_0$. On the other hand, for a noncentral fiber $M\_t$ I cannot find the way to determine the equations of its ($-n+2m$)-section.
I would be grateful for any comments.
| https://mathoverflow.net/users/476418 | Where's the negative section of a deformation of a Hirzebruch surface? | You can take $y\_0 = 0$, $y\_1 = -tx\_1^m$, $y\_2 = x\_0^m$.
EDIT. Here is a simple computation of the normal bundle of the curve
$$
C\_t = \{([x\_0:x\_1],[0:-tx\_1^m:x\_0^m])\} \subset \mathbb{P}^1 \times \mathbb{P}^2
$$
in the surface
$$
S\_t = \{x\_0^ny\_1−x\_1^ny\_0+tx\_0^{n−m}x\_1^my\_2=0\}
$$
for $t \ne 0$. First, the normal bundle fits into the exact sequence
$$
0 \to N\_{C\_t/S\_t} \to N\_{C\_t/\mathbb{P}^1 \times \mathbb{P}^2} \to N\_{S\_t/\mathbb{P}^1 \times \mathbb{P}^2}\vert\_{C\_t} \to 0.
$$
Since $C\_t$ is a graph of a map $\mathbb{P}^1 \to \mathbb{P}^2$ (of degree $m$), its normal bundle is the pullback of the tangent bundle of $\mathbb{P}^2$, hence its degree is
$$
\deg(N\_{C\_t/\mathbb{P}^1 \times \mathbb{P}^2}) = 3m.
$$
On the other hand, $S\_t$ is a divisor of type $(n,1)$ on $\mathbb{P}^1 \times \mathbb{P}^2$, hence its normal bundle restricted to $C\_t$ has the degree
equation which has degree
$$
\deg(N\_{S\_t/\mathbb{P}^1 \times \mathbb{P}^2}\vert\_{C\_t}) = n + m.
$$
Therefore,
$$
\deg(N\_{C\_t/S\_t}) = 3m - (n + m) = 2m - n.
$$
| 6 | https://mathoverflow.net/users/4428 | 415287 | 169,309 |
https://mathoverflow.net/questions/415277 | 2 | Consider the category of chain complexes over a ring $R$.
We can show that $\text{Ext}^1(M, N)$ classifies extensions using the triangulated category structure: the homotopy kernel of a map $N \rightarrow M[1]$, $K$, gives rise to the designated triangle $M \rightarrow K \rightarrow N$. This gives a simple picture of Yoneda's extension theorem for $\text{Ext}^1(M, N)$, which also seems to rely on the "threeness" of a triangulated category.
My question is, is there an extension of this view to $\text{Ext}^n(M, N)$?
The proof of the situation for $\text{Ext}^1$ is quite simple using triangulated categories and it would be nice if the situation could be handled the same way for $\text{Ext}^n$.
| https://mathoverflow.net/users/30211 | Yoneda Ext theorem and extensions | We have that $\mathrm{Ext}^n(M, N) = \mathrm{Hom}(M, N[n])$, in other words, an element of $\mathrm{Ext}^n(M, N)$ is a map
$$
M \longrightarrow I\_N[n]
$$
where $N \to I\_N$ is an injective resolution. From this we obtain a distinguished triangle,
$$
M \longrightarrow I\_N[n] \longrightarrow C \overset{+}\longrightarrow
$$
The homotopical invariant gives an extension of complexes
$$
0 \longrightarrow I\_N[n] \longrightarrow C \longrightarrow M \longrightarrow 0 \
$$
which, plugin-in the beginning of the resolution of $N$, yields the classical $n$-extension
$$
0 \longrightarrow N \longrightarrow I^0\_N \longrightarrow I^1\_N \cdots \longrightarrow I^{n-1}\_N \longrightarrow M \longrightarrow 0
$$
as wanted
| 3 | https://mathoverflow.net/users/6348 | 415295 | 169,311 |
https://mathoverflow.net/questions/412526 | 2 | The following paper is about how a K-flow is produced from a K-induced map, but it is written in Russian. Does someone know where to find its English version? Do some textbooks include this topic?
B. Gurevič, [Certain conditions for the existence of K-decompositions for special flows](http://mi.mathnet.ru/eng/mmo192),
Tr. Mosk. Mat. Obs., 1967, Volume 17, Pages 89–116.
| https://mathoverflow.net/users/124254 | K-flows reference | The Gurevich paper you cite was translated as "Some Existence Conditions for K-Decompositions for Special Flows" in Trans. Moscow Math. Soc. 17 (1967), 99-128 (if you don't have access to a university library I can send a copy to you; you can find my university email address by searching my name). There is also the Totoki paper as mentioned in the [comments](https://mathoverflow.net/questions/412526/k-flows-reference#comment1057360_412526). Totoki also has a book titled *Ergodic Theory*, which includes the argument in the paper without much change it seems. Ornstein and Smorodinsky's paper "[Ergodic flows of positive entropy can be time changed to become $K$-flows](https://doi.org/10.1007/BF03007657)" has an argument very similar to Totoki's argument.
Sinai's two papers "Dynamical systems with countable Lebesgue spectrum " [I](http://dx.doi.org/10.1090/trans2/039/04) and [II](http://dx.doi.org/10.1090/trans2/068/03) are the standard references for $K$-flows.
Katok has a related paper titled "[Smooth Non-Bernoulli $K$-Automorphisms](https://www.personal.psu.edu/axk29/pub/KatokInv1980.pdf)" where he interprets the $K$-property for skew products as a cohomological equation. (This paper uses smooth structures; instead of being purely ergodic theoretical.)
Finally, most ergodic theory books don't include flows at all; one book that does is Cornfeld, Fomin, Sinai's *[Ergodic Theory](https://doi.org/10.1007/978-1-4615-6927-5)* with a special emphasis on the connection between $K$-property and spectral properties (there is also a chapter on special flows; but the connection between $K$-property for special flows and the $K$-property for the base automorphism is not developed). Nadkarni's *[Basic Ergodic Theory](https://doi.org/10.1007/978-93-86279-53-8)* also has a chapter on flows; but it does not have anything on the $K$-property.
| 2 | https://mathoverflow.net/users/66883 | 415315 | 169,323 |
https://mathoverflow.net/questions/415312 | 4 | In my earlier (soft) [MO post](https://mathoverflow.net/questions/415230/on-a-variation-of-the-vandermonde-matrix), an elementary [response](https://mathoverflow.net/questions/415230/on-a-variation-of-the-vandermonde-matrix#comment1064945_415230) was given by Ofir Gorodetsky in regard to the determinant of the symbolic counterpart to the numerical matrix $\mathbf{M}\_n=(i^j-j^i)\_{i,j}^{1,n}$.
I'm now looking at yet another variation. Let $\mathbf{A}\_n=(j^j-i^j)\_{i,j}^{1,n}$ be an $n\times n$-matrix. Denote the (signed) [Stirling numbers of the first kind](https://en.wikipedia.org/wiki/Stirling_numbers_of_the_first_kind) by $s(n,k)$. For clarity, here are some contrasting examples:
$$\mathbf{M}\_3=\begin{pmatrix} 0&-1&-2 \\ 1&0&-1 \\2&1&0 \end{pmatrix}
\qquad \text{and} \qquad \mathbf{A}\_3=\begin{pmatrix}0&3&26 \\-1&0&19 \\ -2&-5&0 \end{pmatrix}.$$
I would like to ask:
>
> **QUESTION.** Is this true? With the convention that $0^0=1$, we have
> $$\det\mathbf{A}\_n=\prod\_{j=1}^{n-1}j!\cdot\sum\_{k=0}^n\,s(n+1,k+1)\cdot k^k.$$
>
>
>
**Remark.** If it helps, since the factor $\prod\_{j=1}^{n-1}j!$ is exactly the determinant of the Vandermonde matrix, $\mathbf{V}\_n=(i^{j-1})\_{i,j}^{1,n}$, we can say
$$\frac{\det\mathbf{A}\_n}{\det \mathbf{V}\_n}=\sum\_{k=0}^n\,s(n+1,k+1)\cdot k^k.$$ This formalism goes in the "spirit" of the (specialization) $s\_{\lambda}(1,\cdots,1)$ of the [*Schur polynomials*](https://en.wikipedia.org/wiki/Schur_polynomial).
| https://mathoverflow.net/users/66131 | Yet, another numerical variant of the Vandermonde matrix | This is true and this is not about numbers $j^j$ at all. Consider a more genral matrix $(c\_j-i^j)\_{1\leqslant i,j\leqslant n}$. Denote $f\_j(x)=c\_j-x^j$, and find the numbers $\beta$, $\alpha\_1,\ldots,\alpha\_{n-1}$ such that $f\_n(x)+\sum\_{j=1}^{n-1}\alpha\_j f\_j(x)=\beta$ for all $x\in \{1,2,\ldots,n\}$. Then the matrix $(f\_j(i))\_{i,j}$ has the same determinant as if we replace the column with $f\_n$ by a column of $\beta$'s, that is seen from operations with columns. Now by operations with columns we may replace each $f\_j$ to $-x^j$, so we get almost a Vandermonde matrix (up to $n-1$ changes of signs in the columns and the cyclic shift, and $\beta$'s instead of 1's), so its determinant equals to the determinant of Vandermonde matrix for the numbers $1,\ldots,n$ times $(-1)^{n-1}(-1)^{n-1}\beta=\beta$. Thus, it appears to find $\beta$. For this, write
$$
f\_n+\sum\_{j=1}^{n-1}\alpha\_jf\_j(x)=\beta-(x-1)(x-2)\ldots(x-n)=
\beta-\sum\_{k=0}^n s(n+1,k+1)x^k.
$$
Equalising the coefficients of $x^k$ we get $\alpha\_k=s(n+1,k+1)$ for all $k=1,\ldots,n$ (where $\alpha\_n=1$ by agreement), thus $\beta-s(n+1,1)=\sum\_{k=1}^n c\_k s(n+1,k+1)$, $\beta=\sum\_{k=0}^n c\_k s(n+1,k+1)$, where $c\_0=1$ by agreement.
| 6 | https://mathoverflow.net/users/4312 | 415317 | 169,325 |
https://mathoverflow.net/questions/415328 | 1 | For the Lie algebra $\frak{sl}\_{n+1}$ we denote its fundamental irreducible representations by $V(\pi\_i)$, with $i=1, \dots, n$. Where can I find a table of the character formula (in other words a list of all the non-zero weight spaces of $V(\pi\_i)$ together with multiplicities)?
P.S. From this answer
[Fundamental representations and weight space dimension](https://mathoverflow.net/questions/314597/fundamental-representations-and-weight-space-dimension)
it looks that the multiplicities are also 1. So my question simplifies to a table of the non-zero weight spaces.
| https://mathoverflow.net/users/378228 | Character formula for the fundamental representations of $\frak{sl}_n$ | These are examples of so called *minuscule* weights which are defined through the property that all the weights are given by $w\lambda$ where $\lambda$ is the highest weight and $w$ is an element of the Weyl group. So, in the $\mathfrak{sl}\_{n+1}$, to get all the weights in "the $\epsilon$ basis" you just take all possible permutations of coordinates of $\lambda$.
| 4 | https://mathoverflow.net/users/6818 | 415329 | 169,327 |
https://mathoverflow.net/questions/415341 | 4 | A positive integer $n$ is called $k$-rough if all of the prime factors of $n$ strictly exceed $k$.
For $k$ fixed and $n$ large, what's the shortest interval $[n,n+t]$ that contains a $k$-rough number ?
| https://mathoverflow.net/users/476460 | Rough numbers in short interval | Being $k$-rough is the same as being relatively prime to the product of primes up to $k$. Hence if $P(k)$ denotes that product, and $j$ denotes the [Jacobsthal function](https://oeis.org/wiki/Jacobsthal_function), then $t=j(P(k))-1$ works for every $n$, while $t=j(P(k))-2$ fails for infinitely many $n$'s. In particular, it follows from a result of [Iwaniec (1971)](http://matwbn.icm.edu.pl/ksiazki/aa/aa19/aa1911.pdf) that $t=ck^2$ works for some absolute constant $c>0$.
| 7 | https://mathoverflow.net/users/11919 | 415342 | 169,332 |
https://mathoverflow.net/questions/415339 | 3 | Let $k = \mathbb{F}\_q$ be a finite field with $q$ elements and $Q\subset\mathbb{P}^n\_k$ a quadric hypersurface defined over $k$.
By the Chevalley-Warning theorem if $n\geq 2$ then $Q$ has a point. Is there a formula for the number of points of $Q$ or at least a bound of the type $\#Q(k)\geq f(q,n)$ where $f$ is a function of $q$ and $n$?
| https://mathoverflow.net/users/14514 | Number of points of a quadric hypersurface over a finite field | Yes. The number of points of a smooth quadric hypersurface is $\frac{q^n-1}{q-1}$ if $n$ is even or $\frac{q^n-1}{q-1} \pm q^{ \frac{n-1}{2}}$ if $n$ is odd.
This can be proven in a number of ways. Here is a (perhaps extravagantly) geometric approach:
Fix a point $x$. A line through $x$ contains one other points of $Q$, unless that line is tangent to $Q$ at $x$ - in other words, is contained in the tangent plane of $Q$ at $x$. If so, then the line contains no other points of $Q$, except if the line is contained in $Q$, in which case it contains $q$ other points. So the number of points is
$$1 + \frac{ q^{n}-1}{q-1} - \frac{ q^{n-1}-1}{q-1} + qN = q^{n-1} +1 + qN$$
where $N$ is the number of lines through $x$ that lie in $Q$, since I have taken the point $x$, added the number of lines, subtracted the number of tangent lines, and added $qN$.
Now the lines through $x$ in the tangent plane are parameterized by an $n-2$-dimensional projective space, and the lines contained in $Q$ form a quadric surface in that projective space. So the number of points on a quadric hypersurface in $\mathbb P^n$ for $n\geq 2$ is $q^{n-1} +1$ plus $q$ times the number of points on a quadric hypersurface in $\mathbb P^{n-2}$. The formulas in above then follow by induction from the easy base cases $n=0, n=1$.
| 8 | https://mathoverflow.net/users/18060 | 415348 | 169,334 |
https://mathoverflow.net/questions/415192 | 0 | Let $(\Omega, \mathcal F, \mu)$ be a standard probability space.
**Question:** For each $f \in L^\infty (\Omega)$, does there exist an ergodic measure preserving transformation $T: \Omega \to \Omega$ such that the following expression is maximised?
$$\lim\_{n \to \infty} \frac{1}{n} \sum\_{k = 1}^n\int\_\Omega |T^k f - T^{k-1} f| \, d\mu$$
Where for $k \geq 1$, $T^k f (x) := f(T^k x)$ and $T^0 f = f$ by convention.
To be unambiguous, the expression is maximized for fixed $f$, over all ergodic measure preserving transformations $T$.
**Remark:** The limit exists due to the pointwise ergodic theorem.
| https://mathoverflow.net/users/173490 | Entropy maximising ergodic transformation | I don't know if you're still interested now that the average collapses, but I think that there is such a maximizing invertible ergodic $T$ as long as you assume that your probability space is Lebesgue.
Under these assumptions, if you normalize so that the median of $f$ is $0$ (i.e. so that $\mu(f < 0) = \mu(f > 0)$), then I think the maximum value of your quantity
$\int |f - Tf|$ is given by $2 \int |f| dx$.
It's clear from the triangle inequality that for every measure-preserving $T$, $\int |f - Tf| \leq \int (|f| + |Tf|) = 2\int |f|$.
Now we want to find $T$ for which $\int |f - Tf| = 2\int |f|$.
Define $Z = \{x \ : \ f(x) = 0\}$, $N = \{x \ : \ f(x) < 0\}$, and $P = \{x \ : \ f(x) > 0\}$. If $Z$ has positive measure, split it arbitrarily into $Z^+$ and $Z^-$ of equal measure (since we assumed $\mu$ is nonatomic). Define $P' = P \cup Z^+$ and $N' = N \cup Z^-$. Then $\mu(P') = \mu(N') = 1/2$, and $f$ is nonnegative on $P'$ and nonpositive on $N'$.
I claim that any measure-preserving $T$ with $T(N') = P'$ and $T(P') = N'$ works.
For such $T$, and for every $x \in N'$, $Tx \in P'$, meaning that $Tf(x) - f(x)$ is nonnegative. Similarly, for every $x \in P'$, $Tf(x) - f(x)$ is nonpositive. Therefore, $\int |f - Tf|$ is
$\int\_{N'} (Tf - f) + \int\_{P'} (f - Tf) = \int\_{N'} Tf - \int\_{N'} f + \int\_{P'} f - \int\_{P'} Tf = 2 (\int\_{P'} f - \int\_{N'} f) = 2 \int |f|$. (The second-to-last equation follows from the fact that $T$ is measure-preserving.)
Now we just need to justify that there exists an invertible ergodic $T$ with $T(P') = N'$ and $T(N') = P'$, but this should be easy. Just find a measure-preserving bijection $S: P' \rightarrow N'$ ($P'$ and $N'$ are positive measure subsets of a Lebesgue space, and so are also Lebesgue and thus isomorphic as probability spaces) and any invertible ergodic map $R: P' \rightarrow P'$. Finally, define
$T(x) = \begin{cases}
S(x) & n \in P' \\
RS^{-1}(x) & n \in N'
\end{cases}$
Since $T^2$ restricted to $P'$ is just $R$, for every positive measure subset $A \subset P'$, $\bigcup\_n T^{2n} A = \bigcup\_n R^n A = P'$ by ergodicity. Ergodicity of $T$ should follow almost immediately.
| 2 | https://mathoverflow.net/users/116357 | 415351 | 169,335 |
https://mathoverflow.net/questions/415125 | 3 | Let $B$ be a standard Brownian motion, and $A$ a process of finite variation on compacts almost surely, not necessarily adapted to the Brownian filtration.
**Question:** Denoting by $\mathcal L$ the Lebesgue measure, is it true that $$\mathcal L(\{t \, | \, B\_t = A\_t \}) = 0$$
almost surely?
| https://mathoverflow.net/users/173490 | Intersection of Brownian motion and finite variation process | Fix a finite interval $I$.It suffices to show that almost surely,
$$\mathcal L(\{t \in I \, | \, B\_t = A\_t \}) = 0 \,.$$
Brownian motion restricted to $\{t \in I \, | \, B\_t = A\_t \}$ has bounded variation, so a positive answer is implied by the following stronger result, a special case of Theorem 1.3 of [1]:
**Theorem[1]** Let $\{B(t): t\in [0,1]\}$ be a standard Brownian motion.
Then, almost surely, for all $S\subset [0,1]$,
if $B|\_{S}$ is of bounded variation, then $\overline{\dim}\_M S\leq 1/2$.
Here $\overline{\dim}\_M$ is the upper Minkowski dimension (a.k.a. upper box dimension.)
[1] Angel, Omer, Richárd Balka, András Máthé, and Yuval Peres. "Restrictions of Hölder continuous functions." Transactions of the American Mathematical Society 370, no. 6 (2018): 4223-4247.
<https://www.ams.org/journals/tran/2018-370-06/S0002-9947-2018-07126-4/S0002-9947-2018-07126-4.pdf>
<http://wrap.warwick.ac.uk/84253/7/WRAP-restrictions-continuous-functions-Peres-2018.pdf>
| 4 | https://mathoverflow.net/users/7691 | 415358 | 169,338 |
https://mathoverflow.net/questions/415381 | 5 | Let $A$ be a finite-dimensional algebra over a field $F$. A representation $M$ of $A$ is called **absolutely irreducible** if $M\otimes\_FE$ is irreducible as a representation of $A\otimes\_FE$ for all field extensions $E$ over $F$.
A field $F$ is said to be a **splitting field** of $A$ if any irreducible representation of $A$ is absolutely irreducible.
There are some results on splitting fields. For example, suppose that $F$ is a splitting field of $A$. Then any field extension $E$ over $F$ is a splitting field, and moreover, irreducible representations of $A$ are in one-to-one correspondence with irreducible representations of $A\otimes\_FE$ via $S\mapsto S\otimes\_FE$.
Conversely, suppose that $A$ is an $F$-algebra and $E$ (over $F$) is a splitting field of $A$. Then $F$ is a splitting field if and only if irreducible representations of $A$ are in one-to-one correspondence with irreducible representations of $A\otimes\_FE$ via $S\mapsto S\otimes\_FE$.
Now suppose that $A$ is an $F$-algebra and $E$ (over $F$) is a splitting field of $A$. My question is that, if we only know that every irreducible representation of $A\otimes\_FE$ is equal to $S\otimes\_FE$ for some irreducible representation $S$ of $A$, can we conclude that $F$ is also a splitting field of $A$? Notice that we do not assume that $S\otimes\_FE$ is an irreducible representation of $A\otimes\_FE$ for every irreducible representation $S$ of $A$.
| https://mathoverflow.net/users/56989 | Absolutely irreducible representation and splitting field | Yes, $F$ is a splitting field of $A$. Put $A\_E = A\otimes\_F E$ for convenience. Let $S'$ be a simple $A$-module. Then $S'\otimes\_F E$ has a simple $A\_E$-submodule, which by hypothesis is of the form $S\otimes\_F E$ for some simple $A$-module $S$. Note that $$0\neq \hom\_{A\_E}(S\otimes\_F E,S'\otimes\_F E)\cong \hom\_A(S,S'\otimes\_F E)$$. But as an $A$-module, $S'\otimes\_F E$ is isomorphic to a direct sum of $[E:F]$ copies of $S'$ (this can be infinite if the field extension is infinite, but that's ok). So $$0\neq \hom\_A(S,S'\otimes\_F E)\leq \hom\_A(S,S')^{[E:F]}$$ (this is equality if $[E:F]<\infty$). So by Schur's lemma, $S\cong S'$ and therefore $S'\otimes\_F E$ is simple. It now follows that $F$ is a splitting field because $E$ is one.
| 3 | https://mathoverflow.net/users/15934 | 415386 | 169,343 |
https://mathoverflow.net/questions/415383 | 2 | MOTIVATION.
Let $f:[0,+\infty)\to \mathbb{R}$.
Solutions to $\partial\_tf(t) = -\lambda f(t)$, $f(0)\neq 0$ approach zero *exactly* as $e^{-\lambda t}$.
This property is preserved if we apply an exponentially decaying perturbation, i.e. we consider
$$\partial\_t f(t) = -\lambda f(t) +R(t) f(t)$$
where $|R(t)|< Ce^{-\delta t}$, $\delta >0$.
To see this we can use that $f(t) = f(0) e^{-\lambda t + \int\_{0}^tR(s) ds}$, and
$$e^{-\lambda t + \int\_{0}^tR(s) ds} \geq e^{-\lambda t + \int\_0^t Ce^{-\delta s}ds}
\geq e^{-\lambda t + C\_1}.$$
Therefore even in the perturbed case $f(t) = \Theta(e^{-\lambda t})$ for $t\to +\infty$.
QUESTION:
>
> what happens in higher dimension? Suppose that $f:[0,+\infty) \to H$, $H$ Hilbert space, solves
> $$ \partial\_t f(t) = - Af + R(t)f(t)$$
> $$f(0) \neq 0 $$
> where $A, R(t) \in \mathcal{L}(H)$ and $||R(t)||< C e^{-\delta t}$ and $A$ is symmetric and has an (ortho)-eigenbasis $\{\psi\_\lambda\}\_{\lambda\in \sigma(A)}$, $\sigma(A) $ discrete, $0\not \in \sigma(A)$.
> Does still hold that $f\_\lambda(t) = \Theta(e^{-\lambda t})$ when not $0$?
>
>
>
Here $f\_\lambda(t) = \langle f(t), \psi\_\lambda\rangle$ is the projection on the subspace generated by the eigenvector $\psi\_\lambda$.
Of course if $[A,R(t)]= 0$ then the result holds as we can reduce to the 1D case.
| https://mathoverflow.net/users/99042 | Decay rate for a small perturbation of a simple linear ODE | $\newcommand\R{\mathbb R}$The answer is no.
E.g., suppose that $H=\R^2$, with the standard basis $(e\_1,e\_2)$, $Ae\_1=3e\_1$, $Ae\_2=e\_2$, $R(t)e\_1=0$, $R(t)e\_2=e^{-t}e\_1$, and $f(0)=(0,1)$. So, $(e\_1,e\_2)$ is an ortho-eigenbasis of $A$, with the corresponding eigenvalues $3$ and $1$.
However, $f(t)\cdot e\_1=e^{-2t}-e^{-3t}\ne\Theta(e^{-3t})$.
| 1 | https://mathoverflow.net/users/36721 | 415398 | 169,346 |
https://mathoverflow.net/questions/415395 | 9 | There are 10 compact flat 3 manifolds up to diffeomorphism, 6 orientable and 4 non orientable. I am looking to better understand how to construct the orientable ones.
The six orientable ones are determined by their holonomy groups
$$
C\_1,C\_2,C\_3,C\_4,C\_6
$$
and
$$
C\_2 \times C\_2
$$
The five with cyclic holonomy all arise as the mapping torus of a mapping class of $ T^2 $ with the corresponding order: 1,2,3,4, or 6. These five Euclidean manifolds with cyclic holonomy can even be constructed as a quotient of the special Euclidean group $ SE\_2 $ by a cocompact lattice constructed as the semidirect product of a lattice in $ \mathbb{R}^2 $ and a finite cyclic subgroup of $ SL\_2(\mathbb{Z}) $ preserving that lattice. For example $ C\_1 $ corresponds to the three torus $ T^3 $.
**I am very curious about the compact flat orientable 3 manifold with holonomy $ C\_2 \times C\_2 $ (known as the Hantzsche-Wendt manifold). It is not a mapping torus of $ T^2 $ like the other five, but perhaps it is a mapping torus of the Klein bottle $ K $?**
| https://mathoverflow.net/users/387190 | Compact flat orientable 3 manifolds and mapping tori | The remaining orientable manifold is called the Hantzsche-Wendt manifold $M^{HW}$, and is not a mapping torus over the Klein bottle. It has first homology $H\_1(M^{HW}, \mathbb{Z}) = \mathbb{Z}\_4 \times \mathbb{Z}\_4$. Any mapping torus $MT(M, f)$ of a manifold $M$ via the map $f: M \to M$ has $\pi\_1(MT(M, f)) \cong \pi\_1(M\_0) \rtimes\_{f\_\*} \mathbb{Z}$. Abelianizing and using Hurewicz's theorem yields that if $M$ is a mapping torus of the Klein bottle, $H\_1(M, \mathbb{Z})$ must be of the form $\mathbb{Z} \times H$, where $H$ is a quotient of $H\_1(K, \mathbb{Z}) = \mathbb{Z} \times \mathbb{Z}\_2$, and hence cannot be $\mathbb{Z}\_4 \times \mathbb{Z}\_4$. It is worth mentioning that the $3$-torus is a normal covering space of all flat compact $3$-manifolds.
The Hantzsche-Wendt Manifold has a nice description as a branched cover of the complement of the Borromean rings, which is described in Zimmerman's paper *[On the Hantzsche-Wendt Manifold](https://eudml.org/doc/178505)*. The standard reference for a classification of compact flat 3-manifolds is J. Wolf's book *Spaces of Constant Curvature,* which gives a classification of them and explicit constructions as quotients of $\mathbb{R}^3$ by isometries.
A resource to develop visual intuition about these manifolds is Jeffrey Weeks' program [Curved Spaces](https://www.geometrygames.org/), which simulates how it would look to "fly around inside of" manifolds and has a number of flat 3 manifolds as pre-built examples, including the Hantzsche-Wendt manifold.
| 4 | https://mathoverflow.net/users/136267 | 415400 | 169,347 |
https://mathoverflow.net/questions/415170 | 2 | Let $R$ be a commutative Noetherian ring and $I$ is a proper ideal of $R$. suppose that $M$ is a f.g. $R$-module.
$\DeclareMathOperator\Ext{Ext}$I'm looking for an example that has this property:
$$\Ext^i\_R(M,R/I)\neq 0$$ for all $i$ and there exists $n$ such that for all $i>n$
$$\Ext^i\_R(R/I,R/I)=0.$$
I don't know how to deal with this question. Any help will be appreciated.
| https://mathoverflow.net/users/66837 | Example of non vanishing Ext | Take $R$ a local artinian non-Gorenstein ring, $I=0$, $M$ the residue field of $R$. The vanishing of $Ext^i\_R(R/I,R/I)=Ext^i\_R(R,R)$ for $i>0(=:n)$ is immediate, while for the non-vanishing of $Ext^i\_R(M,R/I)=Ext^i\_R(k,R)$ see e.g. Bourbaki, Algèbre Commutative, chap. X, $\S$ 3, n. 7, Lemme 2.
| 4 | https://mathoverflow.net/users/92322 | 415405 | 169,351 |
https://mathoverflow.net/questions/403152 | 4 | Let $M\_0^4$ and $M\_1^4$ be two closed smooth 4-manifolds and let $M$ be an $h$-cobordism between them (i.e., a compact smooth 5-manifold with boundary the disjoint union of $M\_0$ and $M\_1$ and with the inclusions of $M\_0$ and $M\_1$ into $M$ both being homotopy equivalences). In the case where $M\_0$ and $M\_1$ are simply-connected there is a result due to [Curtis, Hsiang, Freedman, and Stong](https://link.springer.com/article/10.1007/s002220050031) that says that actually there is a manifold $A \subset M$ that is an h-cobordism between $A\_0 := A \cap M\_0$ and $A\_1 := A \cap M\_1$ (here $A$ is a manifold with corners) such that $A\_0$ and $A\_1$ are contractible manifolds and $M - int(A)$ is a product cobordism. Further conditions where later added on (for example, that $A$ may be chosen so that $M-A$ is simply-connected) - see [here](https://math.berkeley.edu/%7Ekirby/papers/Kirby%20-%20Akbulut%27s%20corks%20and%20h-bordisms%20of%20smooth%2C%20simply%20connected%204-manifolds%20-%20MR1392665.pdf) for further discussion.
I am wondering if a similar result exists in the case of $h$-cobordisms between non-simply-connected 4-manifolds. Does anyone know of such a result or have ideas on why such a result should not exist?
| https://mathoverflow.net/users/99414 | h-cobordisms between non-simply-connected 4-manifolds | Let $X$ be a smooth, closed 4-manifold. Every element of $\operatorname{Wh}(\pi\_1(X))$ can be realised, for some $k \in \mathbb{N}$, as the Whitehead torsion $\tau(W,X \#^k S^2 \times S^2) \in \operatorname{Wh}(\pi\_1(X))$ of a smooth $h$-cobordism $(M;X\#^k S^2 \times S^2,Y)$, for some $Y$. The Whitehead group is nontrivial for many fundamental groups, for example $\pi\_1(X) \cong C\_5$.
If an $h$-cobordism $M$ admits a decomposition as in the statement of the CFHS theorem, then its Whitehead torsion lies in the image of $\operatorname{Wh}(\{1\}) \to\operatorname{Wh}(\pi\_1(X))$, and is therefore trivial since $\operatorname{Wh}(\{1\})=0$.
It follows that the $h$-cobordisms with nontrivial Whitehead torsion do not admit such a contractible submanifold $A$ as in the question.
An alternative version of the question would perhaps begin with the hypothesis of an $s$-cobordism.
| 4 | https://mathoverflow.net/users/102454 | 415407 | 169,352 |
https://mathoverflow.net/questions/415396 | 4 | I first asked this question at [math.stackexchange](https://math.stackexchange.com/questions/4373073/contact-variety-is-equidimensional) with no success, so I decided to repost it here.
I am reading the paper "Weakly Defective Varieties" by L. Chiantini and C. Ciliberto, available [here](https://www.ams.org/journals/tran/2002-354-01/S0002-9947-01-02810-0/S0002-9947-01-02810-0.pdf). In it they define the contact variety as follows.
Let $X$ be a reduced irreducible non-degenerate projective variety in $\mathbb{P}^n\_{\mathbb{C}}$ and $P\_1,\ldots,P\_{k+1}\in X$ be general points. If $H$ is a general hyperplane tangent to $X$ at these $k+1$ points, then the *contact variety* of $H$ is the union $\Sigma$ of the irreducible components of $\text{Sing}(H\cap X)$ containing $P\_1,\ldots,P\_{k+1}.$
Then on page 2 the authors claim:
>
> Since $P\_1,\ldots,P\_{k+1}$ are
> general points, an obvious monodromy argument shows that $\Sigma$ is equidimensional.
>
>
>
Could someone please elucidate this 'obvious' argument? I honestly don't even know what monodromy even means in this context, I am only familiar with it in the context of automorphisms of covering spaces in topology.
| https://mathoverflow.net/users/131868 | Contact variety to projective variety is equidimensional | I think that the monodromy argument they refer to is not entirely obvious. It is presumably related to a version of the *uniform position principle* but in the context of tangent hyperplanes, which I guess involves some intricate technical details. Let me instead sketch a proof of their statement that avoids monodromy.
For any $k \geq 0$, let us denote by $S^k(X)$ the variety:
$$ S^k(X) = \overline{ \{z \in \mathbb{P}^n, \ \textrm{exists a generic $k$-uple} \ (x\_1, \ldots, x\_k) \in X^k, \ \textrm{with} \ z \in \langle x\_1, \ldots, x\_k \rangle \} }, $$
where $\overline{A}$ is the Zariski closure of $A$. The variety $S^k(X)$ is called the variety of $k$-secants $\mathbb{P}^{k-1}$ to $X$. The variety $S^{k}(X)$ is the closure of the projection of a $\mathbb{P}^{k-1}$-fibration over an open dense subset of $X^k$, hence it is irreducible.
Terracini's Lemma asserts that for generic $x\_1, \ldots, x\_k \in X$ and for generic $z \in \langle x\_1, \ldots, x\_k \rangle$, we have:
$$T\_{S^k(X),z} = \langle T\_{X,x\_1}, \ldots, T\_{X,x\_k} \rangle.$$
Consider $X^\*$ the dual variety of $X$ (that is the closure of the set of generic tangent hyperplanes to $X$). The variety $X^\*$ is the closure of the image in $\left(\mathbb{P}^n\right)^\*$ of the projectivization of the normal bundle of $X$ over the dense open subset $X\_{smooth}$. Hence it is irreducible. We also define $X^\*\_k$ to be:
$$ X^\*\_k = \overline{ \{ H \in X^\*,\ \textrm{exists a generic} \ (x\_1,\ldots, x\_k) \in X^k, \ \textrm{with} \ T\_{X,x\_1} \subset H, \ldots, T\_{X,x\_k} \subset H \}}$$
Terracini's lemma implies that $X^\*\_k = S^k(X)^\*$, where $S^k(X)^\*$ is the projective dual to $S^k(X)$. We let
$$p : \overline{\mathbb{P}(N^\*\_{X\_{smooth}/\mathbb{P}^n}(-1))} \longrightarrow \mathbb{P}^n$$
and $$ q : \overline{\mathbb{P}(N^\*\_{X\_{smooth}/\mathbb{P}^n}(-1))} \longrightarrow \left(\mathbb{P}^n \right)^\*$$
the natural projections, where $\overline{\mathbb{P}(N^\*\_{X\_{smooth}/\mathbb{P}^n}(-1))}$ is the closure of the incidence variety *hyperplane/tangent spaces* in $\mathbb{P}^n \times (\mathbb{P}^n)^\*$. Set $I\_{X^\*\_k} = q^{-1}(X^\*\_k)$. By hypothesis, we have:
$$ p(I\_{X\_k^\*}) = X.$$
Furthermore, Terracini's lemma implies that the generic fiber (over $x \in X$) of $p|\_{I\_{X^\*\_k}}$ is $S(T\_{X,x}, S^{k-1}(X))^\*$, which is irreducible, as the projective dual of an irreducible variety. As a consequence, there is only one irreducible component of $I\_{X^\*\_k}$ which dominates $X$. We denote this component by $Z$. Note that $Z$ dominates $X^\*\_k$. Moreover, since $Z$ is the only irreducible component of $I\_{X^\*\_k}$ which dominates $X$, the *contact locus* of $H$ with $X$ as defined in the paper you mention is:
$$\mathrm{Tan}(H,X) = q|\_{Z}^{-1}(H),$$
for generic $H \in X^\*\_k$.
The Theorem of the generic fiber guarantees that the generic fiber of the restriction $q|\_{Z} : Z \longrightarrow X^\*\_k$ is equidimensional.
| 4 | https://mathoverflow.net/users/37214 | 415408 | 169,353 |
https://mathoverflow.net/questions/415441 | 11 | Sheffer polynomials $\{P\_n(x)\}$ have generating function $P(x,t) = \sum\_{n=0}^{\infty}P\_n(x)t^n=A(t)e^{xu(t)}$.
This form reminds me of the Lie group–Lie algebra correspondence. Is there any connection to Lie theory, obvious or not?
(I've deleted my reference to orthogonal polynomials, which was incorrect and irrelevant here.)
Here are some articles on Sheffer polynomials:
* [Daniel Galiffa, Tanya Riston. An elementary approach to characterizing Sheffer A-type 0 orthogonal polynomial sequences. 2015.](https://doi.org/10.2140/involve.2015.8.39)
* [Dongsu Kim, Jiang Zeng. A Combinatorial Formula for the Linearization Coefficients of General Sheffer Polynomials. 2001.](https://doi.org/10.1006/eujc.2000.0459)
| https://mathoverflow.net/users/94182 | How are Sheffer polynomials related to Lie theory? | First, a general set of Sheffer polynomials is not orthogonal with respect to some weight function; for example, the prototypical sequence $p\_n(x) = x^n$, which belongs to both special sub-groups of Sheffer polynomials—the Appell and binomial Sheffer polynomials—with the e.g.f. $e^{xt}$, is not an orthogonal set—neither is the celebrated Bernoulli Appell Sheffer sequence. (Usually the Sheffer polynomials are defined by $A(t) e^{xB(t)} = \sum\_{n\geq 0} p\_n(x) \frac{t^n}{n!}= e^{tp\_\cdot(x)}$. The normalization $P\_n(x)=n!p\_n(x)$ often serves to give integer coefficients.)
However, there are connections to constructs associated to Lie theory:
***1)*** The binomial Sheffer sequences have e.g.f.s of the form $e^{xh(t)}$, where $h(0)=0$ and $h'(0) \neq 0$. A dual sequence, its umbral compositional inverse, has the e.g.f. $e^{x\;h^{(-1)}(t)}$ defined by the compositional inverse function. As formulated by Charles Graves as early as 1853, with $g(z) = \frac{1}{\partial\_z h(z)}$, the action on a function $f(z)$ analytic at $z$ of the exponential map of the infinitesimal generator (infinigen)
$$ g(z)\frac{\partial }{\partial z} = \frac{1}{h'(z)}\frac{\partial }{\partial z} = \frac{\partial }{\partial h(z)} = \frac{\partial }{\partial \omega} $$
is
**(Eqn. 1)**
\begin{align}
& e^{t\;g(z)\frac{\partial }{\partial z}}\; f(z) = e^{t \;\frac{\partial }{\partial \omega}}\; f(h^{(-1)}(\omega)) \\
= {} & f[h^{(-1)}(\omega + t)] =f[h^{(-1)}(h(z)+t)]. \tag{Eqn. 1}
\end{align}
Then
$$e^{t\;g(z)\partial\_z}\;z \Big\rvert\_{z=0}= h^{(-1)}(t),$$
so the e.g.f. for the associated binomial Sheffer sequence is
$$\left. e^{t\;p.(x)}= e^{x\;h^{(-1)}(t)}= e^{t\;g(z)\partial\_z} e^{zx} \;\right\rvert\_{z=0}.$$
Edit Oct. 16, 2022: (Start)
Then
$$p\_n(x) = (g(z)\partial\_z)^n\; e^{zx} \;\Big\rvert\_{z=0}$$
$$= e^{-x} (g(y-1)\partial\_y)^n e^{yx} \; |\_{y=1}= \left. e^{-x} \left[x \;g\left(\frac{u}{x}-1\right)\partial\_u\right]^n e^{u} \; \right|\_{u=x}.$$
Specializing to $g(z)=(1+z)^{m+1}$, allows a general connection between the normal ordering of iterates of the Scherk-Witt-Lie vectors / infinigens $z^{m+1}\partial\_z$ and well-known binomial Sheffer sequences. Then
\begin{align}
p\_n^{(m)}(x) & =((1+z)^{m+1}\partial\_z)^n \;e^{zx}\;\Big|\_{z=0} \\[6pt]
& = e^{-x}\; (y^{m+1}\partial\_y)^n \;e^{yx} \; \Big|\_{y=1} \\[6pt]
& = e^{-x}\left. \left(x \left(\frac{u}{x} \right)^{m+1} \partial\_u\right)^n\; e^u \right|\_{u=x} \\[6pt]
& = e^{-x}\; x^{-mn} \;(u^{m+1}\partial\_u)^n\; e^{u} \; \Big|\_{u=x} \\[6pt]
& = e^{-x}\; x^{-mn} \;(x^{m+1}\partial\_x)\;^n e^x = e^{-x}\; p\_n^{(m)}(:x\partial\_x:)\; e^x
\end{align}
with, by definition, $(:x\partial\_x:)^n := x^n\partial\_x^n$, implying
**(Eqn. 2a)**
\begin{align}
(x^{m+1} \partial\_x)^n & = x^{mn} p\_n^{(m)}(:x\partial\_x:) \\[6pt]
& = x^{mn}\; m^n St1\_n^r \left( \frac{x\partial\_x}{m} \right) \tag{Eqn. 2a}
\end{align}
where the e.g.f. for the $m$-th family of binomial Sheffer polynomials $p\_n^{(m)}(x)$ is
**(Eqn. 2b)**
\begin{align}
& e^{tp.^{(m)}(x)} = e^{h^{(-1)}(t)x} \\[6pt]
= {} & \exp[((1-mt)^{-\frac{1}{m}}-1)x] \tag{Eqn. 2b}
\end{align}
and $St1\_n^{r}(x)$ are the reversed unsigned Stirling polynomials of the first kind of OEIS [A094638](https://oeis.org/A094638) (see also [A008275](https://oeis.org/A008275), [A048994](https://oeis.org/A048994), and [A130534](https://oeis.org/A130534)).
For the special linear Lie algebra $sl\_2$:
For $m =-1$, the infingen $g(z)\partial\_z = g(1+z)\partial\_z= \partial\_z$ has associated $h(z)=h^{(-1)}(z)=z$ giving the **translation** $e^{t\partial\_z}\;f(z) = f(z+t)$, consistent with Eqn. (2) with $p^{(-1)}\_n(z) = z^n$, the iconic Appell Sheffer sequence, which is also a binomial Sheffer sequence.
For $m=0$, the infingen $g(z)\partial\_z = z\partial\_z$ has associated $h(z)=\ln(z)$ and $h^{(-1)}(\omega)=e^\omega$ giving, from Eqn. (1), the **scaling** $e^{tz\partial\_z}f(z) = f[\exp(\ln(z)+t)] = f(e^tz)$. Eqn. (2) associated with $g(z)=(1+z)$, $h(z) =\ln(1+z)$, and $h^{(-1)}(\omega) =e^{\omega}-1$ gives
$$(z\partial\_z)^n = p^{(0)}\_n(:z\partial\_z:) = \operatorname{St2}\_n(:z\partial\_z:), $$
the Bell / Touchard / Scherk / Stirling polynomials of the second kind with the e.g.f. $\exp[(e^t-1)x]$ (cf. [A008277](https://oeis.org/A008277) and [A048993](https://oeis.org/A048993)). This is easy to corroborate using $(z\partial\_z)^n z^k = k^nz^k$.
For $m=1$, the infingen $g(z)\partial\_z = z^2\partial\_z$ has associated $h(x)=-\frac{1}{z}$ and $h^{(-1)}(\omega)=-\frac{1}{\omega}$ giving, from Eqn. (1), the **vertical shearing** $e^{tz^2\partial\_z}f(z) = f\left [-\frac{1}{-\frac{1}{z}+t}\right] = f(\frac{z}{1-tz})$. Eqn. (2) associated with $g(z)=(1+z)^2$, $h(z) =\frac{z}{1+z}$, and $h^{(-1)}(\omega) =\frac{\omega}{1-\omega}$ gives
$(z^2\partial\_z)^n =z^n p^{(1)}\_n(:z\partial\_z:) = z^n Lah\_n(:z\partial\_z:)$,
the shifted Lah polynomials, or shifted, normalized, unsigned Laguerre polynomials of order -1, $n! \operatorname{Lag}\_n^{(-1)}(-z)$, with the e.g.f. $\exp[z \frac{t}{1-t}]$ (cf. [A008297](https://oeis.org/A008297) and [A111596](https://oeis.org/A111596)). This is corroborated by noting
$$(z^2\partial\_z)^n = z\;(z\partial\_zz)^n\;z^{-1} = z\;z^n\partial\_z^n z^n \;z^{-1} = z^n\; z \;(\partial\_z^nz^n)\; z^{-1} = z^n\; n!\; z\; \operatorname{Lag}\_n(:z\partial\_z:)\;z^{-1}$$
from the Rodriguez formula for the associated Laguerre polynomial sequences.
(End)
***2)*** Each Sheffer sequence has a pair of ladder ops—the raising/creation and lowering/destruction/annihilation ops defined by $L \; p\_n(x) = n \; p\_{n-1}(x)$ and $R \; p\_n(x) = p\_{n+1}(x)$—satisfying the Graves–Lie bracket of vector fields, the commutator, relation
$$[L,R] = 1,$$
from which the Graves–Pincherle derivative
$$[f(L),R] = f'(L)$$
follows.
There is also the commutator
$$[(g(z)\partial\_z)^n,u(z)] = n \; (g(z)\partial\_z)^{n-1}$$
for the powers of the Lie derivative / infinitesimal generator for the Scherk-Comtet partition polynomials of [A139605](https://oeis.org/A139605), forming an underlying calculus for the binomial Sheffer sequences.
Edit July 1, 2023: (Start)
The typo $u(z)$ just above should be $h(z)$ so that
$$[(g(z)\partial\_z)^n,h(z)] = n \; (g(z)\partial\_z)^{n-1}.$$
Proof:
$$e^{tg(z)\partial\_z} \;h(z)f(z) - h(z) \;e^{tg(z)\partial\_z}\;f(z)$$
$$= (h(z)+t)\; f(h^{(-1)}(h(z)+t)) -h(z)\; f(h^{(-1)}(h(z)+t))$$
$$= t\; f(h^{(-1)}(h(z)+t)) = t e^{tg(z)\partial\_z} \;f(z)$$
$$=\sum\_{n \geq 0} (g(z)\partial\_z)^{n} \frac{t^{n+1}}{n!} \;f(z) =\sum\_{n \geq 1}n (g(z)\partial\_z)^{n-1} \frac{t^n}{n!}\;f(z) $$
(End)
***3)*** A general Sheffer sequence (a semidirect product of the Appell and binomial Sheffer sequences) is associated with a generalized Lie derivative via
$$e^{t(q(z)+g(z)\partial\_z)} \; e^{xz} |\_{z=0} = A(t) e^{xf^{(-1)}(t)}$$
with $q(z) = \partial\_t \ln(A(t)) \;|\_{t=f(z)}$ and $g(z) = 1/f'(z)$.
I have an extensive set of posts at my web blog and links to numerous OEIS entries and MO-Q&As and MSE-Q&As related to this topic. All the symmetric polynomials/functions--complete, elementary, power, Faber--are related to Sheffer Appell sequences with their related ladder ops. The Faa di Bruno/Bell and cycle index polynomials of the symmetric groups, a.k.a the refined Stirling partition polynomials of the second and first kinds, and other compositional partition polynomials are all Appell Sheffer sequences in a distinguished indeterminate. The closely related sets of Lagrange inversion partition polynomials, including the refined Euler characteristic partition polynomials of the associahedra, all have the raising op $g(z)\partial\_z$, as shown above--one set, OEIS [A134264](https://oeis.org/A134264), is an Appell sequence and is related to free probability, inversion of Laurent series, and characterization of Kac-Schwarz operators related to Heisenberg-Virasoro groups. Multiplicative inversion is intimately bound with the refined Euler characteristic partition polynomials of the permutahedra and the operational (differential/matrix) calculus of Appell Sheffer polynomials. The Bernoulli Appell polynomials are of course related to the BCHD theorem, exponential mappings of the Lie commutator, topology, Todd operator and class of Hirzebruch, the Hurwitz zeta function, and more. Formal group laws (and generalized local Lie groups) are related to the linearization coefficients of products of binomial Sheffer polynomials. The list goes on.
The orthogonal Sheffer sequences such as the associated Laguerre polynomials have a lot of underlying group theory and vector fields associated with them. Vilenkin with "[Special Functions and the Theory of Group Representations](https://doi.org/10.1090/mmono/022)", Talman, Miller, Gilmore, Feinsilver, and many others have written books on the underlying group theory. (None covers every aspect.)
One group of researchers has also written extensively on this topic. See, e.g., "[One-parameter groups and combinatorial physics](https://arxiv.org/abs/quant-ph/0401126)" by
Duchamp, Penson, Solomon, Horzela, and Blasiak (although quite negligent of reffing the OEIS--tribal instincts...sigh...What can you do?) as well as Wolfdieter Lang.
**4)** (Added Oct. 16, 2022) The raising operator of an Appell Sheffer sequence gives the infinitesimal generator, related to the Riemann zeta function, for Heaviside fractional differ-integral operators. See, e.g., my MO-Q "[Riemann zeta function at positive integers and an Appell sequence of polynomials related to fractional calculus](https://mathoverflow.net/questions/111165/riemann-zeta-function-at-positive-integers-and-an-appell-sequence-of-polynomials)" and MO-A to "[What's the matrix of logarithm of derivative operator (lnD)? 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)".
| 13 | https://mathoverflow.net/users/12178 | 415451 | 169,361 |
https://mathoverflow.net/questions/415401 | 1 | Given a (bounded) sequence $\{q\_n\}\_{n\geq 0}$ such that $\lvert q\_n\rvert \leq 1$ for all $n \geq 0$ and $\sum\_{n\geq 0} q\_n = 0$. We can impose the condition that $\sum\_{n\geq 0} \lvert q\_n\rvert \leq 2$ as well. I am wondering whether there exists a fixed constant, independent of the sequence $\{q\_n\}$, such that $$\sum\_{n \geq 0} \lvert q\_n - q\_{n+1}\rvert^2 \geq C \sum\_{n \geq 0} \lvert q\_n\rvert^2$$ holds true?
---
Edit: It seems that $\sum\_{n\geq 0} |q\_n| \leq 2$ might not be enough, I am wondering if higher order moment bound such as $\sum\_{n\geq 0} (1+n)|q\_n| < \infty$ will suffice.
| https://mathoverflow.net/users/163454 | A discrete version of Poincaré's inequality | You can duplicate the usual proof of Hardy type inequalities to the discrete case.
Suppose $\{q\_n\}$ is an eventually 0 sequence (you can weaken this to $\lim\_{n\to \infty} n^{1/2} q\_n = 0$). Then by telescoping you have (all sums are over $n\geq 0$)
$$ \sum (n+1) q\_{n+1}^2 - n q\_{n}^2 = 0 $$
Rewrite as
$$ \sum q\_n^2 = - \sum (n+1) (q\_{n+1} + q\_n) (q\_{n+1} - q\_n) $$
Take absolute values and apply Cauchy-Schwarz on the RHS
$$ \sum q\_n^2 \leq \left( \sum (n+1)^2 |q\_{n+1} - q\_n|^2 \right)^{1/2} \left( \sum (q\_{n+1} + q\_n)^2 \right)^{1/2} $$
the second factor can be bounded by $2 ( \sum q\_n^2)^{1/2}$. Cancelling and you get
$$ \sum q\_n^2 \leq C \sum (n+1)^2 |q\_{n+1} - q\_n|^2 $$
---
Scott Armstrong's comment is similar. As long as $\lim\_{n\to\infty} q\_n = 0$ you have
$$ q\_n = \sum\_{k \geq n} q\_k - q\_{k+1} $$
then
$$ \sup\_{n\geq 0} |q\_n| \leq \sum\_{n\geq 0} |q\_n - q\_{n+1}| $$
---
If you want to look at "Sobolev type" inequalities: they are all essentially based on the fundamental theorem of (discrete) calculus applied in various ways.
---
Finally: note that you can also do a scaling argument.
Let $\lambda$ be a positive integer.
Let $q^{(\lambda)}\_{n}$ be such that
$$ q^{(\lambda)}\_m = q\_n \text{ if } m \in [\lambda n, \lambda (n+1))$$
This scaling preserves the $ \sum |q\_{n+1} - q\_n|^2$ semi-norm, but has $ \sum |q^{(\lambda)}\_n|^2 = \lambda \sum |q\_n|^2$, which immediately falsifies your proposed Poincaré inequality.
You see that the inclusion of weights in Hardy avoids this difficulty.
| 5 | https://mathoverflow.net/users/3948 | 415453 | 169,362 |
https://mathoverflow.net/questions/415476 | 5 | There is the following exercise in [Vershynin's book on High-Dimensional Probability](https://www.math.uci.edu/~rvershyn/papers/HDP-book/HDP-book.html).
**Exercise 3.1.6:**
Let $X = (X\_1, \dots, X\_n) \in \mathbb{R}^n$ be a random vector with independent coordinates $X\_i$ that satisfy $\mathbb{E}[X\_i^2] = 1$ and $\mathbb{E}[X\_i^4] \leq K^4$. Show that $\mathrm{Var}(\|X\|\_2) \leq CK^4$.
What if we remove the condition that the coordinates are independent? Suppose instead that we have only the condition that the coordinates are uncorrelated (i.e., $X$ has covariance $I$), and we additionally have some bound on the fourth moment (e.g., $\mathbb{E} [ \langle X, v\rangle^4]\leq K^4$ for every unit vector $v$). What is the best bound we can prove on $\mathrm{Var}(\|X\|\_2)$?
| https://mathoverflow.net/users/37267 | Variance of the norm of a random variable under finite-moment assumptions | $\newcommand\R{\mathbb R}
\renewcommand{\P}{\operatorname{\mathsf P}}\newcommand\E{\operatorname{\mathsf{E}}}\newcommand\Var{\operatorname{\mathsf{Var}}}$The best bound on $\Var \|X\|\_2$ is about $n$.
Indeed,
\begin{equation\*}
\Var \|X\|\_2\le \E \|X\|\_2^2=\sum\_{i=1}^n \E X\_i^2=n. \tag{1}\label{1}
\end{equation\*}
On the other hand, suppose that
\begin{equation\*}
\mu\_X=q\,\mu\_{aG}+p\,\mu\_{bG},
\end{equation\*}
where $\mu\_Y$ denotes the distribution of a random vector $Y$, $G$ is a standard Gaussian random vector in $\R^n$, and
\begin{equation\*}
a:=0,\quad b\in[1,\infty),\quad p:=\frac1{b^2},\quad q:=1-p.
\end{equation\*}
Then the covariance matrix of $X$ is $pb^2I=I$.
Also, for all unit vectors $v\in\R^n$,
\begin{equation\*}
\E (X\cdot v)^4=pb^4\E (G\cdot v)^4=3b^2=K^4
\end{equation\*}
if $b=K^2/\sqrt3$ (assuming that $K^4\ge3$).
So, all your conditions hold.
However, $\E\|X\|\_2^2=pb^2\E\|G\|\_2^2=n$ and $\E\|X\|\_2=pb\E\|G\|\_2\le pb\sqrt{\E\|G\|\_2^2}=pb\sqrt n=\sqrt n/b$ and hence
\begin{equation\*}
\Var \|X\|\_2=\E\|X\|\_2^2-(\E\|X\|\_2)^2\ge\Big(1-\frac1{b^2}\Big)n=\Big(1-\frac3{K^4}\Big)n; \tag{2}\label{2}
\end{equation\*}
the inequality $\Var \|X\|\_2\ge\big(1-\frac3{K^4}\big)n$ trivially holds if $K^4\not\ge3$.
>
> **Conclusion 1:** The trivial upper bound $n$ on $\Var \|X\|\_2$ in \eqref{1} is optimal, in the sense that it cannot be replaced by $cn$ for any real $c<1$, if $K$ is allowed to be large enough.
>
>
>
---
Nonetheless, the upper bound $n$ on $\Var \|X\|\_2$ in \eqref{1} can be "second-order" improved as follows. Note that
\begin{equation\*}
\E\|X\|\_2^4=\E\Big(\sum\_{i=1}^n X\_i^2\Big)^2=\sum\_{i,j=1}^n \E X\_i^2 X\_j^2
\le\sum\_{i,j=1}^n \frac{\E X\_i^4+\E X\_j^4}2
=n\sum\_{i=1}^n \E X\_i^4\le n^2K^4.
\end{equation\*}
So, by the log-convexity of $\E\|X\|\_2^p$ in $p>0$,
\begin{equation\*}
(\E\|X\|\_2)^2\ge\frac{(\E\|X\|\_2^2)^3}{\E\|X\|\_2^4}\ge\frac{n^3}{n^2K^4}
=\frac n{K^4}.
\end{equation\*}
So,
\begin{equation\*}
\Var \|X\|\_2=\E\|X\|\_2^2-(\E\|X\|\_2)^2\le\Big(1-\frac1{K^4}\Big)n, \tag{3}\label{3}
\end{equation\*}
which is an improvement of \eqref{1}.
>
> **Conclusion 2:** In view of \eqref{2} and \eqref{3}, the best upper bound on $\Var \|X\|\_2$ is of the form
> \begin{equation\*}
> \Big(1-\frac C{K^4}\Big)n
> \end{equation\*}
> for some $C\in[1,3]$.
>
>
>
This refines Conclusion 1.
| 5 | https://mathoverflow.net/users/36721 | 415478 | 169,371 |
https://mathoverflow.net/questions/415477 | 1 | Say we generate an $N \times N$ sparse random matrix $W$, where each element $W\_{ij}$ was independently chosen to be $1$ with probability $p=\frac{a}{N}$, and $0$ with probability $1-p$. We are interested in the case where $N \rightarrow \infty$.
I wonder what happens if we permute the rows and the columns of $W$ so that the non-zero entries are moved as close as possible to the diagonal of the matrix. My intuition is that, if $a=1$ and $N \rightarrow \infty$, we can find a permuted version of the matrix that is exactly an identity matrix with high probability (perhaps with probability 1). Expanding on this intuition, if $a>1$, we will get a permuted matrix that has a band of width $a$ along the diagonal of the matrix. Is this intuition correct? How do I prove/disprove this?
| https://mathoverflow.net/users/173974 | Permute a sparse random matrix to resemble a diagonal matrix as much as possible | When $a=1$, the probability to get a permutation matrix is
$$N! \left(\frac{1}{N}\right)^N \left(1-\frac{1}{N}\right)^{N^2-N} < \frac{N!}{N^N}e^{-(N-1)} < \sqrt{2\pi N}\,e^{-2N + 1 + \frac1{12N}}.$$
So, it's not high, but rather tends to 0 exponentially.
| 2 | https://mathoverflow.net/users/7076 | 415479 | 169,372 |
https://mathoverflow.net/questions/415448 | 4 | I'm interested in whether there exists a well known class of finite groups that generates all finite groups through quotienting.
I will be more formal: does there exist a class of ("nice") well known finite groups $\{G\_n\}$ such that for any finite groups $G$ we have that there exists $G\_n$ and a normal subgroup $N$ of $G\_n$ such that $G\_n/N=G$?
The answer is trivially positive if we remove the condition that $G\_n$ be finite: we can take the free groups of all finite ranks.
| https://mathoverflow.net/users/155306 | Families of finite groups of which every finite group is a quotient | It obviously depends a lot on your definition of "nice", but one class defined by a reasonably nice property, while also relevant in theory, is the class of all complete groups (where complete means that both the center and the outer automorphism group are trivial). That every finite group is a quotient of a complete group (even with some extra properties) is shown in Hartley, Robinson, "On finite complete groups" (1980).
I remember noticing a curious consequence of this fact some years ago: Complete groups have the property that, if they themselves are a normal subgroup of some finite group, then they must also be a quotient of that group. This implies that, if every finite group $G$ is a normal subgroup of some Galois group over, e.g., $\mathbb{Q}$, then (via embedding $G$ as quotient into a complete group and then using the above property), every finite group is indeed (a quotient of a Galois group, and hence itself) a Galois group over $\mathbb{Q}$.
| 6 | https://mathoverflow.net/users/127660 | 415483 | 169,374 |
https://mathoverflow.net/questions/415482 | 3 | Let $k$ be a positive integer. Note that $a/b=ab^{k-1}/b^k$ for any integers $a$ and $b>0$.
If every $n\in\mathbb N=\{0,1,2,\ldots\}$ can be written as $x\_1^k+\cdots+x\_{s}^k$ with $x\_1,\ldots,x\_s\in\mathbb N$, then each $r\in\mathbb Q\_{\ge0}$ can be written as $x\_1^k+\cdots+x\_s^k$ with $x\_1,\ldots,x\_s\in\mathbb Q\_{\ge0}$, where
$$\mathbb Q\_{\ge0}=\{r\in\mathbb Q:\ r\ge0\}.$$
Instead of the classical Waring problem over $\mathbb N$, we may consider Waring's problem over $\mathbb Q\_{\ge0}$. Let $s(k)$ be the least positive integer $s$ such that each $r\in\mathbb Q\_{\ge0}$ can be written as $x\_1^k+\ldots+x\_s^k$ with $x\_1,\ldots,x\_s\in\mathbb Q\_{\ge0}$. Then
$s(k)$ exists and moreover $s(k)\le g(k)$. It is interesting to find the exact value of $s(k)$.
By Theorems 233 and 234 in Hardy and Wright's book *A Introduction to the Theory of Numbers*, we have $s(3)=3$ (the inequality $s(3)\le 3$ was obtained by Richmond in 1923).
**Question.** Can one prove that $s(k)\ge k$ for each integer $k>2$? Is it true that $s(4)=5$?
Your comments are welcoeme!
| https://mathoverflow.net/users/124654 | Waring's problem over $\mathbb Q_{\ge0}$ | I can answer the last question: $s(4)$ equals $15$, not $5$. Variables below will denote integers.
First we show that $s(4)\leq 15$. We use the result of [Davenport (1939)](http://djvu.org/) that there exists $m\geq 1$ such that every positive integer outside $\{16^h k:\text{$h\geq 0$ and $k\leq m$}\}$ is a sum of $15$ integral biquadrates. Now let $a,b\geq 1$ arbitrary, and let $n$ be an odd number such that $n^4>m$. Then $ab^3n^4$ is a sum of $15$ integral biquadrates, hence $a/b$ is a sum of $15$ rational biquadrates.
Now we show that $s(4)\geq 15$. It suffices to show that $15$ is not the sum of $14$ rational biquadrates. Equivalently, $15 n^4$ (for $n\geq 1$) is not the sum of $14$ integral biquadrates. Assume that $15n^4$ is the sum of $14$ integral biquadrates, with $n\geq 1$ minimal. We shall use twice that every integral biquadrate is $\equiv 0,1\pmod{16}$. If $n$ is even, then $15 n^4\equiv 0\pmod{16}$, so every biquadrate in the decomposition is even. That is, $15(n/2)^4$ is also the sum of $14$ integral biquadrates, contradicting the minimality of $n$. If $n$ is odd, then $15 n^4\equiv 15\pmod{16}$, while a sum of $14$ biquadrates is $\equiv 0,\dotsc,14\pmod{16}$. This is again a contradiction.
| 13 | https://mathoverflow.net/users/11919 | 415489 | 169,375 |
https://mathoverflow.net/questions/415508 | 3 | Given 2-tangles $T\_1,T\_2\subset B^3$ with their endpoints at some fixed points NW, NE, SW, SE of $\partial B^3$ we can glue them along $\partial B^3$ to obtain a link $L=T\_1\cup T\_2\subset S^3.$
**Q:** Does that link $L$ together with $T\_1$ determine the homeomorphism class of $(B^3,T\_2)$?
Remarks
1. Note that $L$ and $T\_1$ do not determine the isotopy class of $T\_2$ (within the space of tangles with endpoints at NW, NE, SW, SE, fixed on $\partial B^3$).
2. More generally, any 3-manifolds $M\_1,M\_2$ glued along their boundaries or partial boundaries, $S\hookrightarrow M\_1, S\hookrightarrow M\_2$, form a 3-manifold $N=M\_1\cup\_S M\_2$. However $N$ and $M\_1$ do not determine the homeomorphism class of $M\_2$ in general.
| https://mathoverflow.net/users/23935 | Glueing two 2-tangles | No, $L$ and $T\_1$ together do not determine $T\_2$. Take a non-trivial knot $K$ and suppose that $L$ is $K\#K$, and $T\_1$ is the trivial tangle. Then $T\_2$ can certainly be:
* a 2-tangle with one boundary-parallel component, and one with a boundary-parallel component connected-summed with $L$;
* a 2-tangle with two boundary-parallel components, each connected-sum with a copy of $K$.
I think you can do similar examples when $L$ and $T\_2$ are prime, for instance by cutting a 2-bridge knot in the usual way or in a more "silly" way where all the knotting is in one of the two tangles.
| 2 | https://mathoverflow.net/users/13119 | 415515 | 169,385 |
https://mathoverflow.net/questions/304644 | 9 | The Euler characteristic is an invariant (under homeomorphism) of manifolds that can be computed from a cellulation by (weighted) counting of different kinds of objects, namely
\begin{equation}
\chi=\textrm{#vertices}-\textrm{#edges}+\textrm{#faces}-\ldots
\end{equation}
Are there any other (independent) invariants that are computed in a similar way? I would allow for arbitrary weights (not only $\pm 1$) and arbitrary kinds of "objects" (not only $d$-cells, but also things like "corners", "cycles consisting of $3$ edges", "pairs of a $3$-cell and a $1$-cell that is part of its boundary").
In particular, in odd dimensions where the Euler characteristic is trivial by Poincare duality, is there any sort of replacement for it?
| https://mathoverflow.net/users/115363 | Are there invariants of cell complexes similar to the Euler characteristic? | As pointed out in the comments, every characteristic class in $H^d(BO(d), G)$ provides a $G$-valued locally computable invariant of $d$-manifolds, by pulling back via the classifying map of the tangent bundle. As argued by [Levitt and Rourke](https://www.ams.org/journals/tran/1978-239-00/S0002-9947-1978-0494134-6/S0002-9947-1978-0494134-6.pdf), there are local combinatorial formulas on trianglulations for dual simplicial $d-i$-cycles representing any degree-$i$ characteristic classes, whose values on a $d-i$-simplex only depend on the star of that simplex. To be precise, for this to work we need to add some decorations to the triangulation, namely a local ordering.
So every degree-$d$ characteristic class yields a local combinatorial formula for a simplicial $0$-cycle, whose summation over vertices is an invariant of the type in the question. The "objects" are the different stars of vertices, to which we associate elements of $G$ according to the local formula. Summing up all the $G$-values on the vertices corresponds to evaluating the invariant.
Concrete local formulas are in fact known for many cases. For $G=\mathbb{Z}\_2$, the relevant characteristic classes are generated by degree-$d$ polynomials (with $\mathbb{Z}\_2$ sum and cup product) of Stiefel-Whitney classes. For the latter, combinatorial formulas are given by [Goldstein and Turner](https://www.ams.org/journals/proc/1976-058-01/S0002-9939-1976-0415643-5/S0002-9939-1976-0415643-5.pdf), and a combinatorial cup product for cycles can be easily obtained via its geometric interpretation as intersection. For $G=\mathbb{Z}$ on oriented manifolds, we additionally need the Bockstein homomorphism for the short exact sequence
$$\mathbb{Z}\rightarrow \mathbb{Z}\rightarrow \mathbb{Z}\_2$$
and the Pontryagin classes. The former only involves simplex-wise operations and the co-boundary operator. The Pontryagin classes are the only case for which no satisfactory local formulas exist to date. $\mathbb{Q}$-valued formulas have been described by [Gaifullin](https://www.maths.ed.ac.uk/%7Ev1ranick/papers/gaifullin2.pdf), but not very explicitly.
I'm not aware of any converse argument that every local formula for a $0$-cycle invariant corresponds to a characteristic class, however I don't think there are any known examples for invariants which don't.
| 2 | https://mathoverflow.net/users/115363 | 415526 | 169,390 |
https://mathoverflow.net/questions/415469 | 18 | I am trying to understand some things about [Condensed Mathematics](https://nlab-pages.s3.us-east-2.amazonaws.com/nlab/show/condensed+mathematics) and the [Liquid Tensor Experiment](https://xenaproject.wordpress.com/2021/06/05/half-a-year-of-the-liquid-tensor-experiment-amazing-developments/). The aim of the LTE is to provide a formalised proof of Theorem 9.4 in Scholze's paper [Lectures on Analytic Geometry](https://www.math.uni-bonn.de/people/scholze/Analytic.pdf) (describing joint work with Clausen). Part of the LTE is a [blueprint](https://leanprover-community.github.io/liquid/) of the general approach. Theorem 9.4 is stated for a rather general class of chain complexes, but the blueprint works with a more restricted class which is presumably sufficient for the intended application; I have not yet understood this reduction. Specifically, the blueprint considers complexes defined with the aid of a structure called a Breen-Deligne package (Definition 2.11 in the blueprint). As pointed out in Definition 2.13 of the blueprint, there is a default example of a Breen-Deligne package. My question is: is it sufficient to consider this default example, or is there a real need to consider all possible examples?
If it is sufficient to consider the default example, then it seems to me that the whole proof can be substantially simplified. (However, I am just beginning to get to grips with these ideas, so I could easily be mistaken.)
| https://mathoverflow.net/users/10366 | Breen-Deligne packages and the liquid tensor experiment | The comments have already given the answers, but let me assemble them here with my account of the story.
When Scholze first posted the [Liquid Tensor Experiment](https://xenaproject.wordpress.com/2020/12/05/liquid-tensor-experiment/), it was quickly identified (by both Peter and Reid, somewhat independently I think) that Breen–Deligne resolutions would be the largest input
in terms of prerequisites that needed to be formalised. Back then, I had no idea what these resolutions were, or how to prove that they existed. But they were needed for the statement of Theorem 9.4, which seemed to be a very natural first milestone to aim for.
So I set out to formalise the statement of the existence of Breen–Deligne resolutions, with the expectation that we would never prove the result in Lean, but just assume it as a black box. Let me recalll the statement:
**Theorem (Breen–Deligne).** For an abelian group $A$, there exists a resolution of the form
$$ \dots → \bigoplus\_{j = 0}^{n\_i} ℤ[A^{r\_{i,j}}] → \dots → ℤ[A^2] → ℤ[A] → A → 0 $$
that is functorial in $A$.
*On the proof.* See appendix to Section IV of [Condensed.pdf](https://www.math.uni-bonn.de/people/scholze/Condensed.pdf). As Reid remarked in the comments, the proof relies on the fact that stable homotopy groups of spheres are finitely generated. ∎
(Aside: for the proof of Theorem 9.4, we really need a version that applies to *condensed* abelian groups $A$, so in practice we want to abstract to abelian sheaves.)
In the rest of the Lecture notes, Scholze often uses a somewhat different form of the above resolution, by assuming $n\_i = 1$ and effectively dropping all the $\bigoplus$'s. I think this was done mostly for presentational reasons.
I did the same thing when I axiomatized Breen–Deligne resolutions in Lean. So really, they weren't axiomatized at all. I don't know of any reason to expect that there exists a *resolution* with the property that $n\_i = 1$ for all $i$, but I also don't see any reason why there shouldn't. Anyway, I needed a name for the axioms that I did put into Lean, and I chose [*Breen–Deligne package*](https://leanprover-community.github.io/liquid/sect0001.html#BD_package) for that.
So what is that exactly? Well, a functorial map $ℤ[A^m] → ℤ[A^n]$ is just a formal sum of matrices with coefficients in $ℤ$. So as a first approximation, we record
* the natural numbers $r\_j = r\_{1,j}$ (since $n\_i = 1$);
* for every $j$, a formal sum of $(r\_{j+1}, r\_j)$-matrices with coefficients in $ℤ$.
But we need one more property of Breen–Deligne resolutions: if $C(A)$ denotes the complex, then there are two maps induced by addition. There is the map $σ \colon C(A^2) → C(A)$ that comes from the functoriality of $C$ applied to the addition map $A^2 → A$. But there is also the map $π \colon C(A^2) → C(A)$ that comes from addition in the objects of the complex. (All objects are of the form $ℤ[A^k]$ and we can simply add elements of these free abelian groups.) The final axiom of a Breen–Deligne package is
* there is a functorial homotopy between $σ$ and $π$.
While playing around with the axioms, I noticed that I could write down inductively a somewhat non-trivial [example](https://leanprover-community.github.io/liquid/sect0001.html#BD_eg) of such a package. When I discussed this example with Peter, he suggested that it might in fact be suitable as a replacement for Breen–Deligne resolutions in all applications in his lecture notes so far. Several months later we found out that I had rediscovered MacLane's $Q'$-construction. So, let's denote by $Q'$ the complex corresponding to the example package. The following result is, as far as I know, original:
**Lemma.** Let $A$ and $B$ be (condensed) abelian groups. If $\text{Ext}^i(Q'(A),B) = 0$ for all $i ≥ 0$, then $\text{Ext}^i(A,B) = 0$ for all $i ≥ 0$.
*Proof.* I've written up a proof sketch on Zulip ([public archive of that thread](https://leanprover-community.github.io/archive/stream/267928-condensed-mathematics/topic/breen.20deligne.20lemma.html)). ∎
We are in the process of formalizing the necessary homological algebra to verify the proof in Lean. It's the last major milestone left to complete the challenge in Peter's original blogpost. After that, we mostly need some glue. See [this blogpost](https://leanprover-community.github.io/blog/posts/lte-update/) for an update on the formalisation effort. (Edit: see also <https://math.commelin.net/files/LTE.pdf> for a more precise roadmap of what remains to be done to complete the challange.) Once everything is done, it should all be written up in some paper.
So now, let me turn to the question
>
> is it sufficient to consider this default example, or is there a real need to consider all possible examples?
>
>
>
It is indeed sufficient to consider this default example. The reasons for working with the abstract concept of Breen–Deligne packages are
* historical: I had formalised the statement of Theorem 9.4 and some other material in terms of BD packages before I realised that this default example was indeed sufficient for our purposes.
* practical: as Remy points out in the comments, it is (i) helpful to abstract away concrete details into a conceptual object, and (ii) there is a chance that there might be better examples leading to better constants.
What would be really interesting is an example of a BD package that gives *resolutions* instead of merely functorial complexes. It is known that the $Q'$ construction is not a resolution. Indeed, one can easily show inductively that $H\_i(Q'(ℤ))$ contains a copy of $ℤ^{2^i}$. With more work (relying again on abstract homotopy theory), Peter gave a proof that $Q'(A) ≅ Q'(A)^{⊕2}[-1] ⊕ ℤ ⊗\_{\mathbb S} A$ on Zulip ([public archive](https://leanprover-community.github.io/archive/stream/267928-condensed-mathematics/topic/explicit.20computation.html#243019768)).
| 27 | https://mathoverflow.net/users/21815 | 415540 | 169,396 |
https://mathoverflow.net/questions/415514 | 8 | Cross posting from MSE, where [this question](https://math.stackexchange.com/q/4369960/643521) received no answers.
The following Latin square
$$\begin{bmatrix}
1&2&3&4&5&6&7&8\\
2&1&4&5&6&7&8&3\\
3&4&1&6&2&8&5&7\\
4&3&2&8&7&1&6&5\\
5&6&7&1&8&4&3&2\\
6&5&8&7&3&2&4&1\\
7&8&5&2&4&3&1&6\\
8&7&6&3&1&5&2&4
\end{bmatrix}$$
has the property that for all pairs of two different rows $a$ and $b$, the permutations $ab^{-1}$ have the same cycle type (one 2-cycle and one 6-cycle).
What is known about Latin squares with the property that all $ab^{-1}$ have the same cycle type (where $a$ and $b$ are different rows)? For example, do they have a particular structure, for which cycle types do they exist, are there any infinite families known, do they have a name, etc?
The only example of an infinite family I'm aware of are powers of a single cyclic permutation when $n$ is prime, for example:
$$\begin{bmatrix}
1&2&3&4&5\\
2&3&4&5&1\\
3&4&5&1&2\\
4&5&1&2&3\\
5&1&2&3&4
\end{bmatrix}$$
| https://mathoverflow.net/users/144261 | Latin squares with one cycle type? | One way to achieve the required property is to construct a Latin square whose autotopism group acts transitively on unordered pairs of rows. This can be achieved for orders that are a prime power congruent to 3 mod 4, by means of the quadratic orthomorphism method, described [in this paper](https://www.combinatorics.org/ojs/index.php/eljc/article/view/v12i1r22). The focus of that paper is the atomic squares I mentioned in other comments, but the construction will always achieve the property that each pair of rows produces the same cycle structure. For prime powers that are 1 mod 4, quadratic orthomorphisms will build Latin squares with at most two types of cycle structures formed by pairs of rows.
| 6 | https://mathoverflow.net/users/351290 | 415541 | 169,397 |
https://mathoverflow.net/questions/415532 | 14 | Let $A=\{x^4+y^4:\ x,y\in\mathbb Q\}$. Then
$$A-A:=\{a-b:\ a,b\in A\}=\{u^4+v^4-x^4-y^4:\ u,v,x,y\in\mathbb Q\}.$$
Motivated by [Question 415482](https://mathoverflow.net/questions/415482), here I ask the following question.
**Question.** Is it true that $A-A=\mathbb Q$? Any effective way to approach it？
By my computation, $A-A$ at least contains $0,1,\ldots,562$. For example,
$$248=\left(\frac{95}{28}\right)^4+\left(\frac{135}{14}\right)^4-\left(\frac{13}7\right)^4-\left(\frac{269}{28}\right)^4\in A-A.$$
From the viewpoint of additive combinatorics, the question looks interesting. I guess that it should have a positive answer. Any ideas to solve it?
| https://mathoverflow.net/users/124654 | Does $A-A=\mathbb Q$ hold for $A=\{x^4+y^4:\ x,y\in\mathbb Q\}$? | According to Tito Piezas's website [$x^4+y^4-(z^4+t^4) = N$](https://sites.google.com/site/tpiezas/001b),
There is an identity
$((2a+b)c^3d)^4 + (2ac^4-bd^4)^4 - (2ac^4+bd^4)^4 - ((2a-b)c^3d)^4 = a(2bcd)^4$
where $b = c^8-d^8$, for arbitrary {$a,c,d$}.
| 19 | https://mathoverflow.net/users/150249 | 415560 | 169,403 |
https://mathoverflow.net/questions/415522 | 16 | As the title suggests, I am a physicist and have a question about how to show certain superelliptic curves are also hyperelliptic. The superelliptic Riemann surfaces in question has the form $$w^n = \prod\_{\alpha=1}^N(z-u\_\alpha)(z-v\_\alpha)^{n-1},\quad u\_1<v\_1<\dotsb<u\_N<v\_N$$ and is of interest for certain questions 1+1D CFTs (see [Coser, Tagliacozzo, and Tonni - On Rényi entropies of disjoint intervals in conformal field theory](http://arxiv.org/abs/1309.2189)) that I am working on. For N = 2, there is a change of variables that can be done (see section 6 of [Enolski and Grava - Singular $Z\_N$-curves and the Riemann-Hilbert problem](https://doi.org/10.1155/S1073792804132625)) to bring this into the form $\nu^2 = f(\rho)$ where $f(\rho)$ is a polynomial of degree $2n$. That is, the $N=2$ case is a hyperelliptic curve and there is a change of variables that makes that obvious. My question is: is there a way to prove that for some or all values of $N$, the above curve is also hyperelliptic? While having a way to construct the appropriate change of variables would be nice, if there is a way say whether it is or is not hyperelliptic without constructing such a transformation would be very helpful.
I hope this question isn't too basic for around here.
| https://mathoverflow.net/users/476604 | From a physicist: How do I show certain superelliptic curves are also hyperelliptic? | This curve is not hyperelliptic unless $n=2$ or $N=2$.
First, note that it is more convenient to write the curve as
$$ w^n = \frac{ \prod\_{\alpha=1}^N (z- u\_\alpha )} {\prod\_{\alpha=1}^N (z- v\_\alpha)}$$ after a change of variables dividing $w$ by $\prod\_{\alpha=1}^N (z- v\_\alpha)$.
A straightforward calculation shows that this equation defines a smooth curve in $\mathbb P^1 \times \mathbb P^1$, of bidegree $(n,N$), thus of genus $(n-1)(N-1)$. So, if the curve is hyperelliptic, the hyperelliptic involution must have $2g+2 = 2(n-1)(N-1) +2$ fixed points.
Since hyperelliptic curves have a unique hyperelliptic involution, this involution commutes with all other automorphisms, so it commutes with $w \to e^{ 2\pi i /n} w$.
Thus it sends orbits of $w \to e^{ 2\pi i /n} w$ to orbits. Since these orbits are determined by the value of $z$, the hyperelliptic involution must express the new value of $z$ as a function of the old value of $z$. This must be invertible, so the new $z$ is a rational linear transformation $f(z)$ of the old $z$.
Assume $f$ is not the identity. Then $f$ fixes at most $2$ values of $z$.
Each value of $z$ corresponds to at most $n$ values of $w$, so the hyperelliptic involution has at most $2n$ fixed points, and thus $$2n \leq 2(n-1)(N-1) +2$$ or $$ (n-1) \leq (n-1) (N-1)$$ which implies $N \leq 2$ or $n \leq 1$. The $n \leq 1$ case is clearly genus $0$ and can be discarded, so if $N>2$ then we must have $f(z)=z$.
If $f(z)=z$, then the hyperelliptic involution sends $w$ to another $n$th root of $w^n$. This must be obtained by multiplying $w$ by an $n$th root of unity. Since the hyperelliptic involution is an involution, this must be multiplying $w$ by $-1$, so $n$ must be even and the quotient by the hyperelliptic involution is given by
$$ w^{n/2}= \frac{ \prod\_{\alpha=1}^N(z- u\_\alpha )} {\prod\_{\alpha=1}^N (z- v\_\alpha)}.$$
This is a smooth curve of bidegree $(n/2,N)$ in $\mathbb P^1 \times \mathbb P^1$, and must have genus $0$, so we must have $n/2 \leq 1$ or $N \leq 1$. Again the case $N \leq 1$ is trivial, so we must have $n=2$.
So the only hyperelliptic cases are $n=2$ and $N=2$.
| 14 | https://mathoverflow.net/users/18060 | 415570 | 169,404 |
https://mathoverflow.net/questions/415538 | 1 | I am reading a paper on multiple solutions for boundary value problems of fourth-order differential systems. In the paper, there is a nonlinear term $f\in C\left[(0,1)\times \mathbb{R}^+\times \mathbb{R}, \mathbb{R}\right]$, which is assumed to be semipositone. I found a definition of semipositione online, which says: A (nonlinear) function on the reals is called positone if it is continuous, positive and monotone. A function on the reals is called semipositone if it is positone expect at zero where it is not positive.
I am confused because if the function is negative a zero it will also be negative in a small enough neighbourhood of zero by dint of its continuity. Could you please provide me with a good definition of semipositone?
Thank you in advance.
| https://mathoverflow.net/users/244943 | What are semipositone functions? | The continuous function $f:[0,\infty)\rightarrow\mathbb{R}$ is called *semipositone* if it is monotonically increasing, $f(0)<0$ and $f(u)>0$ for some $u>0$. So indeed, $f$ is negative in a neighborhood of the origin. If instead $f(0)>0$ the function is called *positone*. The case $f(0)=0$ does not have a special name.
The word positone was coined by Keller and Cohen in their [1967 paper](https://authors.library.caltech.edu/100708/1/16087.pdf).
| 4 | https://mathoverflow.net/users/11260 | 415572 | 169,405 |
https://mathoverflow.net/questions/415565 | 17 | Étale cohomology of schemes $X$ is constructed as follows: one associates to $X$ the so-called étale topos of $X$, and then one just takes the sheaf cohomology of that topos.
Is it possible to associate to each smooth manifold $M$ a "de Rham topos of $M$", whose sheaf cohomology yields the de Rham cohomology of $M$?
| https://mathoverflow.net/users/476516 | De Rham via topoi | One can define an analogue of the crystalline topos for smooth manifolds.
This is known as the [de Rham stack](https://ncatlab.org/nlab/show/de+Rham+space) of $M$.
One of the easiest constructions of the de Rham stack
embeds smooth manifolds fully faithfully (using the Yoneda embedding)
into the category of ∞-sheaves on affine smooth loci, the latter being defined as the opposite category of [$\def\Ci{{\rm C}^∞} \Ci$-rings](https://ncatlab.org/nlab/show/C%5E%E2%88%9E-ring) satisfying certain properties.
In this language, the de Rham stack of an ∞-sheaf $F$
is the ∞-sheaf $\def\dR{{\rm dR}} \dR(F)$ defined by $\def\Spec{\mathop{\rm Spec}} \def\red{{\rm red}} \dR(F)(\Spec A)=\dR(F)(\Spec(\red(A))$,
where $\Spec A$ denotes the spectrum of a $\Ci$-ring (defined purely formally in this context) and $\red(A)$ denotes the quotient of $A$ by its ideal of nilpotent elements.
One can then prove that the commutative differential graded algebra of smooth functions on $\dR(M)$ is precisely the de Rham algebra of $M$.
The de Rham stack has other exciting properties: vector bundles (and, more generally, sheaves) on $\dR(M)$ can be identified with D-modules, etc.
The cited nLab article has the relevant pointers to the literature.
The de Rham stack is also closely related to the definition
of the differential graded algebra of differential forms in [synthetic differential geometry](https://ncatlab.org/nlab/show/synthetic%20differential%20geometry)
as the differential graded algebra of infinitesimal smooth singular cochains equipped with the cup product.
See the nLab article [differential forms in synthetic differential geometry](http://nlab-pages.s3.us-east-2.amazonaws.com/nlab/show/differential+forms+in+synthetic+differential+geometry) for further pointers to the literature.
| 13 | https://mathoverflow.net/users/402 | 415581 | 169,406 |
https://mathoverflow.net/questions/415567 | 2 | I'm studying an article but I'm not able to understand one of his statements. I have the following hypotheses:
1. $G$ is a solvable group with trivial center, $J=\phi^{-1}(F(G/F(G)))$ and $J\_2=\phi^{-1}(Z(F (G/F(G)))$ where $\phi$ is the canonical homomorphism of $G$ to $G/F(G)$.
i) $\mathrm{mdc}(|\phi(J/F(G))|,|F(G)|)=1$.
2. $x\_r\in G$ is a prime element such that:
i) $x\_r\in J\_2$;
ii) $C\_G(x\_r)F(G)$ is a Frobenius group with complement $C\_G(x\_r)$ and kernel $F(G)$;
iii) $C\_G(x\_r)\cap F(G)=\{1\_G\}$;
iv) $C\_G(x\_r)$ has odd order;
v) $G=N\_G(\langle x\_r\rangle)F(G)$;
vi) $N\_G(\langle x\_r\rangle)\cap F(G)=\{1\_G\}$;
vii) $G/C\_G(x\_r)F(G)$ is cyclic.
In the article he considers a prime order element $c\in C\_G(x\_r)$ and shows that $c\in Z(F(C\_G(x\_r))$ (this I managed to show) and, because of that, he says that $c\in J\_2$ But I can't understand why $c\in J\_2$ Could someone help me?
The name of the article is: (The commuting graph of a soluble group) and it was written by: (Christopher Parker). My question is on page 845 within the proof of Theorem 1.1. and link is: (arxiv.org/abs/1209.2279).
| https://mathoverflow.net/users/334556 | Element that is in $\phi^{-1}(Z(F (G/F(G)))$ | For simplicity let $\bar G=G/F(G)$ and for any subset or element $x$ of $G$, let $\bar x$ be the image of $x$ in $\bar G$ (the "bar convention"). By (v), $\langle \bar x\_r\rangle\triangleleft \bar G$. In particular, $\bar x\_r\in F(\bar G)$.
It suffices to show that for any $\bar y\in F(\bar G)$ of prime power order $p^a$, $[\bar c,\bar y]=1$. And for this it suffices to show that $\bar y\in C\_{\bar G}(\bar x\_r)$, since then $\bar y\in F(\bar G)\cap C\_{\bar G}(\bar x\_r)\le F(C\_{\bar G}(\bar x\_r))$, in which you have shown $\bar c$ is central. (Note that reduction modulo $F(G)$ induces an isomorphism $\alpha:C\_G(x\_r)\to C\_{\bar G}(\bar x\_r)$. By (iii), $\alpha$ is injective. Since $|x\_r|$ is coprime to $|F(G)|$, a Frattini argument shows that $\alpha$ is surjective.)
Finally, say $\bar x\_r$ has (prime) order $r$. If $p\ne r$, then $\bar y$ and $\bar x\_r$ are of coprime orders in $F(\bar G)$, so $\bar y\in C\_{\bar G}(\bar x\_r)$. If $p=r$, then $\bar y\in C\_{\bar G}(\bar x\_r)$ as $\langle \bar x\_r\rangle\triangleleft \bar G$. QED
| 4 | https://mathoverflow.net/users/99221 | 415588 | 169,409 |
https://mathoverflow.net/questions/415556 | 6 | Suppose that $f \in S\_k(\Gamma\_0(N)) $ be a Hecke eigenform whose Fourier expansion at $ i\infty $ is given by
$$
f(z) = \sum\_{n=1}^{\infty} \lambda(n) n^{\frac{k-1}{2}} \exp(2\pi i n z),
$$
normalized so that $\lambda(1)=1$. In this setting the Ramanujan-Petersson conjecture states that $ |\lambda(n)| \leq d(n) $ the number of divisors of $n$ (for all $ n $ coprime to the level $ N $).
Does the same bound hold if I consider the Fourier expansion of $f$ at some other cusp?
| https://mathoverflow.net/users/148866 | Ramanujan-Petersson conjecture at various cusps | The estimate $|\lambda(n)| \leq C\_N d(n)$ remains valid at all cusps, but $C\_N$ cannot in general be taken independent of $N$. See Remark 3.14 of [this paper](https://mathscinet.ams.org/mathscinet-getitem?mr=3110797) ([arxiv link](https://arxiv.org/abs/1205.5534)), where it is noted that for certain $(N,f,n)$, with $f$ a newform, one can have $\lambda(n) \gg n^{1/4}$ at some cusp. (One can take for $n$ a suitable power of a prime $p$ for which $p^2 | N$.)
| 9 | https://mathoverflow.net/users/29622 | 415589 | 169,410 |
https://mathoverflow.net/questions/415485 | 7 | How can we generate the eigenspace for the Laplace Beltrami operator on SU(2)?
| https://mathoverflow.net/users/475670 | On eigenfunctions of the Laplace Beltrami operator | For $\mathrm{SU}(2)$, with the scale for the biïnvariant metric so that it becomes isometric to the unit $3$-sphere $S^3$ in Euclidean $4$-space, it is well-known what the eigenvalues of the Laplace-Beltrami operator are and, indeed, what the eigenfunctions are as well. The eigenvalues are $\lambda\_k = k(k{+}2)$ for $k=0,1,2,\ldots$, and the $k$-th eigenspace consists of the restrictions to $S^3$ of the harmonic polynomials homogenous of degree $k$ on $\mathbb{R}^4$. Hence, the multiplicity of $\lambda\_k$ is $(k{+}1)^2$. For example, see:
* the classic reference: Berger, M., Gauduchon, P., Mazet, E.: *Le spectre d'une variété riemannienne*, Springer Lecture Notes No.194, 1971.
* these lecture notes from MIT (theorem 2.9): [https://math.mit.edu/~dav/spheres.pdf](https://math.mit.edu/%7Edav/spheres.pdf)
More generally, for the biïnvariant metric on a compact, simply-connected simple Lie group $G$ corresponding to the (negative of the) Cartan-Killing form on the Lie algebra, the computation of the eigenvalues of its Laplace-Beltrami operator and their multiplicities can be reduced to a combinatorial problem by means of the Peter-Weyl Theorem, Cartan's classification of the irreducible representations of $G$, the Casamir operator, and various representation-theoretic formulas such as Weyl's character formula, Steinberg's formula, and formulae of Kostant. It is not a trivial combinatorial problem, though. I would suggest getting a reference such as A. Knapp's *Lie Groups: Beyond an Introduction* and then searching the internet for phrases such as *the spectrum of compact simple groups* to get started.
In principle, one could work out the spectrum for any left-invariant metric on a compact Lie group, but the combinatorial problems that one runs into in carrying this out explicitly can be thorny for all but the simplest cases.
**Addendum:** Because of [IGT's question/comment below from June 24, 2022](https://mathoverflow.net/questions/415485/on-eigenfunctions-of-the-laplace-beltrami-operator#comment1093465_415595), I decided to include the example of a left-invariant metric on $\mathrm{SU}(2)\simeq S^3$ in the first few cases to illustrate what's going on. If we let $g\_0$ be the biïnvariant metric on $\mathrm{SU}(2)$ with sectional curvature $1$, and let $g$ be any other left-invariant metric on $\mathrm{SU}(2)$, then there are left-invariant $1$-forms $\omega\_i$ and positive constants $c\_i$ such that $g\_0 = {\omega\_1}^2+{\omega\_2}^2+{\omega\_3}^2$ while $g = {\omega\_1}^2/c\_1+{\omega\_2}^2/c\_2+{\omega\_3}^2/c\_3$. If $X\_i$ are the dual left-invariant vector fields, then the Laplacian with respect to the metric $g$ is easily seen to be the second-order operator $\Delta = -c\_1\,{X\_1}^2 -c\_2\, {X\_2}^2 -c\_3\, {X\_3}^2$. Since differentiation with respect to $X\_i$ (whose flow is an isometric rotation of the sphere) clearly preserves the $(k{+}1)^2$ dimensional subspace $H\_k$ that consists of restrictions to $S^3$ of the harmonic polynomials homogeneous of degree $k$ on $\mathbb{R}^4$, it follows that $\Delta$ preserves this space as well, i.e., $\Delta(H\_k)=H\_k$ for $k\ge1$.
On $H\_1$, $\Delta$ has a single eigenvalue: $c\_1 + c\_2 + c\_3$, of multiplicity $4$.
on $H\_2$, $\Delta$ has three eigenvalues:
$4(c\_i + c\_j)$ ($i\not=j$), each of multiplicity $3$.
on $H\_3$, $\Delta$ has two eigenvalues: $5(c\_1+c\_2+c\_3)\pm 4\delta$, where
$$
\delta = \sqrt{{c\_1}^{2}+{c\_2}^{2}+{c\_3}^{2}-c\_1c\_2-c\_2c\_3-c\_3c\_1},
$$
and each eigenvalue has multiplicity $8$. (Note that $\delta=0$ if and only if $c\_1=c\_2=c\_3$.)
Finally, on $H\_4$, $\Delta$ has five eigenvalues:
Three of them are $4(c\_1+c\_2+c\_3)+12\,c\_i$ for $i=1,2,3$, and
the other two are $8(c\_1+c\_2+c\_3\pm\delta)$.
Each of these $5$ eigenvalues has multiplicity $5$.
As one can see, the eigenvalues of $g$ need not all be integer multiples of a single value, or even all rational if the $c\_i$ are rational. Also, note that if one of the $c\_i$ is much larger than the other two (for example, when $c\_1 > 3(c\_2{+}c\_3)$), the eigenfunction belonging to the lowest eigenvalue of $\Delta$ may not be the restriction of a linear function on $\mathbb{R}^4$ to $S^3$.
| 11 | https://mathoverflow.net/users/13972 | 415595 | 169,411 |
https://mathoverflow.net/questions/415585 | 8 | 1. What is Deligne's motivation in Appendice 9 of Exposé VI to prove that every coherent topos has enough points? For instance, does that have applications in étale cohomology (or other parts of algebraic geometry)?
2. Which topics are discussed in the Exposés Vbis, VI, VIII, and IX? An answer to the question should provide either buzzwords or *English* literature covering these topics. What role do topoi play in these Exposés?
3. [Here](https://indico.math.cnrs.fr/event/747/contributions/2523/attachments/1818/1962/introductionJoyal.pdf) Joyal writes the following. What could he mean by that?
>
> About half of the topos theory of SGA4 is devoted to categorical generalities. They are now subsumed by the modern theory of (locally) presentable categories.
>
>
>
| https://mathoverflow.net/users/476516 | Questions about SGA 4 | 1. I'm not aware of any applications of Deligne's result to algebraic geometry at the time. It does formally imply that certain topologies (e.g. fppf) have enough points, but this wasn't very useful because the actual points in many well-known topologies were only described by [Gabber–Kelly](https://arxiv.org/abs/1407.5782) and [Schröer](https://arxiv.org/abs/1407.5446) in 2014. On the other hand, points in the étale topology are just geometric points, so those are very useful in the theory, but you don't need Deligne's completeness theorem for this.
Deligne's theorem does have formal consequences to the systematic study of coherent topoi, such as in Barwick–Glasman–Haine's [*Exodromy*](https://arxiv.org/abs/1807.03281), where it is used in the proof of a base change theorem for oriented fibre products of bounded coherent $\infty$-topoi (Thm. 7.1.7).
2. This is many questions in one, but let me say some words.
* V$^{\text{bis}}$ is the origin of hypercoverings and cohomological descent. This is a powerful tool that is used frequently in algebraic geometry, especially in combination with resolution of singularities or alterations to build up the cohomology of a variety from the cohomology of smooth projective varieties. This happens for example in Deligne's [*Hodge III*](http://www.numdam.org/item/PMIHES_1974__44__5_0/) paper, where he uses proper hypercoverings to construct the mixed Hodge structure on a singular or open variety.
One alternative source for the material from V$^{\text{bis}}$ is the [chapter on hypercoverings](https://stacks.math.columbia.edu/tag/01FX) in the Stacks project, but I imagine to a category theorist this might look a bit dated.
Descent with respect to arbitrary hypercoverings is also a topic that comes up in $\infty$-topoi, where this property is called *hyperdescent*. The relation between hypercompleteness and hyperdescent is explained in sections 6.5.3 of Higher Topos Theory, and §6.5.4 makes the case that you do *not* want to impose this as an axiom on an $\infty$-topos. Because hypercompleteness is related to Whitehead's theorem (which holds on the $\infty$-topos of a point), an $\infty$-topos can only have enough points (in the sense of question 1) if it is hypercomplete!
* VI is about coherent topoi and is purely topos-theoretic in nature (although the definitions of *quasi-compact* and *quasi-separated* are inspired by the spaces occurring in algebraic geometry). I don't know secondary references for this, nor is it incredibly clear to me where it is used, so I will leave this to someone else to answer.
* VIII is the first place where some actual computations in the étale cohomology happen. For example, it is explained that the étale site of a field $k$ with absolute Galois group $\Gamma\_k$ is a $B\Gamma\_k$ (at least if you understand $\Gamma\_k$ as a profinite group and interpret $B\Gamma\_k$ in that light).
This material is all specific to the étale site in algebraic geometry, and strays away from general topos theory. Most of the results can be found in the Stacks Project, or Milne's book or notes on *Étale Cohomology*, or the book by Freitag and Kiehl, or ...
* IX is about constructible sheaves and the cohomology on a curve. Constructible sheaves also exist in topology, but in algebraic geometry there are pervasive finiteness hypotheses for everything to work.
More interesting is the computation of the cohomology of a curve. For example, you want to know that étale cohomology with finite coefficients vanishes above dimension $2$. In algebraic geometry, the only way to do this is to show that the Galois group of a function field of a curve has cohomological dimension $1$, and then use the Leray spectral sequence for $\operatorname{Spec} K(C) \hookrightarrow C$. The rest of the computation of the cohomology on a curve is based on its Picard group (or Jacobian).
3. This has been addressed by Tim Campion in the comments. As I noted, schemes don't start coming into SGA 4 until V$^{\text{bis}}$ (a little bit) and really starting at VII. All the material prior to IV (and even some of the material after that) really is general category theory, a substantial part of which consists of putting size bounds on everything to make sure all categories are small or at least locally small. This is exactly the study of accessible categories. The classical reference is probably Adámek–Rosicky's *Locally presentable and accessible categories*, and I suspect SGA 4 would be substantially shortened if one already has access to all that.
| 10 | https://mathoverflow.net/users/82179 | 415598 | 169,413 |
https://mathoverflow.net/questions/415425 | 3 | Recall that the $i$*-weight* of a Tychonoff space $X$, $iw(X)$, denotes the minimal weight of all Tychonoff spaces onto which $X$ can be condensed. A standard fact about this cardinal number is that the relations $\text{log}(X)\leq iw(X) \leq nw(X)$ always hold. Furthermore, if $X$ is an infinite discrete space, then $\text{log}(X)$ and $iw(X)$ are actually the same.
Is it true that all infinite metrizable spaces $X$ satisfiy the equality $\text{log}(X)=iw(X)$?
| https://mathoverflow.net/users/146942 | $i$-weight of a metrizable space | Countable powers of copies of $H(\kappa)$, the hedghog space of spininess $\kappa$, are universal for metrizable spaces of weight $\kappa$ (i.e., any metrizable space of weight $\kappa$ embeds into such a countable power as a subspace). Note first that $H(2^\kappa)$ condenses onto a subspace of $[0,1]^\kappa$ (it is easier to see geometrically how $H(2^\omega)$ condenses onto the unit disc using the usual geometric picture of the hedgehog, but, in general, to map $H(2^\kappa)$ onto $[0,1]^\kappa$: for each $f\in 2^\kappa\setminus \{0\}$, let $I\_f=\{t\cdot f:0\leq t\leq 1\}$ and map the spine $I\_f$ in $H(2^\kappa)$ to the subspace $I\_f$ in the product space $[0,1]^\kappa$). The weight of this subspace is $\leq\kappa$. Now, if a metrizable space has cardinality $2^\kappa$, it embeds into $H(2^\kappa)^\omega$ which in turn condenses onto a subspace of $(I^{\kappa})^\omega$ and so $iw(X)\leq \kappa=Log(X)$ and by the reverse inequality mentioned in your question you get the equality you were asking for.
| 4 | https://mathoverflow.net/users/121994 | 415600 | 169,414 |
https://mathoverflow.net/questions/415610 | 2 | Let $X$ be random vector on the unit-sphere $S\_{n-1}$ in $\mathbb R^n$. We don't assume that the distribution of $X$ is uniform on $S\_{n-1}$
>
> I'm interested in proving the existence of a (deterministic) direction $v \in S\_{n-1}$ such that
> $$
> \mathbb P(|\langle X,v\rangle | \ge \alpha) \ge \beta,
> \tag{1}
> $$
> where $\alpha,\beta \in (0,1]$ are absolute constants.
>
>
>
Thus in a sense, the random vector $X$ "likes" the direction $v$. The larger the constants $\alpha,\beta$ the better. For example, if the support of $X$ is made up of a bounded number of atoms (i.e not dependent on the dimension $n$), then (1) holds.
**Assumption.** Suppose $X$ admits a second-moment matrix $\Sigma := \mathbb E\, X \otimes X \in \mathbb R^{n \times n}$ and largest eigenvalue $s\_1$ of $\Sigma$ is lower-bounded by an absolute constant $c>0$.
Let $v$ be a unit-vector in the corresponding eigenspace. Since $Z:=|\langle X,v\rangle|^2$ takes values in the interval $[0,1]$, it is clear that the variance $\sigma^2$ of $Z$ is at most $1/4$. On the other hand, by linearity of trace and expectation, the expectation of $Z$ is $$
\mathbb E Z = \mathbb E [\mbox{trace}((v\otimes v)(X \otimes X))] = \mbox{trace}((v \otimes v)\Sigma) = v^\top \Sigma v = v^\top (\Sigma v) = s\_1\|v\|^2 = s\_1.
$$
Therefore, by Cantelli's inequality, for any $0<a<s\_1$, we have $$
\mathbb P(Z \ge s\_1-a) \ge 1-\sigma^2/(\sigma^2+a^2)=(1+a^2/\sigma^2)^{-1} \in (0,1).
$$
That is, for any $0 < a < s\_1$, (1) holds with $\alpha = s\_1 - a$ and $\beta=(1+a^2/\sigma^2)^{-1}$.
>
> **Question.** *Is my above reasoning correct ? Is there an alternative / more powerful way to use the above assumptions to obtain a stronger inequality (i.e large $\alpha$ and $\beta$) of the form (1) ?*
>
>
>
| https://mathoverflow.net/users/78539 | Existence of preferred direction for a random vector with arbitrary distribution on sphere, under a condition on its covariance matrix | $\newcommand\R{\mathbb R}\newcommand{\Si}{\Sigma}\newcommand{\si}{\sigma}$The answer is no: in general (and usually) there are no positive absolute constants $a$ and $b$ such that for some unit vector $v$ one has
$$P(|X\cdot v|\ge a)\ge b.$$
Indeed, otherwise one would have $E(X\cdot v)^2\ge c:=ba^2>0$. However, if $X$ is uniformly distributed on the unit sphere in $\mathbb R^n$ and $v$ is a unit vector, then $(X\cdot v)^2$ has the beta distribution with parameters $1/2,(n-1)/2$ and hence $E(X\cdot v)^2=1/n<c$ if $n>1/c$.
---
The OP has rectified the confusion raised by the initial formulation of the their question. The changes invalidate the above answer. Here is an updated answer to the current state of the question.
Let $Y:=|X\cdot v|$, where $v$ is a unit eigenvector corresponding to the eigenvalue $s\_1$. Then, $0\le Y\le1$ and $EY^2=s\_1$. So, for all $a\in(0,1)$ we have the inequality
$$1(Y>a)\ge\frac{Y^2-a^2}{1-a^2},$$
with the equality on the event $\{Y\in\{a,1\}\}$,
and hence taking expectations gives
\begin{equation}
P(|X\cdot v|>a)=P(Y>a)\ge\frac{\max(0,s\_1-a^2)}{1-a^2}. \tag{1}\label{1}
\end{equation}
This lower bound on $P(|X\cdot v|>a)$ is exact: It is attained if
(i) $a^2\le s\_1\le1$ and $(X\cdot v)^2$ only takes values $a^2$ and $1$ (with mean $E(X\cdot v)^2=s\_1\in[a^2,1]$) or if
(ii) $0\le s\_1<a^2$ and $(X\cdot v)^2$ only takes value $s\_1$.
**Addendum 1:** Strictly speaking, to show that the lower bound on $P(|X\cdot v|>a)$ in \eqref{1} is exact, we also need to show that
(I) for any $s\_1\in[a^2,1]$ and at least for some $n$, there exist a random unit vector $X$ in $\R^n$ and a unit vector $v\in\R^n$ such that $(X\cdot v)^2$ only takes values $a^2$ and $1$, with mean $E(X\cdot v)^2=s\_1$, and, moreover, $s\_1$ is the largest eigenvalue of the covariance matrix $\Si$ of $X$;
(II) for any $s\_1\in(0,a^2)$ and at least for some $n$, there exist a random unit vector $X$ in $\R^n$ and a unit vector $v\in\R^n$ such that $(X\cdot v)^2$ only takes value $s\_1$ and, moreover, $s\_1$ is the largest eigenvalue of the covariance matrix $\Si$ of $X$.
To prove (I), do take any $s\_1\in[a^2,1]$, and take any unit vector $v\in\R^n$, where $n\ge2$.
Then let $\mu\_X=p\mu\_V+\frac q2\,\mu\_W+\frac q2\,\mu\_{-W}$, where $\mu\_Y$ denotes the distribution of a random vector $Y$,
\begin{equation}
p:=\frac{s\_1-a^2}{1-a^2},\quad q:=1-p=\frac{1-s\_1}{1-a^2},
\end{equation}
$P(V=v)=P(V=-v)=1/2$, $W:=av+\sqrt{1-a^2}\,U$, and $U$ is uniformly distributed on the unit sphere of the vector space that is the orthogonal complement of $\text{span}(\{v\})$ to $\R^n$. Then $P((X\cdot v)^2=1)=p$, $P((X\cdot v)^2=a^2)=q$, $E(X\cdot v)^2=s\_1$, and the eigenvalues of the covariance matrix $\Si$ of $X$ are $s\_1$ and $\dfrac{1-s\_1}{n-1}$ (the latter one of multiplicity $n-1$). So, if $n\ge1/s\_1$, then $s\_1$ is the largest eigenvalue of $\Si$. Thus, all the desired conditions are satisfied, and (I) is proved.
The proof of (II) is similar, and a bit simpler. Here, we let $\mu\_X=\frac12\,\mu\_T+\frac12\,\mu\_{-T}$, where $T:=\sqrt{s\_1}\,v+\sqrt{1-s\_1}\,U$.
**Addendum 2:** The lower bound in your post depends on the variance $\si^2$ of $Z=Y^2$. You did not fully specify a value of $\si^2$, just noting that $\si^2\le1/4$, since $0\le Z\le1$. The latter bound on $\si^2$ can be improved to the optimal (in terms of $EZ$) bound $EZ(1-EZ)=s\_1(1-s\_1)$, which is attained when $Z$ only takes values $0$ and $1$. It now follows that your lower bound cannot be exact -- because, as seen from above, the exact lower bound is only attained when $Z=(X\cdot v)^2$ takes values in the set $\{a^2,s\_1,1\}$, which does not contain the value $0$. Also, it can probably be shown directly that, in distinction from the lower bound on $P(|X\cdot v|>a)$ in \eqref{1}, the lower bound in your post cannot be exact.
| 1 | https://mathoverflow.net/users/36721 | 415619 | 169,419 |
https://mathoverflow.net/questions/415045 | 0 | I solved many cases for the following dynamical system $\dot{x} = x (1-x-ay)$ and $\dot{y} = c y (1- b x -y)$. However, I reached the case where $c>0$ and $a>1$, $b=1$ and I ended up with the Jacobian of the fixed point $(1,0)$, which is
$$A\_{(1,0)} =
\begin{pmatrix}
-1 & -a\\
0 & 0
\end{pmatrix}$$
It is clear that we have a zero real part eigenvalue, so we need to find the center manifold. Here is the problem, every time i assume $y=h(x)=c\_1 x^2 + c\_2 x^3 +...$ I get all the coefficients zero! What is the problem
| https://mathoverflow.net/users/471464 | Why all the coefficients of the center manifold of this system are zeros? | The center manifold for the first problem is not $y = O(x^2)$. It is
tangent to the center subspace of the linearized operator, which is
not $x$-axis in this case.
Please find an eigenvector associated with zero eigenvalue and change
the coordinates so that the center subspace corresponds one of the
coordinate axes. In the new coordinates, you will get the answer soon.
| 0 | https://mathoverflow.net/users/471464 | 415623 | 169,420 |
https://mathoverflow.net/questions/415361 | 3 | Every continuous vector bundle on a closed smooth manifold $M$ has a smooth structure. On the other hand, every vector bundle $E$ is the image of a trivial bundle $M\times\mathbb{C}^n$ under some projection $p$ in a matrix algebra $M\_n(C(M))$. I am wondering if it possible to approximate this projection by a smooth idempotent, call it $e$, and whether it is possible to preserve certain properties of $p$.
More precisely:
**Question 1:** Take any $\varepsilon>0$. Does there exist an idempotent $e\in M\_n(C^\infty(M))$ such that $\|p(x)-e(x)\|<\varepsilon$ for all $x\in M$?
**Question 2:** Suppose $p$ is Lipschitz in the sense that $\|p(x)-p(y)\|\leq d(x,y)$ for all $x,y\in M$, where $\|\cdot\|$ is the operator norm on $M\_n(\mathbb{C})$ and $d$ is the distance determined by a Riemannian metric on $M$. Take any $\varepsilon>0$. Does there exist an idempotent $e\in M\_n(C^\infty(M))$ that is $(1+\varepsilon)$-Lipschitz and such that $\|p(x)-e(x)\|<\varepsilon$ for all $x\in M$?
I'd be grateful for any references that discuss this.
| https://mathoverflow.net/users/78729 | Approximation of continuous projections on a manifold by smooth idempotents | Well, I learned this was Rieffel's question and was already solved by Hanfeng Li. <https://mathscinet.ams.org/mathscinet-getitem?mr=2994680>
**Obsolete:**
The following is not an answer but maybe useful (see the comments above).
**Lemma.** Let $a,b$ be self-adjoint operators and $0<\epsilon<0.01$. Assume that $\|b-a\|<\epsilon$ and that the spectra of $a$ and $b$ are contained in the $\epsilon$ neighborhood of $\{0,1\}$. Then the spectral projections satisfy $$\|1\_{(1-\epsilon,1+\epsilon)}(b)-1\_{(1-\epsilon,1+\epsilon)}(a)\|\le(1+999\epsilon)\|b-a\|.$$
**Proof.** Put $b(t):=a+t(b-a)$ and $p(t):=1\_{(1-2\epsilon,1+2\epsilon)}(b(t))$.
By Cauchy's integral formula, one has
$$p(t)=\frac{1}{2\pi i}\oint\_\Gamma \frac{dz}{z-b(t)},$$
where $\Gamma:=\{|z-1|=\frac{1}{2}\}$. Take the derivative
$$p'(t)= \frac{1}{2\pi i}\oint\_\Gamma (z-b(t))^{-1}b'(t)(z-b(t))^{-1}\,dz.$$
For $p:=p(0)$, one has $\|(z-b(t))^{-1}-(z-p)^{-1}\|<99\epsilon$ for $z\in\Gamma$,
because $\|(z-b(t))^{-1}\|<3$ and $\|b(t)-p\|\le\|b(t)-a\|+\|a-p\|<2\epsilon$.
Hence
\begin{align\*}
p'(t) &\approx\_{999\epsilon\|b-a\|}\frac{1}{2\pi i}\oint\_\Gamma (z-p)^{-1}(b-a)(z-p)^{-1}\,dz\\
&=p(b-a)p^\perp + p^\perp(b-a)p.
\end{align\*}
Note that $\|p(b-a)p^\perp + p^\perp(b-a)p\|=\|p(b-a)p^\perp\|\le\|b-a\|$.
It follows that
$$\|p(1)-p(0)\|\le(1+999\epsilon)\|b-a\|.$$
| 2 | https://mathoverflow.net/users/7591 | 415625 | 169,421 |
https://mathoverflow.net/questions/415615 | 10 | Littlewood-Richardson coefficients $c\_{\nu\mu}^{\lambda}$ (where $\nu$,$\mu$ and $\lambda$ are integer partitions such that $|\nu| + |\mu| = |\lambda|$) are well-known coefficients appearing in various contexts.
In terms of Schur functions, which form a basis of the symmetric functions, the coefficients are the multiplicative constants, meaning that
$$
s\_{\nu}s\_{\mu} = \displaystyle \sum\_{\lambda} c\_{\nu\mu}^{\lambda} s\_{\lambda}.
$$
This is the context in which the rule was first stated. As Schur functions are the characters of irreducible representations of $\text{GL}\_n$, it means that if we denote $V\_{\lambda}$ for the irreducible representation of $\text{GL}\_n$ associated to $\lambda$, then:
$$
V\_{\nu} \otimes V\_{\mu} \cong \displaystyle \bigoplus\_{\lambda} (V\_{\lambda})^{\oplus c\_{\nu\mu}^{\lambda}}
$$
The transition between those two contexts is relatively easy.
There is also another context closely related to the two above, which is the representations the symmetric group. Via the Schur-Weyl duality, the above results implies that if $S^{\lambda}$ is the Specht module associated to $\lambda$ (which are the irreducible representations of $\mathfrak{S}\_{|\lambda|}$), then
$$
\left( S^{\nu} \otimes S^{\mu} \right) \big\uparrow\_{\mathfrak{S}\_{|\nu|} \times \mathfrak{S}\_{|\mu|}}^{\mathfrak{S}\_{|\nu|+|\mu|}} \cong \bigoplus\_{\lambda} \left( S^{\lambda} \right)^{\oplus c\_{\nu\mu}^{\lambda}}
$$
Using Schur's lemma, the latter means that
$$
\text{dim} \left( \text{Hom}\_{\mathfrak{S}\_{|\nu|+|\mu|}} \left( S^{\lambda}, \left( S^{\nu} \otimes S^{\mu} \right) \big\uparrow\_{\mathfrak{S}\_{|\nu|} \times \mathfrak{S}\_{|\mu|}}^{\mathfrak{S}\_{|\nu|+|\mu|}} \right) \right) = c\_{\nu\mu}^{\lambda}
$$
My question is: is there any construction of the Littlewood-Richardson coefficients which is stated only in terms of the symmetric group and that gives the last equality? For example, Specht modules is the span of certain elements of the symmetric group algebra, called polytabloids. Can we calculate the Littlewood-Richardson coefficients using those?
| https://mathoverflow.net/users/73667 | Littlewood-Richardson coefficients in terms of Specht modules | The answer is Yes. You don't need $GL(n, \mathbb C)$ or symmetric functions to define/understand LR-coefficients. There are numerous references to choose from, which wary how much Young tableau combinatorics you want to see.
1. For more combinatorial explanations, I recommend two similarly titled papers by [Kerov](https://iopscience.iop.org/article/10.1070/RM1984v039n02ABEH003121/meta) and by [Zelevinsky](https://sites.math.washington.edu/%7Ebilley/classes/Hopf.algebra/bulletins/zelevinsky.1981.pdf). Roughly, you want to take pairs of SYTs of shapes $\mu \circ \nu$, apply the jeu-de-taquin and count the number of times a fixed $A\in$SYT$(\lambda)$ appears. When $A$ is lexicographically smallest, these gives one of the standard combinatorial interpretation for $c^\lambda\_{\mu,\nu}$.
2. For a more algebraic interpretation, I recommend a concise [paper by Donin](http://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=dan&paperid=7324&option_lang=eng) which defines LR-coefficients in terms of dimension of certain spaces of homomorphisms between $S\_n$-modules related to Zelevinsky's "pictures".
There are many more references, generalizations, etc. Note that in both cases you need to start with the [Frobenius reciprocity](https://en.wikipedia.org/wiki/Frobenius_reciprocity) formula since it's a bit easier to think about decomposition of large reduced $S\_n$-modules.
| 10 | https://mathoverflow.net/users/4040 | 415636 | 169,422 |
https://mathoverflow.net/questions/415630 | 2 | The Gamma function $\Gamma$ is defined by
\begin{equation\*}
\Gamma(x)=\int\_{0}^\infty t^{x-1}e^{-t} \,\mathrm{d}t,
\end{equation\*}
for $x>0$. It satisfies the well-known functional equation
$$\Gamma(x+1)=x\Gamma(x)\label{1}\tag{i}$$
for all $x>0$. Apparently, the definition of Gamma function can be extended to $\mathbb{R}\setminus \{0,-1,-2,\dots\}$ by using \eqref{1}. With this extended definition, the equation \eqref{1} will also hold for negative real numbers except at non-positive integers.
Now the question is:
>
> Let $A\subseteq \mathbb{R}$ containing $(0,\infty)$, and let $f:A\to\mathbb{R}$. Suppose that you have shown that $f(x)=\Gamma(x)$ for all $x>0$ by using Bohr-Mollerup Theorem. However, if $f(x+1)=xf(x)$ for all $x\in A$, would it imply that $f(x)=\Gamma(x)$ hold for all $x\in A\setminus \{0,-1,-2,\dots,\}$?
>
>
>
The statement of the Bohr-Mollerup Theorem is:
>
> **Bohr-Mollerup Theorem**: Gamma function is the only positive function $f:(0,\infty)\to\mathbb{R}$ that is logarithmically convex, $f(1)=1$ and $f(x+1)=xf(x)$ for all $x>0$.
>
>
>
But in Artin's "the Gamma function", he formulate (probably the translator) the Bohr-Mollerup Theorem a bit different:
>
> **B-M2 Theorem**: If a function $f(x)$ satisfies the following three conditions, then it is identical in its domain of definition with the gamma function:
>
>
> 1. $f(x+1)=xf(x)$
> 2. The domain of definition of $f(x)$ contains all $x>0$, and is logarithmically convex for these x.
> 3. $f(1)=1$.
>
>
>
What Artin did was as follows (the way I understand it): It has been shown that $\Gamma$ satisfies these conditions, so the existence is OK. Let us assume $f$ that satisfies these conditions, and want to check its uniquenessness. First, assume $x\in (0,1]$. After some steps, we end up getting
$$
f(x)=\lim\_{n\to\infty} \frac{n!n^x}{x(x+1)\cdots (x+n)}.
$$
which shows that $f$ is unique on $(0,1]$. But the uniqueness of $f$ also holds on $(1,2]$ by condition (1). Keep doing like this, then $f$ is unique for whole $(0,\infty)$. From here, I am not really sure, how to conclude that $f$ is also unique on whole of its domain, when the domain is not explicitly known, which is why I asked the question.
| https://mathoverflow.net/users/476697 | Gamma function and the somewhat extended version of Bohr-Mollerup theorem | $\newcommand\R{\mathbb R}\newcommand{\Ga}{\Gamma}\newcommand\Z{\mathbb Z}$The answer is yes. Indeed, the conditions $f\colon A\to\R$ and $f(x+1)=xf(x)$ for all $x\in A$ imply that $x+1\in A$ for any $x\in A$ and hence, by induction, $x+k\in A$ for all natural $k$. We also have $(0,\infty)\subseteq A\subseteq\R$ and $f(x)=\Ga(x)$ for all $x>0$.
Now take any $y\in A\cap(-\infty,0]\setminus \{0,-1,-2,\dots,\}$. It remains to show that $f(y)=\Ga(y)$. Take any natural $n$ such that $y+n>0$. Then $\Ga(y+n)\ne0$ and
\begin{equation}
f(y)\prod\_{k=0}^{n-1}(y+k)=f(y+n)=\Ga(y+n)=\Ga(y)\prod\_{k=0}^{n-1}(y+k). \tag{1}
\end{equation}
So, $\prod\_{k=0}^{n-1}(y+k)\ne0$ and $f(y)=\Ga(y)$, as desired. (The conclusion $\prod\_{k=0}^{n-1}(y+k)\ne0$ also follows because $y$ is not an integer and hence $y+k\notin\mathbb Z$ for any integer $k$.)
| 2 | https://mathoverflow.net/users/36721 | 415656 | 169,429 |
https://mathoverflow.net/questions/415384 | 1 | One of the consequences of the well-known Motzkin-Taussky theorem (<https://www.jstor.org/stable/1990825>) is the following : if two complex matrices $A, B$ generate a vector space of diagonalisable matrices, then $A$ and $B$ commutes and in particular are simultaneously diagonalisable.
Does the result hold for nilpotent matrices : Let $A$ and $B$ two (complex) matrices such that $sA+tB$ are nilpotent for all $s,t\in \mathbb{C}$ are they simultaneously triangularisable ?
| https://mathoverflow.net/users/476498 | Nilpotent matrices with (Motzkin-Taussky) property L | The matrices $$A = \begin{pmatrix} 0&1&0 \\ 0&0&1 \\ 0&0&0 \end{pmatrix}, B = \begin{pmatrix} 0&0&0 \\ 1&0&0 \\ 0&-1&0 \end{pmatrix}$$ satisfy $(sA+tB)^3=0$ for all $s,t$, but $$AB = \begin{pmatrix} 1&0&0 \\ 0&-1&0 \\ 0&0&0 \end{pmatrix},$$ which is not nilpotent, so $A$ and $B$ are not simultaneously triangularisable (as strictly upper triangular matrices).
| 1 | https://mathoverflow.net/users/425351 | 415674 | 169,437 |
https://mathoverflow.net/questions/415643 | 15 | There is a folklore in the empirical computer-science literature that, given a tree $(X,d)$, one can find a bi-Lipschitz embedding into a hyperbolic space $\mathbb{H}^n$ and that $n$ is "much smaller" than the smallest dimension of a Euclidean space in which $(X,d)$ can be bi-Lipschitz embedded with similar distortion.
**Question A:** Is there any theoretical grounding to this claim? Namely, can one prove that $(X,d)$ *(where $\# X = n\in\mathbb{N}\_+$)* admits a bi-Lipschitz embedding into some $\mathbb{H}^n$ with:
* distortion strictly less that $O(\log(n))$
* $n<O(\log^2n)$?
**! Edit - (Following Discussion of YCor, WillSawin, and TomTheQuant):** *What can be said if $s=1$ in Equation (1)?*
**Question B (Converse):** For every $n\in \mathbb{N}\_+$ and every $D>0$ does there exist a *finite metric space* $(X,d)$, which don't admit a bi-Lipschitz (resp. possibly [uniform embedding](https://mathoverflow.net/questions/101412/uniform-embedding-into-euclidean-space)) into $\mathbb{H}^n$ with distortion at-most $D$?
**Relevant Definition (For completeness)**
A bi-Lipschitz embedding $f:X\rightarrow \mathbb{H}^n$ of a metric space $(X,d)$ into $\mathbb{H}^n$ with distortion $D>0$ is a Lipschitz homeomorphism $f:X\rightarrow \mathbb{H}^n$ Lipschitz inverse $f^{-1}$ such that there is some $s>0$ satisfying
$$
sd(x\_1,x\_2) \leq d\_{\mathbb{H}^n}(f(x\_1),f(x\_2)) \leq sDd(x\_1,x\_2)
\qquad (1)
$$
for every $x\_1,x\_2\in X$. Here, $d\_{\mathbb{H}^n}$ is the usual geodesic distance on the $n$-dimensional hyperbolic space.
**Some Relevant posts:**
*Hyperbolic embeddings*
* [References to embedding into hyperbolic spaces](https://mathoverflow.net/questions/215550/embedding-graphs-into-hyperbolic-spaces?rq=1)
* [Representability of finite metric spaces](https://mathoverflow.net/questions/12394/representability-of-finite-metric-spaces?rq=1)
*Flat Embeddings*
* [Problem with embedding expanders into "flat" spaces](https://mathoverflow.net/questions/146679/embedding-expanders-in-cat0-spaces?rq=1)
* [Characterizing finite metric spaces which embed into Euclidean space](https://mathoverflow.net/questions/297654/a-characterization-of-metric-spaces-admitting-a-bi-lipschitz-embedding-into-a-hi/297761#297761)
*Uniform Embeddings*
* [Notes on coarse and uniform embeddings](https://arxiv.org/abs/1512.03109)
| https://mathoverflow.net/users/469470 | Are hyperbolic spaces actually better for embedding trees than Euclidean spaces? | I am not sure if the following paper answers your question. The abstract suggests so, but it is written in a computer science style that is less transparent to me in terms of stating a precise theorem. Also: (a) I am not an expert, (b) I am confused by the way that $n$ seems to play two different roles in your question, and (c) I doubt if it is the earliest answer to your question, if indeed it does answer it.
[Low Distortion Delaunay Embedding of Trees in Hyperbolic Plane by Rik Sarkar](https://homepages.inf.ed.ac.uk/rsarkar/papers/HyperbolicDelaunayFull.pdf)
>
> Abstract. **This paper considers the problem of embedding trees into the
> hyperbolic plane.** We show that any tree can be realized as the
> Delaunay graph of its embedded vertices. **Particularly, a weighted tree
> can be embedded such that the weight on each edge is realized as the
> hyperbolic distance between its embedded vertices.** Thus the embedding
> preserves the metric information of the tree along with its topology.
> **Further, the distance distortion between non adjacent vertices can be
> made arbitrarily small – less than a $(1+\epsilon)$ factor for any given $\epsilon$.**
> Existing results on low distortion of embedding discrete metrics into
> trees carry over to hyperbolic metric through this result. The
> Delaunay character implies useful properties such as guaranteed greedy
> routing and realization as minimum spanning trees.
>
>
>
(bolding is mine.)
| 7 | https://mathoverflow.net/users/476736 | 415679 | 169,439 |
https://mathoverflow.net/questions/415617 | 16 | Let $f(x)$ be sufficiently regular (e.g. a smooth function or a formal power series in characteristic 0 etc.). In my research the following recursion made a surprising entrance
$$
f\_1(x) = f(x),\ f\_{n+1}(x) = f(x) f'\_{n}(x)
$$
Thus I would like to understand the sequence
$$
\left(f(x) \frac{d}{dx} \right)^n f(x)
$$
This looks like a classical question that must have been already studied in dynamical systems, or Weyl algebras (say for $f(x) \in R[x]$, $R$ a commutative ring of characteristic $0$, and the derivation $\partial := f(x) \frac{d}{dx} \in A\_1(R)$), or generating functions in combinatorics. But I have been unable to pinpoint it. My question is this:
>
> Is there a known formula in the spirit of the general Leibniz formula which expresses
> $$
> \left(f(x) \frac{d}{dx} \right)^n f(x)
> $$
> in terms of $f$ and its derivatives $f',f'',\dots, f^{(n)}$?
>
>
>
Any references would be also very much appreciated!
| https://mathoverflow.net/users/1849 | A Leibniz-like formula for $(f(x) \frac{d}{dx})^n f(x)$? | Revamped Feb. 12, 2022:
I posted an answer to this (perennial) question in detail in the old MO-Q "[Formula for n-th iteration of dx/dt=B(x)](https://mathoverflow.net/questions/41039/formula-for-n-th-iteration-of-dx-dt-bx)" and pointed out a common conflation of related but distinct number arrays, all related to 'natural growth' of rooted trees (and therefore Lehmer codes, see Adler ref below, p. 12). With decreasing order of refinement, they are
**1)** OEIS [A139002](https://oeis.org/A139002): the Connes-Moscovici weights of the Connes-Moscovici Hopf algebra, enumerating forests of 'naturally-grown' rooted trees (see Hivert et al. in Taylor's answer, eqns. 34 and 36-40)
**2)** OEIS [A139605](https://oeis.org/A139605): coefficients for the Scherk-Comtet partition polynomials, the normal-ordered operator expansion of the diff op $(f(z)\partial\_z)^n$ in terms of the derivatives of $f$ and the derivative operator
**3)** OEIS [A145271](https://oeis.org/A145271): the refined Eulerian numbers, coefficients for the expansion of $(f(z)\partial\_z)^n \; f(z)$, or, equivalently, $(f(z)\partial\_z)^{n+1} \; z$, as a partition polynomial in terms of the derivatives of $f(z)$ (see Hivert et al., eqns. 15-20). ***This is the expansion the OP, M.G., is addressing***.
The paper by Hivert et al. presented by Peter Taylor addresses item 1 and illustrates item 3; however, there is no discussion of a multinomial-type formula for the refined Eulerian numbers--only eqns. 28 and 35 are given, essentially stating that item 3 is a coarsening/reduction of item 1. As far as I can tell, beyond eqn. 40, Hivert et al. deal with trees in relation to generic statistics generated by various codes and don't come back to the illustration of item 3. The Findstat entry created by Hivert that Taylor links to does give examples of the refined Eulerian numbers generated by the Lehmer code algorithm. This is an algorithm, rather than a formula of the multinomial-type (correct me if I'm mistaken)--formulas that are available for item 1 and for specialized Lagrange inversion partition polynomials (see notes III and IV below) generated from item 3, but not for item 3 directly. Giving the algorithm rather than a formula is analogous to saying a number array counts the number of perfect matchings of the vertices of the n-simplices and giving an algorithm that constructs the matchings and then counts them, or saying the number array enumerates a sum over the non-intersecting dissections of the convex polygons and giving a construction algorithm, without giving a numerical formula in terms of the number of vertices of the constructs. A direct formula for the refined Eulerian numbers is still an open question, I believe.
The paper by Domininci that skbmoore (any relation to eecummings?) cites deals with related but distinct expansions, giving as the most general example the associahedra partition polynomials I point out below. Although very important, this is not the expansion the OP is addressing. (Dominici doesn't address the connections to the associahedra nor flow equations nor does he go into any detail on trees or other combinatorial constructs, givibg just a ref. on trees)
Additional refs and further notes:
I) "[Set partitions and integrable hierarchies](https://arxiv.org/abs/1510.02900)" by Adler presents the first few partition polynomials of the expansion the OP desires on p. 11 and discusses them in the context of 'natural growth' sequences $T\_n$. He also alludes to Lehmer codes in the context of $T\_n$. Cayley's rooted trees and the related Connes-Moscovici weights are other manifestations of 'natural growth'.
II) OEIS A145271: the partition polynomials for $(g(z)\partial\_z)^n g(z)$ with $g(z) = 1/f'(z)$ (so just change notation). I called (unaware of the Hivert et al. paper until now) the coefficients of the polynomials the refined Eulerian numbers since they naturally reduce to the Eulerian numbers [A008292](https://oeis.org/A008292), or [A123125](https://oeis.org/A123125). My blog post "[A Creation Op, Scaled Flows, and Operator Identities](https://tcjpn.wordpress.com/2022/02/02/a-creation-op/)" contains detail on related flow functions, p.d.e.s, and more as a prelude to introducing the action of the iterated generalized Lie derivative $q(z)+g(z)\partial\_z$. See also the recent MO-Q "[How are Sheffer polynomials related to Lie theory?](https://mathoverflow.net/questions/415441/how-are-sheffer-polynomials-related-to-lie-theory)" on connections to the Sheffer polynomials, the core polynomials related to the umbral / finite operator calculus.
III) Particular series reps for $f(z)$ in $g(z) = 1/f'(z)$ of the refined Eulerian partition polynomials lead to the classic Lagrange inversion partition polynomials (LIPs) [A134685](https://oeis.org/A134685) for compositional inversion of functions and formal Taylor series / e.g.f.s with $f(z)= a\_1 z + a\_2 \frac{z^2}{2!}+ a\_3 \frac{z^3}{3!}+...$ ; the associahedra LIPs [A111785](https://oeis.org/A111785) (renormalized [A133437](https://oeis.org/A133437)), for formal power series / o.g.f.s with $f(z)= b\_1 z + b\_2 z^2+ b\_3 z^3+...$ ; the LIPs [A133932](https://oeis.org/A133932) with the log rep $f(z) = c\_1 z + c\_2 \frac{z^2}{2}+ c\_3 \frac{z^3}{3}+...$ ; and finally, but not least, the noncrossing partition LIPs / free cumulant partition polynomials [A134264](https://oeis.org/A134264), so prominent in free probability and related quantum theory and random matrix theory, with $f(z) = z/h(z) = z/( h\_1 z + h\_2 z^2+ h\_3 z^3+...)$. {For more info, see, e.g., my answer to the MO-Q "[Important formulas in combinatorics](https://mathoverflow.net/questions/214927/important-formulas-in-combinatorics/215203#215203)" and the recent posts "[Ruling the Inverse Universe, the inviscid Hopf-Burgers evolution equation ...](https://tcjpn.wordpress.com/2022/01/27/ruling-the-inverse-universe-the-inviscid-hopf-burgers-evolution-equation-compositional-inversion-associahedra-diff-ids-integrable-hierarchies-and-translation/)" and "[A Taste of Moonshine in Free Moments](https://tcjpn.wordpress.com/2022/01/29/a-taste-of-moonshine-in-free-moments/)".}
IV) There are multinomial-type expressions for each numerical coefficient of the normal-ordering expansion of the operator $(g(z)\partial\_z)^{n}$ (item 2), which are presented in the MO-Q "[Differential operator power coefficients](https://mathoverflow.net/questions/80828/differential-operator-power-coefficients/80873)". In addition, direct simple multinomial-type expressions exist for the coefficient of any given partition monomial for all the LIPs listed in III. However, as far as I know, no such multinomial-type formula currently exists for the refined Eulerian numbers even though each full partition polynomial can be calculated independently of the others using the corresponding partition polynomials for the other LIPs in III and the permutahedra partition polynomials of [A133314](https://oeis.org/A133314) (or the o.g.f. version [A263633](https://oeis.org/A263633)) or calculated via a matrix computation presented in the MO-Qs "[Сlosed formula for $(g\partial)^n$](https://mathoverflow.net/questions/337766/expansions-of-iterated-or-nested-derivatives-or-vectors-conjectured-matrix-c)" and "[Expansions of iterated, or nested, derivatives, or vectors--conjectured matrix computation](https://mathoverflow.net/questions/337766/expansions-of-iterated-or-nested-derivatives-or-vectors-conjectured-matrix-c)".
V) A noncommutative version of A139605 (link therein) was presented by Kentaro Ihara in "Derivations and automorphisms on non-commutative power series".
VI) Around 1853, when Lie was about ten, Charles Graves published the elegant generalized Taylor shift formula
$$ e^{t \; g(z)\partial\_z} H(z) = H[f^{(-1)}(f(z)+t)]$$
and, in the 1850s also, published the operator commutator
$$[L,R] =LR-RL= 1$$ (e.g., $L=\partial\_z$ and $R=x$) from which the Graves-Lie-Pincherle commutator $$[h(L),R] = \frac{dh(L)}{dL} = h'(L)$$
can be inferred and the dual
$$[L,h(R)] = \frac{dh(R)}{dR} = h'(R).$$
For Sheffer polynomial sequences, $p\_n(z)$, the lowering/annihilation/destruction op $L$ and the raising/creation op $R$ are defined by $L \; p\_n(z) = n \; p\_{n-1}(z)$ and $R \; p\_n(z) = p\_{n+1}(z)$.
Obviously, $R= g(z)\partial\_z$ is the raising op for the partition polynomials formed from $(g(z)\partial\_z)^n g(z)$, and, with $g(z)=1/f'(z)$,
$$[g(z)\partial\_z,f(z)] = 1,$$
implying, similar to the commutator identities above, that
$$[(g(z)\partial\_z)^n,f(z)] = n \; (g(z)\partial\_z)^{n-1}.$$
Then we have the functional identity
$$[(g(z)\partial\_z)^n,f(z)]g(z) =(g(z)\partial\_z)^n f(z)g(z) - f(z)(g(z)\partial\_z)^n g(z) = n \; (g(z)\partial\_z)^{n-1}g(z).$$
Recall the commutator acting on a function as in
$$[\partial\_z,H(z)]K(z) = \partial\_zH(z)K(z) - H(z)\partial\_zK(z)$$
is the Newton-Leibniz formula in disguise
$$\partial\_zH(z)K(z) = H(z)\partial\_zK(z) + [\partial\_z,H(z)]K(z) = H(z)\partial\_zK(z) + K(z)\partial\_zH(z).$$
| 8 | https://mathoverflow.net/users/12178 | 415681 | 169,440 |
https://mathoverflow.net/questions/415678 | 1 | $\newcommand{\Perf}{\operatorname{Perf}}$This is a toy example that I want to understand, I will be grateful for any help. Given a ring $R$ and $A=\Perf(R)$ the category of perfect complexes over $R$ . Suppose that $m\in\Perf(R)$ and $B$ the smallest thick subcategory generated by
$ m $. Let $C$ be the verdier quotient $ C=\Perf(R)/B$.
I want to understand $C$ concretely up to equivalence of triangulated categories.
Is it correct that $C$ Is equivalent to a triangulated subcategory $ D$ of $A$ where $d\in D$ iff the graded abelian groups $$ \operatorname{Hom}\_A^n(m,d)=0$$ for any integer $n.$
| https://mathoverflow.net/users/141114 | Verdier localisation | No, not in general. This is true if you pass to "big" categories, so $\mathrm{Mod}(R)$ instead of Perf, and you take the smallest localizing subcategory containing $m$, but in general wrong at the level of small categories.
So what you can do is look at the full subcategory of $\mathrm{Mod}(R)$ of these objects, and take the compact objects therein (which need not be compact in $\mathrm{Mod}(R)$ !).
There are many counterexamples, such as $R= \mathbb Z$ and $m = \mathbb Z/p$. In that case the quotient is equivalent to $\mathrm{Perf}(\mathbb Z[1/p])$ (and the localization functor is base-change), and you can clearly see that no object of $\mathrm{Perf}(\mathbb Z)$ has $\mathbb Z[1/p]$ as its (graded) endomorphism ring.
| 3 | https://mathoverflow.net/users/102343 | 415695 | 169,446 |
https://mathoverflow.net/questions/415680 | 2 | My setting is the following : let $G$ be a topological group and $X$ be a topological space. I have the head filled with two possible definitions for a continuous action of $G$ on $X$.
The first could be called the separately continuous action and means that every element $g\in G$ acts via a continuous map of $X$ into itself, and that the action $G \rightarrow X, \,\, g\mapsto g \cdot x$ on every element $x\in X$ is continuous. Equivalently, this means that we have a continuous morphism of group from $G$ into $\mathcal{C}^0(X,X)$ where the last group structure is for the composition, which is continuous for the simple convergence topology on $\mathcal{C}^0(X,X)$ (which is not a toplogical group structure in general, but this is not useful).
The second one could be called the joint continuous action. It means a group action of $G$ on $X$ such that the deduced map $G\times X \rightarrow X$ is continuous.
I have seen some conditions for the two definitions to be equivalent (for instance for $X$ with extrastructure in *Locally analytic vectors in representations of locally $p$-adic analytic groups* by M. Emerton). However, seing the lectures of P. Scholze and D. Clausen and especially the definition of the site $G$-$\mathrm{pfsets}$, I came to have the question in the following specific setting : $X$ (resp. $G$) is a profinite set (resp. group). I have also manage to prove that in the case where both limits defining the profinite things can be taken countable, the two conditions are equivalent. I have tried some generalization (but the lack of Mittag-Leffler systems makes it difficult) and counterexamples without success. So my question is the following :
1. Are both definitions equivalent for this profinite setting ?
2. Are they more general conditions on the category indexing the limits so that the two conditions are equivalent ?
| https://mathoverflow.net/users/242651 | Difference between definitions of continuous action, profinite case | Both definitions are equivalent in much greater generality. I believe the profinite case has as more direct proof, perhaps it can be found in the book of Ribes and Zalesski.
There is a very general result by Robert Ellis (Ellis, Robert
Locally compact transformation groups.
Duke Math. J. 24 (1957), 119–125) that covers a much more general situation than yours. If $G$ is a group with a topology for which multiplication is separately continuous in each variable and if the action map $\pi\colon G\times X\to X$ is separately continuous in each variable, and $X$ is locally compact, then $\pi$ is continuous. His paper is written in a slightly hard to digest language and is nontrivial. He views $G$ as a subgroup of the homeomorphism group of $X$ (but the action doesn't really need to be faithful) and expresses the other continuity condition as saying that the topology on $G$ contains the topology of pointwise convergence.
| 2 | https://mathoverflow.net/users/15934 | 415698 | 169,448 |
https://mathoverflow.net/questions/415693 | 0 | Let $a(n)$ be the sequence of composite numbers (starting from $4$). Let
$$b(n)=a(n-1)a(n-2) \operatorname{mod} a(n)$$
Obviously, $b(1)=b(2)=0$.
I conjecture that with the only exception for the $b(3)=0$ all terms are belong to $\left\lbrace2,3,6,8\right\rbrace$.
My second conjecture is that the sequence $b(n)$ can be partitioned into blocks $\left\lbrace8\right\rbrace$ and $\left\lbrace3\underbrace{2\cdots2}\_{2k-1}6\right\rbrace$.
My third conjecture is that for $b(n)=6$ the remainders of the division $a(n) \operatorname{mod} b(n)$ are belong to $\left\lbrace0,2\right\rbrace$.
My fourth conjecture is that we take the sequence $a\_1(n)$ of composite numbers without Sarrus numbers instead (simply as complement of primes and Sarrus numbers), then for $b\_1(n)=6$ we have violation of the rule above somewhere, exactly $a\_1(n) \operatorname{mod} b\_1(n) = 4$. If so, then $b\_1(n)-1$ is the Sarrus numbers (since it is divisible by $3$). The sequence of such Sarrus numbers begins
```
561, 1905, 8481, 18705, 23001, 87249, 154101, 206601, 215265, 289941, 427233, 526593
```
Here similarly
$$b\_1(n)=a\_1(n-1)a\_1(n-2) \operatorname{mod} a\_1(n)$$
and Sarrus numbers is a composite odd numbers $n$ such that $n$ divides $2^n - 2$.
Is there a way to prove all or part of those conjectures?
| https://mathoverflow.net/users/231922 | Property of composite numbers | For any $m$ we have that either $m+1$ or $m+2$ is composite. Therefore a tuple of three consecutive composites must be of one of the following forms:
$$(k-4,k-2,k),(k-3,k-2,k),(k-3,k-1,k),(k-2,k-1,k).$$
Since we have $(k-a)(k-b)\equiv ab\pmod k$, we get the four cases you ask for with $k=a(n)$. The second conjecture follows easily as well - your blocks correspond to blocks of composites between consecutive primes.
For the third conjecture, note that if $b(n)=6$, then the pattern above must be $(k-3,k-2,k)$, meaning $k-1$ is a prime, meaning $k-1\equiv 1$ or $5\pmod 6$, so $k\equiv 2$ or $0\pmod 6$.
For the fourth conjecture, I'm assuming you meant $a\_1(n)-1$ is a Sarrus number. This is true - by a similar analysis as above, we get that if $b\_1(n)=6$, then $a\_1(n)-1$ must be either a prime or a Sarrus number. But $a\_1(n)-1\equiv 3\pmod 6$ under the assumption on $a\_1\pmod b\_1$, so it can't be prime.
| 3 | https://mathoverflow.net/users/30186 | 415699 | 169,449 |
https://mathoverflow.net/questions/415703 | 28 | I am thinking about the Axiom of Choice and I am trying to understand the Axiom with some but a little progress. Many questions are arising in my head. So, I know that there exists a model of ZF set theory in which the set of real numbers, which is provably uncountable, is a countable union of countable sets.
**Question:** does there exist a model of ZF set theory for which there exists a collection $A\_n$, $n\in\mathbb{N}$, of pairwise disjoint **two-element** sets such that their union is not countable?
Some thoughts. Let $A\_n$, $n\in\mathbb{N}$, be a collection of pairwise disjoint two-element sets.
Then for every $n\in\mathbb{N}$ there exists a bijection $f:\{1,2\}\to A\_n$.
But when we want to prove that $\bigcup\_{n\in\mathbb{N}}A\_n$ is countable, we have to choose a countable number of bijections $f\_n:\{1,2\}\to A\_n$, $n\in\mathbb{N}$, at once (simultaneously).
After this we plainly define the bijection $f:\mathbb{N}\to\bigcup\_{n\in\mathbb{N}}A\_n$
by $f(1):=f\_1(1)$, $f(2):=f\_1(2)$, $f(3):=f\_2(1)$, $f(4):= f\_2(2)$, and so on.
Rigorously, we write $f(k)=f\_l(1)$ if $k=2l-1$ and $f(k)=f\_l(2)$ if $k=2l$.
Clearly, $f$ is a bijection and we are done.
But without the Axiom of Countable Choice we can not choose $f\_n$, $n\in\mathbb{N}$, simultaneously and the argument does not work.
It is worth mentioning that if $A\_n$ are subsets of $\mathbb{R}$, then we **can** choose $f\_n$, $n\in\mathbb{N}$, simultaneously.
Indeed, we can define $f\_n(1):=\min A\_n$ and $f\_n(2):=\max A\_n$, $n\in\mathbb{N}$,
and the natural proof given above works.
So if a counterexample exists, the sets $A\_n$, $n\in\mathbb{N}$, have to be "abstract", say pairs of socks.
| https://mathoverflow.net/users/48157 | Can a countable union of two-element sets be uncountable? | Yes, it is possible. This phenomenon is sometimes called Russell's socks, named after an analogy due to Russell about how one can pick out a shoe from an infinite set of pairs of shoes, but not for socks since socks in a pair are indistinguishable.
[Horst Herrlich, Eleftherios Tachtsis, On the number of Russell’s socks or 2 + 2 + 2 + . . . = ?](https://www.emis.de/journals/CMUC/pdf/cmuc0604/herrtach.pdf) is a nice overview which proves some basic properties, including consistency of existence of Russell's socks.
| 39 | https://mathoverflow.net/users/30186 | 415707 | 169,454 |
https://mathoverflow.net/questions/415611 | 4 | Let $d\ge 3$ be a constant. Is there an explicit construction of an infinite family of $d$-regular graphs such that for $G$ in this family with $n$ vertices, every subgraph $H$ of on at most $\alpha n$ vertices where $\alpha$ is a constant depending on $d$ has average degree bounded by, say, $2.1$?
Note that a random $d$-regular graph satisfies this property (see, for example, Section 4.6 of [https://www.cs.huji.ac.il/~nati/PAPERS/expander\_survey.pdf](https://www.cs.huji.ac.il/%7Enati/PAPERS/expander_survey.pdf) for a proof).
Let's also say we aren't picky about achieving this for literally every choice of $d$, and would be happy with obtaining a construction for arbitrarily large $d$. The notion of explicit here is that the graph is constructible in polynomial time in the number of vertices. Let's also say we are happy if the average degree bound of $2.1$ is replaced by any constant $C$ as long as it does not depend on $d$.
Do any of the explicit constructions of Ramanujan graphs, such as the Lubotzky-Phillips-Sarnak/Margulis/Morgenstern constructions, achieve this property? The best bound we know on the average degree is about $\sqrt{d-1}+1$, which only uses the fact that the second largest magnitude eigenvalue is at most $2\sqrt{d-1}$. Is this bound improvable?
| https://mathoverflow.net/users/91432 | Explicit constructions of regular graphs with very sparse induced subgraphs | I think that for Ramanujan graphs the situation is similar to vertex expansion, which got more publicity.
Namely, a variation of the combinatorial construction of Kahale about vertex expansion (<https://www.researchgate.net/publication/2782658_Eigenvalues_and_Expansion_of_Regular_Graphs>, Theorem 7, the same paper which proves the edge expansion result you states) should show that you can't expect to get any better using only the second-largest eigenvalue.
Furthermore, in my recent paper with Tali Kaufman (<https://arxiv.org/abs/2103.04311>, Theorem 1.8 in the current version) we show that you can construct "number theoretic" Ramanujan graphs with plenty of nice properties (high girth, large automorhism group) with a small set (of size square-root of the graph) having exactly the $\sqrt{d-1}+1$ average degree.
On the other hand, as with vertex expansion, the general belief/ hope is that LPS graphs have the property you mentioned, but there is no proof of this in sight.
Two more optimistic comments:
1. In the case when $\sqrt{d-1}$ is not an integer the result can be very slightly improved, essentially because not every vertex can have the average degree.
2. Perhaps you can hope for a construction as in [http://www.cs.tau.ac.il/~nogaa/PDFS/acaproc.pdf](http://www.cs.tau.ac.il/%7Enogaa/PDFS/acaproc.pdf) for this specific question. Namely, the graphs will not have some local structure which can help you. More realistically, maybe this can lead to a construction such that every set of size bounded by $\alpha n$ have a vertex of constant induced degree. I don't even know if this far weaker problem was solved.
| 1 | https://mathoverflow.net/users/450073 | 415712 | 169,456 |
https://mathoverflow.net/questions/414250 | 2 | I know that the weak lower semi-continuity of the KL divergence was proved in [1]. If I remember well, the same property is true for any $f$ divergence (with suitable assumptions on the probability space). I am looking for some reference about it.
[1] Posner, Random Coding Strategies for Minimum Entropy, 1975.
**Edit.**
Here is what I believe a standard definition of $f$ divergences, which includes the case of measures not absolutely continuous to each other. This definition is taken from <http://people.lids.mit.edu/yp/homepage/data/LN_fdiv.pdf>
>
> **Definition 7.1.** Let $f:(0,\infty)\to\mathbb R$ be a convex function with $f(1)=0$. Let $P$ and $Q$ be two probability distributions on a measurable space $(\mathcal X, \mathcal F)$. If $P\ll Q$ then the $f$-divergence is defined as
> $$D\_f(P\|Q)=\mathbb E\_Q[f(dP/dQ)]$$
> where $dP/dQ$ is the Radon-Nikodym derivative and $f(0)=f(0+)$. More generally, let $f'(\infty)=\lim\_{x\to 0}xf(1/x)$. Let $R$ be such that $Q\ll R$ and $P\ll R$ (such an $R$ always exists, for instance take $R=\frac{1}{2}(P+Q)$. Then we have
> $$D\_f(P\|Q) = f'(\infty)P(dQ/dR=0)+\int\_{dQ/dR>0}\frac{dQ}{dR}f\left(\frac{dP/dR}{dQ/dR}\right)dR\,,$$
> with the agreement that if $P(dQ/dR=0)=0$ the last term is taken to be zero regardless of the value of $f'(\infty)$ (which could be infinite).
>
>
>
| https://mathoverflow.net/users/475450 | Looking for a reference: $f$-divergences are lower semicontinuous | I think the most general reference is:
M. Liero, A. Mielke, and G. Savaré. [Optimal entropy-transport problems and a newhellinger–kantorovich distance between positive measures](https://link.springer.com/article/10.1007/s00222-017-0759-8). *Inventiones mathematicae*, **211**, 2018.
[arxiv:1508.07941](https://arxiv.org/abs/1508.07941)
Corollary 2.9 proves the lower semicontinuity wrt to the narrow topology under very general assumptions, and Remark 2.1 shows that the weak topology and the narrow topology coincide if $\mathcal X$ is Polish.
| 0 | https://mathoverflow.net/users/475450 | 415713 | 169,457 |
https://mathoverflow.net/questions/415618 | 3 | I'll first describe my problem in layman's terms. I have a map with $m$ countries and I want to color each country with a different color (this has nothing to do with the 4-color theorem). How do I choose the colors so as to maximize contrasts?
Each color is a 3D vector in the unit cube. I want the minimum distance between any two of these vectors, to be maximum. What is the 3D lattice that accomplishes this goal, that is, maximizing the minimum distances between vertices?
This is related to crystal structures. If the minimum distance between atoms/molecules/ions in a crystal, is maximum, maybe the crystal is more resistant against compression. Maybe diamond crystals achieve this, I don't know. Or you can view it as a graph theory problem: finding a graph where the minimum distances between nodes, is maximum. (as a side question, is this an NP-hard problem?)
| https://mathoverflow.net/users/140356 | Lattice-like structure with maximum spacing between vertices | No proof of optimality, but here are conjectured values for $m \le 10$ based on numerical optimization:
\begin{matrix}
m & \text{maximin?} \\
\hline
2 & \sqrt{3} \\
3 & \sqrt{2} \\
4 & \sqrt{2} \\
5 & \sqrt{5}/2 \\
6 & 3\sqrt{2}/4 \\
7 & 1.0010898245 \\
8 & 1 \\
9 & \sqrt{3}/2 \\
10 & 3/4 \\
\end{matrix}
```
m = 2:
1 1 1
0 0 0
m = 3:
1 0 1
0 1 1
0 0 0
m = 4:
0 1 1
1 1 0
1 0 1
0 0 0
m = 5:
1.0 0.0 0.0
0.0 0.0 0.5
0.5 1.0 0.0
0.0 1.0 1.0
1.0 0.5 1.0
m = 6:
0.25 0.00 0.00
0.00 1.00 0.25
0.75 1.00 1.00
1.00 0.75 0.00
0.00 0.25 1.00
1.00 0.00 0.75
m = 7:
0.292123 1.000000 1.000000
0.033021 1.000000 0.033021
0.000000 0.000000 0.000000
0.033021 0.033021 1.000000
1.000000 0.033021 0.033021
1.000000 0.292123 1.000000
1.000000 1.000000 0.292123
m = 8:
0 0 0
0 1 0
0 1 1
1 0 1
0 0 1
1 1 0
1 1 1
1 0 0
m = 9:
1.0 1.0 1.0
1.0 0.0 1.0
0.0 0.0 0.0
0.0 0.0 1.0
1.0 0.0 0.0
0.0 1.0 0.0
1.0 1.0 0.0
0.0 1.0 1.0
0.5 0.5 0.5
m = 10:
1.00 0.25 1.00
0.50 0.75 0.50
0.00 1.00 0.00
1.00 1.00 0.00
0.00 0.25 1.00
0.00 0.25 0.00
0.50 0.00 0.50
0.00 1.00 1.00
1.00 1.00 1.00
1.00 0.25 0.00
```
| 3 | https://mathoverflow.net/users/141766 | 415716 | 169,458 |
https://mathoverflow.net/questions/415719 | 3 | I'm reading some notes(\*) about arithmetic lattices in $\operatorname{SU}(n,1)$. I'm trying to understand the data that classifies (up to commensurability) the arithmetic lattices of the "first type" in this group. Fix a totally real number field $F$ with a totally imaginary quadratic extension $E$. Then these lattices are defined as the integer points of a unitary group preserving a Hermitian form $H$ of signature $(n,1)$ defined over $E$, i.e. $\operatorname{SU}(H;\mathcal O\_E)$, which is positive- or negative-definite at all other infinite places.
One invariant you might try to pick up from such a construction is the determinant of $H$ (thought of as an $(n+1) \times (n+1)$ Hermitian matrix), which lies in $F^\times$. But this determinant is not actually invariant under a change of basis. It turns out that the relevant invariant lies in $F^\times/N\_{E/F}(E^\times)$. Now here's my confusion: in section 6.6 ("Parity and describing commensurability classes") of the cited notes, the author seems to be claiming that the group $F^\times/N\_{E/F}(E^\times)$ is a group with only two elements. He refers to the "trivial and nontrivial class[es]". I can see that all elements of this group have order 2, but I don't see why the group itself has order 2.
Is it true that $F^\times/N\_{E/F}(E^\times)$ is a group of order 2?
(\*) D.B. McReynolds. [*Arithmetic lattices in $\operatorname{SU}(n,1)$*](https://citeseerx.ist.psu.edu/viewdoc/summary?doi=10.1.1.573.7677).
(I should point out that these notes are clearly unfinished and may not have been intended for publishing. I don't want to call to question the author's work.)
| https://mathoverflow.net/users/151664 | Multiplicative group of number field mod field norms of quadratic extension | The index is definitely not finite. If $F$ is an arbitrary number field and $E$ is a finite extension with $[E:F] > 1$, the norm subgroup ${\rm N}\_{E/F}(E^\times)$ has infinite index in $F^\times$. A proof in general is in a recent stackexchange post [here](https://math.stackexchange.com/questions/4368481/is-f-times-n-l-fl-times-non-trivial-or-infinite-when-l-f-is-an-extensi).
Let's consider the simplest case relevant to what you are reading: $F = \mathbf Q$ and $E = \mathbf Q(i)$. Then $F^\times/{\rm N}\_{E/F}(E^\times) = \mathbf Q^\times/{\rm N}\_{\mathbf Q(i)/\mathbf Q}(\mathbf Q(i)^\times)$, and for primes $p$ and $q$ such that $p, q \equiv 3 \bmod 4$, the ratio $p/q$ is not of the form $a^2 + b^2$ for $a, b \in \mathbf Q$. Thus the infinitely many primes $p$ such that $p \equiv 3 \bmod 4$ are all inequivalent in $\mathbf Q^\times/{\rm N}\_{\mathbf Q(i)/\mathbf Q}(\mathbf Q(i)^\times)$, so they show that quotient group not only size has bigger than $2$, but is in fact infinite.
The standard setting where a quotient group $F^\times/{\rm N}\_{E/F}(E^\times)$ has order $2$ when $[E:F] = 2$ is for a quadratic extension of *local fields*. (In case of local fields of characteristic $2$, I should say quadratic *Galois* extension of local fields, which would be redundant outside of characteristic $2$.) In that case it is true that ${\rm N}\_{E/F}(E^\times)$ has index $2$ in $F^\times$. This is a special case of local class field theory: if $E/F$ is an abelian extension of local fields then ${\rm Gal}(E/F) \cong F^\times/{\rm N}\_{E/F}(E^\times)$. Here "local field" includes archimedean completions, e.g., ${\mathbf R^\times}/{\rm N}\_{\mathbf C/\mathbf R}(\mathbf C^\times) = {\mathbf R}^\times/{\mathbf R\_{>0}} = \{\pm 1\}$ has order $2$. Maybe there is an implicit use of archimedean completions when you refer to quotient groups with norms, since compatible archimedean completions of a totally real number field and a totally imaginary quadratic extension will look like the extension $\mathbf C/\mathbf R$.
| 5 | https://mathoverflow.net/users/3272 | 415721 | 169,459 |
https://mathoverflow.net/questions/346487 | 11 | Gelfand duality states that the functor of continuous functions $C(-)$ from compact Hausdorff topological to commutative $C^\*$-algebras is an equivalence of categories. In other words, all topological properties of such a topological space $X$ are encoded in the algebraic properties (and of course the norm) of $C(X)$.
I have the following related questions:
1. How does one recognize that $X$ is simply connected from the algebraic properties of $C(X)$?
2. How can the fundamental group $\pi\_1(X)$ be defined in terms of the algebraic properties of $C(X)$?
3. If $\tilde{X}$ is a universal cover of $X$, what is the algebraic relation between $C(\tilde{X})$ and $C(X)$?
| https://mathoverflow.net/users/2622 | Fundamental group under Gelfand duality | It seems that this question has been investigated in the literature, where the fundamental group is approached through the notion of a regular covering. For instance, the following reference establishes an equivalence between the categories of regular coverings $\tilde{X} \to X$ and Galois extensions of $C(X)$, meaning that the algebraic analog of a universal cover is a universal Galois extension and the fundamental group is the group associated to that extension:
>
> *Høst-Madsen, Anders*, [**Separable algebras and covering spaces**](http://dx.doi.org/10.7146/math.scand.a-14306), Math. Scand. 87, No. 2, 211-239 (2000). [ZBL1002.54010](https://zbmath.org/?q=an:1002.54010). [MR1795745](https://mathscinet.ams.org/mathscinet-getitem?mr=1795745).
>
>
>
More literature on this topic is cited therein.
| 2 | https://mathoverflow.net/users/2622 | 415729 | 169,461 |
https://mathoverflow.net/questions/415728 | 8 | Given an random variable $Y:\Omega \to \mathbb{R}$ with finite mean $\mu$ and finite, positive variance $\sigma^2$, let $X = \frac{Y-\mu}{\sigma}$ be the renormalization with mean $0$ and variance $1$. what are some general techniques for showing that $Y$ has a normal distribution? That is,
$$P(X\leqslant a) = \frac{1}{\sqrt{2\pi}}\int\_{-\infty}^a e^{-t^2/2}\,dt.$$
The standard technique I know is to compute the moments or [cumulants](https://en.wikipedia.org/wiki/Cumulant) and then use the fact that the normal distribution is characterized by its moments/cumulants. Are there any other general techniques, and what are their advantages and disadvantages?
---
The motivation for this question is number theoretic (as with [this related question](https://mathoverflow.net/questions/102964/convergence-of-moments-implies-convergence-to-normal-distribution)), hence the number theory tags. Specifically, the motivating theorem is Selberg's central limit theorem (first published in [Tsang's thesis](https://mathscinet.ams.org/mathscinet-getitem?mr=2633927), see also [this article](https://mathscinet.ams.org/mathscinet-getitem?mr=3832861) of Radziwill-Soundararajan) which states that for large $T$, the real valued random variable on $[T,2T]$ given by $t \mapsto \log|\zeta(\tfrac12 + it)|$ is approximately normally distributed with mean $0$ and variance $\frac{1}{2}{\log\log T}$.
Both the proofs I know of (Selberg's original proof, and that of Radziwill-Soundararajan) use the method of moments. Morally speaking, the analytic-number-theoretic input goes into showing that the contributions from the zeros of zeta can be controlled, and hence at least for the distributional question with $t \in [T,2T]$,
$$\log|\zeta(\tfrac12 + it)| \simeq \Re\sum\_{p\leqslant T^{o(1)}} \frac{1}{p^{1/2 + it}}.$$
One can then compute the moments of the right hand side and show that as $T \to \infty$, the moments appropriately normalized converge to the moments of a standard Gaussian.
The hope is to see if there's a way to prove Selberg's CLT in a situation where the moments are harder to compute, and so the method of moments may not be tractable.
| https://mathoverflow.net/users/37327 | Ways of proving normal distribution (with a view towards Selberg's central limit theorem) | There are multiple books about ways to characterize the normal distribution. For instance, Bryc’s [book](https://homepages.uc.edu/%7Ebrycwz/probab/charakt/charakt.pdf) starts with Herschel-Maxwell’s theorem:
>
> If $X$ and $Y$ are independent variables whose joint distribution is rotationally invariant, then $X$ and $Y$ are both normal.
>
>
>
He immediately notes that one can strengthen this to Polya’s theorem:
>
> If $X$ and $Y$ are independent variables, and rotations of $\pi/4$ and $\pi/2$ leave the distribution of $X$ invariant, then $X$
> and $Y$ are both normal.
>
>
>
Perhaps somewhere in such books you’ll find a characterization that avoids moments but is number-theoretically tractable.
| 5 | https://mathoverflow.net/users/nan | 415741 | 169,467 |
https://mathoverflow.net/questions/415687 | 6 | Maybe the answer to the following question is known but I am unable to find it in the literature. Anyway, let me begin my question by fixing some notations and terms.
Let $G = (A, \Delta)$ be a compact quantum group in the sense of Woronowicz. For a finite dimensional unitary representation $U \in B(H) \otimes A$ of $G$, the contragredient $U^c$ of $U$ is the representation $(j \otimes \operatorname{Id})(U^\*) \in B\bigl(\overline{H}\bigr) \otimes A$ of $G$, where the carrier space $\overline{H}$ is the conjugate Hilbert space of the finite dimensional Hilbert space $H$, and $j : B(H) \to B\left(\overline{H}\right)$ the $\ast$-anti-isomorphism sending $T$ to $\overline{T^\*}$, the latter denotes the operator $\overline{\xi} \mapsto \overline{T^\* \xi}$ on $\overline{H}$. Define $c(G)$ to be the supremum of all $\| U^c \|$, where $U$ runs through all finite dimensional unitary representations (of course, irreducible ones suffice) of $G$.
It is clear that if $G$ is of Kac-type, then $c(G) = 1$, as $U^c$ in the above is always unitary. With some effort, I can prove that $c(G) = +\infty$ for $G = SU\_q(2)$ with $-1 < q \ne 0 < 1$, but whether $c(G) = +\infty$ for general non-Kac type $G$ seems more delicate, which prompts me into asking the following
**Question.** Does $c(G) < +\infty$ imply $G$ being of Kac type?
As there are already many results concerning characterization of Kac type compact quantum groups, it may well be possible that this question is already settled, in which case, I appreciate a reference to the literature.
| https://mathoverflow.net/users/40789 | Norm of contragredient of unitary representations of compact quantum groups | Yes, $c(G) < +\infty$ implies that $G$ is of Kac type. As far as I know, this result is not in the literature, but can be proven as follows.
Let $h$ be the Haar state. For every irreducible unitary representation $u \in M\_n(\mathbb{C}) \otimes A$, there is a unique positive invertible $Q\_u \in M\_n(\mathbb{C})$ with $\operatorname{Tr}(Q\_u) = \operatorname{Tr}(Q\_u^{-1})$ and giving the orthogonality relations
$$h(u\_{ij} u\_{kl}^\*) = \delta\_{i,k} \, (Q\_u)\_{jl} \, \operatorname{Tr}(Q\_u)^{-1} \; .$$
Note that $G$ is of Kac type if and only if $Q\_u = 1$ for all irreducible unitary representations $u$. This positive $Q\_u$ also unitarizes $u^c$. First defining the matrix $u^c \in M\_n(\mathbb{C}) \otimes A$ by $(u^c)\_{ij} = (u\_{ij})^\*$, we get that $(Q\_u^{1/2} \otimes 1) u^c (Q\_u^{-1/2} \otimes 1)$ is a unitary representation that we denote as $\overline{u}$. This is the unitary contragredient of $u$.
**Proposition.** The following statements are equivalent.
1. $G$ is of Kac type.
2. There exists a $C > 0$ such that $\|Q\_u\| \leq C$ for every irreducible unitary representation $u$.
3. There exists a $C > 0$ such that $\|u^c\| \leq C$ for every irreducible unitary representation $u$.
**Proof.** When $u$ is an arbitrary finite dimensional unitary representation, we can still canonically define $Q\_u$ by decomposing $u$ as a sum of irreducibles. We have $Q\_{u \otimes v} = Q\_u \otimes Q\_v$. Also, $Q\_{\overline{u}} = Q\_u^{-1}$. For every unitary representation $u \in M\_n(\mathbb{C}) \otimes A$, we denote $\dim u = n$ and $\dim\_q u = \operatorname{Tr}(Q\_u)$. The orthogonality relations then say that
$$(\mathord{\text{id}} \otimes h)((u^c)^\* u^c) = \frac{\dim u}{\dim\_q u} \, Q\_u \; .$$
1 $\Rightarrow$ 2 is trivial.
2 $\Rightarrow$ 3. Since $Q\_{\overline{u}} = Q\_u^{-1}$, also $\|Q\_u^{-1}\| \leq C$. By definition, $\|u^c\| \leq C$ for every irreducible unitary representation $u$.
3 $\Rightarrow$ 1. Assume that $G$ is not of Kac type. Fix $C > 0$. We construct an irreducible unitary representation $u$ with $\|u^c\| > C$.
Since $G$ is not of Kac type, we can choose an $n$-dimensional irreducible unitary representation $v$ such that $Q\_v \neq 1$. Define $q\_{\max}$ as the largest eigenvalue of $Q\_v$. Since $Q\_v \neq 1$, we get that $\operatorname{Tr}(Q\_v) < q\_{\max} \, n$. Take an integer $k \geq 1$ such that $(\operatorname{Tr}(Q\_v)^{-1} \, q\_{\max} \, n)^k > C^2$. Define $w$ as the $k$-fold tensor power of $v$. Then,
$$(\mathord{\text{id}} \otimes h)((w^c)^\* w^c) = \frac{\dim w}{\dim\_q w} \, Q\_w = \Bigl(\frac{\dim v}{\dim\_q v}\Bigr)^k \, Q\_v^{\otimes k} \; .$$
By construction, the operator norm of the right hand side is strictly larger than $C^2$. The operator norm of the left hand side is the maximum of $\| (\mathord{\text{id}} \otimes h) ((u^c)^\* u^c)\|$, where $u$ runs through the irreducible subrepresentations of $w$. So, there exists an irreducible $u$ such that $\| (\mathord{\text{id}} \otimes h) ((u^c)^\* u^c)\| > C^2$. It follows that $\|u^c\| > C$.
| 6 | https://mathoverflow.net/users/159170 | 415746 | 169,468 |
https://mathoverflow.net/questions/415688 | 11 | What is a reference for ["the equivalence between geometric theories and frames internal to the free topos"](https://indico.math.cnrs.fr/event/747/contributions/2523/attachments/1818/1962/introductionJoyal.pdf)? [1] This seems to be an extremely interesting theorem.
[1] André Joyal, “A crash course in topos theory: the big picture”, lectures at Topos à l'IHÉS, 2015.
| https://mathoverflow.net/users/476516 | Equivalence between geometric theories and frames internal to the free topos | Simon essentially answered the question already, but I will expand some of the parts that may not be clear to the experts. Sketches of an elephant is a good reference for everything I am going to say.
I will take a bit of a different narrative to Simon's, implicitly assuming a good understanding of classifying topoi, blurring the distinction between a topos and the geometric theory it classifies, $$\mathcal{E} \simeq \text{Set}[\mathbb{T}] .$$
**Def (Prebounds).** Let $\mathcal{E}$ be a topos. A *prebound* $e \in \mathcal{E}$ is an object such that the subobjects of its finite powers $m: a \to e^n$ are a generator for the topos. Such an object always exist and can be obtained by manipulating a generator (or a site).
**Construction (From prebounds to localic geometric morphisms).** Given a couple $(\mathcal{E},e)$ where $\mathcal{E}$ is a topos and $e$ is a prebound, we can construct a localic geometric morphism $$f\_e: \mathcal{E} \to \text{Set}[\mathbb{O}]. $$
Of course, this is the same of a cocontinuous left exact functor $f\_e^\*: \text{Set}[\mathbb{O}] \to \mathcal{E}$, which is the same of a lex functor $\text{Fin}^\circ \to \mathcal{E}$,
$$\text{Topoi}(\mathcal{E}, \text{Set}[\mathbb{O}]) \simeq \text{Cocontlex}( \text{Set}[\mathbb{O}], \mathcal{E}) \simeq \text{Cocontlex}( \text{Set}^{\text{Fin}}, \mathcal{E})\simeq \text{Lex}(\text{Fin}^\circ, \mathcal{E}). $$
The latter, is given by sending $n \mapsto e^n$. The geometric morphism obtained in this way is localic by definition of prebound. This construction appeared for the first time in **Freyd**'s *All topoi are localic*.
*Remark.* If you think about it, I am just spelling out in categorical terms what Simon suggested in somewhat mystical language.
*Remark (Morita-like phenomena)*. Notice that *each prebound* (and we can construct a prebound from any site) gives a different localic morphism, thus we have many localic representation for *the same* topos!
*Remark (Topoi are geometric theories, generators are their presentation)*. Following Thm 2.1.1 in **Caramello**'s *Theories, Sites, Toposes*, we see that a generator, or a site, is essentially the same of a linguistic presentation of the geometric theory classified by the topos.
**Theorem (Internal locales are localic geometric morphisms)**. There is a biequivalence of categories between the $2$-category of internal locales in $\text{Set}[\mathbb{O}]$ and the $2$-category of localic geometric morphisms over $\text{Set}[\mathbb{O}]$, $$\text{Loc}(\text{Set}[\mathbb{O}]) \leftrightarrows \text{Topoi}\_{\text{loc} / \text{Set}[\mathbb{O}]}. $$
*Proof.* Lemma 1.2 in **Johnstone**, *Factorization theorems for geometric morphisms*. Cahiers, 22, no1 (1981)
*Remark (On the emergence of Lawvere-like doctrines).* When one spells out what a locale internal to $ \text{Set}[\mathbb{O}]$ is, one discovers that it is nothing but a functor $$\mathbb{P}: \text{Fin} \to \text{Frames}$$ verifying the Beck-Chevalley condition and Frobenius reciprocity (see Lemma C.1.6.9 and Cor. C.1.6.10 in Sketches of an Elephant). Suddenly we see how doctrine-like objects emerge in the representation of theories! That's beautiful in my opinion. $\text{Fin}$ acts as a fact as a set of variables, while $\mathbb{P}(n)$ gives us the poset (a frame in fact) of formulas on those $n$-variables.
**Def (Well presented topoi).** The $2$-category WTopoi of *well presented topoi* has objects $(\mathcal{E},e)$ where $\mathcal{E}$ is a topos and $e$ is a prebound and morphism geometric morphisms whose left adjoint preserve the prebuound.
*Remark.* This notion does not appear in the literature (to my knowledge), I just need it as an intermediate notion. A good intuition for it is that the topos is specified together with a precise language generator of the geometric theory it classifies. WTopoi is really much more a $2$-category *of sites*, together with a *relational* notion of morphism of sites, rather than a $2$-category *of topoi*.
*Remark (Every topos can be well presented).* Of course, the WTopoi is not the same of Topoi but the forgetful functor $$\mathsf{U}: \text{WTopoi} \to \text{Topoi} $$ is essentially surjective on objects, and on morphisms (!).
**Theorem (Internal locales are well presented topoi and vice versa)**. There is a biequivalence of categories $$\text{Loc}(\text{Set}[\mathbb{O}]) \leftrightarrows \text{Topoi}\_{\text{loc} / \text{Set}[\mathbb{O}]} \leftrightarrows \text{WTopoi}.$$
| 10 | https://mathoverflow.net/users/104432 | 415753 | 169,470 |
https://mathoverflow.net/questions/415584 | 1 | I'm looking for a solution to $$\int x^{-a} \Gamma\left( b, c x^{-d} \right) dx.$$
Mathematica gives me the following solution, but I'd like to know/understand the steps involved in finding it.
$$\int x^{-a} \Gamma\left( b, c x^{-d} \right) dx = \frac{x^{1-a} \left(\left(c x^{-d}\right)^{\frac{1-a}{d}} \Gamma
\left(\frac{a+b d-1}{d},c x^{-d}\right)-\Gamma \left(b,c
x^{-d}\right)\right)}{a-1}.$$
The following solution, to a similar integral, seems to be a starting point. However, I have no idea how to proceed from that.
<https://functions.wolfram.com/GammaBetaErf/Gamma2/21/01/02/01/>
Does someone know how to find this solution?
| https://mathoverflow.net/users/103291 | Solution to $\int x^{-a} \Gamma\left( b, c x^{-d} \right) dx$ | To remove this question from the "unanswered" list; given the indefinite integral
$$\int z^{\alpha-1}\Gamma(b,z)\,\mathrm{d}z=\alpha^{-1}[ z^\alpha \Gamma(b, z)-\Gamma(b + \alpha, z)]$$
change variables to $z=cx^{-d}$, $\mathrm{d}z=-dcx^{-d-1}\mathrm{d}x$, hence
$$-\int dc^\alpha x^{-\alpha d-1}\Gamma(b,cx^{-d})\,\mathrm{d}x=\alpha^{-1}[ c^\alpha x^{-\alpha d} \Gamma(b, cx^{-d})-\Gamma(b + \alpha, cx^{-d})].$$
Now identify $a=\alpha d+1$ to arrive at
$$\int x^{-a}\Gamma(b,cx^{-d})\,\mathrm{d}x=-\frac{1}{a-1}[x^{1-a}\Gamma(b,cx^{-d})-c^{(1-a)/d}\Gamma(b+(a-1)/d,cx^{-d})],$$
which is the desired answer.
| 1 | https://mathoverflow.net/users/11260 | 415761 | 169,474 |
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