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https://mathoverflow.net/questions/407823
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40
|
Recently I was preparing an undergrad-level proof of (a form of) the Jordan Curve Theorem, and I had forgotten just how much work is involved in it. The proof stored my head was just using Alexander duality plus some sanity-checks on the topology of the curve in question, which is a fine approach but does require a bit of algebraic topology machinery my audience didn't have access to. The more elementary proof (or at least the one I landed on) has a straightforward idea behind it, but turning that into a proper argument required slogging through quite a few details about polygons, regular neighborhoods, etc. Similarly, I couldn't help noticing just how much of a pain it is to prove the 2- and 3-dimensional versions of Stokes' theorem without some notion of manifolds, let alone the usual Stokes' theorem in some suitable setting.
Those are both elementary examples, but it got me thinking about the general topic of results that have very rough proofs from more elementary principles but have much clearer and smoother proofs with some more advanced background. Specifically, what are some examples of results from more advanced or narrow topics of mathematics can vastly simplify or explain in retrospect theorems that are encountered and proved laboriously in less specialized or more common areas of math? (If it helps clarify what I'm trying to get at, another example in my mind is May's "Concise Course in Algebraic Topology," which I think of as having the premise of, "So, now that you've gone through the standard intro algebraic topology course, here's what was secretly going on behind the scenes the whole time.")
|
https://mathoverflow.net/users/61829
|
Results with short, advanced proofs or long, elementary proofs
|
The associativity of the group law on an elliptic curve can be proved in an elementary way by explicitly manipulating algebraic expressions, but this is not very enlightening. By using more advanced geometric ideas, one can [prove associativity more conceptually](https://mathoverflow.net/q/6870).
|
43
|
https://mathoverflow.net/users/3106
|
407842
| 167,058 |
https://mathoverflow.net/questions/407839
|
2
|
I am working on an eigenvalue problem whose general solutions involve the associated Legendre functions. Since the goal is to find bounded solutions, my question boils down to understanding the behavior of the following function,
$$f(\theta) = \frac{P^\mu\_{\nu}(\cos \theta)}{\sin^\mu \theta}$$
when $\theta\to 0$ and $\theta \to \pi,$ where $\mu,\nu\in \mathbb{R}$ and $\nu$ is a parameter depending on the eigenvalue.
According to the DLMF entry [here](https://dlmf.nist.gov/14.8), we see that $P^\mu\_{\nu}(\cos \theta) \sim\_{\theta \to 0} c\_{\mu}(1-\cos \theta)^{-\mu/2}$ and so
$$f(\theta)\sim\_{\theta \to 0} c\_{\mu} (1-\cos \theta)^{-\mu}$$
when $\mu \neq 1,2,\cdots.$ This is not quite helpful since I would like to obtain some condition on $\nu$ (which will tell me about the eigenvalues) based on how $f$ behaves at the boundaries, i.e. $\theta \in \{0,\pi\}.$ So, I was wondering if there are any asymptotic expressions for the function $f$ in terms of the parameter $\nu$ as $\theta\to 0$ or $\theta \to \pi?$
|
https://mathoverflow.net/users/68232
|
How to compute this limit involving the associated Legendre function?
|
I find it useful to represent the Legendre function in terms of a hypergeometric function, using a formula from [Wikipedia](https://en.wikipedia.org/wiki/Associated_Legendre_polynomials#Generalization_via_hypergeometric_functions),
$$f(\theta)=\frac{ (1+\cos \theta)^{\mu/2} \, \_2F\_1\left(-\nu,\nu+1;1-\mu;\frac{1}{2} (1-\cos \theta)\right)}{\Gamma (1-\mu)\sin^\mu\theta(1-\cos \theta)^{\mu/2}}.$$
Then Mathematica gives me the small-$\theta$ expansion
$$f(\theta)=\frac{ 4 (1-\mu)+\theta^2 \bigl(\tfrac{1}{3}(1-\mu) \mu- \nu (\nu+1)\bigr)+{\cal O}(\theta^4)}{2^{2-\mu}\theta^{2\mu} \Gamma (2-\mu)}.$$
The leading order term is $\nu$-independent, as noted in the OP, the next order term does depend on $\nu$.
|
3
|
https://mathoverflow.net/users/11260
|
407845
| 167,059 |
https://mathoverflow.net/questions/407619
|
7
|
It is not clear to me the role of the domain and target in the definition of [prederivators](https://webusers.imj-prg.fr/%7Egeorges.maltsiniotis/ps/m.pdf).
For instance, the classical references put the domain as $\mathit{Dia}$, others as $\mathit{Cat}$ itself.
Sometimes $\mathit{Dia}$ comes with some [conditions](http://user.math.uzh.ch/ayoub/PDF-Files/THESE.PDF), and sometimes not.
The target can also vary from the large CAT, or the 2-category of triangulated categories….
Grothendieck uses
$$
D: \mathit{Dia}^\circ \rightarrow \mathit{Cat}.
$$
*On suppose tout au moins que $\mathit{Dia}$ contient les ensembles ordonnés finis, que $\mathit{Dia}$ est stable par limites finies, par sommes finies. Qu'avec toute catégorie $C$ et toute sous-catégorie strictement pleine ouverte (resp. fermée) elle contienne la sous-catégorie complémentaire.*
Wikipedia says that the idea of this domain is to serve as "[indices](https://en.wikipedia.org/wiki/Derivator#Prederivators)".
>
> What prevents us from using not a 2-sub-category of $\mathit{Cat}$ but the complete category as domain and target?
>
>
>
Sorry if this question is too simple for the site. But can someone clean this up explaining the use of $\mathit{Dia}$ here?
|
https://mathoverflow.net/users/429204
|
Indexing categories of derivators
|
The idea is to have more freedom of expression. No one wants to work with a fixed $Dia$ for ever. In particular, $Dia$ is made more general to allow us to change $Dia$ at will, according to our needs, including to make weird technical choices in some proof (for instance, if you want that (pre)derivators collectively form a reasonnable $2$-category without using Grothendieck universes, you could be happy to know that $Dia$ is small, or you might want to avoid (pre)derivators taking values in categories which are not locally small because you love the Yoneda Lemma and you do not know how to prove that your main example has this property for arbitrary shapes of diagrams yet). However, we can define $Cat$-indexed derivators and say that the restriction to some category of diagrams $Dia$ satisfy the axiom of derivators are satisfied, and this is fine most of the time. But there are properties which can proved using only certain shapes of diagrams and it is natural to restrict the theory to the fragment which only deals with them. A nice example is [the universal property of bounded derived categories](https://arxiv.org/abs/1512.02691), mimicking Keller's approach via [towers of triangulated categories](https://webusers.imj-prg.fr/%7Ebernhard.keller/publ/dcp.pdf). The proof only uses finite diagrams. It is formal to deduce an analogous universal property for the $Cat$-indexed derivator associated to an exact category $\mathcal{E}$ (associating to each $C$ the bounded derived categories of $\mathcal{E}$-valued functors defined on $C$). However, this is a typical example where I do not see how to deduce formally that the version for $Cat$-indexed derivators implies the version for derivators indexed by finite partially ordered sets (if you need that the assignment $X\mapsto\mathbb{D}(X\times C)$ is a (stable) derivator on finite partially ordered sets for any small category $C$, it is not easy to produce such a thing from a (stable) derivator derfined on finite partially ordered sets only). Therefore, in this case, the version for finite diagrams only seems to be a more powerful statement.
|
5
|
https://mathoverflow.net/users/1017
|
407847
| 167,061 |
https://mathoverflow.net/questions/407542
|
1
|
Note: This is not intended to be a research level question, but concerns graduate level material.
**Theorem.** *The opposite $\Delta^\mathrm{op}$ of the simplex category $\Delta^\mathrm{op}$ (as usually defined via nondecreasing maps) can be [presented](https://ncatlab.org/nlab/show/presentation+of+a+category+by+generators+and+relations) as follows: it is the category generated by the graph of finite nonempty linearly ordered sets with arrows $\delta\_i\colon [n]\to [n-1]$ and $\sigma\_i\colon [n]\to [n+1]$ for each $0\leq i\leq n$ subject to the simplicial identities.*
Here is a proof sketch using the following well-known fact:
**Fact.** *Any nondecreasing map $f$ in $\Delta^\mathrm{op}$ has a unique representation as
$$f=\sigma^{i\_1}\dots\sigma^{i\_n}\delta^{j\_1}\dots\delta^{j\_m},$$
where $i\_1>\dots > i\_n$ and $j\_1 <\dots < j\_m.$*
Here, the $\sigma^i$ and $\delta^j$ are now regarded as nondecreasing maps, while in the above theorem they denote formal arrows. Sorry for the abuse of notation.
The above theorem means that for any category $\mathcal C$ there is a canonical bijection between functors $\Delta^\mathrm{op}\to\mathcal C$ and simplicial objects in $\mathcal C$ (given by face and degeneracy morphisms satisfying the simplicial identities). This is the universal property of being presented by generators and relations. By forgetting information one can a assign a simplicial object to each functor $\Delta^\mathrm{op}\to\mathcal C$. So the challenge is to construct an inverse to that assignment. Let $C\_\bullet$ be a simplicial object with face maps $d\_i$ and degeneracy maps $s\_i$. We want to define a functor $\Delta^\mathrm{op}\to\mathcal C$. On objects this functor is given by $C\_\bullet$. Now let $f\colon [n]\to [m]$. We need to find an induced morphism
$$C(f)\colon C\_n\to C\_m.$$
To get one, write
$$f=\sigma^{i\_1}\dots\sigma^{i\_n}\delta^{j\_1}\dots\delta^{j\_m},$$
and then set
$$C(f)=s^{i\_1}\dots s^{i\_n}d^{j\_1}\dots d^{j\_m}.$$
One can now check that this defines a functor as desired.
*Question:* Is there a proof of the above theorem (i.e., a construction of the bijection between functors $\Delta^\mathrm{op}\to\mathcal C$ and simplicial objects of $\mathcal C$) that does *not* use the above fact? I wonder because the unique representation of a morphism $f$ in $\Delta^\mathrm{op}$ looks a bit random and ad hoc to me. In other settings when wanting to show that some concrete category can be presented by generators and relations, does one always convulsively have to find a way that each morphism can be written in a unique way as a composition of "generators" (satisfying additional properties, like $i\_1 > \dots > i\_n$ in the above case)?
I feel the theorem should be true nevertheless: just write $f$ as *any* composition of $\delta^i$ and $\sigma^i$s, and then map it to the same pattern where $\delta$ is replaced by $d$ and $\sigma$ is replaced by $s$. However, in this approach one has to check well-definedness. Can the well-definedness be checked without essentially reproducing the above fact?
The only advantage of the above fact seems to be to have a unique well-defined way to define the mapping. But I strongly feel like the theorem should be true even without particular ad hoc representation.
|
https://mathoverflow.net/users/442261
|
Proving that the simplex category is generated by the face and generacy maps
|
This is not an answer to the question as stated, but an elaboration on my comment (the OP asked me to clarify, but this would be too long for a comment).
Let me start by pointing out that this theorem becomes really easy to prove once one observes that the composition of $\delta$'s is injective, and that of $\sigma$'s is surjective. Indeed, given our non-decreasing map $f$, if this decomposition is to exist, it must correspond to the decomposition $i\circ s$ of $f$ as a surjection followed by an injection. This reduces both existence and uniqueness to the case of an injection or a surjection.
But both of these are easy again : if $i$ is an injective non-decreasing map between linearly ordered sets (in fact it must then be increasing), then it is entirely specified by its image, and in fact by the complement of its image. Now in the composition $\delta\_{j\_m}\circ ... \delta\_{j\_1}$ in $\Delta$, if $j\_m>...>j\_1$, then the complement of the image is exactly the subset $\{j\_1,...,j\_m\}$, which proves at once existence and uniqueness.
The argument for surjections is very similar, you just observe that now a surjective nondecreasing map between finite linearly ordered sets is entirely determined by the places where it doesn't "jump".
So first of all, the theorem is not hard and quite conceptual once you think about it. Of course, saying "this theorem is nice, don't ask how to avoid using it" does not answer your question...
The second part of my comment was about why something like this "had" to be done anyways. A "presentation" of an algebraic object is a description of this object in terms of generators and relations. This tells you how to describe morphisms out of it : say where the generators go, and check that their images satisfy these relations. What you want to show amounts (by some abstract nonsense) precisely to finding a presentation for $\Delta$ in terms of these $\delta$'s and $\sigma$'s, with relations the simplicial identities.
Now, when you start with an algebraic structure (an algebra, a module, a category , ...), it always has a *tautological* presentation : take all elements (/objects + arrows) as generators, and all relations as relations. This is of course useless in practice, so one has to do some specific, "ad hoc" things to get useful presentations - like the one here. Our presentation of $\Delta$ has to... depend on $\Delta$ ! So you have to prove something "ad hoc", or "random" (even though the first part of this answer argued that it was not "random", and was actually very conceptual, and the combinatorics that appear there are quite easy)
In general, it suffices to prove things like " every element can be represented in such a way, and I know how to navigate between these representations". "navigate" means different things in different contexts, but in the context of categories say, it means "I know how to represent the composite of two such representations". If you can explain, via some relations, why representations compose the way they do, then you have your presentation by generators and relations !
In other words (this is where "rewriting" comes in), given two representations of my morphisms in terms of my generators in a nice way that I like, I can represent their composite in a nice way too by following my relations, I can *rewrite* it using my relations. Rewriting comes up a lot in proving presentations precisely for this reason. If I can prove that every element of my group is of the form, say $a^ib^j$, it'd be nice to know how I can write $a^ib^ja^kb^l$ in these terms. This can get quite complicated, but if I have, say, a simple rule of the form $ba \to a^2 b^5$ then I know I can pass all my $a$'s past the $b^j$ and I will be done. In particular, I will have that the generators $a,b$ and the relation $ba = a^2b^5$ form a presentation of my object.
I hope that clarifies my comment. Let me now maybe explain why I don't think the answer to your question is "there is such a proof", in a reasonable sense.
You need to prove that $\Delta$ has a presentation given by these $\sigma$'s, $\delta$'s, and the simplicial identities. A thing that you *have* to prove anyways, is that any map can be written as a composite of $\delta$'s and $\sigma$'s. Now, if I'm being honest, I don't know how to do that in a clean way if not for the argument that I gave above.
But let's say you can fiddle around and find a proof of that without that argument. Then you still need to prove that simplicial identities are enough to generate all identities. As Dmitri Pavlov points out in the comments, one way to do so is to show that any two representations of morphisms in terms of the generators can be related by a sequence of "exchanges" allowed by the simplicial identities - he even described such a sequence in general, a "rewriting" scheme if you will. But the way he did it is he rewrote everything to reach the canonical form granted by the theorem !!
I guess the point I'm trying to make is that you can't get away from proving something specific to $\Delta$, and that something will, one way or another, look like the theorem in question. But maybe I'm wrong, and I cannot prove "there is no proof that avoids the theorem", and I'm not well-versed enough in combinatorics to be categorical about this, so don't take this as a definitive answer.
|
5
|
https://mathoverflow.net/users/102343
|
407858
| 167,065 |
https://mathoverflow.net/questions/407859
|
8
|
In [The Fourier Transform of the quartic Gaussian $\exp(-Ax^4)$: Hypergeometric functions, power series, steepest descent asymptotics and hyperasymptotics and extensions to $\exp(-Ax^{2n})$](https://doi.org/10.1016/j.amc.2014.05.001), Boyd derives the asymptotic of the integral
$$\mathcal{A}(x) = \int\_{-\infty}^\infty e^{ikx} e^{-k^4}\,dk$$
for large $|x|$ using the method of steepest descent from complex analysis. The result is that $\mathcal{A}$ is a decaying oscillation with amplitude $\mathcal{O}\left(x^{-1/3}e^{-Cx^{4/3}}\right)$ for some $C>0$. Is there a way to derive this result using real-variable techniques?
|
https://mathoverflow.net/users/152473
|
Is there a real-analytic way to derive the asymptotics of $\int_{-\infty}^\infty e^{ikx} e^{-k^4}\,dk$ as $|x|\to\infty$?
|
A differential equation for ${\cal A} (x) $ can be obtained as follows,
$$
\frac{d^3}{dx^3 } {\cal A} (x) = \int\_{-\infty }^{\infty } dk\, (-ik^3 ) e^{ikx} e^{-k^4 } = \frac{x}{4} \int\_{-\infty }^{\infty } dk\, e^{ikx} e^{-k^4 } = \frac{x}{4} {\cal A} (x)
$$
where integration by parts has been used in the second equality. The leading asymptotic behavior of this differential equation is satisfied by
$$
{\cal A} (x) = \exp (-C x^{4/3} )
$$
with
$$
C = e^{\pm i\pi /3 } \frac{3}{4^{4/3} } = \frac{3}{4^{4/3} } \left( \frac{1}{2} \pm i \frac{\sqrt{3} }{2} \right)
$$
(the third solution, with $C$ real and negative, is excluded since it diverges at large $x$). The two solutions must be suitably linearly combined to construct a real ${\cal A} (x)$. To obtain power corrections, make the ansatz
$$
{\cal A} (x) = f(x) \exp (-C x^{4/3} )
$$
yielding the following differential equation for $f$,
\begin{eqnarray\*}
& f''' - 4C x^{1/3} f'' + 3\left( C^2 \frac{16}{9} x^{2/3} - C\frac{4}{9} x^{-2/3} \right) f' & \\
& + \left( x\left[ -\frac{1}{4} -C^3 \frac{64}{27} \right] + C^2 \frac{16}{9} x^{-1/3} + C\frac{8}{27} x^{-5/3} \right) f=0 &
\end{eqnarray\*}
The leading order (i.e., the term in the square brackets) is of course already satisfied by the above choices of $C$, but now we can also read off the next order: Namely, the terms of order $x^{-1/3} f$ and $x^{2/3} f' $ must be brought to cancel. This requires that $f$ behaves as $f\sim x^{-1/3} $ such that the derivative yields the factor $-1/3$ that is needed to effect the cancellation. Thus, also the $x^{-1/3} $ power correction is clear. Further orders can be obtained systematically.
$$
$$
|
12
|
https://mathoverflow.net/users/134299
|
407861
| 167,066 |
https://mathoverflow.net/questions/407867
|
0
|
Let $u \in C^\infty\_c(\mathbb{\Omega})$ and $\varphi$ be an eigenfunction of the fractional Laplacian $(-\Delta)^s$ in $\Omega$ with eigenvalue $\lambda$. In what sense, if any, is it true that
$$\langle u, \varphi \rangle = \frac{1}{\lambda} \langle u, (-\Delta)^s\varphi \rangle = \frac{1}{\lambda} \langle(-\Delta)^s u, \varphi \rangle = \frac{1}{\lambda^2} \langle (-\Delta)^{2s} u, \varphi \rangle = ........?$$
Here $ \langle \cdot, \cdot \rangle$ denotes the scalar product in $L^2(\Omega)$. In other words, is it true that we can integrate by parts using $(-\Delta)^su$, $(-\Delta)^{2s}u$ and so on even though their support is not compact anymore?
|
https://mathoverflow.net/users/122620
|
Iterated integrations by parts using the fractional Laplacian
|
No, we cannot.
Formally, $\varphi$ is the eigenfunction of the unbounded operator $L\_s$ on $L^2(\Omega)$, defined initially by
$$ L\_s u(x) = (-\Delta)^s u(x) \qquad \text{for } x \in \Omega , $$
where $u \in C\_c^\infty(\Omega)$ (and it is understood that $u(x) = 0$ for $x \notin \Omega$), and then extended to an appropriate domain (e.g. by means of Friedrichs extension).
Now the key observation is that $L\_s L\_s$ is *not* equal to $L\_{2s}$, unless $\Omega = \mathbb R^d$. Indeed, in $\Omega$ we have
$$ L\_s L\_s u = (-\Delta)^s (\mathbb 1\_{\Omega} \times (-\Delta)^s u) , $$
while
$$ L\_{2s} u = (-\Delta)^s (-\Delta)^s u , $$
and due to non-locality of $(-\Delta)^s$, the two are not equal.
For the above reasons, we have
$$ \langle u, \varphi \rangle = \frac{1}{\lambda^2} \langle u, L\_s L\_s \varphi \rangle = \frac{1}{\lambda^2} \langle L\_s L\_s u, \varphi \rangle $$
(provided that $L\_s u$ belongs to the domain of $L\_s$, which is rarely the case!), and in general the right-hand side is not equal to $\lambda^{-2} \langle (-\Delta)^{2s} u, \varphi \rangle$.
|
0
|
https://mathoverflow.net/users/108637
|
407874
| 167,069 |
https://mathoverflow.net/questions/407290
|
2
|
$\DeclareMathOperator\wt{wt}$Let $\wt(n)$ be [A000120](https://oeis.org/A000120), number of $1$'s in binary expansion of $n$ (or the binary weight of $n$).
Let $f(n)$ be [A007814](https://oeis.org/A007814), the exponent of the highest power of $2$ dividing $n$, a.k.a. the binary carry sequence, the ruler sequence, or the $2$-adic valuation of $n$.
Also
$$n=2^{t\_1}(1+2^{t\_2+1}(1+\dotsb(1+2^{t\_{wt(n)}+1}))\dotsb).$$
Then we have an integer sequence given by
$$a(n)=\sum\limits\_{j=0}^{2^{\wt(n)}-1}(-1)^{\wt(n)-\wt(j)}\prod\limits\_{k=0}^{\wt(n)-1}\left(1+f\left(\left\lfloor\frac{j}{2^k}\right\rfloor+1\right)\right)^{t\_{k+1}+1},\quad a(0)=1.$$
Let
$$s(n)=\sum\limits\_{k=0}^{2^n-1}a(k).$$
Then I conjecture that $s(n)$ is [A095989](https://oeis.org/A095989), INVERTi transform applied to the ordered Bell numbers.
I also conjecture that
\begin{align}
a(0)=a(1)&=1\\
a(2n+1) &= a(2n)\\
a(2n)& = a(n)+a(2n-2^{f(n)})+b(n-1)\\
b(2n+1) &= b(n)\\
b(2n) &= a(2n).
\end{align}
In other words
\begin{align}
a(2n) &= c(n)\\
c(0)&=1\\
c(n)& = c\left(\left\lfloor\frac{n}{2}\right\rfloor\right)+c\left(\left\lfloor\frac{2n-2^{f(n)}}{2}\right\rfloor\right)+c(g(n-1))
\end{align}
where $g(n)$ is [A025480](https://oeis.org/A025480), $g(2n) = n$, $g(2n+1) = g(n)$.
Is there a way to prove it? Is it possible to at least get a closed form for $s(n)$?
Similar questions:
* [Sum with Stirling numbers of the second kind](https://mathoverflow.net/questions/406738/sum-with-stirling-numbers-of-the-second-kind)
* [Recurrence for the sum](https://mathoverflow.net/questions/405174/recurrence-for-the-sum)
* [Pair of recurrence relations with $a(2n+1)=a(2f(n))$](https://mathoverflow.net/questions/406902/pair-of-recurrence-relations-with-a2n1-a2fn)
* [Sequences that sums up to second differences of Bell and Catalan numbers](https://mathoverflow.net/questions/407758/sequences-that-sums-up-to-second-differences-of-bell-and-catalan-numbers)
|
https://mathoverflow.net/users/231922
|
Sequence that sums up to INVERTi transform applied to the ordered Bell numbers
|
First, we let $P(j,k):=1+f(\lfloor\tfrac{j}{2^k}\rfloor+1)$ and sum over $n$ of fixed weight $\ell:=\mathrm{wt}(n)$ (like in [this answer](https://mathoverflow.net/q/406817)):
\begin{split}
s(n) &= \sum\_{\ell=0}^n \sum\_{t\_1 + \dots + t\_\ell \leq n-\ell}
\sum\_{j=0}^{2^\ell-1} (-1)^{\ell-\mathrm{wt}(j)} \prod\_{k=0}^{\ell-1} P(j,k)^{t\_k+1} \\
&= [x^{n+1}]\ \sum\_{\ell=0}^n \sum\_{j=0}^{2^\ell-1} (-1)^{\mathrm{wt}(j)+1} \prod\_{k=0}^{\ell} \frac{-P(j,k)x}{1-xP(j,k)}.
\end{split}
Now, we notice that $P(j,k)$ depends on runs of unit bits in $j$. Namely, each run of length $u-1$ contributes the term
$$\prod\_{v=1}^{u} \frac{-vx}{1-vx} = x^{u} (-1)^{u} u! \prod\_{v=1}^{u} \frac1{1-vx}.$$
Hence, introducing the number $z$ of zero bits in $j$ padded with an extra zero at the beginning (and so $\mathrm{wt}(j)=\ell+1-z$), we have
\begin{split}
s(n) &= [x^{n+1}]\ \sum\_{\ell=0}^n [y^{\ell+1}]\ \sum\_{z\geq 0} (-1)^{\ell+2-z} \left(\sum\_{u\geq 1} y^u (-1)^u x^u u! \prod\_{v=1}^u \frac1{1-vx}\right)^z\\
&= -[x^{n+1}]\ \sum\_{\ell=0}^n (-1)^{\ell+1} [y^{\ell+1}]\ \left(1+\sum\_{u\geq 1} y^{u} (-1)^u x^u u! \prod\_{v=1}^u \frac1{1-vx}\right)^{-1}\\
&=- [x^{n+1}]\ \left(1+\sum\_{u\geq 1} x^u u! \prod\_{v=1}^u \frac1{1-vx}\right)^{-1}\\
&=- [x^{n+1}]\ \left(1+\sum\_{u\geq 1} x^u u! \sum\_{m\geq 0} S(m,u) x^{m-u}\right)^{-1}\\
&=- [x^{n+1}]\ \left(1+\sum\_{m\geq 1} x^m B\_m^o\right)^{-1},
\end{split}
which can be recognized as INVERTi transform of ordered Bell numbers $B\_m^o$.
|
2
|
https://mathoverflow.net/users/7076
|
407881
| 167,070 |
https://mathoverflow.net/questions/280223
|
6
|
Given an augmented simplicial object $d\_\bullet:X\_\bullet \to \Delta X\_{-1}$, suppose there's a simplicial map $s\_\bullet :\Delta X\_{-1}\to X\_\bullet$ making $d\_\bullet$ a deformation retract, i.e such that $d\_\bullet$ is both a retract and a homotopy-section of $s\_\bullet$. This is equivalent to providing the augmented simplicial object with an "extra degeneracy" (just an alternative description of the simplicial homotopy axioms in this case). Such an extra degeneracy of an augmented simplicial object will be called a *splitting*.
Let $\mathrm{S}$-$s\mathsf C$ denote the category of split simplicial objects (with fixed splitting) and simplicial arrows between them respecting the simplicial homotopies. This category admits a forgetful functor to simplicial objects $s\mathsf C$.
On [page 20](https://books.google.com.au/books?id=427jnAAACAAJ&lpg=PP1&pg=PA20#v=onepage&q=20&f=false) of Duskin's *Simplicial Methods and the Interpretation of Triple Cohomology* ([AMS page](http://bookstore.ams.org/memo-3-163/)), the author remarks this forgetful functor is a *left* adjoint to a shifting functor defined by deleting the top face map, viewing the top degeneracy as an extra one, and shifting to a lower index. (The simplicial homotopy is defined by $h\_i=s\_0^{n-i}s\_{n+1}d\_0^{n-i}$.)
Moreover, the author writes this shifting functor $s\mathsf C\to \mathrm{S}$-$s\mathsf C$ is monadic. In other words, simplicial objects are monadic over split simplicial objects.
1. What's the intuition behind the fact the shifting functor actually takes values in split simplicial objects? This seems strange to me - as if saying a simplicial object becomes contractible if you forget the top face map. How can that be?
2. What's the intuition behind monadicity? An algebra over a split simplicial object (which is already a structured simplicial object) is an arrow to it, so how can additional structure on an already structured object yield back the original notion of object and arrow?
**Added.** I am looking for naive *geometric* intuition for these facts, namely: 1. that a simplicial object becomes contractible upon merely forgetting a face map and reindexing; 2. simplicial objects are monadic over split ones. Ideally, I would like an example of what the contraction deformation retract actually does to the Décalage of a non-contractible simplicial complex, say the boundary of a tetrahedron.
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https://mathoverflow.net/users/69037
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Why are simplicial objects monadic over split (contractible) simplicial objects?
|
Coming across this old question and noticing it still unanswered, here’s an attempt at the “geometric” answer OP wanted.
First note that a split simplicial complex isn’t in general contractible — it’s just *1-truncated*, i.e. each *connected component* is contractible. Concretely, the augmentation indexes the connected components, and the splitting + homotopy gives a distinguished point + contraction for each component.
Write $R : \mathrm{SSet} \to \mathrm{SplSSet}$ for the desired right adjoint. We have $(RX)\_{-1} = X\_0$, so $X\_0$ indexes the components of $RX$, and then for each $x \in X\_0$, the $n$-simplices of $RX$ in the component over $x$ are the $n+1$-simplices whose $(n+1)$th vertex is $x$ (or $0$th vertex, depending on your indexing convention). In other words the $x$-component of $RX$ is the “out-star” (or “out-cocone”, or “out-path-space”) of $x$, consisting of all simplices whose *source* is $x$. But now each out-star is contractible, contracting into the degenerate 1-simplex on $x$; this is easy to check algebraically, but also easy to see geometrically/topologically — the space of “paths in $X$ with source $x$” is contractible to the constant path on $x$.
Briefly: **Under the right adjoint, a simplicial set falls apart into the collection of out-stars of its vertices; and each out-star is clearly contractible to the reflexivity path on that vertex.**
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3
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https://mathoverflow.net/users/2273
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407887
| 167,072 |
https://mathoverflow.net/questions/407868
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0
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There is an array $a\_1,\dotsc,a\_n$ whose elements are pairwise distinct. We define a reverse order pair to be an ordered pair $(a\_i,a\_j)$ such that $i < j$ and $a\_i > a\_j$. Consider the total number of reverse order pairs $N$.
Assume the array is permuted uniformly and randomly, it is well known that $E[N] = \frac{n(n+1)}{4}$. What is the probability distribution of $N$?
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https://mathoverflow.net/users/117620
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The distribution of number of reverse order pairs in a randomly permuted array
|
First, a quick note on terminology: the standard term for a "reverse order pair" is an *inversion*. Knowing this makes it easier to search for the answer:
The generating function for the number of permutations with $k$ inversions is the $q$-factorial $[n]\_q!$ as shown e.g. [here](https://www.tvhoang.com/articles/2021/03/inversion-stat-distribution-proof).
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2
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https://mathoverflow.net/users/46140
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407889
| 167,073 |
https://mathoverflow.net/questions/407773
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1
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**ASSUMPTION 1**: there exists a continuous random vector $(X,Y,Z)$ such that
$$
\begin{cases}
p\_1=\Pr(X\geq 0, Z\geq 0)\\
p\_2=\Pr(Y\geq 0, Z< 0)\\
p\_3=\Pr(X< 0, Y<0)\\
\end{cases}
$$
where $(p\_1,p\_2,p\_3)\in [0,1]^3$ and $p\_1+p\_2+p\_3=1$. Further, the marginal distribution of each of $X,Y,Z$ are symmetric around 0.
Note that such a random vector $(X,Y,Z)$ may not exist for some values of $(p\_1,p\_2,p\_3)$. This is why here I **assume** that it exists.
**QUESTION**: Does Assumption 1 imply that there exists (i.e., we can construct from $(X,Y,Z)$) a continuous random vector $(W,H,Q)$ such that:
1. it holds that
$$
\begin{cases}
\Pr(W\geq 0, Q\geq 0)=p\_1\\
\Pr(H\geq 0, Q< 0)= p\_2\\
\Pr(W< 0, H<0)= p\_3\\
\end{cases}
$$
2. the marginal distribution of each of $W,H,Q$ are symmetric around 0.
3. $Q=W-H$.
---
Note that the map from $(X,Y,Z)$ to $(W,H,Q)$ does not need to be deterministic. For instance, it could be that $W=X+\epsilon$ where $\epsilon$ is another well defined random variable.
---
**SOME DISCUSSION ON THE MOTIVATION BEHIND THE QUESTION:** I have a problem in statistics/computer science where I need to verify the existence of a 3-d distribution function that satisfies constraints 1-3. However, constraint 3 is computationally intractable to implement because infinite-dimensional. Much simpler is to verify the existence of a 3-d distribution function that satisfies constraints 1-2 and, then, use the construction I'm investigating about (if it exists!) to conclude about the existence of the originally desired distribution.
---
**ATTEMPTED ANSWER (with questions)**:
Let $(X,Y,Z)$ and $(W,H,Q)$ be defined on the same probability space $(\Omega,\mathcal{F}, \Pr)$.
Define $(W,H,Q)$ as follows:
* For each $\omega \in \Omega$ such that $X(\omega)\geq 0$ and $Z(\omega)\geq 0$:
$$\big(W(\omega), H(\omega), Q(\omega)\big)=\big(2X(\omega), X(\omega), X(\omega)\big)$$
* For each $\omega \in \Omega$ such that $Y(\omega)\geq 0$ and $Z(\omega)< 0$:
$$\big(W(\omega), H(\omega), Q(\omega)\big)=\big(Y(\omega), 2Y(\omega), -Y(\omega)\big)$$
* Complete the definition of $(W,H,Q)$ with negative values for $W$ and $H$ in such a way that the marginals are symmetric around zero and that $\Pr(W<0, H<0)=p\_3$. Hence, we will need to have: $-2X, -Y$ for $W$; $-X, -2X$ for $H$; $-X, Y$ for $Q$. I'm not sure we can always do this, though. Can we?
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https://mathoverflow.net/users/nan
|
Construct a random vector as a function of another random vector
|
I will begin with a reformulation of your question which makes it not only more symmetric, by also (at least for me) more natural and interesting. I will pass from your variables $(W,H,Q)$ to new variables $(X,Y,Z)$ (which are **not** your original $X,Y,Z$) by putting $X=W, Y=-Q, Z=-H$. Then your condition (3) becomes just $X+Y+Z=0$. Therefore you are asking about a probability distribution $\pi$ on the plane
$$\mathcal P=\{x+y+z=0\}\subset\mathbb R^3$$ such that all its one-dimensional marginals are symmetric, and you are interested in the possible values of
$$
\begin{aligned}
p\_1=\pi\{x>0, y<0\} \;, \\
p\_2=\pi\{y>0, z<0\} \;, \\
p\_3=\pi\{z>0,x<0\} \;.
\end{aligned}
$$
Although it's not completely clear to me what is the **continuity** of random vectors you are referring to, I will presume that it implies that the probability $\pi$ of the line $\ell\_x=\mathcal P \cap\{x=0\}$ and of the analogous lines $\ell\_y,\ell\_z\subset\mathcal P$ are all 0, so that replacing non-strict inequalities with strict ones doesn't change anything (the general case can also be treated in pretty much the same way). For a better visualization one can draw a picture of the plane $\mathcal P$ and the lines $\ell\_x,\ell\_y,\ell\_z$ in the coordinates $x,y$ (so that $z=-x-y$).
Since all the domains from the definitions of the numbers $p\_i$ are pairwise disjoint, and the complement of their union consists of the zero measure lines $\ell\_x,\ell\_y,\ell\_z$, indeed $\sum p\_1=1$. Moreover, the symmetry condition on the marginals means that all numbers $p\_i$ satisfy the conditions $p\_i\le \frac12$ (and, of course $p\_i\ge 0$).
I claim that, conversely, **any collection $(p\_i)$ like this can be realized by a measure $\pi$ satisfying the above conditions**. Indeed, take for $\pi$ the distribution with two atoms at the points $(1,-2,1)$ and $(-1,2,-1)$ and the weight $\frac12$. Then the corresponding $p$-vector $(p\_1,p\_2,p\_3)$ is
$$
v\_3 = \left(\tfrac12, \tfrac12, 0\right) \;.
$$
If you want to have an absolutely continuous measure $\pi$ with the same property, you can just smoothen the above distribution by taking the normalized sum of the uniform distributions on, say, side $\varepsilon$ squares centered at $(1,-2)$ and $(-1,2)$ in the coordinates $(x,y)$.
By symmetry, there are also measures $\pi$ with the $p$-vectors $v\_2=\left(\tfrac12, 0, \tfrac12\right)$ and $v\_1=\left(0, \tfrac12, \tfrac12\right)$. By taking convex combinations of the associated measures $\pi$ one can realize any $p$-vector from the convex hull of $v\_1,v\_2,v\_3$, which is precisely the set determined by the above conditions on $p\_i$ (for visualization one can draw a picture in the coordinates $p\_1,p\_2,1-p\_1-p\_2$).
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3
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https://mathoverflow.net/users/8588
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407890
| 167,074 |
https://mathoverflow.net/questions/407882
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7
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I’m currently studying basics on étale cohomology, by Fu’s and Milne’s book. The formalization of vanishing cycle and nearby cycle particularly interests me. I realized it may relate with reduction and model problems (for example, Oda’s *A Note on Ramification of the Galois Representation on the Fundamental Group of an Algebraic Curve*.)(However, I’m also a beginner on these problems.) I want to learn more about nearby cycles and vanishing cycles, but the classical reference SGA7 is not available as reading French is quite difficult for me (especially reading something new). I have learned the elementary definition and motivation. So are there any textbook, online notes or videos on this topic, with enough detail?
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https://mathoverflow.net/users/170335
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Reference request: the geometry of vanishing cycle
|
Beyond the definitions in section 9.2 of Lei Fu's book, you also find a finiteness result for $R^q\psi\mathcal{F}$ for a finite type $S$-scheme and constructible $\mathcal{F}$ (9.4.1). Nearby cycles, as well as their finiteness feature in the proof of finiteness for $R^q f\_{\ast}\mathcal{F}$, where $f$ is a morphism of finite type $S$-schemes and $\mathcal{F}$ is torsion constructible, in (9.5.1).
I guess you are aware of this. Then, I would say that this is enough for the algebraic formalism. It is better to see them in action.
Since you presumably want to avoid French, and mention models and reduction, you may want to have a look at this paper by *Takeshi Saito : Vanishing Cycles and Geometry of Curves Over a Discrete Valuation Ring*. There, he proves that a local analogue of the stable reduction theorem (of Grothendieck, Deligne-Mumford): for $X/S$ a normal relative curve with an isolated singularity at a closed point $x$ of the special fiber and $Y$ a regular minimal model of $X$ (defined in the paper), the special fiber of $Y$ is a normal crossing divisor of $Y$ if and only if the action of inertia on the stalk $R\psi^1(\mathbb{Q}\_{\ell})\_x$ is unipotent.
As you can imagine, nearby cycles feature prominently in the proof.
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6
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https://mathoverflow.net/users/69401
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407891
| 167,075 |
https://mathoverflow.net/questions/407879
|
3
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$\DeclareMathOperator\SO{SO}\DeclareMathOperator\Spin{Spin}$What is the exact relationship between the finite dimensional representations of the group $\SO(n)$ and its covering group $\Spin(n)$? More precisely, what are examples of representations of $\Spin(n)$ that do not factor through a representation of $\SO(n)$? Can we give a complete description of such representations?
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https://mathoverflow.net/users/176218
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Relationship between the representation theory of $\operatorname{Spin}(n)$ and $\operatorname{SO}(n)$
|
$\def\Spin{\text{Spin}}$The cover $\Spin(n) \to SO(n)$ is $2$ to $1$, the nontrivial element kernel is a central element of $\Spin(n)$ which I'll call $z$. Since $z$ is central, it acts on any irrep of $\Spin(n)$ by $\pm 1$; if $z$ acts by $1$, then the irrep factors through $SO(n)$, if $z$ acts by $-1$, then it does not.
Irreps of Lie groups are usually described in terms of high weight vectors, which are characters of maximal tori. For notational simplicity, I'll take $n = 2m$ to be even. Then a maximal torus $T$ of $SO(2m)$ looks like
$$\begin{bmatrix}
\cos \theta\_1 & - \sin \theta\_1 &&&&& \\
\sin \theta\_1 & \cos \theta\_1 &&& \\
&& \cos \theta\_2 & - \sin \theta\_2 &&& \\
&& \sin \theta\_2 & \cos \theta\_2 &&& \\
&& && \ddots && \\
&&&&& \cos \theta\_m & \sin \theta\_m \\
&&&&& - \sin \theta\_m & \cos \theta\_m \\
\end{bmatrix}$$
so a character of $T$ looks like $e^{i (k\_1 \theta\_i+ \cdots + k\_m \theta\_m)}$ for some $(k\_1, \ldots, k\_m)$ in $\mathbb{Z}^m$. Such a character is dominant if $k\_1 \geq k\_2 \geq \cdots \geq k\_{m-1} \geq |k\_m|$. So irreps are indexed by $m$-tuples of integers $(k\_1, \ldots, k\_m)$ with $k\_1 \geq k\_2 \geq \cdots \geq k\_{m-1} \geq |k\_m|$.
Letting $\widehat{T}$ be a maximal torus for $\Spin(n)$, we have a short exact sequence
$$1 \to \langle z \rangle \longrightarrow \widehat{T} \longrightarrow T \to 1.$$
Characters of $\widehat{T}$ can be thought of as $m$-tuples $(k\_1, \ldots, k\_m)$ which lie in $\mathbb{Z}(1/2, 1/2, \ldots, 1/2) + \mathbb{Z}^m$, with the factors that have integer entries factoring through $T$ and the ones with half-integer entries not factoring. Again, a character is dominant if $k\_1 \geq k\_2 \geq \cdots \geq k\_{m-1} \geq |k\_m|$.
The simplest example of a representation of $\Spin(n)$ which doesn't factor through $SO(n)$ is the irrep with high weight $(1/2, 1/2, \ldots, 1/2)$; call it $W$. There isn't a simple description of this representation, but one thing that I find helpful is that there is a nonzero map $W \otimes W \longrightarrow \bigwedge^{m} \mathbb{R}^{2m}$. If you compute the matrices for $SO(2m)$ acting on $\bigwedge^{m} \mathbb{R}^{2m}$, you'll see that lots of the matrix entries are determinants of $m \times m$ skew symmetric matrices; the matrix entries for $W$ are the square roots of these determinants, meaning certain Pfaffians of $m \times m$ skew symmetric matrices.
There is no simple construction of $\Spin(n)$, but the usual route is by the Clifford algebra and can be found in books like Fulton and Harris's representation theory.
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8
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https://mathoverflow.net/users/297
|
407895
| 167,076 |
https://mathoverflow.net/questions/407904
|
5
|
Given a projective variety $X$ over a field of any characteristic, consider a line bundle $\mathcal{L}$ over $X$.
* The existence of a line bundle $\mathcal{L}^\prime$ with an isomorphism ${\mathcal{L}^\prime}^{\otimes 2} \simeq \mathcal{L}$ is equivalent to the existence of a simple cyclic cover $Y \rightarrow X$ of degree $2$ which is branched on a divisor $D$ associated to $\mathcal{L}$. Such a cyclic cover depends on the choice of the rational section $s$ of $\mathcal{L}$ such that $D = \text{div}(s)$. See section 3.3 of <https://arxiv.org/abs/2009.01831v2> for a proof. In the related question [divisors and powers of line bundles](https://mathoverflow.net/questions/117443/divisors-and-powers-of-line-bundles), Francesco Polizzi gives an example of a line bundle on a K3 surface which does not admit a square root.
* There is also the root stack construction that gives a Deligne-Mumford stack $f : {}^{2}\sqrt{(X,\mathcal{L})} \rightarrow X$ which is the moduli space of square roots of $\mathcal{L}$.
I am wondering if there exists a finite surjective morphism of projective varieties $g : X^\prime \rightarrow X$ such that $g^\*\mathcal{L}$ admits a square root over $X^\prime$. If such a map exists, it should factor through $f$.
|
https://mathoverflow.net/users/158892
|
Square root of a line bundle up to a finite surjective morphism
|
Assume $\mathcal{L}$ is associated with an effective Cartier divisor $D$. Let $D'$ be another Cartier divisor such that $D + D'$ is divisible by 2 in $\mathrm{Pic}(X)$. Let
$$
g \colon X' \to X
$$
be the double covering branched at $D + D'$. Then $g^{-1}(D) = 2R$ for a Cartier divisor $R$ on $X'$, hence $g^\*\mathcal{L} \cong \mathcal{O}\_{X'}(2R)$ has a square root.
If $\mathcal{L}$ is not associated with an effective divisor, you can replace $\mathcal{L}$ by $\mathcal{L} \otimes \mathcal{O}\_X(2N)$ (where $\mathcal{O}\_X(1)$ is an ample line bundle) with $N \gg 0$, so that this bundle is associated with an effective divisor, and apply the previous construction.
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7
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https://mathoverflow.net/users/4428
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407908
| 167,079 |
https://mathoverflow.net/questions/407910
|
0
|
I ask the question because of the following statement found in Mark Burgin's paper, "Algorithmic complexity of recursive and inductive algorithms", *Theoretical Computer Science* 317 (2004) 31-60 (pg. 34):
>
> It is usually assumed that any finite set is recursively computable and even decidable. When a finite set is given by a list, then this is true. However, this assumption is not valid in a general case when a finite set can be defined by a description [the 'presentation'--my comment]. For instance, let us take the set $X$ of all indices of those Turing machines that have the length of their description less than 1000 and that do not terminate on some input with the length less than 1000. This set is finite, but it is not recursively computable (enumerable).
>
>
>
The problem I see with Prof. Burgin's statement is that computability theory can be (and is usually) formulated in $ZFC$ (the background set theory) in which all sets (regardless of description) can be well-ordered (it should be noted that well-ordered sets, finite or otherwise, can be identified as lists--see Oliver Deiser's paper, "An Axiomatic Theory of Well-Orderings", *The Review of Symbolic Logic* Volume 4, Issue 2, June 2011, pp. 186-204). As regards finite sets, the Diaconescu-Goodman-Myhill theorem (taken from nlab)
>
> The following are equivalent:
>
>
> 1. The principle of excluded middle
> 2. Finitely indexed sets are projective (in fact, it suffices 2-indexed sets to be projective)
> 3. Finite sets are choice (in fact, it suffices for $2$ to be choice)
> (Here, a set $A$ is finite or finitely-indexed (respectively) if, for some natural number *n*, there is a bijection or surjection (respectively) {0,...,*n* $-$ 1} $\rightarrow $$A$. Also, $A$ is projective iff every entire relation from $A$ to $B$ (so that every element of $A$ is related to some element of $B$), for any $B$, contains a function $A$ $\rightarrow$ $B$, while $B$ is choice iff every entire relation from $A$ to $B$, for any $A$, contains a function $A$ $\rightarrow$ $B$.)
>
>
>
suggests that Prof. Burgin's example should lead one to question the principle of excluded middle, much as the following:
>
> The 2$\uparrow$$\uparrow$$\uparrow$6'th digit in the decimal expansion of $\pi$ = 1
>
>
>
might.
Does it (in the sense that Brouwer's argument against excluded middle would cause one to question it)?
|
https://mathoverflow.net/users/20597
|
Can finite sets be non-c.e. depending on how they are presented?
|
Burgin seems to conflate the notion of computability (existence of an algorithm) with a stronger notion such as our knowing an algorithm or existence of a proof that a particular algorithm agrees with the given presentation.
In his example, the finite set $X$ is computable. Indeed, any finite set is computable by a table look-up algorithm. For Burgin's $X$, we don't know the table, so we don't know the algorithm, and we have no effective way to produce the table from the given description of $X$. But that doesn't make $X$ uncomputable; it just says that $X$ doesn't have the stronger properties mentioned above.
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8
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https://mathoverflow.net/users/6794
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407912
| 167,081 |
https://mathoverflow.net/questions/407630
|
1
|
In this question, I'm borrowing the notations from Minguez' paper on unramified representations of unitary groups. Let $F$ be a $p$-adic field and let $G$ be a connected reductive group over $F$. Let $P\_0$ be a minimal $F$-parabolic subgroup and $M\_0$ a Levi factor of $P\_0$ defined over $F$. Let $W := \mathrm{N}\_G(M\_0)/M\_0$ denote the spherical Weyl group of $G$. We denote by $X^{\mathrm{un}}(M\_0)$ the set of unramified characters of $M\_0$, that is those (complex) characters of $M\_0$ which are trivial on the unique maximal compact subgroup of $M\_0$. The Weyl group $W$ acts on $X^{\mathrm{un}}(M\_0)$ through the formula $(w\chi)(m) := \chi(w^{-1}mw)$.
We give ourselves a good special maximal compact subgroup $K$ in $G$. A smooth representation of $G$ is called $K$-spherical if it contains a non-zero $K$-invariant vector. Then the set of isomorphism classes of $K$-spherical representations is in bijection with the quotient set $X^{\mathrm{un}}(M\_0)/W$.
Such a bijection is induced by the following map $X^{\mathrm{un}}(M\_0) \rightarrow \{K\text{-spherical representations}\}$. Let $\chi \in X^{\mathrm{un}}(M\_0)$ and denote by $\iota\_{P\_0}^G\,\chi$ the normalized parabolic induction of $\chi$. The space of $K$-invariant vectors in $\iota\_{P\_0}^G\,\chi$ has dimension $1$ and therefore, there is a unique composition series $\pi\_{K,\chi}$ of $\iota\_{P\_0}^G\,\chi$ which is $K$-spherical. The map is then $\chi \mapsto \pi\_{K,\chi}$.
>
> My question is about what happens if I allow $K$ to vary among all
> good special maximal compact subgroups of $G$. For instance, if $G$ is
> a unitary group then there are two distinct conjugacy classes of such
> subgroups.
>
>
> If $K'$ is another such subgroup, are $K$-spherical and $K'$-spherical
> representations the same thing ? In other words, are the
> representations $\pi\_{K,\chi}$ and $\pi\_{K',\chi}$ the same ? Or is
> there a concrete counterexample ?
>
>
>
Unless I'm mistaken, if $\pi$ is a character of $G$ then it is $K$-spherical if and only if it is $K'$-spherical ; at least it seems to be the case for unitary groups. But what about higher dimensional representations ?
|
https://mathoverflow.net/users/125617
|
Can we compare $K$-spherical representations of $p$-adic groups for varying special maximal subgroups $K$?
|
If I'm not mistaken, $\pi\_{K,\chi}$ and $\pi\_{K',\chi}$ can be different.
Consider a bipartite bi-regular tree, with degrees $q\_1+1<q\_2+1$. There are $p$-adic unitary groups that their building in such a tree with $q\_2=q\_1^3$, see for example <https://arxiv.org/abs/1005.3504>.
Then there are two maximal compact subgroups, corresponding to the stabilizers of the two types of vertices, and I believe that they are both special. Denote them by $K\_1$ and $K\_2$.
Now, if you look at Proposition 3.2 in <https://arxiv.org/abs/1005.3504>, there is a classification of the representations of the corresponding Iwahori-Hecke algebra, which are essentially the same as the possible subquotients of $\iota\_{P\_0}^G\chi$ in the question, and is denoted $X(\nu)$ in that paper.
In this case there are two interesting representations that do not occur in the regular case, which are denoted by $\operatorname{ds}$ and $\operatorname{nt}$ in the said proposition, and are the subquotients of some $\iota\_{P\_0}^G\chi$ for some specific value of $\chi$. It holds that $\operatorname{ds}= \pi\_{K\_1,\chi}$ while $\operatorname{nt}= \pi\_{K\_2,\chi}$.
The reason is that the Iwahori-Hecke algebra is generated by two elements $T\_1,T\_2$, which satisfy the formula $(T\_i+1)(T\_i-q\_i)=0$. The $K\_i$-fixed subspace is the same as the $q\_i$-eigenspace of $T\_i$, and you may check that on $\operatorname{ds}$ the operator $T\_1$ acts by $q\_1$ and $T\_2$ acts by $(-1)$, while on $\operatorname{nt}$ the operator $T\_1$ acts by $(-1)$ and $T\_2$ acts by $q\_2$.
It is also related to the fact that on biregular (and non-regular) trees there is a spherical eigenfunction of the adjacency operator of eigenvalue 0, which is in $L^2$, and hence a discrete series (this is where "ds" comes from). It is supported only on vertices of degree $q\_1+1$. Similarly, there is a "non-tempered" ("nt") eigenfunction of eigenvalue 0, supported on vertices of degree $q\_2+1$.
On the other hand, there is only one $\chi$ where this situation happens, so you can expect it to be quite rare in general. Two special maximal compact subgroups are usually conjugate under an adjoint action (see Tits, "Reductive groups over local fields", Section 2.5), and I don't think that this property can happen in this case.
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3
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https://mathoverflow.net/users/450073
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407919
| 167,082 |
https://mathoverflow.net/questions/407924
|
1
|
Let $B(n, k)$ be the number of $k$-rank subspaces of $\mathbb{Z}\_2^n$. One can establish $B(n, k) = {n \choose k}\_2 = \frac{F(n)}{F(k)F(n - k)}$, where $F(x) = \displaystyle\prod\_{i = 1}^x (2^i - 1)$. This expression clearly implies that $B(n, k)$ is odd for all $k, n \in \mathbb{N}\_0$ such that $k \leq n$.
One can equally easy establish a recurrence $B(n, k) = B(n - 1, k) + 2^{n - k}B(n - 1, k - 1)$, which again provides a straightforward inductive proof for the fact that $B(n, k)$ is odd.
**Question**: is there an elementary, simple, non-enumerative, non-inductive argument for the fact that $B(n, k)$ is odd for all $0 \leq k \leq n$? *E.g. a concise explicit involution on $k$-rank subspaces with a single fixed point would qualify as such.*
|
https://mathoverflow.net/users/106512
|
Number of $k$-rank subspaces of $\mathbb{Z}_2^n$ is odd: easy proof?
|
Consider the involution which complements the ''free'' entries in the reduced row echelon form of the matrix representing a subspace. That is, we leave the pivots as well as things that are necessarily zero unchanged while changing everything else. This involution has a single fixed point, e.g.
$$\begin{bmatrix} 0 & 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 0 & 1\end{bmatrix}$$
for $n = 5$ and $k = 2$.
An example of an obit is
$$\begin{bmatrix} 1 & 0 & 0 & 0 & 1 \\ 0 & 1 & 1 & 1 & 1\end{bmatrix} \leftrightarrow \begin{bmatrix} 1 & 0 & 1 & 1 & 0 \\ 0 & 1 & 0 & 0 & 0\end{bmatrix}$$
again for $n = 5$ and $k = 2$.
|
8
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https://mathoverflow.net/users/51668
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407927
| 167,088 |
https://mathoverflow.net/questions/407826
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7
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Let $J\_C$ be the Jacobian of a smooth projective curve $C$ over $\mathbb{C}$. I would like understand the isomorphism between $H^1(J\_C,\mathbb{C})$ and $H^1(C,\mathbb{C})$. I read in a paper that this isomorphism can be easily achieved by the Hodge-theoretical methods, but they do not give any reference.
Maybe someone can give any reference or explanation about it. Sorry if it is a very basic question, I do not a lot about Hodge theory. So a detailed explanation will be very useful for me.
|
https://mathoverflow.net/users/150116
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Relation between the cohomology group of a curve and the cohomology group of its jacobian
|
$\def\Alb{\text{Alb}}\def\Pic{\text{Pic}}\def\CC{\mathbb{C}}\def\ZZ{\mathbb{Z}}\def\RR{\mathbb{R}}\def\cO{\mathcal{O}}$There are two abelian varieties associated to a smooth projective connected $n$-fold $X$: The Albanese variety $\Alb(X)$ and the Picard variety $\Pic^0(X)$. The Albanese has a natural map $X \to \Alb(X)$ which induces an isomorphism $H^1(\Alb(X)) \to H^1(X)$; the Picard variety parametrizes $(n-1)$-cycles of degree $0$ up to rational equivalence. For a curve, we have $\Alb(X) \cong \Pic^0(X)$, and we call both of these the Jacobian. So we have to understand three things: The Albanese variety, the Picard variety, and the isomorphism.
This causes the ambiguity about whether this question is straightforward or not: It is pretty straightforward that $H^1(X) \cong H^1(\Alb(X))$. But the description in terms of $0$-cycles is a description of $\Pic^0(X)$. What I'm describing in this answer is a modern (or at least, 20-th century version) of the [Abel-Jacobi theorem](https://en.wikipedia.org/wiki/Abel%E2%80%93Jacobi_map#Abel%E2%80%93Jacobi_theorem) which gives a criterion for determining when two $0$-cycles on a curve are rationally equivalent in terms of integrating holomorphic $1$-forms.
**The Albanese variety:** Let's start by just thinking about a smooth connected manifold $M$. We get a map $H\_1(M, \ZZ) \longrightarrow H^1\_{DR}(M, \RR)^{\vee}$ by sending a $1$-cycle $\gamma$ to the linear functional $\omega \mapsto \int\_{\gamma} \omega$. The kernel is the torsion part of $H\_1(M, \ZZ)$, so the image is the torsion free quotient $H\_1(M, \ZZ)\_{tf}$. Define $\Alb(M)$ to be $H^1\_{DR}(M, \RR)^{\vee}/H\_1(M, \ZZ)\_{tf}$. Topologically, this is just a torus whose dimension is the first betti number of $M$. We clearly have a natural isomorphism $H\_1(\Alb(M), \ZZ) \cong H\_1(M, \ZZ)\_{tf}$ and so $H^1(\Alb(M), \ZZ) = \text{Hom}(H\_1(M, \ZZ)\_{tf}, \ZZ) \cong H^1(M, \ZZ)$.
We can get a map $M \to \Alb(M)$ as follows: Choose a vector space $V$ of closed $1$-forms on $M$ which maps isomorphically to $H^1\_{DR}(M)$, and choose a base point $x\_0 \in M$. For any $x \in M$, choose a path $\beta$ from $x\_0$ to $x$. Then $\eta \mapsto \int\_{\beta} \eta$ is a linear functional on $V$. If $\beta'$ is a different path from $x\_0$ to $x$, then $\beta' - \beta = \gamma$ for a $1$-cycle $\gamma$, so $\int\_{\beta'} \eta = \int\_{\beta} \eta + \int\_{\gamma} \eta$. In other words, replacing $\beta$ by $\beta'$ changes the linear functional $\int\_{\beta} (\cdot)$ by an element of $H\_1(X, \ZZ)$. So the class of $\int\_{\beta} (\cdot)$ in $V^{\vee}/H\_1(X, \ZZ)$ depends only on $x$, and we get a map $M \to V^{\vee}/H\_1(X, \ZZ)$ sending $x$ to $\int\_{\beta} (\cdot)$. Since we choose $V$ to be isomrophic to $H^1\_{DR}(M)$, we get a map $M \to \Alb(M)$, and it is easy to check that this map induces the isomorphism $H^1(\Alb(M), \ZZ) \cong H^1(M, \ZZ)$.
Everything becomes more canonical if $M$, which I'll now call $X$, is a connected compact Kahler manifold, for example, a smooth connected complex variety. In that case, Hodge theory tells us that $H^1(X, \CC) = H^{10}(X) \oplus H^{01}(X)$, where $H^{10}(X)$ is the global holomorphic $1$-forms, $H^{10}(X) = H^0(X, \Omega^1)$. Concretely, this isomorphism says that we can take our $V$ to be the real parts and the imaginary parts of holomorphic $1$-forms (or, equivalently, we can take $V$ to be the real harmonic $1$-forms). Thus, $\Alb(X) = H^0(X, \Omega^1)^{\vee}/H\_1(X, \ZZ)\_{tf}$. Now $H^0(X, \Omega^1)^{\vee}$ becomes a complex vector space, so $\Alb(X)$ is not merely a real manifold but a complex manifold.
**The Picard variety** Let $X$ be a smooth complex manifold. Divisors on $X$, modulo rational equivalence, are the same as line bundles on $X$, and are the same as classes in $H^1(X, \cO^{\ast})$. (This should be in most algebraic geometry textbooks.) I am going to work in the analytic world here, so $\cO$ is the sheaf of holomorphic functions; $\cO^{\ast}$ is the sheaf of non-vanishing holomorphic functions and my topology is the analytic topology.
We have the exponential sequence of sheaves $0 \to \underline{\ZZ} \overset{2 \pi i}{\longrightarrow} \cO \overset{\exp}{\longrightarrow} \cO^{\ast} \to 1$, where $\underline{\ZZ}$ is locally constant $\ZZ$-valued functions. So we get a long exact sequence of cohomology which includes
$$H^1(X, \underline{\ZZ}) \longrightarrow H^1(X, \cO) \longrightarrow H^1(X, \cO^{\ast}) \longrightarrow H^2(X, \ZZ).$$
The kernel of the map to $H^2(X, \ZZ)$ are called the cycles of degree $0$ and denoted $\Pic^0(X)$, so we have $\Pic^0(X) \cong H^1(X, \cO) / H^1(X, \underline{\ZZ})$.
Again, things are nicer if $X$ is connected compact Kahler. Then $H^1(X, \cO) = H^{01}(X)$ and the map $H^1(X, \ZZ) \to H^1(X, \cO)$ is the composition of $H^1(X, 2 \pi i \ZZ) \subset H^1(X, \CC) \to H^{01}(X)$ where the second map is the projection onto the second summand of the Hodge decomposition. (I'm going to start getting sloppy about dropping the $2 \pi i$.) In particular, it follows from Hodge theory that the image of $H^1(X, \ZZ)$ is a discrete, cocompact lattice in $H^{1}(X, \cO)$, so the quotient $H^1(X, \cO) / H^1(X, \underline{\ZZ})$ is a compact complex manifold.
**The case of curves** So far, we have two abelian varieties:
$$\Alb(X) = H^0(X, \Omega^1)^{\vee}/H\_1(X, \ZZ)\_{tf} = H^{10}(X)^{\vee}/H\_1(X, \ZZ)\_{tf}$$
and
$$\Pic(X) = H^1(X, \cO)/H^1(X, \ZZ) = H^{01}(X)/H^1(X, \ZZ).$$
But, if $X$ is a curve, then Poincare duality gives an isomorphism $H\_1(X, \ZZ) \cong H^1(X, \ZZ)$ and Serre duality gives an isomorphism $H^{10}(X)^{\vee} \cong H^{01}(X)$. (In fact, Serre duality is just the Poincare duality pairing restricted to the two Hodge summands of $H^1(X, \CC)$.)
After checking enough compatibility of diagrams, this gives an isomorphism $\Alb(X) \cong \Pic^0(X)$, and gives that the map $X \to \Alb(X)$ we defined by integration (using the base point $x\_0$) matches the map $X \to \Pic^0(X)$ sending $x$ to the divisor $[x]-[x\_0]$.
|
14
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https://mathoverflow.net/users/297
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407930
| 167,090 |
https://mathoverflow.net/questions/407915
|
6
|
As a part of the research with which I am involved, I would like to understand how to compute the effect of the Atkin-Lehner operator/Fricke involution $W\_2 = \begin{pmatrix} 0 & 1 \\ -2 & 0 \end{pmatrix}$ on the $q$-expansion modular forms with $\Gamma\_0(2)$ level structure. I know of the Magma function $\operatorname{AtkinLehnerOperator}(f,q)$, but I would like to understand how this works at a theoretical level, both to get at theoretical results and to extend this function to work on modular forms over rings that are not the rationals, (specifically the integers and the 3-adic integers).
If anyone has any sources on or explanations as to how to compute how q-expansions are transformed under the Atkin-Lehner/Fricke involution, it would be greatly appreciated!
|
https://mathoverflow.net/users/122371
|
Explicit computation of the effect of the Atkin-Lehner operator/Fricke involution's effect on $q$-expansion
|
In short, there is no simple formula for the $q$-expansion of the transform of a modular form under an Atkin-Lehner operator $W\_Q$, in terms of the $q$-expansion of the original modular form.
The action of $W\_Q$ on modular symbols is simply $\{\alpha,\beta\} \mapsto \{W\_Q \alpha, W\_Q \beta\}$. This is easy to implement, and gives the matrix of $W\_Q$ in the basis of homology given by (linear combinations of) Manin symbols. If you are interested in the action on a given newform, you just restrict this matrix to the corresponding Hecke module. Otherwise, you compute a linear combination of such transforms.
This works well for $\Gamma\_0(N)$ (trivial character), but not in general, because $W\_Q$ doesn't preserve the character.
There is an implementation in Magma, as you have mentioned, with the restriction to cusp forms with trivial character and integral weight $\geq 2$. In Pari/GP there is a completely general implementation, for modular forms of arbitrary weight (including weight 1 and half-integral weight), level and character. In the difficult cases, this relies on a theorem of Borisov and Gunnells, which asserts that a cusp form of weight $>2$ on $\Gamma\_1(N)$ can be expressed as a linear combination of pairwise products of Eisenstein series, plus an Eisenstein series. The idea being that the action of $W\_Q$ on Eisenstein series is very easy to write down. It should be said, however, that finding such a linear combination of pairwise products, if done exactly, turns out to be very costly. For this reason, the Pari/GP algorithm is numerical; but it works well in practice. Some more details are given in the course [Modular forms in Pari/GP](https://hal.inria.fr/hal-01883565) by Belabas and Cohen.
There are also theoretical results. In your case the character is trivial, so that newforms are eigenvectors for Atkin-Lehner operators. If the newform $f$ satisfies $a\_q(f) \neq 0$ for a prime $q$ dividing the level, then there is a formula for the local eigenvalue at $q$, see for example Atkin, Li, [Twists of newforms and pseudo-eigenvalues of $W$-operators](https://eudml.org/doc/142589), Inv. Math. (1978) 48: 221-244. General theoretical formulas are difficult to get. They seem to be more conveniently stated and proved using the adelic language and the Whittaker model of the newform. The resulting formulas involve $\mathrm{GL}\_2$ Gauss sums. For trivial character, there is a completely explicit formula in Nelson, Pitale, Saha, [Bounds for Rankin-Selberg integrals and quantum unique ergodicity for powerful levels](http://dx.doi.org/10.1090/S0894-0347-2013-00779-1), J. Amer. Math. Soc. (2014), 27: 147-191, see p. 177. I should say here that $\begin{pmatrix} 0 & 1 \\ -2 & 0 \end{pmatrix} = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix} \begin{pmatrix} 2 & 0 \\ 0 & 1 \end{pmatrix}$, so that the problem is reduced to the action of $\sigma = \begin{pmatrix} 0 & 1 \\ -1 & 0 \end{pmatrix}$. I guess that in your case, the formula will be simpler.
|
6
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https://mathoverflow.net/users/6506
|
407937
| 167,091 |
https://mathoverflow.net/questions/407939
|
4
|
Define $A=\sup \limits\_{f} \{m^\*(f[0, 1])\}$, here $f$ are all the functions which satisfy that
$f: [0, 1] \rightarrow [0, 1]$; $\forall x, f(x)-x \in \mathbb{Q}; \forall x, y, x-y \in \mathbb{Q} \rightarrow f(x)=f(y)$.
What is the value of $A$? Obviously, $0<A \leq 1$.
|
https://mathoverflow.net/users/345221
|
What is the value of this supremum?
|
Your conditions may be restated as: $f$ induces a choice function for the partition of $[0,1]$ defined by the equivalence $x\sim y$ iff $x-y\in \mathbb Q$. So the image of $f$ is a Vitali set (a choice of representatives for that partition). Therefore $A=1$ because there are Vitali set of outer measure $1$. Here are
constructions:
<https://math.stackexchange.com/questions/14591/vitali-type-set-with-given-outer-measure>
<https://math.stackexchange.com/questions/157532/vitali-set-of-outer-measure-exactly-1>
|
7
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https://mathoverflow.net/users/6101
|
407942
| 167,093 |
https://mathoverflow.net/questions/407918
|
6
|
I want to simplify the following expression
$$\sum\_{1\leq a,b\leq n}\mathcal{P}(\gcd(a,b,n)),$$ where $\mathcal{P}(x)$ is the number of partitions of $x$.
It turned out this number is the top betti number of the $C\_n\times C\_n$ covering of the Hilbert scheme of $n$ points of the 2-torus. Simplifying this expression may lead to other possible ways of calculating it.
When $n$ is prime, the expression becomes $\mathcal{P}(n)+n^2-1$. The first few terms are 1
5
11
23
31
60
63
109
126
183
176
330
269
420
496
645
585
995
850. I didn't manage to find anything on OEIS.
Many thanks in advance.
|
https://mathoverflow.net/users/111070
|
$\sum_{1\leq a,b\leq n}\mathcal{P}(\gcd(a,b,n))$
|
Introducing $d:=\gcd(a,b,n)$, we get
$$\sum\_{1\leq a,b\leq n} \mathcal{P}(\gcd(a,b,n)) = \sum\_{d\mid n} \mathcal{P}(d) J\_2(\tfrac{n}d),$$
where $J\_2(\cdot)$ is the [Jordan totient function](https://en.wikipedia.org/wiki/Jordan%27s_totient_function) (see also [OEIS A007434](https://oeis.org/A007434)).
In particular, when $n$ is prime, we have the sum of just two terms:
$$\mathcal{P}(1) J\_2(n) + \mathcal{P}(n) J\_2(1) = n^2 - 1 + \mathcal{P}(n)$$
as expected.
|
5
|
https://mathoverflow.net/users/7076
|
407944
| 167,094 |
https://mathoverflow.net/questions/407940
|
12
|
Let $A$ be a subset of $\mathbb{R}^2$ which intersects every straight line in exactly two points.
Is there a such set which is Lebesgue measurable or Borel?
A well-known fact is that there exists such set which is not Lebesgue measurable.
|
https://mathoverflow.net/users/345221
|
Is there a set that intersects every line twice which is Lebesgue measurable or Borel?
|
There is such a set which is Lebesgue measurable, and indeed of Lebesgue measure zero. To see this, start with a subset $S$ of $\mathbb R^2$ such that every line intersects it in continuum many points, for instance $C\times\mathbb R\cup\mathbb R\times C$, where $C$ is the Cantor set. Now repeat your favorite transfinite recursive construction of a set $A$ intersecting each line at exactly two points, modifying it in such a way that we all the points picked belong to $S$. Since $A$ is a subset of a measure zero set $S$, it itself is Lebesgue measurable of measure zero.
It appears that existence of such a set which is Borel is an open problem, see [this MO post](https://mathoverflow.net/a/21864/30186).
|
20
|
https://mathoverflow.net/users/30186
|
407946
| 167,095 |
https://mathoverflow.net/questions/407943
|
8
|
In Voevodsky’s ICM address:
<https://www.uio.no/studier/emner/matnat/math/MAT9580/v18/documents/voevodsky-a1-homotopy-theory-icm-1998.pdf>
In theorem 4.3 it is claimed that given a symmetric monoidal category $(C, \wedge, 1)$ and an object $ X \in C$ in order for $C[X^{-1}]$ be be symmetric monoidal it is enough for $ (123) : X^3 \rightarrow X^3$ to be the identity in $C[X^{-1}]$.
Question: Is there a way to see why this condition on $(123)$ is enough to give us the monoidal structure in an explicit example? Or at least an example of why it is necessary?
Thanks!
|
https://mathoverflow.net/users/374433
|
Inverting objects in a symmetric monoidal category
|
To be clear, this claim refers to a very specific construction of $\mathcal{C}[X^{-1}]$, where you copy the construction of the localization of the ring and defines it as the colimits of:
$$ \mathcal{C} \overset{\\_ \otimes X}{\to} \mathcal{C} \overset{\\_ \otimes X}{\to} \mathcal{C} \overset{\\_ \otimes X}{\to} \mathcal{C} \dots $$
If you are just looking at constructing a $\mathcal{C}[X^{-1}]$ which is universal for adding an inverse to the object $X \in \mathcal{C}$ then this exists without no assumption by the (higher categorical version of the) special adjoint functor theorem.
The condition mentioned there is only important for the localization being computed in the expected way...
For maybe a concrete explanation of why this (123) conditions appears, it is because if $X$ is an invertible object in a symetric monoidal category, then you can show that it satifies the (123) conditions, so when we construct $\mathcal{C}[X^{-1}]$, we are going to have at some point to "kill" the action of (123) on $X^{\otimes 3}$ and the naive iteration presented above doesn't do that at all, so we at least need to assume that this (123) action is already trivial (or more precisely that it is trivial on $X^{\otimes 3} \otimes X^{\otimes n}$ for $n$ large enough).
To give an example let's look at the free symetric monoidal category on on object, i.e. the category $\mathcal{S}$ of finite set and bijection being them with the tensor product being the disjoint union. In this situation Voevodsky's condition isn't satisfied.
If I take the colimit:
$$ \mathcal{S} \overset{ \coprod \{1\}}{\longrightarrow} \mathcal{S} \overset{ \coprod \{1\}}{\longrightarrow} \mathcal{S} \overset{ \coprod \{1\}}{\longrightarrow} \mathcal{S} \overset{ \coprod \{1\}}{\longrightarrow} \mathcal{S} \dots $$
I get a category whose (isomorphisms class of ) objects are indexed by $\mathbb{Z}$, with no non-invertible maps and where each object has as endomorphisms monoids $\Sigma\_\infty = \text{colim } \Sigma\_n$.
This can't be a symetric monoidal category: if it were the unit would have a commutative monoid of endomorphisms and here each object has this non-commutative monoid $\Sigma\_\infty$.
Now $\mathcal{S}[\{1\}^{-1}]$ exists : it is the free symetric monoidal category generated by an invertible object, so it is the free symetric monoidal groupoid on object, so you get the 1-truncation of the sphere spectrum, i.e. a category that has object indexed by $\mathbb{Z}$ (with monoidal structure the addition) and where each object has a $\pi\_{n+1}(S^n) \simeq \mathbb{Z}/2\mathbb{Z}$ of automorphisms, which is involved in the braiding of symetric structure.
Note that, there is a map from the naive iterative construction to the correct localization, that sends each $\Sigma\_\infty$ to $\mathbb{Z}/2\mathbb{Z}$ through the signature of a permutation, and it corresponds to the fact that the localization can be obtained from the naive iteration by impossing a few more relations in the colimit which among other things kills the action of (123) (it is also related to Quillen's $+$ construction).
I have a work in progress with Mathieu Anel where we generalized this condition and which I think make it a bit more clear how it appears and how we can still construct the localization iteratively when it is not satisfied, but it won't be out before a few month.. so I guess I'll come back when it is available.
|
8
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https://mathoverflow.net/users/22131
|
407948
| 167,096 |
https://mathoverflow.net/questions/407872
|
2
|
There are well-described methods of generalizing arbitrary functions to matrices in a natural way.
Basically, if $A=PD\_AP^{-1}$ where $D\_A$ is a diagonal matrix, then $f(A)=Pf(D\_A)P^{-1}$, where the function $f$ is applied to the diagonal matrix element-wise.
This is automatized in some CAS systems, such as Mathematica, so one can apply arbitrary functions to matrices.
For instance, it can be applied to the `Sign` function with `MatrixFunction[Sign, A]`. It should be noted that this function is defined on the complex plane as $z/|z|$ and $\operatorname{sign} 0=0$, which is different from some other definitions ([Higham - What Is the Matrix Sign Function?](https://nhigham.com/2020/12/15/what-is-the-matrix-sign-function/) also talks about generalizing the sign function to matrices, but it seems they use a different definition).
For instance, this method gives
$$\operatorname{sign}\left(
\begin{array}{cc}
1 & -8 \\
1 & 7 \\
\end{array}
\right)=1$$
(e.g., produces an identity matrix) but
$$\operatorname{sign}\left(
\begin{array}{cc}
1 & -8 \\
-1 & 7 \\
\end{array}
\right)=\left(
\begin{array}{cc}
-\frac{3}{\sqrt{17}} & -\frac{8}{\sqrt{17}} \\
-\frac{1}{\sqrt{17}} & \frac{3}{\sqrt{17}} \\
\end{array}
\right).$$
The function can be even applied to some zero divisors.
It also can be applied to hypercomplex numbers represented in matrix form. Now, I noticed that while $\operatorname{sign} z$ can take infinitely-many values on the complex numbers, it can take only 9 values on split-complex numbers: $0$, $1$, $-1$, $j$, $-j$, $1/2+j/2$, $1/2-j/2$, $-1/2+j/2$, $-1/2-j/2$. When applied to dual numbers, it seems to give 5 different values.
The usual rule $\operatorname{sign} (AB)=\operatorname{sign} A\cdot \operatorname{sign} B$ still holds though.
That said, I wonder, whether it has any fundamental importance (telling us about the properties of the ring), whether a ring has finite (as split-complex numbers) or infinite (as complex numbers) set of possible values of the sign function? Can this numerocity be predicted?
What about the $p$-adic rings, can the sign function be generalized there as well?
|
https://mathoverflow.net/users/10059
|
Is the number of values the sign function can take on a ring ("signedness") of any fundamental importance? Can it be predicted?
|
Let $\mathcal{A}$ be a finite-dimensional commutative associative unital $\mathbb{R}$-algebra. (This covers split-complex, hyberbolic and a lot of other number "systems".) We want to extend the function $\operatorname{sign}(x)$ to $\mathcal{A}$ in a meaningful way (to be determined) and understand the number of values in its range.
If $\mathcal{A}$ carries a compatible $\mathbb{C}$-structure, then any extension of $\operatorname{sign}(z) = z/|z|$ takes infinitely many values ($\mathbb{T}$ is in the range) for trivial reasons since $\mathbb{C} \hookrightarrow \mathcal{A}$.
Furthermore, by virtue of finite dimensionality $\mathcal{A}$ is Artinian and hence decomposes into a direct sum of **local** Artinian algebras $\mathcal{A} = \bigoplus\_{i=1}^n (\mathcal{A}\_i, \mathfrak{m\_i})$. Thus it suffices to understand the extension of $\operatorname{sign}(x)$ to such a local Artinian algebra $(\mathcal{A},\mathfrak{m})$. Moreover, by the previous considerations we can assume that $(\mathcal{A},\mathfrak{m})$ does not carry a compatible complex structure, hence $\mathcal{A} / \mathfrak{m} \cong \mathbb{R}$ as algebras and $\mathcal{A} = \mathbb{R} \oplus \mathfrak{m}$ as $\mathbb{R}$-vector spaces.
Write $Z = \lambda \oplus X \in \mathcal{A}$ with $\lambda \in \mathbb{R}$ and $X \in \mathfrak{m}$. Thus, in particular, $r$ is the eigenvalue (multiple) of $Z$ and $X$ is nilpotent. A possible extension of $\operatorname{sign}(x)$ is an element of the $\mathcal{A}$-module $\mathscr{D}'(\mathbb{R})[X]$ via the "Taylor series":
$$
\operatorname{sign}(Z) = \operatorname{sign}(\lambda \oplus X) := \operatorname{sign}(\lambda) + 2 \sum\_{k=1}^{N-1} \frac{\delta^{(k-1)}(\lambda)}{k!} X^k
$$
where $N \in \mathbb{N}$ is the smallest integer such that $\forall X \in \mathfrak{m}: X^N = 0$. Note that $\operatorname{sign}(x) \in L^1\_{\mathrm{loc}}$ and $0\cdot\infty = 0$. Alternatively, one can look at it as a mapping
$$
\mathfrak{m} \to \mathscr{D}'(\mathbb{R}) \otimes\_\mathbb{R} \mathcal{A}
$$
One can have an honest function when $\lambda \neq 0$, namely $\operatorname{sign}(Z) := \operatorname{sign}(\lambda)$, and so $\operatorname{sign}(Z)$ takes exactly two values on $\mathcal{A}^\times$. If $\lambda = 0$, I am currently unsure what would mean to "count the values of its range". In any case, there are infinitely many nilpotents, unless $\mathfrak{m} = 0$. Lastly, if $\mathcal{A} = \bigoplus\_{k=1}^n \mathcal{A}\_k$, we have $\mathcal{A}^\times = \prod\_{k=1}^n \mathcal{A}\_k^\times$ and so the possible-values count on $\mathcal{A}^\times$ is $2^n$.
**Summary:**
1. If $\mathcal{A}$ has a complex structure, then $\operatorname{sign}(z)$ would have infinitely many distinct values in its range. In particular, it contains the circle $\mathbb{T}$.
2. If $\mathcal{A}$ has no complex structure and is not a product of other algebras, then $\operatorname{sign}(x)$ has a boring extension **as a function** to $\operatorname{sign}(\lambda + X) := \operatorname{sign}(\lambda)$, where $\lambda$ is the eigenvalue part and $X$ is the nilpotent part. In this case, its range contains 3 distinct values inherited from $\mathbb{R}$. And if $\mathcal{A}$ is a product of $n$ such algebras without complex structure, then $\operatorname{sign}$ has $3^n$ distinct values in its range.
3. If $\mathcal{A}$ has no complex structure and is not a product of other algebras, then **the distribution** $\operatorname{sign}(x)$ admits an extension as a distribution depending on the nilpotent parameter $X$ and whose integrals are $\mathcal{A}$-valued. In this case, we get an uncountable family of different distributions, parametrized by $X$.
**Remaining Parts:**
4. Find an extension to general zero divisors and general idempotents.
5. Make sense of "continuity" of the distributional extension $\operatorname{sign}(Z)$ over $\mathcal{A}$ (see discussion in the comments below the original question).
|
2
|
https://mathoverflow.net/users/1849
|
407960
| 167,101 |
https://mathoverflow.net/questions/407976
|
1
|
Let $\mathcal M$ be the collection of martingle diffusions starting at zero and ending in $\{-1,1\}$. Equivalently, $X\in \mathcal M$ iff there exists a measurable function $a$ s.t. it holds almost surely
$$X\_t=\int\_0^t a(s,X\_s)dW\_s ~\in~ [-1,1],~ \forall t\ge 0 \quad\mbox{and} \quad X\_{\infty}:=\lim\_{t\to\infty}X\_t\in \{-1,1\}.$$
What is the condition on $a$ so that the above properties of $X$ can be satisfied? My guess is
$$a(t,x)=(1-|x|)^{p}k(t,x) \quad \mbox{or}\quad a(t,x)={\bf 1}\_{\{|x|<1\}}k(t,x)$$
for some $p>0$ and suitable function $k$. The simplest example is given as $a(t,x)={\bf 1}\_{\{|x|<1\}}$ by Mateusz Kwaśnicki ([Martingale representation of a stopped Brownian motion](https://mathoverflow.net/questions/400736/martingale-representation-of-a-stopped-brownian-motion))
Any answer, comments and references are highly appreciated.
PS : For the time-homogeneous case, i.e. $a(t,x)\equiv a(x)$, Mateusz provides a sufficient and necessary condition on $a$ to ensure the properties of $X$. I still look for the condition for general $a$.
|
https://mathoverflow.net/users/261243
|
Characterization of martingale diffusions ending in $\{-1,1\}$
|
This is an extended comment on the time-homogeneous case, when the coefficient $a(t, x)$ does not depend on $t$, so that it can be written as $a(x)$.
Suppose that $a(\pm 1) = 0$ and that $1/(a(x))^2$ is locally integrable over $(-1, 1)$. We claim that in this case the desired process $X\_t$ exists.
---
Define the *speed measure* $m$ by
$$ m(x) = \int \frac{1}{(a(x))^2} \, dx . $$
Then there is a (unique) *Feller diffusion* $X\_t$ on $(-1, 1)$ with speed measure $m$ and scale function $s(x) = x$, and this diffusion is a weak solution of the SDE
$$ dX\_t = a(X\_t) dB\_t \qquad \text{for $t$ less than the hitting time of } \{-1, 1\}; $$
for further information and references, see, for example, [DOI:10.1007/978-94-009-3859-5\_35](https://doi.org/10.1007/978-94-009-3859-5_35). (In this book chapter the author seems to assume that $a$ is locally bounded and bounded away from zero, but I believe the result is quite general. Time permitting, I can look up a better reference.)
This diffusion either hits $\{-1, 1\}$ in finite time $\tau$, in which case we extend it past $\tau$ so that $X\_t = X\_{\tau-}$ for $t \geqslant \tau$, or it stays in $(-1, 1)$ forever, approaching either $1$ or $-1$ as $t \to \infty$. Due to our assumption $a(\pm 1) = 0$, the above extension satisfies the SDE
$$ dX\_t = a(X\_t) dB\_t \qquad \text{for all } t > 0 . $$
Finally, we have already seen that $X\_\infty \in \{-1, 1\}$, as desired.
---
*Edit:* I believe that the above reference does not really cover the case of general $a(x)$ and processes in an interval. It was meant to give an entry point to the abundant literature on the subject.
However, the construction of $X\_t$ is in fact fairly simple. We start with a standard Brownian motion $W\_s$ and its local time $L\_s^x$ (not really needed here), we denote by $\sigma$ the hitting time of $\{-1, 1\}$ for $W\_s$, and we let
$$ T\_s = \int\_0^{s \wedge \sigma} \frac{1}{(a(W\_r))^2} dr = \int\_{-1}^1 \frac{L\_{s \wedge \sigma}^x}{(a(x))^2} dx . $$
Then $T\_s$ is strictly increasing on $[0, \sigma)$ and so it has a continuous inverse function $T\_t^{-1}$, defined on $[0, T\_\sigma)$ (with $T\_\sigma$ possibly infinite). Then we define
$$ X\_t = W\_{T\_t^{-1}} \qquad \text{for } t \in [0, T\_\sigma) $$
and $X\_t = W\_\sigma$ for $t \geqslant T\_\sigma$.
Then clearly $X\_t$ is a continuous diffusion in $(-1, 1)$ — a time-changed Brownian motion — which either reaches $W\_\sigma \in \{-1, 1\}$ in finite time $T\_\sigma$ and stays there forever, or it converges to $W\_\sigma \in \{-1, 1\}$ as $t \to T\_\sigma = \infty$, depending on whether $T\_\sigma$ is finite or not.
It remains to see that $X\_t$ is a weak solution of $dX\_t = a(X\_t) dB\_t$. To this end, oberve that for $t < T\_\sigma$, the quadratic variation of $X\_t$ satisfies
$$ d\langle X \rangle\_t = d\langle W \rangle\_{T\_t^{-1}} = dT\_t^{-1} = (a(W\_{T\_t^{-1}}))^2 dt = (a(X\_t))^2 dt , $$
and so
$$ dB\_t = (a(X\_t))^{-1} dX\_t $$
defines a continuous martingale $B\_t$ with quadratic variation $d\langle B\rangle\_t = dt$, hence a Brownian motion, for $t \in [0, T\_\sigma)$. Thus $X\_t$ indeed solves
$$ dX\_t = a(X\_t) dB\_t $$
for $t \in [0, T\_\sigma)$, and extension to $t \geqslant T\_\sigma$ is now immediate (expand the probability space to accommodate another Brownian motion $\tilde B\_t$, and let $dB\_t = d\tilde B\_t$ for $t \geqslant T\_\sigma$).
---
In fact, this should be essentially an "if and only if" condition, but due to time limitations I did not try to check this.
|
2
|
https://mathoverflow.net/users/108637
|
407980
| 167,106 |
https://mathoverflow.net/questions/407558
|
2
|
Is there a *graph manifold* (<https://en.wikipedia.org/wiki/Graph_manifold>) that doesn't admit an orientation reversing involution? If so, what would be a simple example?
|
https://mathoverflow.net/users/13441
|
A graph manifold without an orientation reversing involution?
|
$\mathbb{N}il^3$-manifolds and $\widetilde{SL}$-manifolds are orientable
and do not have orientation-reversing self homotopy equivalences.
This is most easily seen algebraically.
The fundamental group $\Gamma$ of the $S^1$-bundle over the torus $T$ with Euler class a generator of $H^2(T;\mathbb{Z})$ is a central extension of $\mathbb{Z}^2$ by $\mathbb{Z}$.
(This is the Heisenberg group of upper triangular matrices in $SL(3,\mathbb{Z})$.) Direct calculation of the automorphisms of $\Gamma$ shows that an automorphism which induces $A\in{GL(2,\mathbb{Z})}$ on the central quotient $\mathbb{Z}^2$ acts by $\det{A}$ on the centre, and hence is orientation-preserving on the extension.
In general, the fundamental groups of such 3-manifolds are virtually
central extensions by $\mathbb{Z}$ of orientable surface groups $S$,
with non-zero extension class in $H^2(S;\mathbb{Z})$.
Moreover, the centre of the group is $\mathbb{Z}$,
and so the extension is preserved under any automorphism of the group.
Considering the effect of an automorphism on the centre and the quotient
shows that the extension class is preserved if and only if the induced
automorphisms of the centre and the central quotient $S$ are both orientation-preserving or both orientation reversing.
An analogous situation holds for $\mathbb{S}^3$-manifolds.
However the argument breaks down for some lens spaces (as observed by Ryan), as these admit self-maps which do not preserve the Seifert fibration.
|
6
|
https://mathoverflow.net/users/58488
|
407985
| 167,108 |
https://mathoverflow.net/questions/407933
|
0
|
Consider the following Sturm-Liouville problem,
$$(\sqrt{\sin \theta} Y')' + \lambda \sqrt{\sin \theta} Y =0$$
where $Y(\theta):[0,\pi] \to \mathbb{R}$ with boundary conditions $Y'(0)=Y'(\pi)=0.$
I used maple and got the following explicit solution,
$$Y(\theta) = \sin(\theta)^{1/4} \left(c\_1P^\mu\_{\nu}(\cos \theta) + c\_2Q^{\mu}\_{\nu}(\cos \theta)\right)$$
where $\mu=1/4$ and $\nu = \frac{\sqrt{16\lambda+1}}{4}-\frac{1}{2}.$ When I try to differentiate this expression and plug in the boundary conditions, I get division by zero error. What other ways can I use to compute the eigenvalue of this expression?
**Edit:**
I tried to convert the above ode into the following form $u''+Vu=0$
where
$$u(\theta) = \sin^{1/4}(\theta) Y(\theta)$$
and
$$V(\theta) = \lambda + \frac{1}{4}\csc^2(\theta)-\frac{1}{16}\cot^2(\theta).$$
I am wondering if some property of the potential $V$ can be exploited to find perhaps upper or lower bounds for the eigenvalues.
|
https://mathoverflow.net/users/68232
|
Asymptotic for eigenvalues for the following ode?
|
I asked Mathematica about the boundary behavior. First, the $P^{1/4}\_{\nu } $ solution: We have, for $\epsilon \searrow 0$,
$$
(\sin \epsilon )^{1/4} P^{1/4}\_{\nu } (\cos \epsilon ) = \frac{2^{1/4} }{\Gamma(3/4)} + O(\epsilon^{2} )
$$
and
\begin{eqnarray\*}
(\sin (\pi -\epsilon ))^{1/4} P^{1/4}\_{\nu } (\cos (\pi -\epsilon) ) &=& \frac{2^{3/4} \pi }{\Gamma(3/4)\Gamma(-\nu )\Gamma(1+\nu )} \\
& & -\frac{2^{1/4} \pi }{\Gamma(5/4)\Gamma(-1/4-\nu )\Gamma(3/4+\nu )} \sqrt{\epsilon } \\
& & + O(\epsilon^{2} )
\end{eqnarray\*}
So, the $P^{1/4}\_{\nu } $ solution automatically satisfies the boundary condition at $\theta =0$, whereas at $\theta =\pi $, we have to eliminate the term proportional to $\sqrt{\epsilon } $. That determines the eigenvalues: We need either $-1/4-\nu $ to be a negative integer or 0, or $3/4+\nu $ to be a negative integer or 0.
The specification of $\nu $ in the OP suggests the constraint $\nu \geq -1/2$; this excludes the second alternative, and therefore we obtain the spectrum $\nu = n-1/4$, $n=0,1,2,3,\ldots $.
The $Q^{1/4}\_{\nu } $ solution, on the other hand, exhibits behavior proportional to $\sqrt{\epsilon } $ at $\theta=0$, with coefficient
$$
\frac{\pi^{2} }{2^{1/4} } \frac{\cos ((4\nu +1)\pi /8) \Gamma (-\nu /2 -1/8) \Gamma (\nu /2 +9/8) - \sin ((4\nu +1)\pi /8)\Gamma (-\nu /2 +3/8) \Gamma (\nu /2 +5/8)}{\Gamma (5/4) \Gamma (-\nu /2 -1/8) \Gamma (-\nu /2 +3/8) \Gamma (\nu /2 +3/8)\Gamma (\nu /2 +7/8)}
$$
A plot as a function of $\nu $ suggests that, for $\nu \geq -1/2$, this is positive and monotonically rising (I have not attempted to verify this analytically); the behavior proportional to $\sqrt{\epsilon } $ at $\theta=0$ can therefore not be eliminated, nor can it be compensated by admixture of the $P^{1/4}\_{\nu } $ solution. Thus, there are no further solutions involving $Q^{1/4}\_{\nu } $.
In summary, the complete spectrum is given by $\nu = n-1/4$, $n=0,1,2,3,\ldots $, or, in terms of $\lambda $, $\lambda = n(n+1/2)$, $n=0,1,2,3,\ldots $.
|
1
|
https://mathoverflow.net/users/134299
|
407986
| 167,109 |
https://mathoverflow.net/questions/407977
|
1
|
Is the Implicit Function Theorem in the following form correct:
Let $V\_1,V\_2,W$ be Banach spaces, and $Ω⊂V\_1×V\_2$ an open subset containing $(x\_0,y\_0)$. Let consider a continuously differentiable map $f:Ω→W$ with $f(x\_0,y\_0)=0$ and s.t. the derivative on the second component
$D\_2f(x\_0,y\_0):V\_2\ni y↦Df(x\_0,y\_0)(0,y)∈W$
is one-to-one (**but not necessarly onto**).
Then there exists an open set $Ω\_1×Ω\_2⊂Ω$ around $(x\_0,y\_0)$ and a unique map
$g:Ω\_1→Ω\_2$ s.t $f(x,g(x))=0$ for all $x∈Ω\_1$.
Indeed, the standard strategy would be to consider
$L \equiv D\_2f\_{(x\_0,y\_0)}$ and $
\Phi(x,y)\equiv y - (L^{-1}\circ f)(x,y).
$
However, the map $\Phi$ is not defined everywhere. Indeed, it is defined only on
$$f^{-1}\bigg(L\big(V\_2\big)\cap f(\Omega)\bigg).$$
I look forward to reading your comment, possibly references where I can find a proof if that statement is correct. Thanks in advance
|
https://mathoverflow.net/users/451228
|
Does the Implicit Function Theorem in Banach spaces holds if the differential is only one-to-one (not onto!)?
|
**No, it is not correct.** Simple counterexamples can be obtained as follows. Let $V\_1=W$ and $V\_2$ be any Banach spaces such that the identity is continuous but not surjective $V\_2\to W$. Then $f$ defined by $(x,y)\mapsto x+y$ is even smooth with $\partial\_2 f(x\_0,y\_0)$ injective $V\_2\to W$ and there cannot be any $g,\Omega\_1,\Omega\_2$ as required since we would have $g(x)=-x$ for all $x\in\Omega\_1$ and further $W\subseteq V\_2$.
|
1
|
https://mathoverflow.net/users/12643
|
407992
| 167,111 |
https://mathoverflow.net/questions/407739
|
3
|
Let $(\mathcal{X},d)$ be a Polish (metric) space and let $\{X\_n\}\_{n=1}^{\infty}$ be a sequence of i.i.d. $\mathcal{X}$-valued random elements defined on a common complete (standard) probability space $(\Omega,\mathcal{F},\mathbb{P})$ with $\mathbb{E}[d(X\_1,x)^p]<\infty$ for some $p>1$ and some $x\in \mathcal{X}$.
Are the known concentration inequalities of the form:
$$
\mathbb{P}\left(
\mathcal{W}\_1\left(\frac1{n}\sum\_{k=1}^n \delta\_{X\_k},Law(X\_1)\right)>\delta
\right) \leq I(n,\delta),
$$
for some ``well-behaved'' function $I:\mathbb{N}\times (0,\infty)\rightarrow \infty)$ which is:
* Monotone decreasing in its both its arguments and converges to $0$,
* Upper semi-continuous in its second argument.
**Additional Piece of Information:** I'm not interested in *best/sharp* rates, I only really case about some quantitative rates of the above form.
---
**What I've seen so far:**
I only know of this [2015 PTRF article](https://link.springer.com/article/10.1007/s00440-014-0583-7) which describes sharp rates in the case where $(\mathcal{X},d)$ is a finite-dimensional Euclidean space.
|
https://mathoverflow.net/users/36886
|
Wasserstein-type concentration inequalities for empirical measures on polish spaces
|
Yes, there are various results available in more general settings. The typical route would be to combine an upper bound on the expected distance between the law and the empirical measure (like Theorem 1.1 in [Boissard and Le Gouic](http://www.numdam.org/item/10.1214/12-AIHP517.pdf)) with a concentration estimate around this expectation (like Proposition 20 in [Weed and Bach](https://arxiv.org/abs/1707.00087)). In particular, both of these results work on a general metric space which is totally bounded.
I do not know of a result like the one you ask for that works in an arbitrary Polish space; I expect one would need to impose further regularity assumptions on the law of $X\_1$.
|
3
|
https://mathoverflow.net/users/100163
|
407994
| 167,113 |
https://mathoverflow.net/questions/407987
|
3
|
Let $A$ be a $n\times m$ random matrix, whose elements $a\_{ij}$ are independent standard Gaussian random variables.
I am interested in the case $n=\alpha N\,$, $\,m=(1-\alpha)N$ for $\alpha\in(0,1)$ fixed and $N\to\infty$.
Denote by $\sigma\_\max(N)$ the largest singular value of $\frac{1}{\sqrt{N}}A$, that is the square root of the largest eigenvalue of the symmetric square matrix $\frac{1}{N}AA^T$.
I am looking for a concentration result of type:
$$ \mathbb P(\sigma\_\max(N)\geq K) \,\leq\, e^{-N F(K)}$$
for constants $K$ in a suitable range, a suitable function $F(K)$, and $N$ sufficiently large. I do not need $F$ to be optimal.
In the square matrix case ($\alpha=\frac{1}{2}$), an analogous result holds true for the top eigenvalue $\lambda\_\max(N)$ of the symmetric matrix $\frac{A+A^T}{\sqrt{2N}}$. But I could not find anything about large deviations of the top singular value in the rectangular case. Any reference or suggestion is very welcome.
|
https://mathoverflow.net/users/58793
|
Top singular value of large random matrices: concentration results
|
This largest singular value is the norm of the matrix. You can use a net argument to show that there is a $C$ so that $$\mathbb{P}(\| A \|\_{op} \geq C\sqrt{N}(\sqrt{\alpha} + \sqrt{1 - \alpha} + t)) \leq 2 e^{-Nt^2}$$
for all $\alpha, t$. For a reference, this appears as Theorem 4.4.5 in Vershynin's [High Dimensional Probability book](https://www.math.uci.edu/%7Ervershyn/papers/HDP-book/HDP-book.html).
|
1
|
https://mathoverflow.net/users/69870
|
408014
| 167,116 |
https://mathoverflow.net/questions/408012
|
5
|
Let $\boldsymbol{V}\_{1},\dots,\boldsymbol{V}\_{n}\in\mathbb{R}^{d\times m}$ be $n$ “tall” matrices (where $d\ge m$) with orthonormal columns.
And let $\boldsymbol{P}\_{1},\dots,\boldsymbol{P}\_{n}\in\mathbb{R}^{d\times d}$ be the orthogonal projection matrices defined as $\boldsymbol{P}\_{i}=\boldsymbol{V}\_{i}\boldsymbol{V}\_{i}^{\top}$.
Finally, let $\boldsymbol{T}$ be an operator defined the concatenation of the projection matrices $\boldsymbol{T}=\boldsymbol{P}\_{n}\cdots\boldsymbol{P}\_{1}$.
**Question:** does there exists a sequence $\boldsymbol{V}\_{1},\dots,\boldsymbol{V}\_{n}$ such that the concatenation operator $\boldsymbol{T}$ is nilpotent with index $k\ge3$? That is,
$$\boldsymbol{T}^{k-1}=\left(\boldsymbol{P}\_{n}\cdots\boldsymbol{P}\_{1}\right)^{k-1}\neq\boldsymbol{0}\_{d\times d},\,\,\,\,\,\,\,\boldsymbol{T}^{k}=\left(\boldsymbol{P}\_{n}\cdots\boldsymbol{P}\_{1}\right)^{k}=\boldsymbol{0}\_{d\times d}$$
---
**Special case (example):** when $n=3$ and $d=2$, we can choose projections such that $\boldsymbol{T}$ is a nilpotent operator with an index of $k=2$.
Choose $\boldsymbol{v}\_{1}=\left[0,1\right]^{\top},
\boldsymbol{v}\_{2}=\frac{1}{\sqrt{2}}\left[1,1\right]^{\top},\boldsymbol{v}\_{3}=\left[1,0\right]^{\top}$.
Then, the projection matrices are
$$\boldsymbol{P}\_{1}=\boldsymbol{v}\_{1}\boldsymbol{v}\_{1}^{\top}=\left[\begin{array}{cc}
0 & 0\\
0 & 1
\end{array}\right],\,\,\,\boldsymbol{P}\_{2}=\frac{1}{2}\left[\begin{array}{cc}
1 & 1\\
1 & 1
\end{array}\right],\,\,\,\boldsymbol{P}\_{3}=\left[\begin{array}{cc}
1 & 0\\
0 & 0
\end{array}\right]$$
And the concatenation of these matrices is
$$\boldsymbol{T}=\boldsymbol{P}\_{3}\boldsymbol{P}\_{2}\boldsymbol{P}\_{1}=\frac{1}{2}\left[\begin{array}{cc}
0 & 1\\
0 & 0
\end{array}\right]$$
which is a nilpotent matrix with an index of $k=2$ (i.e., $\boldsymbol{T}^{2}=\boldsymbol{0}\_{d\times d}$).
|
https://mathoverflow.net/users/100796
|
Can the concatenation of projection operators be nilpotent with an index k>=3?
|
I think your example is easily generalisable for any index. For example, let
$$
Q\_1=P\_1\oplus(1),
Q\_2=P\_2\oplus(1),
Q\_3=P\_3\oplus(1)=(1)\oplus P\_1,
Q\_4=(1)\oplus P\_2,
Q\_5=(1)\oplus P\_3.
$$
Then $Q\_5Q\_4Q\_3Q\_2Q\_1$ is nilpotent of index 3, and I think that a similar construction with $2n-1$ matrices of size $n$ gives nilpotence of index $n-1$.
|
5
|
https://mathoverflow.net/users/1306
|
408016
| 167,117 |
https://mathoverflow.net/questions/407594
|
12
|
Let $A, B, C \in \mathbb{R}^{n\times n}$ such that $N = \begin{bmatrix} A & B\\ B^{\top} & C\end{bmatrix}$ is a symmetric positive definite matrix. I'm trying to show that the following matrix
$$
M = \begin{bmatrix} A & B & 0\\ B^{\top} & C & -B^{\top} \\ - A & -B & A\end{bmatrix}
$$
has eigenvalues with positive real part. Numerical tests suggest this is true, but I cannot prove it.
**Edit** It is not true that $M + M^\top$ has positive eigenvalues, i.e. that $\langle x, Mx\rangle \geq 0$ for all $x$, which is (?) a sufficient condition for $M$ to have eigenvalues with positive real parts.
**Edit 2** After further numerical tests, letting $\lambda\_{\min}$ the smallest eigenvalue of $N$, and $v\_{\min}$ the smallest real part of the eigenvalues of $M$, it seems like we should have a bound like
$$
v\_{\min} \geq \rho \cdot \lambda\_{\min}
$$
where the constant $\rho$ is $\simeq 0.963$.
**Edit 3** I expect that we can answer the question by finding the right factorization for $M$. For instance it is easy to show that
$$
\begin{bmatrix} A & B & -A\\ B^{\top} & C & -B^{\top} \\ - A & -B & A\end{bmatrix}
$$
has positive eigenvalues since it equals $ \begin{bmatrix} I & 0 & -I \\ 0& I & 0\end{bmatrix}^{\top}N \begin{bmatrix} I & 0 & -I \\ 0& I & 0\end{bmatrix}$
|
https://mathoverflow.net/users/173967
|
Show that the eigenvalues of a non-symmetric matrix built from positive matrices have positive real parts
|
A simple brute force method worked (even though I'm not happy with this).
Let $\zeta=\xi+i\eta$ be a non-positive eigenvalue of $M$ and
$\left[\begin{matrix} x & y & z \end{matrix}\right]^T$ be
a corresponding eigenvector.
This gives equations
\begin{align}
Ax + By \qquad &= \zeta x \\
B^Tx + Cy - B^Tz &= \zeta y\\
-Ax -By + Az &= \zeta z
\end{align}
From the first and the third, one obtains
$$z = -\zeta(\zeta-A)^{-1}x
\mbox{ and } x-z = (1+\zeta(\zeta-A)^{-1})x.$$
To ease the notation, put $A(\zeta):=1+\zeta(\zeta-A)^{-1}=(2\zeta-A)(\zeta-A)^{-1}$.
From the second,
$$y = (\zeta-C)^{-1}B^T(x-z) = (\zeta-C)^{-1}B^T A(\zeta) x.$$
By combining with the first, one obtains
$$(\zeta-A)x = By = B(\zeta-C)^{-1}B^T A(\zeta)x.$$
Note that the imaginary part of $(\zeta- C)^{-1}$ is
$$\Im\frac{1}{\zeta- C}=\Im\frac{\bar{\zeta}-C}{|\zeta-C|^2}
=-\eta\frac{1}{|\zeta-C|^2}.$$
Take the inner product with $A(\zeta)x$ and look at the imaginary part:
$$\Im \langle (\zeta-A)x,A(\zeta)x\rangle
= -\eta \langle B|\zeta-C|^{-2}B^TA(\zeta)x,A(\zeta)x\rangle.$$
Now, since
\begin{align\*}
A(\bar{\zeta})(\zeta-A)
&= \frac{(2\bar{\zeta}-A)(\zeta-A)^2}{|\zeta-A|^2} \\
&=\frac{2|\zeta|^2\zeta-4|\zeta|^2A+2\bar{\zeta}A^2-\zeta^2A+2\zeta A^2-A^3}{|\zeta-A|^2},
\end{align\*}
one obtains
$$2\eta\langle \frac{|\zeta|^2-\xi A }{|\zeta-A|^2}x,x\rangle
=-\eta \langle B|\zeta-C|^{-2}B^TA(\zeta)x,A(\zeta)x\rangle.$$
Hence, unless $\eta=0$,
$$\xi\geq\frac{|\zeta|^2}{\|A\|}>0.$$
Let's deal with the case $\eta=0$.
Suppose for a contradiction that $\xi\le0$. Then
\begin{align\*}
\langle A(\xi)(-\xi+A)x,x\rangle
&= \langle B(-\xi+C)^{-1}B^T A(\xi)x,A(\xi)x\rangle\\
&\le \langle BC^{-1}B^T A(\xi)x,A(\xi)x\rangle\\
&< \langle A A(\xi)x,A(\xi)x\rangle,
\end{align\*}
but this is in contradiction with the fact that
$A(\xi)\succ0$ and $-\xi + A\succeq A(\xi)A$.
ADDED:
We assume $\left[\begin{smallmatrix}
A\_0 & B\_0 & \\
B\_0^T & C\_0\end{smallmatrix}\right]\succ 2\epsilon$ and will show
that any eigenvalue of
$\left[\begin{smallmatrix}
A\_0 & B\_0 & \\
B\_0^T & C\_0 & -B\_0^T \\
-A\_0 & -B\_0 & A\_0
\end{smallmatrix}\right]$ has its real part at least $\epsilon$.
To ease notation, we consider $A=A\_0-2\epsilon$ and $C=C\_0-2\epsilon$
instead of $A\_0$ and $C\_0$.
By the IVT trick, it suffices to show
there is no solution for
$$\left[\begin{matrix}
A+2\epsilon & B & \\
B^T & C+2\epsilon & -B^T \\
-(A+2\epsilon) & -B & A+2\epsilon
\end{matrix}\right]
\left[\begin{matrix}
x \\ y \\ z
\end{matrix}\right]
= (\epsilon+i\eta) \left[\begin{matrix}
x \\ y \\ z
\end{matrix}\right],$$
with $\left[\begin{smallmatrix}
A & B & \\ B^T & C\end{smallmatrix}\right]\succ 0$,
$\epsilon>0$, $\eta\in\mathbb{R}$, and
$\left[\begin{smallmatrix} x & y & z \end{smallmatrix}\right]\neq 0$.
The first row + the third row: $-\epsilon x + (A+\epsilon) z = i\eta (x+z).$
Hence
$$x-z=(1-\frac{\epsilon+i\eta}{A+\epsilon-i\eta})x
=\frac{A-2i\eta}{A+\epsilon-i\eta}x=:A(\eta)x.$$
Together with the second row:
$$y=-(C+\epsilon-i\eta)^{-1}B^T(x-z)=-(C+\epsilon-i\eta)^{-1}B^TA(\eta)x.$$
It follows that $x\neq0$.
Together with the first row:
\begin{align\*}
(A+\epsilon -i\eta) x
= -By
= B(C+\epsilon-i\eta)^{-1}B^TA(\eta)x
\end{align\*}
and so
\begin{equation}\tag{$\ast$} \langle (A+\epsilon -i\eta) x,A(\eta)x\rangle
= \langle B(C+\epsilon-i\eta)^{-1}B^TA(\eta)x, A(\eta)x\rangle.\end{equation}
Do some calculations:
$$A(\eta)^\*(A+\epsilon -i\eta)
= |A+\epsilon-i\eta|^{-2}(A+\epsilon-i\eta)^2(A+2i\eta),$$
$$(A+\epsilon-i\eta)^2(A+2i\eta)
= A^3 + 2 \epsilon A^2 + \epsilon^2 A + 3 \eta^2 A
+ 4\epsilon \eta^2 + 2i\eta (\epsilon A + \epsilon^2 - \eta^2),$$
$$A(\eta)^\*B(C+\epsilon-i\eta)^{-1}B^TA(\eta)
=A(\eta)^\*B\frac{C+\epsilon +i\eta}{|C+\epsilon-i\eta|^2}B^TA(\eta),$$
$$A(\eta)^\*B\frac{C+\epsilon}{|C+\epsilon-i\eta|^2}B^TA(\eta)
\preceq A(\eta)^\*B C^{-1} B^T A(\eta) \prec A|A(\eta)|^2,$$
$$A|A(\eta)|^2 = |A+\epsilon-i\eta|^{-2}(A^3+4\eta^2A).$$
Look at the real part of $(\ast)$: for $w=|A+\epsilon-i\eta|^{-1}x$,
$$\langle(2\epsilon A^2+\epsilon^2A+4\epsilon\eta^2)w,w\rangle
< \langle \eta^2Aw,w\rangle.$$
Look at the imaginary part of $(\ast)$:
as $\eta\neq0$ from the previous inequality,
$$\langle(\epsilon A + \epsilon^2 - \eta^2)w,w\rangle
=\langle A(\eta)^\*B\frac{1}{|C+\epsilon-i\eta|^2}B^TA(\eta)x,x\rangle/2
\geq0.$$
Combine the last two:
\begin{align\*}
2\epsilon \|Aw\|^2+\epsilon^2\langle Aw,w\rangle +4\epsilon\eta^2\|w\|^2
&< \eta^2\langle Aw,w\rangle\\
&\le (\epsilon \frac{\langle Aw,w\rangle}{\|w\|^2} +\epsilon^2)\langle Aw,w\rangle\\
&\le\epsilon\|Aw\|^2 +\epsilon^2\langle Aw,w\rangle.
\end{align\*}
We arrive at a contradiction.
|
8
|
https://mathoverflow.net/users/7591
|
408017
| 167,118 |
https://mathoverflow.net/questions/408010
|
0
|
My problem is to prove
$$
\left(\cos\frac{m}{2n}\pi\right)^{4n}\ge \left(\cos\frac{m+1}{2n}\pi\right)^{2n-1}
\left(\cos\frac{m-1}{2n}\pi\right)^{2n+1}
$$
holds for any positive integer $n$ and $m = 1, 2, \dots, n-1$.
|
https://mathoverflow.net/users/451922
|
An inequality involving the power of cosine
|
A bit more general inequality is
\begin{equation\*}
g(h):=g\_x(h):=\ln\frac{\cos ^{1+h}(\pi (x-h)) \cos ^{1-h}(\pi (x+h))}{\cos ^2(\pi x) }\le0 \tag{1}
\end{equation\*}
for $x\in(0,1/2)$ and $h\in(0,\min(x,1/2-x))$.
Suppose that indeed $x\in(0,1/2)$ and $h\in(0,\min(x,1/2-x))$, so that $h\in(0,1/4)$.
Note that
\begin{equation\*}
\frac{g''(h)}\pi=f(\pi (x-h))+f(\pi (x+h))-\frac\pi2 c,
\end{equation\*}
where
\begin{equation\*}
f(u):=2 \tan u-\frac\pi2 \sec ^2u=(\sin 2u-\frac\pi2)\sec ^2u
\end{equation\*}
and
\begin{equation\*}
c:=(1+2 h) \sec ^2(\pi (x-h))+(1-2 h) \sec ^2(\pi (x+h)).
\end{equation\*}
Obviously, $f(u)<0$ for $u\in(0,1/2)$ and $c>0$. So, $g''<0$ and hence the function $g$ is concave. Also, $g(0+)=g'(0+)=0$. So, (1) follows.
|
5
|
https://mathoverflow.net/users/36721
|
408018
| 167,119 |
https://mathoverflow.net/questions/408011
|
8
|
Does there exist a topological group which is locally homeomorphic to the Hilbert cube $[0,1]^{\mathbb N}$?
Let me note that Hilbert cube has the fixed point property and thus it is not homeomorphic to a topological group. Also, as a consequence of a recent paper by Arhangelskii and van Mill (Covering Tychonoff cubes by topological groups, Topology and its applications 2020), there is no topological group which is locally homeomorphic to $[0,1]^{\kappa}$, where $\kappa\geq\omega\_1$.
A related question could be: is there a topological group structure on $\mathbb S^1\times [0,1]^{\mathbb N}$?
|
https://mathoverflow.net/users/128723
|
Topological group locally homeomorphic to the Hilbert cube
|
The answer is no.
Since the Hilbert cube is compact and locally contractible, such a group would be a locally contractible locally compact group. And every locally contractible locally compact group is Lie (i.e., locally homeomorphic to $\mathbf{R}^d$ for some integer $d<\infty$).
---
For a reference
>
> Szenthe, J.
> On the topological characterization of transitive Lie group actions.
> Acta Sci. Math. (Szeged) 36 (1974), 323–344. [Link](http://acta.bibl.u-szeged.hu/14497/1/math_036_fasc_003_004_323-344.pdf)
>
>
>
Theorem 3 there: *Let $G$ be a locally compact group and $H$ a closed subgroup such
that the coset space $G/H$ is locally contractible. Then $G/H$ is a free* [=disjoint] *union of manifolds
which are coset spaces of Lie groups.*
(I've seen it attributed, when $H=1$ to earlier work of Gleason and Montgomery-Zippin without precise reference.)
*Edit:* Taras Banakh mentions in a comment that the Szenthe's proof has a gap, and that this gap is fixed independently in:
>
> S. Antonyan, T. Dobrowolski, Locally contractible coset spaces.
> Forum Math. 27 (2015), no. 4, 2157–2175. [DOI link](https://doi.org/10.1515/forum-2013-0033)
>
>
>
>
> K. Hofmann, L. Kramer. Transitive actions of locally compact groups on locally contractible spaces. J. Reine Angew. Math. 702 (2015), 227–243 (+ erratum 245–246). [DOI link](https://doi.org/10.1515/crelle-2013-0036) [ArXiv link](https://arxiv.org/abs/1301.5114)
>
>
>
Also, user Tyrone mentions that a negative solution to the OP's question (not addressing general locally contractible locally compact groups) is the statement of Theorem 3.1 in
>
> A. Fathi, Y. Visetti. Deformation of open embeddings of Q-manifolds. Trans. Amer. Math. Soc. 224 (1976), no. 2, 427–435 (1977). [link at AMS site](https://www.ams.org/journals/tran/1976-224-02/S0002-9947-1976-0420668-4/S0002-9947-1976-0420668-4.pdf) [DOI link](https://doi.org/10.2307/1997487)
>
>
>
|
10
|
https://mathoverflow.net/users/14094
|
408026
| 167,121 |
https://mathoverflow.net/questions/408034
|
1
|
Let $\bar x \in \mathbb R$. Is there a cut-off function such that $\phi\_\epsilon \in C^\infty(\mathbb R)$, $0 \le \phi \le 1$, and
$$\phi\_\epsilon(x) = \begin{cases}
1 &\text{ if } |x - \bar x| \ge \epsilon\\\
0 &\text{ if }|x-\bar x|\le \epsilon/2
\end{cases}
$$
and
$\phi' \le c\_\epsilon \phi$?
|
https://mathoverflow.net/users/122620
|
Construct suitable cutoff function
|
The answer is no.
Indeed, let $f:=\phi=\phi\_\epsilon$. Let $a:=\sup\{x\colon f(x)=0\}$. Then $a$ is real, $f(a)=0$ and $f>0$ on the interval $(a,\infty)$. Without loss of generality, $a=0$, so that $f(0)=0$ and $f>0$ on the interval $(0,\infty)$.
Suppose now that for some real $c>0$ we have $f'\le cf$. Then $(\ln f)'\le c$ on $(0,\infty)$, whence $\ln f(1)-\ln f(0+)\le c$, that is, $0=f(0+)\ge f(1)e^{-c}>0$, a contradiction. $\quad\Box$
|
0
|
https://mathoverflow.net/users/36721
|
408038
| 167,122 |
https://mathoverflow.net/questions/408035
|
9
|
I know the following facts: $\text{SL}\_2(\mathbb{Z})$ is generated by everyone's favorite matrices
\begin{equation\*}
S =
\begin{pmatrix}
0 & -1 \\
1 & 0
\end{pmatrix}
\end{equation\*}
and
\begin{equation\*}
T =
\begin{pmatrix}
1 & 1 \\
0 & 1
\end{pmatrix}
\end{equation\*}
and $\text{SL}\_2(\mathbb{Z})$ acts transitively on $\mathbb{P}^1(\mathbb{Q})$.
I have been told that the answer to the following questions have something to do with continued fraction expansions, but I would like to find a reference/pointers to the particulars.
(1) How do I write a given matrix $A \in \text{SL}\_2(\mathbb{Z})$ in terms of $S$ and $T$?
(2) For a given $p/q \in \mathbb{Q}$, how do I find an element $A \in \text{SL}\_2(\mathbb{Z})$ with $A(\infty) = p/q$?
|
https://mathoverflow.net/users/419791
|
$\text{SL}_2(\mathbb{Z})$ and continued fractions?
|
1. See Remark 2.2 [here](https://kconrad.math.uconn.edu/blurbs/grouptheory/SL(2,Z).pdf).
2. If you expand $p/q$ into a continued fraction then the successive convergents, as columns of a $2 \times 2$ matrix, have determinant $\pm 1$. Provided $p/q$ is in reduced form and $q > 0$, the last convergent $p\_n/q\_n$ in the continued fraction for $p/q$ will have $p\_n = p$ and $q\_n = q$. Let the second to last convergent be $p\_{n-1}/q\_{n-1}$. Then $p\_{n-1}q\_n - q\_{n-1}p\_n = \pm 1$, and it is easy to pass from such an equation to a $2 \times 2$ integral matrix with determinant $1$ and first column $\binom{p}{q}$.
Let's illustrate these algorithms by starting with the second question on the rational number $p/q = 37/11$, which is in reduced form. What is a matrix in ${\rm SL}\_2(\mathbf Z)$ with first column $\binom{37}{11}$? The continued fraction of $37/11$ is $[3,2,1,3]$ and the successive convergents in this continued fraction are $3/1$, $7/2$, $10/3$, and $37/11$. Using the last two convergents, we obtain $\det(\begin{smallmatrix}10&37\\3&11\end{smallmatrix}) = -1$, so $11 \cdot 10 - 37 \cdot 3 = -1$, so $37(3) - 11(10) = 1$. Thus the matrix $A = (\begin{smallmatrix}37&10\\11&3\end{smallmatrix})$ is in ${\rm SL}\_2(\mathbf Z)$ with first column $\binom{37}{11}$, so $A(\infty) = 37/11$.
Next, for the matrix $A = (\begin{smallmatrix}37&10\\11&3\end{smallmatrix})$ in ${\rm SL}\_2(\mathbf Z)$, how can we write $A$ in terms of $S$ and $T$? For this we will use a continued fraction for the first column ratio $37/11$ using nearest integers from above rather than from below: $37/11 = 4 - 1/(2 - 1/(3 - 1/(2 - 1/2)))$. Using the entries $4, 2, 3, 2$, and $2$, form the matrix product $M = T^4ST^2ST^3ST^2ST^2S$. Its first column will be $\binom{37}{11}$ but its second column might not match that of $A$, so $M^{-1}A$ will be a power of $T$. Indeed, $M = (\begin{smallmatrix}37&-27\\11&-8\end{smallmatrix})$ and $M^{-1}A = (\begin{smallmatrix}1&1\\0&1\end{smallmatrix}) = T$, so
$$
A = MT = T^4ST^2ST^3ST^2ST^2ST.
$$
|
13
|
https://mathoverflow.net/users/3272
|
408044
| 167,124 |
https://mathoverflow.net/questions/408043
|
1
|
Let $Z\_n=\sum\_{k=1}^n a\_k X\_k$ with $(a\_k)$ a strictly decreasing sequence of positive real numbers that tend to zero. The random variables $X\_k$ are independent and satisfy $P(X\_k=1) =p\_k, P(X\_k=-1)=1-p\_k$. Here $p\_k=\frac{1}{2}$. The normalized series is defined as $Z^\*\_n=(Z\_n-\mbox{E}[Z\_n])/\sqrt{\mbox{Var}[Z\_n]}$. Note that all odd moments, including $\mbox{E}[Z\_n]$, are zero. In particular, $\mbox{Var}[Z\_n]=\mbox{E}[Z\_n^2]$.
**My questions are**:
* Is my computation of the moments correct (see below)?
* What is the limiting distribution of $Z^\*\_n$ if $a\_k=1/k^s$ with $s=1/2$? Or
if $a\_k=1/2^k$? Or if $a\_k=1/3^k$?
**Progress made so far:**
The characteristic function is $\phi(t)=\prod\_{k=1}^n \cos(a\_k t)$. In particular, if $a\_k=1/2^k$, then $\phi(t)=(\sin t)/t$ when $n=\infty$. If $a\_k=1/3^k$, I expect the resulting distribution to have an awkward support domain, and related to Cantor sets. This is because $(a\_k)$ converges too fast to zero. But if $a\_k=1/k^s$, with $1/2 < s < 1$, the resulting distribution, for $Z\_n$, behaves like any smooth continuous distribution with support domain equal to the set of real numbers. If $s\leq 1/2$, the variance is infinite and we need to use $Z^\*\_n$ rather than $Z\_n$. However, I am wondering what the limiting distribution of $Z\_n^\*$ is. Is it Gaussian? Is the Central Limit Theorem applicable? I think so if $s<1/2$, but what if $s=1/2$ (the critical point)? I would not be surprised if this is connected to the behavior of the Riemann Zeta function (its roots) in the critical strip, especially if you allow $s$ to be a complex number.
Turning to the moment generating function, we have
$$\mu(t)\equiv \mbox{E}[\exp(t Z\_n)] = \prod\_{k=1}^n \cosh(a\_k t).$$
Thus, taking the logarithm and differentiating we get:
$$\mu'(t)=\mu(t)g(t), \mbox{ with } g(t)=\sum\_{k=1}^n a\_k\tanh(a\_k t).$$
We can iteratively differentiate to get $\mu''(t)=\mu'(t)g(t)+\mu(t)g'(t)$ and so forth, and recursively compute the moments $\mbox{E}[Z^m]=\mu^{(m)}(0)$ for $m=1, 2$ and so on. I used Mathematica to compute $m$-th derivative of $\tanh(a\_k t)$ at $t=0$, and this is what I eventually ended up with:
$$\mbox{E}[Z\_n^4]=3\Big(\sum\_{k=1}^n a\_k^2\Big)^2 - 2 \Big(\sum\_{k=1}^n a\_k^4\Big),\\
\mbox{E}[Z\_n^6]=15\Big(\sum\_{k=1}^n a\_k^2\Big)^3 - 30 \Big(\sum\_{k=1}^n a\_k^2\Big)\Big(\sum\_{k=1}^n a\_k^4\Big)+16\Big(\sum\_{k=1}^n a\_k^6\Big).
$$
Of course $\mbox{E}[Z\_n^2]=\sum\_{k=1}^n a\_k^2$ is well-known, but I have never seen the 4-th and 6-th moments published anywhere. It would we nice if my computations could be double-checked, as it may help statisticians using this type of distribution for model fitting, by choosing $(a\_k)$'s that provide a good fit with some empirical moments.
Assuming my computations are correct and using the notation $Z=Z\_\infty$, for the random harmonic series $(a\_k=1/k)$ we would have:
$$\mbox{E}[Z^2]=\frac{\pi^2}{6}, \mbox{ } \mbox{ } \mbox{ } \mbox{E}[Z^4]=\frac{11\pi^4}{180}, \mbox{ } \mbox{ } \mbox{ } \mbox{E}[Z^6]=\frac{233\pi^6}{7560}.$$
The first identity is well-known.
|
https://mathoverflow.net/users/140356
|
Generalized random harmonic series
|
By the [Berry--Esseen inequality](https://en.wikipedia.org/wiki/Berry%E2%80%93Esseen_theorem#Non-identically_distributed_summands), the limit distribution of $Z\_n^\*$ is the standard normal distribution if $a\_k=1/k^{1/2}$.
---
If $a\_k=1/2^k$, then the limit distribution of $Z\_n^\*$ is uniform on the interval $[-\sqrt3,\sqrt3]$. This follows (i) because $(X\_k+1)/2$ may be viewed as the $k$th binary digit of a random number uniformly distributed on the interval $[0,1]$ and (ii) $Var\,Z\_n\to1/3$ as $n\to\infty$.
For the more general case with $a\_k=t^k$ for $t\in(0,1)$, see e.g. [this survey](https://arxiv.org/abs/1608.04210).
---
Using the formula (cf. e.g. [formula 1.411.6 for $\tanh=(\ln\circ\cosh)'$](http://fisica.ciens.ucv.ve/%7Esvincenz/TISPISGIMR.pdf))
$$\ln\cosh x=\sum\_{j=1}^\infty \frac{2^{2 j-1} \left(4^j-1\right) B\_{2 j}}{j (2 j)!}\,x^{2j}$$
for all $x\in(-\pi/2,\pi/2)$, where the $B\_{2 j}$'s are the Bernoulli numbers, we get all the [cumulants](https://en.wikipedia.org/wiki/Cumulant#Definition) and thence the [moments](https://en.wikipedia.org/wiki/Cumulant#Cumulants_and_moments) of $Z\_n$, in terms of [Bell polynomials](https://en.wikipedia.org/wiki/Bell_polynomials#Bell_polynomials).
|
2
|
https://mathoverflow.net/users/36721
|
408046
| 167,125 |
https://mathoverflow.net/questions/408033
|
4
|
Let $[n]\_q=\frac{1-q^n}{1-q}$ with $[0]\_q=0$. Recall the $q$-factorials $[n]\_q!=[1]\_q[2]\_q\cdots[n]\_q$ (with $[0]\_q!=1$) and the $q$-binomials
$$\binom{n}k\_q=\frac{[n]\_q!}{[k]\_q!\,[n-k]\_q!}.$$
Now, consider the polynomials
$$W\_n(q):=\frac{1-q^{3n}}{1-q^{2n}}\binom{2n}n\_q.$$
**Examples.** $W\_1(q)=q^2+q+1$ and $W\_2(q)=q^6 + q^5 + 2q^4 + q^3 + 2q^2 + q + 1$.
>
> **QUESTION.** Is this true? The number of monomials in $W\_n(q)$ equals $n^2+n+1$.
>
>
>
|
https://mathoverflow.net/users/66131
|
A (mild?) question on the number of monomials
|
Using the fact that $1-q^{3n}=(1-q^{2n})+q^{2n}(1-q^n)$, we can write
$$W\_n(q)=\binom{2n}{n}\_q+q^{2n}\binom{2n-1}{n-1}\_q$$
and then the result follows from the fact that the degree of $W\_n(q)$ is $n^2+n$, together with the fact that $q$-binomial coefficients are polynomials in $q$ with positive coefficients (unimodal even).
|
13
|
https://mathoverflow.net/users/2384
|
408049
| 167,126 |
https://mathoverflow.net/questions/408031
|
4
|
The following is motivated by a (now-deleted) MSE-question by @aglearner.
Suppose that $X\subset {\mathbb C}^n$ is an affine subvariety, equipped with the classical (Euclidean) topology. Consider the group $G= {\mathbb C}^\times$, and suppose that $G\times X\to X$ is an algebraic action. Assume, in addition, that each $G$-orbit in $X$ is closed.
**Question 1.** Is the quotient $X/G$ (in the sense of general topology) Hausdorff?
(Note that I am *not* taking the GIT quotient here.)
Hausdorffness of the quotient fails if I relax the assumption on $X$ to that of a *quasi-affine* subvariety in the following standard example: Take $X={\mathbb C}^2\setminus \{(0,0)\}$ and consider the action given by
$$
(t, (x,y)) \mapsto (tx, t^{-1}y), \quad t\in G, (x,y)\in X.
$$
The next question is a complex-analytic version of Question 1:
**Question 2.** Suppose that $X\subset {\mathbb C}^n$ is a Stein submanifold, $G$ is as above and $G\times X\to X$ is a holomorphic action with closed orbits. Is $X/G$ Hausdorff?
I expect answers to both questions to be negative (and counter-examples given by a smooth affine subvariety $X$) but cannot think of any examples.
|
https://mathoverflow.net/users/39654
|
Group actions on affine varieties with closed orbits
|
I think the answer to Question 1 is yes, since under your hypothesis the set-theoretical quotient *is* essentially the GIT quotient. This follows from the fact that $G$-invariant regular functions on $X$ separate orbits (when these are closed).
There is an analytic proof of this fact. Let $V$ and $W$ be two disjoint closed orbits. Then 1 can be written as the sum of two regular functions $f\_V$ and $f\_W$ vanishing respectively on $V$ and $W$ (because $I(V) + I(W) = \mathbb C[X]$). In particular, $f\_V = 0$ on $V$ and $f\_V = 1$ on $W$. After averaging $f\_V$ under the action of the unit circle $U\_1\subset \mathbb C^\times$, one can assume that $f\_V$ is $U\_1$-invariant. Since $f\_V$ is holomorphic, the action of $\mathbb C^\times$ is holomorphic and $U\_1$ is a real form of $\mathbb C^\times$, this implies that $f\_V$ is in fact $\mathbb C^\times$-invariant.
My impression is that one could do the same with holomorphic functions on Stein manifolds. My only concern is about the existence of a holomorphic function which is $0$ on $V$ and $1$ on $W$. (This condition ensures that, in the averaging process, $f\_V$ remains non-zero on $W$.)
|
3
|
https://mathoverflow.net/users/173096
|
408055
| 167,127 |
https://mathoverflow.net/questions/408050
|
4
|
I have been reading Felix E. Browder's [Convergence Theorems for Sequence of Nonlinear Operators in Banach Space](https://link.springer.com/article/10.1007/BF01109805) and I was hoping I could find answers to a couple of questions I have about the paper. Consider Lemma 6.
>
> Let $F$ be a closed convex subspace of a Hilbert space H. Let
> $\{u\_n\}$ be a sequence in $H$ such that
>
>
> 1. For each $f\in F$, $\|u\_n-f\|$ is a non-increasing sequence
> 2. Each weak limit point of the sequence $\{u\_n\}$ lies in $F$.
>
>
> Then $u\_n \to f\_0$ weakly for some point $f\_0$ in $F$.
>
>
>
I can follow most of the proof. However, is the bound $\|u\_n\|\leq \|u\_1\|+\|f\|$ for a fixed $f$ necessarily correct? It seems that it should be $\|u\_n\|\leq \|u\_1\|+2\|f\|$. In any case, I am not sure how one obtains $p(f):=\lim\_{n\to \infty} \|u\_n-f\| \ge \|f\|-M\_0$ (I don't think he mentions what $M\_0$ stands for, but I strongly suspect it is the bound of $\|u\_{n}\|$, for which the inequality would make sense).
Also, how does lower semi-continuous $p$ assumes minimum at some point $f\_0$ of $F$? Is there is a version of the following theorem ([link](https://math.stackexchange.com/questions/3082099/show-that-lower-semi-continuous-function-attains-its-minimum-proof-verificati)) for closed convex subsets of Hilbert spaces and lower semi-continuous $p$?
Further, consider Lemma 7.
>
> Let $X$ be a strictly convex Banach space. $U$ is a non-expansive mapping of a convex subset $C$ of $X$ into $X$. Then the fixed point set of $U$ in $C$ is convex.
>
>
>
The proof starts off by considering two fixed points $f\_0$ and $f\_1$ of $U$ and trying to prove that $y\_t= (1-t)f\_0+tf\_1$ is a fixed point for $0\leq t \leq 1$. I have been able to follow the proof except that I don't see how the remark that $U(y\_t)$ lies on the line segment between $f\_0$ and $f\_1$ follows. Thanks!
|
https://mathoverflow.net/users/nan
|
On some convergence theorems by Felix E. Browder (1967)
|
**Concerning Lemma 6:** Clearly, this lemma is false if $F=\emptyset$ and, say, $u\_n=nu$ for some nonzero $u\in H$. Assume therefore that $F\ne\emptyset$.
The inequality $\|u\_n\|\le\|u\_1\|+\|f\|$ will not always hold in this setting -- consider e.g. the case when $u\_1=0$ and $u\_n=2f\ne0$ for $n\ge2$.
However, since $F\ne\emptyset$, there is some $f\_\*\in F$. Then $\|f\_\*-u\_n\|\le\|f\_\*-u\_1\|$ for all $n$, and hence $\|u\_n\|\le M:=\|f\_\*\|+\|f\_\*-u\_1\|$ and $\|f-u\_n\|\ge\|f\|-\|u\_n\|\ge\|f\|-M$.
So, $p(f)\ge\|f\|-M$, and hence $p(f)>p(0)$ if $\|f\|>r:=M+p(0)$. So, $$\inf\_F p=\inf\_{F\cap B\_r} p,$$
where $B\_r$ is
the closed ball of radius $r$ in $H$ centered at $0$. The ball $B\_r$ in the Hilbert space $H$ is [weakly compact](https://en.wikipedia.org/wiki/Banach%E2%80%93Alaoglu_theorem). The set $F$ is convex and closed and [hence weakly closed](https://math.stackexchange.com/questions/1151148/a-convex-subset-of-a-banach-space-is-closed-if-and-only-if-it-is-weakly-closed). So, $F\cap B\_r$ is weakly compact. Also, the lower semicontinuous **convex** function $p$ is [weakly lower semicontinuous](https://link.springer.com/content/pdf/10.1007%2F978-1-4419-9467-7_9.pdf). Thus, $p$ does attain a [minimum](https://en.wikipedia.org/wiki/Semi-continuity#Properties) on $F\cap B\_r$ and hence on $F$ (provided that $F\ne\emptyset$).
---
**Concerning Lemma 7:** The desired conclusion follows immediately from [third listed property equivalent to the strict convexity of the normed space](https://en.wikipedia.org/wiki/Strictly_convex_space).
|
1
|
https://mathoverflow.net/users/36721
|
408060
| 167,131 |
https://mathoverflow.net/questions/408059
|
2
|
Let $M$ be a compact smooth manifold without boundary. Let $T>0$ and let $g$ be a smooth Riemannian metric on $M$. Given any $f \in L^2(M)$ let $u$ be the unique solution to the equation
$$\partial\_t u -\Delta\_gu=0 \quad \text{on $(0,T)\times M$},$$
subject to initial data $f$ at times $t=0$. Let the map
$$ G: L^2(M) \to L^2(M)$$
be defined via $Gf=u(T,\cdot)$ on $M$.
Is it true that $G$ is compact?
|
https://mathoverflow.net/users/50438
|
Compactness for initial-to-final map for heat equation
|
Parabolic regularity show that $u$ is regular for all positive times; in particular $u(t,\cdot) \in W^{1,2}(M)$ for all $t > 0$. Interior parabolic estimates additionally show that there is a constant $C = C(g,T) > 0$ so that
\begin{equation}
\lvert u(t,\cdot) \rvert\_{W^{1,2}} \leq C \lvert f \rvert\_{L^2}
\quad \text{ for all $t \in [T/2,2T]$.}
\end{equation}
Rellich–Kondrachov implies that the map $G: f \in L^2(M) \mapsto u(T,\cdot) \in L^2(M)$ is compact.
|
6
|
https://mathoverflow.net/users/103792
|
408062
| 167,132 |
https://mathoverflow.net/questions/408008
|
4
|
Here is a statement having a proof that involved the CFSG.
Let $p$ be a prime, and $S$ be a nonabelian finite simple group such that $S$ is isomorphic to a subgroup of $S\_p$ with $p\mid |S|$. Then $\mathrm{Out}(S)$ is a $p'$-group (i.e. $p\nmid |\mathrm{Out}(S)|$).
**Question** Is there a CFSG-free proof for this statement?
|
https://mathoverflow.net/users/44312
|
Outer automorphism of a finite simple group which is isomorphic to a subgroup of $S_p$
|
Let $p$ be a prime, and let $G$ be a non-abelian simple group with $p \mid |G|$. Suppose that $G \leq S\_p$ and that $G$ is transitive.
By a theorem of Burnside a non-solvable transitive group of prime degree is $2$-transitive, so $G$ is $2$-transitive.
Let $B \leq G$ be such that $|B| = p$. Then $G = AB$, where $A$ is a point stabilizer. Note that $A \cap B = 1$, so $A$ is a Hall subgroup.
Now it seems to be a theorem of Wielandt that in this case $Aut(G)/G$ is cyclic of order dividing $p-1$. This is according to the following paper of Cameron (I haven't been able to find the paper by Wielandt), see §3 there.
>
> Cameron, Peter J. On groups with several doubly-transitive permutation representations. Math. Z. 128 (1972), 1–14.
>
>
>
The paper by Wielandt is
>
> Wielandt, Helmut. On automorphisms of doubly transitive permutation groups. 1967 Proc. Internat Conf. Theory of Groups (Canberra, 1965) pp. 389–393 Gordon and Breach, New York
>
>
>
You could also use the proofs given by Cameron, which do not need CFSG.
Denote the image of $N\_{S\_p}(G)$ in $Aut(G)$ by $P(G)$, this is called the group of permutation automorphisms. Then Theorem 3 in Cameron shows that $P(G)$ is normal in $Aut(G)$ and $Aut(G)/P(G)$ is an elementary abelian $2$-group.
Next we can show that $P(G)/Inn(G)$ is cyclic of order dividing $p-1$, which gives you the desired result. For this let $R = N\_{S\_p}(G)$. By the Frattini argument $R = GN\_R(B)$. Then $$\frac{P(G)}{Inn(G)} \cong \frac{R}{GC\_{S\_p}(G)} = \frac{GN\_R(B)}{GC\_{S\_p}(G)} \cong \frac{N\_R(B)}{GC\_{S\_p(G)} \cap N\_R(B)}.$$
Because $B \leq GC\_{S\_p(G)} \cap N\_R(B)$, it follows that $P(G)/Inn(G)$ is isomorphic to a quotient of $N\_R(B)/B$. Now $B$ is self-centralizing in $S\_p$, so $N\_R(B)/B$ is isomorphic to a subgroup of $Aut(B)$, hence cyclic of order dividing $p-1$.
|
8
|
https://mathoverflow.net/users/38068
|
408068
| 167,134 |
https://mathoverflow.net/questions/407707
|
1
|
In [this book](https://books.google.co.kr/books?id=doLoDwAAQBAJ&pg=PA372&lpg=PA372&dq=every%20graph%20with%20m%20edges%20and%20maximum%20degree%20k%20has%20a%20proper%20(k%20%2B%201)-%20edge%20coloring%20in%20which%20each%20color%20is%20used%20%E2%8C%88%20m%20k%2B1%20%E2%8C%89%20or%20%E2%8C%8A%20m%20k%2B1%20%E2%8C%8B%20times&source=bl&ots=Y3LyU9oqRj&sig=ACfU3U2lCEhFOvOnxIveCnrY9G49ffOcLQ&hl=ko&sa=X&ved=2ahUKEwivlcHN9_zzAhXJa94KHVENAbQQ6AF6BAgVEAM#v=onepage&q=every%20graph%20with%20m%20edges%20and%20maximum%20degree%20k%20has%20a%20proper%20(k%20%2B%201)-%20edge%20coloring%20in%20which%20each%20color%20is%20used%20%E2%8C%88%20m%20k%2B1%20%E2%8C%89%20or%20%E2%8C%8A%20m%20k%2B1%20%E2%8C%8B%20times&f=false) (I found it from other references, and it was a nice book to study.), there is an exercise that proving the following two statements.
>
> Every graph $G$ with $m$ edges and maximum degree $k$ has a proper $(k+1)$-edge-coloring with each color used $\lfloor \frac{m}{k+1} \rfloor$ or $\lceil \frac{m}{k+1} \rceil$ times.
>
>
>
>
> Every bipartite graph with maximum degree at least $k$ has a $k$-edge-coloring (*not necessarily proper*) in which at each vertex $v$, each color appears $\lfloor \frac{d(v)}{k} \rfloor$ or $\lceil \frac{d(v)}{k}\rceil$ times.
>
>
>
First one is proved by [Fedor Petrov](https://mathoverflow.net/questions/334036/equitable-edge-coloring-of-graphs) in a very elegant method.
But for second one, it seems that a quite different strategy is needed, and I have yet no idea with how to prove it.
[The book](https://books.google.co.kr/books?id=doLoDwAAQBAJ&pg=PA372&lpg=PA372&dq=every%20graph%20with%20m%20edges%20and%20maximum%20degree%20k%20has%20a%20proper%20(k%20%2B%201)-%20edge%20coloring%20in%20which%20each%20color%20is%20used%20%E2%8C%88%20m%20k%2B1%20%E2%8C%89%20or%20%E2%8C%8A%20m%20k%2B1%20%E2%8C%8B%20times&source=bl&ots=Y3LyU9oqRj&sig=ACfU3U2lCEhFOvOnxIveCnrY9G49ffOcLQ&hl=ko&sa=X&ved=2ahUKEwivlcHN9_zzAhXJa94KHVENAbQQ6AF6BAgVEAM#v=onepage&q=every%20graph%20with%20m%20edges%20and%20maximum%20degree%20k%20has%20a%20proper%20(k%20%2B%201)-%20edge%20coloring%20in%20which%20each%20color%20is%20used%20%E2%8C%88%20m%20k%2B1%20%E2%8C%89%20or%20%E2%8C%8A%20m%20k%2B1%20%E2%8C%8B%20times&f=false) only says 'Use graph transformation.' but no other hints.
Would you help me?
|
https://mathoverflow.net/users/384338
|
An equitable edge-coloring of bipartite graphs
|
There is a proof with a very similar principle than in Petrov's proof.
Take an arbitrary coloring. If the property is not respected, you have a vertex $v$ where one of the colors $c\_1$ appear strictly over $⌈\frac{d(v)}{k}⌉$, and another color $c\_2$ appear less or equal than $⌊\frac{d(v)}{k}⌋$ (or one which appear strictly less than $⌊\frac{d(v)}{k}⌋$ and another which appear more or equal than $⌈\frac{d(v)}{k}⌉$, but the argument is th same).
Take the graph composed only of the edges colored by $c\_1$ and $c\_2$, and orient all $c\_1$ edges in the same direction (from one part of the bipartite graph to the other) such that the arcs are leaving $v$. Same with $c\_2$ but with arcs entering $v$). Follow a path in the graph starting from $v$, you cannont get stuck at $v$ as there are strictly more outcoming edges than incoming edges. Once you get stuck, you have found a node where the number of incoming arcs is strictly more than the number of outcoming arcs. It is easy to show that swapping colors in the path will strictly decrease the distances to average degree.
|
1
|
https://mathoverflow.net/users/381833
|
408076
| 167,136 |
https://mathoverflow.net/questions/408042
|
0
|
Suppose that we have the following PDE
$$\partial\_t \mu\_t = \nabla\cdot \left(\nabla \mu\_t - (b\*\mu\_t)\mu\_t\right), \tag{1}$$
with $\mu\_0$ being a (smooth) probability measure/density on $\mathbb{R}^d$ and with the notation $$b\*\mu(x) := \int\_{\mathbb{R}^d} b(x,z)\,\mu(z)\,\mathrm{d}z.$$ For $N \in \mathbb{N}\_+$ denote $N$ times tensorized product of $\mu$ by $\mu^{\otimes N}$. It is stated (without proof) in page 5 of [this paper](https://www.aimsciences.org/article/doi/10.3934/dcds.2020243) that $\mu^{\otimes N}\_t$ can be seen as the pushed forward of $\mu^{\otimes N}\_0$ by the application $\zeta^{\otimes N}\_{t,0}$ defined as the solution to the following ODE
$$\zeta\_{t,0}(x) = x + \int\_0^t \left(b\*\mu\_s - \nabla \ln \mu\_s\right)\,\mathrm{d}s. \tag{2} $$
However, I have no clue as to deriving $(2)$ from $(1)$. I understand the definition of push forward of measures (see for instance [here](https://math.stackexchange.com/questions/1970164/push-forward-measure-equivalent)) but I really need someone to help me out on this particular problem. Thank you very much for your kind help!
|
https://mathoverflow.net/users/163454
|
Rewriting PDE as "push-forward"
|
Let's focus on $N=1$ only, the case $N>1$ is just a tensorization of the argument below.
The whole argument is actually unrelated to the specific (aggregation-diffusion) PDE or gradient flow: As soon as you have a curve of probability measures $\mu\_t$ satisfying the continuity equation
$$
\partial\_t\mu\_t+\nabla\cdot(v\_t\mu\_t)=0
$$
for some (smooth enough) vector-field $v=v\_t(x)$, then in fact you have the representation
$$
\mu\_t=\Phi\_t\#\mu\_0,
$$
where $\Phi\_t(x)$ is the flow-map associated to the ODE $\begin{cases}\dot X\_t=v\_t(X\_t)\\
X\_0=x\end{cases}$.
This is easy to check formally, and for a precise reference I can suggest reading chapter 8 in [1] (e.g. lemma 8.1.6 or prop. 8.1.8).
---
Now, for your specific context, it is just a matter of rewriting the PDE into a continuity equation. Factoring out and explointing the classical identity $\Delta\mu=\nabla\cdot\nabla\mu=\nabla\cdot(\frac{\nabla\mu}{\mu}\mu)=\nabla\cdot((\nabla\log \mu) \mu)$ you see that $\mu$ solves the continuity equation
$$
\partial\_t\mu\_t=\nabla\cdot(\nabla\mu\_t-(b\ast\mu\_t)\mu\_t)
\quad\Leftrightarrow \quad
\partial\_t\mu\_t+\nabla\Big((\underbrace{b\ast\mu\_t-\nabla\log\mu\_t}\_{v\_t})\mu\_t\Big)=0.
$$
Then the flow-map $\Phi\_t$ can be simply represented from the implicit Duhamel formula as
$$
\begin{cases}\dot X\_t=v\_t(X\_t)\\
X\_0=x\end{cases}
\quad\Leftrightarrow\quad
X\_t=x+\int\_0^t v\_s(X\_s)ds.
$$
Plugging in the explicit formula for the velocity field $v\_t=b\ast\mu\_t-\nabla\log\mu\_t$ gives exactly your expression (2).
---
[1] Ambrosio, L., Gigli, N., & Savaré, G. (2008). Gradient flows: in metric spaces and in the space of probability measures. Springer Science & Business Media.
|
1
|
https://mathoverflow.net/users/33741
|
408080
| 167,137 |
https://mathoverflow.net/questions/408085
|
7
|
It was mentioned in a lecture on Faltings’s proof of Mordell Conjecture that there’s some kind of correspondence between Galois representation (of cohomology, or some complex of constructible sheaves?) , but it hadn’t been explained explicitly. Also, I’m studying étale cohomology and Weil conjectures. The trace formula also seems suggest that correspondence: Frobenius elements form the Galois group of finite field, and the eigenvalues of them are the zeros and poles of Grothendieck’s L-function. But I can’t see the explicit ‘correspondence’(is there?) and I have no idea about the case of other base field.
|
https://mathoverflow.net/users/170335
|
L-functions and Galois representations: What’s the explicit relation?
|
The situation is:
One defines an $L$-function associated to a representation $V$ of the Galois group of a number field $K$ as $$L(V,s) = \prod\_{ \mathfrak p } \frac{ 1}{ \det( 1 -|\mathfrak p|^{-s} \operatorname{Frob}\_{\mathfrak p} , V^{I\_{\mathfrak p}} )}$$ with the product taken over the primes of $K$, where $V^{I\_{\mathfrak p}}$ are the inertia invariants of $p$ (which, for all reasonable Galois representations agree with $V$ away from finitely many primes).
So there is a correspondence between $L$-functions and Galois representations simply by definition. One can check that if two Galois representations over the rational numbers have the same $L$-function, then they have the same characteristic polynomial of Frobenius at each place, and if in addition they are semisimple, then they are isomorphic. So in some cases this is a one-to-one correspondence.
Furthermore one can define, associated to a scheme $X$ over $\mathcal O\_K$, the Hasse-Weil zeta function. This definition doesn't reference Galois representations at all, and just refers to the closed points of $X$. However, one can check using the Lefschetz fixed point formula (applied to the fiber of $X$ over each finite field) that the Hasse-Weil zeta function is
$$ \prod\_i L( H^i (X\_{\overline{K}}, \mathbb Q\_\ell), s)^{(-1)^i} $$ up to changing factors at finitely many primes (the primes where $X$ has singularities and the primes dividing $\ell$).
The zeta function appearing in the Weil conjectures is the Euler factor of the Hasse-Weil zeta function at a particular prime, and the proof of this is almost identical to the proof of the expression of the Weil zeta function in terms of cohomology in the proof of the Weil conjectures.
|
18
|
https://mathoverflow.net/users/18060
|
408086
| 167,140 |
https://mathoverflow.net/questions/408032
|
3
|
I have stumbled upon the following definitions in a paper by Gavin Wraith.
**Definition 1.** Say a ring morphism $A\overset \varphi \to B$ is formally unramified if every $b\in B$ admits:
* $b\_0\in B$ which is a simple root of some $\varphi (p)$ for some monic $p\in A[x]$;
* monics $f,g\in A[x]$ such that $\varphi (g)(b\_0)\in B^\times $ and $b=\tfrac{\varphi(f)(b\_0)}{\varphi(g)(b\_0)}$.
**Definition 2.** Say a ring morphism $A\overset \varphi \to B$ is étale if there exist comaximal (generating the unit ideal) elements $b\_1,\dots ,b\_r\in B$ such that for each $1\leq i\leq r$ there's a $B$ algebra isomorphism $$B[\tfrac 1{b\_i}]\cong \tfrac{A[x\_1,\dots,x\_n][1/g]}{\langle f\_1,\dots ,f\_n\rangle},\;\det\tfrac{\partial f\_i}{\partial x\_j}\mid g.$$
**Question.** When are these definitions equivalent to the modern ones involving lifts and square-zero ideals? (And how to prove the equivalence?)
|
https://mathoverflow.net/users/69037
|
Alternative definitions of étale and formally unramified in Wraith
|
Following a suggestion made in the comments, I post this as an answer.
These are the "local structure theorems" of unramified and étale algebras. They are equivalent to the definitions with liftings and square-zero ideals. The proofs of the equivalence are not straightforward as far as I know: they are elementary but intricate and a bit lengthy. An excellent reference for all this is Michel Raynaud, *Anneaux locaux henseliens*; more precisely, see I. Prop. 8, V. Theorem 1 and Theorem 5.
|
2
|
https://mathoverflow.net/users/69401
|
408087
| 167,141 |
https://mathoverflow.net/questions/408091
|
10
|
Let $G$ be a commutative connected algebraic group over $\mathbb{C}$. A theorem of Serre says that there exists an exact sequence
$$1\to \mathbb{G}\_a^n\times \mathbb{G}\_m^m\to G\to A\to 1,$$
where $A$ is an abelian variety. (See [here](https://icerm.brown.edu/materials/Slides/sp-s12-w3/On_Mordell-Lang_in_algebraic_groups_of_unipotent_rank_1_%5D_Paul_Vojta,_University_of_California,_Berkeley.pdf), for example.)
**I wonder what are some examples of such groups $G$ where this extension is non-trivial. That is, such that $G\ncong \mathbb{G}\_a^n\times \mathbb{G}\_m^m\times A$.**
|
https://mathoverflow.net/users/131975
|
What is a "non-trivial" example of a commutative algebraic group over $\mathbb{C}$?
|
One example is provided by the generalized Jacobian. For a smooth projective curve $C$ and a divisor $D$ on $C$, the generalized Jacobian is defined to be the moduli space parameterizing pairs consisting of a line bundle of degree $0$ on $C$ together with a trivialization of that line bundle over $D$.
This admits a map to the usual Jacobian, whose kernel is a product of $\mathbb G\_m$s and $\mathbb G\_a$s depending on the multiplicity of points of $D$.
If this were trivial, than we could define a section of this map, which would give for an arbitrary line bundle a canonical trivialization at each point of $D$. That would mean that for each point $x\in D$, the line bundle on the usual Jacobian whose fiber at a point $L \in J$ is the fiber of $L$ at $x$ would admit a section and thus be trivial. But from the duality theory of abelian varieties, these line bundles are different for distinct points $x\in C$, so they cannot all be trivial as long as $D$ contains two or more points.
|
17
|
https://mathoverflow.net/users/18060
|
408092
| 167,142 |
https://mathoverflow.net/questions/408045
|
0
|
I wonder if it is possible to solve analytically the following equation
$$
\dot{\alpha}\_t = -\frac{2}{m} \alpha^2\_t + \frac{1}{2m} (\alpha\_t - \alpha\_t^\*)^2
$$
Where $\alpha\_t$ is a complex function, $\alpha\_t^\*$ is its complex conjugate and $\dot{\alpha}\_t$ is the time derivative.
All the best!
|
https://mathoverflow.net/users/191299
|
Solution of this differential equation
|
Yes, this can be integrated explicitly. First, notice that, since $m\not=0$, we can write $\alpha(t) = 2m\bigl(x(t)+iy(t)\bigr)$, in which case, the given equation becomes
$$
\dot x + i\,\dot y = -4(x + iy)^2 + (2iy)^2 = -4 x^2 - i\,(8xy),
$$
so $\dot x = -4x^2$ and $\dot y = -8 xy$. Thus, by standard ODE techniques,
$$
x(t) = \frac{x(0)}{(1+4x(0)\,t)}\quad\text{and}\quad
y(t) = \frac{y(0)}{(1+4x(0)\,t)^2}\,.
$$
|
5
|
https://mathoverflow.net/users/13972
|
408093
| 167,143 |
https://mathoverflow.net/questions/407780
|
2
|
To recall Arrow's theorem:
Suppose we have a finite set $X$ of voters and a finite set $Y$ of candidates.
An **election** is a map $\phi: X \rightarrow T$ where $T$ is the space of total orderings of $Y$. So we are voting via a total ranked list ballot. Let $E$ be the space of all elections.
A **social choice function** is a function $f: E' \rightarrow T$ defined on some subset $E' \subset E$.
Kenneth Arrow asks that a social choice function satisfy some reasonable-looking axioms, namely:
1. **Unanimity** - If every single voter prefers $a > b$, then the outcome should also give $a > b$,
2. **Independence of irrelevant alternatives** - Holding fixed an election $\phi$ and candidate $c$, suppose we alter the ballots by moving around only $c$ in the ranking of each ballot, without disturbing the relative order of all other candidates. Call this altered election $\phi'$. Then the outcome $f(\phi')$ should coincide with $f(\phi)$ except possibly with $c$'s place in the ranking changed. In particular, it should not reverse any other pair order, such as $a > b$ into $b > a$,
3. **Non-monarchy** - The social choice function is not just picking one voter, copying their ballot, and ignoring everyone else,
4. **Unrestricted domain** - The social choice function is defined on all of $E$.
**Arrow's theorem** asserts that *if three or more candidates are running, then no such social choice function exists*.
But I've never understood why we should adopt the axiom of unrestricted domain. It seems a wildly unrealistic thing to ask.
For example, we can easily manufacture (very low probability) election outcomes with symmetric results, such as one-third voting $a > b > c$, one-third voting $b > c > a$, and one third voting $c > a > b$. The only way I see to deal with such edge cases is to restrict our social choice function to not consider such elections.
So I find it no surprise that such a strong hypothesis as unrestricted domain would ruin any attempt to find a satisfactory social choice function. But I would like to think that the essence of Arrow's theorem is something more robust than just exploiting unrestricted domain to get a contradiction.
So my question is: ***is there a weaker, more reasonable substitute for unrestricted domain, and a corresponding version of Arrow's theorem in this more general setting?***
Maybe I should add that I would much prefer a deterministic social choice function.
|
https://mathoverflow.net/users/126543
|
Is there a version of Arrow's theorem without unrestricted domain?
|
There are two possible directions one can take this. One is to look at weaker conditions than full domain that still allow one to obtain ArrowÄs theorem, or one can look at restrictions that allow for more positive results. A classical criterion in the first category is the following:
**Chain Property:** We say $E'$ has the *chain property* if for every two pairs of alternatives $(u,v)$ and $(x,y)$ there is a finite sequence of the form $u,v,y\_1,\ldots,y\_n,x,y$ and for every three consecutive candidates in the sequence, there exist profiles with all possible rankings of these three candidates.
If one replaces the assumption of full domain by the assumption that $E'$ has the chain property, Arrow's theorem still holds. For a proof (and related results), you can look at:
>
> Campbell, Donald E., and Jerry S. Kelly. "Impossibility theorems in
> the Arrovian framework." *Handbook of Social Choice and Welfare 1*
> (2002): 35-94.
>
>
>
On the other hand, there are so-called *domain restrictions* under which one obtains non-dictatorial rules and a large literature studying such restrictions.
|
4
|
https://mathoverflow.net/users/35357
|
408094
| 167,144 |
https://mathoverflow.net/questions/408099
|
3
|
Consider the domain $[0,1] \times [0,T]$ and the uniformly parabolic operator $L -\partial\_t$ with smooth coefficient. Suppose I have $u\_1(x,t) \in C^\infty([0,1] \times [0,T])$ solving
\begin{equation}
\left\{\begin{aligned}
&L u\_1 -\partial\_t u\_1= 0& \hspace{10pt} &\text{for $(x,t) \in (0,1) \times (0,T]$}
;\\
& u\_1(0,t)
=f\_1(t) & \hspace{10pt} &\text{for $t \in \big[0,T\big]$;}\\
& u\_1(1,t)
=g\_1(t) & \hspace{10pt} &\text{for $t \in \big[0,T\big]$;}\\
& u\_1(x,0)
=h(x) & \hspace{10pt} &\text{for $x \in \big(0,1\big)$.}\\
\end{aligned}\right.
\end{equation}
Suppose I also have $u\_2(x,t) \in C^\infty([0,1] \times [0,T])$ solving
\begin{equation}
\left\{\begin{aligned}
&L u\_2 -\partial\_t u\_2 = 0& \hspace{10pt} &\text{for $(x,t) \in (0,1/2) \times (0,T]$}
;\\
& u\_2(0,t)
=f\_2(t) & \hspace{10pt} &\text{for $t \in \big[0,T\big]$;}\\
& u\_2(1/2,t)
=u\_1(1/2,t) & \hspace{10pt} &\text{for $t \in \big[0,T\big]$;}\\
& u\_2(x,0)
=h(x) & \hspace{10pt} &\text{for $x \in \big(0,1\big)$.}\\
\end{aligned}\right.
\end{equation}
Question: Is $\partial\_x u\_1(1/2,t) = \partial\_x u\_2(1/2,t)$?
Reason: I think it is true since $u\_1$ and $u\_2$ solve the same equation. Also, their values at $x=1/2$ are the same. Therefore $u\_3=u\_1 \chi\_{[1/2,1]\times [0,T]} +u\_2 \chi\_{[0,1/2)\times [0,T]}$ is at least Lipschitz in the variable $x$. So $u\_3$ is a smooth unique solution solving the equation with boundary conditions $u(0,t) = f\_2(t)$ and $u(1,t) = g\_1(t)$.
|
https://mathoverflow.net/users/87922
|
Gluing of two solutions to the same parabolic equation
|
Absolutely not! Taking the difference $v=u\_1-u\_2$, you see that $v(x,t)$ solves
$$
\begin{cases}
(L-\partial\_t) v=0 & \mbox{for }(x,t)\in(0,1/2)\times(0,T];
\\
v(0,t)=f(t) & \mbox{for }x=0,\,t\in(0,T]\\
v(1/2,t)=0 & \mbox{for }x=1/2,\,t\in(0,T]\\
v(x,0) =0 & \mbox{for }x\in (0,1/2),\,t=0
\end{cases}
$$
for some left boundary data $f=f\_1-f\_2$ which is a priori non-zero, and your question amounts to asking whether the overdetermined Neumann boundary condition $\partial\_x v(1/2,t)=0$ is satisfied as well at the rightmost endpoint $x=1/2$. Of course there is no reason why this should hold. For a counterexample, just take $f(t)=t$, so that by the maximum principle $v(t,x)\geq 0$ and in fact $v(t,x)>0$ for all $(x,t)\in (0,1/2)\times(0,T)$. Hopf's boundary lemma then guarantees that, since $v$ attains its minimum value $v(1/2,t)=0$ on the right boundary, necessarily $\partial\_x v(1/2,t)<0$ there.
|
5
|
https://mathoverflow.net/users/33741
|
408103
| 167,145 |
https://mathoverflow.net/questions/408121
|
3
|
Consider the number of integer partitions $p(n)$ of $n$ whose (product) generating function reads
$$\sum\_{n\geq0}p(n)\,x^n=\prod\_{k\geq1}\frac1{1-x^k}.$$
There are many congruences for $p(n)$ including those due to Ramanujan: $p(5n+4)\equiv\_50, p(7n+5)\equiv\_70$ and $p(11n+6)\equiv\_{11}0$. Here, I would like to ask:
>
> **QUESTION.** Is this true? For any prime $q$,
> $$\#\{p(n)\,\,\, \text{mod}\,\,\, q: n\in\mathbb{N}\}=q.$$
> That means, do all modular residues appear? In particular, given a prime $q$, does there always exist an $n$ such that $p(n)\,\,\,\text{mod}\,\,\, q=0$?
>
>
>
|
https://mathoverflow.net/users/66131
|
Congruence residues of integer partitions
|
Yes, this is true. In fact something even stronger is known:
For arbitrary positive integers $r,j$, and prime $\ell\geq 5$, there are infinitely many values of $n$ for which
$$p(n)=r\pmod{\ell^j}.$$
This is a special case of a conjecture of Newman and is proved in the paper "[Coefficients of half-integral weight modular forms modulo $\ell^j$](https://doi.org/10.1007/s00208-004-0555-9)" by Scott Ahlgren and Matthew Boylan. I believe it is still unknown whether there are infinitely many $n$ for which $p(n)=0\pmod{3}$.
|
5
|
https://mathoverflow.net/users/2384
|
408124
| 167,147 |
https://mathoverflow.net/questions/408117
|
9
|
Is there a "nice" way to compute the signature of a smooth toric manifold of even complex dimension in terms of the moment polytope? By signature I mean in the sense of topology (see <https://en.wikipedia.org/wiki/Signature_(topology)>).
If one goes through all the machinery it is clear that the signature is encoded. I am asking if there is a some way you can "see" this quantity from the moment polytope, quickly and effortlessly. For example the topological Euler characteristic is the number of vertices. The signature won't be as nice as this, certainly, but that is the spirit of what I am looking for (i.e. the less machinery the better). In complex dimension 2 the signature is also a function of the number of vertices by the minimal model program, but that seems to be a stroke of fortune owing to the low dimension.
|
https://mathoverflow.net/users/99732
|
"Nice" way to compute the signature of a toric manifold?
|
The Hodge index theorem for smooth projective varieties (or compact Kähler manifolds) says that the signature is $$\sum\_{p,q} (-1)^p h^{p,q}(X)$$
For toric varieties, $h^{p,q}=0$ unless $p=q$, and equals the Betti number $h^{2p}(X)$, so the signature is $\sum\_p (-1)^p h^{2p}(X)$.
The Betti number $h^{2p}$ is equal to $\sum\_{i=p}^n (-1)^{i-p} (-1)^{i-p} \binom{i}{p} N\_i$ where $N\_i$ is the number of $i$-dimensional cells in the polytope.
So the signature is $$ \sum\_p(-1)^p \sum\_{i=p}^n (-1)^{i-p} (-1)^{i-p} \binom{i}{p} N\_i = \sum\_i (-1)^i \sum\_{o=0}^i \binom{i}{p} N\_i = \sum\_i (-1)^i2^i N\_i = \sum\_i (-2)^i N\_i.$$
Thus, for a two-dimensional variety, this is the number of vertices minus twice the number of edges plus four times the number of faces. For a polygon, the number of vertices is the number of edges, and the number of faces is one, so this is four minus the number of vertices, which should be your formula.
Similarly, in higher dimensions we can use the [Dehn-Somerville equations](https://en.wikipedia.org/wiki/Dehn%E2%80%93Sommerville_equations) for simplicial polyhedra to eliminate half of the variables, as smooth toric varieties have simplicial moment polytopes, as pointed out by David Speyer in the comments.
|
12
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https://mathoverflow.net/users/18060
|
408134
| 167,148 |
https://mathoverflow.net/questions/408138
|
8
|
A pair of continuous mappings $f \colon X \to Y$ and $g \colon Y \to X$ is called $\pi\_1$-equivalence if they induce mutually inverse isomorphisms of fundamental groups. Spaces are called $\pi\_1$-equivalent if there is $π\_1$-equivalence between them.
Let $X, Y$ be CW-complexes
1. Is it true that if $f \colon X \to Y$ induces an isomorphism of fundamental groups, then $X$ and $Y$ are $π\_1$-equivalent?
2. Is it true that if $\pi\_1(X)$ is isomorphic to $\pi\_1(Y)$, then $X$ and $Y$ are $\pi\_1$-equivalent?
(added later)
3. Is it true that if $\pi\_1(X)$ is isomorphic to $\pi\_1(Y)$, then there is of a mapping $f \colon X \to Y$ inducing an isomorphism or there is of a mapping $g \colon Y \to X$ inducing an isomorphism?
|
https://mathoverflow.net/users/148161
|
Does the isomorphic of the fundamental groups imply the existence of a mapping inducing an isomorphism?
|
No and no. For an explicit counterexample to 1. (which is also a counterexample to 2.) take the map $\mathbb{R}P^2\to \mathbb{R}P^{\infty}$.
|
24
|
https://mathoverflow.net/users/39747
|
408142
| 167,149 |
https://mathoverflow.net/questions/408021
|
3
|
Consider the fractional Sobolev space
$$
W^{k,2}(\mathbb R^n):=\big\{f \in \mathcal S'\,\big|\,(1+\|\xi\|)^k\hat f(\xi)\in L^2(\mathbb R^n)\big\}
$$
for some $k\in\mathbb R$, and let $\mathcal M$ denote the space of Lebesgue-measurable functions on $\mathbb R^n$ (equivalence classes of functions, where two functions are deemed equivalent if they differ on a set of measure zero), equipped with the [topology of local convergence in measure](https://en.wikipedia.org/wiki/Convergence_in_measure#Topology).
For which values of $k\in\mathbb R$ do we have $W^{k,2}(\mathbb R^n)\subset \mathcal M$?
More formally, for which $k\in\mathbb R$ does the identity map $C^\infty\_c(\mathbb R^n)\hookrightarrow \mathcal M$ extend by continuity to a map $W^{k,2}(\mathbb R^n) \to \mathcal M$?
|
https://mathoverflow.net/users/5690
|
Sobolev embedding into measurable functions
|
My question has been answered in this MathStackexchange post: <https://math.stackexchange.com/questions/4033589/sobolev-space-with-negative-index>
For every $k<0$, there exist a measure $\mu\_k$ which is singular with respect to Lebesgue measure, and such that $\mu\_k\in W^{k,2}(\mathbb R^n)$.
So $W^{k,2}(\mathbb R^n)$ does not embed into the space of Lebesgue-measurable functions.
Thank you Raffaele Scandone.
|
0
|
https://mathoverflow.net/users/5690
|
408156
| 167,154 |
https://mathoverflow.net/questions/407832
|
1
|
Let $X$ be a smooth complex projective variety. The Chow variety of degree $d$, $r$ dimensional subvarieties is denoted by $C\_{d,r}(X)$. The Chow variety can have many topologically connected components (The number of connected components can grow to infinity as $d$ goes to infinity). Each connected component can be a reducible variety with irreducible components of different dimensions.
There is a natural monoidal structure on the Chow variety. We call a complex point on $C\_{d,r}(X)$ indecomposable if it does not lie in the image of $C\_{d\_1,r}(X)\times C\_{d\_2,r}(X)$ for any $d\_1>0$ and $d\_2>0$ where $d\_1+d\_2=d$ under this monoidal operation. The indecomposable points form a Zariski open subset of complex points of each irreducible component. For example when $r=0$, for $d>1$ there are no indecomposable points. Indecomposable points exist only when $d=1$ and it is the $X$ itself. I have two questions:
1. Are the Zariski open subvariety of indecomposable points of irreducible components of $C\_{d,r}(X)$ necessarily smooth? If not how bad the singularity can be? Are they local complete intersections?
2. Define the indecomposable dimension of $C\_{d,r}(X)$ as the minimum of dimensions of irreducible components that happen to consist of only indecomposable points, if there are no such components then we set it to infinity (for example when $r=0$ there is only $X$ in degree $1$ and for $d>1$ this dimension is infinite). Does dimension grow to infinity as $d$ goes to infinity?
|
https://mathoverflow.net/users/127776
|
Singularities of Chow varieties
|
I am just posting my comments as an answer.
**Question 1 is false.** Here is a variant of my comment. Consider the dense open subscheme of the Chow variety parameterizing degree-$3$ curves in projective $n$-space that are (geometrically) irreducible and reduced. This has one irreducible component that is a dense open subscheme of the total space of a projective space bundle of relative dimension $9$ over the Grassmannian parameterizing $2$-planes in the projective $n$-space. The fibers parameterize the space of irreducible plane cubics in the corresponding $2$-plane. So the total dimension of this irreducible component equals $3(n-2) + 9 = 3n+3$.
There is a second irreducible component that has a dense open subscheme that is a dense open in the total space of a Zariski locally trivial fiber bundle of relative dimension $12$ over the Grassmannian of $3$-planes. The fibers parameterize linearly nondegenerate, irreducible and reduced, degree-$3$ curves of arithmetic genus-$0$ in the corresponding $3$-plane. Since the Grassmannian parameterizing $3$-planes in projective $n$-space has dimension $4(n-3)$, the total dimension of this irreducible component equals $4(n-3)+12 = 4n$.
These two irreducible components intersect along the locus parameterizing nodal plane cubics. This locus is nonempty, irreducible, and has dimension $3n+2$. Every local complete intersection scheme, indeed every Cohen-Macaulay scheme, is pure-dimensional. Thus, when $n\geq 4$, the Chow variety parameterizing degree-$2$ curves in projective $n$-space is not locally a complete intersection, it is not Gorenstein, it is not Cohen-Macaulay, and it is not even pure-dimensional since it has two intersection components of differing dimension $3n+3$ and $4n$.
**Question 2 is false.** The blowing up of the projective plane along the base locus of a very general pencil of plane cubics is a surface that admits a morphism to the projective line (the "base" of the "pencil") whose geometric fibers are at-worst-nodal, integral curves of arithmetic genus equal to $1$. However, there are infinitely many $(-1)$-curves in this surface. The $9$ base points give $9$ cross-sections of the morphism to the projective line. Since this is a genus-$1$ fibration, we have an action on the smooth locis of the fibers by the degree-$0$ relative Picard. The differences between pairs of the cross-sections gives a free Abelian subgroup of the relative Picard of rank $8$, and this countable group acts on the set of $(-1)$-curves to give a countably infinite collection of $(-1)$-curves. Each of these curves gives an isolated point of the Chow variety of curves (or the Hilbert scheme, or the space of stable maps, etc.). By quasi-compactness of the components of the Chow variety with fixed degree (or the Hilbert scheme with fixed Hilbert polynomial, etc.), only finitely many of these isolated points can lie in a single Chow variety of fixed degree. So there are infinitely many positive degrees such that the corresponding Chow variety has an isolated point.
|
2
|
https://mathoverflow.net/users/13265
|
408162
| 167,157 |
https://mathoverflow.net/questions/408151
|
0
|
Let $X,Y,Z$ be smooth connected manifolds and $f \colon X \times Y \rightarrow Z$ a smooth map. Suppose that we have $H\_{\*}(X \times Y; \mathbb{Z})$ is isomorphic to $\bigoplus\_{p+q=\*}(H\_{p}(X; \mathbb{Z}) \otimes H\_{q}(Y; \mathbb{Z}))$ by the Künneth theorem, which can be assumed if $H\_{i}(Y; \mathbb{Z})$ has no torsion for any $i$.The map $f$ induces a map on homologies, $f\_\* \colon H\_{\*}(X \times Y; \mathbb{Z}) \rightarrow H\_{\*}(Z; \mathbb{Z})$.
My question is whether one can interpret the image of $f\_{\*}$ in terms of $\bigoplus\_{p+q=\*}(H\_{p}(X; \mathbb{Z}) \otimes H\_{q}(Y; \mathbb{Z}))$. More precisely, for $a \otimes b \in H\_{p}(X; \mathbb{Z}) \otimes H\_{q}(Y; \mathbb{Z})$, we have $a \times b \in H\_{p+q}(X \times Y; \mathbb{Z})$, where $\times$ denotes the cross product. Then is it possible to describe $f\_{\*}(a \times b)$ in terms of $f\_{\*}i\_{\*}(a) \in H\_{p}(X \times Y; \mathbb{Z})$ and $f\_{\*}j\_{\*}(b) \in H\_{q}(X \times Y; \mathbb{Z})$. Here $i,j$ are canonical inclusions of $X, Y$ into $X \times Y$, respectively. In particular, I would be very happy if I can say $f(a \times b)=0$ for any $a$ and $b$ with $f\_{\*}j\_{\*}(b)=0$.
Please let me know anything you know about this question.Thank you in advance.
|
https://mathoverflow.net/users/41200
|
Künneth formula and induced map in homologies
|
Here is an example which will not make you very happy. There is a degree one map $f:S^2\times S^1 \to S^3$ which just collapses the complement of an embedded open disk. Take $a\in H\_2(S^2;\mathbb{Z})$ and $b\in H\_1(S^1;\mathbb{Z})$ to be generators. Then $f\_\*(a\times b)\in H\_3(S^3;\mathbb{Z})$ is a generator, while $f\_\*j\_\*(b)\in H\_1(S^3;\mathbb{Z})$ is zero.
This obviously generalizes to a degree one map $X\times Y\to S^n$ whenever $\operatorname{dim}(X) + \operatorname{dim}(Y)=n$ and $\operatorname{dim}(X),\operatorname{dim}(Y)>0$.
|
6
|
https://mathoverflow.net/users/8103
|
408163
| 167,158 |
https://mathoverflow.net/questions/408169
|
4
|
I am interested in when a Du Val surface singularity is smoothable. By Du Val singularity, I mean (the germ of a) isolated double point surface singularity admitting a resolution by blowups of isolated double points over the original. By smoothable, I mean when does there exist a flat family containing the singularity as one of the fibers, and a smooth germ as another fiber.
Any solution/reference would be appreciated.
|
https://mathoverflow.net/users/103594
|
Are Du Val singularities smoothable?
|
Du Val singularities are indeed smoothable and, in fact, more is true: they are *of class* $T$, namely, they are quotient singularities admitting a $1$-parameter $\mathbb{Q}$-Gorenstein smoothing.
See Definition 1.1 and Proposition 1.2 in the paper
M. Manetti: [On the moduli space of diffeomorphic algebraic surfaces](http://dx.doi.org/10.1007/s002220000101), *Invent. Math*. **143**, No. 1, 29-76 (2001). [ZBL1060.14520](https://zbmath.org/?q=an:1060.14520),
where they are called *rational double points*.
|
8
|
https://mathoverflow.net/users/7460
|
408172
| 167,161 |
https://mathoverflow.net/questions/408175
|
8
|
One of the standard examples of a forcing which is distributive but not proper (equiv. not closed) is a club shooting into a stationary co-stationary set.
But that forcing is $S$-proper for the stationary set into which we shoot the club. So it's at least a little bit proper. I want more. I am looking for an example of a forcing that is not $S$-proper for any stationary set $S$.
Of course, we can be silly and take the lottery sum over all possible club shooting, that is, first we choose a stationary set, then we shoot a club into it. Since no stationary set is guaranteed to survive, this is not $S$-proper for any stationary set. But this is silly, since the forcing is "locally $S$-proper for some $S$". That is, every condition can be extended such that below the extension the forcing is $S$-proper for some $S$.
So what I am really looking for is a forcing that is both distributive and not $S$-proper below any condition for any stationary set $S$.
|
https://mathoverflow.net/users/7206
|
Example of a distributive forcing which is entirely improper
|
$\mathbb P$ being $S$-proper means that for all countable $M \prec H\_\theta$ with $M \cap \omega\_1 \in S$, every $p \in \mathbb P \cap M$ can be extended to an $M$-generic condition.
A positive answer to your question is given by [work of Gitik](https://mathscinet.ams.org/mathscinet-getitem?mr=2648158), who showed that relative to a supercompact, there can exist an inaccessible $\kappa$ and a $\kappa^+$-saturated normal ideal concentrating on $\mathrm{cof}(\omega\_1)$. Let $\mathbb P = \mathcal P(\kappa)/I$ for such an ideal $I$.
The reason this $\mathbb P$ doesn't add $\omega$-sequences is that it forces an embedding $j : V \to M \subseteq V[G]$ with $M^\kappa \cap V[G] \subseteq M$. It can't add bounded subsets of $\kappa$, and by the saturation, being $(\omega,\infty)$-distributive is equivalent to being $(\omega,\kappa)$-distributive, which is equivalent to not adding $\omega$-sequences in $\kappa$. Since it forces $\mathrm{cf}(\kappa)=\omega\_1$, adding an $\omega$-sequence in $\kappa$ means adding a bounded one, which it does not.
Now since $\mathbb P$ makes $\kappa$ a singular cardinal, it must kill the stationarity of some $T \subseteq \kappa \cap \mathrm{cof}(\omega)$, since it adds a club of smaller cardinality than some partition into $\kappa$-many stationary sets. Let $\dot C$ be a name for a club in $\kappa$ forced to be disjoint with $T$. Now let $S \subseteq \omega\_1$ be stationary and let $N \prec H\_\theta$ be a countable elementary submodel with $N \cap \omega\_1 \in S$ and $\sup(N\cap\kappa) \in T$, where $\theta \gg \kappa$ and $\dot C \in N$. If $q$ is $N$-generic, then it forces $\sup(N \cap \kappa) \in \dot C$, contradiction. Thus $\mathbb P$ cannot be $S$-proper.
I am curious whether this can be done with less strength or with easier constructions than Gitik's.
**EDIT:** Here’s an example without large cardinals. The argument above shows that if a forcing $\mathbb P$ is $S$-proper for some stationary $S \subseteq \omega\_1$, then it must preserve all stationary $T \subseteq \kappa \cap \mathrm{cof}(\omega)$ for regular $\kappa>\omega\_1$.
There is a two-step iteration $\mathbb P \* \dot{\mathbb Q}$ that is countably closed such that $\mathbb P$ adds a $\square\_{\omega\_1}$-sequence without adding subsets of $\omega\_1$, and then $\dot{\mathbb Q}$ threads it with an ordertype-$\omega\_1$ club in $\omega\_2^V$. So $\dot{\mathbb Q}$ is forced to be countably distributive.
If $\langle C\_\alpha : \alpha < \omega\_2 \rangle$ is the square sequence added by $\mathbb P$, then for unboundedly many $\gamma<\omega\_1$, the set $T\_\gamma = \{ \alpha : \mathrm{ot}(C\_\alpha) = \gamma \}$ is stationary in $\omega\_2$.
Let $D$ be the thread added by $\mathbb P \* \dot{\mathbb Q}$, and let $D’$ be the limit points of $D$. Then $D’$ intersects each $T\_\gamma$ in at most one point. For if $\alpha \in D’ \cap T\_\gamma$, then $D \cap \alpha = C\_\alpha$ and $\mathrm{ot}(D \cap \alpha) = \gamma$, which holds for exactly one point in $D$. Thus the stationarity of some $T\_\gamma$ has been killed by $\mathbb Q$.
|
9
|
https://mathoverflow.net/users/11145
|
408180
| 167,164 |
https://mathoverflow.net/questions/408170
|
3
|
Optimizing the spectral norm of some positive semidefinite matrix $A(x) \in S^{n}$, w.r.t. a list of variables $x \in \mathbb{R}^d$ and semidefinite constraints is, in general, a nonconvex problem (ref.: [Can one maximize the spectral norm of a matrix via semidefinite programming?](https://mathoverflow.net/questions/110496/can-one-maximize-the-spectral-norm-of-a-matrix-via-semidefinite-programming))
Example:
$$
\max \;\;\left\lVert\begin{pmatrix}
1 & x\_1 & x\_2\\
x\_1 & 1 & x\_3 \\
x\_2 & x\_3 & 1
\end{pmatrix}\right\rVert\_2 \\
\text{s.t. } \;\; 0 \leq x\_1, x\_2, x\_3 \leq 1
$$
What other approaches are possible?
For instance, are there relaxations that turn the problem convex (e.g., an upper bounds w.r.t. some quantity that can be optimized). Or can it be modeled as a hierarchy of conditions (as is possible, for example, with polynomial optimization)?
Any reasonably good upper bound would be helpful.
|
https://mathoverflow.net/users/455887
|
Relaxations for the spectral norm maximization problem
|
Minimizing a concave function subject to convex constraints is Concave Programming.
If the constraints of a Concave Programming problem are compact, as in your example, there must be a global optimum at an extreme of the constraints. In this example, if any semidefinite constraints are ignored, the extreme point of the constrains are the 8 vertices having $x\_i = $ 0 or 1. So evaluating the objective function at all 8 points and picking the largest objective value will produce the global optimum. This is a viable method if the number of variables is not too high.
If there are (also) semidefinite constraints, explicit enumeration of all extreme points will not be possible. However, eliminating the semidefinite constraints constitutes a relaxation of the original problem, and therefore provides an upper bound on the optimal objective value of the original problem. If the optimal solution of the relaxation satisfies the semidefinite constraints, then the solution to the relaxed problem is tight, i.e., solves the original unrelaxed problem. If not, this method does not provide information on how tight the upper bound is. In your particular example, the relaxed problem has optimal objective value 3, achieved at $x\_1 = x\_2 = x\_3 =1$, which produces a positive semidefinite matrix, hence the solution to the relaxed problem solves the original unrelaxed problem.
If you can generate a series of linear (affine) inequalities which constitute a relaxation of one or more of the semidefintie constraints, the above method can still be applied, and may provide a tighter upper bound on the original problem.
Note: If the constraints are not compact and convex, there need not be a global optimum at an extreme of the constraints.
|
1
|
https://mathoverflow.net/users/75420
|
408183
| 167,165 |
https://mathoverflow.net/questions/408174
|
2
|
Let $X$ be a scheme. For $Y$ a scheme over $X$, the representable presheaf $h\_Y : U\mapsto \mathrm{Hom}\_X(U,Y)$ on the small étale site $X\_{et}$ is actually a sheaf, and by the Yoneda lemma the mapping $Y\to h\_Y$ is fully faithful when restricted to the category of schemes étale over $X$. Is there a bigger subcategory of the schemes over $X$ such that it is still fully faithful ?
Similarly, what about algebraic spaces over $X$ ? Seing those as sheaves on the big fppf site of $X$, we can restrict them to the small étale site. This is operation is fully faithful when done on algebraic spaces that are étale schemes, can we get a bigger subcategory for which it is still fully faithful ?
|
https://mathoverflow.net/users/138396
|
Extending the domain of the yoneda embedding map from étale schemes to the small étale topos so that it is still fully faithful
|
The good setting to answer this question really is that of algebraic spaces. Then I believe that the answer is no, basically because every sheaf $F$ on the small étale site is representable by an étale $X$-algebraic space. The algebraic space that represents $F$ is constructed as follows: consider the collection of pairs $(V,x)$ with $V$ an affine étale $X$-scheme and $x\in F(V)$. Let $U$ be the disjoint sum of the $V$'s indexed by this collection. If $(V,x)$ and $(V',x')$ are two pairs, the natural morphism $V\times\_FV'\to V\times\_XV'$ is an open immersion. This shows that $U\times\_FU$ is open in $ U\times\_X U$ and hence defines an étale equivalence relation on $U$. The $X$-algebraic space we're looking for is the quotient of this equivalence relation.
This is due to Michael Artin, reference: his book "Théorèmes de représentabilité pour les espaces algébriques", chap. VII. Le théorème de finitude en cohomologie étale, §1. Définition des faisceaux constructibles.
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4
|
https://mathoverflow.net/users/17988
|
408186
| 167,167 |
https://mathoverflow.net/questions/408184
|
2
|
Consider the following wave-type equation,
$$u\_{tt}-\frac{2}{t}u\_t-\Delta u=g(t,x)$$
where $(t,x)\in [\epsilon, 1]\times \mathbb{R}^3$ for some $\epsilon>0.$ Furthermore assume that $(u,u\_{t})=(0,0)$ at $t=\epsilon.$ My goal is to estimate the energy $E(u\_t) = \int (u\_{tt})^2 + |\nabla u\_t|^2.$ I already know that $E(u) \leq C t^6$ and the constant depends on the function $f.$
As usual, we first obtain the equation satisfied by $\tilde{u} =u\_t$
$$\tilde{u}\_{tt} - \frac{2}{t}\tilde{u}\_{t}-\Delta \tilde{u} = g\_t -\frac{2}{t^2}\tilde{u}.$$
Thus the time derivative of the energy can be estimated as follows,
\begin{align\*}
\dot{E} &= 2\int \tilde{u}\_t (\tilde{u}\_{tt}-\Delta \tilde{u}) \\
&= 2\int \tilde{u}\_t (2t^{-1}\tilde{u}\_t+g\_t -2t^{-2}\tilde{u})\\
&\leq \frac{10}{t}E + \frac{t}{2}\|g\_t\|\_{L^2}^2 + \frac{1}{t^3} \|u\_t\|\_{L^2}^2 \\
&\leq \frac{10}{t}E + \frac{t}{2}\|g\_t\|\_{L^2}^2 + C t^{3}.
\end{align\*}
I am not sure how to proceed after this step. Ideally, I would apply Gronwall and get $E(t)\leq C t^{10}$ but the term with $t^{3-10}$ will create singularity as $\epsilon\to 0$.
|
https://mathoverflow.net/users/68232
|
How to estimate higher order regularity for wave type equation with time dependant coefficients?
|
$$ \tilde{u}\_{tt} - \frac{2}{t}\tilde{u}\_{t}-\Delta \tilde{u} = g\_t -\frac{2}{t^2}\tilde{u} $$
$$ \dot{E} = 2 \int \tilde{u}\_t (2 t^{-1} \tilde{u}\_t + g\_t - 2 t^{-2} \tilde{u} ) $$
$$ \dot{E} = 2 \int \tilde{u}\_t (2 t^{-1} \tilde{u}\_t + g\_t) - 2 t^{-2} \frac{d}{dt} \int \tilde{u}^2 $$
$$ \dot{E} + \frac{d}{dt} (2 t^{-2} \int \tilde{u}^2 ) = 2 \int \tilde{u}\_t (2 t^{-1} \tilde{u}\_t + g\_t) - 4 t^{-3} \int \tilde{u}^2 $$
If you are bounding with $\leq$, the final term on the RHS now has a good sign (is negative) and can be dropped. And the first two terms can be bounded by $\frac{6}{t} E + \frac{t}{2} \|g\_t\|\_{L^2}^2 $. Absorb the $\int \tilde{u}^2$ term into $E$ you can then Gronwall that instead.
|
2
|
https://mathoverflow.net/users/3948
|
408189
| 167,168 |
https://mathoverflow.net/questions/408204
|
9
|
Let $L$ be any differential operator (not necessarily linear).
Given initial conditions and boundary conditions (of any type), I am interested in general statements of the form:
Given a boundary value problem on some domain $\Omega$. If $L$, the initial condition, and the boundary are of a certain form, then separation of variables will yield the solution to the boundary value problem.
Do such general statements exist, and if they do, where can I find them? I am interested in statements about strong (classical) solutions and weak (distributional) solutions.
I am aware that questions similar to this one are already posted on this site, however, all of them consider specific differential operators and then they focus on separability of the domain. I am interested in statements about general differential operators.
|
https://mathoverflow.net/users/114299
|
General validity of separation of variables
|
The short answer is No. A major problem is that there is no single universal definition for what it means for an equation to be solvable by separation of variables. This defect in the theory, as well as the lack of general statements of the form that you would like to see, was highlighted a while ago in this BAMS book review:
>
> Tom H. Koornwinder. "Review: Willard Miller, Jr., Symmetry and separation of variables." [Bull. Amer. Math. Soc. (N.S.) 1 (6) 1014 - 1019, November 1979](https://www.projecteuclid.org/journals/bulletin-of-the-american-mathematical-society-new-series/volume-1/issue-6/Review-Willard-Miller-Jr-Symmetry-and-separation-of-variables/bams/1183544920.full).
>
>
>
Unfortunately, the situation has not changed so much since then. Koornwinder himself proposed a general definition for an equation to be *separable* and you can find a few papers that refer to it or extend it by looking up papers that cite this review. But there are no really general results, as far as I am aware.
|
10
|
https://mathoverflow.net/users/2622
|
408206
| 167,170 |
https://mathoverflow.net/questions/408136
|
3
|
Let $\mathrm{С}$ be some class of topological spaces that includes at least all subspaces of $\mathbb{R}^n $. Further we are in the category $\mathrm{С}\_{\*}$ (the category of point spaces; all continuous maps and homotopies preserve the chosen base points). We define $π'\_1(X) $ as the set of homotopy classes of products of injective loops in $X$ with the operation $[a] \cdot [b] = [a \cdot b]$. We say that $\mathrm{D} \subset \mathrm{C}$ is essentially wide if for every $X \in \mathrm{C}$ there exists $Y \in \mathrm{D}$ such that $X$ is homotopy equivalent to $Y$.
Is there a $ D $ essentially wide subclass of $ C $ such that, for $ X \in D $
1. The operation is well defined, i.e. there always exist an injective loop homotopic to $a \cdot b$
2. The natural embedding of $π'\_1 (X) \to π\_1 (X)$ is an isomorphism.
We define $π''\_1 (X)$ similarly, but with homotopies in the class of injective loops.
Is there a $ D $ essentially wide subclass of $ C $ such that, for $X \in D$
3. The operation is well defined, i.e. among the classes of injective homotopy there exists and is uniquely a class of loops freely homotopic $a \cdot b$
4. The natural embedding $π''\_ 1(X) \to π\_1 (X)$ is an isomorphism.
|
https://mathoverflow.net/users/148161
|
Can the loops in the definition of the fundamental group be considered injective?
|
In the [Griffiths Twin Cone](https://wildtopology.com/2014/06/28/the-griffiths-twin-cone/) (or double cone over the shrinking wedge of circles), $G\subseteq \mathbb{R}^3$, all injective loops are null-homotopic yet $\pi\_1(G)$ is uncountable. Hence, injective loops don't generate any of the fundamental group. However, as @WillSawin mentions in the comments, it is possible that you can pass to $G\times C$ where $C$ is contractible. This should mean that your first question has a positive answer at least in the case where $C$ is the set of subsets of $\mathbb{R}^n$.
Specifically, given a space $X$ in your class of spaces, we can consider $Y=X\times [0,1]^2$ which is homotopy equivalent to $X$ by the projection and seems to be in the class you are interested in. Given a loop $\alpha:(S^1,\ast)\to (X,x\_0)$, we can define paths $\beta\_1,\beta\_2:[0,1]\to [0,1]^2$ so that $\beta\_1(t)=(t,0)$ and $\beta\_2:[0,1]\to [0,1]^2$ is some other arc from $(0,0)$ to $(1,0)$ that only meets $[0,1]\times\{0\}$ at its endpoints. Now consider the path $(\alpha(t),\beta\_1(t))$ in $Y$ and follow it by the path $(\alpha(\ast),\beta\_{2}(1-t))$. This concatenation defines an injective loop in $Y$ whose projection to $X$ is homotopic to $\alpha$. Hence, the injective loops in $Y$ generate the entire fundamental group of $Y$.
|
4
|
https://mathoverflow.net/users/5801
|
408211
| 167,173 |
https://mathoverflow.net/questions/408181
|
0
|
It is well known that, for a dynamical system $T$ on a metric space $(X,d)$, the variational principle connects the definition of metric entropy and topological entropy. In other words,
if
$$M(X,T) := \{ \mu\,\, \text{probability measure} : \mu= T\_\*\mu \} $$
is the set of invariant measures for $T$, then
$$h\_\text{top}(
T)= \sup\_{\mu \in M(X,T)} h\_{\mu}(T) $$
where $h\_\text{top}(T)$ is the topological entropy and $h\_{\mu}(T)$ is the metric entropy relative to $\mu$.
I have seen somewhere that, if we denote by $E(X,T) \subset M(X,T)$ the set of invariant ergodic measures for $T$, then
$$h\_\text{top}(T)= \sup\_{\mu \in E(X,T)} h\_{\mu}(T) $$
My questions are: is this true? If it is true, how is it proven?
|
https://mathoverflow.net/users/297294
|
Metric entropy and topological entropy
|
It's true for a very simple reason: the entropy of a dynamical system with respect to a (not necessarily ergodic) invariant measure is the average of the entropies of its ergodic components.
|
2
|
https://mathoverflow.net/users/8588
|
408213
| 167,174 |
https://mathoverflow.net/questions/407731
|
12
|
The Airy differential equation
$$y''(x)\ = \ xy(x)$$
is one of the simplest *irregular* differential equations (so not determined by its monodromy data, there is more structure, the Stokes data). Associated to this, Katz constructs an *Airy sheaf* $\text{Ai}$ on the affine line $\mathbf{A}^1\_{\mathbf{F}\_q}$; $\text{Ai}$ is the Fourier transform of some nonconstant lisse sheaf. However, I don't know anything about Katz's theory, and this question is an attempt to learn more about it, in an example.
**Question:** What is the evidence that this is the correct sheaf analogue of the Airy differential equation? e.g.
1. (How) does this $\ell$-adic sheaf see the Stokes data of the differential equation?
2. You can write the Airy DE as a limit of DE's which have regular singular points. Does the Katz construction work in families i.e. apply it to the family of DE's $P\_ty(x)=0$ where setting $t=0$ gives back Airy?
3. Other properties I might not have thought of.
|
https://mathoverflow.net/users/119012
|
Katz's $\ell$-adic Airy sheaf
|
(1.) The part of the Stokes data we can see is the restriction of the differential equation to the field of formal Laurent series at each point. There is a classification of differential equations over this field, and a classification of Galois representations of a field of formal Laurent series over a finite field, and the two classifications are pretty similar.
When we compare these classifications, the Airy sheaf and Airy differential equation match up at each point. The most important aspect of this is the slopes - the Airy differential equation has slopes $3/2$ at $\infty$, and so does the Airy sheaf, for the suitable definition of slope in each context. Furthermore, both objects have no singularities away from infinity.
(2.) Doesn't this work, by a similar argument, for every differential equation with regular singularities?
Deformations of sheaves are a bit more restricted then deformations of differential equations. In fact, we have two different notions of deformations of sheaves. In one, the deformation space has coordinates in the coefficient field of the sheaf ($\ell$-adic, in this case), while in the other, the deformation space has coordinates in the base field of the variety the sheaf lives on (a field of characteristic $p$, in this case). Only by combining the two can we obtain as much power as deformations of a differential equation, where both relevant fields are $\mathbb C$ and there's a single complex variety parameterizing both types of deformations.
(3.) Here are some additional properties:
a. The Airy differential equation is preserved by the change of variables $x \mapsto \omega x$ where $\omega^3=1$. The Airy sheaf is preserved under the same change of variables.
b. The Fourier transform of the Airy differential equation (i.e. the differential equation satisfied by the Fourier transform of the Airy function) is a differential equation of rank $1$, with no singularities except $\infty$, with slope $3$ at $\infty$. The Fourier-Deligne transform of the Ary sheaf is a sheaf of rank $1$, with no singularities except $\infty$, with slope $3$ at $\infty$.
In fact, properties a and b characterize the Airy differential equation up to scaling $x$ (and changes of variables in $y$), and they also characterize the Airy sheaf up to scaling $x$.
c. The image of the action of the differential Galois group on solutions of the Airy differential equation is $SL\_2$, and the monodromy group of the Airy sheaf (i.e. the Zariski closure of the image of the Galois group on sections of the Airy sheaf) is also $SL\_2$.
|
6
|
https://mathoverflow.net/users/18060
|
408214
| 167,175 |
https://mathoverflow.net/questions/408131
|
2
|
Let $\langle X, X' \rangle$ be a dual pair equipped with the weak and weak\* topologies.
Let $C$ be a weak\* compact subset of $X'$ with nonempty interior. For each $x \in X$, let $M(x)$ be the set of minimisers of $\langle x, \cdot \rangle$ on $C$. Each $M(x)$ is non-empty, convex, and closed.
My question is: What additional conditions on $C$ imply that $M$ admits a continuous selection, i.e. a continuous function $f: X \to X'$ such that $f(x) \in M(x)$?
|
https://mathoverflow.net/users/96899
|
How to choose minimisers in a continuous way
|
I think the answer is: only in the trivial case when each $M(x)$ is a singleton $\{f(x)\}$. Indeed, if $M(x)$ has more than one point, it has at least $2$ extremal points. An extremal point of a convex compact set $C$ of a LCTVS (here, $X’$ with its weak\* topology) is the unique minimiser on $C$ for some continuous linear form (thus, here, an evaluation). So for $i\in\{1,2\}$ there are $x\_i$ in $X$ and $y\_i$ in $M(x)$ such that $M(x\_i)=\{y\_i\}$. Moreover, $M(x\_i)=M(tx\_i+(1-t)x)$ for any proper convex combination of $x\_i$ and $x$. This forbids the existence of a continuous selection for $M$, for any selection has to jump at $x$ from $y\_1$ to $y\_2$.
In fact, for any $x\in X$ and for any selection $f$ of $M$ the whole set $M(x)$ can be recovered as closed convex hull of the discontinuity set of $f$ at $x$:
$$M(x)=\bigcap\_{U\text{ nbd of } x}\overline{\text{co}}f(U)= \overline{\text{co}}\Big( \bigcap \_{U\text{ nbd of } x} f(U)\Big).$$
In particular, there exists a continuous selection of the multi $M$ if and only if there is only one selection, that is $M(x)$ is a singleton for all $x$: in other words, every point of $\partial C$ is extremal, that is $C$ is strictly convex ($\partial C$ contains no non-empty open segment).
|
2
|
https://mathoverflow.net/users/6101
|
408216
| 167,177 |
https://mathoverflow.net/questions/408194
|
13
|
The sequence that is addressed here is resourced from the most useful site OEIS, [listed as A014153](https://oeis.org/A014153), with a generating function
$$\frac1{(1-x)^2}\prod\_{k=1}^{\infty}\frac1{1-x^k}.$$
In particular, look at these two interpretations mentioned there.
$a(n)=$ Number of partitions of $n$ with three kinds of $1$.
**Example.** $a(2)=7$ because we have $2, 1+1, 1+1', 1+1'', 1'+1', 1'+1'', 1''+1''$.
$b(n)=$ Sum of parts, counted without multiplicity, in all partitions of $n$.
**Example.** $b(3)=7$ because the partitions are $3, 2+1, 1+1+1$ and removing repetitions leaves $3, 2+1, 1$. Adding these shows $b(3)=7$.
I could not resist asking:
>
> **QUESTION.** Can you provide a combinatorial proof for $a(n)=b(n+1)$ for all $n$.
>
>
>
|
https://mathoverflow.net/users/66131
|
Two interpretations of a sequence: an opportunity for combinatorics
|
Here is a bijective proof that equates both $a(n)$ and $b(n+1)$ to the quantity
$$p(n)+2p(n-1)+\cdots+np(1)+(n+1)p(0) \tag1$$
where $p(n)$ is the number of partitions of $n$.
For $a(n)$: each partition with three types of $1$'s corresponds to a partition of $k$ together with $(n-k)$ parts from $\{1',1''\}$. This second gadget can be chosen in $(n-k+1)$ ways, so we get $a(n)=\sum\_{k=0}^n p(k)(n-k+1)$.
For $b(n+1)$: Let us count the number of times that a summand "$k$" appears, for $1\le k\le n+1$. It only appears in partitions that have at least one part equal to $k$. The number of such partitions is $p(n+1-k)$. So the total sum calculating $b(n+1)$ is $\sum\_{k=1}^{n+1}kp(n+1-k)$.
**Remark.** The origin of (1) is this: $\frac1{1-x}\prod\_{k\geq1}\frac1{1-x^k}=\sum\_{m\geq0}\sum\_{k=0}^mp(k)x^n$ and hence $\frac1{(1-x)^2}\prod\_{k\geq}\frac1{1-x^k}=\sum\_{n\geq0}\sum\_{m=0}^n\sum\_{k=0}^mp(k)x^n$. Furthermore,
$$\sum\_{m=0}^n\sum\_{k=0}^mp(k)=(1).$$
|
18
|
https://mathoverflow.net/users/2384
|
408219
| 167,178 |
https://mathoverflow.net/questions/408146
|
4
|
I want
\begin{equation}
\int\_a^b f(g(x))dg=\int\_{g(a)}^{g(b)} f(t)dt,\tag{$\heartsuit$}\label{heart}
\end{equation}
where $g$ is a continuous but not necessarily monotone (or bounded variation) function of $[a,b]$, both integrals are Riemann–Stieltjes (that is, $\int\_a^b h(x)dg(x)$ is the limit of Riemann–Stieltjes sums $\sum\_{i=0}^{n-1} h(c\_i)(g(x\_{i+1})-g(x\_i))$ where $a=x\_0<x\_1<\dotsb<x\_n=b$ is a partition of $[a,b]$, $c\_i\in [x\_i,x\_{i+1}]$ for $i=0,\dotsc,n-1$, and $\max \lvert x\_{i+1}-x\_i\rvert$ goes to 0).
If both integrals in \eqref{heart} exist, does it follow that they are equal?
Note that without continuity of $g$ this is not necessarily so, even if $g$ is increasing: say, if $a=0$, $b=1$, $g(x)=\chi\_{[1/2,1]}$, and $f(0)=f(1)=0$, then LHS is 0 but RHS may be non-zero. For increasing continuous $g$ this is less or more obvious. But we sometimes would like to have not monotone change of variables (cf. $\int\_a^b f(x^2)x\,dx=\frac12 \int\_{a^2}^{b^2} f(t)dt$ with $a<0<b$).
If not, what are additional conditions under which \eqref{heart} holds?
|
https://mathoverflow.net/users/4312
|
Change of variables in Riemann–Stieltjes integral
|
Looks like the answer is yes, they are equal, and perhaps you do not really need to assume the existence of the integral in the right-hand side, that is, $\int\_{g(a)}^{g(b)} f(t) dt$. This is proved in the context of Henstock–Kurzweil integrals as Theorem 6.1 in [1].
The key lemma is that if $g(a) = g(b)$, then $\int\_a^b f(g(x)) dg(x) = 0$. This is applied to reduce the problem to the case when $g$ is monotone, which is rather well-known.
Reference:
[1] Michael Bensimhoun, *Change of Variable Theorems for the KH Integral*, Real Analysis Exchange 35(1) (2009–2010): 167–194, [DOI:10.14321/realanalexch.35.1.0167](https://doi.org/10.14321/realanalexch.35.1.0167)
|
5
|
https://mathoverflow.net/users/108637
|
408220
| 167,179 |
https://mathoverflow.net/questions/408193
|
6
|
Disclaimer: This question was originally posted in [math.stackexchange.com](https://math.stackexchange.com/questions/4272335/are-there-a-in-depth-classification-of-branch-points-in-complex-analysis) and, after 30 days with no answers, I followed the instructions of [this topic](https://meta.mathoverflow.net/questions/2637/cross-posts-to-math-se/2638#2638).
In complex analysis we have well known results about isolated singularities. Poles are characterized by ‘nice’ (rational) controlled growth around them and for essential singularities we have the [Great Picard's Theorem](https://en.wikipedia.org/wiki/Picard_theorem).
**Question:** Is there a similar classification for branch points? I mean: a clear list with all possibilities and results that characterizes each case?
For example: if we compare $f(z)=\sqrt z$ and $g(z) = \sin(\ln(z))$, they have very different behavior, one has a well defined limit as we approach $z=0$ in any branch and the other has an accumulation point of zeros. Are there results that characterize the ‘fast oscillations’ of $g(z) = \sin(\ln(z))$ and the ‘calm’ behavior of $f(z)=\sqrt z$? (Maybe in an appropriate Riemann surface.)
|
https://mathoverflow.net/users/424703
|
Is there a, in depth, classification of branch points in complex analysis?
|
Yes, there is a classification. An isolated branch point can be algebraic or logarithmic. If the branch point is at 0, algebraic means that $f(z^n)$ has a pole or removable singularity at 0. It can also have an essential singularity, but this does not have an accepted name. In the case of a logarithmic point
$f(e^z)$ is an (arbitrary) meromorphic function in some left half-plane. Anyway, plugging $z^n$ or $e^z$ (this is called local uniformization) reduces any classification at an isolated branch point to the case of a single valued function.
Literature: on singularities of analytic functions in general,
and their classification, there is a book
P. Dienes,
Leçons sur les singularités des fonctions analytiques; professées à l’université de Budapest,
Paris: Gauthier-Villars, VIII + 172 S. 8∘ (1913),
which is somewhat out of date.
"Isolated branch points" by themselves are
not a subject of a special
study, but "functions whose all singularities are isolated and
may include isolated branch points" play an important role in Ecalle's theory
of "resurgent functions". Jean Ecalle wrote several books on them, but they are very hard to read. Of the many expositions of Ecalle's theory I can mention
C. Mitschi and D. Sauzin,
Divergent series, summability and resurgence I. Monodromy and resurgence.
Lecture Notes in Mathematics 2153, Springer (ISBN 978-3-319-28735-5/pbk; 978-3-319-28736-2/ebook). xxi, 298 p. (2016).
See also arXiv:1405.0356.
|
8
|
https://mathoverflow.net/users/25510
|
408224
| 167,181 |
https://mathoverflow.net/questions/408221
|
5
|
This is a question inspired by T. Amdeberhan's [recent question](https://mathoverflow.net/questions/408194/two-interpretations-of-a-sequence-an-opportunity-for-combinatorics), as well as [another previos MO question](https://mathoverflow.net/questions/127000/partitions-sum-of-divisors-identity).
For an integer partition $\lambda$, and $k\in \mathbb{N}\cup\{\infty\}$, let $|\lambda|\_k$ denote the sum of the parts of $\lambda$, but where we only count each number at most $k$ times. E.g., $|\lambda|\_1$ is the sum of the parts of $\lambda$ after removing repeated parts, and $|\lambda|\_{\infty}=|\lambda|$ is the usual size of the partition.
Define the coefficients $a\_k(n)$ by
$$ \frac{\sum\_{\lambda} |\lambda|\_k \cdot q^{|\lambda|}}{\sum\_{\lambda} q^{|\lambda|}} = \left( \sum\_{\lambda} |\lambda|\_k \cdot q^{\lambda} \right) \cdot \prod\_{i=1}^{n} (1-q^i) = \sum\_{n\geq 0} a\_k(n) q^{n}.$$
From the above-linked questions, we see that $a\_1(n) = n$, while $a\_{\infty}(n) = \sigma(n) = \sum\_{d\mid n} d$, the sum of divisors of $n$. So $a\_k(n)$ give a sequence of numbers which "interpolate" between $n$ and $\sigma(n)$ in some sense.
**Question**: What are these $a\_k(n)$ for arbitrary $k$? Do they have any nice expression in general?
|
https://mathoverflow.net/users/25028
|
Coefficients obtained from ratio with partition number generating function
|
The bijection described by Mark Wildon in the second linked question can be adapted to your generalization. Indeed, $|\lambda|\_k$ counts the number of ways of selecting a box $B$ of $\lambda$ such that, if the row containing $B$ has length $m$, then there are at most $k-1$ other rows of length $m$ above it. By erasing the rows of length $m$ that contain box $B$ or are above the row that contains $B$, we obtain the partition $\mu$. Let $1\le c\le m$ be the column containing $B$. Then for each $m$ we have a bijection $(\lambda, B)\leftrightarrow (\mu, c)$. Which gives
$$\sum\_{\lambda \vdash n}|\lambda|\_k=\sum\_{m=1}^n\sum\_{r=1}^k p(n-rm)m$$
With a little calculation we see that this statement is equivalent to
$$a\_k(n)=n\left(\sum\_{d\le k \,\text{and}\, d|n}\frac{1}{d}\right).$$
In this form, it is easier to see how $a\_k(n)$ interpolates between $n$ and $\sigma(n)$.
|
9
|
https://mathoverflow.net/users/2384
|
408228
| 167,182 |
https://mathoverflow.net/questions/408201
|
9
|
Let $S$ be a set of $2n$ points in $\mathbb{R}^2$. Which is the maximum number of different bi-partitions of $S$ generated by a straight line?
More precisely, which is the maximum number of partitions $S=S\_1 \sqcup S\_2$ such that $|S\_1|=n$ and there exists a line $\ell$ such that $S\_1$ and $S\_2$ and are the intersection of $S$ with the two open half-planes determined by $\ell$.
|
https://mathoverflow.net/users/167834
|
Bi-partitioning $2n$ points on the plane with a straight line
|
These are called halving lines, and we don't know the exact order of their magnitude, just that it is between $\Omega(n2^{\sqrt\log n})$ and $O(n^{4/3})$.
For more information, see <https://jeffe.cs.illinois.edu/open/ksets.html>.
|
11
|
https://mathoverflow.net/users/955
|
408233
| 167,184 |
https://mathoverflow.net/questions/408237
|
4
|
Let $G$ be a reductive group and $X$ a smooth $G$-variety. Then the fixed point subvariety $X^G$ is also smooth (this is theorem 13.1 of Milne's book on algebraic groups). Suppose in addition that the canonical bundle $\omega\_X$ is trivial. Then is $\omega\_{X^G}$ also trivial?
One idea I had was to use the adjunction formula, which gives that $\omega\_{X^G} = \det(I/I^2)^\vee$, where $I$ is the ideal sheaf of $X^G\subset X$. When $X = \text{Spec}(A)$ is affine, we have $I=(g\cdot f - f)\_{g\in G, f\in A}$, but in the general case I'm having trouble giving a sufficiently effective description of $I/I^2$ to calculate its determinant.
|
https://mathoverflow.net/users/334560
|
If $X$ is a smooth $G$-variety with trivial canonical bundle, then does $X^G$ also have trivial canonical bundle?
|
No. Let $C \subset \mathbb{P}^2$ be a smooth sextic curve, let $X$ be the double covering of $\mathbb{P}^2$ ramified over $C$, and let $G = \mathbb{Z}/2$ acting on $X$ be the involution of the double covering. Then $X$ is a K3 surface, hence its canonical bundle is trivial. But $X^G = C$ is a curve of genus $10$, its canonical class is not trivial.
|
14
|
https://mathoverflow.net/users/4428
|
408238
| 167,185 |
https://mathoverflow.net/questions/408208
|
8
|
[This list](http://shpilrain.ccny.cuny.edu/gworld/problems/probhyp.html) of open problems from <http://grouptheory.info/> includes the question:
*"Is every biautomatic group which does not contain any $\mathbb{Z} \times \mathbb{Z}$ subgroups, hyperbolic?"*
It is credited to Gersten but I don't see any mention of it in [the provided reference](https://link.springer.com/chapter/10.1007/978-1-4613-9730-4_10).
The most similar question there seems to be:
*"Let $\phi \in \operatorname{Out}(F)$, where $F$ is a finitely generated free group, and let $G = F \rtimes\_\phi \mathbb{Z}$. Is $G$ automatic?*
*It has also been conjectured that $F \rtimes\_\phi \mathbb{Z}$ is word hyperbolic if it contains no subgroup isomorphic to $\mathbb{Z} \oplus \mathbb{Z}$."*
Which doesn't even seem that related. I haven't been able to find any other mentions of this question or further work.
Since every hyperbolic group is biautomatic, it seems to me to be a natural question to ask when biautomatic groups are hyperbolic.
Is this question still open? Has there been any related work?
What about characterizing when bicombable groups are hyperbolic?
|
https://mathoverflow.net/users/133733
|
When are biautomatic groups hyperbolic?
|
$\DeclareMathOperator\BS{BS}$Since this question goes in several directions, I hope a discursive answer is appropriate.
The question fits into an important family of questions in geometric group theory which look to provide some sort of "algebraic" characterisation of hyperbolic groups. The search for this characterisation starts with two famous properties.
>
> **Proposition:** Let $\Gamma$ be a hyperbolic group. Then:
>
>
> * $\Gamma$ contains no Baumslag–Solitar subgroups; and
> * $\Gamma$ acts properly and cocompactly on an aspherical complex (its Rips complex). In particular, if $\Gamma$ is torsion-free it is of finite type.
>
>
>
(Recall that a Baumslag–Solitar group is given by a presentation $\BS(m,n)=\langle a,b\mid ba^mb^{-1}=a^n\rangle$. In particular, $\BS(1,1)\cong\mathbb{Z}^2$.)
An algebraic characterisation might look like a converse to this proposition. In 1990, the most general form of this question was:
>
> **Question 1:** If $G$ is a finitely presented group with no subgroups isomorphic to $\BS(1,n)$ for any $n$, must $G$ be hyperbolic?
>
>
>
Question 1 was answered in the negative by Noel Brady, who gave an example of a torsion-free finitely presented subgroup $H$ of a hyperbolic group with $b\_3(H)=\infty$; in particular, $H$ couldn't be of finite type. So for the last 20 years, the question has been:
>
> **Question 2:** If $G$ is a group of finite type with no subgroups isomorphic to $\BS(1,n)$ for any $n$, must $G$ be hyperbolic?
>
>
>
This was the first question on Bestvina's famous problem list, and I think there's a good case to be made that it was the most important question in geometric group theory: when $G$ is a 3-manifold group, the answer is "yes" by geometrisation, so it could be thought of as a geometrisation conjecture for groups.
Very excitingly, as Matt Zaremsky mentions in comments, Question 2 has recently been answered in the negative by Italiano–Martelli–Migliorini, who constructed a normal subgroup $K$ of infinite index in a hyperbolic group, which is both of finite type and $4$-dimensional. It follows that $K$ can't be hyperbolic, by deep theorems of Paulin and Rips.
Nevertheless, Question 2 can, and has been, specialised to any of your favourite classes of finite-type groups: CAT(0) groups, (bi)automatic groups, one-relator groups... As far as I know, it remains open in all these cases. However, as Matt again suggests, the important task remains to figure out whether or not Italiano–Martelli–Migliorini's group $K$ has any of these properties, since it may also provide a counterexample to more refined versions of the question.
Let me quickly finish by noting that there are several classes where positive answers to Question 2 are known. These include:
1. $\Gamma$ is a 3-manifold group (Perelman).
2. $\Gamma$ is $F\rtimes\mathbb{Z}$ (Brinkmann).
3. $\Gamma$ is the fundamental group of a special cube complex (Caprace–Haglund).
4. Ascending HNN-extensions of free groups (Mutanguha).
So, in summary, the questions you ask about are open (and important) today, but the recent work of Italiano–Martelli–Migliorini raises the hope that more progress might be possible soon.
(By the way, let me make a quick aside about the attribution of the question. These kinds of questions are indeed sometimes attributed to Gersten – the one-relator version of the question is the one I've heard attributed explicitly to him. But it's clear that many of these famous questions must have been asked verbally in the '80s and '90s, but may not have been written down by the authors they're associated with. For instance, the question of surface subgroups for hyperbolic groups is always attributed to Gromov, but I've never found it explicitly anywhere.)
|
13
|
https://mathoverflow.net/users/1463
|
408239
| 167,186 |
https://mathoverflow.net/questions/408243
|
9
|
I'm looking for a generalization of Nash's embedding theorem (for Riemannian manifolds) to vector bundles with a connection.
Given a smooth manifold $M$ together with a vector bundle $V$ on $M$ equipped with some connection $D$, I want to find an orthogonal bundle $W$ with a *flat* connection $D'$ such that $V\subset W$ is a subbundle and $D$ is induced from $D'$.
Here by "induced" I mean the following: given a section $s$ of $V$ and a vector field, the covariant derivative of $s$ using $D$ should be the composition of the covariant derivative of $s$ (seen as section of $W$) using $D'$ with the projection onto $TV$ using the orthogonal structure of $W$.
A special case is when $M$ is Riemannian. For $V=TM$ and $D$ the Levi-Civita connection, the answer to the question is Nash's embedding theorem: there is an embedding of $M$ into some $\mathbb{R}^N$ such that the connection $D$ is induced by the trivial connection $d$ on $\mathbb{R}^N$.
Is there a result of this kind? Any references?
|
https://mathoverflow.net/users/142627
|
Embedding of a bundle with connection into a bundle with flat connection?
|
The paper “Existence of universal connections” by Narasimhan, M. S.; Ramanan, S. proves that the Grassmanian is universal for connections not just bundles. That is any connection in a U(n) or O(n) bundle is pulled back from the canonical connection in the appropriate grassmanian by a map. Since the canonical connection has the desired form so does so does the original connection. They also estimate the required rank of the complement.
To clarify a bit. The tautological bundle over the Grassmannian
$\gamma\_k\to Gr\_k(\mathbb{R}^N)$ has a complement $\gamma^\bot$ the bundle whose fiber at a subspace $V$ is the ortho-complement of $V$ in $\mathbb{R}^N$. It follows that
$\gamma\oplus\gamma^\bot =Gr\_k(\mathbb{R}^N) \times \mathbb{R}^N$. The connection in this paper is the connection induced from the trivial connection by projection.
|
12
|
https://mathoverflow.net/users/12605
|
408258
| 167,197 |
https://mathoverflow.net/questions/408247
|
5
|
In pseudo-Riemannian geometry it is well known that every three-dimensional Einstein manifold has constant curvature. A proof of this is sketched [here](https://math.stackexchange.com/a/2194463/242708).
**Question**. Does anyone know where in the literature I can find a proof of such result?
|
https://mathoverflow.net/users/74033
|
Proof that every three-dimensional Einstein manifold has constant curvature
|
This is precisely Proposition 1.120 on p.49 in Besse's "Einstein Manifolds" (I am using the reprint of the 1987 edition, so the numbering may be different in the older edition):
>
> A 3-dimensional pseudo-Riemannian manifold is Einstein iff it has constant (sectional) curvature.
>
>
>
Note that the proof is before the proposition.
|
6
|
https://mathoverflow.net/users/1849
|
408259
| 167,198 |
https://mathoverflow.net/questions/408168
|
7
|
The following integral appears naturally within the computation of the Fourier series coefficients of a real analytic $\mathrm{SL}\_2(\mathbb{C})$-Eisenstein series:
\begin{align\*}
\int\_{-\infty}^{\infty}\left(\int\_{-\infty}^{\infty} \frac{(x\_1+ i x\_2)^{\ell}}{(x\_1^2+x\_2^2+a^2)^s} e^{ i x\_1\zeta\_1}dx\_1 \right) e^{ i x\_2\zeta\_2}
dx\_2,
\end{align\*}
where $i=\sqrt{-1}$, $\zeta\_1,\zeta\_2,a\in\mathbb{R}$, $\ell\in\mathbb{Z}\_{\geq 0}$ and $s\in\mathbb{C}$ with $\Re(2s-\ell-1)>1$. Using tables of one dimensional integrals, it is possible to write down this double integral as a "linear combination" of $K$-Bessel functions $K\_{\nu}\left(a\sqrt{\zeta\_1^2+\zeta\_2^2}\right)$ for various order $\nu$ which is very similar to some of the formulas which appear in Friedberg's paper [1], see below.
Here is ma question:
**Q:** Is it possible to express this double integral in a very concise way using an appropriate hypergeometric function (e.g. a Meijer G-function with the appropriate parameters or something alike) ?
>
> [1] S. Friedberg, *On Maass wave forms and the imaginary quadratic Doi-Naganuma lifting*. Math. Ann. 263 (1983), no. 4, 483–508
>
>
>
|
https://mathoverflow.net/users/11765
|
On a certain double integral appearing in the Fourier series coefficients of $\mathrm{SL}_2(\mathbb{C})$-Eisenstein series
|
Following @Abdelmalek's great advice in the comments above it is enough to compute the following triple integral:
\begin{align\*}
I=\frac{\pi^s}{\Gamma(s)}\int\_{0}^{\infty}\left(\int\_{-\infty}^{\infty}\int\_{-\infty}^{\infty}
(x\_1+x\_2 i)^{\ell}\cdot e^{-\pi t(x\_1^2+x\_2^2+a^2)} e^{2\pi i (x\_1\zeta\_1+x\_2\zeta\_2)} dx\_1 dx\_2 \right)
t^s\frac{dt}{t}.
\end{align\*}
So we have
\begin{align\*}
I&=\frac{\pi^s}{\Gamma(s)}\int\_{0}^{\infty}t^{-1}\cdot\left(\frac{i}{t}(\zeta\_1+i\zeta\_2)\right)^{\ell}\cdot
e^{-\frac{\pi}{t}(\zeta\_1^2+\zeta\_2^2)}\cdot e^{-\pi ta^2}t^{s}\frac{dt}{t}\\
&=\frac{(i)^{\ell}\pi^s}{\Gamma(s)}(\zeta\_1+i\zeta\_2)^{\ell}\int\_{0}^{\infty}
e^{-\frac{\pi}{t}(\zeta\_1^2+\zeta\_2^2)-\pi ta^2}\cdot t^{s-\ell-1} \frac{dt}{t} \\
&=\frac{(i)^{\ell}\pi^s}{\Gamma(s)}(\zeta\_1+i\zeta\_2)^{\ell}\cdot 2\left(\frac{|\zeta|}{a}\right)^{s-\ell-1}
K\_{s-\ell-1}\left(2\pi|\zeta|a\right)
\end{align\*}
where $\zeta=\zeta\_1+i\zeta\_2$.
So modulo some minor mistakes made above, out of excitement, it means that the initial messy formulas that I had obtained could be vastly simplified, as what I was hoping for. In any case this new approach is just better and simpler!
I can't wait for somebody to publish a book with tables of double integrals (or even multiple integrals)!
**added** To answer partly to @Abdelmalek's comment below, for the second equality, I'm using the fact that
\begin{align\*}
\widehat{P(x)e^{-\pi t\langle x,x\rangle}}(y)=t^{-n/2}\cdot P(\frac{i}{t}y)e^{-\frac{\pi}{t}\langle y,y\rangle}
\end{align\*}
where $x\in\mathbb{R}^n$ is a length $n$ real vector, $\langle,\rangle$ is the usual inner product on $\mathbb{R}^n$, $P(x)$ is a spherical polynomial with respect to the standard Laplacian and $\widehat{}$ corresponds to the Fourier transform. Of course, using a change a of variable we may obtain a more general formula which applies to any inner product $\langle,\rangle\_Q$ associated to any positive definite quadratic form $Q$.
|
4
|
https://mathoverflow.net/users/11765
|
408260
| 167,199 |
https://mathoverflow.net/questions/408261
|
2
|
Given a finite simple graph $G$ with $n$ vertices, for each 0-1 colouring $\alpha \in \mathbb{Z}\_2^n$ of its vertices consider the subgraph $G\_\alpha$, whose vertices are the $1$-coloured vertices in $\alpha$, created as follows: if two vertices with labels $1$ are adjacent, then add the edge between them.
Is there a way of computing $$\Theta(G) = \sum\_{k = 1}^n (-1)^k \sum\_{\substack{\alpha \in 2^n\\|\alpha|= k}} \# G\_\alpha,$$
where $|\alpha|$ is the sum of the components of $\alpha$ (so the number of vertices with label $1$) and $\#G\_\alpha$ is the number of connected components of $G\_\alpha$.
As an example, for complete graphs we have $\Theta(K\_m)= -1$.
Are the values of $\Theta$ known at least on simple families of graphs, such as the cycle graphs or the $1$-skeletons of the $m$-dimensional cubes?
|
https://mathoverflow.net/users/16507
|
number of components of subgraphs
|
Something to start with.
We have $$\Theta(G)=\sum\_{\alpha} (-1)^{\alpha}\#G\_\alpha=\sum\_{\alpha}(-1)^\alpha\sum\_U \mathbf{1}(U\,\text{is a component of}\,G\_{\alpha}),$$
where $U$ runs over all induced connected subgraphs of $G$. Changing the order of summation, we get
$$
\Theta(G)=\sum\_U \sum\_{\alpha:\, U\,\text{is a component of}\,G\_{\alpha}} (-1)^{\alpha}.
$$
Look at the inner sum. Let $\tilde{U}$ denote the set of vertices $v\in V\setminus U$ (here $V=$set of vertices of $G$), which have a neighbour in $U$, and $U^\*$ denote $V\setminus (U\cup \tilde{U})$. Then $G\_\alpha$ must contain $U$, should not have vertices in $\tilde{U}$ and may contain arbitrary subset of $U^\*$. It yields that the inner sum has $2^{|U^\*|}$ summands, and that it vanishes unless $U^\*=\emptyset$. In the latter case, the inner sum equals $(-1)^{|U|}$. Such $U$ for which $U^\*=\emptyset$ (that is, every vertex either belongs to $U$ or has a neighbour in $U$) are called dominating. Thus
$$
\Theta(G)=\sum\_{U\,\text{is dominating and connected}} (-1)^{|U|}.
$$
If $G=C\_n$ is a cycle of length $n$, then $U$ should be either the whole cycle (1 variant), or a path of length $n-1$ ($n$ variants) or a path of length $n-2$ ($n$ variants), so we get $\Theta(C\_n)=(-1)^n$.
|
2
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https://mathoverflow.net/users/4312
|
408263
| 167,200 |
https://mathoverflow.net/questions/408250
|
8
|
I wonder whether there is a website or a survey collecting all known NP-complete or NP-hard problems on graph theory?
|
https://mathoverflow.net/users/375270
|
Is there a website or a survey collecting all NP-complete problems on graph theory?
|
Here is a [section on graph theory](https://www.csc.kth.se/%7Eviggo/wwwcompendium/node8.html) in *A compendium of NP optimization problems* by P. Crescenzi and V. Kann.
|
7
|
https://mathoverflow.net/users/7076
|
408269
| 167,201 |
https://mathoverflow.net/questions/408264
|
-1
|
Consider the space $S$ of real functions with the norm $$\|f\|^2 = \frac{1}{\sqrt{2 \pi}} \int\_{-\infty}^{\infty} e^{-x^2/2} f^2(x) ~\mathrm{d}x, $$
or any reasonable Euclidean norm such that bounded functions have a finite norm.
Can we construct a continuous function mapping from $S$ to the space of distributions over $\mathbb{R}$ such that periodic functions of period $\tau$ are mapped to $\delta\_{\tau}$?
The motivation is to define some reasonable notion of the fundamental of a signal.
|
https://mathoverflow.net/users/8737
|
Fundamental of a signal
|
Make a sequence of functions of period $2\pi$ whose limit is of period $\pi$. Your map is discontinuous.
|
1
|
https://mathoverflow.net/users/13268
|
408271
| 167,202 |
https://mathoverflow.net/questions/408262
|
11
|
If $0^\sharp$ exists then the $L$-indiscernibles form a proper class of ordinals without any infinite constructible subset, as $0^\sharp$ can be defined from any infinite increasing sequence $\langle \kappa\_i\mid i<\omega\rangle$ of them as
$$0^\sharp = \{\varphi(v\_0,\dots, v\_n)\mid L\_{\sup\_{i<\omega}\kappa\_i}\models\varphi(\kappa\_0,\dots, \kappa\_n)\}$$
On the other hand we, can force a set of ordinals of arbitrary size $\lambda$ without infinite constructible subsets by forcing with finite partial functions $p:\lambda\rightarrow 2$ (which is the same as adding $\lambda$-many Cohen reals). But clearly the class-sized version of this does not preserve $\mathrm{ZFC}$.
My question is:
>
> Is there a model of $\mathrm{ZFC}$ with a definable proper class of ordinals without an infinite constructible subset, yet $0^\sharp$ does not exist in this model?
>
>
>
Such a model cannot be a generic extension (by set-sized forcing) of $L$, as otherwise one condition in the generic filter must force an infinite amount of ordinals into this class and $L$ can see this.
This question is related to [this question](https://mathoverflow.net/q/101821/125703) by Joel Hamkins in the following way: Suppose $V$ has a nonconstructible real and there is a definable embedding $\pi:(V, \in)\rightarrow (L, \in)$ (in the sense of that question). Let $C$ be the range of $\mu\circ\pi$ where $\mu$ sends a set in $L$ to its rank in the canonical global wellorder. Then by the [answer of Joel Hamkins](https://mathoverflow.net/a/147417/125703), $0^\sharp$ does not exist and the arguments in the [answer of Farmer Schlutzenberg](https://mathoverflow.net/a/396181/125703) show that $C$ does not have an infinite constructible subset. It seems to be open whether this scenario is consistent.
*(I have [asked this question](https://math.stackexchange.com/q/4288933/457161) on MSE before but it did not get answered there)*
|
https://mathoverflow.net/users/125703
|
A proper class of ordinals without an infinite constructible subset
|
>
> *Stanley, M. C.*, [**A cardinal preserving immune partition of the ordinals**](http://dx.doi.org/10.4064/fm-148-3-199-221), Fundam. Math. 148, No. 3, 199-221 (1995). [ZBL0843.03028](https://zbmath.org/?q=an:0843.03028).
>
>
>
An infinite set (or class) of ordinals is said to be immune if it neither contains nor is disjoint from any infinite constructible subset of its supremum. The main result of the paper reads as follows: There exists an immune class of ordinals in a cardinal-preserving and GCH-preserving class generic extension of L.
|
14
|
https://mathoverflow.net/users/11115
|
408272
| 167,203 |
https://mathoverflow.net/questions/408276
|
2
|
I posed this question on Math.Stackexchange (see [here](https://math.stackexchange.com/q/4301068/75923)) but until now there was no response. This made me decide to give it a try here.
---
Let $k\subseteq F$ denote an algebraic field extension and let $\alpha\in F$ having $f\in k[x]$ as its minimal polynomial. Further let $G=\mathsf{Aut}\_k(F)$.
My question:
>
> If $\beta\in F$ is a root of $f$ then does there exists some $\sigma\in G$ with $\sigma(\alpha)=\beta$?
>
>
>
In other words: is **every** root of $f$ in $F$ also an element of orbit $G\alpha$?
I know that the answer is "yes" if the extension is normal but am puzzling whether this condition can be dropped.
Thank you in advance for taking notice of this question, and sorry if it is a duplicate (or for some other reason not suitable for MathOverflow).
---
**Edit**: at first hand I forgot to state that $\beta$ is assumed to be an element of $F$. That is repaired now by. Sorry for confusion.
|
https://mathoverflow.net/users/40263
|
If $G=\mathsf{Aut}_k(F)$ acts on field $F$ algebraic over $k$ then do we have: orbit $G\alpha=\text{ roots of minimal polynomial of }\alpha$?
|
Here is a different interpretation of the question, which is hopefully closer to OP's intent:
>
> Let $F/k$ be an algebraic field extension, and let $\alpha\in F$. Does $Aut\_k(F)$ act transitively on the conjugates of $\alpha$ *which are contained in $F$*?
>
>
>
The answer to this question is no in general. For instance, consider $k=\mathbb Q$, $F=\mathbb Q(\sqrt[4]{2})$, and $\alpha=\sqrt{2}$. Then $F$ contains the conjugate $-\sqrt{2}$ of $\alpha$, but there is no automorphism of $F$ carrying $\sqrt{2}$ to $-\sqrt{2}$, because the former is a square in $F$ and the latter is not.
---
There is, however, one important case where the answer *is* positive, specifically when $F=k(\alpha)$. Indeed, in this case, for any conjugate $\beta$ of $\alpha$ contained in $F$, we must have $k(\beta)=F$ as well, since the two have the same degree over $k$ (equal to the degree of the minimal polynomial $\alpha$ and $\beta$). From standard field theory you get an isomorphism from $k(\alpha)$ to $k(\beta)$ fixing $k$ and taking $\alpha$ to $\beta$, which is then an automorphism of $F$.
|
7
|
https://mathoverflow.net/users/30186
|
408283
| 167,207 |
https://mathoverflow.net/questions/408061
|
1
|
I am looking for a reference or derivation for the following question:
Consider a cycle graph $G$ with $N$ vertices (see example [here](https://en.wikipedia.org/wiki/Cycle_graph)). Let two independent continuous-time random walkers$^\star$ start on node $i$ and node $j$. Let $T$ be the time when the two walkers meet for the first time.
What will be the probability density function of $T$?
I am looking for the exact expression of that pdf for finite $N$. I am sure this is a well known question with a precise formula but I have not been able to find a clear reference solving it. This is a new field for me so perhaps I am using the wrong vocabulary? With the hope that this post will help. Thanks!
($^\star$: each walker jumps to a neighboring vertex (chosen at random) with rate one.)
|
https://mathoverflow.net/users/420641
|
Reference: probability distribution of first meeting time of two random walks on a cycle graph
|
Assume that $i<j.$ You can reduce this to the following simpler-looking question: Consider a continuous-time RW on the integers, moving at rate two, started at $k:=j-i$. Let $T$ be the hitting time of $\{0,N\}$ by this walk, also known as as the exit time from $[1,N-1]$. Let $\tau$ denote the hitting time of $\{0,N\}$ by discrete-time random walk. The distribution of $\tau$ can be determined from the arguments in Chapter 21 of [1]. (Spitzer gives the general method, and then provides all the details only for the case $k=N/2$.) Then you can deduce the distribution of $T$ via
$$P\_k[T>q] = \sum\_n P[{\rm Poisson}(2q)=m] \cdot P\_k[\tau>m] \,.$$
See also [2].
[1] Spitzer, Frank. Principles of random walk. GTM Vol. 34. Second edition, Springer.
[2]<https://www.researchgate.net/publication/254211645_Exit_times_for_a_class_of_random_Walks_Exact_distribution_results>
|
0
|
https://mathoverflow.net/users/7691
|
408303
| 167,211 |
https://mathoverflow.net/questions/408301
|
20
|
Are there arbitrarily large sets $\mathcal S=\{a\_1,\ldots,a\_n\}$ of strictly positive integers such that all sums $a\_i+a\_j$ of two distinct elements in $\mathcal S$ are squares?
Considering subsets in $\mathbb Z$ should essentially give the same answer since such a set can contain at most one negative integer.
An example of size $3$ is given by $\{6,19,30\}$.
(Allowing $0$, one gets $\{0,a^2,b^2\}$ in bijection with
Pythagorean triplets $c^2=a^2+b^2$.)
There is no such example with four integers in $\{1,\ldots,1000\}$.
(Accepting $0$, solutions are given by Euler bricks:
$\{0,44^2,117^2,240^2\}$ is the smallest example. I suspect thus that there are strictly positive solutions in $\mathbb N^4$).
An equivalent reformulation: Consider the infinite graph with vertices $1,2,3,\ldots$ and edges $\{i,j\}$ if $i+j$ is a square.
Does this graph contain arbitrarily large complete subgraphs?
(Trivial observation: Every edge $\{a,b\}$ is only contained in finitely many different complete subgraphs.)
*Motivation* This is somehow a variation on question [Generalisation of this circular arrangement of numbers from $1$ to $32$ with two adjacent numbers being perfect squares](https://mathoverflow.net/questions/408235)
|
https://mathoverflow.net/users/4556
|
Size of set of integers with all sums of two distinct elements giving squares
|
The size of such sets is bounded by some (unknown) constant, assuming a big conjecture in arithmetic geometry.
The Bombieri-Lang conjecture (non-trivially via the Uniformity Conjecture, see Stanley Yao Xiao's comment) implies that for any $f(x)\in \mathbb{Z}[x]$ of degree $5$, with no repeated roots, there are at most $B$ many rational numbers $m$ for which $f(m)$ is a square - here $B$ is some absolute constant (so the conjecture goes), completely independent of $f$.
This implies that the size of a set $A$ such that $a+a'$ is a square for any two distinct elements $a,a'\in A$ is at most $B+5$. Indeed, take any $5$ distinct elements $a\_1,\ldots,a\_5\in A$, and consider $f(x) = (x+a\_1)\cdots(x+a\_5)$. For any $m\in A\backslash \{a\_1,\ldots,a\_5\}$, we know that $f(m)$ is a square, and so $\lvert A\rvert-5\leq B$.
I learnt of this kind of argument (and the Uniformity Conjecture) via this paper of Cilleruelo and Granville, which has many similar arguments and applications: <https://arxiv.org/pdf/math/0608109.pdf>.
|
24
|
https://mathoverflow.net/users/385
|
408306
| 167,212 |
https://mathoverflow.net/questions/408287
|
1
|
Let $M$ be a compact smooth manifold with a smooth boundary. Given a smooth Riemannian metric $g$ on $M$, we denote by $\{\phi\_k\}\_{k=1}^{\infty}$ an $L^2(M)$--orthonormal basis consisting of Dirichlet eigenfunctions of $-\Delta\_g$ on $M$.
Now let us denote by $X$ the collection of all pairs $(f,g)$ with $f$ a smooth function on $M$ and $g$ a smooth Riemannian metric on $M$. Is it true that for a generic element in $X$ the following properties are simultaneously satisfies:
1. The spectrum of $(M,g)$ is simple.
2. Given any $k\in \mathbb N$, there holds: $\int\_M f(x) \phi\_k(x)\,dV\_g \neq 0$.
|
https://mathoverflow.net/users/50438
|
A property for generic pairs of functions and metrics
|
I believe that a combination of the following two facts essentially confirms the desired result. I phrased the second point in $C^2(M)$—working directly in $C^\infty(M)$ is a bit awkward because it is not a Banach space—but I think it should hold more widely.
* The generic simplicity of the Laplacian eigenvalues is a result of Uhlenbeck [Theorem 8, 1].
* For the second condition, we fix the metric $g$ andg make the following observation. For all $k \in \mathbf{N}$, the set $U\_k = \{ u \in C^{2}(M) \mid \langle u , \phi\_k \rangle \neq 0 \}$ is open and dense. After all, on the one hand if $\langle u , \phi\_k \rangle \neq 0$ then one can choose $v \in C^2(M)$ so small that $\lvert v \rvert\_{L^2} < \lvert \langle u , \phi\_k \rangle \rvert$. This ensures $\langle u - v, \phi\_k \rangle \neq 0$, confirming that $U\_k$ is open. On the other hand, if instead $\langle u , \phi\_k \rangle = 0$ then obviously $\langle u + t \phi\_k,\phi\_k \rangle \neq 0$ for all $t \neq 0$, no matter how small. This confirms that $U\_k$ is dense. Therefore the countable intersection $\cap\_k U\_k \subset C^2(M)$ is a generic set.
[1] Karen Uhlenbeck. Generic Properties of Eigenfunctions. *American Journal of Mathematics*, 1976, Vol. 98, No. 4, pp. 1059-1078.
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3
|
https://mathoverflow.net/users/103792
|
408312
| 167,214 |
https://mathoverflow.net/questions/406784
|
5
|
Remember, that an incomplete r.e. set $A$ is cuppable if there is an incomplete r.e. set $B$ such that $A\oplus B \equiv\_T 0'$. It's relatively easy to build a low cuppable set but my question is whether for every cuppable set $A$ there is a low r.e. set $B$ such that $A \oplus B \equiv 0'$
I'm pretty sure this must be false (or at least unsolved but I'm betting false) or people wouldn't have had to prove results like the ZBC theorem (one can think of this as the claim that if S is r.e. in 0' then there is an r.e. set A with $A' \equiv 0' \oplus S$ and a $A$ r.e. set $B$ which is low relative to $A$ and cups to $A'$) but maybe someone can point me to a proof.
---
Note that I'm asking about cupping and capping in R (set of re degrees here) not in D to clarify whats going on with the equivalence claim.
Also, the link with ZBC theorem results was mistaken as just because $0' \oplus S \equiv\_T B'$ doesn't guarantee that S itself is re in B. Hence, you can't use low-cuppability to show the ZBC theorem.
|
https://mathoverflow.net/users/23648
|
Does every cuppable r.e. set cup with a low set?
|
It seems that a short comment is not sufficient.
***The answer is negative. I.e. there is a cuppable r.e. set which is no low cuppable.***
By the results from Soare's book, low cuppability is equivalent to prompt simplicity which is equivalent to noncappbability.
Now the fact is that there is an r.e. degree which is both cappable and cuppable. The result was proved by Harrington in 1970's in an unpublished handwritten notes. Yang and I have an improvement of this by showing that there is a cappable degree which does not belong to the ideal generated by the union of nonbounding and noncuppable degrees (see <https://www.jstor.org/stable/pdf/27588357.pdf?refreqid=excelsior%3Aa9865b098fcf6a982c2dfeeb4155a184>).
Certainly you may find earlier published papers that imply the result.
|
3
|
https://mathoverflow.net/users/14340
|
408313
| 167,215 |
https://mathoverflow.net/questions/408230
|
0
|
I'm looking for some continuous functions $\{f\_i(x,t)\}$, here $x=(x\_1,x\_2..., x\_n)$, such that:
* $f\_i(x,t):R^n\times [0, +\infty)\rightarrow R ~~\text{is continuous for each}~~i $
* $f\_i(x,0)=x\_i$
* $\Sigma\_1^n f\_i^2(x,t)\rightarrow \infty ~~\text{as}~~\|(x,t)\|^2=\Sigma\_1^n x\_i+t^2 \rightarrow \infty$
If such functions exist, could you please show me some examples, thanks.
|
https://mathoverflow.net/users/166368
|
Looking for some special functions
|
I assume you want squares on $x\_i$ in the expression for the norm: $\|(x,t)\|^2=x\_1^2+\dots+x\_n^2+t^2$.
Then I believe there are no such functions for topological reasons. Consider the restriction of $f=(f\_1,\dots,f\_n)$ to a half-sphere
$$X=\{(x,t);\,x\_1^2+\dots+x\_n^2+t^2=C,\,t\geq 0\}.$$
Suppose that $f\neq 0$ on $X$. Then, $f/|f|$ would be a continuous map from $X$ to the unit sphere $S^{n-1}$ in $\mathbb R^n$. Moreover, the restriction to the boundary $t=0$ is $f(x,0)=x/|x|$, that is, it is the identity under the natural identification $\partial X\approx S^{n-1}$. My understanding of algebraic topology is near zero, but I believe it follows from homotopy theory that no such map exists. So $f$ has a zero on $X$ for any $C>0$, which contradicts your last assumption.
|
2
|
https://mathoverflow.net/users/10846
|
408319
| 167,217 |
https://mathoverflow.net/questions/322043
|
12
|
On Saturday 4 September 1999, [Vaughan Jones](https://en.wikipedia.org/wiki/Vaughan_Jones) put on arXiv a paper entitled [*Planar algebras, I*](https://arxiv.org/abs/math/9909027).
Until now, this preprint was cited 343 times (according to Google Scholar). It is often cited with the mention "to appear in New Zealand J. Math.", even nowadays after 20 years.
**Question**: Why is this paper not (yet) published? Is it still *under review* in New Zealand J. Math.? What are/were the requests of the referee? Who is/was the referee?
I would like to know the whole truth on this subject.
*Remark*: for people thinking that I just have to ask him, I want to say that I am not asking this question for myself only, I think that this information should be known to anyone interested in planar algebras from far or near. Moreover, I don't want to annoy him with a question that must have been asked to him too many times...
|
https://mathoverflow.net/users/34538
|
Why is Planar algebras I (by Vaughan Jones) not published?
|
This paper is now published in New Zealand Journal of Mathematics Vol. 52 (2021).
<https://doi.org/10.53733/172>
[pdf file](https://nzjmath.org/index.php/NZJMATH/article/view/172/61)
|
11
|
https://mathoverflow.net/users/164194
|
408320
| 167,218 |
https://mathoverflow.net/questions/408266
|
2
|
Is it true the following statement?
>
> Given two Polish spaces $X,Y$ and a Borel function $f:X\rightarrow Y$, there exists a Polish space $Z$ and a Borel function $g:X \rightarrow Y\times Z$ with closed graph such that $f(x) \ = \pi\_Y(g(x))$ for all $x \in X$.
>
>
>
In case it was true, do you have any hint for the proof?
Thanks!
|
https://mathoverflow.net/users/141146
|
Is every Borel function a projection of a Borel function with closed graph?
|
Yes, this is true: by Exercise 13.5 in [Kechris' "Classical Descriptive Set Theory"](https://link.springer.com/book/10.1007/978-1-4612-4190-4), for any Borel function $f:X\to Y$ between Polish spaces there exists a continuous bijective map $i:Z\to X$ from a Polish space $Z$ such that the function $f\circ i:Z\to Y$ is continuous.
Now consider the function $g:X\to Y\times Z$, $g:x\mapsto (f(x),i^{-1}(x))$ and observe that it has closed graph $$\Gamma=\{(x,y,z)\in X\times Y\times Z:y=f(x),\; x=i(z)\}=\{(x,y,z)\in X:y=f\circ i(z),\;x=i(z)\}$$by the continuity of the functions $i$ and $f\circ i$. It is clear that $f=\pi\_Y\circ g$.
|
6
|
https://mathoverflow.net/users/61536
|
408325
| 167,220 |
https://mathoverflow.net/questions/408298
|
0
|
For the Lie algebra $\mathfrak{so}(2n+1, \mathbb{C})$, there is a matrix representation given by the following matrices:
\begin{align}
\left( \begin{matrix} 0 & x & y \\ -y^T & A & B \\ -x^T & C & -A^T \end{matrix} \right),
\end{align}
where $x, y$ are $1 \times n$ matrices, $A,B,C$ are $n \times n$ matrices. Are there similar matrix representations for Lie group $\mathrm{SO}(2n+1, \mathbb{C})$? Thank you very much.
Edit: I forgot to mention that $B,C$ are skew-symmetric, $B=-B^T, C=-C^T$.
|
https://mathoverflow.net/users/11877
|
Matrix representations of Lie groups of type $B_n$
|
The quadratic form whose matrix is $\begin{pmatrix} 1 & 0 & 0 \\ 0 & 0 & I\_n \\ 0 & I\_n & 0 \end{pmatrix}$ gives an embedding of $\operatorname{SO}(2n + 1, \mathbb C)$ in $\operatorname{GL}(2n + 1, \mathbb C)$ whose derivative is your specified embedding $\mathfrak{so}(2n + 1, \mathbb C) \to \mathfrak{gl}(2n + 1, \mathbb C)$. Under this embedding,
* $\left\{\begin{pmatrix} 1 & 0 & 0 \\ 0 & A & 0 \\ 0 & 0 & A^{-\mathsf T} \end{pmatrix}\mathrel: \text{$A$ diagonal}\right\}$ is the image of a maximal (algebraic) torus in $\operatorname{SO}(2n + 1, \mathbb C)$,
* the image of a maximal unipotent subgroup of $\operatorname{SO}(2n + 1, \mathbb C)$ is generated by
+ $\exp\begin{pmatrix} 0 & 0 & 0 \\ 0 & E\_{ij} & 0 \\ 0 & 0 & -E\_{ji} \end{pmatrix}$, where $i$ is less than $j$,
and
+ $\exp\begin{pmatrix} 0 & 0 & y \\ -y^{\mathsf T} & 0 & B \\ 0 & 0 & 0 \end{pmatrix}$, where $B$ is skew-symmetric,
and
* the image of the Tits subgroup of $\operatorname{SO}(2n + 1, \mathbb C)$ is generated by
+ $\begin{pmatrix} 1 & 0 & 0 \\ 0 & A & 0 \\ 0 & 0 & A^{-\mathsf T} \end{pmatrix}$, where $A = I - E\_{ii} - E\_{jj} + E\_{ij} - E\_{ji}$ with $i \ne j$,
and
+ $\begin{pmatrix} 1 & 0 & 0 \\ 0 & D' & D'' \\ 0 & D'' & D' \end{pmatrix}$, where $D'$ and $D''$ are diagonal $(0, 1)$-matrices such that $D' + D'' = I\_n$.
(I originally made a comment to this effect, but I had the wrong quadratic form, and wrongly suggested that you had switched $x^{\mathsf T}$ and $y^{\mathsf T}$ in your embedding.)
|
1
|
https://mathoverflow.net/users/2383
|
408326
| 167,221 |
https://mathoverflow.net/questions/408268
|
1
|
Consider the integral
$$\mathcal{I}=\int\_0^t\left(\frac{Ae^{-\lambda t}-Ae^{-\lambda s}}{2}\right)^{2m+1}e^{-\epsilon(t-s)}ds,\tag{1}$$
for constants $A,\lambda,\epsilon,t\in\mathbb{R}$ and $m\in\mathbb{Z}^+$.
The intention of evaluating $\mathcal{I}$, is to find
\begin{equation}
\begin{split}
\mathcal{S}&=\int\_0^tJ\_1\left(Ae^{-\lambda t}-Ae^{-\lambda s}\right)e^{-\epsilon(t-s)}ds,
\\&=\int\_0^t\sum\_{m=0}^{\infty}\frac{
(-1)^m}{m!(m+1)!}\left(\frac{Ae^{-\lambda t}-Ae^{-\lambda s}}{2}\right)^{2m+1}e^{-\epsilon(t-s)}ds,\\
&=\sum\_{m=0}^\infty \frac{
(-1)^m}{m!(m+1)!}\mathcal{I}\_m
\end{split}\end{equation}
where $J\_1$ denotes the first order Bessel function of first kind. Does a closed form exist for $\mathcal{S}$ (or Taylor series)? Could an asymptotic bound simplify things?
According to [Mathematica](https://www.wolframalpha.com/input/?i=integral+%28a-A+exp%28-lambda+s%29%29%5E%282*m%2B1%29+exp%28-epsilon*%28t-s%29%29ds), we have
\begin{equation}
\mathcal{I}=\frac{(A-Ae^{-\lambda t})^2 (Ae^{-\lambda t}-A)^{2m} 2^{-2m-1}e^{-\epsilon t}\cdot{}\_2F\_1\left(1,\frac{\epsilon+\lambda}{\lambda};\frac{\epsilon}{\lambda}-2m,e^{-\lambda t}\right)}{A\left(\epsilon-\lambda(2m+1)\right)}.
\end{equation}
From this, is it possible to prove that
$2\mathcal{S}\sim A\lambda^2\kappa e^{-\lambda t}-A\lambda e^{-\lambda t}$ as $t\rightarrow\infty$, for some $\kappa$?
|
https://mathoverflow.net/users/167822
|
Integral of $J_1\left(Ae^{-\lambda t}-Ae^{-\lambda s}\right)e^{-\epsilon(t-s)}$ with respect to $s$?
|
If $\lambda>0$ and $\lambda>\epsilon$ one has
$$\lim\_{t\rightarrow\infty}e^{\epsilon t}\mathcal{S}=\int\_0^\infty J\_1\left(-Ae^{-\lambda s}\right)e^{\epsilon s}ds$$
$$\qquad=\frac{A/2}{\epsilon-\lambda} \, \_1F\_2\left(\frac{1}{2}-\frac{\epsilon}{2 \lambda};2,\frac{3}{2}-\frac{\epsilon}{2 \lambda};-A^2/4\right).$$
The asymptotics is different for other ranges of $\lambda,\epsilon$, for example, for $\lambda,\epsilon<0$ one has ${\cal S}\rightarrow -e^{-\epsilon t}J\_1(Ae^{-\lambda t})/\epsilon$.
|
3
|
https://mathoverflow.net/users/11260
|
408335
| 167,223 |
https://mathoverflow.net/questions/408308
|
5
|
In P. E. Kloeden & E. Platen (1995). Numerical Solution of Stochastic Differential Equations.
pg.118, they go over some special cases of nonlinear SDEs $dX\_t=\alpha(t,X\_t)\,dt+\sigma(t,X\_t)\,dB\_t$ that have exact solutions.
I am just wondering if there are any more large lists somewhere that you came across. Thanks.
|
https://mathoverflow.net/users/99863
|
Long list of exactly solvable nonlinear SDEs
|
A follow-up of Kloeden and Platen is C.H. Skiadas, [Exact Solutions of Stochastic Differential Equations](https://link.springer.com/article/10.1007/s11009-009-9145-3) (2010).
|
2
|
https://mathoverflow.net/users/11260
|
408341
| 167,225 |
https://mathoverflow.net/questions/408305
|
3
|
Do there exist non-rational algebraic numbers that belong to $\mathbb Q\_p$ for all prime $p$? If yes, can one characterize them?
I spent several days for the first question, and I found nothing. The second one looks even more diffuclt and surely out of my skills.
|
https://mathoverflow.net/users/33128
|
Algebraic numbers in all $\mathbb Q_p$
|
(cw answer based on Wojowu's link)
>
> The only algebraic extension of $\mathbf{Q}$ that embeds into $\mathbf{Q}\_p$ for all $p$ (or even for a density 1 set of primes) is $\mathbf{Q}$ itself.
>
>
>
For if $P\in\mathbf{Q}[t]$ is an irreducible polynomial of degree $\ge 2$, the set of primes $p$ for which $P$ has no root in $\mathbf{Q}\_p$ has positive density: see [this answer](https://mathoverflow.net/a/7100/30186), which also applies to an arbitrary number field (=finite extension of $\mathbf{Q}$), where primes now mean primes of the number field.
The above statement roughly states that *"the only algebraic number that belong to the intersection of all $\mathbf{Q}\_p$ are rationals"*, which seems to answer your question. As already mentioned in comments, this statement doesn't really make sense, since there is no natural way to identify algebraic elements in different completions. An artificial way for it to makes sense is to mod out $\mathbf{Q}\_p$ by the equivalence relation "having the same minimal polynomial (monic or zero) over $\mathbf{Q}$".
|
9
|
https://mathoverflow.net/users/14094
|
408343
| 167,226 |
https://mathoverflow.net/questions/408344
|
11
|
I'm making this post to ask for a reference about combinatorics: I'm a PhD student in representation theory/algebraic geometry. My background is mostly in algebra and geometry (and also mostly theoretical unfortunately).
Right now I'm interested in the cohomology of quiver and character varieties and their links with cohomological Hall algebras, quantum groups and the character ring of $\operatorname{GL}(n,\mathbb{F}\_q)$. There's a lot of interesting combinatorics going on, but I really know very little about it. Especially regarding MacDonald polynomials etc.
Outside of the classic book by Macdonald "Symmetric functions and Hall polynomials" what could be a good reference to get into this area of combinatorics? Ideally I would like a book or notes with strong link to representation theory/cohomology theories etc
|
https://mathoverflow.net/users/146464
|
Reference for combinatorics with view towards representation theory/algebraic geometry
|
M. Haiman "Notes on Macdonald polynomials and the geometry of the Hilbert scheme of points on $\mathbb{P}^2$". By one of the greatest specialists of interactions between combinatorics and algebraic geometry.
|
12
|
https://mathoverflow.net/users/37214
|
408347
| 167,227 |
https://mathoverflow.net/questions/408364
|
3
|
Consider the [binary partitions](https://oeis.org/A000123) of $2n$ in powers of $2$, denoted by $b(2n)$, with the generating function
$$\sum\_{n\geq0}b(2n)\,x^n=\frac1{1-x}\prod\_{k\geq0}\frac1{1-x^{2^n}}.$$
A result of De Bruijn shows the asymptotic growth
$$\log b(2n)\sim \frac1{\log 4}\log^2\left(\frac{n}{\log n}\right).$$
Suppose $t(3n)$ denote the [ternary partitions](https://oeis.org/A005704) of $3n$ in powers of $3$, with a similar generating function
$$\sum\_{n\geq0}t(3n)\,x^n=\frac1{1-x}\prod\_{k\geq0}\frac1{1-x^{3^n}}.$$
>
> **QUESTION.** What is the asymptotic growth of $t(3n)$?
>
>
>
|
https://mathoverflow.net/users/66131
|
Asymptotic growth of ternary partitions of integers $3n$
|
It's right in the beginning of De Bruijin's paper. More generally,
$\log p(rn) \sim \frac 1 {2 \log r} \log^2 \frac n {\log n}$
where $p(rn)$ is the number of partitions of $rn$ into powers of $r$.
The result is attributed to Kurt Mahler's paper "On a special functional equation", published in Journ. London Math. Soc. in 1940.
|
4
|
https://mathoverflow.net/users/114143
|
408367
| 167,231 |
https://mathoverflow.net/questions/407905
|
0
|
[EDIT] The older exposition of this theory was proved inconsistent by EmilJeřábek (see comments). Here, this is a possible salvage. (the new information over the older post shall be put in square brackets)
Working in mono-sorted FOL with equality and membership, add the following axioms:
Define: $set(y) \iff \exists z: y \in z$
If formula $\phi$ does not use $``x"$, then:
1. **Comprehension:** $(\exists! x \ \forall y \ (y \in x \leftrightarrow set(y) \land \phi))$
Define: $x=V \iff \forall y: set(y) \to y \in x$
2. **Reflection:** $(\phi \to \exists \text { transitive } x \in V: \phi^x)$;
Where $\phi$ is a formula that does not use $``x"$ [having all of its quantifiers bounded by $V$], $\phi^x$ is the formula obtained by merely replacing all bounding occurrences of $V$ in $\phi$ by $x$; "transitive" is defined as a class whose elements are subsets of it.
3. **Subsetting:** $ x \in V \land \forall y \in A (y \subseteq x) \to A \in V$
4. **Global Choice:** For every nonempty class $X$ of pairwise disjoint sets, there is a class having singleton intersection with each nonempty element of $X$.
5. **Foundation**: $\exists y \,(y \in X) \to \exists y \in X \, (y \cap X = \emptyset)$
So the question is:
>
> Is the above theory consistent? Is it equivalent to [$\small \sf MK$](https://en.wikipedia.org/wiki/Morse%E2%80%93Kelley_set_theory)?
>
>
>
|
https://mathoverflow.net/users/95347
|
Is this theory equivalent to MK?
|
This theory is consistent since NBG proves this form of reflection and the rest of axioms are either axioms or theorems of MK, so this theory is a subtheory of MK. Now to establish the other direction of equivalence with MK, we note that Extensionality, Pairing, Boolean union, Set Union, Power, Infinity and Separation all are easily provable using reflection, subsetting, and comprehension, what needs to be proved is Kelley's axiom of [Substitution](https://en.wikipedia.org/wiki/Morse%E2%80%93Kelley_set_theory#The_axioms_in_Kelley%27s_General_Topology).
Now let $F$ be a class function, whose domain is the set $d$, then reflect on the formula $\exists k \in V: k=d \land \forall m \in V (m \in k \to \exists y \in V: y=F(m))$, obviousely from parameters $F,d$; and we get the range of $F$ being a subclass of some transitive set, and by separation it would be a set, thus proving Substitution.
|
0
|
https://mathoverflow.net/users/95347
|
408368
| 167,232 |
https://mathoverflow.net/questions/408294
|
6
|
A subset of a linear space $X$ is called *infinite-dimensional* if it is not contained in a finite-dimensional linear subspace of $X$.
>
> **Problem.** Let $L$ be an infinite-dimensional subset of the linear space $\mathbb R^\omega$. Is there an infinite set $I\subseteq\omega$ such that for every infinite set $J\subseteq I$ the projection of $L$ to $\mathbb R^J$ is infinite-dimensional?
>
>
>
**Remark.** The answer to this problem is affirmative if every function $f\in L$ has finite range.
|
https://mathoverflow.net/users/61536
|
Infinite-dimensional projections of linearly independent sets
|
For $n :=\{0,1,\dots,n-1\}\in\omega$ let’s denote $P\_n:\mathbb{R}^\omega\to \mathbb{R} ^n$ the projection, which is the restriction map $f\mapsto f\_{|n}$. The dimensions of the subspaces $P\_n(L)\subset \mathbb{R} ^n$ can’t be stationary, because that would mean that for some $n\_0$, every function $f\in P\_{n\_0}(L)$ has a unique extension to a function $\tilde f\in L$, in which case $ P\_{n\_0}(L)\ni f\mapsto \tilde f\in L$ is a linear bijection, whereas $L$ was assumed infinite-dimensional. So let $(n\_k)\_k$ be a strictly increasing sequence such that $\dim P\_{n\_k+1}(L)> \dim P\_{n\_k}(L)$. Thus $\ker \big(P\_{n\_k}: P\_{n\_k+1}(L)\to P\_{n\_k}(L) \big)\neq(0)$: there is a function $f\_k\in L$ such that $f\_k|{n\_k}=0$ and $f\_k(n\_k+1)=1$. So if we take $ I:=\{n\_k+1\}\_{k\in\omega}$, for any infinite $J\subset I$ The space $P\_J(L)$ contains the functions $P\_Jf\_k={f\_k}\_{|J}$, which are clearly linearly independent.
|
4
|
https://mathoverflow.net/users/6101
|
408375
| 167,235 |
https://mathoverflow.net/questions/408374
|
21
|
Let $u(i,j)$ denote the number of lattice paths from the origin to a fixed terminal point $(i,j)$ subject only to the condition that each successive lattice point on the path is closer to $(i,j)$ than its predecessor. For example, $u(1,1) = 5$ counts the one-step path $(0,0) \to (1,1)$ and the 4 two-step paths with lone interior point (1,0), (0,1), (2,1), and (1,2) respectively, each of which points is just 1 unit from (1,1) while the origin is at distance $\sqrt{2}$. By symmetry, $u(i,j)=u(j,i)$ and $u(i,j)=u(\pm i, \pm j)$. So we may assume $0\le j \le i$.
The numbers $u(i,j)$ grow rapidly but have only small prime factors. For example,
$u(15,4)=269124680144389687575008665117965469864474632630636414714548567937\\
47381916046142578125
=3^{114}\ 5^{19}\ 13^6\ 17^9.$
Any explanations? Have these paths been considered in the literature?
Here is Mathematica code to generate $u(i,j)$.
```
addListToListOfLists[ls_,lol_] := (ls + #1 &) /@ lol
SelectLatticePtsInDiskCenteredAtO[radius_] := Module[{m},
Flatten[Table[m = Ceiling[Sqrt[radius^2 - i^2]] - 1;
Table[{i,j}, {j,-m,m}], {i,-Floor[radius],Floor[radius]}], 1] ]
SelectLatticePtsInDiskCenteredAtij[{i_,j_}, radius_] :=
addListToListOfLists[{i,j},
SelectLatticePtsInDiskCenteredAtO[radius]]
u[i_,j_] := u[{i,j}];
u[{0,0}] = 1;
u[{i_,j_}] /; j > i := u[{i,j}] = u[{j,i}];
u[{i_,j_}] /; i < 0 := u[{i,j}] = u[{-i,j}];
u[{i_,j_}] /; j < 0 := u[{i,j}] = u[{i,-j}];
u[{i_,j_}] /; 0 <= j <= i := u[{i,j}] =
Module[{cntFromNewOrigin,radius},
radius = Norm[{i,j}];
cntFromNewOrigin[newOrigin_] := u[{i,j} - newOrigin];
Apply[Plus, Map[cntFromNewOrigin,
SelectLatticePtsInDiskCenteredAtij[{i,j}, radius]]]]
In[918]:= Table[ u[{i,j}],{i,0,3},{j,0,i}]
Out[918]= {{1}, {1, 5}, {25, 125, 1125}, {5625, 28125, 253125,
102515625}}
```
|
https://mathoverflow.net/users/29500
|
Why are the numbers counting "ever-closer" lattice paths so round?
|
For any $k\in\mathbb N$ let $a\_k$ be the number of points on the circle of radius $\sqrt{k}$ (this number may be zero). For any path as in question and for any $k$ between $1$ and $i^2+j^2-1$, there is going to be either none or exactly one of the points on the circle of radius $\sqrt{k}$ around $(i,j)$. As the sequence of points determines the path, this tells us that the number of possible such paths is
$$\prod\_{k=1}^{i^2+j^2-1}(1+a\_k).$$
This is a product of comparatively large number of factors, each of which is small: we easily see $a\_k<4k$, but in fact we have tighter bounds, for instance $a\_k\leq 4d(k)$ where $d$ is the divisor counting function (follows e.g. from [this result](https://en.wikipedia.org/wiki/Sum_of_squares_function#k_=_2)), which is $O(k^c)$ for all $c>0$. As all these factors are small, they can only have small prime factors, explaining your observation.
|
41
|
https://mathoverflow.net/users/30186
|
408377
| 167,236 |
https://mathoverflow.net/questions/408362
|
2
|
Let $[x\_1,x\_2,x\_3,x\_4]$ be coordinates of $\mathbb{P}^3$ and $Z\subset \mathbb{P}^3$ the subscheme given by the ideal $$I\_Z=(x\_1,x\_2,x\_3^2) \subset \mathbb{C}[x\_1,x\_2,x\_3,x\_4]$$ i.e. $Z$ is a double point supported on the line $L:(x\_1=x\_2=0)$.
I want to consider the blowup of $\mathbb{P}^3$ along $Z$, i.e. $X=Bl\_Z\mathbb{P}^3$.
In order to do so I try two different approaches:
1. Blowup directly the entire scheme $Z$. In this way I get a singular variety $X$.
2. First blow up the reduced point $p=[0,0,0,1]$, with $p$ equal to the support of $Z$, and then blow up the point $q$ in $E\_p\subset Bl\_p\mathbb{P}^3$ corresponding to the direction of the line $L$ in the exceptional divisor $E\_p$. In this way I obtain a smooth variety $$Y=Bl\_q(Bl\_p\mathbb{P}^3)$$ with the exceptional divisor $E\_q \subset Y$ and the strict transform $E'\_p$ of $E\_p$ with $E'\_p$ isomorphic to the blowup of $\mathbb{P}^2$ at one point.
My questions now are the following:
a) It seems that the output of procedure 1) is not the same as the output of 2), in fact $X$ is singular while $Y$ is smooth. But the resulting scheme that I'm blowing up it seems the same to me, with the only difference that in procedure 1) I'm blowing it up in a single step while in 2) I've done it in two subsequent steps. Why am I obtaining two different varieties?
b) I know that the exceptional divisor of a point in $\mathbb{P}^3$ is isomorphic to $\mathbb{P}^2$. Is this also the case in the second blowup when I'm considering the blowing up of $q \in E\_p$? In other words is still $E\_q$ isomorphic to $\mathbb{P}^2$? I've the feeling that the answer is no, essentially because the line $l=E'\_p \cap E\_q$ seems to have negative self intersection in $E\_q$. So at this point what even is $E\_q$?
Thanks in advance for eventual answers and observations/remarks.
|
https://mathoverflow.net/users/146431
|
Equivalence of sequences of blowups of $\mathbb{P}^3$
|
These are definitely not the same. $Z$ is a complete intersection subscheme, so if you blow up $Z$, then the exceptional divisor (=set) is a single $\mathbb P^2$, while in the second case it's a union of a copy of $\mathbb P^2$ and a copy of a $\mathbb P^2$ blown up at a point (as you point out it is easy to see that it is not irreducible). I don't know how to answer your question of why you are obtaining different varieties, if not by a counter question: why would you get the same? In the second blow up you are blowing up something that does not come from the original variety. They are just not the same.
For your questions in b): If you blow up a non-singular point on any variety, then the exceptional set will be a projective space of one dimension less then what you are blowing up. I doubt that $l$ has negative self-intersection on $E\_q$, because the latter is also a $\mathbb P^2$. (It has negative self-intersection on $E\_p'$, but that makes sense.)
|
1
|
https://mathoverflow.net/users/10076
|
408393
| 167,243 |
https://mathoverflow.net/questions/408396
|
8
|
It is not known if there are infinitely many prime Fibonacci numbers. But can one assert that there is no Fibonacci number >2 that is also highly composite (<https://en.wikipedia.org/wiki/Highly_composite_number>) - or that there are only finitely many such numbers?
**Remarks:** As given in <http://www.maths.surrey.ac.uk/hosted-sites/R.Knott/Fibonacci/fibtable.html>, every Fibonacci number bigger than 1 [except F(6)=8 and F(12)=144] has at least one prime factor that is *not* a factor of any earlier Fibonacci number. So, Fibonacci numbers tend to have large prime factors and it is quite conceivable that none of them are highly composite. However, a few are seen to be *semiprimes* (<https://en.wikipedia.org/wiki/Semiprime>). Not sure if the question of whether there are infinitely many Fibonacci semiprimes has been answered.
|
https://mathoverflow.net/users/142600
|
On Fibonacci numbers that are also highly composite
|
The largest highly composite Fibonacci number is $F\_{3} = 2$.
If $p$ is a prime number, then either $p \mid F\_{p-1}$ (if $p \equiv \pm 1 \pmod{5}$), $p \mid F\_{p}$ (if $p = 5$), or $p \mid F\_{p+1}$ (if $p \equiv \pm 2 \pmod{5}$). It follows that if $n > 12$ and $p$ is a prime that divides $F\_{n}$ and no previous Fibonacci number, then $p \geq n-1$. Assuming $F\_{n}$ is highly composite implies that all primes $\leq n-1$ divide $F\_{n}$. It follows that $F\_{n} \geq \prod\_{p \leq n-1} p$. This will lead to a contradiction for $n$ sufficiently large (which boils down to the fact that $\frac{1+\sqrt{5}}{2} < e$).
Let $\theta(x) = \sum\_{p \leq x} \log(p)$. The prime number theorem is equivalent to $\theta(x) \sim x$ and Rosser and Schoenfeld showed (see page 70 of their [1962 Illinois Journal of Mathematics paper](http://projecteuclid.org/euclid.ijm/1255631807)) that for $x \geq 41$, $\theta(x) \geq x \left(1 - \frac{1}{\log(x)}\right)$. This implies that for $n \geq 42$, we have
$$
\log\left(\frac{1}{\sqrt{5}}\right) + n \log\left(\frac{1 + \sqrt{5}}{2}\right) \geq \log(F\_{n}) \geq \theta(n-1) \geq (n-1) - \frac{n-1}{\log(n-1)}.
$$
For $n \geq 22$, we have $(n-1) - \frac{(n-1)}{\log(n-1)} \geq \frac{2}{3} (n-1)$, which implies that the right hand side of the centered inequality above is greater than the left.
It suffices to check the prime factorization of $F\_{n}$ for $n \leq 42$ to verify that $F\_{n}$ is not highly composite for $4 \leq n \leq 42$. This can be done easily using the table of [Brillhart, Montgomery, and Silverman](https://www.ams.org/journals/mcom/1988-50-181/S0025-5718-1988-0917832-6/).
|
21
|
https://mathoverflow.net/users/48142
|
408421
| 167,250 |
https://mathoverflow.net/questions/408409
|
8
|
What do people in homotopy theory mean when they say "... is a model for spaces / $(\infty,1)$-categories"? And why does one need models?
Is this related to the notion of a model category?
|
https://mathoverflow.net/users/442261
|
"Models" in homotopy theory
|
Yes, this is absolutely related to model categories (although the concept of models for an homotopy theory is much more general): the notion of model category was introduced by Quillen exactly to express the idea of models for homotopy types. I quote Quillen's *Homotopical Algebra* (Chapter I, page 0.3):
>
> The term "model category" is short for "a category of models for a
> homotopy theory" where the homotopy theory associated to a model
> category $\underline C$ is the homotopy category $Ho \, \underline C$
> [...]. The same homotopy theory may have different models e.g.
> ordinary homotopy theory with a basepoint is the homotopy theory of
> the following model categories: 0-connected topological spaces,
> reduced simplicial sets, and simplicial groups.
>
>
>
In fact, Quillen introduced model structures to give sufficient conditions for two model categories to induce equivalent homotopy theories in a way which is compatible with possible extra structures of interest (e.g. suspension, cofiber sequences, etc): the notion of Quillen equivalence.
We need models because we need to define a written formal language to speak of mathematical objects, including $\infty$-categories or $\infty$-groupoids (=spaces or anima...). And a formal language is made of letters, the concatenation of which is strictly associative, and basic induction rules will follow the principle of modus ponens which is a strictly associative process as well. In particular, with so much strictly associative processes to begin with, the formal language we choose will define a $1$-category. Now, there is no unicity of language. For instance, if we could decide to interpret $\infty$-groupoids as CW-complexes (morphisms in $\infty$-groupoids being paths), or as Kan complexes (morphisms in $\infty$-groupoids being $1$-dimensional simplices) we get two interpretations. It is a theorem of Milnor that the homotopy theory of CW-complexes up to homotopy is equivalent to the homotopy theory of Kan complexes up to simplicial homotopy. This can be promoted to a Quillen equivalence between suitable model category structures à la Quillen on topological spaces and on simplicial sets. Having a Quillen equivalence implies that we have in fact an equivalence of $\infty$-categories, out of which one can define all the extra structures that Quillen was thinking about when he wrote his monograph on homotopical algebra. This why, nowadays, we define homotopy theories as $\infty$-categories.
Now, there are infinitely many different models for the homotopy theory of $\infty$-groupoids - and similarly for $\infty$-categories (in fact for any homotopy theory coming from an $\infty$-category...).
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https://mathoverflow.net/users/1017
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408424
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https://mathoverflow.net/questions/408351
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Let $ G $ be a group of $ n \times n $ matrices. Suppose that some subset $ \{ g\_j: 1 \leq j \leq n^2 \} $ of $ G $ is a basis for the space of all $ n \times n $ matrices. Furthermore suppose that the set
$$
\{ \overline{g\_i}: 1 \leq j \leq n^2 \}
$$
is a group in $ \operatorname{PGL}\_n $. What can we say about the size of $ G $? Must it be the case that
$$
\lvert G\rvert\geq n^2 \prod p\_i
$$
where $ \prod p\_i $ is the product of all distinct primes dividing $ n $?
When we take our field to be algebraically closed and characteristic 0 I think I have groups that saturate this bound for all $ n $. Is this bound even true? Can we do better? Mostly interested in doing everything over $ \mathbb{C} $ but could be cool to hear about other fields too.
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https://mathoverflow.net/users/387190
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Bound on the size of group related to a matrix basis
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I consider the situation over $\mathbb{C}$. We can replace $G$ by the subgroup generated by the $g\_j$'s, so assume that $G$ is generated by the $g\_j$'s. That $G$ contains a spanning set of the space of all matrices yields that the center of $G$ only contains scalar matrices, and also that the inclusion $G \hookrightarrow \operatorname{GL}(n,\mathbb{C}) $ is an absolutely irreducible representation.
That the $g\_j$'s modulo the center form a subgroup means that $|G: Z(G)| = n^2$. So $G$ has an irreducible representation of degree $n$ with $n^2 = |G:Z(G)|$. Such groups are called *groups of central type*. DeMeyer and Janusz (*DeMeyer, F. R.; Janusz, G. J.*, [**Finite groups with an irreducible representation of large degree**](http://dx.doi.org/10.1007/BF01114468), Math. Z. 108, 145-153 (1969). [ZBL0169.03502](https://zbmath.org/?q=an:0169.03502), Theorem 2) have shown that a finite group is of central type if and only if each Sylow $p$-subgroup $S$ of $G$ is of central type and $S\cap Z(G) =Z(S)$. Now a $p$-group always has a nontrivial center of order at least $p$. Thus the center of $G$ has an order divisible by all the primes involved in $n$, and your bound follows.
As you say, there are groups attaining the bound: An extraspecial $p$-group of order $p^{2k+1}$ has an irreducible, faithful representation of dimension $p^k$, and the image is a matrix group as in the question with $n=p^k$. For composite $n=p\_1^{k\_1}\dotsm p\_m^{k\_m}$, we take a direct product $P\_1 \times \dotsb \times P\_m$ , where each $P\_i$ is extraspecial of order $p\_i^{2k\_i+1}$.
The situation should be the same over any field of characteristic zero, because the assumptions in the question yield that the given natural representation is absolutely irreducible, and also over fields such that the characteristic does not divide $n$. When the prime $p$ divides $n$, then my guess is that such a matrix group over a field of characteristic $p$ is not possible.
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5
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https://mathoverflow.net/users/10266
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408434
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