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https://mathoverflow.net/questions/402873
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0
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Let $M$ be a $3\times 3$ real invertible diagonal matrix and $H$ a Hilbert space of infinite dimension (for example, we can take $H$ as the space of square integrable functions over a bounded lipschitz domain $\Omega\subset\mathbb{R}^3$).
I want to determine the essential spectrum of $M$ as a multiplication operator acting from $H^3$ to itself.
Does it consist of the eigenvalues of $M$? Or its numerical range ?
Thank you.
|
https://mathoverflow.net/users/152870
|
Essential spectrum of constant invertible diagonal matrix acting on a product of Hilbert spaces
|
$H^3$ is $\mathbb{C}^3 \otimes H$ and the action of $M$ on it is just the usual action on the first tensor factor. It is a completely straighforward exercise to prove that the $\lambda$-eigenspace of $M\otimes\operatorname{id}\_H$ is exactly $\operatorname{Eig}\_\lambda(M) \otimes H$ and similarly for generalized eigenspaces. In particular: The spectrum of $M\otimes\operatorname{id}\_H$ is exactly the same as the spectrum of $M$, i.e. discrete consisting precisely of the eigenvalues.
|
0
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https://mathoverflow.net/users/3041
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402888
| 165,306 |
https://mathoverflow.net/questions/402808
|
3
|
Let $\Omega$ be an algebraically closed field of characteristic $0$, $k$ a subfield such that $\mathrm{tr.deg}(\Omega/k)=\infty$. Let $u\_1,\dots,u\_n,u\_{n+1}\in \Omega$ be algebraically independent over $k$, $P$ be a prime(not maximal) ideal of $k[X\_1,\dots,X\_n]$. Does there exists $x\_1,\dots,x\_n\in \Omega$ such that the following conditions are satisfied:
$$\begin{aligned}&(1)u\_{n+1}=u\_1 x\_1+\cdots +u\_n x\_n,\\
&(2)\text{The image of }k(u\_1,\dots,u\_n)[X]\rightarrow \Omega, X\_i\mapsto x\_i\text{is isomorphic to } k(u\_1,\dots,u\_n)[X]/P
\end{aligned}$$
where we still denote by $P$ the ideal $P k(u\_1,\dots,u\_n)[X] $ which is prime since a purely transcendental extension is regular.
The second condition is easy to be satisfied. Actually, if we forget $(1)$ and set $u\_{n+1}’:=u\_1x\_1+\cdots u\_nx\_n$, then we can prove that $u\_{n+1}’$ is transcendental over $k(u\_1,\dots,u\_n)$(see p212, Introduction to algebraic geometry, S. Lang). But I am not sure whether the converse is true or not. Thanks in advance if anyone could offer some help.
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https://mathoverflow.net/users/313627
|
Existence of generic zeros
|
I found the answer from Serge Lang’s book. As I mentioned in the description of the question, take arbitrary $x\_1,\dots,x\_n\in \Omega$ such that $P$ is the kernel of $k(u\_1,\dots,u\_n)[X]\rightarrow \Omega, X\_i\mapsto x\_i$. Let $u\_{n+1}’:=u\_1x\_1+\cdots+u\_nx\_n$, then $u\_{n+1}’$ is algebraically independent over $k(u\_1,\cdots,u\_n)$. Denote that $k(u’):= k(u\_1,\dots,u\_n,u\_{n+1}’
)(\text{reps. } k(u):= k(u\_1,\dots,u\_n,u\_{n+1}))$. Consider an isomorphism
$$\delta:k(u’)\rightarrow k(u),u\_{n+1}’\mapsto u\_{n+1}.$$
We extend $\delta$ to an isomorphism of $k(u’,x\_1,\dots,x\_n)$, then the image of $x\_1,\dots,x\_n$ is the tuple satisfying our two conditions.
|
0
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https://mathoverflow.net/users/313627
|
402892
| 165,309 |
https://mathoverflow.net/questions/402910
|
6
|
Let $f, g \in L^2(\mathbb{R})$.
Is it true that if both $|f|=|g|$ and $|\hat f|=|\hat g|$ hold, then there exists $\theta \in \mathbb{R}$ such that $f=ge^{i\theta}$?
I am not able to prove it or disprove it. I suspect that this is true. Do you have a reference for this?
|
https://mathoverflow.net/users/94414
|
Absolute values of two functions and absolute values of their Fourier transform coincides
|
This is true. This was proved by Hardy and Littlewood and the proof is reproduced in Zygmund's Trigonometric Series (which I don't have access to at the moment).
(Contradicting my prior "answer")
The answer to this question is negative. Such counterexamples are known as "Pauli partners" and are studied in, among other places, the quantum mechanics literature.
See, for example:
(J. V. Corbett and C. A. Hurst) Are Wave Functions Uniquely Determined by their position and momentum distributions? [https://www.cambridge.org/core/services/aop-cambridge-core/content/view/228E4A34D0B3C63C54B1A01006278C42/S0334270000001569a.pdf/are-wave-functions-uniquely-determined-by-their-position-and-momentum-distributions.pdf]
(P. Jaming) Phase Retrieval Techniques for Radar Ambiguity Problems [https://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.521.4906&rep=rep1&type=pdf]
|
9
|
https://mathoverflow.net/users/630
|
402912
| 165,317 |
https://mathoverflow.net/questions/402919
|
4
|
The question is in the title. It was asked by Paul Erdős, e.g. as part of Section 9 in this [paper](https://www.ime.usp.br/%7Eyoshi/resenhas/abstracts/Erdos.pdf).
|
https://mathoverflow.net/users/174530
|
Does the equation $\varphi(n)=\sigma(m)$ have infinitely many solutions?
|
Yes, this was proved by [Ford, Luca and Pomerance](https://faculty.math.illinois.edu/%7Eford/wwwpapers/phi=sig.pdf) in 2010 (paper in Bulletin of the London Math. Soc.).
|
19
|
https://mathoverflow.net/users/38624
|
402920
| 165,318 |
https://mathoverflow.net/questions/402905
|
0
|
I know that in the literature there are interesting articles involving the sequence of Ramanujan primes, I refer the [*Ramanujan Prime*](https://mathworld.wolfram.com/RamanujanPrime.html) from the online encyclopedia Wolfram MathWorld. This week I wondered what about experimental mathematics concerning this sequence of prime numbers (in the past I've known for example the article $1/f$ *noise in the distribution of prime numbers* by Marek Wolf, Physica A: Statistical Mechanics and its Applications Vol. **241**, Issue 3–4, (1997), pp. 493-499). I have thought the following question inspired by the article *Un nuevo patrón en los números primos* that the author Bartolo Luque refers in his column *Juegos matemáticos* of a scientific journal, the Spanish edition of Scientific American, that is the journal Investigación y Ciencia, pages 91-93 (Julio 2019). He is author of [1].
>
> **Question.** Do the first-digit frecencies of Ramanujan primes satisfy a Benford's law? **Many thanks.**
>
>
>
Thus I'm asking if you know how to deduce, or to disprove it, (showing your computational evidence or reasonings) if first-digit frecuencies of Ramanujan primes obey some Benford's law. I add that Wikipedia has an article dedicated to [*Benford's law.*](https://en.wikipedia.org/wiki/Benford%27s_law)
I hope that my question is interesting, feel free to add comments about it. Also if my question is in the literature answer it as a reference request.
References:
-----------
[1] Bartolo Luque and Lucas Lacasa, *The First-Digit Frequencies of Prime Numbers and Riemann Zeta Zeros*, Proceedings: Mathematical, Physical and Engineering Sciences Vol. 465, No. 2107 (Jul. 8, 2009), pp. 2197-2216 (Royal Society).
|
https://mathoverflow.net/users/142929
|
New experiments involving Ramanujan primes: Benford's law
|
If I am following what is being asked, the answer is no.
Set $R$ to be the set of Ramanujan primes. Let $R\_d$ be the set of Ramanujan primes with lead digit $d$. For a set of positive integers integers $S$, we'll write $S(x)$ to be the number of elements in S which are at most $x$. Then you are asking whether for any $d=1,2, \cdots 9$ we have
$$\lim\_{x \rightarrow \infty}\frac{R\_d(x)}{R(x)} =\log\_{10} \left(\frac{d+1}{d}\right)$$
This statement is false. Let $Ram\_n$ be the $n$th Ramanujan prime. Then by [Sondow's theorem](https://arxiv.org/abs/0907.5232) $Ram\_n$ is asymptotic to the $2n$th prime, which is asymptotic to $2n \log 2n \sim 2n \log n$ by the prime number theorem. So $$R(x) \sim \frac{x}{2\log x}.$$
Now, set $x=(10^t)$ Then we have (neglecting small error terms) $$R\_9(x) \geq R(x)) - R(\frac{9}{10}x) = \frac{x}{2 \log x} - \frac{\frac{9}{10}x}{2 \log x} = \frac{x}{10 (2)\log x} .$$
So for this set of value of $x$ we have $R\_9(x)/R(x)$ is at least about $\frac{1}{10}$. But $\log\_{10} \frac{10}{9}$ is much smaller, a little under $0.046.$
Note that this proof really doesn't use anything deep about the Ramanujan primes other than their asymptotic. In general, for any set of integers $S$ where $$S(x) \sim \frac{cx}{(\log x)^k}$$ for some positive constants $c$ and $k$, it will fail the base $10$ version of Benford's law. And this will apply to any other base $b>2$ by the same reasoning. Base $b$ will always have too many elements starting with $b-1$. (And of course in base $b=2$ Benford's law is trivial.) In order to have a Benford's law distribution one generally needs to be growing at least as slowly as $x^{\alpha}$ for some $\alpha <1$ or not have a good asymptotic at all. Edit: Actually see Will's comment below, one in fact needs a much stronger density requirement than this to have any hope of satisfying Benford's law.
|
4
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https://mathoverflow.net/users/127690
|
402935
| 165,324 |
https://mathoverflow.net/questions/402939
|
5
|
It is known that with $M(x) = \sum\_{n\le x}\mu(n)$, there are infinitely many $x$ s.t. $|M(x)|\ge x^{\frac{1}{2} - \varepsilon}$ (see Chapter 15 of Montgomery-Vaughan, for example). Is there any way to make this result effective in the following sense: show that for large $X$, there exists some $x\in [X, f(X)]$ s.t. $|M(x)|\ge x^{\frac{1}{2} - \varepsilon}$ (or $x^{\sigma - \varepsilon}$ if $\zeta(\sigma + it) = 0$ for some $\sigma > 1/2$), where $f(X)$ is some explicit function in $X$? (Potentially naive) heuristics suggest one can take $f(X) = X + X^{2\sigma}$.
|
https://mathoverflow.net/users/40983
|
Frequency of large values of the Mertens function
|
It was proved by Kaczorowski and Pintz (Acta Math. Hungar. 48 (1986), 173-185, doi: [10.1007/BF01949062](http://dx.doi.org/10.1007/BF01949062)) that there exists $x\in[X,X^{1+o(1)}]$ such that $M(x)\geq x^{\sigma-\varepsilon}$, and there also exists $x\in[X,X^{1+o(1)}]$ such that $M(x)\leq -x^{\sigma-\varepsilon}$. See Corollary 1 in their paper.
|
3
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https://mathoverflow.net/users/11919
|
402971
| 165,333 |
https://mathoverflow.net/questions/402929
|
24
|
To every complex-oriented ring spectrum $E$ there is associated a formal group law, which is a power series $F\_E(x,y)\in E\_\*[[x,y]]$.
Suppose $E$ and $F$ are two complex-oriented ring spectra and suppose I have an isomorphism of coefficient rings $\phi:E\_\*\rightarrow F\_\*$ that carries $F\_E(x,y)$ to $F\_F(x,y)$.
Does this imply that $E$ and $F$ are homotopy equivalent spectra?
Note that if $F\_E(x,y)$ and $F\_F(x,y)$ are "Landweber exact" formal group laws then the answer is yes.
|
https://mathoverflow.net/users/163893
|
Are complex-oriented ring spectra determined by their formal group law?
|
The following is a communal answer from the algebraic topology Discord [[1](http://nodorek.net)], primarily put forward by Irakli Patchkoria (correcting previous half-answers by Tyler Lawson and me).
Kiran suggested it be recorded here to ease future reference.
The idea is to produce two topological realizations $M$, $N$ of a single $MU\_\*$–module by finding two distinct resolutions whose effect on homotopy is the same.
The two associated square-zero extensions then give a counterexample.
We'll reduce complexity first by considering $ku$–modules rather than $MU$–modules, and second by aiming for a $ku$–module whose homotopy cleaves into small even and odd parts, forcing its $ku\_\*$–module structure to trivialize.
$\DeclareMathOperator{\Sq}{Sq}
\newcommand{\F}{\mathbb{F}}
\newcommand{\Z}{\mathbb{Z}}
\newcommand{\HFtwo}{H\F\_2}
\newcommand{\Susp}{\Sigma}
\newcommand{\co}{\colon\thinspace}$
The nice homotopy groups of complex $K$–theory, $ku\_\* = \Z[u]$, can be used to show that its bottom $k$–invariant $\kappa\_{ku}$ is $ku$–linear: in the diagram
$$
\begin{array}{ccccc}
& & \Susp^4 ku \\
& & u \downarrow \\
\Susp^{-1} H\Z & \to & \Susp^2 ku & \to & \Susp^2 H\Z \\
& & u \downarrow \\
& & ku & \to & H\Z,
\end{array}
$$
the vertical maps are multiplication by homotopy elements, hence are $ku$–linear; in turn the horizontal co/fibers are also $ku$–linear; and, finally, the $k$–invariant appears as the middle composite, hence is also $ku$–linear.
Similarly, we can show the $ku$–linearity of the bottom $k$–invariant of $ku/2$ and of the Bockstein map $\beta\co \HFtwo \to \Susp H\Z$ (relying on the $ku$–linearity of $2\co H\Z \to H\Z$).
Stringing some of these together gives a $ku$–linear composite $$\HFtwo \xrightarrow{\kappa\_{ku/2}} \Susp^3 \HFtwo \xrightarrow{\beta} \Susp^4 H\Z \to \Susp^4 \HFtwo.$$
The bottom $k$–invariant $\kappa\_{ku/2}$ of $ku/2$ is given as the Milnor primitive $Q\_2 = \Sq^3 + \Sq^2 \Sq^1$, the composite of the latter two maps is given as $\Sq^1$, and hence the whole composite is the nontrivial Steenrod operation $$\Sq^1 Q\_2 = \Sq^1(\Sq^3 + \Sq^2 \Sq^1) = \Sq^3 \Sq^1.$$
Meanwhile, the homotopy groups of the cofiber $M$ of this composite are $\Susp \F\_2 \oplus \Susp^4 \F\_2$, which splits as a $ku\_\*$–module — hence this $ku\_\*$–module could alternatively be modeled by $N = \Susp H\F\_2 \oplus \Susp^4 \HFtwo$ (i.e., the cofiber of the zero map).
To finish, set $E = ku \oplus M$ and $F = ku \oplus N$.
|
20
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https://mathoverflow.net/users/1094
|
402987
| 165,338 |
https://mathoverflow.net/questions/402986
|
2
|
I recently came across the paper *Les variétés de dimension 4 à signature non nulle dont
la courbure est harmonique sont d’Einstein* by Jean Pierre Bourguignon. What he shows in §8 is that the Weitzenböck curvature operator $\mathfrak{Ric}\_\text{R}$ on $p$-forms is given by $$\mathfrak{Ric}\_\text{R}(\omega)(X\_1,\dots,X\_p) = (\hat{\omega} \circ \hat{R\_p})(X\_1,\dots,X\_p)$$
where
$$R\_p = \left(\tfrac{1}{2(p-1)}\text{Ric} \mathbin{\bigcirc\mspace{-19mu}\wedge\mspace{3mu}} \text{g} - \text{Rm}\right) \mathbin{\bigcirc\mspace{-19mu}\wedge\mspace{3mu}} \text{g}^{p-2}.$$
Here, I just can't figure out what he means by $\hat{\omega} \circ \hat{R\_p}$.
Earlier in the paper he defines (in (2.11)) for a $C \in S^2\Lambda^{2}V$, interpreted as a self-adjoint map $\Lambda^{2}V \to \Lambda^{2}V$, and a form $\eta \in \Lambda^2(V)$, that $$\hat{C}(\eta) = \sum\limits\_{i,j = 1}^n \eta(e\_i,e\_j)C(e\_{i},e\_{j})$$ for an orthonormal basis $(e\_i)\_{1 \le i \le n}$ of $V$.
I see how this can be generalized into:
For a $C \in S^2\Lambda^{p}V$, interpreted as a self-adjoint map $\Lambda^{p}V \to \Lambda^{p}V$, and a form $\eta \in \Lambda^p(V)$, define $$\hat{C}(\eta) = \sum\limits\_{i\_1, \dots, i\_p = 1}^n \eta(e\_{i\_1},\dots,e\_{i\_p})C(e\_{i\_1},\dots,e\_{i\_p})$$ for an orthonormal basis $(e\_i)\_{1 \le i \le n}$ of $V$.
And I am thinking that this is what he means here for $C = R\_k$. But this still doesn't answer what $\hat{\omega}$ would be.
I am thinking that $(\hat{\omega} \circ \hat{R\_p})$ may just mean $\hat{R\_p}(\omega)$; but why wouldn't he have written it like this then? Probably I am missing something central here..
The paper can be found here for free:
[Les variétés de dimension 4 à signature non nulle dont la courbure est harmonique sont d’Einstein by Jean Pierre Bourguignon](https://gdz.sub.uni-goettingen.de/id/PPN356556735_0063?tify=%7B%22pages%25%2022:%5B271%5D%7D)
|
https://mathoverflow.net/users/142232
|
Invariant description of the Weitzenböck curvature operator by Bourguignon
|
I think the idea is to think of $\hat{R}\_p$ as a mapping from $\Lambda^pM$ to itself, and $\hat{\omega}$ as a mapping from $\Lambda^pM$ to $E$ (the vector bundle in which $\omega$ takes values), and then just compose these two operators (to get something $E$ valued in the end).
In the case where $E$ is the trivial (scalar) bundle, it works out to be $\hat{R}\_p(\omega)$ since $R\_p$ is symmetric.
|
1
|
https://mathoverflow.net/users/3948
|
402989
| 165,339 |
https://mathoverflow.net/questions/402762
|
18
|
Let $G=\operatorname{GL}(n,\mathbb{C})$ and $\mathfrak{g}=\operatorname{Mat}(n,\mathbb{C})$ and let us consider the two varieties $X,Y$ defined as $$X=\{(x,y) \in G \times G \ | \ xy=yx\} $$ and $$Y=\{(x,y) \in \mathfrak{g} \times \mathfrak{g} \ | \ xy=yx\} .$$
The group $G$ acts on both of them by conjugation: I'd like to find out what is known in the literature for the $G$-equivariant cohomology of $X,Y$ (an the mixed Hodge structure on it).Moreover, is the cohomology of their GIT quotients $X//G$, $Y//G$ known too? Is there a relation between them?
|
https://mathoverflow.net/users/146464
|
Variety of commuting matrices
|
There's been a good deal of work since the papers Mark Grant cited.
The rational cohomology of $\mathrm{Hom}(\mathbb{Z}^n,K)//K$ for $K$ a compact connected Lie group was computed by Stafa (<https://arxiv.org/abs/1705.01443>). It's a theorem of Florentino and Lawton that if $G$ is a linearly reductive Lie group with maximal compact subgroup $K$, then $\mathrm{Hom}(\mathbb{Z}^n,G)//G$ deformation retracts to $\mathrm{Hom}(\mathbb{Z}^n,K)/K$, so for the general linear group, we can switch to working with the unitary groups instead. Stafa gives a general formula for the Poincare series, in terms of the order of the Weyl group and its action on the (dual of the) Lie algbra of a maximal torus. The formula reduces to $((1+t)^{2n}+(1-t^2)^n)/2$ for $G = U(n)$ (or $GL\_n (\mathbb{C})$). Florentino and Silva (<https://arxiv.org/abs/1711.07909>) computed algebro-geometric refinements of these Poincare series, and their work recovers Stafa's formula.
There's a similar story for the ordinary rational cohomology of $\mathrm{Hom}(\mathbb{Z}^n,G)$, discussed in a paper I wrote with Stafa, <https://arxiv.org/abs/1704.05793>. Since then D. Kishimoto and M. Takeda have made a good deal of progress, including information about the ring structure and torsion (also they gave a much shorter derivation of the Poincare series).
Regarding equivariant cohomology, Baird has some work in the compact case; see Section 4 of his paper <https://arxiv.org/abs/math/0610761>. Note that the inclusion of $\mathrm{Hom}(\mathbb{Z}^n,K)$ into $\mathrm{Hom}(\mathbb{Z}^n,G)$ is $K$-equivariant and a homotopy equivalence by a result of Pettet and Souto (Geom. and Topol. 17, 2013), and the inclusion of $K$ into $G$ is a homotopy equivalence, so $H\_K^\* (\mathrm{Hom}(\mathbb{Z}^n,K) \cong H\_K^\* (\mathrm{Hom}(\mathbb{Z}^n,G)) \cong H\_G^\* \mathrm{Hom}(\mathbb{Z}^n,G)$. It would be quite interesting to know more about the equivariant cohomology.
|
8
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https://mathoverflow.net/users/4042
|
402991
| 165,340 |
https://mathoverflow.net/questions/402992
|
1
|
I have question about a statement from *Etale Cohomology and the Weil Conjecture* by Freitag, Kiehl
at the top of page 16. It seemingly uses the same notations as introduced at the bottom of page 15
and is seemingly a consequence of following two facts.
Let $A$ be a *strict Henselian* ring (i.e. Henselian + residue class field separably algebraically closed).
(This book works with definition 1.17, p 15, a noetherian local rings $A$ is *Henselian* if
every local quasi-finite homomorphism from $A$ to a localization of a finitely generated
$A$-algebra is *finite*).
Next, the authors make a remark and deduce a consequence:
\*1.18 Remark.\*Let A be a strictly Henselian ring. Then every local-etale homomorphism
$A \to B$ is an isomorphism.
As a consequence it can be deduced from this that For every faithfully flat etale homomorphism
$\varphi: A \to B$ of a strictly Henselian ring $A$ into a finitely generated $A$-algebra $B$ there exist
a right inverse; that is a map $\psi:B \to A$ with $\psi \circ \varphi= id\_A$ ( [here an expanation](https://mathoverflow.net/questions/400750/faithfully-flat-etale-morphism-from-strictly-henselian-ring-from-etale-cohomolo)
why this consequence is true)
Now comes a claim I not understand. Next is said that for *any* sheaf of abelian groups $\mathcal{F}$ the
functor
$$ \operatorname{Et}(B) \to \operatorname{(Ab)}, \ C \mapsto \mathcal{F}(C) $$
is exact. Why that's true? (here $\operatorname{Et}(B)$ is the category of etale extensions of $B$
understood as full subcategory of the category of $A$-algebras)
*Question 1:*
My first confusion is that originally in the book the top line defining the functor reads as
$$ \operatorname{Et}(B) \to \operatorname{(Ab)}, \ B \mapsto \mathcal{F}(B) $$
This not make any sense, so obviously the author meant something else. What?
I guess (but not know) that the authors maybe intended to write $A$ instead of $B$, i.e.maybe to work with $\operatorname{Et}(A)$ etale extensions of a strict Henselian base; but I don't know, that's only a guess of mine, which "looks" for me more "natural". Does anybody see what the authors had there in mind?
*Question 2:* Why the statement about the exactness of the functor above is true?
We have to show that for every exact sequence
$$ 0 \to R\_1 \to R\_2 \to R\_3 \to 0 $$
of etale $B$-algebras die Sequenz of abelain groups
$$ 0 \to \mathcal{F}(R\_1) \to \mathcal{F}(R\_2) \to \mathcal{F}(R\_3) \to 0 $$
is exact too. Why? My first guess was that we can tensoring the $R\_i$ by $- \otimes\_B A$ via the right inverse $\psi$ from above and exploit then a splitting property of $A \to B$ for above remark,
in order to show somehow that maybe $R\_1 \to R\_2$ has left inverse (or $R\_2 \to R\_3$ right inverse), but I not know how or if it really holds.
|
https://mathoverflow.net/users/108274
|
Exactness of functor $ Et(B) \to \operatorname{(Ab)}, \ C \mapsto \mathcal{F}(C) $ (Etale Cohomology and the Weil Conjecture by Freitag, Kiehl )
|
I think they meant to say that the functor $\mathcal F \mapsto \mathcal F(A)$ from sheaves on $A$ to abelian groups is exact.
The proof is:
If $\mathcal F\_2\to \mathcal F\_3 \to 0$ is an exact sequence, we need to show $\mathcal F\_2(A) \to \mathcal F\_3(A)$ is surjective.
Let $s \in \mathcal F\_3(A)$ be a section. By definition of a surjective map of sheaves, there exists a covering $B$ of $A$ such that the pullback of $s$ to $B$ lies in the image of $\mathcal F\_2(B) \to \mathcal F\_3(B)$.
By the previous line, we can express $A$ as an open subset of $B$, so the pullback of $s$ to $\mathcal F\_3(A)$, which is $s$ again, lies in the image of $\mathcal F\_2(A) \to \mathcal F\_3(A)$, as desired.
|
4
|
https://mathoverflow.net/users/18060
|
402995
| 165,341 |
https://mathoverflow.net/questions/402994
|
11
|
Often times, we consult resources, like Abramowitz and Stegun's Handbook of Mathematical Functions [https://www.math.ubc.ca/~cbm/aands/](https://www.math.ubc.ca/%7Ecbm/aands/), NIST's database on special functions <https://www.nist.gov/programs-projects/special-functions>, or Mathematica to find identities which aid us with some kind of computation.
However, what if we want to know if we have found a new identity, want to systematically check against the above resources, and want to add to the library in the case the identity is new? Also, are there journals which, even today, still consider mathematical effort toward discovering identities of classical functions?
|
https://mathoverflow.net/users/115325
|
How to determine if you've discovered a new identity for a special function
|
***Q:*** Are there journals which would publish identities of classical functions?
***A:*** Elsevier's *Applied Mathematics and Computation* has published quite a number of papers in that category, see this [search listing](https://www.sciencedirect.com/search?qs=identities&pub=Applied%20Mathematics%20and%20Computation&cid=271588).
It is ranked as a Q1 journal, open access is optional.
|
11
|
https://mathoverflow.net/users/11260
|
402998
| 165,343 |
https://mathoverflow.net/questions/402950
|
2
|
Recall that a ternary $C^\*$-ring is a complex Banach space $X$, equipped with a associative ternary product $[.,.,.]:X^3 \to X$ which is linear in outer variables and conjugate linear in middle variable, $\|[a,a,a]\|= \|a\|^3$ and $$\|[a,b,c]\| \leq \| a \| \| b\|\| c\|$$
>
> Does there exist a unique $C^\*$-algebra corresponding to each ternary $C^\*$-ring?
>
>
>
$3.2$ Proposition of [this](https://reader.elsevier.com/reader/sd/pii/000187088390083X?token=01B8E5AC9D0B68D9B1C3816A5724216F3593567C5DB7B706F1E7859093E031641858701A9AA2ECD58D273F5FFED0E1D4&originRegion=eu-west-1&originCreation=20210901064606) paper seems to answer what i am looking for. The construction given in paper goes as follows: For $y,z \in X$, consider the bounded linear map $D\_{y,z}: X \to X$ defined as $D\_{y,z}(x)= [x,y,z]$. Let $V=$ span $\{D\_{y,z}: y,z \in X \} \subset L(X)$. After defining like this, the author shows that $V$ is pre $ C^\*$-algebra. Finally, author considers the opposite algebra of the norm closure of $V$ to construct the required $C^\*$-algebra.
>
> Can someone please explain me why do we need to consider opposite algebra and the motivation behind this proof?
>
>
>
|
https://mathoverflow.net/users/129638
|
Trying to understand construction of $C^*$-algebra corresponding to a ternary $C^*$-ring from a paper
|
Yes, Proposition 3.2 from the article [MR0700979] that you link to gives you the answer for ternary $\rm C^\*$-rings. The reason for taking the opposite algebra is that the algebra $V$ acts on the right of $X$ rather than on the left. A right action is the same thing as a left action by the opposite algebra. You can see this right action directly in (3) of this proposition.
As a special case, suppose $X$ is a right Hilbert $\rm C^\*$-module over a $\rm C^\*$-algebra $A$. Denoting the $A$-valued inner product on $X$ by $\langle x|y\rangle\_A$ (which is linear on the right), we get a ternary $\rm C^\*$-ring by $[x,y,z] := x\langle y|z\rangle\_A$. In this setting, $D\_{y,z}$ is right multiplication by $\langle y|z\rangle\_A$. All ternary $\rm C^\*$-rings are essentially like this one, except that the $A$-valued form need not be positive. (See the intro of the cited article, bottom of p118 after the displayed equation.)
|
1
|
https://mathoverflow.net/users/351
|
403000
| 165,345 |
https://mathoverflow.net/questions/402993
|
9
|
The permanent $\mathrm{per}(A)$ of a matrix $A$ of size $n\times n$ is defined to be:
$$\mathrm{per}(A)=\sum\_{\tau\in S\_n}\prod\_{j=1}^na\_{j,\tau(j)}.$$
Let
$$A=\left[\tan\pi\frac{j+k}n\right]\_{1\le j,k\le n-1},$$
$$B=\left[\sin\pi\frac{j+k}n\right]\_{1\le j,k\le n-1},$$
$$C=\left[\cos\pi\frac{j+k}n\right]\_{1\le j,k\le n-1},$$
$$D=\left[\sec\pi\frac{j+k}n\right]\_{1\le j,k\le n-1}.$$
Motivated by [Question 402249](https://mathoverflow.net/questions/402249), I found the following
**Conjecture 1.** For any odd integer $n>1,$
$$(-1)^{(n-1)/2}\mathrm{per}(A)=\frac{2(n!!)^2}{n+1}\sum\_{k=0}^{\frac{n-1}{2}}\frac{(-1)^k}{2k+1}. \tag{1}$$
Numerical calculations show that this is correct for $3 \leq n \leq 33$. See [Question 402249](https://mathoverflow.net/questions/402249) for details.
Inspired by [Question 402572](https://mathoverflow.net/questions/402572), I also found the following identities
**Conjecture 2.** For any odd integer $n>1,$
\begin{align}
(-1)^{(n-1)/2}\mathrm{per}(B)&=\frac{n!}{2^{n-2}(n+1)}，\tag{2}
\\ \mathrm{per}(C)&={\frac{(n-1)!}{2^{n-1}}}\sum\_{k=0}^{n-1}\frac{1}{\binom{n-1}{k}},\tag{3}
\\ \mathrm{per}(D)&= (n-2)!!^2\left( (-1)^{\frac{n+1}{2}}+2n\sum\_{k=0}^{\frac{n-1}{2}} {\frac {\left( -1 \right) ^{k}}{2k+1} }
\right) .\tag{4}
\end{align}
Numerical calculations show that it is correct for $3 \leq n \leq 21$.
**Question.** Are these identities correct? How to prove them?
|
https://mathoverflow.net/users/287674
|
Permanent identities
|
Let $\zeta$ be a primitive $n$-th root of unity. Then
$$\prod\_{j=1}^{n-1}(x-\zeta^j)=\frac{x^n-1}{x-1}=1+x+\cdots+x^{n-1}$$
and hence
$$\sigma\_k=\sum\_{1\le i\_1<\cdots<i\_k\le n-1}\zeta^{i\_1+\cdots+i\_k}=(-1)^k$$
for all $k=1,\ldots,n-1$.
Observe that
\begin{align\*}\mathrm{per}[1-\zeta^{j+k}]\_{1\le j,k\le n-1}=&\sum\_{\tau\in S\_{n-1}}\ \prod\_{j=1}^{n-1}(1-\zeta^{j+\tau(j)})
\\=&\sum\_{\tau\in S\_{n-1}}1+\sum\_{\tau\in S\_{n-1}}\ \sum\_{\emptyset\not=J\subseteq\{1,\ldots,n-1\}}(-1)^{|J|}\zeta^{\sum\_{j\in J}\ (j+\tau(j))}
\\=&(n-1)!+\sum\_{\emptyset \not=J\subseteq \{1,\ldots,n-1\}}(-1)^{|J|}\zeta^{\sum\_{j\in J}j}\sum\_{\tau\in S\_{n-1}}\zeta^{\sum\_{j\in J}\ \tau(j)}.
\end{align\*}
For $\emptyset \not=J\subseteq\{1,\ldots,n-1\}$, clearly
\begin{align\*}&\sum\_{\tau\in S\_{n-1}}\zeta^{\sum\_{j\in J}\ \tau(j)}
\\=&\sum\_{1\le i\_1<\cdots<i\_{|J|}\le n-1}\zeta^{i\_1+\cdots+i\_{|J|}}\sum\_{\tau\in S\_{n-1}\atop\{\tau(j):\ j\in J\}
=\{i\_1,\ldots,i\_{|J|}\}}1
\\=&|J|!(n-1-|J|)!\sigma\_{|J|}=(-1)^{|J|}|J|!(n-1-|J|)!.
\end{align\*}
Therefore
\begin{align\*}\mathrm{per}[1-\zeta^{j+k}]\_{1\le j,k\le n-1}=&(n-1)!+\sum\_{\emptyset\not=J\subseteq\{1,\ldots,n-1\}}|J|!(n-1-|J|)!\zeta^{\sum\_{j\in J}\ j}
\\=&(n-1)!\sum\_{k=0}^{n-1}\frac{(-1)^k}{\binom{n-1}k}=(1-(-1)^n)\frac{n!}{n+1}.
\end{align\*}
Then the conjectural identity $(2)$ follows from this since
$$\sin\pi\frac{j+k}n=\frac{e^{-\pi i(j+k)/n}}{2i}\left(e^{2\pi i(j+k)/n}-1\right).$$
The identities $(3)$ can be proved similarly, in fact we have
$$\mathrm{per}[1+\zeta^{j+k}x]\_{1\le j,k\le n-1}=(n-1)!\sum\_{k=0}^{n-1}\frac{x^k}{\binom{n-1}k}.$$ The idea here is slight modification of my way to establish Theorem 1.1 in my preprint *Arithmetic properties of some permanents* available from <http://arxiv.org/abs/2108.07723>.
I admit that the identities $(1)$ and $(4)$ remain open.
|
10
|
https://mathoverflow.net/users/124654
|
403004
| 165,348 |
https://mathoverflow.net/questions/403011
|
12
|
I am looking for a proof of the following claim:
First define the function $\chi(n)$ as follows:
$$\chi(n)=\begin{cases}1, & \text{if }n \equiv \pm 1 \pmod{10} \\
-1, & \text{if }n \equiv \pm 3 \pmod{10} \\ 0, & \text{if otherwise }
\end{cases}$$
Then,
$$\frac{\pi^2}{5\sqrt{5}}=\displaystyle\sum\_{n=1}^{\infty}\frac{\chi(n)}{n^2}$$
The SageMath cell that demonstrates this claim can be found [here](https://sagecell.sagemath.org/?z=eJwrTbI1NQACa67kjEyNPE3b6sw0jTxVQwNbW0OFmhoFCNNSJ9nWUAcuY4yQMQfK6BoCCQNNTeui1JLSojyNZE3rWq5i2-LSXI08oLbSJB2I2foaeXFGQGVcBUWZeSUaxVqGenBOQGackb6GqVZxYVGJhqkmUBUAw8Eq3Q==&lang=gp&interacts=eJyLjgUAARUAuQ==).
|
https://mathoverflow.net/users/88804
|
A conjectural infinite series for $\frac{\pi^2}{5\sqrt{5}}$
|
More generally, if $1\le k\le N-1$ is an integer, where $N$ is a positive interger,
$$S\_{N,k} := \sum\_{n=0}^\infty\biggl( \frac{1}{(N n+N-k)^2} + \frac{1}{(N n+k)^2}
\biggr) = \frac{\pi^2}{N^2\sin^2(\pi k/N)}.$$
Your sum is $S\_{10,1}-S\_{10,3}$.
|
24
|
https://mathoverflow.net/users/9025
|
403016
| 165,349 |
https://mathoverflow.net/questions/397768
|
3
|
Suppose $x$ is a non-zero vector in a Banach space, and $T$ is a fixed operator. Is the following true:
For any $\varepsilon, \delta$, there exists $S$ in the commutant of $T$ such that $1\leq\|S\|<1+\delta$ and $\|Sx\|<\varepsilon$.
This is true for some $T$ (for example the identity), but is it true for *all* $T$? Is there anything known in this direction, even for $\ell\_2$?
|
https://mathoverflow.net/users/69275
|
Operator in the commutant which is small on a given vector
|
Not even true for $2\times 2$ matrices. Let $T$ be the nilpotent $\begin{pmatrix}0&1\\0&0\end{pmatrix}$ and $x$ be the vector $(0,1)$. Then anything in the commutant of $T$ has form $\lambda+\mu T$. So if $S$ is in the commutant of $T$, then $\|Sx\| = \sqrt{\lambda^2 + \mu^2}$, so if $\|Sx\| < \varepsilon$, then $\max(|\lambda|, |\mu|) < \varepsilon$ and $\|S\| < \sqrt{\frac {3 + \sqrt 5} 2}\varepsilon$.
|
5
|
https://mathoverflow.net/users/356618
|
403031
| 165,355 |
https://mathoverflow.net/questions/403020
|
14
|
Let $\mathcal{P}$ be a set of prime numbers of relative density $\kappa \in (0, 1)$, which means that
$$\#\left(\mathcal{P} \cap [1, x]\right) = \kappa \,\pi(x) + E(x) \quad (x \to \infty)$$
for a "suitable" error term $E(x)$ (of course, $E(x) = o(\pi(x))$).
Let $\mathcal{S}$ be the set of natural numbers having no prime factor in $\mathcal{P}$.
**Question:** Do we have an asymptotic formula for $\#\left(\mathcal{S} \cap [1, x]\right)$ ?
For $\kappa \in (0, 1/2)$, Corollary 1.2 of [1] gives that
$$\#\left(\mathcal{S} \cap [1, x]\right) \sim C(\mathcal{P}) \frac{x}{(\log x)^\kappa} \quad (x \to +\infty)$$
where $C(\mathcal{P})$ is a positive constant depending on $\mathcal{P}$.
I guess that something similar holds in the whole range $\kappa \in (0, 1)$, but I could not find such a result. Thank you for any help.
**P.S.** Although I am interested in general set of primes $\mathcal{P}$ of relative density $\kappa$, results concerning *sufficiently general* $\mathcal{P}$ of relative density $\kappa$ are welcome. For example, in light of Daniel Loughran's comment, sets like those of Chebotarev density theorem $$\mathcal{P} = \{p \text{ unramified in $K$}, \mathrm{Frob}\_p \subseteq C\} ,$$
where $K$ is a finite Galois extension of $\mathbb{Q}$ and $C$ is a conjugacy class of its Galois group, are certainly interesting.
*P.S.* This is very related to question: [Natural density of set of numbers not divisible by any prime in an infinite subset](https://mathoverflow.net/q/369092/160943) , as pointed out by user Hhhhhhhhhhh.
[1] Iwaniec and Kowalski, *Analytic Number Theory* (2004)
|
https://mathoverflow.net/users/342938
|
Numbers without prime factors in a set of positive relative density
|
You basically ask about the sum
$$ \sum\_{n \le x} \alpha(n)$$
where $\alpha$ is a completely multiplicative function with $\alpha(p) = \mathbf{1}\_{p \notin \mathcal{P}}$.
This is addressed by Wirsing in his famous paper ``Das asymptotische Verhalten von Summen über multiplikative Funktionen'' (Math. Ann. 143 (1961), 75–102). The only requirement on $E$ is $E(x)=o(\pi(x))$, and it gives the asymptotic result
$$\tag{$\star$} \sum\_{n \le x} \alpha(n) \sim\frac{ e^{-\gamma (1-\kappa)}}{\Gamma(1-\kappa)} \frac{x}{\log x} \prod\_{p \le x,\, p \notin \mathcal{P}}(1-1/p)^{-1},$$
where $\gamma$ is the Euler-Mascheroni constant (appearing also in Mertens' theorem).
Remark 1: Suppose $\sum\_{p \le x, p \in \mathcal{P}} 1/p = \kappa\sum\_{p \le x} 1/p +C + o(1)$, which holds if $E(x)$ is small enough, say $O(x/\log^{1+\varepsilon} x)$ (by partial summation). Then
$$C(\mathcal{P}) :=\frac{ 1}{\Gamma(1-\kappa)} \prod\_{p \notin \mathcal{P}}(1-1/p)^{-1} \prod\_{p}(1-1/p)^{1-\kappa}$$
converges and the last result may be simplified as
$$C(\mathcal{P})\frac{x}{(\log x)^{\kappa}},$$
by Mertens. This should recover the result from Iwaniec and Kowalski.
Remark 2: In Wirsing's sequel to his own paper, ``Das asymptotische Verhalten von Summen über multiplikative Funktionen. II'' (Acta Math. Acad. Sci. Hungar. 18 (1967), 411–467) he relaxes the condition on $\mathcal{P}$ even further, requiring less than positive relative density, while still retaining $(\star)$.
|
20
|
https://mathoverflow.net/users/31469
|
403033
| 165,356 |
https://mathoverflow.net/questions/403036
|
25
|
According to Claude Shannon, von Neumann gave him very useful advice on what to call his measure of information content [1]:
>
> My greatest concern was what to call it. I thought of calling it 'information,' but the word was overly used, so I decided to call it 'uncertainty.' When I discussed it with John von Neumann, he had a better idea. Von Neumann told me, 'You should call it entropy, for two reasons. In the first place your uncertainty function has been used in statistical mechanics under that name, so it already has a name. In the second place, and more important, no one really knows what entropy really is, so in a debate you will always have the advantage.'
>
>
>
What I am curious about is what von Neumann meant with his last point. I find it particularly surprising given that he axiomatised what we now call the von Neumann entropy in Quantum Mechanics around two decades prior to Shannon's development of Classical Information Theory.
Might modern information theorists know of the specific difficulties he had in mind and whether these have been suitably addressed?
References:
-----------
1. McIrvine, Edward C. and Tribus, Myron (1971). Energy and Information Scientific American 225(3): 179-190.
2. von Neumann, John (1932). Mathematische Grundlagen der Quantenmechanik (Mathematical Foundations of Quantum Mechanics) Princeton University Press., . ISBN 978-0-691-02893-4.
3. Shannon, Claude E. (1948). A Mathematical Theory of Communication Bell System Technical Journal 27: 379-423. doi:10.1002/j.1538-7305.1948.tb01338.x.
4. Olivier Rioul. This is IT: A Primer on Shannon’s Entropy and Information. Séminaire Poincaré. 2018.
5. E.T. Jaynes. Information Theory and Statistical Mechanics. The Physical Review. 1957.
6. John A. Wheeler, 1990, “Information, physics, quantum: The search for links” in W. Zurek (ed.) Complexity, Entropy, and the Physics of Information. Redwood City, CA: Addison-Wesley.
|
https://mathoverflow.net/users/56328
|
John von Neumann's remark on entropy
|
An [alternative version](https://en.wikipedia.org/wiki/Talk%3AHistory_of_entropy) of Von Neumann's quote says *"no one understands entropy very well"*. At the intuitive level, this makes sense, it is much harder to explain the concept of entropy to a novice than it is to explain energy.
One debate that existed at the time of Von Neumann$^1$ and is still argued upon$^2$ is whether the entropy $S=-{\rm tr}\,\rho\log\rho$ from information theory equals the physical (thermodynamic) entropy. This may or may not have been what Von Neumann was thinking about when he made the remark to Shannon, but it's one documented source of confusion.
---
$^1$ *"In the 1950's Jaynes told Wigner that physical entropy is a measure of information and Wigner thought that was absurd, because the information one person possesses differs from that of another, whereas entropy can be measured with thermometers and calorimeters."* [[source](https://mathoverflow.net/a/146573/11260)]
$^2$ [A Man Misunderstood: Von Neumann did not claim that his entropy corresponds to the phenomenological thermodynamic entropy](https://arxiv.org/abs/2007.06673) (2007).
|
18
|
https://mathoverflow.net/users/11260
|
403040
| 165,359 |
https://mathoverflow.net/questions/403008
|
10
|
Is there an increasing function on
$[a, b]$ which is differentiable,
but not absolutely continuous？
|
https://mathoverflow.net/users/345221
|
Is there an increasing function on $[a, b]$ which is differentiable, but not absolutely continuous？
|
An elementary non-existence proof may be of interest.
Let $f:[a,b]\to\mathbb{R}$ an increasing, continuous, and not absolutely continuous function: I claim there exists a point $c\in[a,b]$ where the Dini derivative $D^\*f(c):=\limsup\_{x\to c}\frac{f(x)-f(c)}{x-c}$ is infinite. For the proof, we may assume w.l.o.g. that $f$ is *strictly* increasing (we may just consider $x+f(x)$).
By definition, since $f$ is not absolutely continuous, there exists a sequence of sets $J\_n\subset [a,b]$ such that each $J\_n$ is a finite union of intervals, $|J\_n|\to0$, and $|f(J\_n)|$ is bounded away from $0$.
Let $J^\*\_n\subset J\_n$ be the union of all components $I$ of $J\_n$ such that $|f(I)|\ge\frac12\frac {|f(J\_n)|}{|J\_n|}|I|$. Clearly, $|f(J\_n\setminus J\_n^\*)|\le\frac {|f(J\_n)|}2$, so the sets $f(J\_n^\*)$
have length $|f( J\_n^\*)|\ge\frac {|f(J\_n)|}2$ bounded away from $0$ too. This implies ($\dagger$) that some $p \in f([a,b])$ belongs to infinitely many $f( J\_n^\*)$. Therefore $c:=f^{-1}(p)$ belongs to infinitely many $J\_n^\*$; for these indices, let $[\alpha\_n,\beta\_n]$ be the component of $c$ in $J\_n$. So by construction both $\beta\_n$ and $\alpha\_n$ converge to $c$ along a subsequence; and w.r.to this subsequence, $\max\big\{\frac{f(\beta\_n)-f(c)}{\beta\_n-c},\frac{f(c)-f(\alpha\_n)}{c-\alpha\_n}\big\}\ge\frac{f(\beta\_n)-f(\alpha\_n)}{\beta\_n-\alpha\_n}=\frac{|f(I\_n)|}{|I\_n|}\to+\infty$, proving that $D^\*f(c)=+\infty$.
($\dagger$) It's the well-known argument from measure theory: For a finite measure space $(X,\mathcal{S},\mu)$, if for $n\in\mathbb N$ one has $Y\_n\in\mathcal{S}$ and $\mu(Y\_n)\ge \lambda$, then $\overline{Y}:=\limsup\_{n\to\infty}Y\_n:=\cap\_{k\in\mathbb N}\cup\_{n\ge k}Y\_n$ also has $\mu(\overline{Y})\ge \lambda$. Here $Y\_n:=f( J\_n^\*)$.
|
13
|
https://mathoverflow.net/users/6101
|
403041
| 165,360 |
https://mathoverflow.net/questions/403037
|
1
|
$\DeclareMathOperator\PSL{PSL}$Let $U\subset\mathbb C^2$ be an open set, $f:U\to \PSL(2,\mathbb C)$ a holomorphic map. If the image of $f$ is contained in $\operatorname{PSU}(2,\mathbb C)$, I guess that it can only be a constant map. Is it correct ?
|
https://mathoverflow.net/users/356774
|
Holomorphic map to Möbius group
|
Condition $U\subseteq \mathbb{C}^2$ can be replaced by $U\subseteq \mathbb{C}^1$ (just restrict your map
on a little disk in a complex line in $U$. Moreover, the restricted map lifts to a map into $SU(2)$. Then existence of a map to $SU(2)$
essentially means that you have two analytic functions satisfying
$|f|^2+|g|^2=1$. Such functions must be constant. To see this write
$f=u+iv,\; g=u\_1+iv\_1$, differentiate $u^2+v^2+u\_1^2+v\_1^2=1$ twice wrt $x$ and $y$, then add, and you you obtain that sum of squares of all first derivatives is $0$.
|
2
|
https://mathoverflow.net/users/25510
|
403042
| 165,361 |
https://mathoverflow.net/questions/403039
|
2
|
Consider the (complex) geometric realization of the motivic cohomology theory on simplicial presheaves over complex smooth schemes, which is a functorial homomorphism of $R$-modules, where $R$ is the coefficient ring:
\begin{equation}\label{eq}
\varphi: H^{s,t}\_{mot}(X;R)\to H^s(X(\mathbb{C});R).
\end{equation}
When restricted to the Chow ring $\varphi:H^{2\*,\*}\_{mot}(X;\mathbb{Z})\to H^{2\*}(X(\mathbb{C});\mathbb{Z})$ and $X$ a complex algebraic variety, it is a classical result that $\varphi$ is the cycle class map ring homomorphism.
Generally, is the map $\varphi: H^{\*,\*}\_{mot}(X;R)\to H^\*(X(\mathbb{C});R)$ an R-algebra homomorphism? I was not able to find a direct proof of this result in the literature. Or perhaps this is an obvious fact that I am missing.
|
https://mathoverflow.net/users/100553
|
The multiplicativity of the (complex) geometric realization of motivic cohomology
|
Complex realization sends the motivic Eilenberg-MacLane spectrum to the classical Eilenberg-MacLane spectrum [Theorem 5.5 in Marc Levine's "A comparison of motivic and classical stable homotopy theories]. It is also a symmetric monoidal functor, hence preserves ring spectra. The result follows.
|
4
|
https://mathoverflow.net/users/356850
|
403045
| 165,362 |
https://mathoverflow.net/questions/403043
|
3
|
I am currently reading through differential geometry as a mathematics graduate.
Can somebody give me a brief explainer on the purpose of connections?
I could also use explainers on differential forms. Lie derivatives, and the tangent bundle generally.
I understand that you can have these structures, I just don't understand why we want to. What information can we learn from these structures that we couldn't learn about the manifolds otherwise?
|
https://mathoverflow.net/users/356829
|
The purpose of connections in differential geometry
|
If you are interested in local-to-global results, i.e., collecting local info about the manifold and then patch it together to get a global info then you need tools for the patching part of the process. These often come under the guise of some form of integration.
Here is a simple example. If a compact surface admits a metric with negative curvature (local info) then it has nontrivial fundamental group (global info). This is a special case of Gauss-Bonnet that tells you more.
On the other hand, a theorem of Myers shows that if the Ricci curvature of a Riemann manifold is bounded below by a positive constant (local info) then the manifold is compact and its fundamental group is finite (global info).
This is a more sophisticated result and its proof uses the concept of connection.
The revolution in low dimensional topology brought by gauge theory relies in an essential fashion on the concept of connection. The instantons are simply connections with minimal "energy".
As far as Lie derivatives are concerned, I recommend Arnold's "*Mathematical Methods of Classical Mechanics*" to see a geometric interepration and appreciate some of their uses.
|
7
|
https://mathoverflow.net/users/20302
|
403057
| 165,370 |
https://mathoverflow.net/questions/397198
|
13
|
Let $G$ be a group acting highly transitively (and faithfully) on a set $S$. Suppose that $G$ is finitely presented, and that every stabilizer in $G$ of a finite subset of $S$ is finitely generated. I think I can prove that $G$ embeds in a finitely presented simple group, which in particular implies $G$ has decidable word problem, but I'd like a better understanding of why such a $G$ should have decidable word problem. Is there a pre-existing (and/or more direct) reason that a group admitting such an action should have decidable word problem?
(Edit: Here an action of a group $G$ on a set $S$ is called *highly transitive* if for all $n\in\mathbb{N}$ the induced action of $G$ on the set of $n$-tuples of distinct elements of $S$ is transitive.)
|
https://mathoverflow.net/users/164670
|
Decidability of word problem for group admitting certain action
|
Yes. There is such a reason.
I will write a subset of $G$ is RE if the set of those words over the generators for $G$ which represent elements of the subset is recursively enumerable.
As IJL argued, since $G$ is finitely presented the subset $\{1\}$ of $G$ containing only the identity is RE. It remains to show that $G \setminus \{1\}$ is RE.
Fix $s$ and $t$ in $S$ and let $H$ be the stabiliser of $s$. Since $H$ is finitely generated $H$ is RE.
Let $f$ be some element of $G$ which moves $s$ and fixes $t$.
Let $M$ be the set of elements of $G$ which conjugate $f$ into $H$. Note that $M$ is RE and $1 \notin M$ but any element $g$ of $G$ with $(t)g = s$ is in $M$.
Let $N$ be the set of elements of $G$ conjugate to some element of $M$. Note that $N$ is RE.
$G$ acts $2$-transitively on $S$ so $N$ is in the set of elements of $G$ which move at least one point of $S$. Which is to say that $N = G \setminus \{1\}$.
In short: $G \setminus \{1\} = \left\{ g \in G \mid \textrm{there exists } h \in G \textrm{ with } f^{\left(g^h\right)} \in H \right\}$.
|
7
|
https://mathoverflow.net/users/125391
|
403058
| 165,371 |
https://mathoverflow.net/questions/403053
|
1
|
I am working on a research paper where I need to investigate conditions for the existence of probability distributions satisfying certain characteristics. I have already asked a related question ([here](https://mathoverflow.net/questions/401905/conditions-for-existence-of-a-distribution-with-full-support)), whose answers allowed me to frame better in my mind the problems I'm facing. In what follows, I report a close, although different, question. I will highlight the key differences below.
Consider a $6\times 1$ random vector
$$
\eta\equiv (\eta\_1,\eta\_2,..., \eta\_6)
$$
satisfying the following property (hereafter, called Property 1):
**Property 1:** $$
\begin{pmatrix}
\eta\_1\\
\eta\_2\\
\eta\_3
\end{pmatrix} \sim \begin{pmatrix}
\eta\_4\\
-\eta\_2\\
\eta\_5
\end{pmatrix} \sim \begin{pmatrix}
\eta\_6\\
-\eta\_3\\
-\eta\_5
\end{pmatrix} \sim \begin{pmatrix}
-\eta\_1\\
-\eta\_4\\
-\eta\_6
\end{pmatrix} \sim G
$$
where "$\sim$" denotes "distributed as" and $G$ is an absolutely continuous distribution with full support on $\mathbb{R}^3$.
---
**Question A:** Let $\mathcal{G}$ denote the family of absolutely continuous distribution with full support on $\mathbb{R}^3$ and whose marginals are *symmetric around zero and identical*. For each $G\in \mathcal{G}$, does there exists a vector $\eta$ satisfying Property 1?
---
**Question B:** Let $\epsilon$ be a $4\times 1$ random vector
$$
\epsilon\equiv \begin{pmatrix}
\epsilon\_1\\
\epsilon\_2\\
\epsilon\_3\\
\epsilon\_0\\
\end{pmatrix} $$
For each $(G,\eta)$ satisfying Property 1, does there exist $\epsilon$ satisfying Property 2 described below?
**Property 2:**
$$
\begin{pmatrix}
1 & 0 & 0 & -1\\
1 & -1 & 0 & 0\\
1 & 0 & -1 & 0\\
0 & 1 & 0 & -1\\
0 & 1 & -1 & 0\\
0 & 0 & 1 & -1\\
\end{pmatrix}\*\epsilon=\begin{pmatrix}
\eta\_1\\
\eta\_2\\
\eta\_3\\
\eta\_4\\
\eta\_5\\
\eta\_6
\end{pmatrix}
$$
and the distribution $F$ of $\epsilon$ is absolutely continuous with full support on $\mathbb{R}^4$?
---
**My thoughts:**
I believe that the answer to Question A is "Yes": any distribution in $\mathcal{G}$ satisfies Property 1. Certainly, there exist distributions outside $\mathcal{G}$ that can also satisfy Property 1.
I believe that the answer to Question B is "Yes" as well. However, I'm not 100% sure and I would appreciate your help. The answers [here](https://mathoverflow.net/questions/401905/conditions-for-existence-of-a-distribution-with-full-support) suggest that: if $F$ is absolutely continuous with full support on $\mathbb{R}^4$, then $G$ is absolutely continuous with full support on $\mathbb{R}^3$. Here, however, I'm asking something different: if $G$ is absolutely continuous with full support on $\mathbb{R}^3$, can we always find a distribution $F$ for $\epsilon$ that is absolutely continuous with full support on $\mathbb{R}^4$?
|
https://mathoverflow.net/users/42412
|
From probability distribution in $\mathbb{R}^3$ to probability distribution in $\mathbb{R}^4$
|
I think the answer to both questions is negative.
**Question A.**
You ask whether for all $G\in\mathcal{G}$, there exist a random vector $\eta=(\eta\_i)\_{1\le i\le 6}$ such that they satisfy Condition 1 together.
If $G$ were not assumed to be fully supported, the answer would be easily seen to be negative: taking $G$ supported on $\{(x,y,z)\in\mathbb{R}^3 \mid x=y\}$ would force $\eta\_1=\eta\_2$, $\eta\_4=-\eta\_2$ and $-\eta\_1=-\eta\_4$ almost everywhere, which are incompatible.
But you can start from there and change $G$ slightly to be fully supported: simply start with any distribution supported on $\{(x,y,z)\in\mathbb{R}^3 \mid x=y\}$ with equal and centrally symmetric marginals; assume further that the marginals give a mass less than $1/5$ to $[-1,1]$ (**added in edit:** previously was $1/4$, but we need some room because of the perturbation). Define $G$ as a convolution of that distribution with a Gaussian $\sim\mathcal{N}(0,\varepsilon)$ for some small positive $\varepsilon$. Taking $\varepsilon$ small enough, you can ensure $G(\{(x,y,z)\in\mathbb{R}^3 \mid \lvert x-y\rvert>1/4\})<1/4$ and that any marginal of $G$ give a mass less than $1/5$ to $[-1,1]$. If a random vector $\eta$ were to satisfy condition 1 with $G$, you would have
\begin{align\*}
\mathbb{P}(\lvert \eta\_4\rvert\le 1) &<1/4 \\
\mathbb{P}(\lvert \eta\_1-\eta\_2\rvert>1/4) &<1/4 \\
\mathbb{P}(\lvert \eta\_4+\eta\_2\rvert>1/4) &<1/4 \\
\mathbb{P}(\lvert \eta\_1-\eta\_4\rvert>1/4) &<1/4 \\
\end{align\*}
With positive probability, we would thus have $\lvert \eta\_1-\eta\_2\rvert\le 1/4$, $\lvert \eta\_4+\eta\_2\rvert\le 1/4$, $\lvert \eta\_1-\eta\_4\rvert \le1/4$ and $\lvert \eta\_4\rvert > 1$. Now, this is impossible since
$$\lvert \eta\_4+\eta\_4\rvert \le \lvert \eta\_4-\eta\_1\rvert + \lvert\eta\_1-\eta\_2\rvert+\lvert\eta\_2+\eta\_4\rvert. $$
(I took more room than needed, but that does the trick.)
**Question B. (added in edit)**
Simply take $\eta$ a normal vector $\sim\mathcal{N}(0,I\_6)$. Then condition 1 holds with $G$ a normal distribution, but for any $\epsilon$ its image under the matrix is contained in its image vector space, which has dimension $4$ at most since it is a $6\times 4$ matrix. This cannot have a fully supported law, hence cannot equal $\eta$.
|
2
|
https://mathoverflow.net/users/4961
|
403059
| 165,372 |
https://mathoverflow.net/questions/403029
|
14
|
In the setting of complex Stein manifolds $X$ of complex dimension $d$, the theorem of Andreotti--Frankel implies the vanishing of the singular cohomology group $H^i(X,\mathbb Z)=0$ for $i>d$. With complex coefficients, a simple argument for this is to compute the cohomology in terms of the cohomology of the de Rham complex. Their theorem gives a more precise Morse-theoretic statement.
Now let $U\subset X$ be another Stein manifold open in $X$, and assume that the map $\mathcal O(X)\to \mathcal O(U)$ has dense image. (This condition is not automatic, and necessary for the following.)
>
> Conjecture. In top degree, the map $H^d(X,\mathbb Z)\to H^d(U,\mathbb Z)$ is surjective.
>
>
>
Is this known? The same result should also be true with constructible coefficients. With $\mathbb C$-coefficients, it follows from the comparison with de Rham cohomology (at least when cohomology groups are finite-dimensional, which I'm happy to assume). Is there some argument using Morse theory?
The motivation for the question is that the analogue in rigid-analytic geometry is true (but I found it quite surprising); it is essentially equivalent to a version of Artin vanishing for affine schemes over absolutely integrally closed valuation rings stated by Gabber in Oberwolfach last year.
|
https://mathoverflow.net/users/6074
|
Artin vanishing for Stein manifolds and restriction maps
|
The pairs (U,X) are called Runge pairs. The homology version of your statement is proved in the paper of
Andreotti and Narasimhan Annals of Math vol 76 no 3 (1962) 499-509 using Morse
Theory.The title of the paper is "A Topological property of Runge pairs"
The paper by Coltoiu Mihalache titled On the Homology Groups of Stein spaces and Runge pairs, Journal fur reine und angewandte Mathematik volume 371 no 5 pages 215-220 proves the homology statement for Runge pairs of Stein spaces.
|
13
|
https://mathoverflow.net/users/4696
|
403065
| 165,375 |
https://mathoverflow.net/questions/403047
|
7
|
Is there a homeomorphism $f:CP^2\to CP^2$ that has finitely many fixed points and acts by -1 on $H^2$?
|
https://mathoverflow.net/users/356859
|
Homeomorphism with finitely many fixed points acting by -1 on $H^2(CP^2)$
|
Let $a\_1$, $a\_2$ and $a\_3$ be three distinct positive real numbers. Take the map
$$[z\_1 : z\_2 : z\_3] \mapsto [a\_1 \overline{z\_1}: a\_2 \overline{z\_2} : a\_3 \overline{z\_3}]$$
where the bar is complex conjugation. I claim that the only fixed points are $[1:0:0]$, $[0:1:0]$ and $[0:0:1]$.
Suppose that to the contrary that $[z\_1:z\_2:z\_3]$ is fixed and that $z\_i z\_j \neq 0$ for some indices $i \neq j$. Then there is a scalar $\lambda$ such that
$$\lambda = \frac{a\_i \overline{z\_i}}{z\_i} = \frac{a\_j \overline{z\_j}}{z\_j}.$$
But then we have
$$|\lambda| = a\_i = a\_j,$$
contradicting that the $a$'s are distinct.
|
12
|
https://mathoverflow.net/users/297
|
403066
| 165,376 |
https://mathoverflow.net/questions/403074
|
2
|
Let $H$ and $H'$ be two Hopf algebras, and let $\phi:H \to H'$ be an bialgebra map. Then is $\phi$ automatically a Hopf algebra map?
|
https://mathoverflow.net/users/352001
|
Bialgebra maps and Hopf algebra maps
|
Yes it is.
It is actually a standard result, for any hopf algebra, that under the circumstances you are describing any bialgebra map $\phi$ commutes with the antipode, i.e. we have: $$S\_{H'}\circ \phi=\phi\circ S\_H.$$
(The simplest way to show this is to consider $\operatorname{Hom}(H,H')$ as an algebra —under the convolution product— and to show that $\phi$ is convolution invertible with left inverse $S\_{H'}\circ \phi$ and right inverse $\phi\circ S\_H$).
|
3
|
https://mathoverflow.net/users/85967
|
403075
| 165,378 |
https://mathoverflow.net/questions/402681
|
0
|
I am studying properties of the **two-parameter** Mittag-Leffler function.
$$ E\_{\alpha,\beta}(z)=\sum\_{k=0}^\infty \dfrac{z^k}{\Gamma(\alpha k+\beta)}.$$
I am particularly interested in recurrences and relations, such as the duplication formulas.
However, I am seeking some relation in which **a product of two two-parameter Mittag-Leffler functions is a Mittag-Leffler function**, does anyone know something about this? Maybe about powers, but not necessarily.
|
https://mathoverflow.net/users/172600
|
When is a product of two two-parameter Mittag-Leffler functions a Mittag-Leffler function?
|
The product of two MLF is another MLF, only when $\alpha = \beta =1.$ In the other cases we obtain a similar formula but have to introduce generalized binomial coefficients. To show it, it is enough to use the rule used for multiplication of power series. The coefficients of the product are the discrete convolution of the coefficients of both series.
|
2
|
https://mathoverflow.net/users/345734
|
403079
| 165,379 |
https://mathoverflow.net/questions/402842
|
24
|
Let $P$ be a (convex, bounded) polytope with the following property: for every vertex $v$, there are exactly two facets which do not contain $v$. Does it follow that $P$ is (combinatorially) a Cartesian product of two simplices?
Some remarks
* by "facet" I mean "face of codimension $1$"
* a Cartesian product of two simplices has the desired property
* this is trivially true for $\dim P = 2$
* this is true for $\dim P = 3$: by playing with the Euler formula and the list of polyhedra with small $f$-vectors, one checks that only the triangular prism satisfies the condition
* it is a classical fact that a $n$-dimensional **simple** polytope with $n+2$ facets is a Cartesian product of two simplices. So the answer is positive if $P$ is assumed to be simple.
|
https://mathoverflow.net/users/908
|
Polytope where each vertex belongs to all but two facets
|
There are other polytopes with this property that can be obtained via the *free join* construction.
Given two polytopes $P\_1\subset\Bbb R^{d\_1}$ and $P\_2\subset\Bbb R^{d\_2}$, the free join $P\_1\bowtie P\_2$ is obtained by embedding $P\_1$ and $P\_2$ into skew affine subspaces of $\Bbb R^{d\_1+d\_2+1}$ and taking the convex hull.
The claim is that if $P\_1$ and $P\_2$ have your property, then so does $P\_1\bowtie P\_2$.
To see this, note that the facets of $P\_1\bowtie P\_2$ are exactly of the form $P\_1\bowtie f\_2$ and $f\_1\bowtie P\_2$, where $f\_i$ is a facet of $P\_i$. Now, if $v$ is a vertex in, say, $P\_1\subset P\_1\bowtie P\_2$ that is in all facets of $P\_1$ except for $f,f'\subset P\_1$, then $v$ is in all facets of $P\_1\bowtie P\_2$ except for $f\bowtie P\_2$ and $f'\bowtie P\_2$. Equivalently for vertices in $P\_2$.
**Example.** Take the free join of two squares, which is a 5-dimensional polytope with 8 vertices, 8 facets and vertex degree 6. This polytope is not simple and can thus not be the cartesian product of two simplices.
---
This construction yields counterexamples in dimension $d\ge 5$.
Concerning $d=4$, there should not be any counterexamples in this dimension.
Suppose that $P\subset\Bbb R^4$ is a counterexample.
According to [Moritz Firsching's comment](https://mathoverflow.net/questions/402842/polytope-where-each-vertex-belongs-to-all-but-two-facets/403080?noredirect=1#comment1030725_402842) we can assume that it has $n\ge 10$ facets. Each vertex is then contained in exactly $n-2\ge 8$ of them. More generally, each set of $k$ vertices is contained in at least $n-2k$ facets. But let $f\subset P$ be a 2-face of $P$ and $v\_1,v\_2,v\_3\in f$ three affinely independent vertices, then this set of vertices, and thus also $f$, is contained in $n-6\ge 4$ facets. This cannot be, since each 2-face of $P$ is contained in exactly two facets.
I believe this generalizes to show that any polytope with your property must have $\le 2d$ facets.
|
18
|
https://mathoverflow.net/users/108884
|
403080
| 165,380 |
https://mathoverflow.net/questions/403023
|
7
|
Setting
-------
Consider two independent orthogonal matrices, which are decomposed into 4 blocks:
\begin{align}
Q\_{1}
=
\left[\begin{array}{cc}
A\_{1} & B\_{1}\\
C\_{1} & D\_{1}
\end{array}\right]
,
\,Q\_{2}=\left[\begin{array}{cc}
A\_{2} & B\_{2}\\
C\_{2} & D\_{2}
\end{array}\right]\in \mathbb{R}^{d\times d}
\end{align}
where $A\_i \in \mathbb{R}^{r\times r}$ and $D\_i \in \mathbb{C}^{(d-r)\times (d-r)}$ and $d\ge 3r$.
Also notice that
\begin{align}
\forall i=1,2:I\_{d}=\left[\begin{array}{cc}
I\_{r} & 0\\
0 & I\_{d-r}
\end{array}\right]
&=
Q\_{i}^{\top}Q\_{i}=\left[\begin{array}{cc}
A\_{i}^{\top}A\_{i}+C\_{i}^{\top}C\_{i} & A\_{i}^{\top}B\_{i}+C\_{i}^{\top}D\_{i}\\
B\_{i}^{\top}A\_{i}+D\_{i}^{\top}C\_{i} & B\_{i}^{\top}B\_{i}+D\_{i}^{\top}D\_{i}
\end{array}\right]
\\
&=Q\_{i}Q\_{i}^{\top}=\left[\begin{array}{cc}
A\_{i}A\_{i}^{\top}+B\_{i}B\_{i}^{\top} & A\_{i}C\_{i}^{\top}+B\_{i}D\_{i}^{\top}\\
C\_{i}A\_{i}^{\top}+D\_{i}B\_{i}^{\top} & C\_{i}C\_{i}^{\top}+D\_{i}D\_{i}^{\top}
\end{array}\right]
\end{align}
Goal
----
I am trying for a few days to prove (or refute) the following lemma.
In my numerical simulations, it always holds for random $Q$s.
**Prove (or refute)** that if $v\in\mathbb{C}^{d-r}$ is an eigenvector
of $D\_1^\top M$
(with $\lambda=1$), i.e.,
\begin{align}
D\_{1}^{\top}
\underbrace{D\_{2}\left(B\_{2}^{\top}B\_{1}+D\_{2}^{\top}D\_{1}\right)}\_{\triangleq M}v&=v
\end{align}
then it is in the nullspace of $C\_2^\top M$, i.e.,
\begin{align}
C\_{2}^{\top}
\underbrace{D\_{2}\left(B\_{2}^{\top}B\_{1}+D\_{2}^{\top}D\_{1}\right)}\_{=M}v&=0\_{r}
\end{align}
---
Update: Alternative goal
------------------------
If one could show that any eigenvector such that
$D\_{1}^{\top}
D\_{2}\left(B\_{2}^{\top}B\_{1}+D\_{2}^{\top}D\_{1}\right)v=v$
is also in the null space of $B\_1$, i.e., $B\_1 v = 0\_r$, then I know how to prove the rest of the lemma above using it. And again, empirically, this also seems to always hold.
---
I feel like I may be missing some simple trick here.
Any help would be greatly appreciated.
|
https://mathoverflow.net/users/100796
|
Proving a lemma for a decomposition of orthogonal matrices
|
Using the direction Federico Poloni gave, I was finally able to prove it.
Notice that the matrices are real, so I use $A^\top$ and $A^H$ interchangeably.
---
Let $v\in\mathbb{\mathbb{C}}^{d-r}$ be a normalized eigenvector as required.
* **Lemma:** The norm of $v$ is preserved under the following transformations $\left\Vert \left(B\_{2}^{T}B\_{1}+D\_{2}^{T}D\_{1}\right)v\right\Vert \_{2}=\left\Vert D\_{2}^{\top}D\_{1}v\right\Vert \_{2}=\left\Vert D\_{1}v\right\Vert \_{2}$.
**Proof.** We use a simple trick using the Euclidean (operator) norms,
\begin{align}
v&=D\_{1}^{\top}D\_{2}\left(B\_{2}^{\top}B\_{1}+D\_{2}^{\top}D\_{1}\right)v\\v^{H}v&=v^{H}D\_{1}^{H}D\_{2}\left(B\_{2}^{H}B\_{1}+D\_{2}^{H}D\_{1}\right)v\\1&=\left\Vert v^{H}D\_{1}^{H}D\_{2}\left(B\_{2}^{H}B\_{1}+D\_{2}^{H}D\_{1}\right)v\right\Vert \_{2}\\&\le\underbrace{\left\Vert v^{H}D\_{1}^{H}D\_{2}\right\Vert \_{2}}\_{\le1}\underbrace{\left\Vert \left(B\_{2}^{T}B\_{1}+D\_{2}^{T}D\_{1}\right)v\right\Vert \_{2}}\_{\le1}\\&\le\underbrace{\left\Vert v^{H}D\_{1}^{H}\right\Vert \_{2}}\_{\le1}\underbrace{\left\Vert D\_{2}\right\Vert \_{2}}\_{\le1}\underbrace{\left\Vert \left(B\_{2}^{T}B\_{1}+D\_{2}^{T}D\_{1}\right)v\right\Vert \_{2}}\_{\le1}\le1,
\end{align}
where we used the fact that $\left(B\_{2}^{T}B\_{1}+D\_{2}^{T}D\_{1}\right)$ is the bottom-right block of the orthogonal matrix $Q\_{2}^{\top}Q\_{1}$.
As a conclusion, we get the following equalities:
\begin{align}
\left\Vert \left(B\_{2}^{T}B\_{1}+D\_{2}^{T}D\_{1}\right)v\right\Vert \_{2}&=1\\\left\Vert v^{H}D\_{1}^{H}D\_{2}\right\Vert \_{2}=\left\Vert D\_{2}^{H}D\_{1}v\right\Vert \_{2}=\left\Vert D\_{2}^{\top}D\_{1}v\right\Vert \_{2}&=1\\\left\Vert v^{H}D\_{1}^{H}\right\Vert \_{2}=\left\Vert D\_{1}v\right\Vert \_{2}&=1.
\end{align}
* **Lemma:** The vector $v$ is in the null space of $B\_{2}^{\top}B\_{1}$, i.e., $B\_{2}^{\top}B\_{1}v=0\_{r}$.
**Proof.** We get back to the equality above and show
\begin{align}1&=v^{H}D\_{1}^{H}D\_{2}\left(B\_{2}^{H}B\_{1}+D\_{2}^{H}D\_{1}\right)v\\&=v^{H}D\_{1}^{H}D\_{2}B\_{2}^{H}B\_{1}v+v^{H}D\_{1}^{H}D\_{2}D\_{2}^{H}D\_{1}v\\&=v^{H}D\_{1}^{H}D\_{2}B\_{2}^{H}B\_{1}v+\underbrace{\left\Vert D\_{2}^{H}D\_{1}v\right\Vert \_{2}}\_{=1}\\0&=v^{H}D\_{1}^{H}D\_{2}B\_{2}^{H}B\_{1}v.
\end{align}
We use the fact that $\left(B\_{2}^{T}B\_{1}+D\_{2}^{T}D\_{1}\right)v$ is a normalized vector to obtain
\begin{align}
1&=\left\Vert \left(B\_{2}^{\top}B\_{1}+D\_{2}^{\top}D\_{1}\right)v\right\Vert \_{2}^{2}=\left\Vert B\_{2}^{\top}B\_{1}v\right\Vert \_{2}^{2}+\underbrace{2v^{\top}D\_{1}^{\top}D\_{2}B\_{2}^{\top}B\_{1}v}\_{=0}+\underbrace{\left\Vert D\_{2}^{\top}D\_{1}v\right\Vert \_{2}^{2}}\_{=1}\\0&=\left\Vert B\_{2}^{\top}B\_{1}v\right\Vert \_{2}^{2}\\0\_{r}&=B\_{2}^{\top}B\_{1}v.
\end{align}
* **Lemma:** The vector $v$ is in the null space of $C\_{2}^{\top}D\_{1}$.
**Proof.** We use the above results,
\begin{align}
1&=v^{H}D\_{1}^{H}D\_{2}D\_{2}^{H}D\_{1}v=v^{H}D\_{1}^{H}\left(I\_{d-r}-C\_{2}C\_{2}^{H}\right)D\_{1}v
\\&=\underbrace{v^{H}D\_{1}^{H}D\_{1}v}\_{=1}-v^{\top}D\_{1}^{H}C\_{2}C\_{2}^{H}D\_{1}v\\0&=\left\Vert C\_{2}^{\top}D\_{1}v\right\Vert \_{2}^{2}\\0\_{r}&=C\_{2}^{\top}D\_{1}v.
\end{align}
---
Finally, we are ready to conclude the theorem we wanted to prove.
\begin{align}
C\_{2}^{\top}D\_{2}\left(B\_{2}^{\top}B\_{1}+D\_{2}^{\top}D\_{1}\right)v
&=C\_{2}^{\top}D\_{2}D\_{2}^{\top}D\_{1}v
\\&=C\_{2}^{\top}\left(I\_{d-r}-C\_{2}C\_{2}^{T}\right)D\_{1}v
\\
&=\underbrace{C\_{2}^{\top}D\_{1}v}\_{=0\_{r}}-C\_{2}^{\top}C\_{2}\underbrace{C\_{2}^{\top}D\_{1}v}\_{=0\_{r}}=0\_{r}
\end{align}
|
3
|
https://mathoverflow.net/users/100796
|
403096
| 165,385 |
https://mathoverflow.net/questions/403099
|
10
|
In Volume 1 of "Classical Banach Spaces" Lindenstrauss and Tzafriri note that all surjective linear isometries on $\ell\_\infty$ are of the from $(a\_i) \mapsto (\varepsilon\_i a\_{\pi(i)})$ where $\pi$ is a permutation of $\mathbb{N}$ and $\varepsilon\_i \in\{\pm 1\}$. They mention that the proof is a consequence of the more general result on isometries on $C(K)$ spaces that it will appear in Volume 3 (which does not exist). Does anyone know a reference for the proof of this fact?
|
https://mathoverflow.net/users/15388
|
Surjective linear isometries on $\ell_\infty(\mathbb{N})$
|
I don't know a reference (other users will probably provide one). Here's a (quite immediate) proof anyway. I'll replace $\mathbf{N}$ with an arbitrary set $X$ since the integers play no role.
**Fact** The extremal points of the closed 1-ball of $\ell^\infty(X)$ precisely consists of the maps $X\to\{-1,1\}$.
Proof: Given $f,g,u:X\to [-1,1]$ if $u\in [f,g]$ and for some $x\in X$, $u(x)\in\{-1,1\}$, then we see that $f(x)=g(x)=u(x)$. Hence, all $u:X\to\{-1,1\}$ are extremal. Conversely, let $u$ not be of this form. So $-1<u(y)<1$ for some $y$. For a scalar $\varepsilon$, define $u\_\pm(y)=u(y)\pm\varepsilon$ and $u\_\pm(x)=u(x)$ for $x\neq y$. Then $u=\frac12(u\_++u\_-)$ and for $\varepsilon>0$ small enough, both $u\_\pm$ belong to the closed 1-ball. So $u$ is not extremal.
**Corollary:** every bijective self-isometry of $\ell^\infty$ has the required form (call this "standard", and "diagonal-standard" when the permutation is trivial).
Proof: Write $W\_A=1\_A-1\_{A^c}$. Then, by the fact, the isometry group acts on the power set $2^X$ by $g\cdot W\_A=W\_{gA}$.
Given an isometry, after composing with a diagonal-standard isometry, we can suppose that $f(1\_X)=1\_X$, i.e., $f\in G$, where $G$ is the stabilizer of $1\_X$. We have to show that $G$ is reduced to the permutation group.
Now for $A,B,C\subseteq X$, write $F(A,B,C)$ if $W\_A+W\_B-W\_C=1\_X$; since $G$ acts linearly and fixes $1\_X$, this ternary relation is preserved by $G$. Also it is easy to check that $F(A,B,C)$ holds if and only if $C=A\cap B$. Hence, we see that $G$ (acting on $2^X$ by $f(W\_A)=W\_{gA})$ preserves the intersection operation. In particular, it preserves the ordering of $2^X$, and hence is an automorphism of the Boolean algebra $2^X$. Such an automorphism is necessarily induced by a permutation.
Now for $f\in G$, composing with a permutation (which is standard), we can suppose that $f$ pointwise fixes all maps $X\to\{-1,1\}$. By a $\mathbf{Q}$-linear combination, it therefore preserves all characteristic functions (=maps $X\to\{0,1\}$), and in turn, it therefore preserves all simple functions (= maps $X\to\mathbf{R}$ with finite image). Since these form a dense subspace of $\ell^\infty(X)$, the resulting isometry is the identity, which means that the original isometry is standard.
---
Note: an analogous result also holds for all $p\in [1,\infty]\smallsetminus\{2\}$. The case $1$ is similarly easy, but in the other cases this easy argument doesn't work since then the all sphere consists of extremal points.
---
As pointed out by @GiorgioMetafune the initial argument was incorrect and some combinatorial argument was needed.
|
14
|
https://mathoverflow.net/users/14094
|
403100
| 165,386 |
https://mathoverflow.net/questions/403102
|
3
|
Let $M^3$ be a compact $3$-manifold such that $\pi\_1(M)$ contains a normal subgroup isomorphic to $\mathbb Z$.
Can we show either $\pi\_1(M)$ is torsion-free or $\pi\_1(M)=\mathbb Z \oplus \mathbb Z\_2$ or $\mathbb Z\_2 \* \mathbb Z\_2$?
|
https://mathoverflow.net/users/280895
|
$\pi_1(M^3)$ containing a normal infinite cyclic subgroup
|
The answer is "no" (see below for an example) but it is almost "yes". If $M$ does not have any real projective plane boundary components then this follows from Theorem 7 of the paper [A survey on Seifert fibre space conjecture](https://arxiv.org/abs/1202.4142) by Jean-Philippe Préaux.
In general, one has to understand the "Seifert spaces modulo $\mathbb{P}$" introduced by Heil and Whitten. (This is the first time I've had to think about these - life is simpler when we assume orientability!) See the above survey paper for references.
Here is the promised example. Consider the three torus $T = \mathbb{R}^3 / \mathbb{Z}^3$. There is a $\mathbb{Z}\_2$ action via the "antipodal map" $\tau$ that acts as negation on all coordinates. Note that the fixed point set of $\tau$ is $P = \{(0,0,0), (1/2, 0, 0), \ldots, (1/2, 1/2, 1/2)\}$. Let $T' = T / \tau$. So $T'$ is not a three-manifold, due to the orbifold points at $P'$, the image of $P$. If we remove small neighbourhoods of all of the points of $P'$ we obtain a three manifold $T''$ which has fundamental group $\mathbb{Z}^3 \rtimes \mathbb{Z}\_2$. So (sadly), the fundmental group is neither torsion free nor one of the desired groups.
|
6
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https://mathoverflow.net/users/1650
|
403105
| 165,388 |
https://mathoverflow.net/questions/403073
|
5
|
I'm looking for the following:
(1) an example of a $\Pi^1\_1$ class-theoretic sentence that has no known equivalence to a $\Sigma^1\_1$ sentence, even if large cardinal hypotheses or reflection principles are assumed.
(2) an example of a $\Pi^1\_2$ class-theoretic sentence that has no known equivalence to a $\Sigma^1\_2$ sentence, even if large cardinal hypotheses or reflection principles are assumed.
Admttedly, "large cardinal hypothesis" and "reflection principle" are not well-defined, so these questions are rather vague.
I am assuming impredicative Comprehension (Kelley-Morse class theory) and all forms of choice.
The best answer I currently have for (1) is Vopenka's principle ("every proper class of directed graphs has either a member with a nontrivial endomorphism or two distinct members with a homomorphism between them"). But this is often considered a large cardinal hypothesis, so it's not a good answer.
I don't have any answer for (2).
|
https://mathoverflow.net/users/170446
|
Class-theoretic sentences that are $\Pi^1_1$ or $\Pi^1_2$
|
Much as described by @ElliotGlazer above, let's consider the question for models of the form $(V\_\kappa,V\_{\kappa+1})$, assuming ZFC in the background, considering all such where $\kappa$ is inaccessible in the background.
Then *with respect to this particular class of models*, a $\Pi^1\_2$, non-$\Sigma^1\_2$ statement is "There is a wellorder of the proper classes, which $\Pi^1\_1$-definable from some set parameter". (However, the proof that it is not $\Sigma^1\_2$ breaks if we replace "inaccessible" with "weakly compact", and I don't know what the situation is if we do this.)
For $\Pi^1\_2$ it is a direct computation: First, note that set-quantifiers are absorbed by class quantifiers, so we can say the "There is a $\Pi^1\_1$ formula $\varphi$ and set parameter $p$" at the front. Then just say that $\varphi(p,X,Y)$ defines a strict linear order $<^\*$ of the classes, and for every class $C$ coding a sequence $\left<C\_n\right>\_{n<\omega}$ of classes, there is $n<\omega$ such that $\neg(C\_{n+1}<^\*C\_n)$.
If $V=L$ is the background universe, there is such a wellorder of $V\_{\kappa+1}$. For given $X,Y\subseteq V\_\kappa$, just say "For every set $Z\subseteq V\_\kappa$ coding a model $M$ satisfying ZF$^-$ + "$\kappa$ is the largest cardinal" with $V\_\kappa\cup\{X,Y\}\subseteq M$, we have $M\models$"$X<\_LY$". (We automatically get wellfoundedness of $M$, using the inaccessibility of $\kappa$ in $L$.) This generalizes to when $V$ is a short extender premouse which is iterable in some larger universe, so it is consistent with many Woodin cardinals.
Now let us show that the statement is not equivalent to a $\Sigma^1\_2$ (uniformly in the context mentioned above). For suppose it is, with $\Sigma^1\_2$ formula $\psi$. Force over the background universe to add a Cohen subset $X\subseteq\kappa$ (i.e. with conditions of size ${<\kappa}$).
Then $V\_\kappa^{V[X]}=V\_\kappa^V$, and if $A\in V\_{\kappa+1}$ and $\varphi$ is $\Pi^1\_1$
and $(V\_\kappa,V\_{\kappa+1})\models\varphi(A)$, then $(V\_\kappa,V\_{\kappa+1}^{V[X]})\models\varphi(A)$. (This is a standard fact: letting $\neg\varphi(A)$ be $\exists B\subseteq V\_\kappa\ [\varrho(B,A)]$ where $\varrho$ is $\Sigma^1\_0$,
and $\tau$ be a name for a witness, working in $V$, where the forcing is $\kappa$-closed, we can construct a filter $G$ which meets enough sets that $B=\tau\_G\in V$ also witnesses the statement, a contradiction.)
Therefore $\Sigma^1\_2$ truth also goes up to $V[X]$, so
$(V\_\kappa,V\_{\kappa+1}^{V[X]})\models\psi$. So we have a wellorder of $V\_{\kappa+1}^{V[X]}$ which is definable over $V[X]$ from a parameter in $p\in V\_\kappa\subseteq V$, but then by homogeneity of the forcing, we get $X\in V$, a contradiction.
Remark (correcting remark in earlier version): If we started with $V=L$, then $\kappa$ is not weakly compact in $V[X]$.
---
(Edit): For $\Pi^1\_1$: Consider the statement $\psi$, which says "There are stationarily many ordinals $\alpha$ such that $2^{\alpha^+}>\alpha^{++}$". This is $\Pi^1\_1$, but I claim that if a Mahlo cardinal is consistent then it is not $\Sigma^1\_1$ w.r.t. inaccessibility, i.e. in the sense above. (But like in the $\Pi^1\_2$ case, I don't know about w.r.t. higher large cardinal properties for $\kappa$.)
(Remark: An earlier version had a gap in the argument regarding
the value of $2^{\alpha^+}$ in $L[G]$ for singular cardinals $\alpha$.
It is filled in now.)
For suppose $\kappa$ is Mahlo, hence Mahlo in $L$. Force over $L$ to arrange that in $L[G]$, $\kappa$ is Mahlo, $2^{\alpha}\leq\alpha^{++}$ for all $\alpha<\kappa$, the set $T\_{++}$ of inaccessibles $\alpha<\kappa$ such that $2^{\alpha^+}=\alpha^{++}$ is stationary, and the set $T\_{+++}$ of inaccessibles $\alpha<\kappa$ such that $2^{\alpha^+}=\alpha^{+++}$ is also stationary, and $2^{\alpha^+}=\alpha^{++}$ for singular cardinals $\alpha<\kappa$. For this, first partition the inaccessibles $\alpha<\kappa$ into disjoint stationary sets $T\_{++},T\_{+++}$. Then force with Easton support product $\mathbb{P}$ of forcings $\mathbb{P}\_\alpha$ for $\alpha\in T\_{+++}$, where $\mathbb{P}\_\alpha$ adds $\alpha^{+++}$ many Cohen subsets of $\alpha^+$, with conditions of size $\alpha$, in the usual way.
Claim: $\mathbb{P}$ preserves all cardinals and cofinalities,
and in $L[G]$, we have $2^{\alpha^+}=\alpha^{++}$ for all singular
cardinals $\alpha$ and all $\alpha\in T\_{++}$, and $2^{\alpha^+}=\alpha^{+++}$ for all $\alpha\in T\_{+++}$; therefore $L$ and $L[G]$
also have the same inaccessible cardinals $<\kappa$. Also, $\mathbb{P}$ is $\kappa$-cc, and then it follows that for every club $C\subseteq\kappa$
with $C\in L[G]$, there is a sub-club $D\subseteq C$ with $D\in L$,
and therefore $\mathbb{P}$ preserves stationarity for subsets of $\kappa$, and $\kappa$ is Mahlo in $L[G]$.
Proof: These are standard calculations, but here we go: For $\beta<\kappa$, write $\mathbb{P}\upharpoonright\beta$ for the restriction
of the product to indices $\alpha\in T\_{+++}\cap\beta$, and $G\upharpoonright\beta$ for the corresponding generic, and likewise $\mathbb{P}\upharpoonright[\beta,\kappa)$ to indices $\alpha\in T\_{+++}\cap[\beta,\kappa)$, etc.
So $\mathbb{P}\cong(\mathbb{P}\upharpoonright\beta)\times(\mathbb{P}\upharpoonright[\beta,\kappa))$. Note that $\mathbb{P}\upharpoonright[\beta,\kappa)$ is $(\beta+1)$-closed, so does not change $\mathcal{H}\_{\beta^+}$, and in particular does not collapse $\beta$ (or $\beta^+$). Now $L[G]=L[G\upharpoonright[\beta,\kappa)][G\upharpoonright\beta]$. Suppose $\beta$ is regular in $L$. We want to see that
$\beta$ is regular in $L[G]$. If $\beta$ is inaccessible in $L$
then as the product is Easton and by GCH, $\mathbb{P}\upharpoonright\beta$ has cardinality $\beta$ in $L$ and is $\beta$-cc in $L$, and hence also has these properties in $L[G\upharpoonright[\beta,\kappa)]$ (since
these properties only depend on $\mathcal{H}\_{\beta^+}$),
and therefore $\beta$ is still regular in $L[G]$. So suppose $\beta=\gamma^+$ where $\gamma$ is an $L$-cardinal. We may assume that $\mathbb{P}\upharpoonright\beta$ has cardinality $\geq\beta$ in $L$. If there is an $L$-inaccessible $\delta$ such that $\beta<(\delta^{+\omega})^L$ just
factor at $\delta$, using that $\mathbb{P}\upharpoonright\delta$ has cardinality $\delta$ in $L$ and $\Delta$-system calculations to see $\beta$ remains regular. Otherwise, we get that $\gamma$ is a singular limit of inaccessibles in $L$ and $\beta=\gamma^{+L}$ is the cardinality of $\mathbb{P}\upharpoonright\gamma$ in $L$. We may assume that $\beta$ is the least $L$-regular such that $\xi=\mathrm{cof}^{L[G]}(\beta)<\beta$. So $\xi$ is an $L$-regular and $\xi<\gamma<\beta$. But now we can factor $\mathbb{P}$ into $(\mathbb{P}\upharpoonright\xi)\times(\mathbb{P}\upharpoonright[\xi,\kappa))$. Since $\mathbb{P}\upharpoonright[\xi,\kappa)$ is $(\xi+1)$-closed in $L$,
we get $L[G\upharpoonright[\xi,\kappa)]\models$"$\mathrm{cof}(\beta)\neq\xi$",
and also by the minimality of $\beta$, all $L$-regulars $<\gamma$
are still regular in $L[G]$, hence also in $L[G\upharpoonright[\xi,\kappa)]$, and all $L$-inaccessibles $<\gamma$
are still inaccessible in both models, so $\mathbb{P}\upharpoonright\xi$
has cardinality $<\beta$ in $L$ and in $L[G\upharpoonright[\xi,\kappa)]$.
So if $\beta$ is regular in $L[G\upharpoonright[\xi,\kappa)]$,
then it is still regular in $L[G]$, a contradiction.
So there must be $\beta'\in(\xi,\beta)$ such that $\beta$ has cofinality
$\beta'$ in $L[G\upharpoonright[\xi,\kappa)]$. But then $\beta'$
is regular in $L$ and $\mathrm{cof}^{L[G]}(\beta')=\mathrm{cof}^{L[G]}(\beta)=\xi<\beta'$, contradicting the minimality of $\beta$.
So we have preservation of cardinals, cofinalities, and inaccessibles $<\kappa$; also the fact that $\mathbb{P}$ is $\kappa$-cc is as above,
so $\kappa$ remains inaccessible. It follows immediately
that for $\alpha\in T\_{+++}$, $L[G]\models 2^{\alpha^+}\geq\alpha^{+++}$.
To see $L[G]\models$"$2^{\alpha^+}\leq\alpha^{+++}$" for all $\alpha<\kappa$,
and $L[G]\models$"$2^{\alpha^+}=\alpha^{++}$" for $\alpha\in T\_{++}\cup S$,
where $S$ is the set of singular cardinals $<\kappa$,
use cardinality calculations and factoring as above.
(If $\gamma\in T\_{++}\cup S$, then
$\mathbb{P}\upharpoonright\gamma$ has cardinality $\leq\gamma^{+L}$ in $L$, and since $\gamma\notin T\_{+++}$, factoring at $\gamma$ therefore does the job.)
So $V\_{\kappa+1}^{L[G]}\models\psi$. Now suppose $\psi$ is equivalent to a $\Sigma^1\_1$ statement $\varphi$ w.r.t. inaccessibility,
and $\varphi=\exists A\subseteq V\_\kappa\ \varrho(A)$, where $\varrho$ has only set quantifiers.
Since $L[G]\models\psi$ + "$\kappa$ is Mahlo" (hence inaccessible),
we get $V\_{\kappa+1}^{L[G]}\models\varphi$, so fix a witness $A\in L[G]$.
Now force over $L[G]$ to add a club subset of $T\_{++}\cup S$, with conditions $p$ being closed
subsets of $T\_{++}\cup S$, ordered by extension. Note $T\_{+++}$ is disjoint from $T\_{++}\cup S$. It is shown in Jech/Woodin "Saturation of the closed unbounded filter
on the set of regular cardinals" that this forcing is $\kappa$-distributive, so adds no $<\kappa$-sequences (the proof is easy: for each regular cardinal $\lambda<\kappa$, there is a dense
subset of the forcing which is $\lambda$-closed: consider the set of conditions $p$ with $\sup p>\lambda$). Let $C$ be the generic club.
Then we get $V\_\kappa^{L[G,C]}=V\_\kappa^{L[G]}$ and $\kappa$ is inaccessible in $L[G,C]$. However, because $C\subseteq T\_{++}\cup S$,
and since $V\_\kappa$ was preserved, we get $L[G,C]\models$"$2^{\alpha^+}=\alpha^{++}$" for all $\alpha\in C$, and therefore $L[G,C]\models\neg\psi$.
But also because $V\_\kappa$ was preserved,
$\Sigma^1\_1$ truth passes upward from $L[G]$ to $L[G,C]$,
so $L[G,C]\models\varphi$. Since $\kappa$ is inaccessible in $L[G,C]$,
this is a contradiction.
|
4
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https://mathoverflow.net/users/160347
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403108
| 165,389 |
https://mathoverflow.net/questions/403110
|
0
|
Is there a lower bound on the slope (i.e. ratio of degree to rank) of normal bundles of smooth projective curves embedded in smooth projective varieties?
|
https://mathoverflow.net/users/357747
|
Are the slopes of normal bundles of curves bounded from below?
|
No. For instance the Hirzebruch surface $F\_n$ contains a smooth rational curve with normal bundle $\mathcal{O}(-n)$ of slope $-n$.
|
3
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https://mathoverflow.net/users/4428
|
403113
| 165,392 |
https://mathoverflow.net/questions/403118
|
26
|
**Question 1:** Let $C$ be a small category. Does there exist a poset $P$, a set of $W$ of morphisms in $P$, and an equivalence $P[W^{-1}] \simeq C$?
Here $P[W^{-1}]$ is the universal category receiving a functor from $P$ which carries each morphism of $W$ to an isomorphism.
I'm hoping for an affirmative answer. I'm also interested in the following variation:
**Question 2:** Let $C$ be a small category with finite colimits. Does there exist a join-semilattice $P$, a set $W$ of morphisms in $P$, and an equivalence $P[W^{-1}] \simeq C$?
**Version control:**
* There are actually two versions of question 2 -- in one version we require that $P \to P[W^{-1}]$ preserves finite colimits, and in the other we don't. As I'm hoping for an affirmative answer, it should be *easier* to do this without requiring the preservation of the finite colimits, and I'd be happy with an answer to that version.
* On top of that, I am interested in two versions of these questions: the 1-categorical version and the $\infty$-categorical version (the term "poset" means the same thing in both versions).
**Other Notes:**
* I'm thinking a good way to try to construct such a $P$ in general may be via some sort of of subdivision of $C$. But I'm a bit unclear as to when the barycentric subdivision, say, of a category is a poset.
|
https://mathoverflow.net/users/2362
|
Is every category a localization of a poset?
|
Yes, this is true. It follows form the work of Barwick and Kan on relative categories as a model for $\infty$-categories.
The idea is similar to how Thomason's work shows that every homotopy type can be modeled by a poset (by taking subdivisions of the category of simplicies of the simplicial set).
Specifically in [ArXiv:1011.1691](https://arxiv.org/abs/1011.1691) and [ArXiv:1101.0772](https://arxiv.org/abs/1101.0772) they construct a Quillen equivalence between the Joyal model structure and the model structure of "Relative Categories". This latter is a model structure on the category of small ordinary categories equipped with a collection of "weak equivalences". Importantly they show that the cofibrant objects in the latter are **relative posets** - relative categories whose underlying category is a poset.
This means that every $\infty$-category can be modeled by a relative poset.
If you are just interested in ordinary categories, then you don't even need the Quillen equivalence. View the ordinary category $C$ as a relative category with trivial weak equivalences. Then its cofibrant replacement in the Barwick-Kan model structure will be a relative poset $(P,W)$, and it will satisfy $C \simeq P[W^{-1}]$.
**Added for clarification**
From the comments to the OP is seems that people want to see a bit more about how this works. In particular how can we get a cofibrant replacement? Is it functorial?
In the first paper Barwick and Kan construct an adjunction which they then show is a Quillen equivalence:
$$K\_\xi: ssSet \leftrightarrows RelCat: N\_\xi$$
Here $N\_\xi$ is a sort of nerve functor. Claim: for any relative category $C$, $K\_\xi N\_\xi(C) \to C$ is a cofibrant replacement of $C$ (hence a relative poset modeling $C$). This is clearly functorial in $C$ by construction.
*Proof*: $N\_\xi(C)$ will automatically be a cofibrant object in bisimplicial sets in the Reedy(=injective) model structure, and so $K\_\xi$, being a left Quillen functor, will send it to a cofibrant relative category.
We just need to know that (1) the counit $\epsilon\_\xi :K\_\xi N\_\xi C \to C$ is a weak equivalence in Relative categories.
Weak equivalences in the model category of relative categories are detected by $N\_\xi$ (by construction - it is a transferred model structure). Thus (1) will be true if the map $N\_\xi\epsilon\_\xi: N\_\xi K\_\xi N\_\xi C \to N\_\xi C$ is a weak equivalence.
Prop 10.3 in 1011.1691 states that the unit map $\eta\_\xi: id \to N\_\xi K\_\xi$ is always a weak equivalence. Thus $\eta\_\xi N\_\xi$ is a weak equivalence. This is a left inverse of $N\_\xi \epsilon\_\xi$, so by two-out-of-three $N\_\xi \epsilon\_\xi$, and hence $\epsilon\_\xi$ are weak equivalences. $\square$
You can also get an easier description using the functor "$N$" rather than the more cumbersome $N\_\xi$. See section 7-8 of that same paper.
|
23
|
https://mathoverflow.net/users/184
|
403122
| 165,394 |
https://mathoverflow.net/questions/342846
|
1
|
>
> I [know](https://mathoverflow.net/a/136067/143814) we can't formalize the Kunen inconsistency as an assertion in the first-order language of set theory. But
> [the wikipedia say](https://en.wikipedia.org/wiki/New_Foundations#Strong_axioms_of_infinity) "**Con([MK](https://en.wikipedia.org/wiki/Morse%E2%80%93Kelley_set_theory) + The proper class ordinal is a measurable cardinal.)**$\leftrightarrow$ **Con([NFU](https://en.wikipedia.org/wiki/New_Foundations) + Infinity + Large Ordinals + Small Ordinals)**", So I think it's no problem for **NFU**.
>
>
>
**NFU** is a non-well founded set theory. NFU had Universal set **V**, Universal set cardinal and automorphism j, NFU's axiom schema of comprehension is very different from **[MK](https://en.wikipedia.org/wiki/Morse%E2%80%93Kelley_set_theory)/[TG](https://en.wikipedia.org/wiki/Tarski%E2%80%93Grothendieck_set_theory)**, They are very important in Kunen's inconsistency theorem. So a interesting question is:
Can we use Kunen's inconsistency theorem in **NFU + AC**, to proof **NFU + AC + Reinhardt/super Reinhardt/totally Reinhardt/limit club Berkeley cardinal** $\to 0=1$?
|
https://mathoverflow.net/users/143814
|
Can we use Kunen's inconsistency theorem in NFU+AC?
|
In general, NFU + AC is going to have exactly the same possible large cardinals as ordinary set theory. NFU + AC has the same stratified mathematics as ordinary set theory. There will be no interesting results of this kind, unless they involve unstratified assertions in the language of NFU.
I should be slightly more explicit. NFU + AC has the same strength as bounded Zermelo set theory. However, there will be models of NFU + AC associated with any of the usual levels of strength we are used to.
|
0
|
https://mathoverflow.net/users/345616
|
403142
| 165,400 |
https://mathoverflow.net/questions/342915
|
0
|
J. D. Hamkins proved in ["The foundation axiom and elementary self-embeddings of the universe"](http://jdh.hamkins.org/foundation-axiom-and-self-embeddings-of-the-universe/) that, working in $ZFGC^− +BAFA$, there are nontrivial automorphisms and elementary embeddings of the universe **V** into itself.
But, he say "so (BAFA) there are no Reinhardt cardinals here to be found. The embeddings provided by BAFA have no critical points." in this [post](https://mathoverflow.net/a/200544/143814). What does he mean for **"no critical points"**?
(1) Can we use Kunen's inconsistency theorem to proof $ZFGC^- + BAFA + Reinhardt \to 0=1$?
(2) Can Kunen inconsistency make sure there is no "useful" anti-foundational axiom in **[MK](https://en.wikipedia.org/wiki/Morse%E2%80%93Kelley_set_theory)-AF+Reinhardt** and **[TG](https://en.wikipedia.org/wiki/Tarski%E2%80%93Grothendieck_set_theory)-AF+Reinhardt**?
|
https://mathoverflow.net/users/143814
|
Is there no anti-foundational theory exists Reinhardt and hold Global Choice?
|
In general, anti-foundation axioms have nothing to do with issues of large cardinal consistency strength. You can take your universe of set theory with foundation and convert it into a universe with your favorite antifoundation axiom and no large cardinals will be created or destroyed. This process is reversible: large cardinals have nothing to do with anti-foundation axioms.
|
3
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https://mathoverflow.net/users/345616
|
403143
| 165,401 |
https://mathoverflow.net/questions/403139
|
12
|
In algebraic number theory, we constantly make use of the nine-term *Poitou-Tate* sequence: Let $K$ be a number field and $M$ a finite $K$-Galois module. Then we have the nine-term exact sequence
$$
H^0(K, M) \to \prod' H^0(K\_v,M) \to H^2(K, M^\vee)^\vee \mathop{\to}\limits^\delta H^1(K, M) \mathop{\to}^{\operatorname{loc}} \prod' H^1(K\_v,M) \to \cdots
$$
The kernel of the localization map $\operatorname{loc}$, which is also the image of the connecting map $\delta$, is traditionally called $Ш^1(K, M)$. It is frequently remarked (e.g. in Neukirch, Schmidt, and Wingberg's *Cohomology of Number Fields*, Definition 8.6.2), that $Ш^1$ is finite but not necessarily $0$. However, in all the examples I've been able to compute explicitly (e.g. $M$ cyclic of prime order), $Ш^1 = 0$. Is there a ready example of a number field $K$ and a finite module $M$ such that $Ш^1(K, M) \neq 0$?
|
https://mathoverflow.net/users/105625
|
Finite Galois module whose Ш¹ is nonzero?
|
Wang's conterexample to Grunwald's theorem: $K=\mathbb{Q}(\sqrt{7})$ and $M=\mu\_8$. Then $H^1(K,M) \cong K^\times/(K^\times)^8$. Now $16$ is not an $8$-th power in this field but locally an $8$-th power everywhere. Your group is cyclic of order $2$. See [wikipedia](https://en.wikipedia.org/wiki/Grunwald%E2%80%93Wang_theorem).
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12
|
https://mathoverflow.net/users/5015
|
403145
| 165,402 |
https://mathoverflow.net/questions/403146
|
7
|
There is a classification of simple Lie algebras in $\text{Vec}\_{\mathbb{C}}$ given by Dynkin diagrams. We then have 4 families of simple lie algebras, plus some exceptional ones.
**Question**: How about simple Lie algebras in the bigger category $\text{sVec}\_{\mathbb{C}}$ of super vector spaces? Is there a known classification for simple Lie algebras there?
|
https://mathoverflow.net/users/41644
|
Semisimple super Lie algebras
|
Yes there is a complete classification of finite dimensional, simple Lie superalgebras (over $\mathbb{C}$), which -up to a certain extent- goes very much in parallel with the corresponding case of Lie algebras and incorporates the later as a special case. There are significant conceptual differences though (as to the role and uniqueness of Dynkin diagrams, Cartan matrices etc). Historically it has been fully developed by Kac. The original references are:
* V.G. Kac, Lie Superalgebras, Adv. Math. 26 (1977) 8.
* V.G. Kac, A sketch of Lie superalgebra theory, Commun. Math. Phys. 53 (1977) 31.
* V.G. Kac, Representations of classical Lie superalgebras, Lectures Notes in Mathematics 676 (1978) 597; Springer-Verlag, Berlin.
The above provide a self-contained full account of the work. You can also find a succint description of the classification with lots of detailed references at: the [Dictionary of Lie superalgebras, arXiv:hep-th/9607161v1](https://arxiv.org/abs/hep-th/9607161).
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12
|
https://mathoverflow.net/users/85967
|
403150
| 165,403 |
https://mathoverflow.net/questions/403140
|
0
|
If $H=(V,E)$ is a [hypergraph](https://en.wikipedia.org/wiki/Hypergraph) and $\kappa \neq \emptyset$ is a cardinal, then a map $c:V\to\kappa$ is called a *coloring* if the restriction $c\restriction\_e$ is non-constant for all $e\in E$ with $|e|\geq 2$. The smallest cardinal $\kappa$ for which a coloring $c:V\to \kappa$ exists, is said to be the *chromatic number* $\chi(H)$ of $H$.
We say $J\subseteq V$ is *independent* if $|J\cap e| \leq 1$ for all $e\in E$. We say that a collection ${\cal J}$ of independent sets is an *independent covering* of $H$ if $\bigcup {\cal J} = V$. The smallest cardinality of any independent covering of $H$ is said to be the *independent covering number* $j(H)$ of $H$.
It is not difficult to see that if $H$ is a graph, then $j(H) = \chi(H)$.
We say $H$ is $k$-regular for an integer $k\geq 2$ if $|e|=k$ for all $e\in E$.
**Question.** Given an integer $k\geq 3$ is there a positive integer $C\_k$ such that
$$j(H) \leq C\_k\cdot \chi(H)$$ for all finite $k$-regular hypergraphs $H$?
|
https://mathoverflow.net/users/8628
|
Discrepancy of chromatic number and independent covering number for $k$-regular hypergraphs
|
No, this is false already for $k=3$. Let $A$ and $B$ be disjoint sets of size $n$, and let $H$ be the hypergraph with vertex set $A \cup B$, whose hyperedges are all $3$-subsets $e$ of $A \cup B$ such that $|e \cap A| \in \{1, 2\}$. Note that $\chi(H)=2$, since we can colour all vertices in $A$ red and all vertices in $B$ blue. On the other hand, since every subset of size $2$ is contained in some hyperedge, the largest independent set has size $1$. Thus, $j(H)=2n$.
|
2
|
https://mathoverflow.net/users/2233
|
403155
| 165,404 |
https://mathoverflow.net/questions/403101
|
3
|
A $3$-manifold $M$ is called $P^2$-irreducible if it is irreducible and there is no $2$-sided $P^2$ contained in $M$.
Can we show $M$ is $P^2$-irreducible iff $\pi\_2(M)=0$?
Notice that one direction follows directly from the Sphere theorem.
|
https://mathoverflow.net/users/280895
|
$P^2$-irreducibility of a $3$-manifold
|
Yes.
One direction is immediate by the Sphere theorem (projective plane theorem) as pointed out by the OP.
Assume $M$ satisfies $\pi\_2(M)=0$. Note that this condition with the Poincare conjecture means any sphere in $M$ bounds a $3$-ball (see [here](https://math.stackexchange.com/questions/1494524/homotopically-trivial-2-sphere-on-3-manifold)). Therefore, we conclude that $M$ is irreducible.
Suppose $M$ contains a $2$-sided $P^2$. One considers the orientation double cover $M'\overset{p}{\longrightarrow} M$ and assume $p^{-1}(P^2)=S$ which is a sphere (not $P^2$s by our assumption). As above, $S$ bounds a $3$-ball $B$ and we consider $B\overset{p}{\longrightarrow} p(B)$. It is not hard to show this is a double covering map. However, the involution on $B$ contains a fixed point by Brouwer fixed-point theorem. We obtain a contradiction.
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https://mathoverflow.net/users/16323
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403165
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https://mathoverflow.net/questions/403175
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5
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A "randomly chosen" 2-generated dense subgroup
$$
G \ = \ \langle a, b \rangle \ < \
{\rm A}\_5 \times {\rm A}\_6 \times {\rm A}\_7 \times \dots
$$
of the cartesian direct product of the finite simple alternating groups
is "typically" free of rank 2. The "interesting" cases are those where
such group $G$ is *not* free -- in particular those where all upper
composition factors (i.e. composition factors of finite quotients)
are alternating groups.
Choosing a pair of generators $(a,b)$ amounts to choosing pairs $(a\_n,b\_n)$
of generators for each of the direct factors ${\rm A}\_n$.
**Question:** Assume that we choose $a\_n := (1,2,3)$, and
$b\_n := (1,2, \dots, n)$ for $n$ odd and $b\_n := (2,3, \dots, n)$ for $n$
even. Does this ensure that all upper composition factors of $G$
are alternating groups, and that none of them occurs with multiplicity greater than 1?
-- And if not, which other choice of pairs
$(a\_n,b\_n)$ of generators for the ${\rm A}\_n$ serves the purpose
(provided that there is such a choice)?
|
https://mathoverflow.net/users/28104
|
Quotients of a 2-generated dense subgroup of a Cartesian product of infinitely many finite alternating groups
|
A very close variant of this group (the one in the question) was introduced and studied by B.H. Neumann (Neumann, B. H. Some remarks on infinite groups. J. Lond. Math. Soc. 12, 120-127 (1937).).
Namely, it's the same group, but restricting to the product of $A\_n$ for odd; this shouldn't make much difference.
This group has a homomorphism onto $\mathbf{Z}$ which measures, for large $n$, how a given element "eventually" shifts $(1,\dots,n)$. So technically the answer is no since all prime cyclic groups occur as quotient.
This is, however, the only deviation to the expected behavior. Indeed, this group has a characteristic subgroup $W$, which is the set of elements of finite support. Modding out, what the quotient group $H$ is isomorphic to the subgroup of permutations of $\mathbf{Z}$ generated by $(123)$ and the shift $n\mapsto n+1$. This group contains the alternating group $A$ of $\mathbf{Z}$ as normal subgroup and the quotient is infinite cyclic. Since $A$ is (infinite) simple (Onofri 1929 / J.Schreier-Ulam 1934), every finite quotient of $H$ is a quotient of $H/A$, hence cyclic. Also, $W$ is just the direct sum $\bigoplus\_{2n+1\ge 5}A\_{2n+1}$. All this is in Neumann's paper.
Hence, if $F$ is a finite quotient of $G$, and $M$ is the image of $W$ in $F$, we deduce that $M\simeq\prod\_{n\in I}A\_n$ for some finite subset $I$ of $\mathbf{N}\_{\ge 5}$ and $F/M$ is cyclic. In particular, the composition factors of $F$ are abelian, or alternating, and the alternating ones have multiplicity $\le 1$.
---
If you allow even $n$ to exactly fit the original question, probably the conclusion is the same: the only minor modification is probably just to check that the cycles $(1,2,3)$ and $(2,3,\dots,n)$ generate $A\_n$ for even $n$, which certainly holds for large $n$ and should be checkable by hand anyway (and with computer, say for $n=6,8$).
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6
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https://mathoverflow.net/users/14094
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403176
| 165,408 |
https://mathoverflow.net/questions/402985
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20
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I came across this integral that seems related to the Riemann zeta function $\zeta(2n)$ evaluated at even integers $2n \in 2\mathbb{Z}$. Letting $n$ be an even integer, define the multiple integral over $(2n+1)$ variables $u\_1 \cdots u\_{2n+1}$
\begin{equation}
\mathcal{I}\_{2n} = \int\_0^1 du\_1 \cdots \int\_0^1 du\_{2n+1} \frac{1}{1+u\_1} \frac{1}{u\_1+u\_2} \frac{1}{u\_2+u\_3} \cdots \frac{1}{u\_{2n} + u\_{2n+1}} \frac{1}{u\_{2n+1}+1}.
\end{equation}
For example, the case $2n=4$ has integrand $\frac{1}{1+u\_1} \frac{1}{u\_1+u\_2} \frac{1}{u\_2+u\_3} \frac{1}{u\_{3} + u\_{4}} \frac{1}{u\_{4}+u\_5} \frac{1}{u\_{5}+1}$.
Below are listed some exact and numerical results for the integrals for the first few values of $2n$ that could be numerically evaluated on my laptop. The $2n=0$ case is easy, and the $2n=2$ case could evaluated exactly by Mathematica in terms of a complicated expression of polylogarithms (although below, I show an alternate way to explicitly derive this).
* $\mathcal{I}\_0 = \frac{1}{2} = - \zeta(0)$
* $\mathcal{I}\_2 = \frac{\pi^2}{6} = \zeta(2)$
* $\mathcal{I}\_4 \approx 8 \frac{\pi^4}{90} = 8 \cdot \zeta(4)$
* $\mathcal{I}\_6 \approx 54 \cdot \zeta(6)$
* $\mathcal{I}\_8 \approx 384 \cdot \zeta(8)$
* $\mathcal{I}\_{10} \approx 2880 \cdot \zeta(10)$
These integrals are almost exactly integer multiples of the zeta function at even integers, up to the small errors given by Mathematica. The series of integers $1,8,54,384,2880$ don't appear in the [OEIS](https://oeis.org/), although the first terms $1,8,54,384$ go as $k^2 \cdot k!$, and $2880$ isn't far off from $5^2 \cdot 5! = 3000$.
This integral seems closely related to the one [considered in this question](https://mathoverflow.net/questions/129955/evaluation-of-an-n-dimensional-integral). However, their method for their integral seemed quite magical and I was not able to generalize it. While I'm specifically interested in this integral, it would be nice to know if there's a general method to deal with these integrals.
Below are alternate formulations of the integral that may help, as well as showing the result for $2n=2$. First, one can change variables to $u\_i = \frac{1-v\_i}{1+v\_i}$ to rewrite it as
\begin{equation}
\mathcal{I}\_{2n} = \frac{1}{2} \int\_0^1 dv\_1 \cdots \int\_0^1 dv\_{2n+1} \frac{1}{1 - v\_1 v\_2} \frac{1}{1 - v\_2 v\_3} \cdots \frac{1}{1 - v\_{2n} v\_{2n+1}},
\end{equation}
which is similar to a form found in the other MathOverflow question linked above. As such, the same substitutions used there done backwards yield other expressions
\begin{equation}
\begin{split}
\mathcal{I}\_{2n} &= \frac{1}{2} \int\_0^1 dy\_{2n+2} \int\_0^{y\_{2n+2}} dy\_{2n+1} \cdots \int\_{0}^{y\_3} dy\_{2} \frac{1}{y\_3} \frac{1}{y\_4-y\_2} \cdots \frac{1}{y\_{2n+2}-y\_{2n+1}} \frac{1}{1-y\_{2n+1}} \\
&= \frac{1}{2} \int\_0^1 d \tilde{u}\_{1} \cdots \int\_0^1 d \tilde{u}\_{2n+2} \frac{\delta(1-\tilde{u}\_1 - \cdots - \tilde{u}\_{2n+2})}{(\tilde{u}\_1 + \tilde{u}\_2)(\tilde{u}\_2 + \tilde{u}\_3)\cdots(\tilde{u}\_{2n+1} + \tilde{u}\_{2n+2})}
\end{split}
\end{equation}
which almost match the integral considered in linked question with $2n+2$ variables, except missing the last factor $\frac{1}{\tilde{u}\_{2n+2} + \tilde{u}\_{1}}$
For $2n=2$, one can use [Feynman Parameterization](https://en.wikipedia.org/wiki/Feynman_parametrization) to write
\begin{equation}
\begin{split}
\mathcal{I}\_{2} &= \frac{1}{2} \int\_0^1 dv\_1 \int\_0^1 dv\_2 \int\_0^1 dv\_3 \frac{1}{1 - v\_1 v\_2} \frac{1}{1 - v\_2 v\_3} \\
&= \frac{1}{2} \int\_0^1 du \int\_0^1 dv\_1 \int\_0^1 dv\_2 \int\_0^1 dv\_3 \frac{1}{\left[(1 - v\_1 v\_2)u + (1 - v\_2 v\_3)(1-u)\right]^2} \\
&= \frac{1}{2} \int\_0^1 du \left( \frac{\log(1-u)}{u} + \frac{\log(u)}{1-u} \right) = \int\_0^1 du \frac{\log(1-u)}{u} = \sum\_{k=1}^{\infty} \frac{1}{k^2} = \frac{\pi^2}{6},
\end{split}
\end{equation}
where the second-to-last equality follows from term-by-term integration of the series expansion.
I haven't found a way to generalize the method for any of the other cases and still don't have analytical proof for the next equalities.
---
EDIT: See below for a beautiful answer! It turns out that it's not related to the $\zeta$ function except for the first few values of $n$.
|
https://mathoverflow.net/users/106463
|
A multiple integral that seems related to the $\zeta$ function at even integers
|
We have $$I\_{2n}=\frac{(2n)!!}{(2n+1)!!}\cdot \frac1{2n+2}\cdot \pi^{2n}.$$
To see this, we follow the suggestion by Terry Tao in the comments and apply the diagonalization of the integral operator with the kernel $1/(x+y)$ on $[0,1]$. Change the variable to $1/x\in [1,\infty)$ and use (1.18) [here](https://www.ams.org/journals/spmj/2019-30-01/S1061-0022-2018-01534-X/) (this is Mehler integral operator, as I understand) to diagonalize. Substituting the value of Legendre functions at 1, we get $$I\_{2n}=\pi^{2n}\int\_0^\infty x\tanh x/\cosh^{2n+2} x dx.$$
(The above part is suggested by Vladimir Petrov, I understand nothing about all this special functions and integral transforms stuff. But the answer for small $n$ coincides, so I guess everything is ok:) Well, you are free to ask for more details if necessary.)
For evaluating these integrals, integrate by parts noting that $$\tanh x/\cosh^{2n+2} x dx=\tanh x\cosh^{-2n} xd\tanh x=\frac12(1-\tanh^2 x)^nd\tanh^2 x\\=\frac{-1}{2(n+1)}(1-\tanh^2x)^{n+1},$$
thus $$I\_{2n}=\pi^{2n}\cdot \frac1{2(n+1)}\int\_0^\infty (1-\tanh^2x)^{n+1}dx=\pi^{2n}\cdot \frac1{2(n+1)}\int\_0^1 (1-t^2)^{n}dt\\=\frac{\pi^{2n}}{2(n+1)}\cdot \frac {(2n)!!}{(2n+1)!!}$$
(here $t=\tanh^2 x$, the last integral is well known and may be calculated by induction or using Beta function or how do you prefer.)
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12
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https://mathoverflow.net/users/4312
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403179
| 165,411 |
https://mathoverflow.net/questions/401964
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6
|
First, consider group extensions with non-abelian kernel
$$1\to K\to G \to Q \to 1$$
It is well-known that these are classified by certain cohomological objects, specifically: Any such extension induces an outer action, i.e. a group homomorphism $\omega: Q\to\operatorname{Out}(K)$ (which turns $Z(K)$ into a $\mathbb{Z}Q$-module) and to any outer action we can associate an obstruction in $H^3(Q,Z(K))$ that vanishes iff any extension with the given outer action exists. And if they exist, they are in bijection with the non-abelian cohomology $H^2(Q,K,\omega)$.
Second, consider $Q$-graded, crossed $k$-algebras, i.e. $k$-algebras with a decomposition $A=\bigoplus\_{q\in Q} A\_q$ such that $1\in A\_1$, $A\_{q\_1} \cdot A\_{q\_2} \subseteq A\_{q\_1 q\_2}$, and all $A\_q$ contain a unit. Again, this situation induces an outer action $\omega: Q\to\operatorname{Out}(A\_1)$ (and $Z(A\_1)$ becomes a $kQ$-module), for any such outer action there is an obstruction in $H^3(Q,Z(A\_1)^\times)$ that vanishes iff any $Q$-graded, crossed $k$-algebra with the given outer action exists. And if they exists, they are classified by non-abelian cohomology $H^2(Q,A\_1^\times,\omega)$.
These statements are so similar that it is natural to ask:
**Question**: What is a natural, common generalization of both statements?
There are natural constructions relating the two: Given a group extension, $A:=k[G]$ is $Q$-graded and crossed with $A\_q:=k[qK]$. Conversely, given a crossed algebra, the group of homogeneous units fits into a natural extension $1\to A\_1^\times \to (A^\times)\_{homog.} \to Q\to 1$
But these constructions do not immediately give implications from one theorem to the other: The group extension version does not give you the algebra version, because $\operatorname{Out}(A\_1)$ may be different from $\operatorname{Out}(A\_1^\times)$ and $Z(A\_1)^\times$ may have little to do with $Z(A\_1^\times)$.
Conversely, the algebra version does not give you the group version, $A\_1=k[K]$, again because $\operatorname{Out}(K)$ and $\operatorname{Out}(k[K])$ can be very different.
|
https://mathoverflow.net/users/3041
|
Connection between the classifications of group extensions and group-graded algebras in terms of non-abelian cohomology
|
In the meantime I found the right nlab pages to read... The answer seems to be [2-groups](https://ncatlab.org/nlab/show/2-group)! Specifically [automorphism 2-groups](https://ncatlab.org/nlab/show/2-group#automorphism_2groups). I will write up what I have come to understand so far (though I have not checked every last detail of this):
1. What's a 2-group? Like an ordinary group is a category with 1 object in which every morphism is invertible, a (coherent) 2-group is a 2-category with only one object in which every 1- and 2-morphism is invertible (and we have picked a weak inverse of every 1-morphism).
* Generic Example: For every 2-category $\mathscr{C}$ and every object $\mathcal{C}\in\mathscr{C}$ there is a automorphism-2-group $\operatorname{AUT}\_\mathscr{C}(\mathcal{C})$ which has the self-equivalences $\mathcal{C}\to\mathcal{C}$ as 1-morphisms and natural isomorphisms between those as 2-morphisms.
* This includes both the case of groups by letting $\mathscr{C}$ be the 2-category of categories and the case of $k$-algebra by letting $\mathscr{C}$ be the 2-category of $k$-linear categories.
A group $K$ is a category with one object and $\operatorname{AUT}\_\mathsf{cat}(K)$ is a 2-group with $\operatorname{Aut}\_{\mathsf{Grp}}(K)$ as its set of 1-morphisms and for every $h\in K$ a 2-morphism $\alpha \to \kappa\_h\circ\alpha$ where $\kappa\_h$ is the conjugation with $h$.
A $k$-algebra is a $k$-linear category with one object and we get its automorphism-2-group $\operatorname{AUT}\_{k\mathsf{-cat}}(A)$ which similary consists of $\operatorname{Aut}\_{k\mathsf{-Alg}}(A)$ and $A^\times$ as its sets of 1- and 2-morphisms respectively.
* Other example: Any ordinary group $Q$ can "upgraded" to a 2-group $\mathscr{Q}$ by letting 1-morphisms be elements of $Q$ just as usual and letting 2-morphisms be only identities.
2. Conversely, any 2-group $\mathscr{G}$ naturally defines an ordinary group $\pi\_1(\mathscr{G})$ by considering 1-morphisms up to equivalence.
In our examples
* $\pi\_1(\operatorname{AUT}\_{\mathsf{cat}}(K)) = \operatorname{Out}(K)$,
* $\pi\_1(\operatorname{AUT}\_{k\mathsf{-cat}}(A)) = \operatorname{Out}(A)$,
* $\pi\_1(\mathscr{Q}) = Q$
I will write $\operatorname{Out}\_\mathscr{C}(\mathcal{C})$ as shorthand for $\pi\_1(\operatorname{AUT}\_\mathscr{C}(\mathcal{C}))$
3. In addition, any 2-group $\mathscr{G}$ also defines an abelian group $\pi\_2(\mathscr{G})$ consisting of the 2-isomorphisms of the identity 1-morphism (which is abelian by an Eckmann-Hilton argument) and $\pi\_1$ naturally acts on $\pi\_2$ by conjugation.
* In the generic example, $\pi\_2(\operatorname{AUT}\_\mathscr{C}(\mathcal{C})))$ is the unit group of the commutative monoid $Z(\mathcal{C})$ on which $\operatorname{Out}\_\mathscr{C}(\mathcal{C})$ acts by "conjugation"
* $\pi\_2(\operatorname{AUT}\_\mathsf{cat}(K)) = Z(K)$ with the natural action by $\operatorname{Out}(K)$.
* Similary, $\pi\_2(\operatorname{AUT}\_{k\mathsf{-cat}}(A)) = Z(A)^\times$ with the natural action by $\operatorname{Out}(A)$.
* And trivially $\pi\_2(\mathscr{Q}) = 1$
4. Where do extensions and gradings come in? A common generalisation of both theorem in my question is the following:
>
> Let $Q$ be a group,$(\mathcal{V},\oplus,\otimes)$ be a [bimonoidal category](https://ncatlab.org/nlab/show/rig+category) and $\mathscr{C}:=\mathcal{V}\mathsf{-cat}$ be the 2-category of categories enriched in $(\mathcal{V},\otimes)$.
>
>
> Then: For every group homomorphism $\omega: Q\to\operatorname{Out}\_\mathscr{C}(\mathcal{C}\_1)$, the non-abelian cohomology $H^2(Q,\operatorname{AUT}\_\mathscr{C}(\mathcal{C}\_1),\omega)$ is in canonical bijection with "crossed, $Q$-graded categories" $\mathcal{C}$ that have $\mathcal{C}\_1$ as their degree-1-part and induced outer action $\omega$.
>
>
> Furthermore: These two sets are non-empty iff a certain obstruction $o(\omega)\in H^3(Q,Z(\mathcal{C}\_1)^\times)$ vanishes.
>
>
>
5. This begs the question: What's a "graded category"? A *graded category* is a category $\mathcal{C}$ enriched in $(\mathcal{V},\otimes)$ that is equipped with a decomposition $\mathcal{C}(x,y) = \bigoplus\_{q\in Q} \mathcal{C}\_q(x,y)$ such that $\operatorname{id}\_x \in \mathcal{C}\_1(x,x)$ and composition of morphisms decomposes accordingly as $\circ: \mathcal{C}\_p \otimes \mathcal{C}\_q \to \mathcal{C}\_{pq}$.
In particular: Considering only the morphisms in $\mathcal{C}\_1$ gives us another category (with the same set of objects).
A graded category is *crossed* if every $\mathcal{C}\_q(x,x)$ contains an isomorphism.
* Every group extension $1\to K\to G\to Q\to 1$ gives a $Q$-grading on the one-object category $G$ if we let $(\mathcal{V},\oplus,\otimes)$ be $(\mathsf{Set},\sqcup,\times)$, namely the decomposition into cosets of $K$. These gradings are always crossed.
* A graded $k$-algebra $A=\bigoplus\_{q\in Q}$ gives a $Q$-grading on the one-object category $A$ if we let $(\mathcal{V},\oplus,\otimes)$ be $(k\mathsf{-mod},\oplus,\otimes)$. These gradings are crossed iff they are crossed in the usual sense, i.e. if all $A\_q$ contain at least one unit.A crossed $Q$-graded category induces an outer action $Q\to\operatorname{Out}\_{\mathcal{V}\mathsf{-cat}}(\mathcal{C}\_1)$ by choosing 1-isomorphisms $u\_{q,x}\in \mathcal{C}\_q(x,x)$ for every $x\in Ob(\mathcal{C})$ and $q\in Q$ and mapping $q$ to the equivalence class $[\alpha\_q]\in\operatorname{Out}(\mathcal{C}\_1)$ of the "conjugation"-functor $\alpha\_q: \mathcal{C}\_1\to\mathcal{C}\_1, (x\xleftarrow{f}y) \mapsto (x \xleftarrow{u\_{q,x} \circ f \circ u\_{q,y}^{-1}} y)$.
* For group extensions $1\to K\to G\to Q\to 1$ this is exactly the outer action $Q\to\operatorname{Out}(K)$ induced by conjugation.
* For crossed, $Q$-graded algebras this is also the outer action $Q\to\operatorname{Out}(A\_1)$ induced by conjugation.
6. Now, what precisely does the theorem do? For every fixed group homomorphism $\omega: Q \to \pi\_1(\operatorname{AUT}\_\mathscr{C}(\mathcal{C}\_1))$ it establishes really two pairs bijection between the following three sets:
* $Q$-graded, crossed categories extending $\mathcal{C}\_1$ that induce $\omega$
$$\uparrow \downarrow$$
* 2-group morphisms $\mathscr{Q}\to\operatorname{AUT}\_\mathscr{C}(\mathcal{C}\_1)$ up to equivalence that induce $\omega$ on the $\pi\_1$-groups, where $\mathscr{Q}$ is the 2-group upgrade of $Q$.
$$\uparrow \downarrow$$
* non-abelian cohomology $H^2(Q,\operatorname{AUT}\_\mathscr{C}(\mathcal{C}\_1),\omega)$, i.e. 2-cocycles $(\alpha,\chi)$ up to 2-coboundaries
+ $(\alpha,\chi)$ being a 2-cocycle means $\alpha: Q \to \{1\text{-morphisms}\}, \chi: Q^2 \to \{2\text{-morphisms}\}$ with $\chi\_{p,q}: \alpha\_{p} \alpha\_{q} \Rightarrow \alpha\_{pq}$, $\alpha\_q$ is in the equivalence class $\omega(q)$, and the 2-coycle condition holds: $\chi\_{xy,z}\circ\chi\_{x,y} = \chi\_{x,yz}\circ(\alpha\_x(\chi\_{y,z}))\circ\text{associator}$ as 2-morphisms $(\alpha\_x\circ\alpha\_y)\circ\alpha\_z \Rightarrow \alpha\_{xyz}$
+ A 2-coboundary $(\alpha,\chi) \to (\alpha',\chi')$ is map $\lambda: Q\to\{2\text{-morphisms}\}$ such that $\lambda\_x: \alpha\_x \Rightarrow \alpha'\_x$ with $\lambda\_{xy}\circ\chi\_{x,y} = \chi'\_{x,y}\circ(\lambda\_x\lambda\_y)$ as 2-morphism $\alpha\_x \circ \alpha\_x \Rightarrow \alpha'\_{xy}$.
7. And finally: How does one prove all of that? If I have not made any huge mistakes, the classical proofs from Schreier theory all carry over if one replaces all equations by the appropriate commutative diagrams.
There is even a regular action of the group $H^2(Q,Z(\mathcal{C}\_1)^\times)$ on the set $H^2(Q,\operatorname{AUT}\_\mathscr{C}(\mathcal{C}\_1),\omega)$ if it is non-empty given by $[\zeta] . [\alpha,\chi] := [\alpha, \zeta.\!\chi]$ where $\zeta.\!\chi$ means the 2-morphism $\alpha\_p\circ\alpha\_q \overset{\chi\_{p,q}}{\Longrightarrow} \alpha\_{pq} \overset{1\_{\alpha\_{pq}} \ast \zeta\_{p,q}}{\Longrightarrow} \alpha\_{pq}$
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1
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https://mathoverflow.net/users/3041
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403180
| 165,412 |
https://mathoverflow.net/questions/403181
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1
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Let $\Omega\_1$, $\Omega\_2$ be bounded,convex, open domains with smooth boundary in $\mathbb{R}^2$ and $\overline\Omega\_1\subset\Omega\_2$. Suppose we are given a $C^1$ function $f:\overline\Omega\_1\cup(\mathbb{R}^2\setminus\Omega\_2)\rightarrow\mathbb{R}$ satisfies the following properties:
(1)$f\equiv 1$ on $\partial\Omega\_1$ and $f\equiv 0$ on $\partial\Omega\_2$.
(2)$\nabla f\cdot \nu\_k< 0$ on $\partial\Omega\_k$, where $\nu\_k$ is the unit outward normal vector to $\Omega\_k$, ($k=1,2$).
Now the question is that can we extend $f$ to be $C^1$ in $\mathbb{R}^2$ such that $0\leq f\leq 1$ and $|\nabla f|\neq 0$ in $\Omega\_2\setminus\overline\Omega\_1$?
I think the answer must be yes because I can imagine its figure as a frustum of a cone. I have read some references about the Whitney's extension theorem but they do not match. I would be very appreciate if anyone can provide the proof or references.
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https://mathoverflow.net/users/157939
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Looking for a reference for an extension problem of function
|
The boundary of a convex domain is a Jordan curve (this is the only property of
the boundaries that will be used). Then $\Omega\_2\backslash\Omega\_1$ is a topological ring, and by the well known theorem, there is a conformal map
$\phi:\Omega\_2\backslash\Omega\_1\to A$, where $A=\{ z:r<|z|<1\}$, for some $r\in(0,1)$. This map is continuous on the boundary. Now take
$f=-\log|\phi|/\log r$ as your extesion. Since $\phi'(z)\neq 0$, you have $\nabla f(z)\neq 0$. You only have to be careful when you use the normal vector, since you did not state any smoothness condition of $\partial\Omega\_j$.
Remark. Your condition 2 is redundant, since you did not say that you want the extended $f$ to be $C^1$ in $\Omega\_2$. But this also can be achieved, if desired, by the same method, which reduce the question
to the round ring. You only have to be careful when you mention the normal vector since you did not state any smoothness conditions of $\partial\Omega\_j$.
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1
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https://mathoverflow.net/users/25510
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403182
| 165,413 |
https://mathoverflow.net/questions/403184
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63
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A (non-mathematical) friend recently asked me the following question:
>
> Does the golden ratio play any role in contemporary mathematics?
>
>
>
I immediately replied that I never come across any mention of the golden ratio in my daily work, and would guess that this is the same for almost every other mathematician working today.
I then began to wonder if I were totally correct in this statement . . . which has led me to ask this question.
My apologies is this question is unsuitable for Mathoverflow. If it is, then please feel free to close it.
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https://mathoverflow.net/users/352001
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Golden ratio in contemporary mathematics
|
The "Cleary group" $F\_\tau$ is a version of Thompson's group $F$, introduced by Sean Cleary, that is defined using the golden ratio, and it's definitely of interest in the world of Thompson's groups. See *[An Irrational-slope Thompson's Group](https://arxiv.org/abs/1806.00108)* ( Publ. Mat. 65(2): 809-839 (2021). DOI: 10.5565/PUBLMAT6522112 ). Very roughly, where $F$ arises by "cutting things in half", $F\_\tau$ arises in an analogous way by "cutting things using the golden ratio". There are lots of similarities between $F\_\tau$ and $F$, but also plenty of mysteries, for example I believe it's still open whether $F\_\tau$ embeds into $F$ (i.e., whether there exists a subgroup of $F$ isomorphic to $F\_\tau$).
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47
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https://mathoverflow.net/users/164670
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403186
| 165,415 |
https://mathoverflow.net/questions/403169
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2
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Assume that $G$ is a (finite) abelian group and $M$ is a matroid whose ground set is $G$. Let $X$ and $Y$ be subsets of $G$, and $H$ is the stabilizer of $X+Y$. That is $X+Y+H=X+Y$. We denote the rank function of $M$ by $r$. Then can we say that $r(X+Y)\geq$ $r(X)+r(Y)-r(H)$? Or under what condition this can be true? I am fine to assume $M$ to be a partition matroid(or even a uniform matroid) and also assume $G$ is cyclic of prime order.
(A similar result holds for the cardinality of sumsets of a given group, well-known as Kneser's theorem)
Note: To make the statement plausible, adding the condition of $X+Y$ to be independent won't hurt.
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https://mathoverflow.net/users/165074
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Rank of sumsets in matroids
|
Without any condition on the matroid structure, there is really no reason for your inequality to hold.
For example, take $X=\{a\},Y=\{b\}$ so that $X+Y=\{a+b\}$ where $a,b$ are any nonzero elements of $G$.
Then take any matroid structure such that $a+b$ is a loop but $a,b$ and $0$ are not. We have $0=r(X+Y)<r(X)+r(Y)-r(H)=1$.
Maybe you want to add some kind of compatibility condition between the group and matroid structures to make your question more interesting?
EDIT: Even in a very simple and and well-behaved setting (say $G$ cyclic and $M$ uniform), the inequality fails.
For example, let $G=\mathbb{Z}/N\mathbb{Z}$ with the $n$-uniform matroid structure, where $n<N/2$. Let $X\subset [0, N/2)$ be any subset of size $n$.
Then $r(X+X)=r(X)=r(Y)=n$ and $H$ is trivial so $r(H)=1$. Hence your inequality fails whenever $n>1$.
|
2
|
https://mathoverflow.net/users/160416
|
403192
| 165,417 |
https://mathoverflow.net/questions/403172
|
6
|
$\newcommand{\IR}{\mathbb R}$
$\newcommand{\IT}{\mathbb T}$
$\newcommand{\w}{\omega}$
$\newcommand{\e}{\varepsilon}$
[Taras Banakh](https://mathoverflow.net/users/61536/taras-banakh) and me proceed a long quest answering a [question](https://math.stackexchange.com/questions/1203722/approximate-vanishing-in-pontryagin-dual) of ougao at Mathematics.SE. Recently we encountered a notion of a remote sequence. We are interested whether it was studied before and want to know a solution of a problem below.
Recall that a circle $\mathbb T=\{z\in\mathbb C:|z|=1\}$, endowed with the operation of multiplication of complex numbers and the topology inherited from $\mathbb C$ is a topological group. Let $(r\_n)\_{n\in\w}$ be an increasing sequence of natural numbers. The sequence $(r\_n)\_{n\in\w}$ is called *remote* if there exists $z\in\IT$ such that $\inf\_{n\in\w}|z^{r\_n}-1|>0$. A *minimum growth rate* of $(r\_n)\_{n\in\w}$ is a number $\inf\_{n\in\w} r\_{n+1}/r\_{n}$.
**Problem.** Find the maximal set $M\subseteq [1,\infty)$ such that if the minimum growth rate of $(r\_n)\_{n\in\w}$ belongs to $M$ then $(r\_n)\_{n\in\w}$ is remote.
*Our try.* It is easy to see that $1\not\in M$. On the other hand, we can show that any real number $m>2$ belongs to $M$ as follows. Let $m$ be the minimum growth rate of $(r\_n)\_{n\in\w}$. Find $\e>0$ such that $2+\frac{\e}{1-\e}\le m$. Then $\frac{r\_{n+1}}{r\_n}\ge 2+\frac\e{1-\e}=\frac{2-\e}{1-\e}$ and hence $\frac{r\_n}{r\_{n+1}}\cdot\frac{2-\e}{1-\e}\le 1$ for every $n\in\w$. Let $W=\{e^{it}:|t|<\pi \e\}$ and for every $n\in\w$ consider the set $U\_n=\{z\in\IT:z^{r\_n}\in W\}$. Consider the exponential map $\exp:\IR\to \IT$, $\exp:t\mapsto e^{2\pi t i}$, and observe that for every $n\in\w$ any connected component of the set $\exp^{-1}[U\_n]$ is an open interval of length $\frac{\e}{r\_n}$ and every connected component of the set $\IR\setminus \exp^{-1}[U\_n]$ is a closed interval of length $\frac{1-\e}{r\_n}$. Since $\frac{r\_n}{r\_{n+1}}\cdot\frac{2-\e}{1-\e}\le 1$ and $$\frac{1-\e}{r\_{n+1}}+\frac\e{r\_{n+1}}+\frac{1-\e}{r\_{n+1}}=\frac{2-\e}{r\_{n+1}}=\frac{1-\e}{r\_n}\cdot\frac{r\_n}{r\_{n+1}}\cdot\frac{2-\e}{1-\e}\le \frac{1-\e}{r\_n},$$every connected component of the set $\IR\setminus \exp^{-1}[U\_n]$ contains a connected component of the set $\IR\setminus\exp^{-1}[U\_{n+1}]$. Then for every $n\in\w$ we can choose a connected component $I\_n$ of the set $\IR\setminus\exp^{-1}[U\_n]$ such that $I\_{n+1}\subseteq I\_n$. By the compactness of the set $I\_0$, the intersection $\bigcap\_{n\in\w}I\_n$ contains some real number $t$. Then the point $z=\exp(t)$ does not belong to $\bigcup\_{n\in\w}U\_n$, which implies that $z^{r\_n}\notin W$ for every $n\in\w$. The definition of the neighborhood $W$ ensures that $\inf\_{n\in\w}|z^{r\_n}-1|>0$, that is the
sequence $(r\_n)\_{n\in\w}$ is remote.
Thanks.
|
https://mathoverflow.net/users/43954
|
A property of rapid sequences of natural numbers
|
A sequence is called *lacunary* if, in your terminology, its minimum growth rate is strictly greater than $1$. The following articles prove that every lacunary sequence is remote. If I understand your question correctly, this means that the (only) maximal $M$ you seek is the interval $(1,\infty)$.
*Pollington, Andrew D.*, On the density of sequences $\{n\_k\xi\}$, Ill. J. Math. 23, 511-515 (1979). [ZBL0401.10059](https://zbmath.org/?q=an:0401.10059).
*de Mathan, B.*, [**Numbers contravening a condition in density modulo 1**](http://dx.doi.org/10.1007/BF01898138), Acta Math. Acad. Sci. Hung. 36, 237-241 (1980). [ZBL0465.10040](https://zbmath.org/?q=an:0465.10040).
|
4
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https://mathoverflow.net/users/10457
|
403197
| 165,419 |
https://mathoverflow.net/questions/403187
|
3
|
Define a structure made of objects $A, B, C, \dots$ and morphisms $f, g, \dots$. Each morphism has *a collection of* domain objects and codomain objects. For simplicity we consider the domains and codomains as unordered sets. We require that
* For each $A$ there exists an $f : A \to A$.
* For every $f:B\_1, B\_2, \dots \to C, D\_1, D\_2,\dots$ and $g: C, F\_1, F\_2, \dots \to H\_1, H\_2,\dots$, there exists a morphism $g \circ f : B\_1, B\_2, F\_1, F\_2 \dots \to D\_1, D\_2, H\_1, H\_2\dots$.
* For composable $f, g, h$, $(f\circ g) \circ h = f \circ (g \circ h)$.
Note that this is richer than sequent logics, because it is sort of proof relevant; the same sequent have different proofs. If we impose that for each $f,g:\Gamma \to\Delta$ we have $f=g$, then we recover the usual sequent logic, without any connectives. (Of course, this makes axiom three trivial.)
We can then define connectives with universal properties, just as we define products, coproducts and equalizers etc.
My question is: is there any existing literature on such a structure? What's the name for it? Where can I read more about it?
|
https://mathoverflow.net/users/136535
|
What's the terminology for a sequent-like variant of category?
|
It sounds like you want the notion of a [polycategory](https://ncatlab.org/nlab/show/polycategory).
|
6
|
https://mathoverflow.net/users/126667
|
403204
| 165,422 |
https://mathoverflow.net/questions/403166
|
5
|
>
> $\textbf{Theorem}.1$ (The first Korn inequality) Suppose that $ \Omega $ is a bounded domain in $ \mathbb{R}^d $ with Lipschitz boundary. Then\
> \begin{eqnarray}
> \sqrt{2}\left\|\triangledown u\right\|\_{L^2(\Omega)}\leq \left\|\triangledown u+(\triangledown u)^T\right\|\_{L^2(\Omega)}
> \end{eqnarray}
> for any $ u\in H\_{0}^{1}(\Omega;\mathbb{R}^d) $, where $ (\triangledown u)^T $ denotes the transpose of $ \triangledown u $.
>
>
>
>
> $\textbf{Theorem}.2$ (The second Korn inequality) Suppose that $ \Omega $ is a bounded domain in $ \mathbb{R}^d $ with Lipschitz boundary. If $ u\in H^{1}(\Omega,\mathbb{R}^d) $ is a function with the property that $ u\perp R $ in $ H^{1}(\Omega;\mathbb{R}^d) $, then
>
> \begin{eqnarray}
> \int\_{\Omega}|\triangledown u|^2dx\leq C\int\_{\Omega}|\triangledown u+(\triangledown u)^T|^2dx
> \end{eqnarray}
> where $ R=\left\{\phi=Bx+b:B\in\mathbb{R}^{d\times d} \text{ is skew-symmetric and }b\in\mathbb{R}^d\right\} $ and $ C $ is a constant.
>
>
>
I recently see the two theorems in a book about elliptic equations. I tried to get the estimate for the second inequality by direct computation which works in the proof of the first Korn inequality, but for this inequality, I cannot combine the condition $ u\perp R $ with the final results. Can you give me some hints or references?
|
https://mathoverflow.net/users/241460
|
How to prove the second Korn inequality?
|
You can find a full proof (to my knowledge the simpler one currently known) in the paper [1] and in the book [2], chapter I, §2.1 pp. 14-21. The original proof of Arthur Korn is so long and involved that K.O. Friedrichs, who gave a much simpler yet sophisticated proof, had doubts on his validity: starting from the work of Friedrichs, several authors gave their (in general quite complex) proofs, until Olga Oleĭnik gave a much shorter and simpler one (despite being still not elementary).
**New edit**. While ordering my library, I noted reference [1b]: in this paper Oleĭnik an Kondratiev prove the classical second Korn inequality for bounded domains satisfying the cone condition (theorem 1, a three page proof) and for certain classes of unbounded domains. They also prove that the constants in the inequality are sharp in some precise sense.
**References**
[1] Vladimir Alexandrovitch Kondratiev, Olga Arsenievna Oleĭnik,
"[On Korn’s inequalities](http://gallica.bnf.fr/ark:/12148/bpt6k6236782n/f497.image)" (English), Comptes Rendus de l’Académie des Sciences, Série I, 308, No. 16, pp. 483-487 (1989), [MR0995908](https://mathscinet.ams.org/mathscinet-getitem?mr=MR0995908), [Zbl 0698.35067](https://www.zbmath.org/?q=an%3A0698.35067).
[1b] Vladimir Alexandrovitch Kondrat’ev, Olga Arsenievna Oleĭnik, "Hardy’s and Korn’s type inequalities and their applications". (English) Rendiconti di Matematica e delle sue Applicazioni, VII Serie 10, No. 3, 641-666 (1990), [MR1080319](http://www.ams.org/mathscinet-getitem?mr=MR1080319), [Zbl 0767.35020](https://zbmath.org/0767.35020), also found in the commemorative book *Scritti matematici. Dedicati a Maria Adelaide Sneider*, Università "La Sapienza", 415-440 (1990).
[2] Olga Arsenievna Oleĭnik, Alexei Stanislavovich Shamaev, Grigorii Andronikovich Yosifian, *Mathematical problems in elasticity and homogenization*. (English) Studies in Mathematics and its Applications. 26. Amsterdam-London-New York-Tokyo: North- Holland, pp. xiii+398 (1992), ISBN: 0-444-88441-6, [MR1195131](https://mathscinet.ams.org/mathscinet-getitem?mr=MR1195131), [Zbl 0768.73003](https://www.zbmath.org/?q=an%3A0768.73003).
|
5
|
https://mathoverflow.net/users/113756
|
403211
| 165,425 |
https://mathoverflow.net/questions/403218
|
8
|
Is there a known explicit description of the [abelian $2$-group](https://en.wikipedia.org/wiki/Abelian_2-group) $\mathsf{Ho}(\mathbb{S})\overset{\mathrm{def}}{=}\mathsf{Ho}(QS^0)\cong\Pi\_{\leq1}(QS^0)$?
|
https://mathoverflow.net/users/130058
|
What is the homotopy category of the sphere spectrum?
|
This is the groupoid given by the 1-truncation $\tau\_{\leq 1}(QS^0)$. This groupoid has $\mathbb Z$-many objects (since $\pi\_0^s = \mathbb Z$), and each one has automorphism group $C\_2$ (since $\pi\_1^s = C\_2$). The tensor product on objects is given by addition in $\mathbb Z$, and on morphisms by addition in $C\_2$. One way to see this is to consider the universal functor $\Sigma \to QS^0$ given by the Barratt-Priddy-Quillen theorem (i.e. the fact that $K(\Sigma) = QS^0$; here $\Sigma$ is the groupoid of finite sets with the disjoint union monoidal structure), and to postcompose with the truncation functor $QS^0 \to \tau\_{\leq 1} (QS^0)$; the fact that this functor is symmetric monoidal yields this description of the category. This perspective is discussed a bit more [here](https://mathoverflow.net/questions/386111/is-every-otimes-invertible-object-coherently-sym-central).
From the description I've given, I suppose it follows that $\tau\_{\leq 1} (QS^0)$ splits symmetric monoidally as $\tau\_{\leq 1} (QS^0) = \mathbb Z \times BC\_2$, (where $\mathbb Z$ is a discrete symmetric monoidal groupoid and $BC\_2$ is a 1-object symmetric monoidal groupoid), which is maybe a little surprising. This is not to say that $\tau\_{\leq 1} \mathbb S$ splits...
|
12
|
https://mathoverflow.net/users/2362
|
403221
| 165,430 |
https://mathoverflow.net/questions/403222
|
-3
|
I was looking for a natural power of 3 that could be written like
>
> Binary format:
>
>
> 11..(N times)..11011..(M times)..11
>
>
> Example: 1111110111111111111111 (...isn't a power of 3)
>
>
>
Or could also be written like
>
> 3^x = 2^a - 2^b - 1
>
>
> (x is arbitrary, "a" and "b" are natural numbers, a = N-M-1, b = M, and the single zero in binary format is a must)
>
>
>
But couldn't find any, so I thought there might be some proof that there's no such numbers (altho that would contradict intuition) or maybe it can be proven that there might be such numbers?
|
https://mathoverflow.net/users/359116
|
Prove that the equation $2^a - 2^b - 1=3^c$ has no integral solution with $a,b\geq 3$
|
The equation $2^a-2^b-1=3^c$ has no integral solution with $a,b\geq 3$. Indeed, in this case the left-hand side is $\equiv 7\pmod{8}$, while the right-hand side is either $\equiv 1\pmod{8}$ or $\equiv 3\pmod{8}$.
|
3
|
https://mathoverflow.net/users/11919
|
403226
| 165,433 |
https://mathoverflow.net/questions/402936
|
2
|
I wish to have a proof for the following result:
Let $U\_n$ be an $n\times n$ upper [shift matrix](https://en.wikipedia.org/wiki/Shift_matrix), and $L\_n = U\_n^T$ be a lower shift matrix. For example,
$$
U\_5 = \begin{pmatrix}
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 & 0 \\
0 & 0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 & 0
\end{pmatrix},
\quad
L\_5 = \begin{pmatrix}
0 & 0 & 0 & 0 & 0 \\
1 & 0 & 0 & 0 & 0 \\
0 & 1 & 0 & 0 & 0 \\
0 & 0 & 1 & 0 & 0 \\
0 & 0 & 0 & 1 & 0
\end{pmatrix}.
$$
Then the Lie algebra generated by $U\_n$ and $L\_n$ over $\mathbb{C}$ is isomorphic to $\mathfrak{so}(n, \mathbb{C})$ when $n$ is odd, and is isomorphic to $\mathfrak{sp}(n, \mathbb{C})$ when $n$ is even.
I tested this numerically with `GAP` over the field $\mathbb{Q}$: the Lie algebra generated by $U\_{2n+1}$ and $L\_{2n+1}$ over $\mathbb{Q}$ is of type $B\_n$, while the Lie algebra generated by $U\_{2n}$ and $L\_{2n}$ over $\mathbb{Q}$ is of type $C\_n$.
|
https://mathoverflow.net/users/113258
|
Prove: Lie algebra generated by two $n\times n$ shift matrices is $\mathfrak{so}(n,\mathbb{C})$ ($n$ odd) or $\mathfrak{sp}(n,\mathbb{C})$ ($n$ even)
|
Let ${\mathfrak g} $ be the Lie algebra generated by $U\_n$ and $L\_n$. It's easy to check that $U\_n$ and $L\_n$ preserve the bilinear form determined by the matrix with $1,-1,1,\ldots$ down the antidiagonal and zero elsewhere. So ${\mathfrak g}$ is contained in (a Lie algebra isomorphic to) $\mathfrak{so} \_n$ for odd $n$, $\mathfrak{sp} \_n$ for even $n$. Let's assume $n$ is odd; the argument is the same for even $n$. Clearly ${\mathfrak g}=\mathfrak{so} \_3$ if $n=3$. In the general case, $[U\_n, L\_n]=H\_n$ is a diagonal matrix with just two non-zero entries $\pm 1$ and $U\_n-[H\_n, U\_n]$, resp. $L\_n+[H\_n, L\_n]$ equals $U\_{n-2}$, resp. $L\_{n-2}$ in the "middle" matrix subalgebra. By induction, ${\mathfrak g}$ contains $\mathfrak{so} \_{n-2}$. Now it's easy to see that $[H\_n, U\_n]$, resp. $[H\_n, L\_n]$ is a multiple of the "missing" positive, resp. negative root element, so in fact ${\mathfrak g}$ equals $\mathfrak{so} \_n$.
|
4
|
https://mathoverflow.net/users/26635
|
403240
| 165,443 |
https://mathoverflow.net/questions/403242
|
4
|
Let $G$ be a group quasi-isometric to the fundamental group of a genus 2 surface group $H$. It is well known that $G$ is quasi-isometrically rigid, i.e. $G$ and $H$ are virtually isomorphic. Does the stronger property, that $G$ and $H$ are commensurable, also hold?
If so is there a reference for this?
|
https://mathoverflow.net/users/121307
|
Quasi-isometric rigidity of surface groups and commensurability
|
Yes (I assume that by "virtually isomorphic" you mean commensurable modulo finite kernels, which is a nonstandard misleading use of "virtually"). This is because surface groups have Serre's property D$\_2$ meaning that each 2-cohomology class (in a finite abelian group with trivial action) it trivial on some finite index subgroup.
(Also a side remark: the known result on QI rigidity is stronger, since it says that every group QI to this surface group is, modulo a finite kernel, isomorphic to a cocompact lattice in the isometry group of the hyperbolic plane. I.e., there's a structural statement which does not require passing to a finite index subgroup.)
|
8
|
https://mathoverflow.net/users/14094
|
403243
| 165,444 |
https://mathoverflow.net/questions/379869
|
4
|
$\DeclareMathOperator\GL{GL}$If $G$ is a simple Lie group, and $\rho: G \to \GL(V)$ is a representation, then by Schur's lemma, the group of automorphisms of $\rho$ is a reductive subgroup of $\GL(V)$. I'm wondering whether this generalizes to the case where $\GL(V)$ is replaced by an arbitrary reductive group?
More generally: if $G$ is a semisimple algebraic group (or even reductive), $H$ is a reductive algebraic group, and $\rho : G \to H$ is a homomorphism, is the centralizer of the image of $\rho$ in $H$, $C\_{H}(\rho(G))$, a reductive subgroup of $H$?
Edits:
1. The case I'm most interested is when the field is $\mathbb{C}$, but I would be interested in hearing about other cases as well.
2. As pointed out in the comments, in the case $G = H$, the centralizer is the centre, which is reductive but not necessarily connected. So the centralizer may not be connected in the general case as well.
|
https://mathoverflow.net/users/69713
|
Centralizers of semisimple subgroups
|
In characteristic zero, the connected closed subgroups of $G$ are in 1-1 correspondence with the Lie subalgebras of ${\rm Lie}(G)={\mathfrak g}$, and the Killing form a suitably chosen symmetric bilinear $G$-equivariant form on ${\mathfrak g}$ is non-degenerate. *In fact we can choose this form to be the trace form for a rational representation for $G$.* Since the Killing form this form is $G$-equivariant, it is $H$-equivariant, and irreducible summands for non-isomorphic $H$-submodules are orthogonal. Hence the Killing form on ${\mathfrak g}$ restricts to a non-degenerate form on ${\mathfrak g}^H={\rm Lie}(Z\_G(H)^\circ)$.
I now claim that if $K$ is a connected algebraic group with a non-degenerate trace form $\kappa$ arising from a rational representation $\rho:K\rightarrow {\rm GL}(V)$, then $K$ is reductive. Indeed, if the unipotent radical $R\_u(K)$ is non-trivial then $\rho(R\_u(K))$ is a unipotent subgroup of ${\rm GL}(V)$, so after conjugation is contained in the subgroup of upper-triangular unipotent matrices, so the restriction of $\kappa$ to ${\mathfrak u}={\rm Lie}(R\_u(K))$ is zero. Since ${\mathfrak u}$ is an ideal of ${\rm Lie}(K)$, this contradicts the non-degeneracy of the form. In particular, $Z\_G(H)^\circ$ is reductive.
|
3
|
https://mathoverflow.net/users/26635
|
403246
| 165,447 |
https://mathoverflow.net/questions/402257
|
7
|
Simplicial sets are presheaves on the simplex category $\Delta$, while augmented simplicial sets are presheaves on $\Delta\_+$, the augmented simplex category. Because Day convolution allows us to lift monoidal structures on a category $\mathcal{C}$ to its category of presheaves $\mathrm{Sets}^{\Delta^\circ}$, it is therefore of interest to find monoidal structures on $\Delta$ and $\Delta\_+$, as these then provide "natural" monoidal structures on simplicial sets.
The only monoidal structure I know of is the ordinal sum of $\Delta\_+$ (which is not braided), whose Day convolution gives the join of simplicial sets, and whose internal hom is given by
$$[X,Y]\_n=\mathrm{hom}\_{\mathrm{Sets}^{\Delta^\circ\_+}}(X,\mathrm{Dec}^{n+1}Y)$$
Is this the only monoidal structure on $\Delta\_+$? If not, what other monoidal structures are there on $\Delta\_+$, and what are there on $\Delta$?
|
https://mathoverflow.net/users/130058
|
Is $\oplus$ the only monoidal structure on the simplex category?
|
Here is half of a classification. Let $\otimes$ be a monoidal structure on $\Delta\_+$. As I mentioned in a [comment](https://mathoverflow.net/questions/402257/is-oplus-the-only-monoidal-structure-on-the-simplex-category#comment1032524_402257), the monoidal unit must be $[-1]$ or $[0]$ because these are the only objects with commutative endomorphism monoids.
Suppose that the monoidal unit is $[0]$. Let us consider $[1] \otimes [1]$. We have that $[0]$ is a retract of $[1]$ in 2 ways, and as a result we obtain 4 retracts of $[1] \otimes [1]$ with support $[0] \otimes [0] = [0]$. Consider the induced linear ordering on these 4 points. We also have 4 ways that $[1] = [1] \otimes [0] = [0] \otimes [1]$ is a retract of $[1] \otimes [1]$, and from this we can deduce most of the ordering. It must have the following relations
$\require{AMScd} \begin{CD} 0 \otimes 0 @>>> 0 \otimes 1\\ @VVV @VVV\\ 1 \otimes 0 @>>> 1 \otimes 1 \end{CD}$
To complete this to a linear order, without loss of generality we must have $0 \otimes 1 \leq 1 \otimes 0$. But now, one of our 4 projections onto $[1]$ is the coordinate projection onto the right column of the above square. The fact that this projection is order-preserving implies that $1 \otimes 0 = 0 \otimes 1 = 1 \otimes 1$. This contradicts the fact that the right column exhibits $[1]$ as a retract of this subset of $[1] \otimes [1]$.
Therefore the monoidal unit is not $[0]$; it must be $[-1]$.
I also think I'm ready to conjecture that $\oplus$, $\oplus^{rev}$ (as mentioned [by Peter](https://mathoverflow.net/questions/402257/is-oplus-the-only-monoidal-structure-on-the-simplex-category?noredirect=1#comment1032530_402257)) and the [degenerate monoidal structure](https://mathoverflow.net/questions/402257/is-oplus-the-only-monoidal-structure-on-the-simplex-category?noredirect=1#comment1032531_402257) are probably the only ones. You could imagine a classification starting as follows. Consider the maps $[0] = [0] \otimes [-1] \to [0] \otimes [0]$ and $[0] = [-1] \otimes [0] \to [0] \otimes [0]$. If these are the same, then we should have the degenerate monoidal structure. If they are different, then one is less than the other, and those two cases should correspond to $\oplus$ and $\oplus^{rev}$.
|
6
|
https://mathoverflow.net/users/2362
|
403250
| 165,448 |
https://mathoverflow.net/questions/403217
|
8
|
$\mathbb{Z}$-graded rings play an important role in algebra and algebraic geometry, so when moving to derived algebra and spectral algebraic geometry, it's natural to ask about ring spectra graded in the sphere spectrum $\mathbb{S}$.
As discussed in Section 2 of [Bunke–Nikolaus's *Twisted differential cohomology*](https://arxiv.org/abs/1406.3231), such an object may be defined as a symmetric lax monoidal functor $\mathbb{S}\to\mathsf{Sp}$, where $\mathbb{S}$ is considered as a monoidal $\infty$-groupoid.
In particular, given an $\mathbb{S}$-graded ring spectrum $R$, it follows that the ring $\pi\_0(R)$ is $\mathsf{Ho}(\mathbb{S})$-graded. [By the description here](https://mathoverflow.net/a/403221), such a $\tau\_{\leq1}\mathbb{S}$-graded ring consists of a pair $(R\_\bullet,\{\sigma\_k\}\_{k\in\mathbb{Z}})$ with
* $R\_\bullet$ a $\mathbb{Z}$-graded ring;
* $\{\sigma\_k\colon R\_k\to R\_k\}\_{k\in\mathbb{Z}}$ a family of order $2$ automorphisms, one for each $R\_k$.
Moreover, if $R$ is an $\mathbb{E}\_{\infty}$-ring, then $\pi\_0(R)$ is "$\mathsf{Ho}(\mathbb{S})$-graded commutative", in that we additionally have
$$
ab
=
\begin{cases}
ba &\text{if $\deg(a)\deg(b)$ is even,}\\
\sigma\_{\deg(a)+\deg(b)}(ab) &\text{if $\deg(a)\deg(b)$ is odd}
\end{cases}
$$
for each $a,b\in R\_\bullet$.
This includes in particular $\mathbb{Z}$-graded commutative algebras by picking $\sigma\_k(a)\overset{\mathrm{def}}{=}-a$ for each $k\in\mathbb{Z}$, as in that case the above condition becomes
$$ab=(-1)^{\deg(a)\deg(b)}ba.$$
---
**Question.** So, are there any interesting/non-trivial "in nature" examples of $\mathbb{S}$-graded ring spectra?
|
https://mathoverflow.net/users/130058
|
Grading ring spectra over the sphere spectrum
|
One of the default examples of ordinary graded commutative rings is the polynomial ring $\mathbf Z[t]$. Let us first examine the analogue of that, and then see where else that leads!
**1. $S$-grading on $S\{t\}$**
For the sake of clarity, allow me to denote the underlying infinite loop space (equivalently: grouplike $\mathbb E\_\infty$-space) of the sphere spectrum $S$ by $\Omega^\infty(S)$.
Recall that, by the Barrat-Priddy-Quillen Theorem, the ininite loop space $\Omega^\infty(S)$ can be described as the group completion of the groupoid of finite sets and isomorphisms $\mathcal F\mathrm{in}^\simeq$. Consider the constant functor $\mathcal{F}\mathrm{in}^\simeq \to\mathrm{Sp}$ with value $S$ - since the latter is the monoidal unit, this is a symmetric monoidal functor. Left Kan extension along the canonical map $\mathcal F\mathrm{in}^\simeq \to (\mathcal F\mathrm{in}^\simeq)^\mathrm{gp}\simeq \Omega^\infty(S)$ produces a symmetric monoidal functor $\Omega^\infty(S)\to \mathrm{Sp}$, and as such exhibits an $S$-graded $\mathbb E\_\infty$-ring.
But which one? To figure out which one, note that the passage to the "underlying $\mathbb E\_\infty$-ring" of an $S$-graded $\mathbb E\_\infty$-ring is given by passage to the colimit. Thus the underlying $\mathbb E\_\infty$-ring, which we have just adorned with an $S$-grading, is
$$
\varinjlim\_{\Omega^\infty(S)}\mathrm{LKan}^{\Omega^\infty(S)}\_{\mathcal{F}\mathrm{in}^\simeq}(S)\simeq \varinjlim\_{\mathcal F\mathrm{in}^\simeq}S,
$$
where we used that the left Kan extension of a functor does not change its colimit. Now we can use the explicit description of the groupoid of finite sets $\mathcal F\mathrm{in}^\simeq \simeq \coprod\_{n\ge 0}\mathrm B\Sigma\_n$ to obtain
$$
\varinjlim\_{\mathcal F\mathrm{in}^\simeq} S\simeq \bigoplus\_{n\ge 0}S\_{h\Sigma\_n}.
$$
We may recognize this as the free $\mathbb E\_\infty$-ring on a single generator $S\{t\}$, corresponding in terms of spectral algebraic geometry to the smooth affine line $\mathbf A^1$ (as opposed to the flat affine line $\mathbf A^1\_\flat = \mathrm{Spec}(S[t])$ for the polynomial $\mathbb E\_\infty$-ring $S[t]\simeq \bigoplus\_{n\ge 0}S$).
**2. $S$-grading on symmetric algebras**
The previous example can be easily generalized by observing that $\mathcal F\mathrm{in}^\simeq$ is the free $\mathbb E\_\infty$-space ( = symmetric monoidal $\infty$-groupoid) on a single generator. That means that a symmetric monoidal functor $\mathcal F\mathrm{in}^\simeq \to \mathrm{Sp}$ (which always factors through the maximal subgroupoid $\mathrm{Sp}^\simeq\subseteq\mathrm{Sp}$) is equivalent to the data of a spectrum $M\in \mathrm{Sp}$ (the image of the singleton set). The functor is given by sending a finite set $I$ to the smash product $M^{\otimes I}$, and is evidently symmetric monoidal. The same Kan extension game as before now gives rise to an $S$-graded $\mathbb E\_\infty$-ring spectrum, this time with underlying $\mathbb E\_\infty$-ring
$$
\mathrm{Sym}^\*(M)\simeq \bigoplus\_{n\ge 0}M^{\otimes n}\_{h\Sigma\_n},
$$
the free $\mathbb E\_\infty$-ring generated by $M$. We recover the prior situation by setting $M=S$. For $M = S^{\oplus n}$, we obtain an $S$-grading on $S\{t\_1, \ldots, t\_n\}$, corresponding to the spectral-algebro-geometric smooth affine $n$-space $\mathbf A^n$.
**3. $S$-grading on $S\{t^{\pm 1}\}$**
Another way to generalize the example of the $S$-grading on $S\{t\}$ is to take directly the constant functor $\Omega^\infty(S)\to \mathrm{Sp}$ with value $S$, instead of starting with a constant functor on $\mathcal F\mathrm{in}^\simeq$ and Kan-extending it. This produces a perfectly good $S$-graded $\mathbb E\_\infty$-ring, which let us denote $S\{t^{\pm 1}\}$.
From the group-completion relationship between $\Omega^\infty(S)$ and $\mathcal F\mathrm{in}^\simeq$, it may be deduced that $S\{t^{\pm 1}\}$ and $S\{t\}$ are related in terms of $\mathbb E\_\infty$-ring localization as $S\{t^{\pm 1}\}\simeq S\{t\}[t^{-1}]$, justifying our notation. Here we are localizing $S\{t\}$ at the element $t\in \mathbf Z[t] = \pi\_0(S\{t\})$. In terms of spectral algebraic geometry, this is encoding the spectral scheme $\mathrm{GL}\_1$, the smooth punctured line.
**4. Remark on non-negative grading**
In algebraic geometry, we often prefer to think about non-negatively graded commutative rings than graded commutative rings. Just as the latter are equivalent to lax symmetric monoidal functor $\mathbf Z\to\mathrm{Ab}$, so are the former equivalent to lax symmetric monoidal functors $\mathbf Z\_{\ge 0}\to \mathrm{Ab}$.
By analogy, the "non-negatively $S$-graded $\mathbb E\_\infty$-rings" are lax symmetric monoidal functors $\mathcal{F}\mathrm{in}^\simeq \to\mathrm{Sp}$. Indeed, just as $\mathbf Z\_{\ge 0}$ is the free commutative monoid on one generator, so is $\mathcal F\mathrm{in}^\simeq$ the free $\mathbb E\_\infty$-space on one generator. That is the reason why we were encountering such functors above (and Kan extending them along group completion, as we would to view as $\mathbf Z\_{\ge 0}$-grading as a special case of a $\mathbf Z$-grading).
**5. Some actual "non-tautological" examples though**
So far, a not-completely-unreasonable complaint might be that all the examples of $S$-graded $\mathbb E\_\infty$-rings were sort of tautological.
For a very non-tautological example, see the main result of [this paper of Hadrian Heine](https://arxiv.org/abs/1712.00521). It shows that there exists an $S$-graded $\mathbb E\_\infty$-ring spectrum such that its $S$-graded modules are equivalent to the $\infty$-category of cellular motivic spectra. In fact, much more is proved: this situation is very common, and under some not-too-harsh compact-dualizable-generation assumptions, a symmetric monoidal stable $\infty$-category will be equivalent to $S$-graded modules over some $S$-graded $\mathbb E\_\infty$-ring. So you may just as well view this result as a wellspring of potentially interesting examples of $S$-gradings "occurring in nature"! :)
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10
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https://mathoverflow.net/users/39713
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403254
| 165,450 |
https://mathoverflow.net/questions/403235
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3
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>
> Is there any integral expression for $\log (X + Y) - \log (X)$ if $X$ and $Y$ are positive definite matrices?
>
>
>
Could anyone give some suggestion as to how to find such an integral expression if there is any?
Thanks a bunch.
|
https://mathoverflow.net/users/352167
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Question on integral expression of positive definite matrices
|
The [formula](https://math.stackexchange.com/a/2085547/87355)
$$\frac{d}{ds}\log Z(s) = \int\_0^1 [(1-t)I+tZ(s)]^{-1}Z'(s) [(1-t)I+tZ(s)]^{-1}\, dt,$$
with $Z(s)=X+sY$, gives upon integration of
$$\int\_0^1 \frac{d}{ds}\log Z(s)\,ds=\log Z(1)-\log Z(0)$$
an integral expression for
$$\log(X+Y)-\log X=\int\_0^1\int\_0^1 [(1-t)I+t(X+sY)]^{-1} Y [(1-t)I+t(X+sY)]^{-1}\, dt\,ds.$$
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3
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https://mathoverflow.net/users/11260
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403258
| 165,452 |
https://mathoverflow.net/questions/403272
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7
|
Given a ring spectrum $R$ and an $R$-module $E$, we have the [spectral symmetric algebra](https://ncatlab.org/nlab/show/spectral+symmetric+algebra) $\mathrm{Sym}\_R(E)$ of $E$ over $R$, defined by
$$
\begin{align\*}
\mathrm{Sym}\_R(E) &\overset{\mathrm{def}}{=} \mathrm{colim}\_{\mathbb{F}}(\Delta\_{E})\\
&\cong \bigoplus\_{n\in\mathbb{N}}E^{\otimes\_\mathbb{S}n}\_{\mathsf{h}\Sigma\_{n}},
\end{align\*}
$$
where $\mathbb{F}\overset{\mathrm{def}}{=}\mathsf{FinSets}^\simeq$ is the groupoid of finite sets and permutations. As [A Rock and a Hard Place showed here](https://mathoverflow.net/a/403254), the $\mathbb{E}\_\infty$-ring $\mathrm{Sym}\_R(E)$ comes with a natural grading by the sphere spectrum, inducing a $\mathbb{Z}$-grading on $\pi\_0(\mathrm{Sym}\_R(E))\cong\mathrm{Sym}\_{\pi\_0(R)}(\pi\_0(E))$. So e.g. picking $R=E=\mathbb{S}$, gives
$$
\begin{align\*}
\pi\_0(\mathrm{Sym}\_{\mathbb{S}}(\mathbb{S})) &\overset{\mathrm{def}}{=} \pi\_0(\mathbb{S}\{t\})\\
&\cong \mathbb{Z}[t],
\end{align\*}
$$
which carries the natural $\mathbb{Z}$-grading.
However, the $\pi\_0$ of an $\mathbb{S}$-graded ring can be [more complicated](https://mathoverflow.net/q/403217) than just a commutative $\mathbb{Z}$-grading, and for instance allows for the multiplication on the $\pi\_0$ to be supercommutative, satisfying $ab=(-1)^{\deg(a)\deg(b)}ba$. This led me to the following pair of questions:
1. Is there an "spectral exterior algebra" construction $\bigwedge\_RE$, whose $\pi\_0$ is the $\mathbb{Z}$-graded supercommutative exterior algebra $\bigwedge\_{\pi\_0(R)}\pi\_0(E)$? If so, does it come with an $\mathbb{S}$-grading?
2. [One of the more homotopy-theoretic points of view](https://arxiv.org/abs/1110.4753) on symmetric and exterior algebras is that the passage from the former to the latter corresponds to considering a larger portion of the sphere spectrum. More generally, do we have an $\mathbb{N}$-indexed sequence of "higher exterior algebra" constructions $\mathrm{Sym}\_R(E)$, $\bigwedge\_R(E)$, $\bigwedge^{\mathbf{2}}\_R(E)$, $\ldots$?
|
https://mathoverflow.net/users/130058
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Is there a "spectral exterior algebra" construction in higher algebra?
|
Interesting question! I can't give a real answer, but here are some idle musings:
Note that one way to encode exterior powers is as $\Lambda^i\_R(E) = \Sigma^{-i}(\mathrm{Sym}^i\_R(\Sigma(E)))$ (since placing the generators in degree one switches, by virtue of the Koszul sign rule from the usual $\Sigma\_n$-action to the alternating one). So the symmetric algebra in SAG or DAG on a $1$-shifted module looks like a sheared version of the exterior algebra. That is, if $R$ is an ordinary commutative ring and $M$ an $R$-module, then
$$
\mathrm{Sym}^\*\_R(\Sigma (M)) \simeq \Sigma^{\*}(\Lambda\_R^\*(M))
$$
(though here both symmetric and exterior algebras are understood in the spectral or derived sense, and may not agree with their usual algebraic counterparts unless $M$ is flat). In the DAG context, we similarly have
$$
\mathrm{Sym}^\*\_R(\Sigma^2(M)) \simeq \Sigma^{2\*}(\Gamma\_R^\*(M)),
$$
where $\Gamma^\*\_R(M)$ is the free divided power $R$-algebra generated by $M$. See Section 25.2 of [the SAG book](https://www.math.ias.edu/%7Elurie/papers/SAG-rootfile.pdf) for the proofs.
Therefore, perhaps the candidates for your degree $n$ higher exterior algebras might be some kind of $n\*$-fold desuspensions of the symmetric algebra $\mathrm{Sym}^\*\_R(\Sigma^n(R))$. For $n=0$, that would recover (flatness issues notwithstanding) the usual symmetric algebra, for $n=1$ the exterior algebra, for $n=2$ the free divided power algebra, and for $n\ge 3$, who knows? :)
Operating on this provisional understanding of what the higher exterior algebras are, let us discuss the $S$-grading. We see from the construction that the "unshearing" of it is the symmetric algebra $\mathrm{Sym}^\*\_R(\Sigma^n(R))$, which does indeed carry an $S$-grading. The question about whether the higher exterior algebras themselves carry a canonical $S$-grading now becomes a question about how, if at all, some kind of a "shearing" functor $(M\_n)\mapsto (\Sigma^n M\_n)$ interacts with $S$-gradings. I believe this going to be a very non-trivial question.
Even if we were not talking about the much-more-complicated $S$-gradings, but rather $\mathbf Z$-gradings, the shearing functor $\Sigma^\*$ is indeed an automorphism of the $\infty$-category of $\mathbf Z$-graded spectra $\mathrm{Fun}(\mathbf Z,\mathrm{Sp})$, but I don't think it is symmetric monoidal. Indeed, consider rather its inverse squared $\Sigma^{-2\*}$. If it were symmetric monoidal, it would send the polynomial $\mathbb E\_\infty$-algebra $S[t]\simeq \bigoplus\_{n\ge 0}S$, with its usual $\mathbf Z$-grading, into the "$2$-shifted polynomial $\mathbb E\_\infty$-algebra" $S[\beta] \simeq \bigoplus\_{n\ge 0}\Sigma^{-2n}S$. The latter thing does indeed exist as an $\mathbb E\_2$-ring, studied in Section 2.8 of [this paper by Lurie](https://people.math.harvard.edu/%7Elurie/papers/Waldhaus.pdf). But I have been told before that, due to some obstructions one can calculate, it can be shown that said $\mathbb E\_2$-ring does not support an $\mathbb E\_\infty$-structure. That would seem to imply that the shearing functor $\Sigma^{-2\*}$ (and consequently $\Sigma^\*$) can not be symmetric monoidal.
And that was just in the $\mathbf Z$-graded world, let alone all the additional complications that the $S$-graded setting might bring!
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6
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https://mathoverflow.net/users/39713
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403274
| 165,460 |
https://mathoverflow.net/questions/403275
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29
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Let $\mathrm{SO}(3)$ be the group of rotations of $\mathbb{R}^3$ and let $S\_\infty$ be the group of all permutations of $\mathbb{N}$. Is $\mathrm{SO}(3)$ isomorphic to a subgroup of $S\_\infty$?
This question is due to Ulam. It is discussed in V.2 of Ulam's "A collection of mathematical problems". It is also discussed in de le Harpe's "Topics in Geometric Group Theory", Appendix III.B. According to de la Harpe it was open as of 2003. Is it still open?
Ulam also asks the more general question of whether every Lie group is isomorphic to a subgroup of $S\_\infty$. De la Harpe constructs homomorphic embeddings $\mathbb{R} \to S\_\infty$ and $\mathbb{R}/\mathbb{Z} \to S\_\infty$. There is an obvious embedding $S\_\infty \times S\_\infty \to S\_\infty$, it follows that any connected abelian Lie group is a subgroup of $S\_\infty$. So it is natural to look at $\mathrm{SO}(3)$ or other low dimensional non-abelian Lie groups.
It is worth pointing out that we cannot hope to realize $\mathrm{SO}(3)$ as a subgroup of $S\_\infty$ without using a lot of choice. Suppose that $G$ is a compact lie group and $f : G \to S\_\infty$ is a homomorphism constructed without choice. Then $f$ is Baire measurable, so by the Pettis lemma $f$ is continuous, hence $f(G)$ is compact and connected. Finally, any compact connected subgroup of $S\_\infty$ is trivial. (In general closed subgroups of $S\_\infty$ are totally disconnected.) So a faithful action of $\mathrm{SO}(3)$ would be rather different than the usual kinds of actions.
|
https://mathoverflow.net/users/152899
|
Does $\mathrm{SO}(3)$ act faithfully on a countable set?
|
From MathSciNet:
Thomas, Simon Infinite products of finite simple groups. II.
J. Group Theory 2 (1999), no. 4, 401–434.
Summary: "(...) In the course of our classification proof, we also show that if $K$ is a field of cardinality $2^\omega$ and $G$ is a non-trivial linear group over $K$, then there exists a subgroup $H$ of $G$ such that $1<[G:H] \le \omega$."
[This is almost the same since enlarging $G$, we can always assume $G$ is simple, say $G=\mathrm{PSL}\_m(K)$, and then the induced homomorphism $G\to \mathrm{Sym}(G/H)$ is injective. The proof uses countable valuations as well.]
---
Ershov, Yu. L. ; Churkin, V. A. On a problem of Ulam. (Russian)
Dokl. Akad. Nauk 399 (2004), no. 3, 307-309.
Theorem 1. Every linear group over a field with continuum cardinality (in particular over the field $\mathbf{R}$ of all real numbers or over the field $\mathbf{C}$ of all complex numbers) can be embedded (as an abstract group) in the permutation group of a countable set.
(Unlike S. Thomas, they were aware of Ulam's question. On the other hand they were visibly not aware of S. Thomas paper.)
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16
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https://mathoverflow.net/users/14094
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403299
| 165,469 |
https://mathoverflow.net/questions/403294
|
4
|
$\DeclareMathOperator\Gr{Gr}$Let $E$ be a real, finite dimensional vector space of dimension $n$. Let $\Gr(k)$ be the set of linear subspaces of dimension $k$ of $E$. I am wondering what structures the manifold $\Gr(k)$ inherits from $E$, beyond differentiability. When $E$ is Euclidean, $\Gr(k)$ gets a Riemannian metric. When $E$ is not, I guess $\Gr(k)$ should still inherit a structure from the affine structure of $E$ but I can't clarify my ideas.
In particular, I am wondering if there is a notion of straight path between subspaces.
Can anyone help? Thanks.
|
https://mathoverflow.net/users/176470
|
Grassmannians on a vector space without metric
|
$\DeclareMathOperator\Gr{Gr}$I discussed this in my thesis.
**Lemma:** Every tangent space of the Grassmannian is a tensor product $T\_P \Gr(k)=P^\* \otimes (E/P)$; these isomorphisms are invariant under the general linear group $\operatorname{GL}\_E$.
**Proof:** Take a smooth path $P(t)\in \Gr(k)$. Write each $P(t)$ as the kernel of a linear map $\varphi(t)$, smoothly varying in $t$, say $\varphi(t)\colon E\to Q$, for fixed vector spaces $E,Q$, with $\varphi(t)$ onto. This $\varphi(t)$ is uniquely determined by its kernel $P(t)$ up to changing to $g(t)\varphi(t)$, where $g(t)$ is a path in $\operatorname{GL}\_Q$. Let $\bar\varphi(t)$ be the quotient isomorphism $E/P(t)\cong Q$ by quotienting out the kernel of $\varphi(t)$. Then check easily that $\eta(t)=\left.\bar\varphi(t)^{-1}\varphi'(t)\right|\_{P(t)}$ is well defined (i.e. unchanged if we replace $\varphi(t)$ by $g(t)\varphi(t)$). Check that to each tangent vector $v=P'(0)\in T\Gr(k)$, there is a unique well defined value of $\eta(0) \in P(0)^\*\otimes (E/P(0))$, so that $v\mapsto \eta(0)$ is a linear isomorphism $T\_P \Gr(k)\cong P^\*\otimes (E/P)$. $\Box$
In particular each tangent space contains a cone of tangent vectors which are rank one, i.e. pure tensors $\xi\otimes v$ for some $\xi\in P^\*$ and $v\in E/P$, called the bad cone; it was named by Gluck, Warner and Yang.
>
> H. Gluck, F. Warner, C. T. Yang, *Division algebras, fibrations of spheres by great spheres and the topological determination of a space by the gross behaviour of its geodesics*, **Duke Mathematical Journal**, vol. 50, no. 4, December 1983.
>
>
>
You can recover the Grassmannian from this field of cones in some sense; see papers of Hwang and Mok.
>
> J.M. Hwang, N. Mok, *Uniruled projective manifolds with irreducible reductive $G$-structures*, **Journal fur die reine und angewandte Mathematik**, vol. 490, 1997, pp. 55-64.
>
>
>
A tensor product structure on each tangent space is equivalent to that field of cones. A tensor product structure has local invariants, which one can uncover using Cartan's method of equivalence, and the case of maximal dimensional Lie algebra of local symmetry vector field is that of open subsets of the Grassmannian. In particular, a compact and simply connected manifold with a tensor product structure in its tangent spaces, with maximal dimensional symmetry Lie algebra, is diffeomorphic to the universal covering space of the Grassmannian with its usual tensor product structure. The proof is involved: you use the tensor product structure to reduce structure group of the frame bundle, and then arrive at the Cartan geometry of the Grassmannian. The complete proof, for a much broader class of geometric structures on manifolds (in the real and complex cases), appears in lemma 12 of my paper
>
> B. McKay, Holomorphic geometric structures on Kaehler-Einstein manifolds, **Manuscripta Math.**, vol 153 (no. 1-2), 2017, pp.1--34.
>
>
>
To answer your more recent question: The automorphism group of the Grassmannian does not preserve a projective connection or a path geometry, as I proved in
>
> McKay, Benjamin, *Complete complex parabolic geometries.* **Int. Math. Res. Not.** 2006.
>
>
>
In other words there is no family of paths on the Grassmannian which are, in local coordinates, precisely the solutions of a second order system of ordinary differential equations, and which is invariant under the action of the projective general linear group. But I suspect that there is an invariant family of paths everywhere tangent to the bad cone and defined by some ordinary differential equations of second or third order, so that is what I would suggest you might look for. The idea is to look at the representation theory of the projective group: in the Lie algebra $\mathfrak{sl}\_{n+1}$ of the projective group, take the noncompact simple root $\alpha$, and the root vectors $e\_{\alpha},e\_{-\alpha}$ and their bracket; together these three vectors span a copy of $\mathfrak{sl}\_2$ inside $\mathfrak{sl}\_{n+1}$. This $\mathfrak{sl}\_2$ has an orbit in the Grassmannian, which is a rational curve tangent to the bad cone. These curves form a small dimensional family, so solve some lower order ordinary differential equation. See
>
> McKay, Benjamin *Rigid geometry on projective varieties.* **Math. Z.** 272 (2012), no. 3-4, 761–791
>
>
>
for a discussion of these on arbitrary flag manifolds and on manifolds with parabolic geometries.
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14
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https://mathoverflow.net/users/13268
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403301
| 165,471 |
https://mathoverflow.net/questions/403304
|
2
|
Let $a \ge 0$, $b,c>0$ be fixed constants, and let $X$ be an $m \times d$ random matrix with entries drawn iid from $N(0,1/d)$. Consider the random psd matrix $S := a 1\_m 1\_m^\top + b XX^\top + c I\_m$.
>
> **Question 1.** In the limit $m,d \to \infty$ with $m/d \to \rho \in (0,\infty)$, what does $d^{-1}\mbox{trace}(S^{-2})$ converge to ?
>
>
>
**Observation.** If $a = 0$, then one computes
$$
d^{-1}\mbox{trace}(S^{-2}) = \frac{1}{d}\sum\_{i=1}^n \frac{1}{(bXX^\top +c)^2} \overset{a.s}{\to} b^{-2}m'\_{MP(\rho)}(-c/b),
$$
where $m\_{MP(\rho)}$ is the Stieltjes transform of the Marchenko-Pastur distribution with parameter $\rho$.
---
Update
------
I've often heard that
>
> "*Finite-rank perturbations don't change limiting empirical spectral distribution of random matrices.*"
>
>
>
Unfortunately, I can't find a definitive reference for this statement.
>
> **Question 2.** In view of the previous remark, is it true that $d^{-1}\mbox{trace}(S^{-2}) \overset{a.s}{\to} b^{-2}m'\_{MP(\rho)}(-c/b)$ for all $a \ge 0$ and $b,c>0$ ?
>
>
>
|
https://mathoverflow.net/users/78539
|
Compute the limit of trace of inverse of square of rank-1 perturbation of Wishart matrix
|
Let me try to answer the updated question. If $A$ and $B$ are Hermitian $m\times m$ matrices with $B$ of rank $r$ then the two sequences of eigenvalues $\lambda\_k(A+B)$ of $A+B$ and $\lambda\_k(A)$ of $A$, each sorted in ascending order, are related by
$$\lambda\_k(A)\leq\lambda\_{k+r}(A+B)\leq\lambda\_{k+2r}(A),\;\;1\leq k\leq m-2r.$$
For a proof see Theorem 8 of [these notes.](https://www.math.drexel.edu/~foucart/TeachingFiles/F12/M504Lect5.pdf)
In this case $A=S(a)$ differs from $A+B=S(0)$ by a rank-one matrix, $r=1$, so the eigenvalues of $S(a)$ are interlaced with those of $S(0)$. Their spectral densities in the large-$m$ limit must coincide and Question 2 can be answered in the affirmative.
|
2
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https://mathoverflow.net/users/11260
|
403312
| 165,475 |
https://mathoverflow.net/questions/403313
|
3
|
In an exercise of Voisin book, says:
Let $j:C\rightarrow S$ the inclusion of a smooth curve on a smooth connected projective surface. Set
$H=ker(j\_\*:H^1(C,\mathbb{Z})\rightarrow H^3(S,\mathbb{Z}))$.
We also write $A\subset J(C)$ for the Abelian subvariety corresponding to the Hodge substructure $H$.
I do not understand how the correspondence between Abelian subvarieties and Hodge substructures goes, so I will be grateful if someone can suggest to me a reference to learn it.
Thank you
|
https://mathoverflow.net/users/150116
|
Abelian varieties corresponding to Hodge substructures
|
Hodge substructures of $H^1(C,\mathbb Z)$ and Abelian subvarieties of $J(C)$ ar essentially the same thing:
In general, let $V,\omega$ be a free $\mathbb Z$-module of even rank with an integral symplectic form. A *polarized Hodge structure of weight 1* on $V$ is the data of a decomposition
$V\_{\mathbb C} := V\otimes\_{\mathbb Z} \mathbb C = V^{1,0} \oplus V^{0,1}$
with $V^{0,1} = \overline{V^{1,0}}$ and a positivity condition with respect to $\omega$. This is canonically equivalent to the data of a linear complex structure $I$ on $V\_{\mathbb R} := V\otimes\_{\mathbb Z} \mathbb R$ for which $\omega$ is of positive. The associated abelian variety is $V\_{\mathbb R} / V\_{\mathbb Z}$, with this linear complex structure.
A Hodge substructure is a $\mathbb Z$-submodule $W\subset V$ such that $W\otimes \mathbb C$ is compatible with the Hodge decomposition, i.e.
$$W\_\mathbb C:= W \otimes \mathbb C = W\_\mathbb C \cap V^{1,0} \oplus W \_\mathbb C \cap V^{0,1}~.$$
Equivalently, $W\_\mathbb R := W\otimes \mathbb R$ is $I$-invariant. Then $W\_\mathbb R/W$ is an abelian subvariety of $V\_\mathbb R / V$. Conversely, every abelian subvariety is a the quotient of a $I$-invariant subspace.
In your case, $V= H^1(C,\mathbb Z)$, $V^{1,0} = H^{1,0}(C)$, the associated abelian variety is $J(C)$ and $H= W$. What remains to understand is that $H$ is a Hodge substructure, and this is because $j\_\*$ is compatible with the Hodge decomposition (it maps $H^{1,0}$ to $H^{2,1}$ and $H^{0,1}$ to $H^{1,2}$).
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5
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https://mathoverflow.net/users/173096
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403317
| 165,477 |
https://mathoverflow.net/questions/402940
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3
|
The Auslander correspondence said there exists a bijection between the set of Morita-equivalence classes of representation-finite finite-dimensional algebras $\Lambda$ and that of finite-dimensional algebras $\Gamma$ with gl.dim $\Gamma \leq 2 \leq$ dom.dim $\Gamma$. It is given by $\Lambda \mapsto \Gamma:=\operatorname{End}\_{\Lambda}(M)$ for an additive generator $M$ of $\bmod \Lambda$.
I don't understand why $\Lambda \mapsto \Gamma:=\operatorname{End}\_{\Lambda}(M)$ is a bijection between the set of Morita-equivalence classes. And what is the importance of Morita-equivalence classes in representation theory of Artin algebra?
Thank you.
|
https://mathoverflow.net/users/118028
|
Why is Auslander correspondence a bijection between the set of Morita-equivalence classes?
|
Let me answer your second question about the importance of Morita-equivalence classes first. Two rings $R$ and $S$ are called Morita equivalent if $\operatorname{Mod} R$ and $\operatorname{Mod} S$ are equivalent as categories. If one considers representation theory as the study modules over a ring, then in some sense $R$ and $S$ are indistinguishable, so it makes sense to study rings up to Morita equivalence. (Of course, there are many other aspects of representation theory that people are insterested in, which differ a lot if one changes the Morita representative: monoidal structure, dimension of simple modules, existence of subrings, just to name a few).
Now with your particular question, you said that the map is defined by sending $\Lambda$ to $\operatorname{End}\_\Lambda(M)$ where $M$ is a additive generator of $\operatorname{mod} \Lambda$. If you would just define that on isomorphism classes, instead of Morita equivalence classes, then this would not be well-defined, as there are many different such additive generators, yielding non-isomorphic algebras. Just to give a trivial example: If $R=\Bbbk$ is the ground field, then $\Bbbk^2$ is an additive generator of $\operatorname{mod}\Lambda$ and $\operatorname{End}\_\Bbbk (\Bbbk^2)\cong \operatorname{Mat}\_{2\times 2}(\Bbbk)$ is Morita equivalent, but not isomorphic to $\Bbbk\cong \operatorname{End}\_\Bbbk(\Bbbk)$. A different way to fix this uses the basic representative on each side. Then you would send (a basic representation-finite) $\Lambda$ to $\operatorname{End}\_\Lambda(M)$ where $M$ has precisely one summand from each isomorphism class of indecomposable modules.
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2
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https://mathoverflow.net/users/15887
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403320
| 165,479 |
https://mathoverflow.net/questions/403310
|
9
|
In the context of [another MO question](https://mathoverflow.net/q/400714), the following question arose: Does there exist any software for detecting Brauer–Manin obstructions to the existence of integer solutions to a single polynomial equation?
Failing that, has anyone written down a detailed description of such an algorithm?
It may be that one (ahem) obstruction to the existence of such software is that the standard accounts of the Brauer–Manin obstruction are written in the language of modern algebraic geometry, which is unfamiliar to many people who might otherwise have the right skills to write the software. To some extent, such language is unavoidable, but it would be nice to have an account that is as elementary as possible. Perhaps some of the building blocks have already been implemented in (say) Sage, and it is not too hard to explain what is needed to put them together.
|
https://mathoverflow.net/users/3106
|
Software for detecting Brauer-Manin obstructions?
|
I strongly disagree with the assertion that "the language of modern algebraic geometry [...] is unfamiliar to many people who might otherwise have the right skills to write the software". You are denying the existence of a flourishing research field, computational arithmetic geometry! See e.g. this paper:
*Bright, M. J.; Bruin, N.; Flynn, E. V.; Logan, A.*, [**The Brauer-Manin obstruction and $\text{Ш}[2]$**](http://dx.doi.org/10.1112/S1461157000001455), LMS J. Comput. Math. 10, 354-377 (2007). [ZBL1222.11084](https://zbmath.org/?q=an:1222.11084).
From the abstract: "We discuss the Brauer-Manin obstruction on del Pezzo surfaces of degree 4. We outline a detailed algorithm for computing the obstruction and provide associated programs in MAGMA."
This isn't exactly the same question as you asked, since the relevant surfaces are intersections of two quadrics in $\mathbf{P}^4$, so not defined by a single equation; but it should serve to demonstrate that there is a substantial literature focussing on explicit algorithmic computations of the Brauer--Manin obstruction -- maybe you can find something matching your question more precisely among papers citing (or cited by) this one.
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11
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https://mathoverflow.net/users/2481
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403323
| 165,480 |
https://mathoverflow.net/questions/204790
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17
|
So this question probably shows my inner model theoretic ignorance, but:
In "Two remarks on elementary embeddings of the universe" (<http://projecteuclid.org/download/pdf_1/euclid.pjm/1102969567>), Jech defines - for $j: V\rightarrow M$ a definable-with-parameters elementary embedding - an inner model $L(j)$, and proves that $L(j)$ is the smallest inner model which *admits* $j$ (in the sense that the inner model thinks $j$ is an elementary embedding as well).
This is a neat, short argument, and the definition of $L(j)$ is very straightforward. Naively, I would suspect that properties of the inner model $L(j)$ would tell us interesting information about the embedding $j$, and similarly for models $L(\overline{j})$ defined for sequences of embeddings $\overline{j}$.
For instance, here's a question which seems natural to me: given an inner model $M\subset V$ and a family of embeddings $j\_\eta: V\rightarrow M$ $(\eta\in \kappa)$, it's reasonable to ask for which $I\subset \kappa$ is there an inner model $N$ admitting precisely the $j\_\eta$ with $\eta\in I$? In particular, given a left distributive algebra of embeddings, which sub-left distributive algebras can be captured this way? Naively, if I want to capture $\{j\_\eta: \eta\in I\}$, my first guess would be to look at $L(\{j\_\eta: \eta\in I\})$.
*NOTE: this isn't my actual question, I'm just including it to motivate interest in the $L(j)$s.*
However, I've never run across these models before, and I can't seem to find any recent reference to them. So, my question is:
>
> Are the $L(j)$s interesting at all from the point of view of modern inner model theory? If not, why not, and if so, why doesn't there seem to be any modern literature about them (maybe they are extremely hard to work with, or maybe there is such literature which I just haven't found yet)?
>
>
>
(For my purposes, "modern" means "since 1990.")
|
https://mathoverflow.net/users/8133
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Whatever happened to $L(j)$?
|
I recently noticed that Mitchell returned to $L[j]$ in ["Applications of the covering lemma for sequences of measures,"](https://www.ams.org/journals/tran/1987-299-01/S0002-9947-1987-0869398-2/S0002-9947-1987-0869398-2.pdf) where he shows that the model is quite a bit larger than one might expect. The thing is, $L[j]$ *isn't* $L[E\_j]$ where $E\_j$ is the short extender of $j$, it's $L[P]$ where $P = \{(x,A) : x\in j(A)\}$ is the entire *proper class long extender* of $j$. Assuming there is no inner model with a measurable $\kappa$ of order $\kappa^{++}$, Mitchell shows that if $j$ is nontrivial, $L[j]$ contains an iterate of the core model. Moreover, under the same assumption, he proves that if $j$ is the ultrapower embedding associated to a normal ultrafilter, then $L[j]$ satisfies $V = K$. So $L[j]$ contains many measurable cardinals of high order if $K$ does. As far as I know, no one has worked on the question at the level of strong/Woodin cardinals.
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6
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https://mathoverflow.net/users/102684
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403329
| 165,481 |
https://mathoverflow.net/questions/403321
|
1
|
Let $$r\_k(x)=\prod\_{j=1}^k {(\frac{x+j}{j}})^{\min(j,k-j)}.$$
Computations suggest that $$r\_{2k}(x)=\sum\_{j=0}^{(k-1)^2}a(2k,j)\binom{k^2+x-j}{k^2}$$ and $$r\_{2k+1}(x)=\sum\_{j=0}^{k^2-k}a(2k+1,j)\binom{k^2+k+x-j}{k^2+k},$$ where the coefficients $a(k,j)$ are positive, palindromic and gamma positive.
Is there an elementary proof?
|
https://mathoverflow.net/users/5585
|
Worpitzky-like identities?
|
Upgrading my comment to an answer. It is not hard to see that $r\_k(x)$ is the same as MacMahon's famous formula for the number of [plane partitions](https://en.wikipedia.org/wiki/Plane_partition) in a $\lfloor k/2 \rfloor \times \lceil k/2 \rceil \times x$ box. Then the same argument as in [my previous answer](https://mathoverflow.net/questions/402767/generating-functions-for-hankel-determinants-of-catalan-numbers/402795#402795) (see in particular Section 3.15.2 of Stanley's EC1, which explains that $a(k,j)$ is the number of linear extensions of $[\lfloor k/2 \rfloor]\times [\lceil k/2 \rceil]$ with $j$ descents) implies the $\gamma$-positivity you are interested in.
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3
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https://mathoverflow.net/users/25028
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403330
| 165,482 |
https://mathoverflow.net/questions/300151
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5
|
**Problem.** Let $K$ be a compact subset of the plane such that the projection of $K$ on each line has non-empty interior in the line. Has $K+K$ or $K-K$ non-empty interior in the plane?
**Remark.** The results of [this paper](https://arxiv.org/pdf/1805.01997) imply the affirmative answer to this problem for compact subsets $K$ of positive dimension in the plane.
|
https://mathoverflow.net/users/61536
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A compact set in the plane with small sum-set and large projections
|
The answer is negative. The key point is the following
**Lemma:** For every $L>0$ and $x\in\mathbb R^2, x\ne 0$, there exist $\ell>0$ and a closed origin-symmetric set $F\subset\mathbb R^2$ such that every interval $I$ of length $L$ on the plane contains a subinterval $J\subset I\cap F$ of length $\ell$ while $F\cap(F+x)=\varnothing$.
**Proof:** WLOG, $x=(0,1)$. Now choose a huge integer $N=N(L)$ and consider the sinusoidal strip $S=\{(x,y):|y-\sin x|\le\frac 1{10(2N+1)}\}$. Take $F=\cup\_{k\in\mathbb Z}(S+(0,\frac{2k}{2N+1}))$.
Now just enumerate some dense set $x\_1,x\_2,\dots$ on the plane, take $L\_1=1$ and use $x\_k$ and $L\_k$ to construct $F\_k$ and $L\_{k+1}$. Put $K=\cap\_k F\_k\cap \bar D(0,10)$, say (note that each finite intersection of $F\_k$ intersects every interval of length $1$, so this property will survive for the full intersection $\cap\_k F\_k$).
|
3
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https://mathoverflow.net/users/1131
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403358
| 165,488 |
https://mathoverflow.net/questions/403348
|
18
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I am reading through Clausen's and Scholze's Lectures on condensed mathematics. I am struggling to understand the concept of solid abelian groups so I am looking for some examples.
Is the underlying condensed abelian group of a finitely generated module (with the unique induced topology) over a Banach ring solid? Do you know of other examples?
|
https://mathoverflow.net/users/118220
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Examples of solid abelian groups
|
Here's a rule of thumb: As long as the construction is nonarchimedean and does not involve noncompleted tensor products, it's solid.
More precisely, anything you can build from discrete abelian groups by repeatedly forming limits (in particular, kernels), colimits (in particular, cokernels), extensions, and internal Hom's, is solid. This includes any nonarchimedean Banach ring: These are (usually) of the form $A=A\_0[\tfrac 1t]$ where $A\_0\subset A$ is the unit ball and $t\in A$ is some pseudouniformizer, and $A\_0$ is $t$-adically complete, $A\_0=\varprojlim\_n A\_0/t^n$. Here, all $A\_0/t^n$ are discrete abelian groups, thus solid; thus the limit $A\_0$ is solid; thus the colimit $A=A\_0[\tfrac 1t]$ is solid.
Now if $M$ is a finitely generated $A$-module, you can write it as a cokernel of a map $f: A^n\to A^m$, and this makes $M$ naturally into a solid $A$-module, again by the above principles. (Beware that if you first do this topologically, endowing $M$ with the quotient topology, you may run into the issue that the quotient topology may not be separated. In that case, the corresponding condensed $A$-module may not agree with the above cokernel of $f$ taken in condensed $A$-modules. But I'd simply argue that here the condensed perspective gives you a more sensible thing to do, namely just take the cokernel of $f$ in condensed $A$-modules.)
Also beware that if $A$ is a Banach ring over the real numbers, then it's not solid (unless $A=0$).
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31
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https://mathoverflow.net/users/6074
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403371
| 165,491 |
https://mathoverflow.net/questions/313956
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5
|
In two important papers of Wirsing, namely ["Das asymptotische Verhalten von Summen über multiplikative Funktionen"](https://link.springer.com/article/10.1007/BF01351892) (1961) and its [follow up](https://akademiai.com/doi/abs/10.1007/BF02280301?pubCode=mobile&journalCode=10473) (1967), several results on mean values of multiplicative functions are obtained. Specifically, Satz 1.1 and Satz 1.2.2 in the latter paper say the following. Suppose $f$ is a non-negative multiplicative function, satisfying
$$\sum\_{p \le x} \frac{\log p}{p} f(p) \sim \tau \log x$$
for some positive $\tau$ (here the sum is over primes). Suppose further there exists $C>0$ with $f(p)\le C$ for all primes $p$. Under some restrictions on the growth of $f(p^k)$ (which I shall not write down), one has
$$(\*) \sum\_{n \le x} f(n) =\frac{e^{-\gamma \tau}}{\Gamma(\tau)} \frac{x}{\log x} \prod\_{p \le x} ( \sum\_{k \ge 0} f(p^k)/p^k) (1+o(1)),$$
where $\gamma$ is the Euler-Masscheroni constant. This is Satz 1.1. If furthermore we have a multiplicative, real function $g$ with $|g(n)|\le f(n)$ for all $n$, then Satz 1.2.2 says
$$(\*\*) \frac{\sum\_{n \le x} g(n)}{\sum\_{n \le x} f(n)} = \prod\_{p} \frac{\sum\_{ k\ge 0} g(p^k)/p^k}{\sum\_{k \ge 0} f(p^k)/p^k} (1+o(1))$$
>
> Q1: Is there any result making the dependence on $f$ and $g$ of the error term in $(\*\*)$ explicit (perhaps under additional restrictions)?
>
>
>
For the applications I have in mind, I do not need a bound on the error term itself, but something weaker:
>
> Q2: Is there a sense in which $(\*\*)$ is uniform in $g$? Concretely, do we have
> $$\sup\_{|g| \le f} \left| \frac{\sum\_{n \le x} g(n)}{\sum\_{n \le x} f(n)} / \prod\_{p} \frac{\sum\_{ k\ge 0} g(p^k)/p^k}{\sum\_{k \ge 0} f(p^k)/p^k} - 1 \right| \to 0,$$
> where the supremum is taken over all real multiplicative functions $g$ with $|g(n)| \le f(n)$? If not, is there some natural subset of these functions for which we do have uniformity?
>
>
>
|
https://mathoverflow.net/users/31469
|
Uniformity in Wirsing's Mean Value Theorems
|
Effective versions of Wirsing's theorems are worked out in a recent paper of G. Tenebaum: ``Moyennes effectives de fonctions multiplicatives complexes'', published in Ramanujan J. 44 (2017), no. 3, 641–701. Questions 1 and 2 are resolved by his Théorème 1.3. He shows that error term is uniform in the sense that it depends only on certain (simple) invariants of the functions at hand.
|
0
|
https://mathoverflow.net/users/31469
|
403388
| 165,499 |
https://mathoverflow.net/questions/403389
|
9
|
Formula (12) in the paper
* Bauer, M., Chetrite, R., Ebrahimi-Fard, K., & Patras, F. (2013).
Time-ordering and a generalized Magnus expansion. Letters in
Mathematical Physics, 103(3), 331-350.
expresses the derivative of an exponential of a parameter-dependent Lie algebra element. The authors call it 'Duhamel's formula', without giving a reference.
Where and in which context did Duhamel prove this formula? (I know the formula as Kubo's formula, from his work in statistical mechanics.)
|
https://mathoverflow.net/users/56920
|
Duhamel's formula
|
Duhamel's formula for the derivative of the exponent of a matrix refers to [Jean-Marie-Constant Duhamel](https://en.wikipedia.org/wiki/Jean-Marie_Duhamel), who described it in [Eléments de calcul infinitésimal](http://catalogue.bnf.fr/ark:/12148/cb3037024) (volume 2, 1856; page 36) as a method to obtain solutions to inhomogeneous linear differential equations in terms of the solution to the homogenous equation.
In the context of Lie groups the [formula for the derivative of the exponential map](https://en.wikipedia.org/wiki/Derivative_of_the_exponential_map) is attributed to Schur (1891).
A modern proof is in these [notes.](https://personal.math.ubc.ca/~feldman/m428/duhamel.pdf) or alternatively [here](http://scipp.ucsc.edu/~haber/webpage/MatrixExpLog.pdf) (theorem 3). A formal proof is given by
\begin{align}
\frac{d}{dt}e^{A(t)}
&= \lim\_{N \to \infty}\frac{d}{dt}\left(1 + \frac{A(t)}{N}\right)^N\\
&= \frac{1}{N}\lim\_{N \to \infty}\sum\_{k=1}^N\left(1 + \frac{A(t)}{N}\right)^{k-1}A'(t)\left(1 + \frac{A(t)}{N}\right)^{N-k}~\\
&=\int\_0^1 e^{sA(t)}A'(t)e^{(1-s)A(t)}\,ds.
\end{align}
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11
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https://mathoverflow.net/users/11260
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403404
| 165,504 |
https://mathoverflow.net/questions/403381
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4
|
Let $C$ be a connected simplicial 2-complex, and $f: C \to \mathbb{S}^3$ an embedding in the 3-sphere. Assume that each of the link graphs of $C$ is connected, and that $f$ nice, e.g. locally flat or piecewise linear.
Is it true that there is a *simply connected* 2-complex $C'$, containing $C$ as a topological subspace, and an embedding $f': C' \to \mathbb{S}^3$ that coincides with $f$ when restricted to $C \subseteq C'$?
(If we relax the condition that $f'$ extends $f$, then the statement is proved by Carmesin in Theorem 7.1 of <https://arxiv.org/pdf/1709.04643>. Carmesin obtains an $f’$ with a rotation system that coincides with that of $f$ when appropriately restricted, but this is weaker than what I am asking for.)
|
https://mathoverflow.net/users/69681
|
Extending a 2-complex embedded in $\mathbb{S}^3$ into a simply connected one
|
Yes. [The tameness of $f$ makes our life *much* easier.]
It will be convenient for the proof to assume that $C$ is *pure*: that is, every vertex and edge of $C$ lies in some triangle of $C$. If this is not the case, then glue triangles on until it is the case.
Take $D = f(C)$; since $f$ is piecewise linear $D$ is tame and we can work with $D$ instead of $C$. Let $N$ be a very small regular neighbourhood of $D$. Note that $D$ is a *spine* for $N$ in the sense that $N - D \cong S \times (0, 1]$ (PL homeo). Let $\tau\_D : N \to D$ be the resulting deformation retraction. (In fact, we can here get away with using nearest point projection.) So $N \subset S^3$ is a PL sub-three-manifold and $S = \partial N$ is a separating surface. Note that $S$ is need not be connected and may have two-sphere components.
Define $M$ to be the closure of $S^3 - N$. Note that $S = \partial M$.
Let $E$ be a special spine for $M$. Again, we will assume that $E$ is pure. (Note that this means that we will have use Bing's house with two rooms, or similar, as special spines for three-ball components of $M$.) Thus $M - E \cong S \times (0, 1]$. Again let $\tau\_E : M \to E$ be the resulting deformation retraction. We can no longer use closest point projection, but we still arrange that $\tau\_E$ is PL.
Let $\Gamma$ be a spine for the surface $S$; that is, $S - \Gamma$ is a collection of disks. Again we assume that $\Gamma$ is pure - so while $\Gamma$ may have vertices of degree one, it does not have isolated vertices. Also, we isotope $\Gamma$ slightly so that it is transverse to $(\tau\_D|S)^{-1}(D^{(1)})$ and also to $(\tau\_E|S)^{-1}(E^{(1)})$.
We build a two-complex from $$D \,\, \sqcup \,\, \Gamma \times [-1, 1] \,\, \sqcup \,\, E$$ by attaching $\Gamma \times \{-1\}$ to $D$ via $\tau\_D$ and attaching $\Gamma \times \{1\}$ to $E$ via $\tau\_E$. This is the desired complex $C'$; the map $f'$ is inclusion. Note that $S^3 - C'$ is a collection of (open) three-balls. That is, $C'$ is a spine for $S^3$. Thus $C'$ is simply connected.
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5
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https://mathoverflow.net/users/1650
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403405
| 165,505 |
https://mathoverflow.net/questions/403345
|
3
|
Fix a field $k$ of characteristic zero, and let $G$ be a connected reductive algebraic $k$-group of isotropic rank $\ge 1$. Fix a maximal $k$-split torus $S$, and let $\Phi\_k$ be the relative root system of $G$ with respect to $S$. Assume that $\Phi\_k$ is reduced and irreducible.
By Theorem 2 of Petrov and Stavrova, for every root $\alpha \in \Phi\_k$ there is an embedding $X\_\alpha:V\_\alpha \to G$ where $V\_\alpha$ is a vector group (EDIT: vector group scheme), and the image of $X\_\alpha$ is the root subgroup $U\_\alpha$. Given $v \in V\_\alpha(k)$, $v \neq 0$, associated to $X\_\alpha(v) \in U\_\alpha(k)$ is a "Weyl group element" $w\_\alpha(v)$, using Lemma 1.3 of Deodhar. If $G$ is a Chevalley group, these are the elements denoted $w\_\alpha(t)$ in Steinberg, and $w\_\alpha(1)$ is literally a representative of a Weyl group element. Steinberg lists some relations, of interest here is (R7), which says that for any two roots $\alpha,\beta \in \Phi\_k$ and $v \in k$,
$$
w\_\alpha(1) \cdot X\_\beta(v) \cdot w\_\alpha(1)^{-1} = X\_{w\_\alpha \beta} \Big( c\_{\alpha,\beta} v\Big)
$$
This says that the action of the (lifts of) Weyl group elements on $G(k)$ by conjugation corresponds to the action of the Weyl group on $\Phi\_k$; conjugating the root subgroup $U\_\beta$ by a Weyl group representative $w\_\alpha(1)$ takes it to the root subgroup $U\_{w\_\alpha \beta}$. The coefficient $c\_{\alpha,\beta}$ is just a sign $\pm 1$ depending on $\alpha$ and $\beta$.
I have done a variety of explicit computations in quasi-split groups, and found in every case that
$$
w\_\alpha(u) \cdot X\_\beta(v) \cdot w\_\alpha(u)^{-1} = X\_{w\_\alpha \beta} \Big( c\_{\alpha,\beta}(u,v) \Big)
$$
for some function $c\_{\alpha,\beta}:V\_\alpha(k) \times V\_\beta(k) \to V\_{w\_\alpha \beta}(k)$. Steinberg's (R7) says that in the split case $c\_{\alpha,\beta}(1,v) = \pm v$. In the preprint (reference 3, equation 3) we give a version of this for a class of quasi-split special unitary groups, where $c\_{\alpha,\beta}(1,v) = \pm v$ or $\pm \overline{v}$, with the bar denoting a Galois automorphism of order 2. I have also done computations in a quasi-split special orthogonal group, where some more complicated functions $c\_{\alpha,\beta}$ arise.
**Question 1**: Why does the conjugation on the LHS above always end up in $U\_{w\_\alpha \beta}$, even in non-split cases? I suspect this is an obvious consequence of a Bruhat decomposition, but I don't understand that as well as I should.
**Question 2**: Is there a known generalization of this relation to reductive isotropic groups, or perhaps just for quasi-split groups?
1. *Deodhar, Vinay V.*, [**On central extensions of rational points of algebraic groups**](http://dx.doi.org/10.2307/2373853), Am. J. Math. 100, 303-386 (1978). [ZBL0392.20027](https://zbmath.org/?q=an:0392.20027).2.
2. *Petrov, V.; Stavrova, A.*, [**Elementary subgroups of isotropic reductive groups.**](http://dx.doi.org/10.1090/S1061-0022-09-01064-4), St. Petersbg. Math. J. 20, No. 4, 625-644 (2009); translation from Algebra Anal. 20, No. 4, 160-188 (2008). [ZBL1206.20053](https://zbmath.org/?q=an:1206.20053).
3. *Rapinchuk, I.; Ruiter, J.*, On abstract homomorphisms of some special unitary groups, arXiv:2107.07351, preprint.
4. *Steinberg, Robert*, [**Lectures on Chevalley groups**](http://dx.doi.org/10.1090/ulect/066), University Lecture Series 66. Providence, RI: American Mathematical Society (AMS) (ISBN 978-1-4704-3105-1/pbk; 978-1-4704-3631-5/ebook). xi, 160 p. (2016). [ZBL1361.20003](https://zbmath.org/?q=an:1361.20003).
|
https://mathoverflow.net/users/361094
|
Conjugation of root subgroups by the Weyl group
|
Let me use $a$ and $b$ for relative roots, so that I can later switch to $\alpha$ and $\beta$ for absolute roots.
If $b$ is a non-multipliable root, then, as you have [said](https://mathoverflow.net/questions/403345/conjugation-of-root-subgroups-by-the-weyl-group#comment1032860_403345), $V\_b$ is the $b$-root space, and $X\_b$ is an exponential-type map. Specifically, it is the unique group homomorphism $V\_b \to G$ whose derivative at the identity is the inclusion of the $b$-root subspace of $\operatorname{Lie}(G)$. It can be described ‘explicitly’, for small values of explicitly, as $v \mapsto \prod X\_\beta(v\_\beta)$, where $\beta$ runs over the absolute roots whose restriction to $S$ is $b$, and $v = \sum v\_\beta$.
If $a$ is also non-multipliable, then we have that $w\_a(u)$ is the *commuting* product $\prod w\_\alpha(u\_\alpha)$, where $\alpha$ runs over the absolute roots whose restriction to $S$ is $a$, and $u = \sum u\_\alpha$. (Actually, now that I think about it, I might already be assuming $G$ quasi-split here.) In particular, $w\_a(u)w\_a(1)^{-1}$ equals $\prod \alpha^\vee(u\_\alpha)$ (or maybe the inverse of this, depending how things are normalised; I didn't check).
You have already observed that $\operatorname{Int}(w\_a(1))\bigl(X\_b(v)\bigr)$ equals $\prod X\_{w\_a\beta}\bigl(c\_{a\beta}\operatorname{Ad}(w\_a(1))v\_\beta\bigr)$, where $c\_{a\beta} = \prod c\_{\alpha\beta}$. Now suppose that $G$ is quasi-split. Then the set of absolute roots $\beta$ restricting to $b$ is a Galois orbit, and it is clear that $\beta \mapsto c\_{a\beta}$ is constant on Galois orbits, so $\operatorname{Int}(w\_a(1))\bigl(X\_b(v)\bigr)$ equals $X\_{w\_a b}\bigl(c\_{a b}\operatorname{Ad}(w\_a(1)v)\bigr)$, where $c\_{a b}$ is the common value of $c\_{a\beta}$. Now just conjugate by $\prod \alpha^\vee(u\_\alpha)$ to finish.
|
1
|
https://mathoverflow.net/users/2383
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403409
| 165,506 |
https://mathoverflow.net/questions/192252
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4
|
I hope that this is the appropriate place for asking about a step I don't understand in a proof which I think is due to a lack of knowledge. This is a step in Drinfeld-Simpson's paper: ``*$B$ structures on $G$-bundles and local triviality*" in which they showed that under some nice conditions, every $G$-bundle on a curve admits a $B$-reduction. Let me get to the specific statement.
Suppose that $X$ is a smooth projective curve over an algebraically closed field $k$. Let $\mathcal{P}$ be a $G$-bundle on $X$ where $G$ is a linear algebraic group over $k$.
Now, some notation: let $B$ be a Borel of $G$, $\alpha\_i: H \rightarrow G\_m$ a simple root. If $\mathcal{E}$ is a $B$-bundle then I can produce the $G\_m$-bundle associated to $\mathcal{E}$ and call its degree $\deg\_i(\mathcal{E})$ just as a line bundle over a smooth projective curve.
The problem set up: suppose I now have another $G$-bundle $\mathcal{P}'$ and (1) I have a trivialization of both $\mathcal{P}$ and $\mathcal{P}'$ on an open subscheme of $X$ and I have chosen $h$ an isomorphism between them and (2) a $B$-reduction on $\mathcal{P}$ (with corresponding $B$-bundle $\mathcal{E}$) which induces a $B$-reduction on $\mathcal{P}'$ with corresponding $B$-bundle $\mathcal{E}'$.
**Question**:
Why does there exist a number $c$ such that $-c < \deg\_i(\mathcal{E}) - \deg\_i(\mathcal{E}') < c$ depending on the singularities of $h$? (as claimed)
I suppose I don't see how these numerical data relate to one another.
|
https://mathoverflow.net/users/24706
|
A step in the proof of the Drinfeld-Simpson theorem
|
Fix two principal $G$-bundles $P\_G$ and $P'\_G$ on $X$ along with an identification $\beta$ away from a divisor $D$. Also fix a simple root $\alpha\_i$.
Let $P\_B$ and $P\_B'$ be arbitrary $B$-reductions (of $P\_G$ and $P\_G'$ respectively) compatible with $\beta$. We write
$$
L = P\_B \overset{B}{\times} \alpha\_i \qquad E = P\_G \overset{G}{\times} V\_{\alpha\_i},
$$
and similarly for $L'$ and $E'$. So $P\_B$ exhibits $L$ (resp. $L'$) as a "highest weight" line subbundle of $E$ (resp. $E'$).
By use of $\beta$, we may write
$$
E(-m \cdot D) \subset E' \subset E(n \cdot D)
$$
for some integers $m$ and $n$ independent of $P\_B$ and $P'\_B$. It follows that
$$
L(-m \cdot D) \subset L' \subset L(n \cdot D).
$$
Taking degrees gives a bound of the desired type.
|
1
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https://mathoverflow.net/users/362094
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403411
| 165,507 |
https://mathoverflow.net/questions/403133
|
18
|
**Question:** Is there a polynomial $f \in \mathbb{Z}[x]$ with $\deg(f) \geq 7$ such that
>
> 1. all roots of $f$ are distinct integers; and
> 2. all roots of $f'$ are distinct integers?
>
>
>
**Background:**
I asked a related question about four years ago on MSE ([**2312516**](https://math.stackexchange.com/questions/2312516/polynomials-f-and-f-with-all-roots-distinct-integers)) that led to the largest known examples having degree six. Shortly after, I suggested an investigation into this problem in the Polymath Proposal big-list on MO ([**219638**](https://mathoverflow.net/questions/219638/proposals-for-polymath-projects/271653#271653)).
Since I haven't seen any subsequent traction post-2017, I would be interested as to whether anyone knows of progress on this problem or ways of dealing with it in this particular direction. (A *different* direction could be e.g. requiring that the distinct integer roots are a feature of all derivatives – not just the first derivative.)
As a clarifying example, here is one of the degree six polynomials:
$$f(x) = (x^2−3130^2)(x^2−3590^2)(x^2−10322^2)$$
has distinct integer roots at $x = \pm 3130, \pm 3590, \pm 10322$, and [**differentiating**](https://www.wolframalpha.com/input/?i=d%2Fdx+%28x%5E2%E2%88%923130%5E2%29%28x%5E2%E2%88%923590%5E2%29%28x%5E2%E2%88%9210322%5E2%29) yields:
$$f'(x) = 6x(x^2 - 3366^2)(x^2 - 8650^2)$$
which has distinct integer roots at $x = 0, \pm 3366, \pm 8650$.
(So, it is straightforward to verify examples but seemingly difficult to generate them!)
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https://mathoverflow.net/users/22971
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Distinct integer roots for a degree 7+ polynomial and its derivative
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Suppose one wishes to solve a system $P\_1(n\_1,\dots,n\_k) = \dots = P\_m(n\_1,\dots,n\_k)=0$ of diophantine equations involving polynomials $P\_1,\dots,P\_m$ of various degrees. We have the following probabilistic heuristic (discussed for instance in [this blog post of mine](https://terrytao.wordpress.com/2012/09/18/the-probabilistic-heuristic-justification-of-the-abc-conjecture/)): if we pick $n\_1,\dots,n\_k$ to be random integers of size $N$ (here we make the assumption that the dominant scenario is one in which the $n\_1,\dots,n\_k$ all have comparable magnitude), then each equation $P\_j(n\_1,\dots,n\_k)=0$ heuristically has a probability $\sim N^{-\mathrm{deg} P\_j}$ to hold, so (assuming no unusual correlations) the whole system should be satisfied with probability $\sim N^{-\mathrm{deg} P\_1 - \dots - \mathrm{deg} P\_m}$. On the other hand, we have $\sim N^k$ choices of candidates $(n\_1,\dots,n\_k)$ here. Thus, if we define the *degree surplus* (or *degree-of-freedom deficit*) of this system to be $\mathrm{deg} P\_1 + \dots + \mathrm{deg} P\_m - k$, we arrive at the following naive predictions:
* For a negative surplus, solutions should be quite abundant (unless there are some obvious local obstructions, or maybe some less obvious obstructions such as [Brauer-Manin obstructions](https://en.wikipedia.org/wiki/Manin_obstruction)), and locatable by brute force search.
* For a positive surplus, one would typically only expect a small number of sporadic or degenerate solutions, or no solutions at all, unless there is some algebraic miracle that allows one to lower the surplus, for instance through some birational embedding. If the surplus is close to zero, e.g., if it is equal to $1$, then it might not be too much to ask for such a miracle, but when the surplus is large then this begins to look unreasonable.
* For zero surplus, the situation is delicate and could go either way.
Very roughly speaking, this surplus is a crude measure of how many "algebraic miracles" would be needed to be present in this problem in order to produce a large number of non-trivial integer solutions.
This heuristic has to be taken with several grains of salt - for instance, one needs the polynomials $P\_1,\dots,P\_m$ to be "independent" in some sense for this prediction to be accurate, one is implicitly assuming that the $n\_1,\dots,n\_k$ are probably of comparable magnitude, and this notion of surplus is not a birational invariant, in contrast to more intrinsic notions such as the genus of an algebraic curve - but it can still serve as a starting point for making initial guesses. Some examples:
* The Fermat equation $a^n + b^n = c^n$ has surplus $n-3$ and so one expects a lot of integer solutions for $n=2$, almost no non-trivial solutions for $n>3$, and no definitive conclusion for $n=3$. (Of course, as it turns out, there are no non-trivial solutions for $n=3$, but the story for say $a^3 + b^3 = c^3 + h$ for other constants $h$, e.g., $h=33$, can be different; see <https://en.wikipedia.org/wiki/Sums_of_three_cubes> .)
* Rational points on elliptic curves $y^2 = x^3 + ax + b$ in Weierstrass form, after clearing out denominators, becomes a cubic equation in three integer variables with a surplus of $3-3=0$, which is consistent with the fact that some elliptic curves have positive rank and thus infinitely many rational points, and others have rank zero and only finitely many rational points (or none at all).
* An elliptic equation in quartic form $y^2 = a\_4 x^4 + \dots + a\_0$ has a surplus of $4-3=1$ so one would naively expect very few rational points here; but (assuming the existence of a rational point) one can use a (rational) Mobius transformation (moving the rational point to infinity in a suitable fashion) to convert the equation into Weierstrass form, thus lowering the degree surplus down to $0$.
The question here is that of finding integer solutions $a\_1,\dots,a\_d,b\_1,\dots,b\_{d-1}$ to the equation
$$ \frac{d}{dx} (x-a\_1) \dots (x-a\_d) = d (x-b\_1) \dots (x-b\_{d-1}).$$
Comparing coefficients, the naive surplus here is
$$ 1 + 2 + \dots + d-1 - (d + d-1) = \frac{d^2-5d+2}{2}.$$
As noted in comments, this is negative for $d=1,2,3,4$, suggesting plentiful solutions in those cases. For $d=5$ the surplus is $1$, suggesting a minor miracle is needed to find solutions. For $d>5$ the surplus is quite large and one would not expect many solutions. However, with symmetry reductions one can do a bit better. For $d$ even one can look at the symmetrised equation
$$ \frac{d}{dx} (x^2-a^2\_1) \dots (x^2-a^2\_{d/2}) = d x(x^2-b^2\_1) \dots (x^2-b^2\_{d/2-1})$$
and now the surplus is
$$ 2 + 4 + \dots + (d-2) - (\frac{d}{2} + \frac{d}{2}-1) = \frac{d^2-6d+4}{4}.$$
In particular for $d=6$ the surplus is reduced from $4$ to $1$, so one is only "one miracle away" from finding many solutions in some sense. But it gets rapidly worse, for instance for $d=8$ this symmetrised problem has surplus $5$ (though this still improves substantially from the non-symmetric surplus of $13$).
In the [paper of Choudry](https://www.sciencedirect.com/science/article/pii/S0022314X15000633) linked by the OP, some ad hoc substitutions and birational transformations are used in the non-symmetric $d=5$ case and the symmetric $d=6$ case (both of which have a surplus of $1$, as noted before) to extract a subfamily of solutions parameterized by a specific elliptic curve in quartic form; the surplus is still $1$, but as discussed previously one can then use a Mobius transform to send the surplus to $0$, giving hope that solutions exist. (One still has to be lucky in that the rank of the elliptic curve is positive, but this happens to be the case for both of the elliptic curves Choudry ends up at, so that paper in fact succeeds in obtaining an infinite family of solutions in both cases.)
Thus I would be doubtful of any non-degenerate solutions (beyond perhaps some isolated sporadic ones) for $d \geq 7$, even if one imposes symmetry; one would need quite a long sequence of miraculous surplus-lowering transformations to be available, and absent some remarkable algebraic structure in this problem (beyond the affine and permutation symmetries) I don't see why one should expect such a sequence to be present. Establishing this rigorously (or even conditionally on conjectures such as the [Bombieri-Lang conjecture](https://en.wikipedia.org/wiki/Bombieri%E2%80%93Lang_conjecture)) looks quite hard though. Being homogeneous, the problem is one of finding rational points to a certain projective variety (avoiding certain subvarieties corresponding to degenerate solutions); if one could somehow show this variety to be of general type, Bombieri-Lang would then say that one should only expect solutions that are trapped inside lower dimensional subvarieties, but a complete classification of such subvarieties looks extremely tedious at best given the relatively high dimension and degree. (And in view of the undecidability of Hilbert's tenth problem, which already shows up at relatively low choices of degree and dimension, there is little chance of a systematic way to resolve these questions even with the aid of conjectures such as Bombieri-Lang.) Still I would very much doubt that there are any non-trivial infinite families of solutions for $d>6$.
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https://mathoverflow.net/users/766
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403425
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https://mathoverflow.net/questions/352051
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10
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Every monoidal category $(\mathcal{C},\otimes)$ can be seen as a one-object bicategory: the morphisms are the objects of $\mathcal{C}$, and the $2$-morphisms are the morphisms of $\mathcal{C}$. In every bicategory, we can speak of left/right Kan extensions. Specifically, if $P$ is an object of $\mathcal{C}$ (the one "along which" we extend) and $F$ is another object of $\mathcal{C}$, then $\mathrm{Lan}\_P(F)$ is an object of $\mathcal{C}$ equipped with a morphism $\alpha : F \to \mathrm{Lan}\_P(F) \otimes P$, such that the evident universal property is satisfied: If $G$ is another object with a morphism $\beta : F \to G \otimes P$, then there is a unique morphism $\gamma :\mathrm{Lan}\_P(F) \to G$ with $(\gamma \otimes P) \circ \alpha = \beta$. The notation $\mathrm{Lan}\_P(F) = F/P$ makes sense, I would say: the left Kan extension tries to approximate this "quotient object".
Is this notion of a Kan extension in a monoidal category already known under a different name? Has it been studied, at least in some examples? I think that the *right* Kan extension is just the internal hom $[P,F]$. So by dualization, the left Kan extension is a kind of internal "co-hom".
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https://mathoverflow.net/users/2841
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Kan extensions inside a monoidal category
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It is certainly the case that the duals of internal homs have appeared significantly less in the categorical literature. I've included a few more references below, but I am not sure this is a satisfying answer; I have not found any reference that lists many naturally-occurring examples.
In the cocartesian monoidal setting, these are known as *coexponential objects*, and are studied in papers such as Filinski's [Declarative Continuations and Categorical Duality](https://citeseerx.ist.psu.edu/viewdoc/download?doi=10.1.1.43.8729&rep=rep1&type=pdf), and Niefield–Wood's [Coexponentiability and Projectivity: Rigs, Rings, and Quantales](http://www.tac.mta.ca/tac/volumes/32/36/32-36abs.html).
More generally, monoidal categories with internal cohoms are usually called *coclosed*. Examples are studied in Wedhorn's [On Tannakian duality over valuation rings](https://www.sciencedirect.com/science/article/pii/S0021869304004107), Chikhladze–Lack–Street's [Hopf monoidal comonads](http://web.science.mq.edu.au/%7Estreet/ChLaSt.pdf), and Silantyev's [Quantum Representation Theory and Manin matrices I: finite-dimensional case](https://arxiv.org/abs/2108.00269) (see §4.4.1 and Proposition 5.16 for instance), which refers to earlier work of Manin; and are made essential use of in Lyubinin's [Tannaka duality, coclosed categories and reconstruction for nonarchimedean bialgebras](https://arxiv.org/abs/1411.3183). However, in several of the explicit examples in the literature, coclosure follows from closure and duality, which makes for less interesting examples.
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6
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https://mathoverflow.net/users/152679
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403432
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https://mathoverflow.net/questions/403415
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6
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I am mainly interested in the $3$-dimensional case. It is a well-known fact, following from the work of E. E. Moise and R. H. Bing in the 1950s, that every $3$-dimensional topological manifold (with or without boundary) admits a triangulation, i.e. its homeomorphic to (the geometric realization of) an abstract simplicial complex. Furthermore, it is a well known fact that a manifold is piecewise-linear if and only if it admits a combinatorial triangulation, i.e. a triangulation in which the link of each simplex is Pl-homeomorphic to a sphere, and that in $d\leq 4$, every triangulation of a manifold is combinatorial. In other words, every $3$-manifold admits a PL-structure.
I am interested in the other way round: Is there a bunch of properties an abstract simplicial complex has to have in order to define a topological manifold? Clearly, not all $3$-dimensional simplicial complexes which one can draw give rise to a manifold. The complex should be at least pure and non-branching, I guess. Is it maybe enough to assume that a complex is combinatorial?
In the literature, I also have found the notions of ''pseudo-manifolds'', which are abstract simplicial complexes, which are pure, non-branching and strongly-connected. How is this related to my question?
Any help is appreciated. If someone could provide some reference, I would be happy too.
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https://mathoverflow.net/users/259525
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Properties a triangulation must have in order to describe a manifold
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From the comments:
1. Suppose that $T$ is a triangulation. If all vertex links are PL $(n-1)$-dimensional spheres then the realisation space of $T$ is a PL manifold and thus a topological manifold. (In the compact case this is equivalent to the usual definition.)
2. There are triangulations of topological manifolds (in fact, of $S^5$) that do not have this property. Examples come from the [double suspension theorem](https://en.wikipedia.org/wiki/Double_suspension_theorem) of Cannon and also Edwards.
Furthermore:
3. There are triangulations of topological manifolds that have no PL structure. This is (essentially) the failure of the [Hauptvermutung](https://en.wikipedia.org/wiki/Hauptvermutung).
4. There are closed topological manifolds that do not admit any triangulation. This is the failure of the Triangulation Conjecture, and is obtained by Casson in dimension four and to [Manolescu](https://arxiv.org/abs/1607.08163) in all higher dimensions.
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6
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https://mathoverflow.net/users/1650
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403435
| 165,512 |
https://mathoverflow.net/questions/403429
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1
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Let $H=(V,E)$ be a [hypergraph](https://en.wikipedia.org/wiki/Hypergraph). For $v\in V$ we let $v^\* = \{e\in E:v\in e\}$. We define the *dual* of $H$ by $H^\*= (E, V^\*)$ where $V^\* = \{v^\*: v\in V\}$. We say that a $H$ is *self-dual* if $H \cong H^\*$.
Is there a self-dual hypergraph $H = (\omega, E)$ with the following property?
>
> Whenever $e\_0\neq e\_1 \in E$ we have both $e\_0 \not\subseteq e\_1$ and $|e\_0| \neq |e\_1|$.
>
>
>
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https://mathoverflow.net/users/8628
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Self-dual hypergraph on $\omega$
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Expanding my comments into an answer: by rephrasing this as a question of constructing a symmetric $\omega\times\omega$ matrix with the same properties, we can give an explicit construction. For convenience's sake I'm going to start numbering at 0. Let $e\_0 = \{v\_0, v\_1\}$ and for $n\geq 1$ let $e\_n = \{v\_i\} \cup \{v\_j: f(n)\lt j\leq f(n+1)\}$, where $f(n)$ is the sum-of-integers function $f(n) = \frac12n(n+1)$ and $i$ is the unique number such that $f(i)\lt n\leq f(i+1)$ — in other words, $i$ is chosen to make the generated matrix symmetric.
This hypergraph has the nice property of being connected: for any two vertices there's some chain of edges that connects them to each other. In fact, it's a tree; for any two vertices there's exactly one chain of edges connecting them, or from a different perspective for any edge $e\_i$ there's exactly one edge $e\_j$ with $j\lt i$ and $e\_i\cap e\_j\neq\emptyset$. By allowing for more overlap in one half of the matrix, we can get 'greater' connectivity: for instance, take $e\_0=\{v\_0, v\_1, v\_2\}$, $e\_1 = \{v\_0, v\_2, v\_3, v\_4\}$, $e\_2 = \{v\_0, v\_1, v\_4, v\_5, v\_6, v\_7\}$, etc.
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2
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https://mathoverflow.net/users/7092
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403436
| 165,513 |
https://mathoverflow.net/questions/403441
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7
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$\DeclareMathOperator\Diff{Diff}$Suppose for simplicity that $X$ is affine, it is then possible to define $\Diff(X)$ — the ring of Grothendieck differential operators. When $X$ is smooth, then
>
> **Definition.** the category of $D$-modules on $X$ is defined to be modules over $\Diff(X)$. (Category **1**)
>
>
>
However, when $X$ is singular, this is not the right category to consider. One usually follows Kashiwara's approach:
>
> **Definition.** choose a closed embedding $X\hookrightarrow V$ and define $D$-modules be to modules over $\Diff(V)$ such that are (set-theoretically) supported $X$. (Category **2**)
>
>
>
The usual reason I heard for why to consider the second category is that $\Diff(X)$ behaves badly when $X$ is singular, and specifically people will point out that $\Diff(X)$ is not Noetherian. (Noetherian = left + right.) For example, this is the case when $X$ is the 'cubic cone' [BGG72].
However, I am no longer satisfied with this answer because of the following:
>
> (1), when $X$ is a curve then $\Diff(X)$ is Noetherian. [SS88]
>
> (2), when $X=V/G$ a quotient singularity then $\Diff(X)$ is Noetherian.
>
>
>
But in these cases, one still considers Category **2** for these $X$. So it has to be the case that, in general and in these cases, $\Diff(X)$ is bad not just because it is not Noetherian, it is also bad for other reasons. So my question is:
>
> **Question:** Why do we work in category **2** in the situations above. Or a better questions, what is **bad** about $\Diff(X)$ besides not being Noetherian.
>
>
>
Note my question is not how to work in category **2**, but why it fails badly if we work in category **1** in situations (1) and (2).
It is worth to remark that:
>
> in (1), if the curve is cuspidal then category **1** $\cong$ category **2**. [SS88] generalised in [BZN04]
>
> in (2), if the $X=\mathbb{C}^2/(\mathbb{Z}/2\mathbb{Z})$ then category **1** $\cong$ category **2** (I think this is true, but do please correct me if I am wrong.)
>
>
>
[BGG72] I. N. Bernˇste ̆ın, I. M. Gel’fand, and S. I. Gel’fand. Differential operators on a cubic cone. Uspehi Mat. Nauk, 27(1(163)):185–190, 1972.
[BZN04] David Ben-Zvi and Thomas Nevins. Cusps and D-modules. Journal of the American Mathematical Society, 17.1:155–179, 2004
[SS88] S. P. Smith and J. T. Stafford. Differential operators on an affine curve. Proc. London Math. Soc. (3), 56(2):229–259, 1988.
**Noted later:** actually it is not true that on $X=\mathbb{C}^2/(\mathbb{Z}/2\mathbb{Z})$ then category **1** $\cong$ category **2**, sorry for the confusion.
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https://mathoverflow.net/users/111070
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Why is the ring of Grothendieck differential operators bad when $X$ is singular?
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One thing that's wrong with the Grothendieck definition on singular varieties is the same thing that's wrong with defining the tangent space at a singular point rather than the tangent complex - it's not sufficiently derived (ie it's a strange truncated notion of the "true" derived object), which is why several different attempts to define D-modules give different answers there (the one quoted by OP, the notion of modules over vector fields - ie the algebra of differential operators generated by first order ones, which needn't be the full algebra in the singular case, and crystals - which amounts to the "option 2" one).
If you define D-modules in a derived fashion, as is done e.g. in the book of Gaitsgory-Rozenblyum, then order is restored: all the different natural notions you might come up with agree [here it's important to be in characteristic zero, a whole lot more interesting stuff happens in positive characteristic]. If you replace the naive notion of the sheaf vector fields by the tangent complex and consider modules for it, or define the Grothendieck differential operators derivedly (as a groupoid algebra for the de Rham groupoid of X), or define crystals (option 2), you get the same notion.
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9
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https://mathoverflow.net/users/582
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403445
| 165,516 |
https://mathoverflow.net/questions/403418
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1
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I am working on a paper of Elie Aidekon : [*‘Speed of the biased random walk on a Galton–Watson tree’*](https://doi.org/10.1007/s00440-013-0515-y) and have a question about one transformation in a proof:
\begin{align}
& 1+\frac{1}{1-\lambda}+\mathbb{E} \biggl[ \frac{\beta\_n(I) \mathbf{1}\_{\{ \beta(j)=0 \ \forall j\neq I\}}}{\lambda-1+\sum\_{i=1}^{\nu(e)}\beta\_n(i)}\biggl] \\
\tag{1}
\label{1}
={} & \frac{\lambda}{1-\lambda}+\mathbb{E} \biggl[ \frac{\beta\_n(I) \mathbf{1}\_{\{ \beta(j)=0 \ \forall j\neq I\}}}{\lambda-1+\beta\_n(I)}\biggl] \\
\tag{2}
\label{2}
={} &\frac{\lambda}{1-\lambda}+\mathbb{E} [\nu q^{\nu-1}] \mathbb{E} \biggl[ \frac{\beta\_{n-1}(e)}{\lambda-1+\beta\_{n-1}(e)}\biggl].
\end{align}
Some additional informations about it:
* It holds that $\lambda<1$,
* $\beta(e)=\mathcal{P}^{e}(\tau\_{e\_{\*}}=\infty)$ is the probability that the random walk start in the root $e$ of a tree and never hits the parent $e\_\*$,
* $\beta\_n(e)=\mathcal{P}^{e}\bigl(\tau^{(n)}<\tau\_{e\_\*}\bigl)$ is the probability that the random walk start in the root $e$ of a tree and hit level $n$ before it hit the parent $e\_\*$.
I absolutely don’t know where the $\frac{\lambda}{1-\lambda}$ in the \eqref{1} equation comes from and how the denominator short up to $\lambda-1+\beta\_n(I)$. Have someone a tip for me?
In the \eqref{2} equation I know that the *branching property* is used, but I don’t see this. I just know the branching property defined by ‘lines’ and can’t see how to take this definition for markov chains (without lines).
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https://mathoverflow.net/users/362153
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Random walks on Galton–Watson trees
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It seems no one checked this calculation prior to the publication of this paper. The factor $\frac{\lambda}{1-\lambda}$ in (1) seems mistaken, it should be $\frac{2-\lambda}{1-\lambda}$. This has no effect on the finiteness claimed in the lemma. The denominator in (1) indeed takes the given form because all the other summands in that denominator vanish due to the indicator in the numerator.
To go from (1) to a variant of (2) observe that
$$\beta\_n(I) \mathbf{1}\_{\{ \beta(j)=0 \ \forall j\neq I\}}=\sum\_{i=1}^{\nu(e)}\mathbf{1}\_{\{I=i\}} \beta\_n(i) \mathbf{1}\_{\{ \beta(j)=0 \ \forall j\neq I\}} \quad (\*)\,.$$
Next we wish to take conditional expectations given $\nu(e)$. For each $i,j \le \nu(e)$ we have
$$ \mathbb{E} \biggl[ \frac{\beta\_n(i) \mathbf{1}\_{\{I=i\}}}{\lambda-1+\beta\_n(i)} \,\bigg| \, \nu(e)\biggl] =\frac{\beta\_{n-1}(e)\mathbf{1}\_S} {\lambda-1+\beta\_n(e)}
$$
by the branching property, and $P[\beta(j)=0 |\nu(e)] =q$ since $\lambda<1$. Thus using independence of the subtrees determined by the offspring of $e$, we get
$$\mathbb{E} \biggl[ \frac{\beta\_n(I) \mathbf{1}\_{\{ \beta(j)=0 \ \forall j\neq I\}}}{\lambda-1+\sum\_{i=1}^{\nu(e)}\beta\_n(i)} \,\bigg| \, \nu(e)\biggl] \le \nu(e) q^{\nu(e)-1} \frac{\beta\_{n-1}(e)\mathbf{1}\_S} {\lambda-1+\beta\_n(e)} \,. $$
Now take another expectation.
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1
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https://mathoverflow.net/users/7691
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403455
| 165,520 |
https://mathoverflow.net/questions/403356
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14
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$\DeclareMathOperator\GL{GL}$Let $R$ be a commutative ring, let $R[n] := R[M\_d^{\oplus n}]$ be the polynomial ring on $nd^2$ variables corresponding to the coordinates of $n$-many $d\times d$ matrices. Let these matrix variables be $X\_1,\ldots,X\_n$. For an $d\times d$ matrix $A$, let $c\_k(A)$ be the coefficient of $T^k$ in the characteristic polynomial $\det(A-TI)$.
The group scheme $\GL\_{d,R}$ acts on $R[n]$ by simultaneous conjugation on the $d$ matrices. Clearly for any product $X\_{i\_1}X\_{i\_2}\cdots X\_{i\_r}$ (where $i\_j\in[1\ldots n], r\ge 1$), the function $c\_k(X\_{i\_1}\cdots X\_{i\_r})\in R[n]$ is invariant under $\GL\_{d,R}$.
Is it true that for any commutative ring $R$, $R[n]^{\GL\_{d,R}}$ is generated as an $R$-algebra by the functions $c\_k(X\_{i\_1}\cdots X\_{i\_r})$?
Does anyone have a reference for this?
Remark - The statement over $\mathbb{C}$ is a classical result of Sibirski and Procesi. This was later extended to the case $R = \mathbb{Z}$ and $R$ any algebraically closed field by Donkin in [*Invariants of several matrices*](https://www.researchgate.net/profile/Stephen-Donkin/publication/226194761_Invariants_of_several_matrices/links/00b4951a7231dd4ed0000000/Invariants-of-several-matrices.pdf). In Concini-Procesi's [*The invariant theory of matrices*](https://bookstore.ams.org/ulect-69), they also seem to obtain the result when $R$ is an infinite field.
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https://mathoverflow.net/users/88840
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Invariants of matrices (by simultaneous $\mathrm{GL}_n$ conjugation) over arbitrary rings
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It is true. The standard reference is the Book by Jantzen, Representations of Algebraic Groups, Second edition. In particular we need the Appendix `Chapter B', and the base change Proposition in part I, 4.18. By Donkin the $G\_{\mathbb Z}$ module $M={\mathbb Z}[n]$ has good filtration. So $H^1(G\_{\mathbb Z},M)$ vanishes. And $H^0(G\_{\mathbb Z},M)$ just means $M^{G\_{\mathbb Z}}$. The base change Proposition tells that the map $M^{G\_{\mathbb Z}}\otimes R\to (M\otimes R)^{G\_{R}}$ is an isomorphism. The result follows.
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13
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https://mathoverflow.net/users/4794
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403460
| 165,521 |
https://mathoverflow.net/questions/403473
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7
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I came to know that the statement below could be proved using *Stallings' binding tie* argument, though I have no reference article proving the statement by the binding tie argument. Can anyone help me telling something related to this? Another question, except this, does there exist any other application of binding tie argument, maybe in other dimensions?
*Given a $\pi\_1$-surjective map $f\colon S\to S'$ between two closed orientable surfaces and a smoothly embedded circle $C$ in $S$, we can homotope $f$ to make it transverse to $C$ so that $f^{-1}(C)$ is either empty or exactly one circle.*
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https://mathoverflow.net/users/363264
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Stallings' binding tie
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Several of the papers citing Stallings paper *A topological proof of Grushko's theorem on free products* are using the binding tie argument. Here are the ones I found using a Google search. More can probably be found using MathSciNet.
1. Jaco [1968] *Constructing 3-manifolds from group homomorphisms*
2. Heil [1972] *On Kneser’s conjecture for bounded 3-manifolds*
3. Feustel [1972] *A splitting theorem for closed orientable 3-manifolds*
4. Bowditch [1999] *Connectedness properties of limit sets*
5. Bellettini, Paolini, Wang [2021] *A complete invariant for closed surfaces in the three-sphere*
Several searches failed to find the precise statement you mention. However, Lemma 3.2 in Jaco's paper comes close (he maps a surface to a graph, pulls back midpoints, and so on).
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6
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https://mathoverflow.net/users/1650
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403481
| 165,529 |
https://mathoverflow.net/questions/403048
|
2
|
Once-punctured torus bundles are a well-studied class of hyperbolic 3-manifolds, but unfortunately I have been unable to find out whether they always have integral traces (in the sense of [1], Def. 5.2.1). Is this already known?
In several dozens of examples that I have calculated with the help of Snappy or the approach in [2], this seems to be the case. Moreover, for bundles with monodromy $LR^n$ this seems to follow from the calculations in [2].
Bass's Theorem ([3]; [1] Thm. 5.2.2) sometimes helps in this situation, but I am not enough of a topologist to tell (and am inclined to say it does not). See my question (and attempt of answer) on MSE: <https://math.stackexchange.com/q/4237616/478924>
When calculating the trace field with the approach of [2], there is a finite set of algebraic solutions one of which belongs to the geometric structure of the manifold. It might be worthwhile to mention that in all my examples, also the other, non-geometric solutions define a (non-discrete) representation with integral traces.
**Related question** discussing closedness in Bass's Theorem: <https://math.stackexchange.com/q/4243973/478924>
[1] MacLachlan, Reid: The Arithmetic of Hyperbolic 3-Manifolds.
[2] Rehmann, Vinberg: On a Phenomenon Discovered By Heinz Helling. Transformation Groups 22, 259–265 (2017).
[3] Bass, Hyman: Chapter VI Finitely Generated Subgroups of Gl2. Pure and Applied Mathematics Vol. 112, 1984, pp 127-136.
|
https://mathoverflow.net/users/129446
|
Do once-punctured torus bundles have integral traces?
|
Floyd and Hatcher give a classification of essential surfaces in once-punctured torus bundles. In particular, there are no closed incompressible surfaces (other than the boundary torus), if there were they would persist in high parameter Dehn fillings, contradicting Corollary 1.2:
*Floyd, W.; Hatcher, Allen E.*, [**Incompressible surfaces in punctured-torus bundles**](http://dx.doi.org/10.1016/0166-8641(82)90035-9), Topology Appl. 13, 263-282 (1982). [ZBL0493.57004](https://zbmath.org/?q=an:0493.57004).
Then the version of Bass's theorem you cite (see Theorem 5.2.2 of the reference below aka [1] above) which shows that for hyperbolic manifolds with non-integral traces contain a closed incompressible surface. Together, these results confirm your computations more broadly. Hyperbolic once-punctured torus bundles have integral traces.
*Maclachlan, Colin; Reid, Alan W.*, The arithmetic of hyperbolic 3-manifolds, Graduate Texts in Mathematics. 219. New York, NY: Springer. xiii, 463 p. (2003). [ZBL1025.57001](https://zbmath.org/?q=an:1025.57001).
|
2
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https://mathoverflow.net/users/27453
|
403488
| 165,531 |
https://mathoverflow.net/questions/403469
|
3
|
An interesting representation of the Euler-Mascheroni constant
$$ \gamma~=~ \lim \limits\_{n\to \infty} \sum \limits\_{k,s=1}^n \frac{s-k}{k\left( s\,n +k\right)},\label{1}\tag{$\*$}$$
can be proved using the relatively recent identity of Macys (Mat. Zametki, 94/5, 695–701, 2013), but it also can be reformulated to
almost mimic a Riemann sum over the unit square $[0,1]^2$
$$ \gamma~=~ \lim \limits\_{n\to \infty} \sum \limits\_{x,y=1/n\,(step: 1/n)}^1 \frac{x-y}{y\,\left( x +y/n\right)} \,\left(\frac1n\right)^2,$$
and there are known similar representations as double integral like
$$ \gamma~=~ \int \limits\_0^1 \int \limits\_0^1 \frac{1-x}{(1-x\, y ) (-\log x\,y)}\,dx\,dy. $$
Is there a valid way to prove \eqref{1} by converting it into an integral representation ?
|
https://mathoverflow.net/users/20804
|
An Euler-Mascheroni double sum
|
$\newcommand\ga\gamma$As usual, let $H\_n$ denote the $n$th harmonic number. Note that
$$\frac{s-k}{k (s n+k)}=\frac1n\frac1k-\frac{n+1}n\frac1{n s+k}$$
and hence
$$
\begin{aligned}
&\sum\_{k,s=1}^n\frac{s-k}{k (s n+k)} \\
&=\frac1n\,\sum\_{k,s=1}^n\frac1k-\frac{n+1}n\,\sum\_{k,s=1}^n\frac1{n s+k} \\
&=H\_n-\frac{n+1}n\,\sum\_{s=1}^n\sum\_{k=1}^n\frac1{n s+k} \\
&=H\_n-\frac{n+1}n\,\sum\_{s=1}^n (H\_{n(s+1)}-H\_{ns}) \\
&=H\_n-\frac{n+1}n\,(H\_{n(n+1)}-H\_n) \\
&=\frac{2n+1}n\,H\_n-\frac{n+1}n\,H\_{n(n+1)}\\
&=\frac{2n+1}n\,(\ga+\ln n+o(1)) \\
&-\frac{n+1}n\,(\ga+\ln(n(n+1))+o(1)) \\
&=\ga+o(1)
\end{aligned}$$
(as $n\to\infty$), which proves your formula $(\*)$.
Of course, $H\_n$ can be represented as an integral; for instance, we have [Euler's representation](https://en.wikipedia.org/wiki/Harmonic_number#Calculation)
$$H\_n=\int \_{0}^{1}{\frac {1-x^{n}}{1-x}}\,dx.$$
---
On the other hand, apparently, your Riemann-sum argument cannot work -- because the boundary effect here is nonnegligible. In particular, summing over the set $\{2,\dots,n\}\times\{2,\dots,n\}$ rather than over the set $\{1,\dots,n\}\times\{1,\dots,n\}$, we get
$$\lim\_{n\to\infty}\sum\_{k,s=\color{red}{\mathbf{2}}}^n\frac{s-k}{k (s n+k)}=\ga-1+\ln2\ne\ga.$$
|
2
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https://mathoverflow.net/users/36721
|
403495
| 165,533 |
https://mathoverflow.net/questions/403489
|
4
|
Let $(C, J)$ be a small site and let $\mathsf{Sh}\_{(2, 1)}(C, J)$ be the $(2, 1)$-sheaf topos of sheaves of (small) groupoids on $(C, J)$. Let $G$ be a sheaf of groups on $(C, J)$, and let $\mathbf{Bun}\_G$ be the hom-stack $[-, \mathbf{B}G]$, which is typically known as the moduli stack of principal $G$-bundles on $(C, J)$. [On the nLab](https://ncatlab.org/nlab/show/principal+bundle#in_terms_of_fiber_sequences), it is stated that (up to homotopy equivalences), principal $G$-bundles over a given base space $X \in (C, J)$, (i.e. objects of $\mathbf{Bun}\_G(X)$) are homotopy pullbacks of the following form:
$\require{AMScd}$
\begin{CD}
P @>>> \*\\
@V V V @VV V\\
X @>>> \mathbf{B}G
\end{CD}
Would anyone mind explaining to me why this is the case, and moreover, how one might obtain "local trivialisations" out of the above pullback square ? The nLab article I linked does go into these topics, but their explanation is a bit too abstract non-sensical for me to be able to cut through.
|
https://mathoverflow.net/users/143390
|
Internal principal $G$-bundles
|
The easiest way to see local trivializations is to compute the homotopy pullback using the local projective model structure.
For differential geometry, we can $C$ to be the category of cartesian spaces and smooth maps, whereas $J$ is the usual topology of open covers.
(Other sites work in the same manner.)
Then maps $\def\B{{\bf B}} X→\B G$ in the homotopy category
correspond to maps $Č(U)→\B G$ of presheaves of groupoids,
where $U$ is a good cover of $X$ and $Č(U)$ denotes the Čech nerve of U.
Unfolding this construction, we see that a map $Č(U)→\B G$
* sends each $U\_i$ to the only object of $\B G(U\_i)$;
* sends each intersection $U\_i ∩ U\_j$ to a morphism in $\B G(U\_i∩U\_j)$, i.e., a smooth map $t\_{i,j}\colon U\_i∩U\_j→G$;
* for each triple intersection $U\_i∩U\_j∩U\_k$, it enforces a cocycle condition $t\_{j,k}t\_{i,j}=t\_{i,k}$.
This data is precisely the traditional description of principal $G$-bundles in terms of cocycles.
Next, we can compute the homotopy pullback by replacing the map $\*→\B G$ with a fibration, namely, $\def\E{{\bf E}} \E G→\B G$,
where $\E G(V)$ is the nerve of the contractible groupoid with its objects being smooth maps $V→G$.
Let's now see what a smooth map $V→P$, i.e., an element of $P(V)$ is in concrete terms.
By definition of a pullback, an element of $P(V)$ can be described
as a pair $(b,f)$, where $b∈Č(U)(V)$, i.e., a smooth map $V→U\_i$ for some $i$, whereas $f∈\E G(V)$, i.e., a smooth map $V→G$.
The pullback compatibility condition for objects is trivial because $\B G(V)$ has a single object.
Thus, restricting our attention to a single $U\_i⊂X$, we see that
maps $V→P$ are pair of smooth maps $(b\colon V→U\_i,f\colon V→G)$,
or, equivalently, a smooth map $τ\colon V→U\_i⨯G$.
The manifold $U\_i⨯G$ is precisely the total space
of the trivial principal $G$-bundle $U\_i⨯G→U\_i$ over $U\_i$.
Next, let's examine a generic morphism of the form $$(b\colon V→U\_i,f\colon V→G)→(b'\colon V→U\_i,f'\colon V→G).$$
According to the definition of a pullback, such a morphism is given by a compatible pair of morphisms $b→b'$ and $f→f'$.
By definition of $\E G(V)$, there is a unique morphism $f→f'$,
and its image in $\B G(V)$ is $f'f^{-1}$.
By definition of $Č(U)$, a morphism $b→b'$ exists if and only if
$b$ and $b'$ factor through the intersection $U\_i∩U\_j$,
in which case such a morphism is unique and its image
in $\B G(V)$ is $t\_{i,j}$.
Finally, the compatibility condition amounts to saying $f'f^{-1}=t\_{i,j}$.
Collecting all the pieces together, we see that two maps
$(b,f)\colon V→U\_i⨯G$ and $(b',f')\colon V→U\_j⨯G$ are identified if they satisfy the relation $f'f^{-1}=t\_{i,j}$.
This is precisely how the total space of a principal bundle
is glued from individual pieces given by trivial principal bundles over $U\_i$ and $U\_j$.
Thus, a map $c\colon X→\B G$ picks out principal $G$-bundles,
and for such a map, the homotopy pullback is precisely the total space of the principal $G$-bundle classified by $c$.
|
5
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https://mathoverflow.net/users/402
|
403499
| 165,536 |
https://mathoverflow.net/questions/403414
|
45
|
The following simple-looking inequality for complex numbers in the unit disk generalizes [Problem B5 on the Putnam contest 2020](https://www.maa.org/sites/default/files/pdf/Putnam/2020/2020%20Putnam%20Session%20B%20Solutions.pdf):
>
> **Theorem 1.** Let $z\_1, z\_2, \ldots, z\_n$ be $n$ complex numbers such that $\left|z\_i\right| \leq 1$ for each $i \in \left\{1,2,\ldots,n\right\}$. Prove that
> \begin{align}
> \left| z\_{1} + z\_{2} + \cdots + z\_n - n \right|
> \geq
> \left| z\_{1} z\_{2} \cdots z\_n - 1 \right| ,
> \end{align}
> and equality holds only if at least $n-1$ of the $n$ numbers $z\_1, z\_2, \ldots, z\_n$ equal $1$.
>
>
>
In the particular case when $n = 4$, the theorem can be proved using stereographic projection onto a line, followed by a longish computation. This is how [both proposed solutions](https://www.maa.org/sites/default/files/pdf/Putnam/2020/2020%20Putnam%20Session%20B%20Solutions.pdf) go.
On the other hand, in the general case, the only elementary solution I know [was given by @mela\_20-15 on AoPS (spread over several posts)](https://artofproblemsolving.com/community/c7t1129634f7h2461230). It has some beautiful parts (Cauchy induction), but also some messy ones (tweaking the points to lie on the unit circle in the induction step). There might also be a heavily analysis-based proof in [Kiran Kedlaya's solutions](https://kskedlaya.org/putnam-archive/2020s.pdf) (not sure if Theorem 1 is proved in full there).
>
> **Question.** What is the "proof from the book" for Theorem 1?
>
>
>
Someone suggested to me to try to interpolate expressions of the form $\dbinom{n-1}{k-1}^{-1} \left|\sum\limits\_{i\_1 < i\_2 < \cdots < i\_k} z\_{i\_1} z\_{i\_2} \cdots z\_{i\_k} - \dbinom{n}{k}\right|$ between the left and the right hand sides in Theorem 1; but this does not work. For example, the inequality $\dfrac{1}{2} \left| z\_1 z\_2 + z\_2 z\_3 + z\_1 z\_3 - 3\right| \geq \left|z\_1 z\_2 z\_3 - 1\right|$ fails [quite often even on the unit circle](https://sagecell.sagemath.org/?z=eJxLyy9SiFfIzFMoSsxLT9UwNDAw0LTiUgCCRB2FJB2FZAVbhejUigINI62CTK1MLaCyFLhSTX0QoZCGaoaxZizYgIKizLwSjaAgjcSkYo1ErSQFbYUkrWQgmQgkdRWMNYGEkRZUEixkqKmpCQBasyZS&lang=sage&interacts=eJyLjgUAARUAuQ==).
A **warning**: Inequalities like Theorem 1 are rather hard to check numerically. Choosing the $z\_i$ uniformly will rarely hit close to the equality case; usually the left hand side will be much larger than the right. Near the equality case, on the other hand, it is hard to tell whether the answer comes out right legitimately or whether accumulated errors have flipped the sign.
|
https://mathoverflow.net/users/2530
|
Putnam 2020 inequality for complex numbers in the unit circle
|
Here is a detailed and self-contained proof for general $n$, which also covers the "equality" case. It is based on Fedja's post, but it only uses (a variant of) the Gauss-Lucas theorem once.
Let $a\_0,\dotsc,a\_{n-1}\in\mathbb{C}$ be coefficients such that
$$p(z):=(z-z\_1)\dotsb(z-z\_n)=(z-1)^n+\sum\_{k=0}^{n-1}a\_k z^k,$$
and assume that $(z-1)^{n-1}$ does not divide the left-hand side. Then, the $k$-sum on the right-hand side is a polynomial of degree $n-1$, because
$$\Re(a\_{n-1}) = \Re\left(n-\sum\_{j=1}^n z\_j\right)=n-\sum\_{j=1}^n\Re(z\_j)>0,$$
and we need to prove that $|a\_{n-1}|>|a\_0|$.
Introducing the differential operator
$$Df(z):=(1-z)f'(z)+nf(z),$$
we claim that every root of the polynomial $Dp(z)$ lies in the closed unit disk. Indeed, assume for a contradiction that $Dp(z)=0$ and $|z|>1$.
Then $p'(z)/p(z)=n/(z-1)$, that is,
$$\frac{1}{n}\sum\_{j=1}^n \frac{1}{z-z\_j}=\frac{1}{z-1}.$$
Let $K$ be the image of the closed unit disk under the Möbius transformation $s\mapsto 1/(z-s)$. Using $|z|>1$, we see that $K$ is a closed disk containing $1/(z-1)$ on its boundary. By the previous display, this boundary point is a convex linear combination of the points $1/(z-z\_j)$, which also lie in $K$. This forces that all the $z\_j$'s are equal to $1$, contrary to our assumption. So the claim is proved.
Now observe that the polynomial $Dp(z)$ is the same as
$$D\left(\sum\_{k=0}^{n-1}a\_k z^k\right)=a\_{n-1}z^{n-1}+\sum\_{k=0}^{n-2}((n-k)a\_k+(k+1)a\_{k+1})z^k.$$
We proved that this polynomial has all its roots in the closed unit disk, therefore
$$|(n-k)a\_k+(k+1)a\_{k+1}|\leq\binom{n-1}{k}|a\_{n-1}|,\qquad k\in\{0,\dotsc,n-2\}.$$
Applying the triangle inequality and re-arranging, we get that
$$\frac{n-k}{k+1}|a\_k|\leq|a\_{k+1}|+\frac{1}{k+1}\binom{n-1}{k}|a\_{n-1}|.\tag{$\ast$}$$
From here we derive by induction that
$$\binom{n}{k}|a\_0|\leq |a\_k|+\binom{n-1}{k-1}|a\_{n-1}|,\qquad k\in\{1,\dotsc,n-1\}.$$
In particular, the special case $k=n-1$ tells us that $n|a\_0|\leq n|a\_{n-1}|$.
To finish the proof, we only need to show that some of our inequalities are strict, so that the final inequality is strict as well. Let us assume that equality holds in each of our inequalities. Then the roots of $Dp(z)$ are equal to a single $w$ on the unit circle, so that
$$(n-k)a\_k+(k+1)a\_{k+1}=\binom{n-1}{k}a\_{n-1}(-w)^{n-1-k},\qquad k\in\{0,\dotsc,n-2\}.$$
In particular, the case $k=n-2$ yields
$$\frac{2}{1-n}a\_{n-2}=(1+w)a\_{n-1}.$$
However, we have equality in $(\ast)$ for $k=n-2$, whence $|1+w|=2$, which forces $w=1$. But then the recursion yields that
$$a\_k=\binom{n-1}{k}(-1)^{n-1-k}a\_{n-1},$$
i.e.,
$$p(z)=(z-1)^n+a\_{n-1}(z-1)^{n-1}.$$
This contradicts our initial assumption, and we are done.
**Added.** As Malkoun pointed out in the comments, $(\ast)$ also implies the inequalities
$$|a\_k|\leq\binom{n-1}{k}|a\_{n-1}|,\qquad k\in\{0,\dotsc,n-1\}.$$
Hence, renaming $k$ to $n-k$, we get the following generalization of the original inequality:
$$\left|\sum\_{1\leq j\_1<\dotsb< j\_k\leq n}z\_{j\_1}\dotsb z\_{j\_k}-\binom{n}{k}\right|\leq\binom{n-1}{k-1}|z\_1+\dotsb+z\_n-n|,\qquad k\in\{1,\dotsc,n\}.$$
|
24
|
https://mathoverflow.net/users/11919
|
403506
| 165,538 |
https://mathoverflow.net/questions/403520
|
2
|
I suspect that
$$ \lim\_{n\rightarrow\infty}\frac{1}{n}\sum\_{i=1}^n\sum\_{j=1}^n \frac{1}{\sqrt{i^2+j^2}} =a\approx 1.76$$
$$ \lim\_{n\rightarrow\infty}\frac{1}{n}\sum\_{i=1}^n\sum\_{j=1}^n \frac{1}{\sqrt{n^2+(i-j)^2}} =b\approx 0.934$$
Is there an analytic expression for $a$ or $b$?
|
https://mathoverflow.net/users/94200
|
Limiting behavior of lattice sums
|
You seem to have a typo in your value of $a$ - evaluating the sum as is yields something more like 1.76. Converting to an integral,
$$
\lim\_{n\rightarrow \infty } \frac{1}{n^2 } \sum\_{i=1}^{n} \sum\_{j=1}^{n} \frac{1}{\sqrt{(i/n)^2 + (j/n)^2 } } =\\ \int\_{0}^{1} dx \int\_{0}^{1} dy\, \frac{1}{\sqrt{x^2 +y^2 } } = 2\ln (1+\sqrt{2} )
$$
On the other hand,
$$
\lim\_{n\rightarrow \infty } \frac{1}{n^2 } \sum\_{i=1}^{n} \sum\_{j=1}^{n} \frac{1}{\sqrt{1+((i/n) - (j/n))^2 } } =\\ \int\_{0}^{1} dx \int\_{0}^{1} dy\, \frac{1}{\sqrt{1+(x-y)^2 } } = 2\ln (1+\sqrt{2} ) +2-2\sqrt{2}
$$
|
9
|
https://mathoverflow.net/users/134299
|
403522
| 165,541 |
https://mathoverflow.net/questions/403547
|
4
|
Consider the Hirzebruch surface $\mathbb{F}\_n = \mathbb{P}(\mathcal{O}\_{\mathbb{P}^1}\oplus \mathcal{O}\_{\mathbb{P}^1}(n))\rightarrow\mathbb{P}^1$. The Picard group of $\mathbb{F}\_n$ is generated by the class of the negative section $\Gamma$, such that $\Gamma^2 = -n$, and the class of a fiber $F$ of the morphism onto $\mathbb{P}^1$. Furthermore, $\Gamma$ and $F$ generate the effective cone of $\mathbb{F}\_n$ so that all divisors of the form $D\_{a,b} = a\Gamma + bF$ with $a,b\in \mathbb{N}$ have sections.
Does there exist a closed formula for the dimension of the vector space $H^0(\mathbb{F}\_n,D\_{a,b})$?
Thank you very much.
|
https://mathoverflow.net/users/nan
|
Cohomology of divisors on Hirzebruch surfaces
|
Sure. To compute $H^0$ one can first pushforward to the base $\mathbb{P}^1$. If $a \ge 0$ one obtains
$$
p\_\*\mathcal{O}(a\Gamma + bF) \cong
p\_\*\mathcal{O}(a\Gamma) \otimes \mathcal{O}(b) \cong
S^a(\mathcal{O} \oplus \mathcal{O}(-n)) \otimes \mathcal{O}(b) \cong
\bigoplus\_{i=0}^a \mathcal{O}(b-in),
$$
and if $a < 0$ the pushforward is zero. Therefore,
$$
\dim H^0(\mathbb{F}\_n, \mathcal{O}(D\_{a,b})) =
\begin{cases}
\sum\_{i = 0}^{\min(a,\lfloor b/n \rfloor)} (b - in + 1),
& \text{if $a \ge 0$},\\
0, & \text{if $a < 0$}.
\end{cases}
$$
|
6
|
https://mathoverflow.net/users/4428
|
403548
| 165,547 |
https://mathoverflow.net/questions/403546
|
10
|
In [1] (section 3), C. Scott introduces the following concept of **regular atlas** for closed $C^\infty$-smooth Riemannian manifolds. He says:
>
> When referring to a coordinate system $(U,\phi)$ as **regular**, we shall mean that there is another system $(V,\psi)$ with $\overline{U}$ compact, $\overline{U} \subset V$ and $\psi\vert\_{U} = \phi$.
>
>
>
His motivation to introduce such concept comes from the problem of defining Sobolev spaces of differential forms. In self-explanatory notation, he defines ($1 \leq l \leq n = \dim M$)
$$\mathscr{W}^{1,p}\_{\mathscr{A}}\left( \bigwedge^l M \right) := \left\{ \omega \in \left( \bigwedge^l M \right) : \omega, |\nabla{\omega}| \in L^p \right\},$$
endowed with the norm $\|\omega\|\_p + \|\nabla \omega \|\_p$. Here,
$$|\nabla \omega(x)|^2 := \sum\_{U \in \mathscr{A}} |\nabla\_U \omega(x) |^2 = \sum\_{U \in \mathscr{A}} \sum\_{I,k} \left| \frac{\partial \omega\_I}{\partial x^k}(x) \right|^2,$$
where $I = { 1 \leq i\_1 < \dots < i\_l \leq n}$. Then, he observes
>
> Simple examples demonstrate that it is possible to choose, in perfectly reasonable ways, two atlases which yield Sobolev spaces that are not equivalent as normed linear spaces. [...] From here on, classical Sobolev space refers to one constructed as above using a regular atlas. This is all fairly familiar and once again, Morrey [*Multiple integrals in the calculus of variations*] is a fine reference.
>
>
>
Such observation is not proved nor references to the literature are provided for the counterexamples. Morever, I have not found a similar definition in Morrey's book (Morrey deals with *admissible cooordinate systems*, see pp. 288 and 300, which are defined differently and it appears that the only requirement about them is the standard compatibility between new and old coordinates). In addition, it seems to me (but I'm quite new to this topic) that this problem is not considered by other authors in this area.
Exceptions are [2] and [3]. In [2] (section 2.2), Iwanienc, Scott and Stroffolini, dealing now with smooth Riemannian manifolds with boundary, give the following definition of regular atlas. If $\mathscr{R}$ is a $C^\infty$-smooth, closed Riemannian manifold,
>
> a *regular open region* $M \subset \mathscr{R}$ is one for which there exists a finite atlas $\mathscr{A}$ on the reference manifold $\mathscr{R}$ consisting entirely of coordinate charts $(U,\kappa) \in \mathscr{A}$ so that $\kappa$ is a $C^\infty$-diffeomorphism *onto* $\mathbb{R}^n$ and $\kappa(U \cap M) = \mathbb{R}^n\_+$ whenever $U$ meets $\partial M$.
>
>
>
In [3], section 2.1.1, the authors are concerned only with closed manifolds. The requirement is the same but obviously there is no need for the second part (as $\partial M = \varnothing$).
It seems to me that no one of these requirements is common in standard differential geometry. For instance, I guess that the sphere $\mathbb{S}^n$ with the most common atlas consisting of the two coordinate patches associated with the stereographic projections from the north and south poles is not regular in the sense of Scott, and this is perhaps quite weird.
Summarizing, these are my questions (they are multiple but strictly tied):
* Is it always possible to introduce such atlases?
* Are the two definitions in [1] and [3] equivalent?
* Are they necessary to give a meaningful definition of Sobolev spaces of differential forms? What are the easy counterexamples mentioned above?
* Are they necessary to prove the $L^p$-version of Gaffney's inequality and Hodge-de Rham-Kodaira decomposition, which are essentially the main results of [1] and [2]?
---
[1] C. Scott, *$L^p$ theory of differential forms on manifolds*, Trans. Amer. Math. Soc. 347 (6), 1995.
[2] T. Iwaniec, C. Scott, B. Stroffolini, *Nonlinear Hodge theory on manifolds*, Ann. Mat. Pura Appl. (IV), CLXXVII (1999), 37-115.
[3] P. Hajlasz, T. Iwaniec, J. Maly, J. Onninen, *Weakly differentiable mappings between manifolds*, AMS.
|
https://mathoverflow.net/users/364344
|
Sobolev spaces of differential forms and regular atlases
|
Ok this is already quite a mouth full, so let me try to give answers to some of your questions:
The main issue is that Sobolev mappings are defined via a boundedness concept (you ask for $L^p$-integrability conditions) and boundedness is not an intrinsic concept on a manifold. This means that definitions in charts usually blow up when boundedness is concerned (as I can always compose with an arbitrary diffeomorphism to make new charts and by choosing badly I can blow up every such condition). This is often glossed over in the older Sobolev literature on manifolds (or they define things chart independent in which case the problem does not exist). For most of what follows now, I refer to the great memoir by Inci, Kappeler and Topalov: On the regularity of the composition of diffeomorphisms (arxiv version is here [1](https://arxiv.org/abs/1107.0488)) where these things are worked out.
1. Atlases with the desired property always exist. Indeed one can even strengthen the conditions to make the atlases more convenient to work with (e.g. Lipschitz boundary so restrictions of your Sobolev maps behave nicely). Atlases with the additional properties are called fine cover in [1](https://arxiv.org/abs/1107.0488) (see Section 3.1 in [1](https://arxiv.org/abs/1107.0488) for discussion and an existence proof)
2. These atlases are necessary to talk about Sobolev functions on a manifold **in charts**. There are chart independent definitions which usually require you to pick a Riemannian metric as an additional structure to make sense of integration and boundedness. For example the classical book by Palais: Foundations of Global Non-Linear Analysis, [2](http://vmm.math.uci.edu/PalaisPapers/FoundationsOfGlobalNonlinearAnalysis.pdf) does it this way. See the Appendix of [3](https://arxiv.org/abs/1909.09982) for a comparison of the notion (even in the boundary case).
3. For the counterexample we can look at [1](https://arxiv.org/abs/1107.0488) p.43 (which I am quoting here):
Consider the torus $M= \mathbb{R}/\mathbb{Z}$ and let $f\colon (−1/2, 1/2) \rightarrow \mathbb{R}$ be the function
$$f(x) := \begin{cases} x^{2/3} &, x \in [0, 1/2)\\
(−x)^{2/3} &, x \in [−1/2, 0).
\end{cases}$$
Extending $f$ periodically to $\mathbb{R}$ we get a function on $M$ that we denote by the same letter. It is not hard to see that $f \in H^1(M, \mathbb{R})$. Now, introduce a new coordinate $y = x^2$ on the open set $(0,1/2) \subseteq M$. Then $\tilde{f}(y) := f(x(y)) = y^{1/3},\ y \in (0, 1/4)$. We have, $\tilde{f}'(y) = 1/(3y^{2/3})$, and hence, $\tilde{f}′ \not\in L^2((0, 1/4), R)$. This shows that $\tilde{f} \not \in H^1((0, 1/4), R)$ and thus provides an example showing that a function may fail to be Sobolev in certain coordinate charts (and the boundedness conditions remedy that). Upshot: Sobolev functiobs are not stable under arbitrary change of charts. Similar ideas will also produce differential forms which are $H^s$ in one but not the other coordinate system.
This answers half of your questions (since I am no expert on Sobolev spaces I will leave the other questions to somebody who is more in the know).
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11
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https://mathoverflow.net/users/46510
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403551
| 165,548 |
https://mathoverflow.net/questions/403545
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4
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Let $R$ be a (commutative, otherwise the answer is easy, see the comment below) ring and let $M$ be a finitely generated $R$-module. Is it possible that $M$ admits an infinite linearly independent set? [Cardinality of maximal linearly independent subset](https://mathoverflow.net/questions/30066/cardinality-of-maximal-linearly-independent-subset) seems relevant, but it does not give an answer to my question. Thank you!
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https://mathoverflow.net/users/7845
|
Infinite linearly independent set in finitely generated module
|
In a commutative ring $R$ this does not exist. Better for any $n\ge 0$, if $M$ is an $R$-module generated by $n$ elements, then $R^{n+1}$ doesn't embed into $M$.
Indeed, lifting if necessary, we can suppose $M=R^n$. So we get an $n\times (n+1)$ matrix $u$ over $R$ defining an injective operator $R^{n+1}\to R^n$. Let $B$ be the unital subring generated by entries of $u$. Then $B$ is finitely generated commutative, hence noetherian, and $u$ defines an injective operator $B^{n+1}\to B^n$, contradiction.
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7
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https://mathoverflow.net/users/14094
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403552
| 165,549 |
https://mathoverflow.net/questions/403554
|
3
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Lately I have been studying reflection groups, and there is a particular example of a complex reflection group that has been very good for guiding my intuition. I would like to know if there is an analogue over the real numbers. To keep this post self-contained, let me start by stating my definition of a reflection group.
**Definition:** Let $k = \mathbb{R}$ or $\mathbb{C}$, and let $V$ be a finite-dimensional vector space over $k$, equipped with an inner product. A *reflection* is an operator $s \in \text{GL}(V)$ such that
1. $s(\alpha)= \zeta \alpha$ for some nonzero vector $\alpha \in V$ and nontrivial root of unity $\zeta \neq 1$, and
2. $s$ fixes the codimension 1 orthogonal complement of the line $k\alpha$.
A reflection group is a subgroup $G \subseteq \text{GL}(V)$ generated by reflections.
At least in my experience, many counterexamples when studying reflection groups can be found by looking at either infinite or abelian reflection groups, so it seems natural to ask if there is a reflection group which is both. We can construct such a complex reflection group as follows:
**Over $\mathbb{C}$:** For each integer $n \geq 2$, pick a primitive $n^{th}$ root of unity $\zeta\_n \in \mathbb{C}$. We can then consider the rank 1 reflection group $W = \langle \zeta\_n : n \geq 2 \rangle$ which is clearly both infinite and abelian.
This construction doesn't work over $\mathbb{R}$ because the only nontrivial root of unity is $\zeta = -1$. This leads me to wonder:
**Question:** Do there exist infinite abelian real reflection groups?
In case it helps to either construct an example or prove nonexistence, let me give the arguments for or against that arose out of my attempts to answer this question.
**Arguments for:**
1. The existence of such an example over $\mathbb{C}$ is encouraging.
2. We can construct finite abelian real reflection groups of arbitrarily large order as follows: for each $n \geq 1$, let $V = \mathbb{R}^n$ and let $\{e\_i\}\_{i=1}^n$ be the standard basis vectors. Then the operators $s\_{e\_i}$ have order 2, commute and there are no other relations between them, so the reflection group they generate is isomorphic to $(\mathbb{Z}/2)^n$. I would like to do something like "take the limit" of this construction, but my definition requires that $V$ be finite-dimensional.
**Arguments against:**
Such a group would have to be infinitely-generated (in line with our $\mathbb{C}$ example). If $W$ were a finitely-generated abelian reflection group then, by the standard classification theorem, we would have $$ W \cong \mathbb{Z}^n \oplus \text{torsion}.$$
For $W$ to be infinite, the exponent $n$ needs to be positive. But $\mathbb{Z}$ does not contain any element of order 2 and, in particular, cannot be generated by a reflection.
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https://mathoverflow.net/users/175051
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Are there infinite abelian real reflection groups?
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I think that you can prove by induction on the dimension of $V$ that there is no infinite Abelian (finite-dimensional) real reflection group $G$. Suppose that $G \subseteq {\rm GL}(V)$ is such an Abelian infinite real reflection group with ${\rm dim}\_{\mathbb{R}}(V)$ minimal. Then $V$ is certainly not $1$-dimensional. Let $t$ be one of the generating reflections of $G$. Then $V = U\_{t} \oplus W\_{t}$ is a $G$-invariant direct sum decomposition of $V$, where $U\_{t}$ is the (one-dimensional) $-1$-eigenspace of $t$ on $V$ and $W\_{t}$ is the $1$-eigenspace of $t$.
Furthermore, every element of $G$ acts as $\pm 1$ on $U\_{t}$, and no non-identity of $G$ acts as the identity on both $U\_{t}$ and $W\_{t}.$ The elements of $G$ which act as $1$ on $U\_{t}$ form a subgroup of $G$ of index $2$, and every generating reflection of $G$ (other than $t$) acts as a reflection on $W\_{t}.$ It follows that $G/\langle t \rangle$ is isomorphic to an infinite Abelian reflection subgroup of ${\rm GL}(W\_{t})$, contrary to the minimality of the dimension of $V$.
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3
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https://mathoverflow.net/users/14450
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403557
| 165,551 |
https://mathoverflow.net/questions/403533
|
2
|
From the standard literature it is well known that for sequences of random variables $X\_{1, n} \stackrel{P}{\rightarrow} X\_1$ and $X\_{2, n} \stackrel{P}{\rightarrow} X\_2$ as $n \rightarrow \infty$ it holds that $(X\_{1, n}, X\_{2, n}) \stackrel{P}{\rightarrow} (X\_1, X\_2)$ for $n \rightarrow \infty$. Using a continuous mapping theorem argument this can be used to establish that $X\_{1, n} + X\_{2, n} \stackrel{P}{\rightarrow} X\_1 + X\_2$ for $n \rightarrow \infty$.
Question in general case
------------------------
Under which conditions can the arguments of the standard literature be extended from a simple sum of two sequences to a sum of sequences whose number of summands also increases to infinity as $n$ goes to infinity. Specifically, given a sequences $X\_{l, n}$, $l \in \mathbb{N}$, with $X\_{l, n} \stackrel{P}{\rightarrow} X\_l$ $(n \rightarrow \infty)$ for all $l\in \mathbb{N}$ under which conditions does it holds that
$$
\lim\_{n \rightarrow \infty} \sum\_{l = 1}^n X\_{l, n} \stackrel{P}{\rightarrow} \sum\_{l = 1}^\infty X\_{l}
$$
as $n \rightarrow \infty$?
Question in special case
------------------------
In my particular use case something more specific would also suffice. Though, I think the more general question is also interesting. If we know that
$X\_{l, n}$, $l \in \mathbb{N}$, are nonnegative random sequences, i.e. each sequence element is a nonnegative random variable, and $X\_{l, n} \stackrel{P}{\rightarrow} 0$ $(n \rightarrow \infty)$ for all $l \in \mathbb{N}$ under which conditions does it holds that
$$
\lim\_{n \rightarrow \infty} \sum\_{l = 1}^n X\_{l, n} \stackrel{P}{\rightarrow} 0
$$
as $n \rightarrow \infty$?
Possibly, what would also be interesting and might be easier to answer is when the sample mean converges to zero, i.e.,
$$
\lim\_{n \rightarrow \infty} \frac{1}{n} \sum\_{l = 1}^n X\_{l, n} \stackrel{P}{\rightarrow} 0
$$
as $n \rightarrow \infty$
|
https://mathoverflow.net/users/302666
|
Convergence in probability of series of random variables
|
$\newcommand\ep\varepsilon\newcommand\de\delta\newcommand{\P}[1]{\overset P{\underset{#1}\longrightarrow}}$What you need is the uniform summability (in probability).
Here are details: Let $Y\_{l,n}:=X\_{l,n}-X\_l$, so that $$Y\_{l,n}\P{n\to\infty}0 \tag{0}$$
for each $l$. We want to have
$$S\_{n,n}\P{n\to\infty}0,\tag{1}$$
where
$$S\_{m,n}:=\sum\_{l=1}^m Y\_{l,n}.$$
The mentioned sufficient uniform summability condition is that
$$S\_{n,n}-S\_{m,n}\P{n\ge m\to\infty}0. \tag{2}$$
Indeed, take any real $\de>0$ and $\ep>0$. Then, by (2), for some natural $m\_1$ we have the following implication:
$$n\ge m\ge m\_1\implies P(|S\_{n,n}-S\_{m,n}|>\ep/2)\le\de/2. \tag{3}$$
Also, (0) implies $S\_{m\_1,n}\P{n\to\infty}0$, so that for some natural $m\_2$ we have the following implication:
$$n\ge m\_2\implies P(|S\_{m\_1,n}|>\ep/2)\le\de/2. \tag{4}$$
Letting now $m\_3:=\max(m\_1,m\_2)$, by (3) and (4) we have
$$n\ge m\_3\implies
P(|S\_{n,n}|>\ep)\le
P(|S\_{n,n}-S\_{m\_1,n}|>\ep/2)+P(|S\_{m\_1,n}|>\ep/2)\le\de. $$
So, (1) holds.
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3
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https://mathoverflow.net/users/36721
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403561
| 165,553 |
https://mathoverflow.net/questions/403517
|
20
|
Is there a non-enumerative proof that, in average, less than 50% of tiles in domino tiling of 2-by-n rectangle are vertical?
It is a nice exercise with rational generating functions (or equivalently, linear recurrence relations) to show that for a random domino tiling of a $2\times n$ rectangle, with $n$ large, we can expect about $\frac{1}{\sqrt{5}}\approx 44.7\%$ of the tiles to be vertical. In the spirit of [Non-enumerative proof that there are many derangements?](https://mathoverflow.net/questions/86118/non-enumerative-proof-that-there-are-many-derangements), I wonder if there is an easy way to see, without computing this fraction, that this average must be some constant < 50%.
**EDIT**: Maybe to sharpen the request, is there any heuristic which works here and also carries over to the case of a $k\times n$ rectangle (with say $k$ fixed and $n$ large).
|
https://mathoverflow.net/users/25028
|
Non-enumerative proof that, in average, less than 50% of tiles in domino tiling of 2-by-n rectangle are vertical?
|
Here is at least a heuristic argument for why we should expect fewer vertical dominoes than horizontal dominoes. As we increase the length of the strip (to the right, say), let us think about how the rightmost four squares are covered. There are three cases: (1) two horizontal dominoes; (2) two vertical dominoes; (3) the last two squares are covered by a vertical domino and the preceding two squares are covered by horizontal dominoes (that "stick out" to the left and cover two additional squares).
Imagine I'm trying to build one copy of every tiling of the $2\times n$ strip and I'm trying to gauge whether I'll need to buy more horizontal dominoes from Harry or more vertical dominoes from Victoria. For every tiling of a $2\times (n-2)$ strip I'll need two horizontal dominoes for case (1) and two vertical dominoes for case (2); this is a tie. But for case (3), I need two horizontal dominoes and only one vertical domino for each tiling of a $2\times (n-3)$ strip. So I'm going to need to buy more dominoes from Harry than from Victoria.
|
16
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https://mathoverflow.net/users/3106
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403563
| 165,554 |
https://mathoverflow.net/questions/403536
|
3
|
I’m reading a note on higher regularity boundary Harnack inequality by D. DE SILVA AND O. SAVIN and I’m kind of confused of the case k=1.
In the paper they used the Hopf lemma to show that $u\_\nu>c>0$, but, as the boundary regularity is just $C^{1, \alpha}$, I don’t think that we can directly use Hopf lemma.
I tried to use the transformation of coordinates to do make a better regularity of boundary, but it only works in the divergence form of equations. I have no clue to the non-divergence form. Is there any way to do this estimate?
|
https://mathoverflow.net/users/348579
|
About the proof of higher regularity boundary Harnack inequality
|
*Edit:* The result is fine: Hopf's lemma was proved in
* G. Giraud, *Problèmes de valeurs à la frontière relatifs à certaines donn ás discontinues*, Bull. de la Soc. Math. de France, 61 (1933), 1–54
Below is my incorrect answer (which I keep for reference), where I mistakenly assumed that $C^{1,Dini}$ is stronger than $C^{1,\alpha}$.
---
*Old answer:*
Looks like you are right: the regularity assumption on the boundary is insufficient, although I did not check this very carefully.
You may have a look at the paper *A counterexample to the Hopf-Oleinik lemma (elliptic case)* by D. E. Apushkinskaya and A. I. Nazarov, [DOI:10.2140/apde.2016.9.439](https://doi.org/10.2140/apde.2016.9.439), [arXiv:1503.02179](https://arxiv.org/abs/1503.02179). Let me quote from p. 2 of this paper:
>
> The reduction of the assumptions on the boundary of $\Omega$ up to $C^{1,Dini}$-regularity was realized for various elliptic operators in the papers [Wid67], [Him70] and [Lie85] (see also [Saf08]). A weakened form of the Hopf-Oleinik lemma (the existence of a boundary point $x\_1$ in any neighborhood of $x\_0$ and a direction $\ell$ such that $\frac{\partial u}{\partial \ell}(x\_1) \ne 0$) was proved in [Nad83] for a much wider
> class of domains including all Lipschitz ones. We mention also the paper
> [Swe97] where the behavior of superharmonic functions near the boundary
> of 2-dimensional domains with corners is described in terms of the main
> eigenfunction of the Dirichlet Laplacian.
>
>
> The sharpness of some requirements was confirmed by corresponding
> counterexamples constructed in [Wid67], [Him70], [KH75], [Saf08], [ABM$^+$11] and [Naz12]. In particular, the counterexamples from [Wid67], [Him70] and
> [Saf08] show that the Hopf-Oleinik result fails for domains lying entirely in
> non-Dini paraboloids.
>
>
>
Later, the paper shows that lack of "Dini condition" on the boundary invalidates the assertion of Hopf's lemma.
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3
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https://mathoverflow.net/users/108637
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403564
| 165,555 |
https://mathoverflow.net/questions/403560
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3
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Let $G$ be a reductive group over a field $k$ of characteristic zero with maximal split torus $T$ and Borel $B \supset T$ defining a set of simple roots $\Delta$. Additionally let $\rho$ be the half sum of all positive roots with respect to $B$. For a character $\lambda \in X^\*(T)$ we define a line bundle $L\_\lambda$ on $G/B$ by
$L\_\lambda(U)=\{f: \pi^{-1}(U) \rightarrow k \mid f(gb)=\lambda(b)f(g) \text{ for all }g \in \pi^{-1}(U)(\bar{k}), b \in B(\bar{k})\}$
for $U \subset G/B$ open and $\pi:G\rightarrow G/B$ the natural projection.
Now I'm interested which character $\lambda$ defines the canonical bundle $\omega\_{G/B}$ on $G/B$. I figured by several sources out that its either $-2\rho$ or $2\rho$. But as already mentioned in <https://math.stackexchange.com/questions/654793/confused-about-borel-weil-theorem>, the literature seems somehow inconsistent and I got stuck.
I will greatly appreciate help with this.
|
https://mathoverflow.net/users/135674
|
Character which defines canonical bundle on flag variety
|
I think the most direct way comes from the following fact. There is a natural $G$-equivariant isomorphism $$ T^\*(G/B) \cong G \times\_{B} \mathfrak{b}^{\bot}$$ where $\mathfrak{b}^{\bot}$ is the sub-Lie algebra of $\mathfrak{g}^\*$ given by the annihilator of $\mathfrak{b}$. This can be found in " Representation theory and complex geometry" by Ginzburg Chriss,chapter 1, page 45-46.
This is true in general for Lie groups. Now, in the reductive group context you have an explicit way of describing $\mathfrak{b}^{\bot}$. You start from the decomposition $$\mathfrak{g}=\mathfrak{t} \bigoplus\_{\alpha \in \Delta\_+}\mathfrak{g}\_{\alpha} \bigoplus\_{\alpha \in \Delta\_{-}} \mathfrak{g}\_{\alpha} .$$
Here $\Delta\_{+}$ is the set of positive roots chosen so that $$\mathfrak{b}= \mathfrak{t} \bigoplus\_{\alpha \in \Delta\_+} \mathfrak{g}\_{\alpha}.$$ In this way you can see that $$\mathfrak{b}^{\bot} \cong \bigoplus\_{\alpha \in \Delta\_{+}} \mathfrak{g}\_{\alpha}$$ in such a way that $$T^\*(G/B) \cong G \times\_{B} \mathfrak{b}^{\bot} \cong \bigoplus\_{\alpha \in \Delta\_+} \mathcal{L}(\alpha) .$$
Taking determinants now you should find that $$\omega\_{G/B}=-2 \rho $$ I think (hoping not havinh some signs messing up...)
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2
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https://mathoverflow.net/users/146464
|
403574
| 165,557 |
https://mathoverflow.net/questions/403512
|
1
|
Consider the definition of "lost melody" given by Merlin Carl in his arXiv preprint, "[The Lost Melody Phenomenon](https://arxiv.org/abs/1407.3624v5)" (arXiv: 1407.3624v5 [math.LO] 16 Mar 2015):
>
> A lost melody is a real number $x \subseteq \omega$ which is recognizable, i.e., for some ITTM program $P$, the the computation of $P$ with $y$ on the input tape (which plays the role of a real oracle for an ITTM) is defined for all $y$ and outputs 1 iff $y = x$ and otherwise outputs 0, but not computable, i.e., no program computes the characteristic function of $x$.
>
>
>
Furthermore, Prof. Carl understands the notion of ITTM as follows:
>
> These machines, which are basically classical Turing machines equipped with transfinite running time….
>
>
>
Hamkins and Lewis define the conditions under which ITTMs can be considered "ordinary Turing machines" and in such manner show that classical computability theory can be deemed a subtheory of ITTM-computability theory:
>
> Thus, restricting to the case of finite input and time, a function $f: 2^{\lt {\omega}} \rightarrow 2^{\lt {\omega}}$ is $\omega$-computable exactly when it is computable in the Turing machine sense.
>
>
>
With this in mind I will rewrite Prof. Carl's definition without mentioning ITTM's:
>
> A lost melody is a real number $x \subseteq \omega$ which is recognizable, i.e., from some Turing machine program $P$, the computation of $P$ with $y$ on the input tape (which plays the role of a real oracle for a Turing machine) is defined for all $y$ and outputs 1 iff $y = x$ and otherwise outputs 0 but not computable, i.e., no Turing machine program computes the characteristic function of $x$ [note that such oracles are standard in classical computability (recursion) theory—my comment (see Hartley Rogers Jr's text, chapter 9.2, "Turing Machines with Auxiliary Tapes", pg. 130)].
>
>
>
All that remains is to note that both $\{x \mathrel| \text{$\varphi\_{x}$ is total}\}$ and its complement are both productive sets, and not recursively (computably) enumerable (which means, to me at least, not writable by any recursive function).
So can the singleton $\{\{x \mathrel| \text{$\varphi\_{x}$ is total}\}\}$ be decidable by Prof. Carl's oracle (restricted to classical recursion theory)?
I believe it can for the following reason: i) if one assumes that the oracle in question has multiple oracle tapes—$2^{\aleph\_0}$ of them in fact (one tape for each set of natural numbers)—when one compares the elements of $x$ with the elements of $y$ by the oracle, it will, in fact, show that $y \neq x$ in a finite number of steps, but only show that $y = x$ if the oracle machine does not halt. Now if one wishes to consider the oracle machine an ITTM (for the reason that if $y = x$ the the oracle machine does not halt), by the limsup rule the oracle machine must output a 1, but this is a trivial use of an ITTM.
|
https://mathoverflow.net/users/20597
|
Can $\{x \mathrel| \text{$\varphi_{x}$ total}\}$ be deemed a "lost melody" relative to classical recursion theory?
|
If we disregard the model of computation we want to use, a *lost melody* is a decidable singleton $\{x\}$ such that the point $x$ is non-computable. Classical computability theory cannot admit any lost melodies in $2^\omega$ because there are no decidable singletons in the first place, and it cannot admit lost melodies in $\mathbb{N}$ because there are no non-computable points.
We could weaken the requirement (as you may be doing in your question), and only ask for $\{x\}$ to be $\Pi^0\_1$, meaning that we can recognize the complement of $\{x\}$ but not necessarily $\{x\}$ itself. With $2^\omega$ as ambient space, this still doesn't give us any examples. However, if use $\omega^\omega$, then we see that every hyperarithmetical Turing degree has a representative $x$ such that $\{x\}$ is $\Pi^0\_1$.
Another way to get lost melodies without using a more powerful model of computation would be to move towards computable analysis, and to use general *represented spaces* as ambient spaces. This allows us trivial constructions such as considering any singleton $\{p\}$ for $p \in 2^\omega$ as ambient space in its own right, and since $\{p\}$ is trivially a decidable subset of $\{p\}$, we have plenty of lost melodies here. On the other hand, "nice" represented spaces would generally be computably separable, which immediately implies that every computably open (ie $\Sigma^0\_1$) singleton consists of a computable point.
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2
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https://mathoverflow.net/users/15002
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403578
| 165,560 |
https://mathoverflow.net/questions/403572
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1
|
In the question [here](https://mathoverflow.net/questions/403523/explicit-eigenvalues-of-matrix)
the author asks for the eigenvalues of an operator
$$A = \begin{pmatrix} x & -\partial\_x \\ \partial\_x & -x \end{pmatrix}.$$
Here I would like to ask if one can extend this idea to the operator
$$A = \begin{pmatrix} x & -\partial\_x +c\\ \partial\_x+c & -x \end{pmatrix},$$
where $c$ is a real constant. It seems to me that this is a non-trivial change in the operator.
|
https://mathoverflow.net/users/150549
|
Eigenvalues of operator
|
The extended operator can be treated along similar lines as the $c=0$ case. One merely has to modify the algebra a little. Again, denote the standard harmonic oscillator eigenfunctions (i.e., the eigenfunctions of $-\partial\_{x}^{2} +x^2 $ with eigenvalues $\lambda\_{n} =2n+1$) as $\psi\_{n} (x)$. Introduce also the standard raising and lowering operators $a^{\dagger } $, $a$, in terms of which $x=(a^{\dagger } +a)/\sqrt{2} $ and $-\partial\_{x} =(a^{\dagger } -a)/\sqrt{2} $. Acting on the $\psi\_{n} $, these act as $a\psi\_{n} = \sqrt{n} \psi\_{n-1} $, $a^{\dagger } \psi\_{n} = \sqrt{n+1} \psi\_{n+1} $. Now, $A$ takes the form
$$
A=\frac{1}{\sqrt{2} } \begin{pmatrix} a^{\dagger } +a & a^{\dagger } -a \\ -a^{\dagger } +a & -a^{\dagger } -a \end{pmatrix}
+ \begin{pmatrix} 0 & c \\ c & 0 \end{pmatrix}
$$
One has the following eigenvectors/eigenvalues:
$$
\begin{pmatrix} \psi\_{0} \\ -\psi\_{0} \end{pmatrix} \mbox{ with eigenvalue } \ \ \ -c
$$
$$
\left[ \frac{c\pm \sqrt{c^2 +2n+2} }{\sqrt{2n+2} } \begin{pmatrix}
\psi\_{n} \\ \psi\_{n} \end{pmatrix} + \begin{pmatrix} \psi\_{n+1} \\ -\psi\_{n+1} \end{pmatrix} \right] \ \ \ \mbox{with eigenvalue} \ \ \
\pm \sqrt{c^2 + 2n+2}
$$
For $c=0$, the spectrum reduces to the one given in the answer to the previous question.
|
3
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https://mathoverflow.net/users/134299
|
403584
| 165,562 |
https://mathoverflow.net/questions/403587
|
4
|
In his pioneering paper *An extension of Schwarz's lemma*, Ahlfors proves the lower bound on the Bloch constant $B \geq \frac{\sqrt{3}}{4}$. The proof of this lower bound proceeds as follows:
Let $W$ be a Riemann surface. For an arbitrary point $\mathfrak{m}$ of $W$, let $\rho(\mathfrak{m})$ denote the radius of the largest simple circle of centre $\mathfrak{m}$ contained in $W$. It is clear that $\rho(\mathfrak{m})$ is continuous, and is zero only at the branch points. We introduce the metric $ds = \lambda | dw |$, with $$\lambda = \frac{A}{2 \sqrt{\rho} (A^2-\rho)},$$ where $\rho = \rho(\mathfrak{m})$ and $w$ denotes the variable on the function plane (not the uniformizing variable), and $A^2>B(f)$ (the Bloch constant of $f$).
If $\mathfrak{a}$ is a branch point, then $\rho = | w - \mathfrak{a} |$. Let $n$ be the multiplicity of $\mathfrak{a}$. Then $w\_1 = (w-\mathfrak{a})^{\frac{1}{n}}$ is a uniformizing variable the corresponding $\lambda\_1$ is determined from $\lambda\_1 | dw\_1 | = \lambda | dw |$. Explicitly, $$\lambda\_1 = n \rho^{\frac{1}{2} - \frac{1}{n}} / 2(A^2 - \rho).$$ For $n=2$ the metric is regular, and for $n>2$, the metric is zero.
>
> Ahlfors gives no explanation for why the $\lambda$ above is chosen. Is it simply constructed so that, with respect to the uniformizing variable, we obtain a continuous pseudo metric? Can one choose another metric to start with?
>
>
>
|
https://mathoverflow.net/users/172177
|
Ahlfors' proof of Bloch's theorem
|
He explains his choice in lines 8-9 on p. 364 of the paper:
"This metric has the curvature $-4$ for it is obtained from the hyperbolic metric by the transformation $w'=w^{1/2}$ ."
Remarks. By more sophisticated choices of metrics later authors,
M. Bonk (MR0979048) and Chen and Gauthier (MR1428103) were able to improve Ahlfors's estimate of the Bloch constant.
|
8
|
https://mathoverflow.net/users/25510
|
403589
| 165,565 |
https://mathoverflow.net/questions/403562
|
3
|
Let $X\subset \mathbb{P}\_{\mathbb{C}}^3$ be a projective singular cubic surface with two singular points. Is the rationalcohomology of such objects known? As an example of the type of surfaces I'd be interested in one can take the hypersurface defined by the homogenous cubic $$x^2w+y^2w+z^2w+xyz-zw^2-w^3=0 .$$
In this case for example the singular points are given by $$([2: -\sqrt{3} :\sqrt{3} :1] , [2:\sqrt{3}:-\sqrt{3}:1] .$$
I think that the cohomology should be known for projective smooth cubic surfaces which should be realized as blowup of $\mathbb{P}^2\_{\mathbb{C}}$ in six points but I'm not sure about singular ones.
|
https://mathoverflow.net/users/146464
|
Cohomology of singular projective cubic surface
|
By the classification theorem of cubic surfaces (p.6 in [this paper](https://arxiv.org/pdf/1305.0178.pdf)), a cubic surface belongs to the following classes
1. Has at worst ADE singularities.
2. Has an elliptic singularity, i.e., the surface is cone over a smooth cubic curve.
3. Non-normal or non-integral, and singular along a curve.
So if $X$ has two singularities, $X$ belongs to case 1 and all singularities are rational. We can compute the cohomology by the minimal resolution $\tilde{X}\to X$. The surface $\tilde{X}$ is called a weak del Pezzo surface, which is still blowup of 6 points on $\mathbb P^2$, but in less general positions, so $H^{2}(\tilde{X})=\mathbb Z^7$.
Now let's do some topology: Let $E$ be the exceptional divisor. Then the long exact sequence of the pair $(\tilde{X},E)$ reads
$$H^1(E)\to H^2(X)\to H^2(\tilde{X})\xrightarrow{r} H^2(E),$$
where we used $H^\*(\tilde{X},E)\cong H^\*(X)$ because $\tilde{X}/E\cong X$ as CW complex.
$E$ is the disjoint union of two bunches of rational curves over the two singularities, so $H^2(E)$ has rank $\mu\_1+\mu\_2$, where $\mu\_i$ is the Milnor number of the singularity. Also, $H^1(E)=0$ and $r$ is surjective, so
$$H^2(X)=\mathbb Z^{7-\mu\_1-\mu\_2}.$$
Cohomologies at other degrees are easy to compute.
|
6
|
https://mathoverflow.net/users/74322
|
403592
| 165,567 |
https://mathoverflow.net/questions/403437
|
1
|
Let $\overline{\widehat{Z}\_i} = \frac{E\_i\left[ \int\_{t\_i}^{t\_{i+1}}\widehat{Z}\_sds\right] }{\Delta t\_i}$ with $\widehat{Z}$ a square integrable process, $\Delta t\_i := t\_{i+1} - t\_i$, and $E\_i$ denotes the conditional expectation w.r.t. $F\_{t\_i}$, with standard probability space/filtration.
Why is then $E\_i\left[ \int\_{t\_i}^{t\_{i+1}}(\widehat{Z}\_s -\overline{\widehat{Z}\_i})ds\right] =0$?
More details can be found in <https://arxiv.org/pdf/2006.01496.pdf> , Page 17, equation 5.8
|
https://mathoverflow.net/users/358044
|
Stochastic Integral + conditional expectation
|
Breaking the integral into two terms, the first term is simply $E\_i\left[ \int\_{t\_i}^{t\_{i+1}}\widehat{Z}\_sds\right]$.
The second term is $E\_i \left[\int\_{t\_i}^{t\_{i+1}} \overline{\widehat{Z}\_i} ds\right]$. The term in the expectation is $\mathcal F\_{t\_i}$ measurable, and so the second term is just $\int\_{t\_i}^{t\_{i+1}} \overline{\widehat{Z}\_i} ds.$
The integrand being independent of $s$, this is just $\nabla t\_i \overline{\widehat{Z}\_i}$, which is $E\_i\left[ \int\_{t\_i}^{t\_{i+1}}\widehat{Z}\_sds\right]$.
So the terms cancel and we get $0$ as required.
|
1
|
https://mathoverflow.net/users/173490
|
403609
| 165,571 |
https://mathoverflow.net/questions/403604
|
2
|
Let $\sigma: \mathbb R \times \mathbb R \to \mathbb R$ be a Lipschitz continuous function bounded below by some $M > 0$.
Let $W$ be a standard Brownian motion, and let $X$ be the solution to the SDE
$$dX\_t = \sigma(t, X\_t) dW\_t$$
with $X\_0 = 0$.
**Question:** Fix $T > 0$. Does there exist, for every $\varepsilon, h > 0$ a $\delta > 0$ such that
$$\mathbb P\left(\int\_{0}^T \mathbf 1\_{[-\delta, \delta]} (X\_s) ds > h\right) < \varepsilon?$$
|
https://mathoverflow.net/users/173490
|
A bound for the occupation time of a diffusion
|
I think so. The martingale $X\_t$ is a time-changed Brownian motion: $X\_t = B\_{A\_t}$, where $$A\_T = \int\_0^T \sigma^2(t, X\_t) dt$$ and $B\_t$ is some Brownian motion. Now write
$$ \int\_0^T \mathbb 1\_{[-\delta,\delta]}(X\_t) dt = \int\_0^{A\_T} \mathbb 1\_{[-\delta,\delta]}(B\_s) dA^{-1}\_s = \int\_0^{A\_T} \frac{\mathbb 1\_{[-\delta,\delta]}(B\_s)}{\sigma^2(A^{-1}\_s, B\_s)} ds \leqslant \int\_0^{A\_T} \frac{\mathbb 1\_{[-\delta,\delta]}(B\_s)}{M^2} ds . $$
If $\sigma(t, x)$ is bounded above by some number $N$, then we get
$$ \int\_0^T \mathbb 1\_{[-\delta,\delta]}(X\_t) dt \leqslant \int\_0^{N^2 T} \frac{\mathbb 1\_{[-\delta,\delta]}(B\_s)}{M^2} ds , $$
and the desired result follows directly from the corresponding statement for the Brownian motion. The general case requires a more delicate bound on $A\_T$. I do not have time now to work out the details, but if need be, I will get back to this when I am more available.
|
3
|
https://mathoverflow.net/users/108637
|
403610
| 165,572 |
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