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2025-03-21T14:48:30.434562
| 2020-04-28T17:41:17 |
358792
|
{
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"Paata Ivanishvili",
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"leo monsaingeon",
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|
Stack Exchange
|
Approximation of smooth function by harmonic function
Let $u(x,y)$ be a smooth function on the square $S:=[0, 1] \times [0, 1]$ (see, for example, Wiki for the definitions) and $\varepsilon > 0$.
Is it possible to approximate $u(x,y)$ by a function $h(x,y)$ harmonic on $S$ s.t.
$$\max_{(x,y)\in S} |u(x,y)-h(x,y)|\le \varepsilon ? $$
No. If it approximates smooth function within $\varepsilon$, take a sphere, say centered at $(1/2, 1/2)$ of radius $1/3$ completely inside the square. Then mean value equality (over spheres) implies that the value of $h(1/2)$ must be within $\varepsilon$ of $u$'s integral average over the sphere whereas $u(1/2)$ can be any arbitrary number, contradiction.
also, uniform limits of harmonic functions are harmonic functions!
@Paata Ivanishvili: Sorry, I don't see any contradiction. Can you elaborate your comment as an answer?
@leo monsaingeon: Can you elaborate your comment as an answer, giving, in particular, a reference to us?
@Paata Ivanishvili: I got it. In order to accept it, can you elaborate your comment as an answer, constructing a counter-example? TIA.
No, this is not possible because (locally) uniform limits of harmonic functions are harmonic so the limit $u=\lim\limits_{\epsilon\to 0} h_\epsilon$ would have to be harmonic to start with.
Well, I guess it depends on what one means by "harmonic", but at least in the flat Euclidean setting there is no amibguity. In this basic setting the fact that uniform limits of harmonic functions are harmonic immediately follows from the standard characterization of harmonic functions by the mean-value property (which is trivially stable under uniform limits).
A counterexample is therefore given by any non-harmonic smooth function. (I will not insult MO's readership by giving an explicit such function!)
Think of it like this: In dimension 1 harmonic functions are affine, and clearly it is impossible to approximate an arbitrary smooth function uniformly (or in any reasonable topology, for that matters) by sequences of affine functions. Going to higher dimensions does not help with the "approximability".
|
2025-03-21T14:48:30.434711
| 2020-04-28T18:15:46 |
358795
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"authors": [
"Yuval Filmus",
"https://mathoverflow.net/users/156535",
"https://mathoverflow.net/users/7732",
"mkultra"
],
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/358795"
}
|
Stack Exchange
|
An attempt to find expected value of clique number of special random graph
Let $G(n)=(V,\mathcal{E})$ be a random graph definded as follows:
$V=[n]=\{1,2, ... ,n\}$ and for all $i,j\in V$ so that $i\ne j$ we have $\{i,j\}\in\mathcal{E}$ with probability $p$. Where $p\in[0,1]$
So i was trying to find expected value of a clique number of a $G(n)$:
$X(n,p)=E[\omega(G(n))]$, where $\omega(G)$ is a clique number of a graph $G$
Here is my attempt:
Lets call $G(n)$ a previous graph, we create a new graph by attaching a star to the previous graph.
$\omega(G(n+1))-\omega(G(n))=\begin{cases} 1, & \mbox{when additional star is connected with a clique in previous graph} \\ 0, & \mbox{otherwise} \end{cases}$
Lets denote the first case as a $A$.
So if we put expected value we will get the following: $X(n+1,p)=X(n,p)+P(A)$
The problem is to find $P(A)$.
Here is my idea:
We can associate to the graph $G(n)$ a clique matrix, which is matrix that columns corresponds to vertices of a graph, and rows corresponds to cliques. All the rows has exactly $\omega(G(n))$ ones and $n-\omega(G(n))$ zeros. If we pick some columns (this is attaching our star) then we would choose some rows of this matrix with some probability. This probability is exactly $P(A)$
Any ideas how to find $P(A)$?
Regards.
Your random graph is not special at all. It's the standard $G(n,p)$ model. You can find the answer in textbooks.
I can't find any exact results, only estimations.
The estimates are pretty good, the error is $O(1)$ or perhaps even smaller. Asking for more is too much to hope for.
Not an “answer” in that this post has no original thought, but too long for a comment.
As Yuval Filmus notes, this is in fact the standard (and most studied) model of random graphs, and it’s called the Erd\H{o}s-R\’enyi model. If you can’t find information on the clique number, then search for the independence number of the Erdos-Renyi graph (which is distributed exactly the same as the clique number of $G(n, 1-p)$). That might google better for you?
These parameters are now very well understood. In short, let $k_0$ be the largest value of $k$ for which ${n \choose k} p^{{k \choose 2}} \geq 1.$
Then the clique number of $G(n,p)$ is essentially equal to $k_0$ (so its expected value is also essentially equal to $k_0$).
By “essentially equal to,” I mean this in a very strong sense. For instance, if $p=1/2$, then with very high probability, the independence number of $G(n,1/2)$ is between $k_0 -3$ and $k_0$. (So it is very usually equal to one of those four values! How cool!!! And in fact, for “most” values of $n$, this result can actually be improved even more, and this has to do with the number theoretic properties of $n$.)
The clique number is also tightly concentrated about its mean for other ranges of $p$.
The above statement is actually relatively easy to prove if you already have a strong background in probability. See for instance these lecture notes on the subject:
https://people.eecs.berkeley.edu/~sinclair/cs271/n19.pdf
You can also find all of this in Alon and Spencer (The Probabilistic Method), which is one of the most commonly used texts on random graphs. It’s pretty good for beginners in the subject, but you may want to skip around and also supplement it with lecture notes you find online.
|
2025-03-21T14:48:30.434953
| 2020-04-28T18:45:34 |
358801
|
{
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"Iosif Pinelis",
"https://mathoverflow.net/users/144000",
"https://mathoverflow.net/users/36721",
"republic"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/358801"
}
|
Stack Exchange
|
Concavity of a function along a path
Suppose that $f(x,y)$ is a continuously differentiable function
and $g(x,y) =xy-f(x,y)$. I know that $g$ is concave
if and only if $(-f_{xx})(-f_{yy}) -(1-f_{xy}) ^{2}>0$ and $f_{xx}>0$.
Now suppose that I "travel" along the function $g$ on a path that satisfies
$f_{x}=y$. Thus, along this path $f_{xy}=1$.
Is it correct that along the path the function $g$ is concave if and only if $
(-f_{xx}) (-f_{yy}) >0$ and $f_{xx}>0$ ? An explanation is much appreciated.
What do you mean by "along the path the function $g$ is concave"?
I mean the following. Suppose that $\left( x_{1},y_{1}\right) ,\left(
x_{2},y_{2}\right) $ and $\left( x_{3},y_{3}\right) $ satisfy $f_{x}\left(
x_{1},y_{1}\right) =y_{1}$, $f_{x}\left( x_{2},y_{2}\right) =y_{2}$,$
f_{x}\left( x_{3},y_{3}\right) =y_{3}$ and $x_{2}=\lambda x_{1}+\left(
1-\lambda \right) x_{3}$ for some $\lambda \in \left( 0,1\right) $. Then $
f\left( x_{2},y_{2}\right) >\lambda f\left( x_{1},y_{1}\right) +\left(
1-\lambda \right) f\left( x_{3},y_{3}\right) $. Thanks!
Of course, your "if and only if" statement is incorrect. E.g., let
$$f(x,y):=x^2/2+y^2/4.$$
Then your conditions $(-f_{xx})(-f_{yy})>0$ and $f_{xx}>0$ obviously hold. Further, here
$f_x(x,y)=y$ means $y=x$, and hence the concavity of $g$ along the path means that $g(x,x)$ is concave in $x$. However, $g(x,x)=x^2/4$ is strictly convex, and thus not concave, in $x$.
As for your reasoning, if you try to formalize it, avoiding terms such as "along the path" and "travel", perhaps you will see where you made mistakes. Hints: (i) $f_x(x,y(x))=y(x)$ does not imply $f_{xy}(x,y(x))=1$ (you are confusing independent and dependent variables here) and (ii) the concavity of $g(x,y)$ in $(x,y)$ means that the restriction of $g$ to every straight line is concave, and this concavity does not in general imply any kind of concavity over curved lines.
|
2025-03-21T14:48:30.435088
| 2020-04-28T19:29:51 |
358805
|
{
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"https://mathoverflow.net/users/102390",
"https://mathoverflow.net/users/134242",
"pupshaw",
"skd"
],
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"url": "https://mathoverflow.net/questions/358805"
}
|
Stack Exchange
|
commutative "subalgebras" of associative ring spectra
A bit of context: for any ordinary associative algebra $A$ and element $x \in A$, the subalgebra spanned by the powers of $x$ is commutative. In the universal example, this says that the free associative algebra on one element is commutative. Or if we prefer: any associative algebra is the union of its commutative subalgebras.
I expect any analogue of these statements to be false for ring spectra, and moreover I no longer have a clear idea of what a "subalgebra" ought to be. But I haven't been able to produce an example of what goes wrong.
So here's a tentative definition. Let $A$ be an $A_\infty$ ring spectrum. We'll say that $x \in \pi_*(A)$ is of commutative origin in case there exists an $E_\infty$ ring spectrum $B$, equipped with an $A_\infty$ map $f: B \to A$ such that $x \in \text{image}(\pi_*(f)).$
My question is: is it easy to produce an example of an associative ring spectrum $A$ and an element $x \in \pi_*(A)$ which is not of commutative origin in this sense?
The question arose in thinking about whether one could reduce nilpotence questions in $A$, which are controlled by $MU$ according to the theorems of Devinatz-Hopkins-Smith, to nilpotence questions in a lift to $B$, where one has the simpler(?) May nilpotence machinery which only requires integral homology.
(May's nilpotence conjecture was affirmed by Mathew-Naumann-Noel in https://arxiv.org/abs/1403.2023)
Let $A$ be an $\mathbf{E}_1$-ring, and let $x\in \pi_n A$. There are two distinct cases to consider. First, if $n = 0$, then the answer to your question is that $x$ is in the image of an $\mathbf{E}_1$-map from an $\mathbf{E}_\infty$-ring. Indeed, then $x:S^0\to A$ extends to a map $\Sigma^\infty_+ \mathbf{Z}_{\geq 0}\to A$, essentially because $B\mathbf{Z}_{\geq 0} = S^1$ (and so is evidently in the image of this map). Now observe that $\Sigma^\infty_+ \mathbf{Z}_{\geq 0}$ admits the structure of an $\mathbf{E}_\infty$-ring (where the commutative monoid $\mathbf{Z}_{\geq 0}$ is regarded as an $\mathbf{E}_\infty$-space).
Next, if $n>0$, then $x$ is not necessarily in the image of an $\mathbf{E}_1$-map from an $\mathbf{E}_\infty$-ring. Consider the universal case, when $A = \Omega S^{n+1}_+$ is the free $\mathbf{E}_1$-ring on one generator in degree $n$, and $x\in \pi_n(A)$ is given by the suspension $E:S^n\to \Omega S^{n+1}$. The question, then, is whether one can find an $\mathbf{E}_\infty$-ring $B$ and an $\mathbf{E}_1$-map $f:B\to \Omega S^{n+1}_+$ such that $E$ lifts to $B$. If there was a lift $\widetilde{E}:S^n\to B$ of $E$, then there must be an $\mathbf{E}_k$-map $\widetilde{E}:\Omega^k S^{n+k}_+\to B$ for every $k\geq 0$. Moreover, the $\mathbf{E}_1$-composite
$$\Omega S^{n+1}_+\to \Omega^k S^{n+k}_+\xrightarrow{\widetilde{E}} B\xrightarrow{f} \Omega S^{n+1}_+$$
would be the identity on $\Omega S^{n+1}_+$. In particular, $\Omega S^{n+1}_+$ would be an $\mathbf{E}_1$-summand of $\Omega^k S^{n+k}_+$. The James splitting gives an equivalence $\Omega S^{n+1}_+ = \bigvee_{k\geq 0} S^{nk}$, and so $S^n$ would be a summand of $\Omega^k S^{n+k}_+$. I think this fails, for instance, once $k\geq 2$ (recall that $n>0$). (By the way: the proof of May nilpotence relies on the nilpotence theorem. I can talk more about this over email if you're interested.)
The essence of this answer is that $S^1$ is the only sphere which admits the structure of an infinite loop space.
One other thing to mention that might be of interest is the following. Although it's very hard (and often impossible) to lift to $\mathbf{E}_\infty$-rings, it's sometimes possible to prove liftings to $\mathbf{E}_k$-rings for finite $k$. For instance, let us work $p$-locally, where $p$ is an odd prime (slight variants also work at $p=2$). A hard theorem of Cohen, Moore, and Neisendorfer says that the map $p:S^{2n-1}\to S^{2n-1}$ factors through the double suspension $E^2:S^{2n-1}\to \Omega^2 S^{2n+1}$. In particular, there's a map $\Omega^2 S^{2n+1}\to S^{2n-1}$ which is degree $p$ on the bottom cell. Looping, we get that $p\cdot E\in \pi_{2n-2} \Omega S^{2n-1}$ factors through an $\mathbf{E}_1$-map $\Omega^3 S^{2n+1} \to \Omega S^{2n-1}$. Iterating, you find that $p^k\cdot E\in \pi_{2n-2} \Omega S^{2n-1}$ factors through an $\mathbf{E}_1$-map $\Omega^{2k+1} S^{2n+2k-1} \to \Omega S^{2n-1}$. Stabilizing, you get (for each $k$) an example of an element in the homotopy of an $\mathbf{E}_1$-ring which lifts to an $\mathbf{E}_k$-ring.
this is very helpful, though I wonder if you could say a bit more about the James splitting. and absolutely, did not mean to imply this could have given a new proof of nilpotence, just was wondering if it might produce an easier-to-check condition. so much for that!
@pupshaw Edited, and added another comment.
|
2025-03-21T14:48:30.435500
| 2020-04-28T19:31:34 |
358806
|
{
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"authors": [
"Johannes Hahn",
"LSpice",
"Richard Stanley",
"YCor",
"Yambo",
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}
|
Stack Exchange
|
Groups whose non-linear irreducible characters are all non-faithful
I am interested in knowing if there is any literature that describes finite solvable groups whose non-linear complex irreducible characters are all non-faithful.
One reference is https://arxiv.org/pdf/1504.01343.pdf.
what's a "non-linear" character?
@YCor, character of a non-1-dimensional representation.
@RichardStanley, I took the question to be about faithful characters (meaning injective?). Is that the same as characters of faithful representations?
@LSpice: A (complex) character is usually called faithful if the associated representations are faithful which is equivalent to $\chi(g)=\chi(1) \implies g=1$
OK; If $G$ is non-abelian the 1-dimensional characters are non-faithful, so the question is about non-abelian groups all of whose characters are non-faithful (but the restriction to non-abelian is artificial).
A necessary and sufficient condition for a finite group $G$ to have a faithful complex irreducible character is given by a Theorem of W. Gaschutz, which states that a finite group $G$ has a faithful irreducible complex character if and only if the socle $S$ of $G$ is generated by the $G$-conjugates of a single element of $S$. Here, the socle of $G$ is the product of all minimal normal subgroups of $G$.
If $G$ is solvable, then all minimal normal subgroups are Abelian of prime-power order, and the socle of $G$ is Abelian of square-free exponent. Note that a non-Abelian solvable group $G$ has no faithful linear character. Hence a non-Abelian solvable group $G$ has no faithful (non-linear) complex irreducible character if and only if its socle $S$ can not be generated by the $G$-conjugates of any single element of $S$.
An extension characterizing countable discrete groups $G$ having a faithful irreducible unitary representation, can be found in Corollary 1.9 of PE Caprace, P. de la Harpe, "Groups with irreducibly unfaithful subsets for unitary representations". The condition is that finite normal subgroups of $G$ satisfy a certain property (and in particular is satisfied if there is no nontrivial abelian finite normal subgroup). See on arXiv.
Thank you, I have something to work with.
|
2025-03-21T14:48:30.435661
| 2020-04-28T19:41:49 |
358807
|
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"ABIM",
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}
|
Stack Exchange
|
Bounding the total variation metric between Gaussian mixtures
Let $\mathcal{P}(\mathbb{R}^d)$ the space of probability measures on $\mathbb{R}^d$ with total variation metric $\delta$, fix $k \in \mathbb{N}$, and let $\mathcal{P}'\subset \mathcal{P}(\mathbb{R}^d)$ be the metric subspace of all probability measures $\nu$ of the form
$$
\nu : =\sum_{i=1}^k \alpha_i \nu(\mu_i,\Gamma_i),
$$
where $\sum_{i=1}^k \alpha_i =1, \, 0\leq \alpha_i\leq 1$, $\mu_i \in \mathbb{R}^d$, and $\Gamma_i$ is a symmetric positive semi-definite matrix.
Can $\delta(\nu_1,\nu_2)$ be bounded above in terms of the distance between the Euclidean distance between their parameters (i.e.: $\alpha_i,\mu_i, \Gamma_i$)?
When $k=1$ bounds are obtained here; but what about in general?
The bound is actually simpler than it looks. Extend your convex combinations to linear ones and then you obtain a fininte-dimensional (Banach) subspace of the Banach space of all finite signed Borel measures. Since all norms are equivalent on finite-dimensional Banach spaces then you're done (after restricting back to the ball intersect the positive orthant).
|
2025-03-21T14:48:30.435766
| 2020-04-28T21:04:01 |
358814
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"authors": [
"Iosif Pinelis",
"Karim KHAN",
"LSpice",
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|
Stack Exchange
|
$ \|u_k-v_k\|_2\leq \min \bigg(\inf_{n\geq k}{\|f_n1_{\{|f_n|\leq k\}}\|_2},\frac{\epsilon_{k-1}}{4k}\bigg) $
Let $(E,\mathcal{A},\mu)$ be a finite measure space and $\{f_n\}_n$ be a sequence of simple functions such that:
\begin{align*}
f_n1_{\{\lvert f_n\rvert\leq k\}}\overset{\sigma(L^2,L^2)}{\underset{n}{\longrightarrow}} u_k, &\qquad\forall k\geq 1 \\
\|u_k\|_2\leq 2\|f_n1_{\{\lvert f_n\rvert\leq k\}}\|_2,&\qquad \forall n\geq k.
\end{align*}
Put $\epsilon_k=\frac{1}{2^k}$ $(k\geq 1)$. Why does there exist, for each $k\geq 1$, a simple function $v_k$ such that:
$$
\|u_k-v_k\|_2\leq \min \bigg(\inf_{n\geq k}{\|f_n1_{\{\lvert f_n\rvert\leq k\}}\|_2},\frac{\epsilon_{k-1}}{4k}\bigg)
$$
The specificity of this, including that $\epsilon_{k - 1}/(4k)$ (which one would expect to come out of the details of a proof, not to be a goal in itself) suggests that this is something from an exercise, or possibly from a paper you are reading. If the former, then it doesn't belong here. If the latter, then what paper?
$\newcommand\de{\delta}$ $\newcommand\ep{\epsilon}$ $\newcommand\al{\alpha}$
Fix any natural $k$. Let $$\de_k:=\inf_{n\ge k}\|f_n 1_{\{|f_n|\le k\}}\|_2$$ and $$\eta_k:=\ep_{k-1}/(4k).$$
We have
$$\|u_k\|_2\le2\de_k \tag{1}$$
and
$$\eta_k>0.$$
We want to show that there exists a simple function $v_k$ such that
$$
\|u_k-v_k\|_2\le\al_k:=\min(\de_k,\eta_k). \tag{2}
$$
If $\de_k=0$, then, by (1), $\|u_k\|_2=0$, and hence we may let $v_k:=0$, to have (2).
Otherwise, $\al_k>0$. By the condition $f_n 1_{\{|f_n|\le k\}}
\overset{\sigma(L^2,L^2)}{\underset n\longrightarrow} u_k$, we have $|u_k|\le k$ $\mu$-almost everywhere ($\mu$-a.e.). So, we can find a simple function $v_k$ such that $|u_k-v_k|\le\al_k/\sqrt{1+\mu(E)}$ $\mu$-a.e. Then we will have the inequality in (2), as desired.
@losif Can we say that there exist q simple function $v_k$ such that $|v_k|\leq k$ $\mu-a.e$ and $|u_k-v_k|\leq \frac{\alpha_k}{\sqrt{1+\mu(E)}}$ $\mu-a.e$?
@KarimKHAN : Yes, we can do that, by the standard procedure.
@losif Can you prove it to me? If you have the time.
@KarimKHAN : I suggest you post this question on Math SE.
|
2025-03-21T14:48:30.435918
| 2020-04-28T21:05:23 |
358815
|
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|
Stack Exchange
|
"Probability" for a partitioned matrix to be singular
Let $A,B\in\mathbb{R}^{n\times n}$ be two nonsingular matrices with $A\ne B$, and consider the following partitioned matrix
$$
M:=\begin{bmatrix}AA^\top + BB^\top & A^\top \Delta_1 A + B^\top \Delta_2 B \\
A^\top \Delta_1 A + B^\top \Delta_2 B & A^\top A + B^\top B \end{bmatrix}
$$
where $\Delta_1$ and $\Delta_2$ are diagonal matrices with $0$-$1$ diagonal entries.
It is quite easy to see that if the diagonal entries of $\Delta_1$ and $\Delta_2$ are $1$s, then $M$ is singular.
However, numerical simulations with randomly generated $\Delta_i$ (whose diagonal entries take values $0$ or $1$ with probability $1/2$) suggests that, in general, it is very "unlikely" that $M$ is singular. I'd like to understand if there is a way to formalize this fact, so my question is:
Given all possibile choices of matrices $\Delta_1$ and $\Delta_2$, how is it "likely" that matrix $M$ is singular?
I'm aware that my question is not very rigorous, but I hope it's understandable.
Thanks in advance for any help.
For each $n$ the answer is rational number with power of two in the denominator. Did you try to search for the sequence of numerators in OEIS?
@VítTuček: No, I did not. Could you please elaborate a little more on this?
My understanding of your question is that you have a set of $2^{n+1}$ diagonal matrices with zeros and ones on the diagonal and you ask how many there are such that $\det(M(\Delta_1, \Delta_2)) = 0$ identicaly as a polynomial in $A$ and $B$ entries. I am suggesting you calculate first few of these numbers and then look for this sequence on https://oeis.org/ If you find something, great -- it can suggest possible way of attacking the problem. If you don't find anything, well... it seems that the sequence is new and hence we don't get any clues for free.
|
2025-03-21T14:48:30.436065
| 2020-04-28T22:01:10 |
358819
|
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"Nate",
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}
|
Stack Exchange
|
Doubly periodic 4 color theorem?
Let $G$ be a graph embedded (without crossings) on a torus $T$. It's fairly well known that this implies the chromatic number of $G$ is at most 7. If I lift $G$ to the universal cover of $T$, we get a doubly periodic planar graph $\tilde{G}$ and of course the four color theorem tells us there is a four coloring of $\tilde{G}$.
With a little work I can improve this slightly to say that for any such $G$ there is a finite cover $\widehat{T}$ such that the corresponding cover $\widehat{G}$ is four colorable. My question is: Can this be done uniformly in $G$? If so, how small can we take the cover?
Concretely: Does there exist a covering map $T' \to T$ such the pull back to $T'$ of any graph embedded on $T$ can be properly four colored? Which covers work and what is the minimal degree of such a cover?
I was especially interested in the case where $T = \mathbb{R}^2/\mathbb{Z}^2$ and $T'$ was the 4-fold cover $\mathbb{R}^2/(2\mathbb{Z})^2$ but would be interested in hearing about any case.
EDIT: Since I thought this was a fun question I thought about it more and did some more searching through literature. Here are my current best partial results:
1) For a surface $\Sigma$ of genus $g$ there exists a degree $36^g$ cover such that any graph embedded on $\Sigma$ becomes $6$-colorable when pulled back to the cover.
2) For genus 1, any graph embedded on a torus becomes $5$-colorable when pulled back to the $\mathbb{Z}/2\mathbb{Z} \times \mathbb{Z}/2\mathbb{Z}$ cover described above.
Can you illustrate how you get your $\widehat{G}$? I'm almost certainly missing something, but if we take $G$ to be $K_7$, then I'm not sure what finite cover of $T$ four-colors $\widehat{G}$...
There is a pigeon hole principal argument - take your favorite essential curve missing the vertices of $G$ and take the infinite cyclic corresponding to this curve. It embeds in the plane and so has a 4-coloring. But the finite chunk of the graph in one of the lifts has only so many 4-colorings, so looking at the different lifts they eventually repeat. That repetition gives you a finite cover. Ian Agol observed this here. I would be interested in Nate's question for arbitrary genus also.
Indeed, that pigeonhole argument sounds exactly like what I came up with. $K_7$ (at least the embedding I tried this afternoon, which might be unique?) was 4-colorable when lifted to the 4 coloring I mentioned above.
@Nate Apparently that embedding is in fact unique - Negami says that any 6-connected toroidal graph is uniquely embeddable in a torus.
I thought a bit about the higher genus case. I believe I can at least show for any genus $g$ there exists a finite cover $\Sigma_h \to \Sigma_g$ such that the pull back of any graph on $\Sigma_g$ is $6$ colorable. Seems plausible such a cover could exist for 4 colors too.
As Michael Klug points out in the comments, I've thought about related questions before. I'll make a few comments on the question.
Firstly, the usual reduction allows one to consider triangulations on a surface: if a graph $G$ does not induce a triangulation of $\Sigma$, then we can complete it to a triangulation $G'$ so that if $G'$ (or a cover $\hat{G'}$ induced by a cover $\hat{\Sigma}$) is 4-colorable then so is $G$ (or $\hat{G}$).
So let's assume that $G$ induces a triangulation of $\Sigma$. Then the dual graph $G^*$ (with respect to the embedding in $\Sigma$) is a cubic graph. If $G^*$ is 3 edge-colorable (i.e. has a Tait coloring), then one can see that a $\mathbb{Z}/2\times \mathbb{Z}/2$-cover $\hat{\Sigma}\to \Sigma$ will give a lift of $G$ which is 4-colorable. To prove this, identify the three colors with the non-zero elements of the Klein 4-group $V=\mathbb{Z}/2\times \mathbb{Z}/2$. Then coloring the vertices of $G$ corresponds to coloring the faces of $G^* \subset \Sigma$. If we color one face of $G^*$ by $0\in V$, then each time we cross an edge of $G$, we change the color by adding the element of $V$ corresponding to the edge coloring. This is locally well-defined near a vertex, but globally might have holonomy in $V$. So passing to a 4-fold cover $\hat{\Sigma}\to \Sigma$ induced by this holonomy, we get a pulled-back graph $\hat{G}$ which is 4-colorable. (In the planar case, there is no holonomy, and hence Tait's observation that Tait colorings suffice).
Thus it suffices to consider 3-edge colorings of cubic graphs in $\Sigma$. The Snark theorem implies that if the graph $G^*$ is not 3-edge colorable, then there is a Petersen minor (that is, a copy of the Petersen graph embedded topologically in $G^*$). The Petersen graph is non-planar, so must be embedded in an essential way in $\Sigma$ (not isotopic into a disk). Hence any Petersen subgraph of $G^*$ will not lift to some 2-fold cover of $\Sigma$. However, passing to a cover to which no Petersen subgraph lifts, there may be new Petersen subgraphs of $\hat{G^*}$ created. Nevertheless, one can ask if there is a finite cover $\hat{\Sigma}\to \Sigma$ such that the preimage of any embedded cubic graph in $\Sigma$ is not a Snark? Seems implausible, but it is a natural question to ask when thinking about virtual Tait coloring.
One can weaken the condition of Tait coloring, allowing passage to a finite-sheeted cover. If a cubic graph $G^*$ has a perfect matching (also called a 1-factor, a degree 1 regular subgraph spanning the vertices), then the complementary subgraph is a 2-factor, i.e. a regular subgraph of degree 2 containing every vertex, homeomorphic to a union of circles, each component a cycle graph . If the 2-factor is also bipartite (2-colorable, r every component has an even number of edges), then we may 2-color the 2-factor and use a third color for the 1-factor to get a Tait coloring of $G^*$. Then we can look for a 2-factor $C\subset G^* \subset \Sigma$ such that every non-bipartite component of $C$ is a non-trivial curve on $\Sigma$. In this case, we can pass to a $2^{2g}$-fold cover in which ever non-separating curve has each component of the preimage an even-index cover, and every separating essential curve has preimage components non-separating, and repeat, to get a finite cover for which the preimage of every essential curve is an even index cover on each component. Then the preimage of a 2-factor with the above properties will be a bipartite 2-factor, and hence the preimage graph will be 3-colorable (and a further 4-fold cover will give a 4-colorable dual triangulation).
One knows that every bridgeless cubic graph has a perfect matching (or 1-factor, and hence a 2-factor), known as Petersen's theorem. One could try to modify the proof to try to show that a graph $G^*\subset \Sigma$ has a 2-factor with odd cycles all essential. But I didn't see how to do this. In any case, it seems possibly easier to find a controlled cover of $\Sigma$ where the preimage of every cubic graph has a 2-factor with essential odd cycles.
Another special case is triangulations of even degree. Then we can try to 3-color the vertices. Once one 3-colors the vertices of a triangle, there is a unique way to continue the coloring, well-definied locally around a vertex because of the even degree hypothesis. This may have non-trivial holonomy, but passing to an $S_3$-cover (of index 6), we get a preimage which is a 3-colorable graph. This works e.g. for $K_7\subset T^2$.
Ultimately, this problem ought to be as hard as the 4-color theorem itself. Given a large graph embedded in a disk, one ought to be able to insert it into a disk on a surface $\Sigma$ of genus $>0$ as a subgraph. Coloring the graph larger graph in a finite-sheeted cover will induce a coloring of the planar graph. So I think one will likely have to use the 4-color theorem or parts of its proof as an essential ingredient in resolving this question.
One reduction I've contemplated is to make the surace the boundary of a handlebody, and pass to the universal cover of the handlebody. The preimage of the boundary is a planar surface, so the preimage of the graph $\tilde{G}$ is 4-colorable. The
space of 4-colorings of $\tilde{G}$ is a closed subset of the Cantor set $4^\tilde{V}$, where $\tilde{V}$ is the vertex set of $\tilde{G}$. The covering translations form a rank $g$ free group. If there is a probability measure on the space of colorings which is invariant under the free group action, then I can show that there is a finite-sheeted cover (induced by a cover of the handlebody) which is 4-colorable, using a theorem of Lewis Bowen. However, I haven't been able to show the existence of such a probability measure (again, this may require non-trivial input from the proof of the 4-color theorem). One could do a similar thing with 2-factors of cubic graphs, where every contractible cycle is bipartite, and ask for an invariant probability measure on these. This approach, if it worked, would likely not give a uniform finite-sheeted cover.
Thanks for detailed response! This is very similar to how I was thinking about the problem.
|
2025-03-21T14:48:30.436624
| 2020-04-28T22:04:12 |
358820
|
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"Ekaterina Bogdanova",
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|
Stack Exchange
|
Is the morphism of sheaves $(R \mapsto GL(R((h)))) \rightarrow (R \mapsto PGL(R((h))))$ surjective in Zariski topology?
Consider two functors given by $R \mapsto GL(R((h)))$ and $R \mapsto PGL(R((h)))$ for a ring $R$. It is easy to see that these functors are sheaves in Zariski topology (in fact for any affine variety $Y$ the functor $R \mapsto Y(R((h)))$ is a sheaf). Is the morphism of sheaves $(R \mapsto GL(R((h)))) \rightarrow (R \mapsto PGL(R((h))))$ surjective in Zariski topology?
Welcome new contributor. No, that morphism of Zariski sheaves is not surjective. There is a discussion of this in Serre's "Galois cohomology".
@JasonStarr thanks a lot! But could you be a little more specific? In which chapter can I find this discussion?
I do not have a copy of the book with me, but it should be prior to the long exact sequence of non-Abelian cohomology. For instance, if $R$ equals $\mathbb{C}[x,y]/\langle y^2-x^2(x-1)\rangle$, the $n$-torsion elements in the Picard group give counterexamples.
Dear @JasonStarr, I am confused. I've looked it up in the Serre`s book and have not find information concerning this question. To this point I am not sure that I formulated it clear enough. My question is the following. For any ring $R$
an element in $PGL(R((h)))$ defines a line bundle $L$ on $Spec(R((h)))$. I wonder whether it is true that there exists an open covering $Spec R = \cup Spec R_{f_i}$ such that pullbacks of $L$ to $Spec(R_{f_i}((h)))$ are trivial.
I don't see why your ring provides a counterexample, so I will be grateful if you explain it in more details.
You are correct, and I am incorrect. I misread the question. I thought you were just asking about surjectivity of the group homomorphism from $\textbf{GL}(R((h))$ to $\textbf{PGL}(R((h))$. Now I see that is not what you are asking. Sorry for the confusion.
|
2025-03-21T14:48:30.436774
| 2020-04-28T22:35:35 |
358824
|
{
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"EquivKun",
"Phil Tosteson",
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|
Stack Exchange
|
Reference for equivariant derived Künneth formula
I'm looking for a reference for the following statement in as much generality as possible, assuming it is correct.
Let's $X$ and $Y$ be "spaces" with a $G$-action. We can take the $G$-product defined by
$$
X \times_G Y = X \times Y / \sim
$$
where $(x,y) \sim (g x, yg^{-1})$.
Let $K$ be a complex of constructible sheaves on $X \times_G Y$. Let $f: X \times Y \to X \times_G Y$ be the canonical map. Suppose $G$ has cohomology supported in degree 0 e.g. it is homotopic to a 0-dimensional space. Then is it true in reasonable cases that
$$
R\Gamma_c(X \times_G Y, K) = R\Gamma_c(X, f^*K|_X) \otimes_G^L R\Gamma_c(Y,f^*K|_Y)
$$
where the right side is derived $\otimes_G$, perhaps up to a shift in dimension? In particular, if $Y$ is a point, and $G$ acts freely on $X$, I'm hoping that $R\Gamma_c(X/G, K)$ will be the derived coinvariants of $R\Gamma_c(X, f^*K)$ up to a shift.
If anyone has some references regarding relationships between any of the above complexes, I'd appreciate it. Thanks.
As stated, this is too strong. Consider $G = e$ and $K$ a complex on $X \times Y$ that is not an external tensor product of complexes on $X$ and $Y$. E.g. $X = Y$ and $K = \Delta_* \mathbb Z_X$.
@PhilTosteson Yes, that's right. Your example shows it's too strong as stated. Perhaps I want a condition like $f^*K$ is the external tensor product of two complexes on $X$ and $Y$... Do you know of a good way to salvage the statement to get something generally true?
You could take $G$-equivariant sheaves $\mathcal F_1$ on $X$ and $\mathcal F_2$ on $Y$. And use them to build a sheaf $\mathcal F_1 \boxtimes_G \mathcal F_2$ on $X \times_G Y$. Alternately you could just consider a single space $T$ with a $G$ equivariant sheaf on it. This isn't actually less general, because you can take $T = X \times Y$.
@PhilTosteson That sounds quite promising. Do you think you could develop your comment into an answer with either a reference or a short proof if you know how to show that statement involving the derived coinvariants? I would prefer if it's possible to use as much "standard formalism" (e.g. six functors) as possible. Thanks.
|
2025-03-21T14:48:30.436937
| 2020-04-28T23:34:41 |
358828
|
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|
Stack Exchange
|
improved regularization for $\lambda$-convex gradient flows
It is well-known that gradient-flows of convex functionals are "parabolic" in some vague sense, and accordingly solutions tend to regularize instataneously. In the abstract context of gradient flows in metric spaces one has
Theorem Let $(X,d)$ be a "nice" metric space, and $\Phi:X\to\mathbb R\cup\{\infty\}$ a convex function. If $t\mapsto u_t$ is a solution to the gradient-flow
$$
"\dot u_t=-\nabla\Phi(u_t)"
$$
then for any $w\in X$ there holds
$$
|\partial\Phi|^2(u_t) \leq \frac{1}{t^2} d^2(u_0,w)+|\partial\Phi|^2(w).
$$
In particular if $\Phi$ attains a minimum at a point $w$ (with $|\partial\Phi|(w)=0$) then this quantifies an instantaneous regularization: The slope decreases at $t=0$ at least as
\begin{equation}
\label{eq:regularization}
|\partial\Phi|^2(u_t) \leq \frac{C}{t^2}
\tag{R1}
\end{equation}
wih $C=d^2(u_0,w)$ depending only on the initial datum.
Moreover if $\Phi$ has modulus of convexity $\lambda>0$ then
\begin{equation}
\label{eq:regularization2}
|\partial\Phi|^2(u_t) \leq e^{-2\lambda t} |\partial\Phi|^2(u_0)
\tag{R2}
\end{equation}
Here $|\partial\Phi|(v)$ denotes the metric slope at a point $v\in X$, and the ODE $"\dot u_t=-\nabla\Phi(u_t)"$ should be understood in the abstract metric sense, for example in the sense of Energy-Dissipation-Inequality - see [AGS]. Please allow me to remain sloppy here and dispense from the technical details/assumptions in order to avoid burdening the exposition (see e.g. [AGS, Theorem 11.2.1] for a rigorous statement in a precise context). This is easy to prove for example in the framework of Hilbert spaces.
Question:
The instantaneous regularization \eqref{eq:regularization} holds if
the driving functional is merely convex, i-e has a modulus of
convexity $0$. Can we strengthen this estimate under the stronger hypothesis that $\Phi$
has a modulus of convexity $\lambda>0$? Typically one would expect a
better rate
$$
|\partial\Phi|^2(u_t)\leq \frac{C}{t}\qquad ???
$$
for some $C>0$ possibly depending on $u_0,\lambda$. Of course
\eqref{eq:regularization2} is already much stronger (and optimal), but it requires the initial regularity $|\partial \Phi|(u_0)<\infty$ to make sense. What if $u_0\not\in D(\partial\Phi)$?
Actually any improvement of the form $|\partial\Phi|^2(u_t)\leq \frac{C}{t^{2-\epsilon}}$ with $\epsilon>0$ would suffice for my purpose.
For a non-trivial example of this improved regularity, take the (quadratic) Wasserstein space $(\mathcal P(\mathbb T^d),\mathcal W_2)$ on the flat torus, and take as a driving functional the entropy $\Phi(u)=H(u)=\int_{\mathbb T^d} u\log u\,dx$ (with the convention that $H(u)+\infty$ whenever $u$ is not absolutely continuous w.r.t the Lebesgue measure $dx$). The the gradient flow is the linear heat equation $\partial_t u=\Delta u$, and the (squared) slope is nothing but the Fisher information $|\partial H|^2(u)= F(u)=\int_{\mathbb T^d} \frac{|\nabla u|^2}{u}=\int_{\mathbb T^d}|\nabla\log u|^2 u$. From \eqref{eq:regularization} we already know that $F(u_t)\leq \frac{C}{t^2}$. However from the Li-Yau inequality [2] (see my previous post) it is know that the Fisher information decays at the BETTER universal rate $F(u_t)\leq \frac{C}{t}$ and in this case \eqref{eq:regularization} is clearly suboptimal.
I would be happy with an aswer even in Hilbert spaces (the statement is probably vacuous in finite dimension, though)
[AGS] Ambrosio, L., Gigli, N., & Savaré, G. (2008). Gradient flows: in metric spaces and in the space of probability measures. Springer Science & Business Media.
[LY] Li, P., Yau, S.T.: On the parabolic kernel of the Schrödinger operator. Acta Mathematica
156(1), 153–201 (1986)
|
2025-03-21T14:48:30.437206
| 2020-04-28T23:44:47 |
358829
|
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|
Stack Exchange
|
Existence of continuous map on real numbers with dense orbit?
Does there exist a continuous map $f:\mathbb{R}\rightarrow \mathbb{R}$ such that the forward orbit of 0 is dense in $\mathbb{R}$?
Existence of a point with the dense orbit is one of the definitions of topological transitivity. To see some examples of such maps on $\mathbb{R}$, google "transitive map on R"
I found a topologically transitive map on \mathbb{R} with backward orbit of every point dense on \mathbb{R}, but not the forward orbits.(Reference:- https://www.researchgate.net/publication/236026813_Some_classes_of_transitive_maps_on_R)
@erz not exactly definition, but characterization (Degirmenci, Kocak, Acta Math. Hungar. 99(3):185-187): for a nonempty and complete metrizable space with no isolated point, topologically transitive (for all $U,V$ nonempty open there exists $n\ge 1$ such that $f^n(U)\cap V$ is nonempty) is equivalent to having a dense forward orbit. (Nonempty is missing in their Prop 2.) In particular this applies to $\mathbf{R}$.
@yogamat therefore the paper you link, + the one I provide (Proposition 2) seems to answer your question, except that the proof doesn't tell you which point has a dense forward orbit: it just tells you that every point in a dense $G_\delta$-dense subset $B$ has a dense orbit. This set $B$ is explicit but exhibiting a point in it doesn't seem easy in such an example as the piecewise affine map from your link.
@YCor from my understanding there are 3 nonequivalent definitions of transitivity: a dense orbit, the one you provide, and the same as the one you provide but with $n\ge 0$. The latter two are equivalent unless the space is irreducible. I don't think it's the only case when historically there are multiple nonequivalent definitions.
Here is another paper that discusses this stuff: https://projecteuclid.org/euclid.rmjm/1181072815
@erz sure, but usage can hopefully converge to a coherent definition. So far my impression is that at places where they were careful about the issue of non-equivalence, they tend to choose the $f^n(U)\cap V$ definition as "topologically transitive".
@YCor, i could get a topologically transitive map from the paper, and it is same as having a dense orbit on $\mathbb{R}$, even though the exact point whose orbit is dense is not known
https://mathoverflow.net/a/363010/53155
Does this answer your question? Existence of topologically transitive map on Euclidean space
A more challenging question: does there exist a Lipschitz map on ${\bf R}$ with dense orbit?
As in this previous answer of mine, emulating random walks works pretty well for this kind of question.
Consider the map $T : \mathbb{S}_1 \to \mathbb{S}_1$ defined by
$$T(x) = 5x [1] \ \text{ if } \ 1/5 \leq x < 4/5,$$
$$T(x) = -5x [1] \ \text{ otherwise.}$$
Its graph is as follows:
The map $T$ is continuous, preserves the Lebesgue measure, is ergodic, and much more.
Now, let me introduce the $\mathbb{Z}$-extension $\widetilde{T} : \mathbb{S}_1 \times \mathbb{Z} \to \mathbb{S}_1 \times \mathbb{Z}$ of $T$ defined by
$$\widetilde{T} (x, p) := (T(x), p+F(x)),$$
where $F(x) = -1$ for $x \in [0,2/5)$, then $F(x) = 0$ for $x \in [2/5,3/5)$ and $F(x) = +1$ for $x \in [3/5,1)$. Note that $\widetilde{T}$ preserves the uniform ($\sigma$-finite) measure on $\mathbb{S}_1 \times \mathbb{Z}$.
The second coordinate of $\widetilde{T}^n (x, p)$ is $p+S_n F(x) := p+\sum_{k=0}^{n-1} F (T^k (x))$. Under the Lebesgue measure on $\mathbb{S}_1$, the sequence $(F \circ T^k)_{k \geq 0}$ is a sequence of i.i.d. random variables of symmetric distribution $2/5\cdot \delta_{-1} + 1/5 \cdot \delta_0+ 2/5\cdot \delta_{+1}$, so that the process $(S_n F)_{n \geq 0}$ is ergodic and recurrent.
A bit more work (but not that much, given the simplicity of the model) gives that $\widetilde{T}$ is ergodic and recurrent for the uniform measure on $\mathbb{S}_1 \times \mathbb{Z}$. This is very much folklore, although I have to admit it can get annoying to pinpoint the best reference and fill the gaps. As a consequence, almost every point has a dense orbit.
Now, everything is on $\mathbb{S}_1 \times \mathbb{Z}$; however, is we identify $\mathbb{S}_1$ with $[0,1)$ and then $\mathbb{S}_1 \times \mathbb{Z}$ with $\mathbb{R}$, we get a map $S$ from $\mathbb{R}$ to $\mathbb{R}$. The specific choice of $T$ gives that $S$ is continuous, actually 5-Lipschitz, with a sawtooth-like graph (in black, the line with equation $y=x$):
Again, Lebesgue-almost every point has a dense orbit. Here is the picture of an orbit from a random (uniform in $[0,1]$) starting point:
The orbit of $0$ is not dense, as it is a fixed point; however, conjugating by a Lebesgue-generic translation gives the map we want.
The same construction work on $\mathbb{R}^2$ (just use $S \times S$), where Lebesgue almost every orbit will be dense. Things get more annoying in higher dimension, since the random walks are no longer recurrent. This can be solved by getting a tweak $R$ of $S$ favouring orbits recurring quickly to zero, ensuring that $R$ preserves a unique absolutely continuous invariant measure, with respect to which it is mixing; then $(R, R, \ldots, R)$ is mixing with respect to the product measure on $\mathbb{R}^n$, and from there topologically mixing.
@Matt F.: No. If $x < 1/5$, then $T(x) = 1-5x$. Going back to the real number, we get $S(x) =1-5{x}+ \lfloor x \rfloor + F({x}) = \lfloor x \rfloor + 1-5{x}-1 = \lfloor x \rfloor -5{x}$ whenever ${x} < 1/5$, so the sequence $(x_n)$ can decrease.
Basically, the graph of this function starts from $(0,0)$, goes down to $(1/5, -1)$, goes up to $(4,/5, 2)$, goes down to $(6/5, 0)$, goes up to $(9/5, 3)$, etc. I will post a graph tomorrow to make the construction more intuitive (because it really is) ; it's a situation where a picture is worth a thousand words.
@Matt F.: The images are now there.
|
2025-03-21T14:48:30.437692
| 2020-04-28T23:54:47 |
358830
|
{
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|
Stack Exchange
|
Conjugacy in the quaternion group
Let $G$ be a non-commutative group, and suppose we are given two elements $x, y \in G$ which are conjugate, i.e. we know there exists some $z \in G$ such that $zxz^{-1} = y$. Can we find $z$ given $x$ and $y$? When $G$ is the set of $n \times n$ invertible matrices, finding $z$ is simply a matter of finding a change of basis which takes $x$ to $y$. I'm interested in the more general case, where $G$ is an arbitrary non-commutative group. I am especially interested in the case where $G$ is the group of quaternions of unit norm.
I'm not sure whether "can we find" makes sense in general, but if $G$ is a computable group, one has an algorithm whose input is a pair $(x,y)$, and: if $x,y$ are conjugate, outputs some $z$ with $zxz^{-1}=y$ (and if they're not conjugate possibly doesn't answer): just enumerate all $z$, test $zx=yz$ and stop when it's true. A better algorithm is when the algorithm can be improved to stop and say no (no pun intended!) when they're not conjugate. This is called solvable conjugacy problem, is well documented (and not always true for f.p. groups with solvable word problem).
You seem to be just asking whether a group $G$ has a solvable conjugacy problem, which is one of the three fundamental decidability problems for groups posed by Dehn in 1911.
@DerekHolt no, because OP takes as an assumption that the elements are conjugate, and asks to output a conjugating element. As you know, plenty of algorithmic problems (e.g., isomorphism problem for hyperbolic groups, etc) have this kind of scheme (input X, output something about X, requiring the program to work if X satisfies some property P and possibly not halt otherwise— where usually property P is not decidable).
@YCor OK, so this problem is solvable in theory for finitely (or even countably) generated groups with solvable word problem. But we seem to see a lot of questions of the type "Can we do this?" where it is not clear whether this means is this problem decidable or is there some practical algorithm to solve it. This one seems to be a mixture of both!
@DerekHolt (Yes, and even for recursive presentations.) Actually I don't even think that the OP considered the problem under an algorithmic angle (as in this recent question). Another example where the "algorithm" is only in principle: if we have two conjugate elements $g,h$ in a symmetric group (finite or infinite), we "choose" a length-preserving bijection between the sets of cycles of $g$ and $h$, we choose an element in each cycle, and then can induce a conjugating element. (...)
(...) For the infinite symmetric group $S_\omega$ (of all permutations of the set $\omega={0,1,\dots}$) with $g,h$ computable this can be done algorithmically if the function mapping $n$ to the (possibly infinite) length of the $g$-cycle of $n$ is computable (and the same for $h$) and in addition being on the same cycle is decidable, which I think is not always the case.
Also here's a way to make the question meaningful, typically for Lie groups: assume that $G$ is a Polish group. Then we can ask whether there exists a Borel subset $X$ of $G^3$ such that for all conjugate $x,y$ in $G$, there exists a unique $z\in G$ such that $(x,y,z)\in X$, and moreover it satisfies $zxz^{-1}=y$. (Clearly we can always suppose that $X\subset{(x,y,z)\in G^3:zxz^{-1}=y}$.) Possibly there are natural variants of this, replacing the "Borel" condition by a weaker or stronger one, I haven't thought enough.
For the case you are especially interested in, the answer is yes! The equation $zxz^{-1}=y$ is equivalent to $zx=yz$ and $z \neq 0$. For fixed unit quaternions $x,y \in \mathbb{H}^1$, this is linear in $z$ so can be solved by linear algebra. Explicitly, write $z=z_0 + z_1 i + z_2 j + z_3 ij$ with $z_i \in \mathbb{R}$ the unknowns, multiply on the right by $x$ and on the left by $y$, and setting these equal you get $4$ linear equations (in terms of the coefficients of $x,y$) in each coordinate.
If there is a nonzero solution, then the real dimension of the solution space is $2$ whenever $x,y$ are nonscalar: you can always post-conjugate by any nonzero element of $\mathbb{R}[y]$, which of course commutes with $y$, and the Skolem--Noether theorem says this all you can do.
From this solution space, you can pick your elements of norm $1$ by rescaling any nonzero solution; the total set of solutions looks like a circle.
This is a positive answer to the case of unit quaternions.
Edited to indicate this
|
2025-03-21T14:48:30.438020
| 2020-04-28T23:57:30 |
358831
|
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|
Stack Exchange
|
What is the dirichlet series of $f(n)=\sum_{d | n}(\log d) / d$ function?
My opinion is ;
We may use id(d)=d arithmetic function and log*id dirichlet convolution in the question.
i thought that ; when we multiply and divide n with $(\log d) / d$ we obtain
$F(S)=\sum_{n=1}^{\infty} \frac{1}{n^{s+1}} \cdot\left(\log*id\right)$
so
$F(S)=D(log,s+1).D(id,s+1)$
So we get
$F(S)=-\zeta^{\prime}(s+1). \zeta(s)$
i am curious that my solution is right or not.You may write your solutions and different ideas. thanks for your helps.
We can verify this by direct calculation. By definition,
$$F(s)=\sum_{n=1}^\infty\frac{f(n)}{n^s}=\sum_{n=1}^\infty\frac{1}{n^s}\sum_{d\mid n}\frac{\log d}{d}.$$
We rearrange the double sum so that $d$ comes first, and then write $n=dm$ to separate variables. We get
\begin{align*}
F(s)&=\sum_{d=1}^\infty\frac{\log d}{d}\sum_{m=1}^\infty\frac{1}{(dm)^s}=\left(\sum_{d=1}^\infty\frac{\log d}{d^{s+1}}\right)\left(\sum_{m=1}^\infty\frac{1}{m^s}\right).
\end{align*}
On the right hand side, the first factor is $-\zeta'(s+1)$, while the second factor is $\zeta(s)$, so
$$F(s)=-\zeta'(s+1)\zeta(s).$$
P.S. Your question is not of research level, but I felt like answering it.
it's nice solution , i understand better it now. thanks:)
|
2025-03-21T14:48:30.438386
| 2020-04-29T00:52:56 |
358840
|
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|
Stack Exchange
|
Injectivity of a locally strictly expanding map on a compact space
Prove that any locally strictly expanding map on an infinite compact metric space is non-injective.
Could you include the definition if "locally strictly expanding"? Do you assume $f$ continuous? does it mean that every $x$ has a neighborhood $V$ such that $d(f(y),f(y'))>d(y,y')$ for all distinct $y,y'\in V$?
f is not continuous, yes the definition you wrote is the one!
I have a suspicion that this is a homework question, because it appears as Lemma 13.1.3 in the book Brucks, Bruin - Topics from One-Dimensional Dynamics. However, the definition of a "locally strictly expanding map" there is different from the one suggested by YCor in the comments. Hence, I will modify the proof from the aforementioned book to show the following:
Let $f$ be a map on a compact metric space $X$ such that for every $x\in X$ there is an open neighborhood $U_x$ of $x$ such that $d(f(u),f(v))>d(u,v)$, for every $u,v\in U_x$. If additionally $f$ is injective, then $X$ is finite.
First, since $f$ is a continuous injection from a compact metric space, it is a homeomorphism onto its image $Y=f(X)$. Hence, $\{f(U_x), x\in X\}$ is an open cover of $Y$. Since $Y$ is compact, by Lebesgue theorem there is $\delta>0$ such that for every $y\in Y$ there is $x\in X$ such that $B(y,\delta)\subset f(U_x)$. In particular, if $d(y,z)<\delta$, then $d(g(y),g(z))<d(y,z)$, where $y,z\in Y$ and $g:Y\to X$ is the inverse of $f$.
It is enough to show that $Y$ is finite, since $g$ is a surjection. From compactness, $Y$ can be covered by $N$ sets of diameter at most $\frac{\delta}{2}$. We will show that $Y$ is finite by showing inductivly that it can be covered by $N$ sets of diameter at most $\frac{\delta}{2^n}$, for every $n\in\mathbb{N}$. The case $n=1$ is done. Assume that $Y=U_1\cup...\cup U_N$, where $U_i$ are of diameter at most $\frac{\delta}{2^n}$.
The function $K(x,y)=\frac{d(g(y),g(z))}{d(y,z)}$ is continuous on the compact set $\{(y,z),\frac{\delta}{2^{n+1}}\le d(y,z) \le \frac{\delta}{2^n}\}$, from where there is $\alpha\in(0,1)$ such that $\frac{\delta}{2^{n+1}}\le d(y,z) \le \frac{\delta}{2^n}$ $\Rightarrow$ $d(g(y),g(z))\le \alpha d(y,z)$. Therefore, $ d(y,z) \le \frac{\delta}{2^n}$ $\Rightarrow$ $d(g(y),g(z))\le \max\{\frac{\delta}{2^{n+1}},\alpha d(y,z)\}$, and so
$diam(g(U_i))\le \max\{\frac{\delta}{2^{n+1}},\alpha diam(U_i)\}$. Since $g$ is a surjection, and $\{U_i\}$ cover $Y$, it follows that $\{V^1_i\}$ cover $Y$, where $V^1_i=g(U_i)\cap Y$.
Define also $V^{k+1}_i=g(V^k_i)\cap Y$. Applying the same argument we get that $\{V^k_i\}$ cover $Y$, and $diam(V^k_i)\le \max\{\frac{\delta}{2^{n+1}},\alpha^k diam(U_i)\}$. If $m$ is such that $\alpha^m\le \frac{1}{2}$, we get that $\{V^m_i\}$ cover $X$, and $diam(V^m_i))\le \frac{\delta}{2^{n+1}}$. This completes the proof.
Edit: it turns out that the two definitions of locally expanding maps are equivalent, as shown in this article. However, the proof is not trivial.
Thank you, for providing a detailed solution! Does the map $f$ have to be continuous function?
@yogamat yes, probably there is a counterexample without this assumption
Why do you claim that your definition of "locally strictly expanding" differs from the one I suggested? the only difference I can see is that you don't assume $u\neq v$, which makes the condition absurd (unless $X$ is empty).
@YCor i claim that your definition is different from the one in the book (if i am not mistaken), and my proof is based on your definition
OK, I got it, thanks.
|
2025-03-21T14:48:30.438616
| 2020-04-29T03:08:05 |
358845
|
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|
Stack Exchange
|
Čech for $\ell$-adic sheaves
On the étale site of a scheme $X$, there is a spectral sequence associated to the data of an étale hypercover $K$ of $X$ and an abelian étale sheaf $\mathcal F$ on $X$:
$$E_2^{p,q}=\check H^p(K,\underline H^q(\mathcal F))\Rightarrow H^{p+q}(X,\mathcal F).$$
(See https://stacks.math.columbia.edu/tag/01GY.)
Here $\underline H^q(\mathcal F)$ is the étale presheaf with value $\underline H^q(\mathcal F)(U)=H^q(U,\mathcal F)$ on $U\to X$ étale.
Does the same spectral sequence exist for $\ell$-adic cohomology ($\mathbf{Z}_\ell$ or $\mathbf{Q}_\ell$ sheaves), with $\underline H^q(\mathcal F)$ a presheaf of $\mathbf{Z}_\ell$-modules or $\mathbf{Q}_\ell$-vector spaces? (With sufficient finiteness hypotheses on $X$ such as being of finite type over a field.) If so, is there a reference? (The closest I have been able to find to something mentioning this are the notes of Brian Conrad titled ‘Cohomological Descent.’)
Thank you.
If you use the pro-étale site of Bhatt and Scholze, the same tag applies. (Small warning: $\mathbf Z_\ell$ and $\mathbf Q_\ell$ are not constant sheaves; rather they carry a topology that affect the sections over a pro-étale cover.)
Thank you very much. As for your warning, if I start with an $\mathbf{Q}\ell$ sheaf $\mathcal F$ on, say, a variety over an algebraically closed field, and consider an étale hypercover of this variety, then $\underline H^q(\mathcal F)$ will be a presheaf on the pro-étale site. If I restrict this presheaf to the étale site, can its sections still be computed naïvely without mentioning the word ‘pro-étale?’ i.e. are sections of $\underline H^0(\mathbf{Q}\ell)$ over a connected $U\to X$ étale still $\mathbf{Q}_\ell$ (even if the sections over a weakly étale $U\to X$ are something else)?
Yes: the sections over $U$ are continuous maps $U \to \mathbf Q_\ell$, so when $U$ has finitely many components (e.g. $U$ is of finite type over $X$) then this is the usual definition of a locally constant function (i.e. constant on each component).
|
2025-03-21T14:48:30.438779
| 2020-04-29T03:17:28 |
358846
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/358846"
}
|
Stack Exchange
|
Is being a deg 0 vector field equivalent to being locally non-surjective?
Let $X:\mathbb R^n\to\mathbb R^n$ be a $C^1$ (or smooth)-vector field, such that $X(0)=0$ is an isolated zero. So we can talk about the mapping of $0$ for $X$.
For convenience assume $0$ be the only zero of $X$. My question is, are the following equivalent?
$X$ has mapping degree 0 at the origin.
There is a $\epsilon_0>0$ such that $X/|X|:\{0<|x|<\epsilon_0\}\setminus\{0\}\to\mathbb S^{n-1}$ is not surjective
or 2'. There is a $\epsilon_0>0$ such that $X/|X|:\{x:\lvert x\rvert =r\}\to\mathbb S^{n-1}$ is not surjective for all $0<r<\epsilon_0$.
I am not sure if there is any counterexample for either direction.
Is the domain in your (2) meant to be ${x : 0 < \lvert x\lvert < \epsilon_0}$? If not, what does "not surjective for $0 < \lvert x\rvert < \epsilon_0$" mean?
@LSpice Thank you, you are right.
I think it can be degree zero and the map you describe in (2') is surjective for any $r$. If the vector field has a transverse zero, then both your (2) and (2') are true, but if it isn't transverse both (2) and (2') are false, in general.
But if the zero is transversal should the vector field has nonzero degree?@RyanBudney
We agree that for every $0<r<+\infty$ one has the self-mapping of the unit sphere
$$f_r:S^{n-1}\to S^{n-1}:x\mapsto X(rx)/\vert X(xr)\vert$$
Since the maps $f_r$ are all two by two homotopic, they have the same degree $d$; this is what you call the "mapping degree of $X$ at the origin".
In particular, if $f_r$ is not onto for at least one $r$, then $d=0$.
But the converse does not hold, here is a counterexample. It is easy to make a continuous
and even smooth ($C^\infty$) self-mapping $f$ of $S^{n-1}$ which is onto but of degree $0$: first flatten the sphere to the unit disk $D^{n-1}$ (by the linear projection
$R^n\to R^{n-1}$) and then wrap $D^{n-1}$ around $S^{n-1}$ (like a hankerchief around a ball). Then, choose a smooth nonnegative real function $u$ on $[0,+\infty)$ which is positive but at $0$, and such that all its derivates at $0$ vanish (e.g. $u(r)=e^{-1/r^2}$). The smooth vector field
$X(x)=u(\vert x\vert)f(x/\vert x\vert)$ on $R^n$ is a counterexample.
|
2025-03-21T14:48:30.438945
| 2020-04-29T04:23:16 |
358848
|
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|
Stack Exchange
|
Is this a positive definite kernel?
Under which conditions on the function :
\begin{array}{l|rcl}
K : & \mathbb R^+ & \longrightarrow & (0, 1)\\
&t & \longmapsto & K(t) \end{array}
is the symmetric function :
\begin{array}{l|rcl}
C : & (\mathbb R^+)^2 & \longrightarrow & \mathbb R\\
& (t,s) & \longmapsto & C(t, s) =\frac{K(t) +K(s)} {2\Gamma(1-\frac{K(t)+K(s)}{2})} \end{array}
a positive definite Kernel ? .Are there references to help me solve this problem ?
NB: $\Gamma$ is the well known Gamma function.
Hi, and welcome to MO. You can edit your own question even after it is posted, instead of adding comments later on. Also, if you are looking for a reference you may want to use the [reference-request] tag
|
2025-03-21T14:48:30.439025
| 2020-04-29T04:38:03 |
358849
|
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"DSM",
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"smapers"
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"url": "https://mathoverflow.net/questions/358849"
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|
Stack Exchange
|
Minimum number of swaps needed to 'group' a sequence?
Let a finite sequence $s=\{s_1,\dots,s_N\}$ (the characters of which are chosen from a finite set $\{c_1, \cdots, c_M\}$) be called "grouped" if for any $s_i=s_j$, $i<j$, we have $s_k=s_i=s_j$ for any $i<k<j$. For example, $\{a,a,c,c,c,b,b\}$ is grouped, where as $\{a,b,a,c,c,c,b\}$ is not.
What is the minimum number of swaps needed to make a given sequence $s$ 'grouped'? Is the computational complexity (in terms of the length of the string and the size of character set) of this problem known in literature?
A practical motivation might stem from (not completely sure though) the process of "disk de-fragmentation", where 'pieces' of the same file are grouped together in contiguous memory segments. .
The answers so far aren’t really addressing the question as asked. This is mainly a reformulation and a few simple observations.
First consider , for a fixed $N,k$ a graph with $\binom{N}k$ vertices labelled by the strings made of $k$ a’s and $N-k$ b’s. Each vertex has degree $k(N-k)$ with an edge to each string which results from swapping an a and a b. There are two distinguished vertices, $k$ a’s then $N-k$ b’s and the reverse.
Q: Given a particular vertex find the closest distinguished vertex and a path to it. The computational complexity matters but swaps are expensive. So it is really a question of finding the absolute minimum number of swaps.
In this case the distance between two strings is half the number of places they differ, i.e. half the Hamming distance. So find which of the two is closest and an optimal sequence of swaps should be clear.
I suppose the sequence $aaabbbaaa$ takes $3$ swaps and generalizes to $\frac{N}3$ swaps. Is that the worst case?
In case $N=2k$ is $ababab\cdots$ taking about $\frac{N}4$ swaps worst?
I general we have all length $N$ sequences resulting from a certain multiset with $M$ distinct characters. The number of these is given by a certain multinomial coefficient. There are $M!$ sorted sequences so we might not want to consider them all. The Hamming distance gives some bound on the distances but it might not be as simple as in the $M=2$ case.
Well, obviously any sorting algorithm can achieve what you want, since if the entries in the sequence are sorted (under whatever arbitrary linear order you impose) they have the grouping property.
Given $N,$ there are sorting algorithms with complexity $O(N \log N).$ These algorithms make no assumptions about the values they are sorting.
However, you can do better, depending on the relationship of the alphabet size $M$ to list length $N$. In particular, you can map your alphabet to $\{1,2,\ldots,M\}$ and thus assume that the values to be sorted are all nonnegative and bounded by $M.$
In that case, the counting sort algorithm can sort your sequence with time complexity $O(N+M).$ Alternatively, you can use radix sort with complexity $O(w N),$ where $w$ is the number of bits required to store the values you are sorting, so $w=O(\log K).$
The counting sort does not output a sequence of swaps though, does it?
This is a sorting problem. The sorting algorithm that performs the minimum possible number of swaps in the worst-case scenario is selection sort, with $n-1$ swaps. Its time complexity is $O(n^2)$.
There are of course sorting algorithms that have lower time complexity, such as counting sort and merge sort, but they are irrelevant to the question because they do not rely on swapping items.
There are $M!$ ways of ordering the characters, and post that one might want to do selection sort to figure out which worked the best. But the $M!$ factor is not very encouraging. Any workaround for that? And also, I would be interested in computing the minimum swaps for a given string (not the worst case scenario).
Grouping the sequence might be easier than sorting it, though.
@DSM, the minimum swap for a given string is likely to be very intractable.
@kodlu, thanks for the comment. That's my hunch as well. Not sure if there are good approximations. For example, say the string consists of ${l_1,\cdots,l_M}$ number of characters. Then if I choose an ordering (for selection sort) of the alphabets greedily based on the smallest Hamming distance between constant strings of character $c_k$ length $l_k$ and the sub-string ${s_{{l_{j_1}+...+{l_{j_{k-1}}},..., s_{l_{j_1}+...+{l_{j_{k-1}}+l_k-1}}$, how good would this approximation be?
Not sure I get your suggestion right but wouldn't your proposal require computations for a large number of the $M!$ possible arbitrary orderings? (unless you have in mind a greedy approach with a Nemhauser-type guarantee on how close the outcome of a greedy approach would be to the true optimum)
|
2025-03-21T14:48:30.439373
| 2020-04-29T05:05:51 |
358852
|
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|
Stack Exchange
|
$C^2$-control using orthonormal frame on a Riemannian manifold
Let $(M,g)$ be a Riemannian manifold. Let $E=(E_1,\dots,E_n)$ be an orthonormal frame for $M$.
So for $M$ itself we have a natural $C^k$-norm $\|f\|_{C^k_g(M)}:=\max\limits_{1\le m\le k}\sup\limits_{x\in M}|\nabla^mf(x)|$ where $|\nabla^mf(x)|=\max\limits_{v_i\in T_xM,|v_i|=1}\langle\nabla^m f(x),v_1\otimes\dots v_m\rangle_g$.
On the other hand we can define norm structures along the frame $E$ by viewing $E_j$ as differential operator $\|f\|_{C^k_E(M)}:=\max\limits_{1\le m\le k}\{\|E_{i_1}\dots E_{i_m}f\|_{C^0(M)}:1\le i_j\le n\}$.
My question is: Is there a universal constant $C=C(n)>0$ such that $\|f\|_{C^2_g(M)}\le C\|f\|_{C^2_E(M)}$ for all $(M,g,E)$ and $f$?
More specifically, can we have $\|(\nabla_{E_i}E_j)f\|_{C^0}\lesssim_n\|E_iE_jf\|_{C^0}+\|E_jE_if\|_{C^0}$?
Note that for $C^1$, $|\nabla f(x)|=\max\limits_{v\in\mathbb R^n,|v|=1}df_x(v_1E_1+\dots+v_nE_n)=\sqrt{\sum_{i=1}^n|E_if(x)|^2}$. So the analogically control holds with $C_1=1$.
More generally, does the analogy holds for $C^k$?
|
2025-03-21T14:48:30.439474
| 2020-04-29T05:57:26 |
358854
|
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|
Stack Exchange
|
Are Conway's combinatorial games the "monster model" of any familiar theory?
This is related to this question about a "mother of all" groups, and so seemed like it'd fit in better at MO than MSE.
If I understand the answer to that question correctly, the surreal numbers have a nice characterization as being the "monster model" of the theory of ordered fields (and I think also the real-closed fields), which means that every ordered field embeds into the surreal numbers. In the answer to the question above, Joel David Hamkins gave an interesting example of what the monster model of the theory of groups would look like, which has the property that every possible group is a subgroup of this group (which caused it to be dubbed the "Hamkins' All-Encompassing Group-Like Thing," or I suppose HAEGLT, in the comments).
This question, then, is about Conway's formalization of combinatorial games, of which the surreal numbers are embedded. Conway's games are much more general than the surreal numbers, and have (among other things) the following structure:
There is a commutative sum of two games (which agrees with the sum on surreal numbers)
For any game, there is an additive inverse (so we have an abelian group)
There is a partial order on the games
There are nilpotent games, such as the star $\{*|*\}$ game of order 2, as seen in Conway's analysis of Nim
My question is, are the Conway games the monster model of the theory of... well, anything familiar related to the above? Abelian groups? Partially ordered abelian groups? Something else?
To my precise, I am sure there is probably some way to devise some artificial theory that the games are technically a monster model of. What I am wondering is if they are a monster model of some familiar algebraic theory that people use all the time, or perhaps some such theory with just a bit of added structure. Since they generalize the surreals in a fairly "natural" way, it seems intuitive that they might be a monster model of some equally "natural" theory that is more general than that of ordered fields.
EDIT: I previously wrote that the surreal multiplication can also be extended to a commutative product on the entire theory of games, as shown on (page 412 of this book). However, this is apparently not entirely true, as written in the comment below, as there is some subtlety with the equality relation.
In general, multiplication of games doesn't respect equality (although it does on the surreals). Indeed there can't be an associative and distributive multiplication on games with $1$ as the unit, because then we would have $0 =0\times\frac{1}{2} = (+)\times\frac{1}{2} = *\times(1+1)\times\frac{1}{2} = *$.
Thanks, that is a good point - I will edit
In On a conjecture of Conway (Illinois J. Math. 46 (2002), no. 2, 497–506), Jacob Lurie
proved Conway's conjecture that the class $G$ of games together with Conway's addition defined thereon is (up to isomorphism) the unique "universally embedding" partially ordered abelian group, i.e.
for each such subgroup $A$ of $G$ whose universe is a set and any such extension $B$ of $A$, there is an isomorphism $f:B\rightarrow G$ that is an extension of the identity on $A$. The terminology "universally embedding", which is due to Conway, is unfortunate since it is sometimes confused with "universal". For partially ordered abelian groups "universally embedding" implies "universal", but I haven't checked if they are equivalent (though I suspect they're not). For ordered fields the notions are not equivalent; whereas $\mathbf{No}$ is up to isomorphism the unique "universally embedding" ordered field, it is not up to isomorphism the unique universal ordered field (though it is of course universal). I point this out in my paper Number systems with simplicity hierarchies: a generalization of Conway's theory of surreal numbers (J. Symbolic Logic 66 (2001), no. 3, 1231–1258). In that paper I further suggest the terminology universally extending in place of universally embedding. As an example of the potential for confusion using Conway's terminology, I point out (p. 1240) that Conway's terminology led Dales and Wooden (Super-real ordered fields, Clarendon Press, Oxford. 1996, p. 58) to mistakenly claim that $\mathbf{No}$ is up to isomorphism the unique universal ordered field.
Edit (4/30/20): For the sake of completeness it is perhaps worth adding that while Lurie's result is the deeper of the two, David Moews proved that Conway's class of games with addition (but without the order relation) is (up to isomorphism) the unique universally embedding Abelian group. See Moews's The abstract structure of the group of games in Richard J Nowakowski (ed.) More games of no chance, MSRI Publications no. 42, Cambridge University Press, Cambridge, 2002, pp. 49-57.
Category theorists would probably say that a "universally embedding" partially ordered abelian group is a partially ordered abelian group which is injective with respect to embeddings.
If I understand the statement correctly, then I think model theorists would say that (the elementary theory of) $\mathbf{No}$ is the model completion of the theory of partially ordered abelian groups. But since the theory of partially ordered abelian groups is not complete, I don't think model theorists would call $\mathbf{No}$ "the monster model" of this theory -- an incomplete theory has one monster model for each completion.
Tim<EMAIL_ADDRESS>I'm not sure I understand your point. The question was not about $\mathbf{No}$ but rather about the more general class of games. Moews's result like Lurie's is about the class of games.
@TimCampion This is stronger than injectivity (otherwise uniqueness would be problematic). This is because homomorphisms required in the statement satisfy $x\leqslant y$ iff $f(x)\leqslant f(y)$; in particular they are embeddings.
Tim<EMAIL_ADDRESS>I still am not clear as to what point you are trying to make. Are you questioning my answer to the question. If so, on the basis of what?
@PhilipEhrlich I'm not questioning your answer. When I wrote $\mathbf{No}$ that was a brain-o for the class of all games. I'm just noting that the use of the term "monster model", which really goes back to the question statement, clashes with the standard usage in model theory.
@მამუკაჯიბლაძე Because being an embedding is a stronger condition than being a monomorphism, injectivity-with-respect-to-embeddings is also in a sense weaker than injectivity-with-respect-to-monomorphisms. Really we're just talking about injectivity in the category whose morphisms are embeddings.
Tim<EMAIL_ADDRESS>Thank you for the clarification.
Thanks @PhilipEhrlich. So if the games are the "universally embedding" partially ordered abelian group, and given that every abelian group can be trivially made into a poset, does that mean it is also the universally embedding abelian group? Likewise, would the surreals be the universally embedding free abelian group?
@PhilipEhrlich, just saw in your edit that my question about the games being the universally embedding abelian group was answered in the affirmative in your edit. So I guess there's the same question for the surreals then, if they are the universally embedding free abelian group...
And, if they are also the universally embedding totally ordered free abelian group if you also count $\lt$...
@TimCampion But in any case, while there are many partially ordered abelian groups that are injective for not-necessarily-embedding morphisms, the one injective for embeddings is unique up to isomorphism (and forcibly a proper class). And I believe it is also injective for not-necessarily-embeddings, no?
<EMAIL_ADDRESS>These are interesting questions to which I have not given any thought.
<EMAIL_ADDRESS>I suspect you are right.
|
2025-03-21T14:48:30.440114
| 2020-04-29T06:44:52 |
358857
|
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|
Stack Exchange
|
Mirror symmetry for $C^*$
The Liouville manifold $T^*S^1$ is said to be "mirror" to the complex variety $C^*$. (see for instance lecture 7 here: http://math.columbia.edu/~topology/Eilenberg_lectures_fall_2016)
This is manifested in the fact that the wrapped Fukaya category of $T^*S^1$ is quasi-equivalent to (a dg enhancement of) the category of coherent sheaves on $C^*.
Given a fiber $F_x \subset T^*S^1$, one can compute that $\operatorname{Hom}_{\operatorname{Fuk}}(F_x, F_x)= \mathbb{C}[x, x^{-1}] = \operatorname{Ext}(\mathcal{O}_{\mathbb{C}^*},\mathcal{O}_{\mathbb{C}^*})$.
So we say that the fibers are "mirror" to the structure sheaf.
Two (related) questions that have been bothering me:
(i) is there a systematic/functorial way to construct a functor from $\operatorname{Fuk}(T^*S^1) \to \operatorname{Coh}(C^*)$? My understanding is that the "Family Floer theory" program constructs this in the setting of compact toric varieties, but this story does not seem to apply here on the nose (e.g. it involves rigid analytic geometry and does not involve wrapped Fukaya categories, at least not in the papers I found).
(ii) the above story seems to rely a lot on the fact that the Fukaya category is defined over $\mathbb{C}^*$. In general, the Fukaya category is defined over a Novikov ring; since $T^*S^1$ is exact, one can also use $\mathbb{C}$-coefficients using a rescaling "trick" described on the top of p7 of these notes: https://arxiv.org/pdf/1301.7056.pdf
Does the story still work with Novikov coefficients? What changes?
|
2025-03-21T14:48:30.440252
| 2020-04-29T07:24:05 |
358861
|
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"so-called friend Don"
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|
Stack Exchange
|
How to prove there is infinite prime numbers of form $5n+3$ without Dirichlet theorem?
Is there a nice elementary way to prove there is infinite prime numbers of form $5n+3$ (also for $5n+2$) with $n\in \mathbb{N}$?
I know how to do it for primes of form $pn+1$ for any prime $p\geq 3$ but not in this case.
I'm also familiar with the theorems of Schur and Murty.
This would be a good question for math.stackexchange.
The question https://mathoverflow.net/questions/16735/is-a-non-analytic-proof-of-dirichlets-theorem-on-primes-known-or-possible is a more general version of this (and to my opinion suitable for this forum). It refers to https://kconrad.math.uconn.edu/blurbs/gradnumthy/dirichleteuclid.pdf for the results of Schur and Murty and more. The question https://mathoverflow.net/questions/15220/is-there-an-elementary-proof-of-the-infinitude-of-completely-split-primes is also relevant.
Why? Is it so trivial to answer this and not worth of a research? @LaurentMoret-Bailly Anyway, I did post it on MSE.
To add to the previous comment, here is link to the post on [math.se]: How to prove there is infinite prime numbers of form $5n+3$ without Dirichlet theorem?
In my opinion, this question would clearly be on-topic here, if it weren't for that Konstantinos's answer in Chris's link also answers this question. It is clearly not trivial with $5$ replaced with an arbitrary integer, evidenced by the fact that no such proof has appeared in the linked question (despite having 60 upvotes) and I personally consider the $m=5$ proof there also quite non-trivial. The question also demonstrates awareness of relevant literature, and I would say is of general interest to mathematicians (or at least to me)... but it is indeed arguably a duplicate.
I wouldn't necessarily consider this a research-level question, but it's certainly not trivial. One can certainly adapt the proof of Dirichlet's Theorem to this special case (checking $L(\chi,s)\neq 0$ should be straightforward in this specific case), but this is probably not the kind of answer OP wants.
To confirm what Wojowu says: For the quadratic character $\chi$, $\sum_{n=1}^{\infty} \tfrac{\chi(n)}{n} = \sum_{m=0}^{\infty} \left( \tfrac{1}{5m+1} - \tfrac{1}{5m+2} - \tfrac{1}{5m+3} + \tfrac{1}{5m+4} \right)= \sum_{m=0}^{\infty} \tfrac{10 (2m+1)}{(5m+1)(5m+2)(5m+3)(5m+4)}$ and all summands are positive.
Since you are familiar with Schur and Murty's theorems, you seem to be asking for an elementary proof that doesn't use a Euclidean polynomial argument. Let me ask then slightly more generally, do you know of any elementary proof of this sort for any residue class that doesn't rely on a polynomial in that way? I don't know of any, but it would seem that trying to find one in a simple case would be an obvious starting point.
For a nonreal character $\epsilon$, we have $\sum_{n=1}^{\infty} \tfrac{\epsilon(n)}{n} = \sum_{m=0}^{\infty} \left( \tfrac{1}{5m+1} \pm \tfrac{i}{5m+2} \mp \tfrac{i}{5m+3} - \tfrac{1}{5m+4} \right) = \sum_{m=0}^{\infty} \tfrac{3}{(5m+1)(5m+4)} \pm i \sum_{m=0}^{\infty} \tfrac{1}{(5m+2)(5m+3)}$, and the summands are similarly positive.
No, I also don't know any such proof and would like to see it if exists. @JoshuaZ I was thinking to present my hischool students this Euclidian type of proofs and then I wonder if there are also some others types. Now it seem that I should dive into the proof of Dirichlet theorem itself in these special cases.
Other question threads on this topic have been linked to above. In one of those, I mention in a comment that there are proofs of Bang, Ricci, Roux, and Erdos that prove Dirichlet's theorem for certain special progressions by methods analogous to those used by Chebyshev to study the distribution of primes. Erdos's argument, for instance, is certainly "elementary" in the technical sense but ... I don't think any of the mentioned proofs are "simple". For your purposes it might be better to just specialize Dirichlet's proof in the ways suggested above.
I incorrectly believed that Euclid's argument would work in this case, but it doesn't. I don't know of a proof that's materially different than Dirichlet's proof.
Ajjaja @LaurentMoret-Bailly Is that only contribution to this question, to vote for close down? I'm very disappointed for such a manner.
|
2025-03-21T14:48:30.440561
| 2020-04-29T09:48:30 |
358869
|
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|
Stack Exchange
|
Does "tensoring" with a fixed field preserve Galois extensions of finite fields?
Let $K$ be a (possibly infinite) field of characteristic $p$, and $L$ be a finite field extension of $\mathbb{F}_p$, so that $L$ is finite and $L/\mathbb{F}_p$ is Galois. Suppose $K \otimes_{\mathbb{F}_p} L$ is a field, can we conclude that $(K \otimes_{\mathbb{F}_p} L) / K$ is Galois? If not, what are some sufficient conditions for this conclusion to hold?
The tensor product of two fields will not in general be a field.
Yes you are right. Let me make an edit.
Yes. The action of the Galois group $G$ on $L$ extends to an action on $K\otimes _{\mathbb{F}_p}L$, and since the functor $K\otimes _{\mathbb{F}p}-$ is exact, the $G$-invariant subfield is $K\otimes{\mathbb{F}_p}\mathbb{F}p=K$. Thus $(K\otimes{\mathbb{F}_p}L)/K$ is Galois with group $G$.
Your argument seems to not require any assumption of finiteness of the fields or of the extension. Is that correct?
|
2025-03-21T14:48:30.440656
| 2020-04-29T10:02:02 |
358871
|
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|
Stack Exchange
|
Choice of finite element spaces in plasticity
I am planning to run numerical simulations in metal elastoplasticity (von-Mises yield condition with and without isotropic hardening). However, I am completely new to this subject and I am unsure about the choice of appropriate finite element spaces.
I read that a problem called “volumetric locking” may occur, which is related to the fact that standard finite element functions are not flexible enough to deal with the incompressible plastic deformations and that mixed formulations may be necessary.
I found some 2D simulations on the internet where it is pointed out that in this case quadratic triangles for the displacement field in a non-mixed formulation are sufficient. However, I am mainly interested in AXISYMMETRIC situations (also in the general 3D case) and I wasn’t able to find precise information on this topic in the literature.
|
2025-03-21T14:48:30.440745
| 2020-04-29T10:52:02 |
358875
|
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|
Stack Exchange
|
Are the tangle functors based off Khovanov homology braided monoidal functors?
I was wondering if the tangle functors constructed in
"A functor-valued invariant of tangles"
https://arxiv.org/pdf/math/0103190.pdf
"An invariant of tangle cobordisms via subquotients of arc rings"
https://arxiv.org/abs/math/0610054
are braided monoidal?
For example in the Reshetekhin and Turaev set up, they have a braided monoidal functor from tangles to $Rep(U_q(sl_2))$. Sending points/tangles to representations/intertwiners, and one can take tensor products of points/tangles to tensor products of representations/intertwiners.
Are there any examples of tangle invariants coming from a braided monoidal functor but give something on the level of khovanov homology (or something that "decategorifies" to the Jones Polynomial)? Namely where points/tangles are assigned to categories/functors, tensor products of points/tangles carry over to some sort of tensor product of categories/functors, and finally also "decategorifies" to the Jones polynomial?
Should such a thing exist?
I have also seen some of the foam constructions and I think they relate to braided monoidal 2-categories, but it seems the points/tangles are not assigned to categories/functors.
I am quite a beginner and have not fully digested the content or understand a bigger picture in the literature.
Any help or comments would be greatly appreciated
It definitely doesn't comes from a braided monoidal functor. We hope it comes from a braided monoidal 2-functor, but before asking this question you'd need a braided monoidal 2-category structure on the target, and AFAIK (I'm not an expert on this) we're not there yet, although there has been a lot of work in that direction.
@Adrien Thanks for the Help!
I have actually be trying to learn your work "Integrating Quantum Groups over Surfaces" with Ben-Zvi and Jordan.
I was wondering if given a braided monoidal 2-category, would one be able to get an $E_n$-algebra? And use the factorization homology framework?
If the answer is in the positive, which $n$? What would be the analogue of $Pr_c$?
Yes, any reasonable definition of braided monoidal 2-category should coincide with the notion of $E_2$-algebra in the (3,1)-category of 2-categories. For Khovanov homology I suppose you want something like a symmetric monoidal $(\infty,1)$-category of presentable dg-2-categories which I have no clue how to define (but I'm sure some people have thought about it). Then indeed you can use the framework of factorization homology to attach 2-categories to (oriented if you have a framed $E_2$-algebra) surfaces. What's more tricky, AFAIK, is how to define the correct version of ribbon 2-category.
@Adrien Thanks a bunch!
|
2025-03-21T14:48:30.440933
| 2020-04-29T11:02:24 |
358877
|
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"Iosif Pinelis",
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|
Stack Exchange
|
Prove $x\mapsto\frac{E[f(X)1(f(X)\geq x X)]}{1+E[X1(f(X)\geq x X)]}$ is Lipschitz-continuous
Let $X$ be a random variable on $[0,A]$, and $f:[0,A]\to[-B_1,B_2]$ be a continuous function.
Let $$g(x) = \frac{g_1(x)}{g_2(x)}$$ where $g_1(x) = E[f(X)\mathbf{1}_{\{f(X)}]$ and $g_2(x) = 1+E[X\mathbf{1}_{\{f(X)\geq 1\}}]$.
My goal is to prove that $g$ is Lipschitz-continous on $[0,E[f(X)]]$.
So far I have, for $\epsilon$
\begin{align}
g(x+\epsilon) = \frac{g_1(x) - E[f(X)\mathbf{1}_{\{xX\leq f(X)\leq 1\}}]}{g_2(x) - E[X\mathbf{1}_{\{xX\leq f(X)\leq 1\}}]}.
\end{align}
In general, the function $g$ is not even continuous. E.g., let $A=4$ and $f(x)=x$ for all $x$. Let $X$ be uniformly distributed on $[0,A]=[0,4]$. Then $Ef(X)=2$ and
$$g(x)=\frac{1(x\le1)}{1+1(x\le1)},$$
so that $g$ is discontinuous at $1\in[0,2]=[0,Ef(X)]$.
Thank you for your answer, however I forgot to explicit the domain of $g$, which is $[0,E[f(X)]]$. Hence in you example $g$ is constant.
@Iques I think you can take Iosif's example with larger $A$ to make $E[f(x)]$ larger.
@Iques : As Nik Weaver noted, even after your change of the question the function $g$ can still be discontinuous, as is now detailed in my answer. More importantly, if your posted question is not what you actually meant, then you (and not the answerer) should take the responsibility for your mistake. That is, you should not change your question so as to invalidate a valid answer. Rather, you may want to post the amended question separately.
@Iques: to echo that, you are a new contributer and wouldn't know this, but it is considered bad form on Mathoverflow to modify a question after the original version has been answered. Nothing wrong with posting the modified question separately.
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2025-03-21T14:48:30.441082
| 2020-04-29T11:04:54 |
358878
|
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|
Stack Exchange
|
Representaility of morphism of stacks for schemes
I have seen two definitions of representability of a morphism of stacks, which should be at least compatible with the definition of a morphism of categories fibered in groupoids.
(Representable morphism of categories fibred in groupoids). Let $\mathcal C$ be a category. Let $\mathcal X,\mathcal Y$ be categories fibred in groupoids over $\mathcal C$. Let $F:\mathcal X\to \mathcal Y$ be a $1$-morphism. We say $F$ is representable, if for every $S\in Ob(\mathcal C)$ and any $G:\mathcal C/S\to\mathcal Y$ the category fibred in groupoids $(\mathcal C/S)\times_{\mathcal Y}S\to S$ is representable.
(Representable morphism of stacks). We say that a morphism of stacks $ f : \mathcal X \to \mathcal Y $ is representable if for all S-schemes $ U $ and morphisms $ y : U \to \mathcal Y $, the fiber product $\mathcal X \times_{\mathcal Y} U $ is an algebraic space.
(Representable morphism of stacks). We say that a morphism of stacks $ f : \mathcal X \to \mathcal Y $ is representable if for all $S$-schemes $ U $ and morphisms $ y : U \to \mathcal Y $, the fiber product $ \mathcal X \times_{\mathcal Y} U $ is a scheme.
How can we reconcile these 3 definitions?
Thank you for your help.
and 3. are genuinely different in general. Just take $f: \mathcal{X} \to \mathrm{Spec} k$ an algebraic space over a field $k$ which is not a scheme. Then this is representable in the sense of 2., but not 3. For stacks definition 2. is the right one in general; sometimes authors refer to 3. as "strongly representable" or "schematic" to differentiate.
This is not an answer, just too long for a comment. So, writing as an answer. It turns out that, one may not be able to see the correspondence between these three definitions as one of them is stated wrongly.
You have mentioned in the question the following:
(Representable morphism of categories fibred in groupoids). Let $\mathcal C$ be a category. Let $\mathcal X,\mathcal Y$ be categories fibred in groupoids over $\mathcal C$. Let $F:\mathcal X\to \mathcal Y$ be a $1$-morphism. We say $F$ is representable, if for every $S\in Ob(\mathcal C)$ and any morphism of categories fibered in groupoids $G:\mathcal C/S\to\mathcal Y$, the $2$-fiber product $(\mathcal C/S)\times_{\mathcal Y}S\to S$ is representable.
This definition wrongly stated (I am hoping it was a typo). Observe that your definition of representability of $F$ depends only with the category $\mathcal{Y}$ and nothing do with neither the category $\mathcal{X}$ or the map $F:\mathcal{X}\rightarrow \mathcal{Y}$
Correct definition of representable morphism of categories fibered in groupoids is the following:
Let $\mathcal C$ be a category. Let $\mathcal X,\mathcal Y$ be categories fibred in groupoids over $\mathcal C$. Let $F:\mathcal X\to \mathcal Y$ be a $1$-morphism. We say $F$ is representable, if for every $S\in Ob(\mathcal C)$ and any morphism $G:\mathcal C/S\to\mathcal Y$ of categories fibered in groupoids, the projection map $(\mathcal C/S)\times_{G,\mathcal Y,F}\mathcal{X}\to (\mathcal C/S)$ is representable.
By "$(\mathcal C/S)\times_{G,\mathcal Y,F}\mathcal{X}\to (\mathcal C/S)$ is representable", it means that the $2$-fibered product $(\mathcal C/S)\times_{G,\mathcal Y,F}\mathcal{X}$ is representable by an object of $\mathcal{C}$; that is $\mathcal C/S)\times_{G,\mathcal Y,F}\mathcal{X}\cong (\mathcal{C}/T)$ for some $T\in \text{Ob}(\mathcal{C})$ and that the morphism $(\mathcal{C}/T)\rightarrow (\mathcal{C}/S)$ is induced from an arrow $T\rightarrow S$ (or $S\rightarrow T$??, make a guess) in $\mathcal{C}$.
I hope you will be able to prove this definition of $(1)$ is related the definition of $(3)$. I will give more details about this if there is still some difficulty.
There are two notions of representability:
Given a category $\mathcal{C}$ and an object $S$ of $\mathcal{C}$, one can associate a category fibered in groupoiods over $\mathcal{C}$, namely the functor $(\mathcal{C}/S)\rightarrow \mathcal{C}$. Now, an arbitrary category fibered in groupoids over $\mathcal{C}$, say $\mathcal{F}\rightarrow \mathcal{C}$ is said to be representable by an object of $\mathcal{C}$ if there is an object $S$ of $\mathcal{C}$ and an isomrophism $\mathcal{F}\cong (\mathcal{C}/S)$. This is the situtation of your point $3$.
It turns out that there are other interesting categories fibered in groupoids over $\mathcal{C}$. If $\mathcal{C}$ is $Sch/S$, we have algebraic spaces. If $\mathcal{C}$ is the category $\text{Man}$, we have what are called Lie groupoids. Given a Lie groupoid $\mathcal{G}$, one can assocaite a category fibered in groupoids $B\mathcal{G}\rightarrow \mathcal{C}$. Now, given a category fibered in groupoids $\mathcal{F}\rightarrow \mathcal{C}$, we can ask if this $F$ is representably by such a special category fibered in groupoids, that is a Lie groupoid $\mathcal{G}$ and an isomorphism $\mathcal{F}\cong B\mathcal{G}$. This is the situtation of your point $2$.
|
2025-03-21T14:48:30.441329
| 2020-04-29T11:06:56 |
358879
|
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|
Stack Exchange
|
Comparing $X+Y$ and $X-Y$ for independent random variables with values in an abelian locally compact group
Let $G$ be an abelian locally (separable?) compact group with Haar measure $\mu$. Inspired by the interesting proof of A sum of two binomial random variables :
Let $X$ and $Y$ be $G$-valued independent random variables with $\mu$-density $f_X$ resp. $f_Y$. Let $f_{X \pm Y}$ be the $\mu$-density of $X \pm Y$. Is it always true that
$$\int f_{X-Y}^2 d\mu = \int f_{X+Y}^2 d\mu \ ?$$
This holds true for $G = \mathbb{Z}$ (here both integrals are finite)
.
The answer is yes. Indeed, let $dx:=\mu(dx)$ for brevity. We have
\begin{align}
I_{X,Y}&:=\iiint f_X(x'+y'-y)f_Y(y)f_Y(y')f_X(x')dx'\,dy\,dy' \\
&=\iint f_{X+Y}(x'+y')f_Y(y')f_X(x')dx'\,dy' \\
&=\iint f_{X+Y}(t)f_Y(t-x')f_X(x')dt\,dx' \\
&=\int f_{X+Y}(t)^2dt.
\end{align}
Also,
\begin{align}
I_{X,Y}&=\iiint f_X(x'-u'+u)f_Y(-u)f_Y(-u')f_X(x')dx'\,du\,du' \\
&=\iiint f_X(x'+u-u')f_{-Y}(u)f_{-Y}(u')f_X(x')dx'\,du\,du' \\
&=\iiint f_X(x'+u'-u)f_{-Y}(u)f_{-Y}(u')f_X(x')dx'\,du\,du' \\
&=I_{X,-Y} =\int f_{X-Y}(t)^2dt,
\end{align}
by what was shown in the previous display.
Thus,
$$\int f_{X+Y}(t)^2dt=\int f_{X-Y}(t)^2dt,$$
as desired.
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2025-03-21T14:48:30.441421
| 2020-04-29T11:46:00 |
358880
|
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|
Stack Exchange
|
Divisors of shifted geometric progressions
For integers $a,b,k$ with $a \geq 1$ and $k\geq 2$, consider the shifted geometric progression $n_i = ak^i + b$. I would like to understand the set of integers (prime or otherwise) that divide at least one of the $n_i$'s. More precisely, let
$$ D = D_{a,b,k} = \{ d \in \mathbb{N} \ : \ \text{there exists } i \geq 1 \text{ with } d \mid n_i \}, $$
$$ P = P_{a,b,k} = \{ p \ : \ p \text{ is prime and there exists } i \geq 1 \text{ with } p \mid n_i \}. $$
Questions:
Is it known the set $P$ has positive relative density inside the primes?
Barring a positive answer to 1., what can be said about the sizes of $D$ and $P$?
(It is certainly possible that $D$ has asymptotic density $0$, e.g. by letting $a,b$ and $k$ be squares. If the answer to the first question is positive then $\# D \cap [1,x] \geq \# P \cap [1,x] \gg x/\log x$. In the second question I'm hoping that perhaps a weaker asymptotic is true, such as $\# D \cap [1,x] \gg x^{1-\varepsilon}$.)
Context: It would follow from Artin's conjecture on primitive roots that $k$ is a primitive root for a positive proportion of the primes, and for each such prime $p$ there exists $i$ with $k^i \equiv -b/a$, whence $p \mid n_i$ (at least, as long as $p$ does not divide $ab$). Hence, if we believe Artin's conjecture then $P$ has positive density inside the primes.
One can also show by purely elementary methods that for each $C$ there exists $i$ such that $n_i$ has $>C$ prime divisors. Hence, $P$ is infinite.
Here is a sketch: Assume, as we may do without loss of generality, that $b$ is coprime to $k$ (else, replace $a$ with $ak^{i_0}/d$, $b$ with $b/d$, and $i$ with $i-i_0$ where $d = \gcd(k^{i_0},b)$ for some large integer $i_0$). This assumption ensures that $n_i$ ($i \geq 1$) are coprime to $k$. Construct a sequence $i_j$ where $i_1$ is arbitrary, and $i_{j+1} = i_j + \varphi(n_{i_j}^2)$, so that $n_{i_{j+1}} \equiv n_{i_j} \bmod n_{i_j}^2$. This is set up so that $n_{i_{j+1}}/n_{i_j}$ is an integer coprime to $n_{i_j}$, and consequently $n_{i_j}$ has at least $j$ distinct prime factors.
You seem to be asking about the "two-variable Artin conjecture". Unfortunately the known (unconditional) results seem much weaker than what you are hoping for; I think the state of the art is https://arxiv.org/abs/1711.06410
@so-calledfriendDon: Thank you! Yes, it seems like what I'm asking is very closely related: If I understand correctly, the case of my question where $b/a \in \mathbb{Z}$ and we only care about prime divisors is corresponds precisely to Theorem 1.1 in the preprint you attached, while Theorem 2.1 is more general than my question (since $ak^i + b$ satisfies a binary recurrence).
This does not fully answer the question, but gives several links to some related literature.
Most papers below study if the order of an element mod $p$ is odd or even,
(or more general). Therefore the links below study the cases $a=1, b=1$ or $b=-1$ in your notation.
Shparlinski's paper is more general, but primarily studies related sequences with more than 2 summands.
a) The following two papers by Hasse give at least some special cases
such as prime divisors of the sequence $k^n+1$, ($k$ is fixed, $n \in \mathbf{N}$).
In particular Hasse proved that the Dirichlet density of primes dividing an integer of the type $2^n+1$ is
$17/24$.
Note: if $p|(2^n+1)$, then $2^n\equiv -1 \bmod p$, and the order of $2$ modulo $p$ is even.
H. HASSE, Über die Dichte der Primzahlen p, für die eine vorgegebene ganzrationale Zahl
$a\neq 0$ von gerader bzw. ungerader Ordnung mod p ist. Math. Ann. 166 (1966), 19-23.
http://www.digizeitschriften.de/dms/img/?PID=GDZPPN002296616
H. HASSE, Über die Dichte der Primzahlen p, für die eine vorgegebene rationale Zahl $a\neq 0$
von durch eine vorgegebene Primzahl $l \neq 2$ teilbarer bzw. unteilbarer Ordnung mod $p$ ist.
Math. Ann. 162 (1965), 74-76
http://www.digizeitschriften.de/dms/img/?PID=GDZPPN002295253
b) Odoni proved corresponding results for natural density.
Journal of Number Theory
Volume 13, Issue 3, August 1981, Pages 303-319
A conjecture of Krishnamurthy on decimal periods and some allied problems,
https://www.sciencedirect.com/science/article/pii/0022314X81900160
In some cases he got positive relative prime density, in other cases 0-density.
c) On RH (related to Artin's conjecture
Stephens, P. J.
Prime divisors of second-order linear recurrences. I.
J. Number Theory 8 (1976), no. 3, 313–332.
d)
Shparlinski, Igor E.
Prime divisors of sparse integers.
Period. Math. Hungar. 46 (2003), no. 2, 215–222.
https://link.springer.com/article/10.1023%2FA%3A1025996312037
e) Hooley's book "Application of sieve methods to the theory of numbers" has some information on prime factors of $2^n+b$.
f) Prime divisors of certain recurrent sequences have been studied e.g. by Ballot, and Moree.
Finally, I believe that the case $|b| \neq 1$ is more difficult,
as there is less algebraic structure (such as order of an element).
Thank you! That's very relevant - and the problem at the level of generality I was hoping for appears to be much harder than I anticipated.
|
2025-03-21T14:48:30.441850
| 2020-04-29T12:55:26 |
358885
|
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|
Stack Exchange
|
Concentration of norm of linearly transformed normal random vector as dimension go to infinity
Earlier asked on MSE, but didn't get an answer, so posting here:
Let $X=(X_1 \dots X_n) \in \mathbb{R}^n, X_i\sim N(0,1), iid.$ Let $B: \mathbb{R}^n \to \mathbb{R}^n $ be the diagonal linear map: $Bx_k:= x_k/ {k}, 1 \le k \le n.$ Then $||B||_F^2= \sum_{k=1}^{n}\frac{1}{k^2}$. Then is: $lim_{n \to \infty}|E||BX|| - ||B||_F |=0?$. How do we compute $E||BX||$ anyway? Note that, if $B$ were $I_n,$ the answer would be yes even for cases $X_i$ non Gaussian, c.f. this question on MO.
Motivation for this question (not needed to answer the question): concentration
Note that: $E[||BX||^2]=||B||_F^2, ||.||_F$ denoting the Frobenius norm. For those familiar with concentration of measure OR Hanson-Wright inequality for concentration of quadratic forms, we could expect that $||BX||^2$ should be concentrated around $E[||BX||^2]=||B||_F^2.$ My question is: is the concentration asymptotically tight when dimension goes to infinity?
Following motivation from the fact: $lim_{n \to \infty}|E||X|| - \sqrt{n} |=0,$ I wonder: does $lim_{n \to \infty}|E||BX|| - ||B||_F |=0?$ Or if not, could we at least have: $\frac{||BX||}{||B||_F } \to _{p} 1$ in probability, as dimension $n \to \infty?$
I purposefully chose $B$ above so that the ratio of Frobenius norm to operator norm of $B$, i.e. $\frac{||B||_F}{||B||}$ does not go to $\infty.$ If it does go to infinity as $n \to \infty$, then we do have: $\frac{||BX||}{||B||_F } \to _{p} 1$, which follows from Hanson-Wright inequality. See the top/first equation from P.144 from this book.
The answer is no. Indeed, first of all, to make sense of the question, we need to deal with an infinite sequence of iid $N(0,1)$ random variables (r.v.'s) $X_1,X_2,\dots$. Next, for $n=1,2,\dots,\infty$, let
$$Y_n:=\sqrt{\sum_1^n\frac{X_k^2}{k^2}}.$$
Your question can then be stated thus: Is it true that
$$EY_n\underset{n\to\infty}\longrightarrow\sqrt{EY_\infty^2}\,?\tag{1}$$
To answer this question, note first that, by the uniform integrability (see e.g. Corollary 12.8 and Proposition 12.9),
$$EY_n\underset{n\to\infty}\longrightarrow EY_\infty.$$
So, the question becomes whether $\sqrt{EY_\infty^2}=EY_\infty$. But the latter equality may occur only if $P(Y_\infty=c)=1$ for some $c\in[0,\infty)$, which implies, in particular, that the r.v. $Y_\infty$ is discrete. In fact, however, the r.v. $Y_\infty$ is clearly absolutely continuous. Thus, the answer to question (1) is indeed no.
Iosif: thank you and much appreciate your answer, upvoted! I'll go through it in detail tomorrow and might comment here if I've further questions. Yes I agree that it's indeed an infinite sequence!
|
2025-03-21T14:48:30.442048
| 2020-04-29T14:33:59 |
358892
|
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"Abdelmalek Abdesselam",
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|
Stack Exchange
|
History of the notion of irreducible representation
I am looking for the earliest references where the study of irreducible representations appears. There has been many articles and books on the history of representation theory. A fundamental feature of this theory, is that in good situations where one is dealing with a semisimple category, one can decompose objects into simple ones, or here, irreducible representations. My understanding is that the introduction of this circle of ideas is usually credited to Frobenius around the end of the 19th century.
However, the decomposition of tensor products of irreducible representations of $SL_2$ can be found in the article by Paul Gordan "Beweis, dass jede Covariante und Invariante einer binären Form eine ganze Function mit numerischen Coefficienten einer endlichen Anzahl solcher Formen ist." in J. reine angew. Math. 323 (1868), 323-354.
It is written in old fashioned language that can be difficult to decipher, but the Clebsch-Gordan decomposition for $SL_2$ is basically there in Section 2 of that article. One could also ask: when was it realized that it was important and very useful to decompose general representations in terms of irreducibles? Reading the proof in that article, one can only conclude that Gordan was very well aware of that.
Also note that for the irreducible representations for $SL_n$, one can find a description already in the article by Alfred Clebsch "Ueber eine Fundamenthalaufgabe der Invariantentheorie", in
Abhandlungen der Königlichen Gesellschaft der Wissenschaften in Göttingen 17 (1872), 3-62, and its shorter follow-up "Ueber eine Fundamentalaufgabe der Invariantentheorie", Math. Ann. 5 (1872), 427-434.
Are there earlier references about irreducible representations?
You have done well already with this. A related question is how did characters of representations first arise? Back in the 1980's when Mike Hopkins and I were working on our generalized character theories applied to complex oriented cohomology theories, Mike was looking at the 19th century representation theory literature, to see when it was realized that characters were telling you about G-modules. G for us was finite, and 40 years later, we still don't really know what an irreducible module in Morava E-theory means, even though we know what the characters look like.
Thanks Nick. Do you remember what Mike excavated from his archeological dig in the 19th century literature?
@FrancoisZiegler: Nice! Indeed, a covariant of degree $n$ and order $k$ of a binary form of degree $d$, in modern language is the same as an element of ${\rm Hom}_{SL_2}(S_k, {\rm Sym}^n(S_d))$ or intertwiner. Here $S_p$ is the $p+1$-dimensional irreducible representation of $SL_2$. So I guess the earlier work of Cayley, etc. on covariants secretly involves irreducibles.
Perhaps this should be migrated to https://hsm.stackexchange.com/
@user347489 Research-level questions belong here.
@FrancoisZiegler as you know HSM has some fantastic Q and A for all type of questions related to history of mathematics, including some that could be considered research level. Has this already been discussed in the meta? I personally believe this belongs there, but I also understand how this falls into a gray area.
@user347489 I don’t know that.
@FrancoisZiegler Thanks for your reply.
@user347489: I also subscribed and contributed to hsm.stackexchange but I asked the question on MO for a reason. I am primarily interested in the input of professional mathematicians. I would not mind having the question cross-posted on hsm but I am against simply migrating the question to hsm.
@FrancoisZiegler: You raised some very good points. But it goes in a direction which is different (more analytic) from how I thought about the question (more combinatorial and category theoretic in flavor). I would have to amend my post to reflect that, but I can't do it now. Too much work.
@FrancoisZiegler: I just saw you added the "harmonic-analysis" tag (which is totally fine by me) but that's what I was trying to say in my previous comment: I would prefer the point of view of representation theory rather than harmonic analysis...if one could separate the two...BTW I think you should collect what you said in the comments and write an answer which would certainly deserve some upvotes.
I convert my comments to an answer per Abdelmalek’s request:
Dieudonné attributes the classification of irreducible $sl_2$-modules to Cayley (1856).
Also the theory of spherical and cylindrical harmonics should qualify as prehistory — told in e.g. Heine (1878, pp. 1–10), Burkhardt (1902–1903, Chap. V).
The words “irreducible” and “degree” hint at another root: if $G$ is finite, decomposing its regular representation $L$ on $\mathbf C[G]$ (elements $x=\smash{\sum x_g\delta^g}$, product $\smash{\delta^g\cdot\delta^h}=\smash{\delta^{gh}}$, $L(x)y=x\cdot y$) amounts to factoring the “group determinant” $\det(L(x))$ into irreducible polynomials in the $x_g$.
(“When was it realized that it was important and very useful to decompose general representations in terms of irreducibles?” inadvertently evokes a whole other question involving the origin of Fourier analysis, going back to at least D. Bernoulli, not to mention celestial epicycles, Pythagorean ideas on musical harmony; or Lang’s Algebra’s casting of Jordan form as “Representation Theory of One Endomorphism” (or the monoid algebra $k[\mathrm X]$ of $(\mathbf N,+)$, with irreducibles etc.) — but I understand the intent was to keep it non-commutative.)
Great!...thank you.
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2025-03-21T14:48:30.442432
| 2020-04-29T14:51:44 |
358893
|
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|
Stack Exchange
|
$S^3$ as a Sasakian Manifold
Reading about Sasakian manifolds one come across two slogans:
A) "A Sasakian manifold is an odd-dimensional analogue of a Kahler manifold."
B) "A Sasakian manifold sits between two Kahler manifolds - one above and one below."
I would like to understand the second slogan for the motivating example
of the three sphere $S^3$. What are the two Kahler manifolds that it sits
between? I would guess that below is the projective line $\mathbb{CP}^1$. But
I cannot guess what lies above.
I will answer your question for $S^{2n+1}$, since there is no difference between the case $n=1$ and the case of general $n$.
Let $(M^{2n+1},g,\theta)$ be a Sasakian manifold. One definition of a Sasakian manifold is that its metric cone is Kähler; this is the "one above". Here the metric cone is the manifold $M\times(0,\infty)$ with metric $dt^2+t^2g$. Thus in the case of $S^{2n+1}$, the metric cone is $\mathbb{C}^{n+1}\setminus\{0\}$ with the flat metric (written in spherical coordinates).
Taking the quotient by the $S^1$-action on $(M^{2n+1},\theta)$ determined by the Reeb vector field gives the "Kähler manifold below". For $S^{2n+1}$, the $S^1$-action is scalar multiplication by $e^{i\phi}$ (regarding $S^{2n+1}$ as the unit sphere in $\mathbb{C}^{n+1}$), so the quotient is $\mathbb{C}P^n$.
Will the metric cone be non-compact in general?
Does this not work for n =0? S1 x R does not seem to be homeomorphic to C.
@Asvin I think it should be $[0,\infty)$, with $M \times 0$ quotiented out. This part goes to the origin.
I see. That makes more sense, thanks!
The usual definition does not include $0$ in the second factor, the reason being that $(M\times[0,\infty)/\sim,dt^2+t^2g)$ is not smooth at the origin unless $M$ is a round sphere (here $\sim$ identifies $M\times{0}$ to a point). I have edited the original answer to reflect that the origin is missing. Also, the metric cone is always noncompact, because of the second factor.
|
2025-03-21T14:48:30.442591
| 2020-04-29T15:22:31 |
358896
|
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|
Stack Exchange
|
How to obtain envelope equation for oscillating functon?
Some oscillating function is given. How can I obtain its envelope? For example, for $ \sin x$ I should get $\pm 1$.
Particularly, I am interested in envelope for $$\begin{equation}\frac{(1-x \cot (2 \kappa a x))^{2}}{\left(1-x \cot (2 \kappa a x)-x^{2} / 2\right)^{2}+x^{2}(1-(x / 2) \cot (2 \kappa a x))^{2}}\end{equation}$$
UPD
by envelope I mean line which lies over maximum of function, for instance
this
Could you define "envelope"?
In your example a good "envelope" appear to be $4(1+x^2)/Q $ with $Q=2+2x^2+x^4-\sqrt{(x^2-1)^2+16x^2}$.
Is the "envelope" uniquely defined for any given function? If not, what characteristics are you looking for?
The maxima are at the poles of the cotangent, so for an envelope just retain only the cotangents
$$\frac{(1-x \cot (2 \kappa a x))^{2}}{\left(1-x \cot (2 \kappa a x)-x^{2} / 2\right)^{2}+x^{2}(1-(x / 2) \cot (2 \kappa a x))^{2}}\mapsto
\frac{(x \cot (2 \kappa a x))^{2}}{\left(x \cot (2 \kappa a x)\right)^{2}+x^{2}((x / 2) \cot (2 \kappa a x))^{2}}=\frac{1}{1+x^2/4}.$$
Here is a comparison for $\kappa a=1$.
|
2025-03-21T14:48:30.442683
| 2020-04-29T15:29:04 |
358897
|
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"Iosif Pinelis",
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|
Stack Exchange
|
Density of $C(X,\operatorname{co}\{\delta_y\}_{y \in Y})$ in $C(X,\mathcal{P}(Y))$
Let $X,Y$ be locally-compact Polish spaces, equip the set $\mathcal{P}(Y)$ of probability measures on $Y$ with the weak$^{\star}$ topology (topology of convergence in distribution), and equip $C(X,\mathcal{P}(Y))$ with the compact-open topology. Let $\operatorname{co}(\{\delta_y\}_{y \in Y})$ denote the set of finite convex combinations of point-masses on $Y$.
(When) is the subset $C(X,co(\{\delta_y\}_{y \in Y}))$ dense in $C(X,\mathcal{P}(Y))$?
$\newcommand{\ep}{\varepsilon}
\newcommand{\de}{\delta}
\newcommand{\B}{\mathcal B}
\newcommand{\K}{\mathcal K}
\newcommand{\NN}{\mathcal N}
\newcommand{\PP}{\mathcal P}
\newcommand{\supp}{\operatorname{supp}}$
The answer to your question is yes.
Indeed, recall that the weak$^{\star}$ topology on $\PP(Y)$ is metrized by the Lévy--Prokhorov metric $L$ defined by the condition: for any real $h>0$ and any $P,Q$ in $\PP(Y)$,
\begin{equation}
L(P,Q)\le h\iff\forall B\in\B(Y)\ P(B)\le Q(B_h)+h,
\end{equation}
where $\B(Y)$ is the Borel $\sigma$-algebra over $Y$ and $B_h$ is the $h$-neighborhood of $B$.
Let $\supp P$ denote the support of $P$. Let $\PP(K):=\{Q\in\PP(Y)\colon\supp Q\subseteq K\}$.
Lemma 1: Take any compact $\K\subseteq\PP(Y)$ and any real $\ep>0$. Then there is a compact $K\subseteq Y$ such that for each $P\in\K$ there is some $Q_P\in\PP(K)$ such that
\begin{equation}
\sup_{P\in\K}L(P,Q_P)\le\ep.
\end{equation}
Proof: By Prokhorov's theorem, for each real $\de\in(0,1)$ there is a compact $K\subseteq Y$ such that for each $P\in\K$ we have $P(K)\ge1-\de$. For all $B\in\B(Y)$, let then
\begin{equation}
Q_P(B):=\frac{P(B\cap K)}{P(K)}.
\end{equation}
Then for all $P\in\K$ and all $B\in\B(Y)$ we have
\begin{equation}
Q_P(B)\le\frac{P(B)}{1-\de}\le P(B)+\frac\de{1-\de}=P(B)+\ep
\end{equation}
if $\de=\ep/(1+\ep)$. So, indeed $L(P,Q_P)\le\ep$ for all $P\in\K$. $\Box$
Let $\PP_{fin}(Y)$ denote the set of all $Q\in\PP(Y)$ with a finite support.
Lemma 2: Take any compact $\K\subseteq\PP(Y)$ and any real $\ep>0$. Then for each $P\in\K$ there is some $R_P\in\PP_{fin}(Y)$ such that
\begin{equation}
\sup_{P\in\K}L(P,R_P)\le2\ep.
\end{equation}
Proof: Let $K\subseteq Y$ and $Q_P\in\PP(K)$ for $P\in\K$ be as in Lemma 1. Since $K$ is compact, there is a finite Borel-measurable partition $(C_{\ep;j})$ of $K$ such that for each $j$ there is a point $y_j\in Y$ such that $C_{\ep;j}$ is contained in the ball $B_\ep(y_j)$ of radius $\ep$ centered at $y_j$.
For any $P\in\PP(Y)$ and any $B\in\B(Y)$, let
\begin{equation}
Q_P^\ep(B):=\sum_j Q_P(C_{\ep;j})1_{y_j\in B}=Q_P(C_{\ep;B}),
\end{equation}
where
\begin{equation}
C_{\ep;B}:=\bigcup_{j\colon y_j\in B}C_{\ep;j}.
\end{equation}
Then $Q_P^\ep\in\PP_{fin}(Y)$. Moreover,
$$L(Q_P^\ep,Q_P)\le\ep$$
for all $P\in\PP(Y)$, because $C_{\ep;B}\subseteq B_\ep$ for all $B\in\B(Y)$. Letting now $R_P:=Q_P^\ep$, we see that Lemma 2 follows by Lemma 1. $\Box$
Finally, take any function $P_\cdot\in C(X,\PP(Y))$, which is a continuous map $X\ni x\mapsto P_x\in\PP(Y)$. Take also any compact $K_X\subseteq X$. Then the image $\K:=\K_{K_X}:=P_\cdot(K_X)$ of the compact $K_X$ under the continuous map $P_\cdot$ is compact. So, by Lemma 2, for each real $\ep>0$ and each $x\in K_X$ there is some $R_{\ep;x}\in\PP_{fin}(Y)$ such that
\begin{equation}
\sup_{x\in K_X}L(P_x,R_{\ep;x})\le2\ep.
\end{equation}
This means that indeed $C(X,\PP_{fin}(Y))$ is dense in $C(X,\PP(Y))$, in the required sense.
@JohnZornSu : I don't think you can make $f_k$ and $g$ continuous. You may want to ask this question in a separate post.
|
2025-03-21T14:48:30.442898
| 2020-04-29T15:39:52 |
358900
|
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|
Stack Exchange
|
Realization of limit of discrete series using Dirac operators
I wonder if there is a geometric realization of limit of discrete series in the flavor of Atiyah-Schmid or Parthasarathy realizing discrete series using Dirac operators on G/K. I know you can see limit of discrete series in Dolbeault cohomology of G/T after you slightly leave the $L^2$-world, so I wonder a similar thing could be done in G/K picture.
|
2025-03-21T14:48:30.442954
| 2020-04-29T16:17:43 |
358902
|
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|
Stack Exchange
|
If all 2-faces of a polytope are $2n$-gons, is the edge-graph bipartite?
This question on MSE has not received a satisfying answer. It can be summarized as follows:
Question: Is is true that the edge-graph of a (convex) polytope is bipartite if and only if all 2-faces are $2n$-gons?
This is motivated by the fact that a graph is bipartite if and only if all its cycles are of even length.
The hope is then that for polytopes it suffices to check the facial cycles instead of all cycles.
I would like to ask this question more generally for polytopal complexes.
Question: Is is true that a simply connected polytopal complex has a bipartite edge-graph if and only if all its 2-faces are $2n$-gons?
Simply connected is necessary, because there are counterexamples otherwise (see the image below; this one was already mentioned in the original post on MSE).
Even more general, I have the feeling this might be true for all polytopal complexes with a trivial first homology group.
The idea is as follows:
Fix a vertex $v$ of the complex.
For a vertex $w$, consider two paths $P_1$ and $P_2$ from $v$ to $w$.
Somehow use the fact that the union of these paths is a boundary (by trivial first homology group) to show that the paths have the same parity (that is, the same length mod 2).
Define a bipartition of the edge graph by assigning to each vertex $w$ the partiy of the paths from $v$ to $w$.
Apparently there are some gaps in the third step that I do not know how to fill in.
Yes your idea works, but you should be taking mod 2 (co)homology. Let $X$ be your manifold, $C_*(X, \mathbb{Z}/2)$ its mod 2 cellular homology complex. Define a map $\eta: C_1(X, \mathbb{Z}/2)\to \mathbb{Z}/2$ (a cochain) by counting the number of edges modulo $2$. Your condition on even edges implies this is a cocycle. Call the corresponding homology element $\alpha$. If you had an odd cycle $C$ then $C$ would define a class in $H_1$ that has nonzero pairing with $\alpha,$ implying $H_1$ with coefficients in $\mathbb{Z}/2$ (hence also in $\mathbb{Z}$) is nontrivial.
How do you define the 2-faces of a polytopal complex? Are you only counting 2-faces on the boundary? Do you want to also consider complexes with no boundary?
@HarryRichman The 2-faces of the complex are the 2-faces of the polytopes the complex is made from. Unbounded complexes are definitely a thing I would like to include if possible, but I don't think they make the task much harder, or do they?
In the given example, do the triangles count as 2-faces? If not, what is the polytopal structure?
@HarryRichman No, the triangles are not 2-faces, otherwise it would not be a counterexample in this context. If you want so, you can understand a polytopal complex as a set of actual convex polytopes together with a description of how they are connected facet to facet. More abstractly, the polytopes could also just be given via their face lattice. The boundary of a convex polytope is an example polytopal complex. The triangles in the toric example are not the 2-face of any of the involved convex polytopes (which in this case are just polygons themselve).
Ok, thanks for clarifying. I got confused thinking of it as a 3-dimensional complex rather than 2-dimensional.
|
2025-03-21T14:48:30.443195
| 2020-04-29T16:44:26 |
358903
|
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|
Stack Exchange
|
A stronger generalized Jordan curve theorem
The generalized Jordan curve theorem is usually stated as such:
Given $X\subseteq S^n$ such that $X$ is homeomorphic to $S^k$, $$\tilde{H}_i(S^n\setminus X)\cong\begin{cases}\mathbb{Z},\quad i=n-k-1 \\ 0,\quad \text{otherwise}.\end{cases}$$
If we relax the hypothesis that $X$ is homeomorphic to $S^k$ and instead only assume that $X$ is homotopy equivalent to $S^k$, does the theorem still hold true? I cannot visualize a counterexample, but I suspect a pathological example may exist. I attempted to prove it using the Mayer-Vietoris sequence like one would for the usual generalized Jordan curve theorem, but ran into an issue in the inductive step:
Let $X\subset S^n$ such that $X\simeq S^k$ and let $f:X\to S^k$ be a homotopy equivalence. Let $A,B\subseteq S^k$ be closed subsets such that $A\cup B=S^k$ and $A\cap B$ is homeomorphic to $S^{k-1}$ (e.g. $A$ and $B$ can closed hemispheres). Then $S^n\setminus f^{-1}(A)$ and $S^n\setminus f^{-1}(B)$ are open subsets of $S^n$ such that $(S^n\setminus f^{-1}(A))\cap(S^n\setminus f^{-1}(B))=S^n\setminus X$. From here, we can use the Mayer-Vietoris sequence, but in order to use induction on $k$, we need to know that $f^{-1}(A\cap B)$ is homotopy equivalent to $S^{k-1}$ and I am not sure whether it is. Also, to make use of the Mayer-Vietoris sequence, we would like $S^n\setminus f^{-1}(A)$ and $S^n\setminus f^{-1}(B)$ to have trivial reduced homology, but it is not obvious to me that this should be the case.
There is a standard argument using Poincare Duality that describes the homology of the compliment in terms of the homology of the submanifold.
@RyanBudney But that probably requires assuming $X$ is a manifold, right? Perhaps there is a counterexample when $X$ is not a manifold.
Not really. It just needs a nice neighbourhood. For example, it works for CW-complexes. What kind of spaces $X$ do you want to prove this for? Look up Alexander-Spanier duality in Bredon's text.
@RyanBudney I am looking at arbitrary subspaces of $S^n$ which are homotopy equivalent to $S^k$, no additional structure.
If you assume that $X$ is compact, you get the desired conclusion, otherwise, I am doubtful.
@MoisheKohan What about compactness implies the desired conclusion?
In this case: Use Alexander duality for Chech cohomology and the fact that Chech cohomology is homotopy invariant.
|
2025-03-21T14:48:30.443361
| 2020-04-29T17:01:00 |
358904
|
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|
Stack Exchange
|
Compatible systems of $l$-adic Galois representations as sheaves
I have a simple translation question:
Suppose we fix embeddings $\iota_l: \mathbb{Q} → \mathbb{Q}_l$ for every rational prime $l$, and take a finite-dimensional vector space $V$ over a number field $E$.
Let $\mathcal{R} = \{R_{l,\iota_l} : Gal(\mathbb{\bar{Q}}/\mathbb{Q}) → GL(V ⊗_{\mathbb{Q},\iota_l} \mathbb{Q}_l)\}$ be a weakly compatible system of Galois representations.
Can we think of $\mathcal{R}$ as a certain kind of sheaf over Spec $\mathbb{Z}$? If yes, how would the properties of $\mathcal{R}$ being strictly compatible/pure/geometric be expressed in terms of properties of this sheaf?
There are conjectures that every compatible system is represented (in an appropriate sense) by a unique object in some motivic category but I don't think it makes sense to think of this category structure as interesting, precisely because of the "weak" adjective. Even if these conjectures were true, the category you get would be analogous to "the category of $X$-graded vector spaces with graded components isomorphic to the stalks of a sheaf" for $X$ a topological space. If you could somehow get rid of the adjective "weakly" (for example by using motives) then your question is more meaningful.
|
2025-03-21T14:48:30.443591
| 2020-04-29T17:06:24 |
358907
|
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|
Stack Exchange
|
closed ideals in L(L_1)
Denote $L_1=L_1[0,1]$ The lattice of closed ideals in $\mathcal{L}(L_1)$ includes the chain
$$
\{0\}\subsetneq\mathcal{K}(L_1)\subsetneq\mathcal{FS}(L_1)
\subsetneq\mathcal{J}_{\ell_1}(L_1)\subsetneq\mathcal{S}_{L_1}(L_1)
\subsetneq\mathcal{L}(L_1).
$$
I believe it's still an open question whether the lattice contains any more distinct elements than the above listed.
One strategy for producing many closed ideals very quickly is to use Schreier-Rosenthal operator ideals, and this has worked, for instance, in the cases $\mathcal{L}(\ell_1\oplus\ell_\infty)$. In particular, given a countable ordinal $\xi$, we could seek some special operator $A\in\mathcal{R}_\zeta(L_1)\setminus\mathcal{R}_\xi(L_1)$ for some $\zeta>\xi$.
Here's how it worked for $\mathcal{L}(\ell_1\oplus\ell_\infty)$. Let $T_\xi$ denote the $\xi$-Tsirelson space, and let $I_{1,\xi}:\ell_1\to T_\xi$ be the formal identity. Of course we can find an embedding $J_{\xi,\infty}:T_\xi\to\ell_\infty$. Then $J_{\xi,\infty}\circ I_{1,\xi}$ is class $\mathcal{R}$ but not class $\mathcal{R}_\xi$. By a result of Beanland/Freeman we can find $\zeta>\xi$ such that it is class $\mathcal{R}_\zeta$ as well. Then use transfinite induction.
Adapting this approach for $\mathcal{L}(L_1)$ presents serious difficulties. Of course $L_1$ contains a complemented copy of $\ell_1$, so we can still use a formal identity map from $\ell_1$ onto any space with an unconditional basis. But we can't use $T_\xi$ because it doesn't embed into $L_1$. (We know this because every infinite-dimensional subspace of $L_1$ contains a copy of $\ell_p$ for some $1\leq p\leq 2$.)
But perhaps there is a way around this. For any scalar sequence $(a_n)$ we can define
$$\|(a_n)\|_\xi=\|(a_n)\|_2\vee\sup_{F\in\mathcal{S}_\xi}\|(a_n)_{n\in F}\|_1$$
This generates a 1-unconditional basic sequence whose closed linear span we denote $X_\xi$. Is it possible that $X_\xi$ (or something like it) embeds into $L_1$? We would also need to make sure that $\ell_1$ doesn't embed into $X_\xi$, but that seems fairly likely given the latter's construction.
In general, to make this work we need to find a space $X_\xi$ with a basis which is $\xi$-equivalent to $\ell_1$ and embeds into $L_1$, but doesn't contain a copy of $\ell_1$. Frankly, I believe this is asking for too much. Nevertheless, I thought it would be worth seeing what y'all thought before I give up on the idea. And if there is no such space, perhaps that would be worth proving anyway.
Thanks guys!
You could have dropped me an email. I keep an eye on this stuff :-)
Your post is awfully technical for MO, IMO.
The lattice of closed ideals in $L(L_1)$ contains at least a continuum of elements:
https://arxiv.org/abs/1811.06571
Oh wow cool, nice! Geez louise, is there anything you can't do?
Among zillions of things I would like to do but cannot, I do not know how to construct new large ideals in $L(L^1)$; in particular, I do not know how to construct new complemented subspaces of $L^1$. The latter is something I have wanted to do for 50 years (or prove that there are only the two known ones).
|
2025-03-21T14:48:30.444069
| 2020-04-29T17:16:48 |
358911
|
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|
Stack Exchange
|
Exponential towers of $i$'s
It's well known that the expression $i^i$ takes on an infinite set of values if we understand $w^z$ to mean any number of the form $\exp (z (\ln w + 2 \pi i n))$ where $\ln$ is a branch of the natural logarithm function.
Since all values of $i^i$ are real, all values of $i^{i^i}$ (by which I mean $i^{\left( i^i \right)}$) are on the unit circle, and in fact they form a countable dense subset of the unit circle.
I can't figure out what's going on with $i^{i^{i^i}}$, though. I've posted a pdf version of Mathematica notebook at https://jamespropp.org/iiii.pdf containing images starting on page 2. Each image shows the points in the set of values of $i^{i^{i^i}}$ lying in an annulus whose inner and outer radii differ by a factor of 10. For instance, here's what we see in the annulus whose inner and outer radii are 100 and 1000 respectively.
Can anyone see what's going on?
Also, what happens for taller exponential towers of $i$'s? Does the set of values of $$i^{i^{i^{{\ \kern3mu\raise1mu{.}\kern3mu\raise6mu{.}\kern3mu\raise12mu{.}\ }^i}}}$$ become dense in the complex plane once the tower is tall enough? The following picture makes me think that for a tower of height five, the set of values is everywhere dense.
Possibly relevant: Convergence of $i^{i^{i \ldots}}$ by Archibald James Macintyre [Proc. AMS 17 #1 (February 1966), p. 67].
When you have multiple (nested) exponentials do you mean to pick a common branch of log once for all before you apply any of them, or to chose a (possibly different) branch for each exponentiation?
I mean to choose a possibly different branch.
For sets of complex numbers $A$ and $B$, let $A^B$ denote the set of all complex numbers of the form $\exp( z (\ln w + 2 \pi i n))$ wtih $w$ in $A$, $z$ in $B$, and $n$ an integer. Then I am asking about $A^{A^{A^A}}$ where $A = {i}$ (as well as taller towers of this form).
on the wiki page List of mathematical constants the constant goes under the name infinite tetration $^\infty i$ of $i$
Hi James, it seems your site is down. I couldn't find the file on the Internet Archive but it did point to your faculty page. For those interested: http://faculty.uml.edu/jpropp/iiii.pdf
Let $f : \mathcal{P}(\mathbb{C}) \rightarrow \mathcal{P}(\mathbb{C})$ be the function on the powerset of the complex numbers defined by:
$$ f(A) = \left\{ \exp\left( \frac{\pi i}{2}(4n + 1)z \right) : z \in A, n \in \mathbb{Z} \right\} $$
Then your question is asking about $f^k(\{1\})$ and its (topological) closure.
One thing to note is that, for any set $A$, we have $\overline{f(A)} = \overline{f\left(\overline{A}\right)}$, where the overline denotes closure. This follows from the fact that the expression $\exp\left( \frac{\pi i}{2}(4n + 1)z \right)$ is continuous as a function from $\mathbb{C} \times \mathbb{Z}$ to $\mathbb{C}$. So, if we want to study the closure of $f^n(\{1\})$, we are allowed to take the closure after each application of $f$.
The OP remarked that $\overline{f^3(\{1\})}$ is the unit circle. Then the set:
$$ B :=\left\{ \frac{\pi i}{2}(4n + 1)z : z \in \overline{f^3(\{1\})}, n \in \mathbb{Z} \right\} $$
is precisely the union of a circle of radius $\frac{\pi i}{2} k$ for each odd integer $k$. Next, we need to take the image of this set $B$ under $\exp$. As the function $\exp$ is many-to-one, this is best understood by first reducing $B$ modulo $2 \pi i$ so that it lies in the strip with imaginary part $[-\pi, \pi)$.
Here's a picture of the first six circles in $B \mod 2 \pi i$:
The intersection of $B \mod 2 \pi i$ with a line of fixed real part is a set of points which contains two limit points (the intersection with the light grey lines in the above plot). The closure of $B \mod 2 \pi i$ therefore contains all of these circles together with the two light grey lines (and nothing else).
$\overline{f^4(\{1\})}$ is therefore just the image of this set under $\exp$ (viewed as a bijective function from the strip to the punctured complex plane) together with the origin.
Now let's consider $\overline{f^5(\{1\})}$. Recall that the first step is to take the union of lots of homothetic copies of $\overline{f^4(\{1\})}$:
$$ B' :=\left\{ \frac{\pi i}{2}(4n + 1)z : z \in \overline{f^4(\{1\})}, n \in \mathbb{Z} \right\} $$
If we pull this back through the inverse of the $\exp$ map (so we're working on the strip instead of the complex plane), this corresponds to taking the union of lots of translates of $B \mod 2 \pi i$. Specifically, we are interested in the set of points:
$$ \log(B') :=\left\{ w + \log(i) + \log\left(\frac{\pi}{2}(4n + 1)\right) : w \in (B \mod 2 \pi i), n \in \mathbb{Z} \right\} $$
Now, the set $\{ \log\left(\frac{\pi}{2}(4n + 1)\right) : n \in \mathbb{N} \}$ has the very desirable property that the set is unbounded to the right and the gaps between the points become arbitrarily small. Also, the intersection of every horizontal line with the original set $B \mod 2 \pi i$ (the one in the picture above) is unbounded to the left. As such, it follows that their convolution is dense in every horizontal line, and therefore dense in the whole strip.
Consequently, $\overline{f^5(\{1\})}$ is the entire complex plane as you suspected.
The single-valued case of the exponential tower of $i$ was studied by Peter Lynch. The tower converges along three spirals to the fixed point $Q=i^Q$, given in terms of the Lambert W-function by
$$Q=\frac{2i}{\pi}W\left(\frac{\pi}{2i}\right)=0.438\cdots+i\,0.361\cdots,$$
see image.
I'd guess this amazing constant is not linked to any other known constant?
it's basically the Lambert W-function at $\pi/2i$, I don't know if it appears in other contexts.
I'd be very surprised if that evaluation of $W$ was exactly as you've written. Would not an $\approx$ be more appropriate?
you mean the numerical value of $W$ ? I only recorded the first few digits, more decimals are at https://oeis.org/A077589 and https://oeis.org/A077590 --- I added trailing $\cdots$
|
2025-03-21T14:48:30.444481
| 2020-04-29T17:43:31 |
358913
|
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|
Stack Exchange
|
On the equation that involves the Dedekind psi function $\psi(x)=n$ with unique solution $x$, for a fixed integer $n\geq 1$
The Dedekind psi function is defined for a positive integer $m>1$ as
$$\psi(m)=m\prod_{\substack{p\mid m\\p\text{ prime}}}\left(1+\frac{1}{p}\right)\tag{1}$$
with the definition $\psi(1)=1$. See the Wikipedia Dedekind psi function (I add also the reference [1]). I wondered (while I was studying elementary problems similar than some problems that are in the literature of the Euler's totient function) if the following advanced problem is in the literature.
Problem. Given an integer $X\geq 1$ we define the set
$$\{\text{integers } 1\leq n\leq X:\text{ the equation }\psi(x)=n\text{ has a unique solution } x\}\tag{2}$$
and we denote the cardinality of this set as $$f(X):=\#\{ 1\leq n\leq X:\psi(x)=n\text{ has a unique integer solution } x\geq 1\}.$$ Determine what is the asymptotic or how grows $f(X)$ as $X\to\infty$.
Question. What work can be done to get a not obvious statement $$f(X)=\text{main term}+\text{error term}\tag{3}$$
as $X\to\infty$? I'm asking a statement at research level if isn't in the literature*, you can provide your statement about this asymptotic behaviour $(3)$ of $f(X)$ using Landau notation or other notation. Many thanks.
*I don't know if this problem is in the literature, if this problem is in the literature answer my question as a reference request and I try to search and read the statement from the literature.
A related sequence is A203444 from the OEIS, but I believe that the sequence of $n$'s such that $\psi(x)=n$ has a unique solution $x$, isn't in the OEIS.
I've calculated with a Pari/GP script the first positive integers $n$ such that $\psi(x)=n$ has a unique solution, its solution $x$. For very small integers $X$ it seems that $f(X)\ll X$ in Vinogradov notation (let's say $f(X)\approx \frac{X}{10}$).
References:
[1] Tom M. Apostol, Introduction to analytic number theory, Undergraduate Texts in Mathematics, New York-Heidelberg: Springer-Verlag (1976).
As was said if this problem is in the literature feel free to refer that this is a well known problem. I suspect that this kind of problems should be difficult since other similar that are in the literature for the Euler's totient function were very difficult. Any case I'm asking about what work can be done to get a statement $(3)$ at research level, for very large $X$ (if you can deduce the corresponding proposition for $X\to\infty$ it is better, of course).
Dedekind psi tabulated at http://oeis.org/A001615 – I think the sequence under discussion is very like http://oeis.org/A291109
Now I'm confused, maybe the problem is well known, if the Problem in the body of the post is known refer the literature in comments or answering the Question as a reference request. Many thanks @GerryMyerson
Sorry, I'm the one who is confused. 291109 resembles numbers not in the range of psi, rather than numbers appearing exactly once as values of psi.
|
2025-03-21T14:48:30.444698
| 2020-04-29T17:52:51 |
358914
|
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|
Stack Exchange
|
Turing machines with all runs decidable
$\DeclareMathOperator\Comp{\mathit{Comp}}
\DeclareMathOperator\succ{\mathit{succ}}$Let $(\Phi_e)_{e\in\omega}$ be your favorite enumeration of Turing machines. For $e,n\in\omega$ there is a structure $\Comp(e,n)$ naturally associated to the run of $\Phi_e$ on input $n$. Intuitively, $\Comp(e,n)$ is an $\omega\times\omega$-array whose $(a,b)$th entry codes the state of the computation $\Phi_e(n)$ at stage $n$.
More formally:
The domain of $\Comp(e,n)$ is $\omega^2$, with "$(a,b)$" representing cell $b$ at time $a$.
The language of $\Comp(e,n)$ consists of: unary function symbols $\succ_t$ and $\succ_s$ (for moving in time and space respectively; a unary relation symbol $S_q$ for each of the finitely many states $q$; a unary relation symbol $A_x$ for each of the finitely many alphabet symbols $x$; and a unary symbol $H$ for the tape head.
We set $\succ_t((a,b))=(a+1,b)$, $\succ_s((a,b))=(a,b+1)$.
We set $S_q((a,b))$ iff at stage $a$ the computation is in state $q$.
We set $A_x((a,b))$ iff at stage $a$ the $b$th cell on the tape has symbol $x$.
We set $H((a,b))$ iff at stage $a$ the head of the Turing machine is at cell $a$.
We use the convention that if the computation halts, we just copy the final configuration forever.
(Note that the actual program, i.e. the transition function, is not coded into the structure; $\Comp(e,n)$ only tells us what happens, not why it happens. That said, this isn't an essential point, and any reasonable tweak will result in an appropriately-equivalent structure.)
My question is about the logical complexity of these structures. Trivially, if $\Phi_e(n)\downarrow$ then $\Comp(e,n)$ is decidable. Broadly, I'm interested in what general tools we might use to prove decidability of $\Comp(e,n)$ when $\Phi_e(n)\uparrow$.
More specifically, say that $\Phi_e$ is run-decidable if $\Comp(e,n)$ is decidable for each $n$. Note that we consider each run of $\Phi_e$ individually here, and do not even demand that $\Comp(e,n)$ be decidable uniformly in $n$. My question is:
Is every c.e. set the domain of some run-decidable machine (at least up to Turing degree)?
(I'm happy to shift attention to the analogue of $\Comp(e,n)$ for other models of computation if that would help.)
What does it mean for a structure to be decidable, that the FO theory is?
@VilleSalo Exactly.
People study Turing machines as dynamical systems, and one thing they do is think about how complex the spacetime diagrams (these are your structures) are. That might or might not be relevant, I certainly don't see the answer though.
@VilleSalo Yes, I took a glance at the subshift literature especially in this context, but I didn't find an answer there (I'm not very familiar with it, though).
I think I must not understand the notion of arity. Aren't your $\operatorname{succ}$, $S$, and $A$ all binary (taking a pair $(a, b)$)? Or is it unary because we regard pairs $(a, b)$ as atomic within the structure, even though they look non-atomic externally?
@LSpice I like \mathit - I didn't know that was an option! I'll definitely use it in the future. (For my eyes, the mathoperator mode blurs into normal text. You're absolutely right about the kerning issue.) I've also thrown in extra parentheses to address the arity issue.
What do you mean by 'up to Turing degree' in your question at the end? Do you mean 'is every c.e. set Turing equivalent to the domain of a run-decidable machine'?
@JamesHanson Yes, that's exactly right. It would be slightly more proper to say 'up to Turing equivalence,' but the phrasing above seems more prevalent in the literature. (An interesting example of how 'up to degree' can matter a lot: you can show that every $\Delta^0_2$ set is c.e. up to degree over some low set, but trivially the complement of the halting problem can't literally be c.e. over any incomplete $\Delta^0_2$ set.)
I was just about to leave for my biquarantinely jog when you asked this nice question, sorry about the quick comments which if anything have lead you to a wild goose hunt. I think the answer is yes, with a much easier trick than the ones I suggested.
First of all, my understanding is that with just successors and first-order logic, all you can do is count finite patterns. If you can do more than that, then the following may not be enough.
The idea is to simulate counter machines with our Turing machine: the machine reads the input, and verifies that it is of the form $ 0^n 12\#0^\omega$ (I assume the input is finite and you have an end-marker $\#$, and you are promised the rest is $0$s). If it's not, just halt. Otherwise, erase the end-marker, go back to the origin of the tape, and start simulating a universal two-counter machine by zig-zagging between the origin and the $2$, moving $1$ and $2$ (keeping them in this order).
The invariant linking the Turing machine run and the counter machine is that when the head returns to the origin for the $m$th time, the tape looks like $q 0^k 1 0^\ell 2 0^\omega$ where $(q,k,\ell)$ is the state of the counter machine after $m$ steps.
Now, assuming I understand what decidability of first-order logic amounts to with your definitions, all you have to be able to decide about the spacetime diagram is counting, i.e. it is sufficient, given a pattern $P$ and a number $k \in \mathbb{N}$, to be able to decide whether $P$ appears at least $k$ times in the spacetime diagram.
Now, the point is that any particular diagram has a finite amount of this type of counting information, because all you have to remember is in which states the head traverses from $1$ to $2$ (or $2$ to $1$, or origin to $1$, or $1$ to origin) infinitely many times, and whether $1$ and $2$ meet infinitely many times and in which states.
Two-counter machines can accept the set $\{(2^n,0) \;|\; n \in S\}$ for any $\Sigma^0_1$ set $S$, so also the set of $0^n12$ on which this machines halts is undecidable if we pick a suitable counter machine to simulate. (Note that on other inputs the machine halts and certainly the spacetime diagram is decidable since it's Presburger.)
edit 5th May 2020
Here's some additional observations and details. I have fixed the argument above a bit as well, as I realized there was a small mistake in what I was counting (I decided exact counts of occurrences of patterns, but we want to decide lower bounds on number of occurrences instead).
Preprocessing: any $\Sigma^0_1$ set can be accepted by a run-computable machine
Obviously any decidable preprocessing can be done to the input, as this only adds finitely many new initial rows of the grid for each input, which does not affect decidability by an easy argument. So we can make our Turing machine turn an arbitrary input $w \in \{0,1,2\}^* \# 0^\omega$ into the form $0^{2^{n(w)}}120^\omega$ where $n : \{0,1,2\}^* \to \mathbb{N}$ is any computable injection (the alphabet doesn't matter either, but I used a ternary one plus marker in the original construction so I'll keep that).
How does the two-counter universality work, again?
Let me recall the outline of the classical argument that we can compute absolutely anything about $n$ if the input is given as $0^{2^n} 12$ and we can only simulate a two-counter machine as I described, just because I can't bother finding a reference.
The way two-counter counter machine universality is usually proved is in two steps, first we simulate Turing machines with three counters, then three with two. So first assuming the Church-Turing thesis, we can compute anything using a Turing machine. A Turing machine's configurations are of the form $u q v$ where $u$ and $v$ are finite words, say over alphabet $\{0,1\}$. We simply replace them with the numbers they present in binary, and we can simulate a Turing machine using just two counters, as long as we can do the following operations to the counters:
Check the parity of a counter (to read a bit).
Divide/multiply a counter by two (to move the head).
With a basic counter machine, we assume we are only allowed to shift the counter values by one and check for zero. So we cannot do the above. Instead, we add a third counter, and now we can check parity of counter number $1$ by moving its contents to counter number $3$ two values at a time, and at the end we see the parity, and move everything back. Multiplying and dividing works the same. It follows that with three counters we can compute anything with two inputs given in the first two counters, assuming the third counter initially contains $0$.
Now to simulate $k$ counters with two, take $k$ distinct primes $p_1, ..., p_k$, and the correspondence is that $k$ counters containing values $(v_1, ..., v_k)$ is replaced by having the first counter contain $p_1^{v_1} p_2^{v_2} \cdots p_k^{v_k}$ and the second contain $0$. You can increment the simulated counter $i$ by dividing the first counter value by $p_i$, and similarly we can divide and check its parity (checking parity means checking whether $p_i$ divides the counter value an odd or an even number of times). Picking $p_1 = 2$, we see that if the input is $2^n$ we can think of the first simulated counter as containing $n$, and the others (however many auxiliary counters we want to use) contain $0$.
What's a pattern and what's "appearing"?
I should maybe explain what I mean by patterns appearing, since this is different terminology than what the asker used. I think of the structure as being an element $x \in A^{\omega^2}$ where $A$ is a finite alphabet containing the information about whether the head is in the current position and what the tape symbol is. I call this the spacetime diagram. A pattern is an element of $A^D$ for finite $D \subset \omega^2$ and appears means $\sigma^{v}(x)_D = P$ for some $v \in \omega^2$, where $\sigma^v$ is the shift, defined by $\sigma^v(x)_u = x_{u+v}$. I say $P$ then appears at $v$ and we say $P$ appears $k$ times if it appears at $v_1, v_2, ..., v_k$ for distinct vectors $v_i \in \omega^2$, and does not appear at any $v \notin \{v_1,v_2,...,v_k\}$.
What's the finite information we need to ask from an oracle, exactly?
Some more details (though still just a sketch) on why we only need a finite amount of information about the spacetime diagram to be able to decide every FO query, for now believing it amounts to counting how many times a given pattern appears. For this it is helpful (though not strictly necessary) to slightly modify the two-counter machine we simulate without changing its halting on any input: Let's assume the two-counter machine behaves as described above, and simulates a $k$-counter machine as described. Let $p$ be a prime it does not yet use to simulate the counters, and have it multiply the first counter value by $p$ between every two steps. Then we have that given $n$ we can compute $m$ such that after $m$ steps at least one of the two simulated counters will always have value at least $m$.
Now, having done that preprocessing, let me explain how you can decide whether a pattern $P \in A^D$ appears in the spacetime diagram $x$ at least $n$ times (and we'll see what information I need to query), assuming the computation never halts (if it halts, it's a Presburger/semilinear spacetime diagram anyway, so trivially FO queries would be decidable).
First, as discussed above, changing finitely many initial rows of $x$ does not change anything (there is a decidable procedure that modifies an FO query to take this into account). So we may assume that in the spacetime diagram $x$, at least one counter value is always larger than the maximal distance between any two elements of $D$. Now, observe that the only sort of thing $P$ may contain is
in $P$ we see just static stuff, no head and at most one counter values. Anything like this will appear infinitely many times (assuming the machine does not halt) so we actually need no information to answer such a query. (Anything where the contents of $P$ does not change from row to row that is, otherwise obviously it does not appear in any spacetime diagram since the configuration is only modified when the machine head moves on it.)
in $P$ we see a lone Turing machine head traveling on a bunch of zeros. This type of queries are easy, we just need to know in which state cycles the head moves over large zero areas infinitely many times.
in $P$ we see shows a Turing machine head hitting one of the counters (and possibly moving it). There are a few different cases here, let's concentrate on the case where the head of the Turing machine hits the $1$-counter (= leftmost of the two) in some state $q$, and we see $m$ many $0$s to the left of that counter in our picture. Now, we need to be a bit careful: the $m$ many zeroes to the left mean that the counter value has to be at least $m$ at this point. Obviously we cannot remember, for each distinct $m$, whether the head hits the $1$ counter with such particular $m$. So a crucial observation is that if we hit the counter at some point so that the counter value is at least $m$, then our simulation is in fact currently in a cycle where it is continually decreasing or increasing the first counter value and decreasing the other. So actually either before or after this step, we see the exact same pattern after a constant number of steps, with a smaller $m$. This is why all we need to actually know is the pictures we see for small $m$. (The preprocessing where we keep multiplying by $p$ removes the cases where both counters contain a small value.)
Why is FO just pattern counting?
First-order definability (for this particular structure!) is the same as threshold-counting, see e.g. https://www.sciencedirect.com/science/article/pii/S0890540196900188 for such a result. (I have not actually read this paper, and it's probably not the optimal reference; it's about finite pictures rather than infinite ones; but anyway same idea.)
What that means is, if you have an FO formula $\phi$, then you can effectively find a finite list $P_1, P_2,... , P_k$ of patterns, and a subset $S \subset \omega^k$ which for some $n_0$ satisfies $t \in S \iff t + e_i \in S$ whenever $t_i > n_0$, where $e_i = (0,0,...,1,0,0,...,0)$ is the $i$th standard generator of $\omega^k$ as a monoid; such that $\phi$ is true for $x \in A^{\omega^2}$ if and only if $(t_1, ..., t_k) \in S$, where $t_i$ is defined by
$$ t_i = |\{\mbox{number of times } P_i \mbox{ appears at in } x\}|. $$
And vice versa you can go from $P_1, P_2 ,... ,P_k$ and such $S$ to an FO formula.
If we just want to decide an FO formula, you only need to be able to check for a single pattern $P$ and $n \in \mathbb{N}$ whether $P$ appears at least $n$ times: given a formula $\phi$, turn it into $P_1, P_2 ,..., P_k$ and $S \in \omega^k$ as described above, and use that decidability result finitely many times as a subroutine to check whether $t \in S$.
Independence of accepted language and run-decidability
As mentioned, any $\Sigma^0_1$ set can be accepted by a run-decidable machine. But of course any $\Sigma^0_1$ set is also accepted by a strongly run-undecidable machine, namely one whose spacetime diagram is undecidable for all non-halting inputs: simply make the machine, while doing its main computation, also simulate the computation for some $\Sigma^0_1$-complete language $L \subset \omega$, and write $10^n1q$ somewhere in the spacetime diagram for all $n \in L$, $q$ marking the position of the head when it's written. (This cannot appear in the original input so if we take some care it does not interfere with whatever the actual language is we want to accept.)
Can we make an even more run-decidable Turing machine?
One may ask if we can do more than decidability of FO logic, say monadic second-order logic. Good question, I don't know the answer!
Biquarantinely = this will be your second one?
Or my first one, and I'm hopeful.
I realized that there's a lot of tricks here that you may need to have a computer science degree to know. I tried writing some of it out in the edit just now, but perhaps it just looks more incomplete now since there's now more places for missing details :) Best I can do for now, I'm happy to try to explain some of the steps if something is unclear.
Oh, I didn't notice that the original question asks about every c.e. set. Today's edit also clarifies that this follows from the original construction. (Not just up to degree, but exactly.)
It will take me a bit of time to accept this answer, but thanks (and it's certainly worth the bounty!).
Thanks for the meaningful internet points! Feel free to ask for clarifications.
This looks great - thanks again! (Of course I'll bother you if I wind up having questions later on.)
I decided to give some thought to my last comment, i.e. monadic second-order logic. Twenty seconds in, I'm not sure it's such a great question after all. The structure is an $\omega \times \omega$ array, so ignoring its contents the tiling problem of $\mathbb{Z}^2$ reduces to its EMSO, and all runs are undecidable.
|
2025-03-21T14:48:30.445922
| 2020-04-29T18:13:17 |
358915
|
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|
Stack Exchange
|
How wealthy are canonical inner models?
One of the way a person shows their wealth is by having many diamonds. The same can be said about models of $\sf ZFC$. We can add generic diamond sequences, while preserving the old ones, so in some sense, some models have more diamonds than others. And some have none at all.
Recall that a diamond sequence is a sequence $\langle A_\alpha\subseteq\alpha\mid\alpha<\omega_1\rangle$ such that $A_\alpha\subseteq\alpha$, and for any $A\subseteq\omega_1$, $\{\alpha\mid A\cap\alpha=A_\alpha\}$ is stationary.
Very very clearly, we can replace any one particular $A_\alpha$, in fact any countably many of them, and in fact any non-stationary set of them, and the sequence is still a diamond sequence. By that trivial observation alone, if there is one diamond sequence, there are $2^{\aleph_1}$ of them.
So the "plain counting argument" is the wrong one. But we can talk about equivalence up to a non-stationary set. That is, two sequence are equivalent, if they differ on a non-stationary set.
Okay, but maybe you can find a permutation of $\omega_1$ which maps stationary sets to stationary sets. And maybe that permutation is not the identity (mod. non-stationary, that is), so maybe that gives you a new sequence, where in fact you just moved that diamond from an earring to a necklace.
So it is unclear to me that agreeing on a club is the right notion here. But maybe it is.
Questions:
Is there a better notion of making diamond sequences distinct?
How wealthy are canonical inner models? In particular does $L$ have only one diamond sequence (up to a reasonable notion of equivalence)?
The reason to ask is that normally we argue that a canonical inner model has a diamond sequence by constructing said sequence from some fine structural properties of the model. But that gives us a very concrete sequence, which may hint at the relative poverty in which canonical inner models live.
I just want to clarify that diamonds in canonical inner models are ethical diamonds, i.e. diamonds that were not obtained by forcing.
I think that equivalence up to a club is still too strong (you still have $2^{\aleph_1}$ many such diamond sequences from one). Take $X \subseteq \omega_1$ and ask the diamond to guess some coding of the pair (A, X), where $A \subseteq \omega_1$. By taking the guess for the A-part, in ordinals in which we guessed the $X$ part correctly, we get a diamond sequence. For any $X$ you get a different diamond sequence and the non-trivial parts of those sequences can agree only on a bounded piece.
Well, I suspected that much. But then, what would be a good way to identify "essentially the same" diamonds?
Let's start with: Is there a candidate (forcing) model where there is some evidence that one should reasonably expect wealth?
@Andrés: Correct me if I'm wrong, but we can add a generic diamond sequence by forcing with its initial segments. Now iterate... Or take a side-by-side product, I suppose, for at least $\aleph_2$ different diamonds.
(Note that this is predicated on a notion of "distinct" that is satisfied by pairwise generic sequences.)
@Asaf I fear that we need to clarify the notion first before much can be said. I was hoping that wealth could be expected because of combinatorial consequences: in this model we can build this and that diamond sequences, the first gives us A, the second B, and "there does not seem to be a reasonable way" of getting B from the first or A from the second.
@Andrés: Indeed that was my thinking. It might be that somehow I'm wrong. Maybe any reasonable notion of sameness results in either "everything is equivalent" or "there is a maximal number of diamonds".
@Andrés: One way is naturally to consider the relative constructibility order on diamond sequences. But then trivially we get that $L$ admits one class. Maybe that's okay...
Since you asked about the situation in $L$, I'll conjecture that, for any "reasonable" notion of equivalence, $L$ will have $\aleph_2$ inequivalent diamond-sequences and that this should be provable by tweaking standard constructions of one diamond sequence in $L$. (Here "reasonable" is intended to mean something like "combinatorial" and to exclude relative constructibility.)
@Andreas: That's interesting, I can see why that would be true. Maybe someone has a suggestion of a reasonable notion of equivalence, and then we can settle this...
A $\diamondsuit$-sequence gives you an almost disjoint family of stationary subsets of $\omega_1$ in a natural way. Does that family capture any essential information about the $\diamondsuit$ sequence? Do you get any mileage out of declaring two $\diamondsuit$ sequences equivalent if those a.d. families generate the same ideal, for example? Just random late night thoughts..
@Todd: I imagine that this family is somehow related to the "agree modulo permutations on a club"?
Another random thought. Replace your diamond sequences by allowing countably guesses at each $\delta$ instead of one. Then you can distinguish diamond sequences by whether or not they are "somewhere" a $\diamond^$-sequence, that is, whether there is a stationary set $S$ where any set is guessed "almost always". Once you've got a $\diamondsuit^(S)$ sequence, any run-of-the-mill single-guessing $\diamondsuit$ sequence on $S$ is built from it in some sense. This would push towards $L$ having only a single type of $\diamondsuit$ sequence, built from the hierarchy in standard way.
@Todd: Aha, that's nice. Does that also satisfy the "generic is different" criteria, so that generic diamonds are different?
|
2025-03-21T14:48:30.446357
| 2020-04-29T18:22:41 |
358916
|
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|
Stack Exchange
|
Is "$\mathbb{E}(T_n|X)\rightarrow 0 $ a.s." equivalent to a statement that does not involve the Radon–Nikodym derivative as a black box?
Let $\{T_n\}_n$ be a sequence of random variables, and let $X$ be another random variable.
Each $\mathbb{E}(T_n|X)$ is a random variable, therefore the statement "$\mathbb{E}(T_n|X)\rightarrow 0$ almost surely" makes sense in the sense of almost sure convergence of random variables.
What makes this difficult to unpack is that almost sure convergence is phrased in terms of individual $\omega$'s in the probability space, but the definition of $\mathbb{E}(T_n|X)$ is via the existence of the Radon-Nikodym derivative, and so is non-constructive...
The question is whether you can have an equivalent definition of this statement that does not use the Radon-Nikodym derivative as a black box. For example, is it equivalent to the statement: "for every measurable set $U$ in the co-domain of $X$, the mean of $T_n$ over $X^{-1}(U)$ converges to $0$"?
At the level of logic, note that it's only the existence of the conditional expectation that relies on the Radon-Nikodym theorem; the definition itself does not. You should be able to formulate, and perhaps prove, a statement like "every conditional expectation of $T_n$ given $X$ converges a.s. to 0", without any reference to Radon-Nikodym.
You are asking: Is the statement "$E(T_n|X)\to0$ almost surely [a.s.]" equivalent to the statement "for every measurable set $U$ in the co-domain of $X$, the mean of $T_n$ over $X^{-1}(U)$ converges to $0$"?
The answer is no in general. First here, to talk about the mean of $T_n$ over $X^{-1}(U)$, you need to assume that $P(X^{-1}(U))>0$. Taking this into account, we can restate your question as follows: Are the following two statements equivalent:
(i) $Y_n\to0$ a.s.
(ii) $EY_n\,1_{X\in U}\to0$ for all measurable sets $U$ in the co-domain of $X$,
where $Y_n:=E(T_n|X)$. Neither one of the implications (i)$\implies$(ii) or (ii)$\implies$(i) holds in general.
Indeed, let $X$ be uniformly distributed on $[0,1]$, and then let $T_n:=nX1_{X<1/n}$, so that $Y_n=T_n$ and, for $U=[0,1]$, $EY_n\,1_{X\in U}=ET_n=1\not\to0$, whereas $Y_n=T_n\to0$ a.s. So, the implication (i)$\implies$(ii) fails to hold.
On the other hand, let again $X$ be uniformly distributed on $[0,1]$, and then let $T_n:=X1_{X\in\delta_n}$, where $\delta_1,\delta_2,\dots$ are subintervals of $[0,1]$ such that $T_n\to0$ in probability but not a.s. Then again $Y_n=T_n$ and hence (i) fails to hold, whereas (ii) holds by dominated convergence. So, the implication (ii)$\implies$(i) fails to hold.
Mmm, you're right of course. Can you see a different way of unpacking $E(T_n|X)\rightarrow 0$ a.s. constructively that would work?
@AndrewNC : I'd surprised if such unpacking is possible. Also, it is unclear what you mean by "does not use the Radon-Nikodym derivative as a black box".
|
2025-03-21T14:48:30.446580
| 2020-04-29T18:23:24 |
358917
|
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|
Stack Exchange
|
Necessary condition to extend a morphism of schemes
Consider two schemes $X,Y$ over a locally noetherian scheme $S$. Let $p \in X$ and assume that $X$ is irreducible and not affine spectrum of a semilocal ring.
We assume moreover we have a morphism $f: \operatorname{Spec}O_{X,p} \to Y$. One may ask what are the neccessary conditions to extend $f$ to an open subscheme $U \subset X$ which contains $p$, that is to extend to a morphism $h: U \to Y$ which restricts under composition with $ \operatorname{Spec}O_{X,p} \to U$ to $f$.
Certainly, a well known sufficient condition is if we assume that $Y$ is locally of finite type over $S$. Then we can choose an affine subscheme $\operatorname{Spec} \ R= S_0 \subset S$ and open subscheme $\operatorname{Spec} \ T= Y_0 \subset Y$. Since $Y$ locally of finite type $T= R[x_1,x_2,..., x_n]/I$. Since $R$ is noetherian, $R[x_1,x_2,..., x_n]$ is also noetherian and the ideal $I$ is finitely generated. Let $\operatorname{Spec} \ A\subset X$ be an affine open neighbourhood of $p \in X$.
Then we consider a morphism $\phi:R[x_1,\dots,x_n]/I\rightarrow \mathscr{O}_ {X,p}$. Now, write $\phi(x_i)=a_i/r$ for some $r \in A$ not vanishing at $p$ and let $s \in A$ not vanishing at $p$ be such that $s \cdot g(\phi(x_1),\dots,\phi(x_n))=0$ for all $g \in I$. Here it is crucial that $I$ is finitely generated ideal, otherwise such $s$ might not exist. then the open set $D(sf)$ makes the job, where $l:V→Y$ corresponds to the morphism $\psi:k[x_1,\dots,x_n]/I\rightarrow A[(sf)^{-1}]$ mapping $x_i$ to $a_i/r$.
Now I ask if to require $Y$ is locally of finite type over $S$ around $y=f(x)$ is also a neccessary condition to obtain the extension above. I guess so but I haven't found a conterexample.
If $X$ is a point then the extension property is vacuously satisfied without any assumptions on $Y$. Do you maybe mean something like "if $Y$ is locally of finite type over $S$ near $y$, then for every $S$-scheme $X$ with a morphism $\operatorname{Spec} \mathcal O_{X,x} \to Y$ taking $x$ to $y$ there exists an extension on an open"?
No, you are right, I just overlooked this trivial case if $X$ is finite or more general discrete. But, no I not meant the "if $Y$ is locally of finite type over $S$ near $y$, then for every $S$-scheme $X$ with a morphism $Spec \ O_{X,x} \to Y$ taking $x$ to $y$ there exists an extension on an open" since we know that in this case it is always true, see my sketch above. I'm interested in the case if we assume that $X$ is irreducible of dimension $\ge 1$. Is it neccessary to require that $Y$ is locally of finite type over $S$ near $y$? I will improve my question...
In other words I'm asking if the statement that the solubility of the extension problem above (with $X$ irreducible of dim $\ge 1$ is equivalent to $Y$ is locally of finite type over $S$ near $y$. One implicaition I solved. For the other I conjecture that it also true but I need a conterexample in order to argue by contraposition
Oh right, I wrote the converse of what I meant; my question was really about varying $X$. But 'irreducible of dimension $\geq 1$' will not help much, because my objection still exists for (semi)local schemes.
Yes, I see the issue. The also obvious semilocal case we have also to exclude. see update
I think you should assume every affine open containing $p$ contains a point which doesn't specialize to $p$ (this is a very mild condition). Then if $Y = \text{Spec}(\mathcal{O}_{X, p})$ and $f$ is the identity you obviously can't extend.
@Johan: This argument I not understand. Assume $X$ affine and contains such point $q$ which doesn't specialize to $p$. That means by definition $q \not \in \overline{{p}}$ or equivalently $p$ is not a prime ideal in $\mathcal{O}{X,q}$. Well, why we cannot extend the identity morphism $f:=id: \text{Spec}(\mathcal{O}{X, p}) \to Y:=\text{Spec}(\mathcal{O}{X, p})$? I not understand. If for example $\mathcal{O}{X, p}$ is finite type over $\mathcal{O}_{S, s}$ ($s$ is the image of $p$ by structure map $X \to S$),
that is $\mathcal{O}{X, p}= \mathcal{O}{S, s}[x_1,\dots,x_n]/I$ with finite generated $I$ since $\mathcal{O}_{S, s}$ by assumption Noetherian. That is I see no obstructions to continue the same argument as in fourth paragraph in my question. or do I oversee a detail?
probably I misunderstood your example, could you explain my thinking error in the previous comments which "propose" that it seems to be possible to extend your morphism?
The English phrase "q doesn't specializes to p" is equivalent to "p isn't a specialization of q" is equivalent to "p isn't in the closure of q" is equivalent to "q is not a point in the spectrum of the local ring at p".
@Johan: Sorry for late response. You are right, I confused the $p$ and $q$. But I still not understand why your proposed morphism $f:=id: \text{Spec}(\mathcal{O}{X, p}) \to Y:=\text{Spec}(\mathcal{O}{X, p})$ is not extendable in the sense I explaned above, i.e. to a morphism $\text{Spec}(\mathcal{O}{X, p}) \subset U \to Y$? Read again my last comments. Essentially, they tell that if $\mathcal{O}{X, p}$ is finite type over $\mathcal{O}_{S, s}$, then we can argue like in the fourth paragraph of my original question.
But this construction whould tell that your $f$ is extendable, or not? Or does your example contain a detail which makes such argument not applicable here?
|
2025-03-21T14:48:30.446935
| 2020-04-29T19:17:42 |
358920
|
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|
Stack Exchange
|
Horizontal lift of fundamental vector field
Suppose $\theta\colon G\times M\to M$ is a transitive smooth left action of a compact Lie group $G$ on a manifold $M$ and $\pi\colon G\to M\cong G/K$ the corresponding smooth submersion for some closed subgroup $K$. Then $\pi$ is equivariant with respect to $\theta$ and left multiplication on $G$, i.e. for all $g,h\in G$
$$\theta(h,\pi(g)) = \pi \circ L_h(g)\qquad \Big(=\pi \circ R_g(h)\Big).$$
If $\langle\cdot\ , \cdot\rangle$ is a bi-invariant Riemannian metric on $G$ then $\pi$ induces a $G$-invariant metric $\gamma$ on $M$, turning $\pi$ into a Riemannian submersion. The vertical bundle $V$ (where each fibre is the nullspace of $d\pi_g$) and the metric $\langle\cdot\ , \cdot\rangle$ define a horizontal bundle $H$ by letting $H_g = V_g^\perp$ with respect to $\langle\cdot\ , \cdot\rangle_g$.
Given $X\in \mathrm{Lie}(G)$ (the set of left-invariant vector fields), consider the induced fundamental vector field $\mathcal{K}(X)\in\mathfrak{X}(M)$ defined for $p=\pi(g)\in M$ by
$$\mathcal{K}(X)|_p= d_t|_{t=0} \theta(\exp^G(tX),p)=d\pi_g\big(d(R_g)_e(X_e)\big).$$
Question: Is it correct that the horizontal lift of $\mathcal{K}(X)$ is the vector field $\tilde{Z}\in\mathfrak{X}(G)$ given by $$\tilde{Z}_g = \big(d(R_g)_e(X_e)\big)^H$$
i.e., the horizontal part of the right (!) translation of the tangent vector $X_e\in T_e G$?
Initially, I thought it should be the horizontal part of $X$ itself which is only true at $g=e$ it seems.
Even though taking the horizontal part commutes with left translation (by construction of the induced metric on $M$), the vector field $\tilde{Z}$ is not left-invariant (i.e. not in $\mathrm{Lie}(G)$) because in general $L_h\circ R_g\neq R_{hg}$. Neither is $\tilde{Z}$ right-invariant because taking the horizontal part does not commute with right translations.
This somehow feels strange to me.
The main point (I think) is that the left action of $\mathfrak g=\operatorname {Lie}(G)$ on $G$ itself induces right invariant vector fields.
The left action "left multiplication on G", i.e. $L\colon G\times G\to G, (g,h)\mapsto L_g(h)$ induces the right-invariant fundamental vector fields $\mathcal{K}^L(X)|_g= \frac{d}{dt} L(\exp^G(tX),g) = d(R_g)_e(X_e)$. Is this what you mean? I do see how that is related to my question.
However, what I find so strange is that in my case above I start with a left-invariant vector field on $G$, calculate the fundamental vector field downstairs on $M$, lift it back upstairs to $G$ and get something that is neither left- nor right-invariant. Is my calculation for $\tilde{Z}$ correct at least?
The action of $K$ is from the right, so the subbundle of $TG$ generated by this action is left invariant, same for its complement. Projecting something right-invariant onto a left invariant subbundle can give something with no invariance at all. Or try to look at it from a geometric point of view: the horizontal lift of the Killing field has the same norm as the Killing field itself. If the horizontal lift was somehow invariant, it would have constant norm. But there are examples $G/K$ with nonzero Euler characteristic, so each vector field must vanish somewhere.
The first part of your comment is what confused me. Thanks for clearing this up for me. I like your geometric argument! Do you want to turn this comment into an answer so I can accept it?
|
2025-03-21T14:48:30.447195
| 2020-04-29T20:58:33 |
358929
|
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|
Stack Exchange
|
Heat flow, decay of the Fisher information, and $\lambda$-displacement convexity
In the whole post I will work in the flat torus $\mathbb T^d=\mathbb R^d/\mathbb Z^d$ and $\rho$ will stand for any probability measure $\mathcal P(\mathbb T^d)$. This question is strongly related to two of my previous posts, universal-decay-rate-of-the-fisher-information-along-the-heat-flow and improved-regularization-for-lambda-convex-gradient-flows.
Fact 0: the quadratic Wasserstein distance $W_2$ induces a (formal) Riemannian structure on the space of probability measures, which gives a meaning to Wasserstein gradients $\operatorname{grad}_{W_2}F(\rho)$ of a functional $F:\mathcal P(\mathbb T^d)\to\mathbb R$ at a point $\rho$
Fact 1: the heat flow $\partial_t\rho_t=\Delta\rho_t$
is the Wasserstein gradient flow
$$
\dot\rho_t=-\operatorname{grad}_{W_2}H(\rho_t)
$$
of the Boltzmann entropy
$$
H(\rho)=\int_{\mathbb T^d}\rho\log\rho
$$
Fact 2: the Boltzmann entropy is $\lambda$-(displacement) convex for some $\lambda$.
Its dissipation functional is the Fisher information,
$$
F(\rho):=\|\operatorname{grad}_{W_2} H(\rho)\|^2_{\rho}=\int _{\mathbb T^d}|\nabla\log\rho|^2 \rho
$$
Fact 3: for abstract metric gradient flows (in the sense of [AGS]) and $\lambda$-convex functionals $\Phi:X\to\mathbb R\cup\{\infty\}$ one expects a smoothing effect for gradient flows $\dot x_t=-\operatorname{grad}\Phi(x_t)$ in the form
\begin{equation}
|\nabla\Phi(x_t)|^2\leq \frac{C_\lambda}{t} \Big[\Phi(x_0)-\inf_X\Phi\Big]
\tag{R}
\end{equation}
at least for small times, where $C_\lambda$ depends only on $\lambda$ but not on $x_0$ see e.g. [AG, Proposition 3.22 (iii)].
Fact 3': with the same notation as in Fact 3, an alternative regularization can be stated as
\begin{equation}
|\nabla\Phi(x_t)|^2 \leq \frac{1}{2e^{\lambda t}-1}|\nabla\Phi(y)|^2 +\frac{1}{(\int_0^te^{\lambda s}ds)^2} dist^2(x_0,y),
\,\,
\forall y\in X
\tag{R'}
\end{equation}
Fact 4: in the Torus the Fisher information decays at a universal rate, i-e there is $C=C_d$ depending on the dimension only such that, for all $\rho_0\in \mathcal P(\mathbb T^d)$ and $t>0$, the solution $\rho_t$ of the heat flow emanating from $\rho_0$ satisfies
\begin{equation}
F(\rho_t)\leq \frac{C}{t}
\tag{*}
\end{equation}
This follows from the Li-Yau inequality [LY], see this post of mine and F. Baudoin's answer.
Question: is there more to ($*$) than just the convexity of the Boltzmann functional? If the driving functional were upper-bounded $\Phi(x_0)\leq C$ (for all $x_0\in X$) in the regularization estimate (R) then we would immediately get the universal decay $|\nabla \Phi(x_t)|^2\leq \frac{C}{t}$.
However, in the specific context of Facts 0-2 it is clearly not true that the Boltzmann entropy is upper-bounded. In fact there are many probability measures with infinite entropy, take e.g. any Dirac mass. Since (R) is optimal, I guess that one cannot simply deduce (*) from general $\lambda$-convexity arguments, and there is more than meets the eyes. But is there any connection? Note that both the Li-Yau inequality and the displacement convexity of the Boltzmann entropy strongly rely on the nonnegative Ricci curvature of the underlying torus.
I tried desperately to use any modified regularization estimate (e.g. R' and variants thereof instead of R), but to no avail so far. I am starting to believe that there is no direct implication, and that the work of Li-Yau is really profoundly ad-hoc (don't get me wrong, I just mean that their results cannot be generalized for abstract gradient-flows, and that their result/proof really leverages the specific structure and setting of the heat flow in Riemannian manifolds, not just any gradient flow). I would immensely appreciate any input or insight!
[AG] Ambrosio, L., & Gigli, N. (2013). A user’s guide to optimal transport. In Modelling and optimisation of flows on networks (pp. 1-155). Springer, Berlin, Heidelberg.
[AGS] Ambrosio, L., Gigli, N., & Savaré, G. (2008). Gradient flows: in metric spaces and in the space of probability measures. Springer Science & Business Media.
[LY] Li, P., & Yau, S. T. (1986). On the parabolic kernel of the Schrödinger operator. Acta Mathematica, 156, 153-201.
There is a generalized notion of $(K,N)$-convexity of the entropy in Wasserstein space, which takes not only the lower Ricci curvature bound into account, but also the (upper bound on the) dimension. See this article by Erbar, Kuwada and Sturm: https://arxiv.org/abs/1303.4382 The Li-Yau inequality is also valid on RCD($K,N$) spaces, but I never looked carefully into that, so I don't know if a proof using $(K,N)$-convexity of the entropy (and not the Bakry-Émery criterion) is known.
Maybe a small addition to the comment of MaoWao: The mentioned paper of Bochner inequalities in RCK(K,N) spaces lead to many follow-up results, which contain Li-Yau estimates. One, which contains the universal bound (*) as Theorem 1.1 is arxiv.org/abs/1306.0494 by Garofalo-Mondino. So the missing ingredient in your Facts 1-3 is the dimension N and also in the classical result by Li-Yau the constant C depends on the dimension of the space. Hence, a general $\infty$-dimensional theory for gradient flows cannot give you the desired bound.
That's a really good point! Thank you André.
Hi @leo, do you have any updated news on obtaining $(*)$ with the machinery of gradient flows?
@Akira no, I moved on and completely left this topic aside
@leomonsaingeon thank you so much for your elaboration!
I won't say that it is impossible, but I don't see how to obtain $(\ast)$ using only general theory. There might be a different strategy that works, but I can tell you why I don't think the Li-Yau estimate can be proven using general properties of convexity. In particular, Li-Yau relies on some careful estimates involving the Laplace-Beltrami operator (and some hard analysis), which I don't think general theory can "see".
For a detailed write up of the Li-Yau estimate, I recommend Lectures on Differential Geometry by Schoen and Yau, which was very helpful for me. From a high level overview, the idea is to let $u$ be a non-negative solution to the heat equation, consider $\log (u + \epsilon)$ and try to bound its derivative. To do this, you consider the point which maximizes $ | \nabla \log (u + \epsilon) |^2$ and use the Bochner formula. Bochner's formula has a correction term due to the curvature, but when the manifold is Ricci positive, this has a favorable sign and we can ignore it (or use something like a barrier function to sharpen the estimate). The key insight is actually a clever use of the Cauchy-Schwarz inequality to eke out a little bit extra from the second derivative terms. It's elementary, but also a stroke of genius, and allows everything else to work.
If you read proofs of Li-Yau, the logarithm tends to appear near the end. However, it was helpful for my intuition to realize that this is not ad hoc; there was always going to be a logarithm because we are using the maximum principle applied to the function $\dfrac{|\nabla u|^2}{(u+\epsilon)^2} = | \nabla \log(u+\epsilon)|^2$.
The fact that $\nabla u$ and $u$ are raised to the same power here is crucial. When the power of $\nabla u$ is less than $u$, integrating out the resulting inequality gives a bounded function (which is significantly less useful). There's this really delicate balancing act in order for everything to work, and logs play an essential role. As a brief aside, I suspect you get different powers of $\nabla u$ and $u$ if you try the Li-Yau strategy with the porous media equation (I'm not entirely sure of this though).
So back to your question about whether this can be done using general properties of gradient flows. It might be a lack of imagination on my part, but it's hard for me to see how this would work. There's several essential steps that rely on hard analysis. For instance, you really need the Cauchy-Schwarz step to work and the resulting function that you get from integrating out should be unbounded. Furthermore, while it's possible to sharpen the estimate, the original version is already fairly sharp, in that there is not a whole lot of wiggle room. As such, while it's possible to adapt the argument to elliptic operators or to include lower order terms, it does seem like there is genuinely more here than the general theory.
Thanks @Gabe for your input. I'm fully aware that Li-Yau relies on hard analysis (a clever comparison principle, really). One might argue that the heat flow is not the only flow with this "universal regularization" property (take e.g. any Fokker-Planck). Of course the heat flow is not only a "good" gradient flow but also has super extra-nice properties, which of course everyone will agree upon. So maybe you're right. However I'm still not fully convinced: This whole gradient-flow machinery actually allows to conclude in any sub-levelset of the entropy, so there is something going on here.
That's true. It's definitely possible that there's some deeper fact that allows this to go through. If you can find a way to prove $(\ast )$ without using Li-Yau it would be very neat and likely have some interesting consequences.
Well, truth be told I'm pretty sure I can't do that. What I'd like to understand is: Which $\epsilon$-improvement is still missing to conclude? OK, maybe I'm still missing $10\epsilon$ or $\sqrt\epsilon$, but I'm pretty sure that any result in this direction (thus using mainly quantitative convexity arguments) would give a whole new perspective on the problem! Also, understanding what is missing would help identifying some hidden structural properties, I feel. In a nutshell: How much better is the heat flow than just any $\lambda$-convex gradient flow? It seems to me that not too much...
To complement MaoWao and André Schlichting's comment: In the general setting of your Facts 3 and 3' (plus some regularity conditions), the instantaneous regularization property
$$|\nabla \Phi(x_t)|^2 \leq \frac{n}{2t}$$
is implied by $(0,n)$-convexity of $\Phi$ (in the sense of Erbar, Kuwada and Sturm (2013)), i.e., $\mathrm{Hess} \Phi \geq \frac{1}{n} (\nabla \Phi \otimes \nabla \Phi)$.
Here is a proof of the implication in the case $X = \mathbb{R}^d$: suppose $\nabla^2 \Phi(x) \succeq \frac{1}{n} \nabla \Phi(x) \nabla \Phi(x)^\top$ for all $x$ and consider $(x_t)_{t \geq 0}$ such that $\frac{d}{dt} x_t = -\nabla \Phi(x_t)$. Consider the potential function $V(t) = \frac{1}{|\nabla \Phi(x_t)|^2}$. We have by direct computations
$$\frac{d}{dt} V(t) = \frac{2 \nabla \Phi(x_t)^\top \nabla^2 \Phi(x_t) \nabla \Phi(x_t)}{|\nabla \Phi(x_t)|^4} \geq \frac{2}{n},$$
so $V(t) \geq \frac{2}{n} t + V(0) \geq \frac{2}{n} t$ and $|\nabla \Phi(x_t)|^2 \leq \frac{n}{2t}$.
For more general $X$, I would guess that the same implication is true though I don't know how this could be proved. (Of course with enough regularity the proof above will apply.)
For $X$ being the Wasserstein space, this gives a heuristic "proof" of your Fact 4 starting from the CD(0,d) property of $\mathbb{T}^d$, that doesn't go via the Li-Yau inequality.
Edit: just to add some context, this choice of potential function $V(t)$ is directly inspired from the proof of $O(1/t)$ convergence of gradient flow for convex objectives which poses $W(t) = \frac{1}{\Phi(x_t) - \min \Phi}$ (see e.g. this blog post)
Thanks Guillaume for this nice insight, I didn't know this trick of $(0,n)$ convexity. Say hi to Lénaïc!
|
2025-03-21T14:48:30.448075
| 2020-04-29T21:48:39 |
358934
|
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|
Stack Exchange
|
Sum and Product game
Two perfect logicians Steve and Pete, who have never met, before are imprisoned by an eccentric villain. "I have two positive integer numbers x and y" he says to them. "I will tell Steve the sum x+y, and Pete the product xy. If at any point either of you know the two numbers x and y, you may shout them out, and I will release you both. You may take turns saying statements giving the other information, but you may only communicate about your state of knowledge. If any part of your statement contains any information about the numbers themselves, you'll both be shot. If any of you tries to communicate any other information not directly implied by the statements, you'll both be shot."
Can the logicians always figure out the numbers?
As an example, if the two numbers are 1 and 6, a potential conversation could go:
P: I don't know the numbers
S: I don't know the numbers
P: I don't know the numbers
S: I don't know the numbers
P: I know the numbers! They are 1 and 6! (S would have known if the numbers were 2 and 3)
Or another example, if the numbers are 2 and 4:
P: I don't know the numbers
S: I didn't know whether you knew until you said that (could have been 1 and 5, and P would obviously have known, as 5 can only be factorized as 1x5)
P: I know the numbers! They are 2 and 4! (S couldn't have said that if the numbers were 1 and 8)
Is there always some strategy like this? (Note that the logicians are not able to communicate beforehand, and each logician must attempt to give useful information based on what they know)
Edit: The question has been raised whether just saying "I don't know" until someone figures it out would work for all possible answers. It doesn't. In order to show this, all one needs to do is construct a "cycle" of possibilities, with each consecutive pair alternatively having the same sum or the same product. Those pairs would be indistinguishable for S and P for all time unless more information is given than "I don't know the numbers"
A simple cycle is given by (2,5)->(3,4)->(1,12)->(3,10)->(5,6)->(1,10)->(2,5).
One could argue that simply eliminating the possibility of having 1 as one of the numbers would solve this problem. But by merely doubling the cycle, this is shown to be false: (4,10)->(6,8)->(2,24)->(6,20)->(10,12)->(2,20)->(4,10). Thus, removing 1 only delays the problem.
Edit 2: As the condition about only talking about states of knowledge is ambiguous, I have decided that a new version is: The logicians can only say either they know/don't know the numbers, or that they knew/didn't know what the previous person just said (from which the other logician can infer that they don't know the numbers, because otherwise they would have said that)
For example, if the numbers are 3 and 4, a conversation could go:
P: I don't know the numbers (could be 1 and 12 or 2 and 6)
S: I knew you didn't (even if the numbers were 2 and 5 or 3 and 4, P wouldn't have known)
P: I didn't know that (S couldn't have said that if the numbers were 2 and 6. However, this doesn't yet eliminate 1 and 12)
S: I know the numbers! They are 3 and 4! (the previous statement eliminated the other options)
Edit 3: 02/05/2020: It's starting to seem like it still might take the logicians quite a long time to figure out the numbers, and the logic gets increasingly difficult to follow. I wouldn't be surprised if for some pairs of numbers they might not be able to figure it out. Here's a particularly interesting (solvable!) case (shortened to aid readability). The numbers are 1 and 10:
P: I know you know that I don't know the numbers
S: I know you know that I didn't know that
P: I know the numbers now! They are 1 and 10!
Presumably, the positive numbers are positive integers?
See discussion at https://artofproblemsolving.com/community/c146h150971
Much discussion at https://www.math.uni-bielefeld.de/~sillke/PUZZLES/sp.txt
@LSpice, yes, positive integers
Those seem to be about very specific cases, where there are restrictions on the sum, and the reader has to figure out the numbers. I'm asking if, given any two numbers, the logicians can figure it out.
So at the beginning the logicians only know that $x,y \in \mathbb{N}$? The solution to this puzzle depends strongly on the given domain of $x,y$. But here you're asking a different question, namely about the strategy of the logicians. However, how do you define "state of knowledge"? What is the difference with "information about the numbers themselves"?
Let's restrict things a bit: Suppose they only keep saying "I don't know" until one of them finds out the numbers. Is there a number which we can construct where they'll never figure it out? My guess is that for any starting pair of positive integers, eventually they'll get it; this is based on the fact that products of two distinct primes have a pretty high density.
@FrançoisBrunault, with regards to state of knowledge. Let's say that the two logicians can only say things like "I know/don't know the numbers", and "I knew/didn't know what you just said before you said it".
@JoshuaZ I think that is too strong of a limitation on the conversation. I'll edit my question to show what I mean, in particular, to show that even (3,4) is impossible to figure out using that method.
Are the logicians allowed to agree on a strategy before the conversation?
@FrançoisBrunault No, they have never met before.
Then I don't understand precisely your question "Is there always some strategy like this?". Do you mean to ask whether the logician S (or P) has a winning strategy regardless of what the other logician will say? That's different from finding a common strategy.
Okay, what I'm asking is this: Given the constraints of the problem, can the logicians figure out the numbers? No strategies, no coded information, only logic.
Sorry but I still don't get it. Do you mean to ask whether there exists at least one sequence of declarations which enables the logicians to find the numbers?
Yeah that is a good description of it
|
2025-03-21T14:48:30.448493
| 2020-04-29T22:34:08 |
358935
|
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|
Stack Exchange
|
Pyramids whose volume can be computed by simple cutting and glueing
Since this question remained without answers even after a bounty, I thought it might be time to ask it here.
For which pyramid can you compute the volume from simple cut-and-glue processes? The Dehn invariant naturally gives the answer, but I failed to turn this in an algorithm. Here are the pyramids, I know of, whose volume is computable by elementary operations:
take a cube and divide into six pyramids from its center (or three pyramids given the 120° symmetry along a diagonal). These pyramids can be subdivided and glued into further pyramids, but there are not so many possibilities.
take a trigonal trapezohedron whose faces are all rhombic. This is not a right prism, but oblique prisms can also be cut-and-glued into a right prism (and hence their volume can be computed). Because of his symmetry group you can cut it into three pyramidal pieces (oblique pyramids). Since the small angle of the rhombus can be $\in ]0, \pi/2[$ (at $\pi/2$ it's just a cube), this gives an infinite family of pyramids. (These oblique pyramids have a symmetry: you can cut them further in half.) EDIT: Note that when the acute angle is $\in ]\pi/3, \pi/2[$ there are actually two ways to make a trigonal trapezohedron out of this rhombus (depending on whether the angle at the polar vertices is acute or obtuse).
But the that's all I could find. Are there any other? and if so, what is the cut-and-glue process?
Some background:
The volume of pyramids is discussed here in Proposition 3 to 5 of Euclid's book
The fact there was no "simple" proof irritated Gauß and later Hilbert. It was purportedly the inspiration for his 3rd Problem, which was solved by Dehn (using his invariant)
There is an earlier (and largely ignored) solution of Hilbert's 3rd problem by a certain Birkenmayer, see this paper.
In two dimensions the Wallace–Bolyai–Gerwien theorem answers the question completely. Beside the Dehn invariant (which shows this is not the case in three dimensions), the Banach-Tarski paradox is also a reminder that volume in three dimensions are tricky.
[EDIT: To clarify simple cute-and-glue (as asked in the comments). One starts with the class $(C)$ of solids (at the beginning it only includes rectangular prisms). You can add another polyhedron $P$ to $(C)$ if:
a number $k \neq 0$ of copies of $P$ can be obtained by cutting solids $\{S_i\}_{i=1,\ldots,s}$ from $(C)$ [possibly many copies of the same solid] into finitely many polyhedral pieces and reassembling these pieces together into the $k$ copies $P$ and a possibly empty collection of solids $\{T_i\}$ with $T_i$ belonging to $(C)$. In that case, $\mathrm{vol}(P) = \tfrac{1}{k} \big( \sum \mathrm{vol}(S_i) - \sum \mathrm{vol}(T_i) \big)$, and so $P$ can be added to $(C)$.
Reassembling means apply isometries of $\mathbb{R}^3$ (rotations, reflections and translations); two solids are congruent if one is the image of the other under such transformations and a copy is also the image of a solid under such transformations. This definition is closely related to scissors-congruence.
Could you please define what constitutes a "simple cut-and-glue process"? Thanks.
Sure, it's essentially scissors-congruence. You have a class of solids whose volume is computable call it (C); at the beginning it includes only rectangular [right] prisms. Then you extend this class by using scissors-congruence (cutting in finitely many pieces). You have the right to say that two congruent (by rotation or reclection or transaltion) solids have the same volume; you have the right to combine many solids together. As an example, here is how you go to triangular [right]-prisms. First, by cutting rectangular prisms in two you get right-angled-triangular [right]-prisms....
Since any triangle can be split in two right-angled-triangles, you can extend (C) to all triangular [right]-prisms. Next, since any polygon in the plane can be cut in triangles, you can extend (C) to all polygonal [right]-prisms. Somewhat trickier is the step to oblique prisms. If I remember correctly you have to cut them in planes which are perpendicular to the basis and reassemble. It's easier to see in the case of a rectangular [oblique] prism which has only two sides which are parallelogramms. Then you can cut it (as you would in the plane) to reassemble it as an oblique prism.
Problems starts when you get to pyramids. Apart the one I described above (which explicitly use the fact that they come from cutting a solid from (C) in congruent pieces), I know of no examples. In any case, you don't have the right to cut in pieces which are not polyhedral (e.g. spheres, cylinders, ...). Hopefully it is clearer?
Thanks, clearer.
|
2025-03-21T14:48:30.448833
| 2020-04-29T23:57:19 |
358938
|
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|
Stack Exchange
|
Shklyarov's Euler class in geometrical setting
In paper HIRZEBRUCH-RIEMANN-ROCH THEOREM FOR DG ALGEBRAS Shklyarov provides construction which for a given dg-algebra (over field $k$, proper, $k$-smooth) $A$ and perfect $A$-module $M$ associate en element $\operatorname{eu}(M)$ in $HH_0(A)$, but the construction is not very straightforward: we should represent our perfect $A$-module as twisted dg $A$-module at first and after that, we have to make some hard computations. Could we simplify the construction in a purely geometric case? I mean, suppose we have some smooth proper $n$-dimensional algebraic variety $X$ over a field $k$ and coherent sheaf $\mathcal F$ on it, interpreted as an element of $\mathcal{Perf}-X$ (with some dg-enhancement), is there exists a more straightforward description of $\operatorname{eu}(\mathcal{F})$ as the element of
$$HH_0(X) := \bigoplus_{p=0}^n H^p(X, \Omega_X^p)$$
? For example, can we compute $\operatorname{eu}(\mathcal{F})$ in the case when $X=\mathbb P^n, \mathcal F = \mathcal{O}(k)$?
Useful remarks, thank you!
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2025-03-21T14:48:30.448922
| 2020-04-30T01:10:48 |
358939
|
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|
Stack Exchange
|
Local contractibility of group of symplectomorphisms for open manifolds
It is well know that for a closed symplectic manifold $(M, \omega)$ the group of symplectomorphisms in locally contractible. The gist of this proof goes as follows.
Given a $\psi \in \operatorname{Symp}(M,\omega)$ that is sufficiently close to the identity in $C^1_{loc}$ topology, we look at the graph of $\psi$ (denoted by $\Gamma(\psi)$) in $M \times M$. And using Lagrangian neighbourhood theorem we can identify a neighbourhood of the diagonal in $M \times M$ with a neighbourhood of the 0-section in $T^*M$. Then using that $\psi$ is sufficiently close to the identity and the following facts:
Fact 1: If $\alpha : M \rightarrow T^*M$ is an embedding sufficiently close to the 0-section then $\alpha$ is the graph of a 1-form.
Fact2: If $\beta : M \rightarrow M \times M$ is an embedding sufficiently close to the diagonal then $\beta$ is the graph of a diffeomorphism.
we get a correspondence between symplectomorphisms in a small enough neighbourhood of the identity and a neighbourhood of 0 in the vector space of closed forms thus giving us local contractibility.
Unfortunately the proofs of both Fact 1 and Fact 2 that I know of seem to rely on compactness of $M$. So
1) Does local contractibility of group of symplectomorphisms still hold for open symplectic manifolds? How about for manifolds with boundary?
2) Is the proof similar to the one above i.e., are Fact 1 and 2 true in the non-compact case as well or do we need new ideas to prove this?
Edit
The question as it was originally asked had a small mistake. I shouldn’t have considered $C^1 $topology on a non-compact manifolds as this would not make sense. Instead The question should be phrased in the $C^1_{loc}$ topology i.e $\phi_n \rightarrow \phi$ in $C^1_{loc}$ iff on every compact set $K \subset M$ $\phi_n \rightarrow \phi$ in the $C^1$ sense.
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2025-03-21T14:48:30.449069
| 2020-04-30T01:16:37 |
358940
|
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|
Stack Exchange
|
Effective radius of section of a convex set compared to that of the convex itself
The effective radius $er(A)$ of a $n$-solid $A$, is defined by Schramm (see the question by Gil Kalai
Volumes of Sets of Constant Width in High Dimensions)
to be the radius of the $n$-ball that has the same volume.
Question(s) :
Let $A$ be a $n$-convex (convex set in the $n$ dimentionnal euclidian affine space $\mathbb A_n$) and $H$ a hyperplane of $\mathbb A^n$ such that $B=A\cap H$ has maximal volume (resp. such that the boundary of $B'=A\cap H$ is maximal in volume)
a) Is $er(B)\geq er(A) $ ?
b) Is $er(B')\geq er(A)$ ?
I think the answer (especially the b) question) if positive implies the resolution of the question in
Volumes of Sets of Constant Width in High Dimensions podted by Gil Kalai.
Let me try an heuristic for the first inequality, in $3$ dimensions (I write vol($A$) as well as for a volume or an area, as soon as there is no ambiguities) :
Take $t\mapsto H_t$ a path of plans parallele to $H$ that go throught $C$ and that intersect $C$ in $B_t$.
Now we associate to $B_t$ a disk $D_t$ that radius is $er(B_t)$ such that its center is $t\in T$ and that $T$ is a line. Consider $E=\bigcup_{t\in T} D_t$.
It is clear that $Vol (E) = Vol (C)$.
It is also clear that $E$ is the $\pi$- rotation solid generated by , say $E_0$ : union of coplanar "effectives diameters"
The question a) for dim 3 is then equivalent to :
Is $vol(E_0)$ smaller then $vol(B)$ ?
This would be automatically the case if $E_0$ can be isometrically embeded in $C$ whom a greater section (in volume, well in this case "area") is $B$ - it appears that the question that translate the possibility of this transversal embedding is : can we put the line $T$ inside $C$ and take in each "efficient disk" an "efficient diametre" inside $C$ in such a way that they are co-planar.. we could imagine that there is some torsion around $T$ that make this operation impossible, despite convexity condiderations... never the less, this intermediary request is stronger then the one of the (here part a) but also wrt part b)) question...
The problem of this heuristic is not only the torsions phenomenons that I think they might not occure in 3 dim (and that we can probably get rid of, considering that the case that we are dealing with are well behaving, especially throught heredity) but the fact that I'm not sure it can be adapted to greater dimensions, because $E_0$ is 2-dimensional in any case... that does not go against the relevance of the questions but that does not give at least to me a clear hope or a clue, well maybe this discussion can be adapted to the general case... or not.
Maybe I have been a bit pessimistic and the a) 3d argument seems to work simmilary (up to the torsion problem)... Another obstruction than the "torsion" is that a) might not directy be applied to the Kalai- Schramm suestion, because we'd have to get that $K_n$ is a maximal volum section, but we can maybe, as i said into the question, get rid of these two problem by considering a sequence convexes set that behave nicely wrt a) conditions ans that contains the $K_i$ in Gil Kalai's question...
I'm thinking about the intersection of $n+1$ $n$-balls of radius , whom centers would be the verteces of a regular simplex of edge 1
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2025-03-21T14:48:30.449355
| 2020-04-30T01:36:16 |
358941
|
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|
Stack Exchange
|
What are the necessary conditions on $f$ if $f(x)=f(\sin(\pi x)+x)\iff x\in\Bbb{Z}$?
I am aware that the statement:
$$f(x)=f(\sin(\pi x)+x)\iff x\in\Bbb{Z}$$
is not true for all $f$. For example, $f$ can be $x$ to any constant power or any constant to the $x$th power but it cannot be the gamma function $\Gamma(x)$ or $\sin(x)$ or $x^x$. I have been told that it is important to note whether or not $f$ is injective. However, $f(x)=x^2$ is not injective, yet it satisfies the statement. If being injective is only a sufficient condition as opposed to a necessary condition, what exactly do we know about the class of functions that makes this statement true?
Thanks in advance!
Here is a condition that $f$ must satisfy if it happens to be periodic: The period must be larger than $1$. Otherwise, the function $f$ satisfies $f(x+T)=f(x)$ for all $x$ where $T\in (0,1]$. But then there exists $x_0\in\Bbb{R}-\Bbb{Z}$ with $\sin(\pi x_0)=T$ for which $f(x_0+\sin(\pi x_0))=f(x_0)$.
Another necessary condition for a continuous $f$ would be: $f$ should not have a unique global maximum or global minimum at any $0<a<1$ over $[0,t]$ where $t=\arg\max_{0\leq x\leq 1}(x + \sin(\pi x)]$. Also, let $g(x) = f(x+\sin(\pi x))$.
Suppose it does have a global maximum over $[0,t]$ at some $0<a<1$. Let $b$ be the smallest positive real such that $b+\sin(\pi b)=a$. Certainly, $0<b<a$. Now, $g(b)=f(a)>f(b)$, since $f(a)$ is the maximum. Similarly, $g(a)<f(a)$. This implies that $f$ and $g$ must intersect somewhere within $[b,a]$, a contradiction. A similar argument hold is there existed a global minimum.
This condition can be applied to $[2m,2m+t]$, for any integer $m$. Hope this helps.
Thanks for the answer! I tried to test this theory on Desmos, and, unless I am mistaken, $f(x)=\sin(x)$ does not have a max/min at any $a$ over $[0,t]$. But I am certain that this $f$ does not satisfy the statement. I calculated $t=-\frac{\cos^{-1}\left(-\frac{1}{\pi}\right)}{\pi}$, which means $[0,t]$ should actually be written $[t,0]$. Is this a case of me not understanding your answer or of an error in the theory?
Sorry about the oversight, and thanks for catching that. It is $t=\arg\max_{0\leq x\leq 1}(x + \sin(\pi x)]$ instead of $t=\arg\min_{0\leq x\leq 1}(x + \sin(\pi x)]$. Hope it helps.
WARNING: INCORRECT see comments for why
If f is continuous on some connected sets it must be monotonic on them. Say it is not. Then there is some point $x^*$ such that in some of its neighborhood V such that $x^*=\sup_V f$. Then it can be shown easily that there are $x_1,x_2$ in some neighborhood of $x^*$ such that $f(x_1)=f(x_2)$ and $d(x_1,x_2)\lt 1$ causing a contradiction.
note: here monotone is defined as that there is no $x^*$ such that it is the superior of all $f(x)$ in some of its neighborhood, which can be shown to be equivalent to the concept of monotone in real valued functions of one real variable.
Why the difference between $x_1$ and $x_2$ should be the same as $\sin \pi x_1$?
@KhashF well I think that I have managed to mess up... I just read your answer and thought, emm, there mustn't be two x with their distance less than 1 and f(x) the same. Would this be a fixable issue (or non fixable which probably means that I need to delete this answer)?
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2025-03-21T14:48:30.449874
| 2020-04-30T02:08:50 |
358943
|
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|
Stack Exchange
|
Metrics on torus without closed contractible geodesics
It is easy to see that any closed geodesic on a flat 2-torus is noncontractible.
Further the same holds true for a torus of revolution.
Indeed either a closed geodesic is a meridian and therefore noncontractible, or it intersects all the meridians transversally and noncontractible as well.
The same argument shows that if a torus admits a geodesic foliation, then it has no contractible closed geodesic.
Are there other examples?
Are there examples of Riemannian metrics on $\mathbb{T}^2$ without closed null-homotopic geodesics and without a geodesic foliation?
A commnet of Dima Burago: It is impossible to find an example by taking a small perturbation of a flat metric. Indeed, by KAM theory any such perturbation has a geodesic foliation.
Make a little hole and put small negative curvature outside the hole, and positive inside. It seems that there are no null-homotopic closed geodesics.
@AlexandreEremenko, maybe, but why --- if you pass to the universal cover then behavior of geodesic reminds random walk --- why not to have a closed trajectory?
You have a mistake in your edit. You mean "..., then it has no contractible closed geodesic."
@RobertBryant, fixed --- thank you.
This is a comment, not an answer, but it's too long to fit into a comment window.
I don't know of a 'generic' condition on metrics on the torus that would guarantee that there are no null-homotopic closed geodesics, but then I'm not sure what 'generic' is supposed to mean.
For example, would it suffice to find some sort of 'open' condition (for example, that a metric be 'sufficiently close' in some appropriate sense to being flat) that would imply that there are no null-homotopic closed geodesics, or are we supposed to be trying to decide whether, given any metric on the torus, there is another metric arbitrarily 'close' to it that has no null-homotopic closed geodesics? (This second meaning is what I take to be the literal meaning of 'generic', which I think is unlikely to hold.)
Meanwhile here is an interesting, 'nongeneric' example that has no nontrivial symmetries (which is, itself, a 'generic' condition in a sense) and has no null-homotopic closed geodesics:
Let $a$ and $b$ be 'generic' positive smooth functions on $\mathbb{R}$, periodic of period $2\pi$, and consider the metric
$$
g = \bigl(a(x)+b(y)\bigr)(\mathrm{d}x^2 + \mathrm{d}y^2).
$$
which is doubly periodic and hence defines a metric on the torus $\mathbb{T} = \mathbb{R}/(2\pi\mathbb{Z}^2)$.
I will show that $g$ has no closed geodesics in $\mathbb{R}^2$.
To do this, note that this is a metric of Liouville type and hence all the unit speed geodesics satisfy the equations (i.e., 'conservation laws' or 'first integrals')
$$
\bigl(a(x)+b(y)\bigr)(\dot x^2+\dot y^2) = 1
\quad\text{and}\quad
\bigl(a(x)+b(y)\bigr)\bigl(b(y)\,\dot x^2 - a(x)\,\dot y^2) = c
$$
for some constant $c$. Given such a geodesic, let $\theta$ be the function on the geodesic that satisfies $\cos\theta = \dot x\sqrt{a(x)+b(y)}$ and $\sin\theta = \dot y\sqrt{a(x)+b(y)}$. This $\theta$ is well-defined up to an integer multiple of $2\pi$ and gives the 'slope' of the geodesic in the standard $xy$-coordinates. Then we have
$$
b(y)\,\cos^2\theta - a(x)\,\sin^2\theta = c.
$$
Notice that, because $a$ and $b$ are strictly positive, this equation implies that, for any given $c$, there is an open set of directions $\phi$ such that $\theta$ cannot attain either value $\phi$ or $\phi+\pi$, an impossibility for a closed curve in the $xy$-plane, since a closed curve has to have a tangent perpendicular to any given direction.
Thus, $g$ has no closed geodesics in $\mathbb{R}^2$ and hence the induced metric on $\mathbb{T}$ has no closed null-homotopic geodesics.
Of course, you can object that Liouville metrics are not 'generic'. But it might be that there is some 'open' condition on a lattice-periodic metric on the plane such that no geodesic satisfying this condition can ever 'turn all the way around', as in this Liouville case. If one can find such a condition, then this would, I take it, be an answer to Anton's question.
Moreover, there are `higher' Liouville metrics, i.e., for which the geodesic flow has higher degree polynomial conservation laws, and, as far as I know, there is no limit on how high the degree of such a conservation law might be. It's possible that one could find examples of arbitrary complexity, so that these metrics are as 'generic' as you could possibly want.
Thank you. Maybe generic means nonintegrable?
@AntonPetrunin: I guess better would be to say 'generic' means `having no geodesic foliation', since that is sufficient to make the argument work that you presumably used for the case of a foliation by closed geodesics. Perhaps 'integrable' would imply that there is a geodesic foliation anyway. Certainly, in the Liouville case I discussed above, there is a geodesic foliation, for example, the one calibrated by $\phi = \sqrt{a(x)},\mathrm{d}x + \sqrt{b(y)},\mathrm{d}y$. Meanwhile, the generic metric on $\mathbb{T}$ that admits a geodesic foliation is not integrable.
|
2025-03-21T14:48:30.450244
| 2020-04-30T03:59:46 |
358949
|
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"A. Gupta",
"Dan Petersen",
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|
Stack Exchange
|
action of symmetric group on the second exterior power
Let $e_i \wedge e_j \ (i < j)$ be a basis for the $\mathbb Z$-module $\wedge^2 \Gamma$, where $\Gamma = \mathbb Z^n$.
Clearly $S_n$ acts on the module $\wedge^2 \Gamma$ via
$$\pi(e_i \wedge e_j) = e_{\pi(i)} \wedge e_{\pi(j)} \ \ \ \forall \pi \in S_n.$$ By restriction this induces an action on the subset $\bar B =
\{ \epsilon e_i \wedge e_j \ (i < j), \ \epsilon \in \{-1, 1\} \}$.
Which (non-trivial) cyclic subgroups of $S_n$ have maximal number of orbits in this action on $\bar B$.
The answer seems to be the subgroups generated by transpositions $\pi = (ij)$. But can there be other permutations $\pi$ that are not transpositions but with the same number of orbits?
Please, note a modification in the question.
Yes, we mean exactly the restriction to $\bar B$ ("latter action" is now made explicit in the question).
The only exception is when $n=4$ in which case a product of two disjoint transposition also works.
Please note: For $n = 4$ there are $7$ orbits for the group $\langle (12) \rangle$, namely, ${\pm e_1\wedge e_2}$, ${e_1 \wedge e_3, e_2 \wedge e_3 }$, ${-e_1 \wedge e_3, -e_2 \wedge e_3 }$, ${e_1 \wedge e_4, e_2 \wedge e_4 }$, ${-e_1 \wedge e_4, -e_2 \wedge e_4 }$, ${e_3 \wedge e_4 }$ and $-{e_3 \wedge e_4}$.
For the action of $\langle (12)(34) \rangle$ there are six orbits only: ${e_1 \wedge e_2, -e_1 \wedge e_2 }$, ${e_3 \wedge e_4, -e_3 \wedge e_4}$, ${e_1 \wedge e_3, e_2 \wedge e_4 }$, ${-e_1 \wedge e_3, -e_2 \wedge e_4 }$, ${e_1 \wedge e_4, e_2 \wedge e_3 }$ and ${-e_1 \wedge e_4, -e_2 \wedge e_3 }$. The reason behind this is that $\langle (12) \rangle$ fixes more points in $\bar B$ (on average).
By the Lemma that is not Burnside's, the number of orbits is the average number of fixed points. An element fixes $\epsilon e_i \wedge e_j$ iff it fixes both $i,j$ (because if it swaps them it reverses the sign, and otherwise it won't even preserve the span $\mathbb{Z} e_i \wedge e_j$. It follows that $\sigma \in S_n$ will have $2\binom{\#\mathrm{Fix}(\sigma)}{2}$ fixed points on $\bar{B}$ because we need to choose pairs $(i,j)$ in its fixed point set.
Now a transposition has the largest number of fixed points of any non-identity element. Accordingly let $G < S_n$ be any non-trivial subgroup and let $C<S_n$ be the subgroup generated by a transposition. Then we have
$ \# \bar{B}/G = \frac{1}{\# G} \sum_{\sigma\in G} 2\binom{\#\mathrm{Fix}(\sigma)}{2} \leq \frac{1}{\# G} n(n-1) + \left(1-\frac{1}{\# G}\right)(n-2)(n-3) = (n-2)(n-3) + \frac{1}{\# G} (4n-6) \leq (n-2)(n-3) + \frac{1}{\# C} (4n-6) = \# \bar{B}/C \,.$
It follows that the number of orbits of $C$ is maximal, with equality iff $G$ is conjugate to $C$.
|
2025-03-21T14:48:30.450551
| 2020-04-30T04:36:44 |
358951
|
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|
Stack Exchange
|
Are there extensions of Hilb's and Laplace's formulas to Jacobi polynomials with $\alpha,\beta\le-1$?
In Szegő's Orthogonal Polynomials book, he gives two interesting asymptotic formulas for Jacobi polynomials with $\alpha,\beta>-1$. The first (Theorem 8.21.12, page 197 is a generalization of Hilb's formula for Legendre polynomials:
Let $\alpha>-1$ and let $\beta$ be arbitrary. Then we have
\begin{align*}
\left(\sin\frac{\theta}{2}\right)^{\alpha}\left(\cos\frac{\theta}{2}\right)^{\beta}P_n^{(\alpha,\beta)}(\cos\theta)&=N^{-\alpha}\frac{\Gamma(n+\alpha+1)}{n!}(\theta/\sin\theta)^{1/2}J_{\alpha}(N\theta)\\
&+\begin{cases}
\theta^{1/2}O(n^{-3/2}) & if & cn^{-1}\leq\theta\leq\pi-\epsilon,\\
\theta^{\alpha+2}O(n^\alpha) & if & 0<\theta\leq cn^{-1},
\end{cases}
\end{align*}
where $N=n+\frac{1}{2}(\alpha+\beta+1)$ and $c$ and $\epsilon$ are fixed positive numbers.
A generalization of Laplace's formula for Legendre polynomials is also stated (Theorem 8.21.13, page 197):
Let $\alpha>-1, \beta>-1$. We have
\begin{align*}
P_n^{(\alpha,\beta)}(\cos\theta)=n^{-1/2}k(\theta)\{\cos(N\theta+\gamma)+(n\sin\theta)^{-1}O(1)\},
\end{align*}
for $cn^{-1}\leq \theta\leq \pi-cn^{-1}$. Here $c$ is a fixed positive number, $k(\theta)=\pi^{-1/2}\left(\sin\frac{\theta}{2}\right)^{-\alpha-\frac{1}{2}}\left(\cos\frac{\theta}{2}\right)^{-\beta-\frac{1}{2}}$ and $\gamma=-(\alpha+\frac{1}{2})\frac{\pi}{2}$.
Szegő seems to suggest that the restrictions of $\alpha,\beta>-1$ are unnecessary after the statement of the latter result, but I haven't been able to find a reference. Are similar theorems known for $\alpha,\beta\leq -1$? Can they perhaps be derived easily with proofs similar to the case of $\alpha,\beta>-1$?
|
2025-03-21T14:48:30.450677
| 2020-04-30T05:56:06 |
358955
|
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"Han Bao",
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"url": "https://mathoverflow.net/questions/358955"
}
|
Stack Exchange
|
How to calculate Fourier transformation of eigenstates in CV quantum information
The position $\hat{q}$ and momentum $\hat{p}$ has $[\hat{q},\hat{p}]=i$.
And we set there eigenstates as $|s\rangle_q$ and $|s\rangle_p$ with eigenvalue s.
In the paper [Phy Rev A. 79, 062318 (2009)], the eigenstates of $\hat{q}$ and $\hat{p}$ can be transfered as
\begin{equation}
\begin{array}{c}|s\rangle_{p}=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} d r e^{i r s}|r\rangle_{q}=F|s\rangle_{q} \\ |s\rangle_{q}=\frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} d r e^{-i r s}|r\rangle_{p}=F^{\dagger}|s\rangle_{p}\end{array}
\end{equation}
Here $F$ is the Fourier transformation and has the form
\begin{equation}
F=e^{i \pi\left(\hat{q}^{2}+\hat{p}^{2}\right) / 4}
\end{equation}
Also $F$ is a special case for the rotation operator
\begin{equation}
R(\theta)=e^{i \theta\left(\hat{q}^{2}+\hat{p}^{2}\right) / 2}
\end{equation}
It rotates a state counter-clockwise in phase space by an angle $\theta$.
My question is how to prove
\begin{equation}
|s\rangle_{p}=F|s\rangle_{q}
\end{equation}
Also, how to calculate the state rotation
\begin{equation}
R(\theta)|s\rangle_{q}
\end{equation}
Many thanks.
This is easiest to understand in terms of the annihilation operator $a$, related to $q$ and $p$ by $a=(q+ip)/\sqrt 2$. The operator $R(\theta)$ is then given by
$$
R(\theta)=e^{i \theta(q^2+p^{2}) / 2}=e^{i\theta/2}e^{i\theta a^\dagger a}.$$
Application of the commutation relation $[a,a^\dagger]=1$ shows that
$$R^\dagger(\theta)aR(\theta)=e^{i\theta}a.$$
Hence $R(-\pi/2)$ transforms $a$ into $-ia$, which means that $q=(a^\dagger+a)/\sqrt 2$ is transformed into
$$R^\dagger(-\pi/2)qR(-\pi/2)=(ia^\dagger-ia)/\sqrt 2=p.$$
The corresponding eigenstates are transformed as $|s\rangle_p=R(\pi/2)|s\rangle_q$.
A general $R(\theta)$ transforms $q$ into $q\cos\theta-p\sin\theta$ and it transforms $p$ into $q\sin\theta+p\cos\theta$.
Many thanks. This really helps me a lot. I want to ask further more. Is it possible to calculate <s|R(theta)|s>. This is the obstacle in my current theory derivation. Or it's too complex to have an expression？
Specially, for θ=0, <s|R(0)|s>=1. This doesn't agree with your fomula above. Do I misunderstand sth?
Also, could you please also give the off-diagnal element <s|R(θ)|t>. Or could you give a hint about how to calculate that. I have no idea about this.
There is still some contradiction. There should be $\langle s|R(\pi)|s\rangle=0$ and $\langle s|R(0)|t\rangle=\delta(s,t) $
|
2025-03-21T14:48:30.450850
| 2020-04-30T07:55:24 |
358961
|
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"authors": [
"Piotr Achinger",
"Praphulla Koushik",
"https://mathoverflow.net/users/118688",
"https://mathoverflow.net/users/3847",
"https://mathoverflow.net/users/68893",
"wkf"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/358961"
}
|
Stack Exchange
|
Higher gerbes and cohomologies
Suppose $X$ is a scheme over a field $k$ and $G$ a group scheme over $k$ (may assume commutative), one can talk about $G$-torsors and $G$-gerbes over $X$ and their isomorphism classes are given by $H^1(X,G)$ and $H^2(X,G)$ (calculated in suitable Grothendieck topologies) respectively. These have been well written in the literature. I think the similar results for higher gerbes and higher cohomologies also hold but I could not find any reference. Could someone suggest a reference for it? Thanks.
Intro + Section 7.2.2 in Lurie "Higher Topos Theory"?
Corollary $<IP_ADDRESS>$ in above reference. Let $\mathfrak{X}$ be an $\infty$-topos, and $n \geq 2$, and $A$ an abelian group object of $Disc(\mathfrak{X})$. There is a canonical bijection of $H^{n+1} (\mathfrak{X}; A)$ with the set of equivalence classes of $n$-gerbes on $\mathfrak{X}$ banded by $A$..
@PiotrAchinger, thanks for mentioning Lurie's paper. I was kinda aware of the paper but unfortunately I'm not very familiar with the language of the infinity topos....
|
2025-03-21T14:48:30.450948
| 2020-04-30T08:20:37 |
358964
|
{
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"Robert Bryant",
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|
Stack Exchange
|
Can a polynomial family of curves be identified by a polynomial satisfied by its derivatives?
Suppose we have a 1-parameter family of curves that foliate a subset of the plane, defined by a function that is a polynomial in the coordinates $(x,y)$ and a parameter that identifies the individual curves.
For example, the family of curves defined by:
$$P(x,y,a) = 16 a^2 \left(x^2+y^2-a^2\right)-\left(2 x_f x +y_f\left(2 y-y_f\right) -x_f^2\right) \left(8 a^2+2 x_f x+2 y_f y-x_f^2-y_f^2\right) = 0$$
is the set of confocal ellipses with foci at $(0,0)$ and $\left(x_f,y_f\right)$, with the parameter $a$ giving the semi-major axis of each ellipse.
Associated with this family of curves is a field of tangents to the curves: at each point $(x,y)$ a unique curve passes through the point, and if we choose an orientation we can assign a unit vector to each point that is tangent to that curve.
We can derive a polynomial equation satisfied by the components of these tangent vectors, $(x',y')$ along with the coordinates $(x,y)$, by supposing we have parameterised each curve as $\left(x_a(t),y_a(t)\right)$ then taking the derivative of $P(x_a(t),y_a(t),a)$ with respect to $t$, to produce a polynomial $D(x,y,x',y',a)$. In our example, we have:
$$D(x,y,x',y',a)=16 a^2 \left(2 x x'+2 y y'-x_f x'-y_f y'\right)-4 \left(2 x_f x+2 y_f y-x_f^2-y_f^2\right) \left(x_f x'+y_f y'\right)$$
When $P$ and $D$ are both zero their resultant with respect to $a$ will be zero, and we have:
$$Res_{a}(P,D) = \left(y'\right)^4 \left(2 x^3 x_f \left(2 y-y_f\right) y_f+x^2 \left(x_f^2 \left(-2 y y_f+y_f^2-4 y^2\right)+y_f^2 \left(y_f-2 y\right){}^2\right)+2 x y^2 x_f \left(2 x_f^2+y_f \left(y_f-2 y\right)\right)-y^2 x_f^2
\left(x_f^2+y_f \left(y_f-2 y\right)\right)\right){}^2+\left(x'\right)^4 \left(2 x^3 x_f \left(2 y-y_f\right) y_f+x^2 \left(x_f^2 \left(-2 y y_f+y_f^2-4 y^2\right)+y_f^2 \left(y_f-2 y\right){}^2\right)+2 x y^2 x_f
\left(2 x_f^2+y_f \left(y_f-2 y\right)\right)-y^2 x_f^2 \left(x_f^2+y_f \left(y_f-2 y\right)\right)\right){}^2-4 \left(x'\right)^3 y' \left(x^3 \left(2 y-y_f\right)+x^2 x_f \left(y_f-3 y\right)+x y \left(x_f^2+3 y
y_f-y_f^2-2 y^2\right)+y^2 x_f \left(y-y_f\right)\right) \left(2 x^2 x_f y_f-x \left(x_f^2 \left(y_f+2 y\right)+y_f^2 \left(y_f-2 y\right)\right)+y x_f \left(x_f^2+y_f \left(y_f-2 y\right)\right)\right){}^2+4 x'
\left(y'\right)^3 \left(x^3 \left(2 y-y_f\right)+x^2 x_f \left(y_f-3 y\right)+x y \left(x_f^2+3 y y_f-y_f^2-2 y^2\right)+y^2 x_f \left(y-y_f\right)\right) \left(2 x^2 x_f y_f-x \left(x_f^2 \left(y_f+2 y\right)+y_f^2
\left(y_f-2 y\right)\right)+y x_f \left(x_f^2+y_f \left(y_f-2 y\right)\right)\right){}^2+2 \left(x'\right)^2 \left(y'\right)^2 \left(-4 x^3 x_f+x^2 \left(2 x_f^2+8 y y_f-y_f^2-8 y^2\right)+y^2 \left(2
\left(y-y_f\right){}^2-x_f^2\right)+2 x y x_f \left(4 y-3 y_f\right)+2 x^4\right) \left(2 x^2 x_f y_f-x \left(x_f^2 \left(y_f+2 y\right)+y_f^2 \left(y_f-2 y\right)\right)+y x_f \left(x_f^2+y_f \left(y_f-2
y\right)\right)\right){}^2$$
So we have eliminated the parameter $a$ to obtain a homogenous polynomial in $(x',y')$ that will be zero for any field of tangents to the curves, since any multiple of $(x',y')$ will also yield zero.
My question is: when and how can this construction be reversed? Given a polynomial $R(x,y,x',y')$ like this that is satisfied by a field of tangents, is there a test that can determine whether it originated from a polynomial family of curves? And can the defining polynomial be reconstructed?
If we write down a polynomial $P$ of a sufficiently high degree, with unknown coefficients, we can carry through this whole construction to produce a resultant polynomial, which we could then attempt to match with our given $R(x,y,x',y')$ by solving a potentially very large number of polynomial equations in the unknown coefficients.
So my question is: can we do better than that brute-force approach?
For the specific case I’m trying to solve, I have a polynomial satisfied by the tangent vectors that is of degree 8 in the derivatives and degree 6 in the coordinates.
Note that there is a symmetry between the coefficients of the derivatives in the resultant in the example: if the terms are given by $C_{i,j}(x')^i(y')^j$, we have
$$C_{i,j} = (-1)^{i}C_{j,i}$$
The $R(x,y,x',y')$ that I wish to integrate also has this symmetry.
This is not a property of the resultant obtained from a generic polynomial $P(x,y,a)$, or even from a generic foliation; for example, $P(x,y,a)=(x^2+1)a-y$ foliates the plane with parabolas, but yields the resultant $R(x,y,x',y')=2 x y x' - (x^2+1) y'$.
Edited to add: The symmetry in the coefficients of the resultant for the family of confocal ellipses can be explained by the well-known fact that the same family extends to the family of confocal hyperbolas, with both an ellipse and a hyperbola passing through every point in the plane, with orthogonal tangents.
Since we have eliminated the parameter $a$ from the resultant, it must be zero for both parameter values at each point, and the symmetry in the coefficients yields:
$R(x,y,-y',x') = \pm R(x,y,x',y')$
which is sufficient to ensure that if $R$ is zero for one tangent, it is also zero for the orthogonal tangent.
There is not much one can say. For example, consider the simplest case: A linear relation between the differentials with polynomial coefficients, i.e., $$\omega = P(x,y),\mathrm{d}x + Q(x,y),\mathrm{d}y = 0.$$ You want a test for when there is an algebraic first integral, i.e, when there is a nonzero algebraic $M(x,y)$ such that $\mathrm{d}\bigl(M(x,y)\omega\bigr) = 0$, i.e., an algebraic integrating factor. There are certainly necessary conditions at the singular points (i.e., points where both $P$ and $Q$ vanish), but I'm not aware of any general sufficient conditions.
|
2025-03-21T14:48:30.451246
| 2020-04-30T10:02:37 |
358965
|
{
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|
Stack Exchange
|
Canonical connection on $\mathcal{A}\times X$
Let $E\rightarrow X$ be a vector bundle and let $\mathcal{A}$ denote the space of connections on $E$. Pulling back $E$ by the second projection we obtain a vector bundle $\mathbb{E}=p_2^*E\rightarrow X$ over $\mathcal{A}\times X$.
There exists a canonical connection $\mathbb{A}$ on $\mathbb{E}$ which is flat in the $\mathcal{A}$ direction and equal to $A$ on the slice $\{A\}\times X$. We obtain the following curvature:
$$
\begin{cases}
\mathbb{A}^2(v,w)= R_A(v,w) & \text{for $v,w\in T_xX$} \\
\mathbb{A}^2(\alpha,v)=\alpha(v) & \text{for $\alpha\in T_A\mathcal{A}$,$v\in T_xX$ } \\
\mathbb{A}^2(\alpha,\beta)=0 & \text{for $\alpha,\beta \in T_A\mathcal{A} $}.
\end{cases}
$$
These identities can be found in Donaldson's "INFINITE DETERMINANTS, STABLE BUNDLES AND CURVATURE" p236 and Itoh and Nakajima's "Yang-Mills Connections and Einstein-Hermitian Metrics" p451.
I do not understand the pairing of the middle line. How does it follow from the definition?
A vector field $v$ on $X$ and a vector field $\alpha$ on $\mathcal A$ give rise to two commuting vector fields on $\mathcal A\times X$, denoted by the same letters. Then we may regard $\alpha$ as (the pullback of) an element of $\Omega^1(X;\operatorname{End}E)$ as well. Write the connection as a covariant derivative $\nabla$, choose a section $s$ of $E\to X$ and compute for its pullback (still denoted $s$) that
$$\underbrace{\nabla_\alpha\nabla_v}_{=\alpha(v)\in\operatorname{End}E}s-\nabla_v\underbrace{\nabla_\alpha s}_{=0} -\nabla_{\underbrace{[\alpha,v]}_{=0}}s=\alpha(v)s\;.$$
Here we have used the interpretation of $\alpha$ as a variation of $\nabla$, and we have used that $\nabla$ is trivial in the $\mathcal A$-direction.
|
2025-03-21T14:48:30.451389
| 2020-04-30T10:25:01 |
358968
|
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"Aaron Bergman",
"Markus Zetto",
"amathematician",
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|
Stack Exchange
|
B-model and Hochschild cohomology
In "On the Classification of Topological Field Theories" in Example 1.4.1, Lurie introduces the B-model with target an (even dimensional) Calabi-Yau variety $X$: The Hochschild cohomology $\operatorname{HH}^*(X)$, together with the "canonical trace map" $\operatorname{HH}^*(X) \rightarrow k$, form a graded commutative Frobenius algebra; and by a classical theorem that can, as he states, analogously applied to the graded setting, this defines a 2-dimensional Topological Quantum Field Theory $Z$ with values in complexes such that $Z(S^1)=\operatorname{HH}^*(X)$.
While I can roughly follow his reasoning (although it would be helpful if someone could provide a short explanation on how this trace map occurs), as far as I knew the states of the B-model on $X$ were related to the (bounded) derived category $\operatorname{D}^b(\operatorname{Coh(X)})$, and I don't really see why Lurie's definition should describe the B-model as I know it, or why Hochschild cohomology occurs here. As Lurie only uses this statement as a motivation for introducing extended TQFTs with values in chain complexes, he doesn't give any further explanation or references, so I wondered if anyone here could. Thanks in advance and greetings,
Markus Zetto
The boundary conditions of the B-model, ie, the D-branes, are the objects in $\mathcal{D}^b(X)$. A little bit of playing with pictures gives that the space of closed string states must be in the center of the algebra of open strings for any given boundary condition. For a category $\mathcal{C}$, this gives a map to $\mathcal{Nat}(id_\mathcal{C},id_\mathcal{C})$. Extending this to the graded situation and using the description of natural transformations as bimodules, you get
$$
HH^\bullet(\mathcal{D}^b(X)) = \mathrm{Ext}^\bullet_{X \times X}(\mathcal{O}_\Delta,\mathcal{O}_\Delta)
$$
where $\Delta$ is the diagonal. There's a version of the HKR theorem that relates this to cohomology of exterior powers of the tangent bundle, the usual space of states in the B-model.
A lot of this is explained in Moore and Segal. I also explained some (with no claim to originality) in the beginning of my old paper Deformations and D-branes.
Thanks, these both seem like very good references! I'll still have a bit of reading to do to figure out what exactly is going on here and how exactly this factors into Lurie's definition (I figure the fact that $Z(S^1)=\operatorname{HH}^*(X)$ just tells me exactly what you are saying, i.e. what the space of states for a closed string is), but this makes it very clear why Hochschild cohomology apears.
You might also want to check out Kevin Costello's work, which focuses only on the 2D case so might be easier to understand: https://arxiv.org/pdf/math/0412149.pdf
For correct attribution, one should at least mention the paper which the preprint of Moore and Segal itself quotes as the source for the particular case of the algebraic description of open-closed TFTs which they give without proof. The following paper was published about 5 years before the preprint of Moore and Segal:
C. I. Lazaroiu, On the structure of open-closed topological field theory in two dimensions, Nucl.Phys.B603 (2001) 497-530, https://arxiv.org/abs/hep-th/0010269
Notice that the spaces of boundary states in the B-model with CY manifold target X are Z-graded, since the physically correct category of B-type topological D-branes is not D^b(X) but rather its shift completion. The Z/2Z grading in Lazaroiu's axioms is the mod 2 reduction of that natural Z-grading -- and this Z/2Z grading is ignored by Moore and Segal. Shift completions were explained in:
C. I. Lazaroiu, Graded D-branes and skew-categories, JHEP 0708 (2007) 088, https://arxiv.org/abs/hep-th/0612041
(which also explains the twist-completion that occurs, for example, in B-type orbifold LG models).
One should also note that the argument using HH(D^b(X)) and the holomorphic HKR isomorphism is not due to Lurie but to Kontsevich. Furthermore this argument (while very nice and natural) is not needed for the B-model, since the closed string state space of the B-model is known independently from localization of the path integral in the bulk sector, which produces the space of polyvector fields endowed with its (Dolbeault) differential and Serre trace. This goes back to Witten and to the paper of BCOV in the early 1990's. The advantage of the polyvector approach is that it provides an off-shell model which also appears in the BCOV theory and has a dual description using a BV bracket.
Without digging too deep into this particular issue, I will note that the Moore and Segal results had been circulating and were available on Greg’s webpage many years before the release of the preprint.
@Aaron What I have heard is that the only thing available (as per Greg Moore himself) were transparencies of a talk that had been unknown to Lazaroiu and which did not contain Lazaroiu's results but only a preliminary attempt at the problem. Lazaroiu (who was a first year post-doc at the time) mentions this in an added note and quotes those transparencies in the second version of his preprint, since he received a very strongly worded email from Greg Moore attaching those transparencies. It is dangerous to make claims about priority which are not supported by documentary evidence.
|
2025-03-21T14:48:30.451747
| 2020-04-30T10:29:50 |
358969
|
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|
Stack Exchange
|
Jordan isomorphisms of type I von Neumann algebras
Let $\mathcal M$ and $\mathcal N$ be two von Neumann algebras. A linear map $J:\mathcal M\to\mathcal N$ is said to be a Jordan isomorphism if $J$ is bijective, $*$-preserving and $J(xy+yx)=J(x)J(y)+J(y)J(x)$ for all $x,y \in \mathcal M.$
Is there any nice classification of type I von Neumann algebras up to Jordan isomorphism? Also is there is any classification of type I factors up to Jordan isomorphism?
I saw in the literature two notion of Jordan isomorphism: one is the condition $J(xy+yx)=J(x)J(y)+J(y)J(x)$. The second is the condition $J(x^2)=J(x)^2$ (which is equivalent to the other one by polarization, as soon as 2 is invertible). This doesn't fit your definition, which forces $J(xy-yx)$ to vanish and hence is excluded if $\mathcal{M}$ is non-commutative. You seem to confuse with the trace condition.
@ YCor!Sorry! Edited!
I have not read through all the details carefully, but I think your questions can probably be answered using the results in Kadison's 1951 paper "On isometries of operator algebras". That paper actually has a discussion of Jordan $\ast$-isomorphisms between ${\rm C}^\ast$-algebras (confusingly, Kadison calls these "${\rm C}^\ast$-isomorphisms") and in particular:
Theorem 10 (with terminology updated) a Jordan $\ast$-isomorphism from a von Neumann algebra onto a ${\rm C}^\ast$-algebra is the direct sum of a $\ast$-isomorphism and a $\ast$-anti-isomorphism.
As for your question about the factor case, this is handled by
Corollary 11: any Jordan $\ast$-isomorphism of a factor is either a $\ast$-isomorphism or a $\ast$-anti-isomorphism.
Yes. And because every type I von Neumann algebra is $$-isomorphic to its opposite algebra, this means that Jordan isomorphic $\Rightarrow$ $$-isomorphic,
@ Ni and Choi. Actually, I was looking for a more concrete kind of classification. For example can it be seen to be a von Neumann algebra by gluing concretely known type von Neumann algebras e.g. $B(\mathcal H)$ ($\mathcal H$ is a Hilbert space) etc.?
I can't make any sense of this comment.
Are you just asking for a classification of type I von Neumann algebras? That is achieved through direct integral decomposition.
Alright! I now understand.
I believe two type I von neumann algebras are Jordan isomorphic if and only if they are $*$-isomorphic. Jordan isomorphism preserve order, so if two von Neumann algebras are Jordan isomorphic they are order isomorphic and hence have the same normal state space. And one should be able to recover any type I von Neumann algebra from its normal state space. (I don't have the reference handy, but this is probably in State Spaces of Operator Algebras by Alfsen and Schulz.)
BTW, this will not be true outside the type I setting. There is a von Neumann algebra (even a factor) that is not $*$-isomorphic to its opposite algebra. But the two are Jordan isomorphic via the identity map.
|
2025-03-21T14:48:30.451952
| 2020-04-30T11:03:55 |
358971
|
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|
Stack Exchange
|
Coordinates for Laminations: geometric versus shear
Let $S$ be an orientable surface with a triangulation T.
A lamination $\ell$ is a simple closed curve on $S$, up to isotopy. We will assume that $\ell$ is drawn in such a way that it intersects the edges of $T$ in the minimal number of points. We can associate to $\ell$ two sets of "coordinates".
The geometric coordinates are defined to be the $|T|$-tuple of numbers $geom(\ell,t)=$ {number of points in $\ell \cap t$}, where $t \in T$ is an edge of the triangulation.
The shear coordinates are another $|T|$-tuple of numbers $shear(\ell,t)$ defined as follows. For every edge $t \in T$ consider the quadrilateral $q$ formed by the two triangles with edge $t$, and the collection of arcs $\ell \cap q$. For each such arc associate a number -1,0,1 depending on how the arc crosses $q$: if it starts and ends at consecutive edges of $q$ associate 0, otherwise if it starts and ends at opposite edge that forms with $t$ a "S" (resp. a "Z") associate -1 (resp. "1"). A more precise definition can be found e.g. in (https://arxiv.org/abs/math/0510312).
My question is whether and how it's possible to move from the two sets of "coordinates", e.g. how to build the $|T|$-tuple $shear(\ell,t)$ from $geom(\ell,t)$.
Yes, it is possible to go back and forth between the two coordinate systems (and a good thing too, as otherwise they would not be coordinates!). My solution goes through a third coordinate system called "normal coordinates".
Suppose that $\gamma$ is a collection of properly embedded arcs in a triangle $T$ which avoids the corners $x, y, z$ of $T$. Suppose that each arc of $\gamma$ meet each edge of $T$ in at most one point. (Thus $\gamma$ forms no bigons with the edges of $T$.) Then we can partition the arcs of $\gamma$ into three collections $\gamma_x, \gamma_y, \gamma_z$ where an arc in $\gamma_x$ (say) separates $x$ from $y$ and $z$. The three numbers $|\gamma_x|, |\gamma_y|, |\gamma_z|$ are called the normal coordinates of $\gamma$ in $T$.
Suppose that $S$ is a surface and $\Delta$ is a triangulation of $S$. Suppose that $\alpha$ is a simple closed multi-curve in $S$ transverse to the skeleta $\Delta^{(k)}$. Suppose that $\alpha$ meets the edges of $\Delta^{(1)}$ minimally, up to isotopies that do not cross the vertices of $\Delta^{(0)}$. (That is, $\alpha$ has no bigons with the edges of $\Delta^{(1)}$.) Then for every triangle $T \in \Delta^{(2)}$, we can compute the normal coordinates of $\alpha$ in $T$.
Here is a sequence of hints for moving between the various coordinate systems.
To go from normal to intersection coordinates - we simply add. That this is well-defined (independent of which triangle you use) is called the matching equalities for normal coordinates.
To go from intersection to normal coordinates - this can be done triangle by triangle. Doing this discovers the mod two condition on intersection coordinates as well as the triangle inequality.
To go from normal to shear coordinates - this is done edge by edge. Suppose that $e \in \Delta^{(1)}$ is an edge. Let $T$ and $T'$ be the adjacent triangles. Let $Q = Q(e)$ be their union; this is the quadrilateral about $e$. If $T = T'$ then we learn that shear coordinates require the surface $S$ to be orientable (unlike the other coordinate systems) and do not allow peripheral curves (that is, components of $\alpha$ homotopic into a neighbourhood of a vertex).
To go from shear to normal coordinates - this is done vertex by vertex. That is, for a vertex $v$, let $\{T_i\}$ be the triangles in cyclic order about $v$. Note that a triangle may appear as many as three times in this list, in different rotations. Let $P = P(v)$ be the union of these triangles; this is the polygon about $v$. From the sequence of signed shears about $v$ we build a weighted train track, with stops, in $P$. This gives the normal coordinates for the $T_i$ at the corner $v$. Doing this discovers the lack of spiralling in shear coordinates as well as a rule of signs - about every vertex (with some shearing) both signs appear.
|
2025-03-21T14:48:30.452211
| 2020-04-30T11:17:53 |
358972
|
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|
Stack Exchange
|
Slopes of classical newforms
Let $f$ be a newform in $S_k(\Gamma_1(Np^r))$ with $r\geq 1$, and let $a_p$ be the $U_p$ eigenvalue of $f$.
If $f$ is in $S_k(\Gamma_0(Np^r))$, it seems to be a well known consequence of Atkin-Lehner theory that the slope of $a_p$ (the $p$-adic valuation of $a_p$) is equal to $k/2-1$ if $r=1$ and is $\infty$ (i.e. $a_p=0$) if $r\geq 2$.
What happens for general newforms $f \in S_k(\Gamma_1(Np^r),\chi)?$ It is no longer true that if $r\geq 2$ then $a_p=0$. However, does there exist a simple description as above of the slope of $a_p$ only in terms of $r$ and the conductor of $\chi$? For example, it seems to be suggested by examples in the LMFDB that if $r\geq 2$, the slope is going to be infinite unless the order of $\chi$ is divisible by $p^{r-1}$.
Thanks!
Theorem 3 in Li, "Newforms and functional equations" gives you some information about $a_p$, in particular its vanishing. Also, if $a_p$ is nonzero then you can relate it with the pseudo-eigenvalue of $f$ for the Atkin-Lehner operator at $p$, see Theorem 2.1 in Atkin--Li, "Twists of Newforms and Pseudo-Eigenvalues of W-Operators". The theorem you mention follows directly from this.
Thanks! To summarize for the benefit of anyone interested, Theorem 3 of Li's article specializes to the following. If $f\in S_k(\Gamma_1(Np^r),\chi)$ with $N$ coprime to $p$ and $r\geq 1$, then the slope is determined by the conductor $cond(\chi)$ of $\chi$ and by $r$. If the $cond(\chi)\nmid Np^{r-1}$, then the slope is $(k-1)/2$. Otherwise, assume $cond(\chi) \mid Np^{r-1}$. If $r=1$ then the slope is $k/2-1$. If $r\geq 2$ then the slope is $\infty$.
In the case $\chi$ is $p$-primitive, I don't think you can deduce the slope: one only knows the archimedean absolute value. In the particular case $a_p$ is real, we can conclude. Note that pseudo-eigenvalues of $W_p$ are algebraic numbers but not necessarily $\pm 1$, if I recall correctly they may have non-trivial $p$-adic valuations.
Sorry, you are right of course (I was just about to write a correction...). In the case $cond(\chi)\nmid Np^{r-1}$ and $r=1$ you can conclude that the absolute value of $a_p$ as a complex number is $p^{(k-1)/2}$, but that's not the same as saying the slope is $(k-1)/2$. Thanks for the correction
Another reference: Mazur, Tate, Teitelbaum, On p-adic analogues of BSD, Section 12
|
2025-03-21T14:48:30.452516
| 2020-04-30T11:35:49 |
358973
|
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|
Stack Exchange
|
Powers of the Euler class, torsion free subgroup of Homeo($S^1$)
For any subgroup $G$ of $\text{Homeo}(S^1)$, we have the Euler class $\chi$ in the group cohomology $H^2(G;\mathbb{Z})$. One can think of this class as the pullback of the generator of $H^2(\mathrm{B}\text{Homeo}(S^1);\mathbb{Z})$. I have two related questions regarding the (non)vanishing of the powers of Euler class for different subgroups of $\text{Homeo}(S^1)$.
Does there exist a torsion-free, finitely generated group $G\subset\operatorname{Homeo}(S^1)$ such that $\chi^k$ for all $k$ are nonzero in $H^{2k}(G;\mathbb{Z})$?
If we let the group $G$ be the group of piecewise linear homeomorphisms of the circle $\mathbb{R}/\mathbb{Z}$ which send $\mathbb{Q}_2/\mathbb{Z}$
onto itself and have singular points only on $\mathbb{Q}_2/\mathbb{Z}$, where $\mathbb{Q}_2$ is the ring of dyadic numbers. Ghys and Sergiescu proved, in this case, $\chi^k$ for all $k$ are nonzero in $H^{2k}(G;\mathbb{Z})$ but $G$, which is one of the variants of Thompson groups, has many torsion elements.
Another example related to this question is due to Solomon Jekel. Let $\Gamma_{g}^1$ be the mapping class group of a surface of genus $g$ with a marked point. It is a subgroup of $\text{Homeo}(S^1)$. Jekel showed that in this case $\chi^{g-1}$ is nonzero in $H^{2g-2}(\Gamma_{g}^1;\mathbb{Q})$. Now, one could let $G$ be a finite index torsion-free subgroup of $\Gamma_{g}^1$. But the power $g-1$ is a threshold for the non-vanishing of powers of the Euler class.
In the direction of the second example, let $\text{Tor}_g^1$ be the Torelli group of the surface of genus $g$ and a marked point. What is the threshold $k$ for which $\chi^k$ is nonzero in $H^{2k}(\text{Tor}_g^1;\mathbb{Q})$? Is it less than $g-1$?
In 1994 ("Topology, Geometry and Field Theory - Proceedings of the 31st International Taniguchi Symposium", p. 105), Morita wrote
However at present even the non-triviality of $e^2 \in H^4(Tor_g^1)$
seems to be unknown.
So I imagine not much is known in general.
Kupers and I recently (accidentally) answered this: $e^2$ is nontrivial, even on $Tor^1_{g, 1}$. This is Corollary 8.3 of "On the cohomology of Torelli groups".
|
2025-03-21T14:48:30.452683
| 2020-04-30T13:02:16 |
358978
|
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|
Stack Exchange
|
For which classes of metric spaces can we prove that quasi-isometry is an equivalence relation in ZF?
Given two metric spaces $(M_1, d_1)$ and $(M_2, d_2)$, a map $\phi \colon (M_1, d_1) \to (M_2, d_2)$ is a large-scale Lipschitz essentially surjective map if there exist constants $A \geq 1, B \geq 0$, and $C \geq 0$ such that
For all $x, y \in M_1$, we have $\frac{1}{A} d_1(x, y) - B \leq
d_2(f(x), f(y)) \leq A \: d_1(x, y) + B$.
For all $z \in M_2$, there exists $x \in M_1$ such that $d_2(z, f(x)) \leq C$.
We define the relation $\mathcal{R}$ on metric spaces by $M_1 \mathcal{R} M_2$ iff there exists a large-scale Lipschitz essentially surjective map $\phi \colon (M_1, d_1) \to (M_2, d_2)$. In ZFC, $\mathcal{R}$ is a symmetric relation (even an equivalence relation) and we normally say that $M_1$ and $M_2$ are quasi-isometric spaces.
However, it is known (see arXiv:1609.01353 for the below cited results) that while $\mathcal{R}$ is always reflexive and transitive, it turns out that symmetry of $\mathcal{R}$ implies the axiom of choice, and hence does not follow from ZF alone (Corollary 2).
It is therefore natural to ask whether if one restricts $\mathcal{R}$ to certain classes of metric spaces one may still have symmetry follow from ZF alone. Even in this case, things are messy: for example, the symmetry of $\mathcal{R}$ on hyperbolic metric spaces fails in ZF (Theorem 1). However, if one restricts one's attention to the class of $\mathbb{R}$-trees, one can do away with choice, and can show in ZF that $\mathcal{R}$ on $\mathbb{R}$-trees is an equivalence relation (Theorem 3). Thus, my question:
For what (interesting) classes of metric spaces, other than $\mathbb{R}$-trees, does symmetry of $\mathcal{R}$ follow from ZF?
Any references and/or thoughts would be appreciated (especially from those set theorists on this site who like to do away with the axiom of choice...).
Clearly, "be quasi-isometric" is not an appropriate term if it fails to be an equivalence relation. You should better refer to it as the relation $X,\mathcal{R},Y$ "there exists a large-scale Lipschitz essentially surjective map $X\to Y$". Corson says "the QI-relation", which is OK.
Also your definition (even in ZFC) is not the usual one: the identity map $(\mathbf{R},d)\to(\mathbf{R},\sqrt{d})$ is coarsely Lipschitz (even large-scale Lipschitz), coarsely surjective (= essentially surjective), but not a quasi-isometry. Either you want to stick to coarse equivalence, or restrict to spaces for which these notions are equivalent (that is to say, large-scale geodesic spaces). I note that the definition used by Corson in the paper you link is the usual one.
Well, interesting is subjective. So I'd daresay "none of them"? :P
@YCor Thank you, you are right of course. I have edited the question to reflect this.
I have no idea how much you can get in ZF, but you should be able to go quite far with weak forms of choice. I would guess that symmetry of $\mathcal{R}$ over separable metric spaces follows from a weaker form of choice, possibly countable choice (CC). CC doesn't imply the pathologies of full choice, in particular it is consistent with the axiom of determinacy (AD rules out all choice pathologies). And ZF + CC is equivalent to ZF + "countable products of compact pseudometric spaces are Baire", so it looks like CC is necessary to develop metric geometry over separable spaces.
In general CC is needed to develop analysis. The result on Baire category is in https://core.ac.uk/download/pdf/82313129.pdf
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2025-03-21T14:48:30.452930
| 2020-04-30T13:34:14 |
358980
|
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|
Stack Exchange
|
Let $U$ be a simply connected open subset of ${\Bbb S}^2$, is the complement of $U$ also simply connected?
I was looking into particular cases for the Poincaré-Bendixson theorem and I came across a topological problem about simply connectivity.
If $\gamma$ is a Jordan curve in ${\Bbb S}^2$ then using Jordan-Schoenflies, we have that ${\Bbb S}^2\setminus \gamma = U\sqcup V$ with $U$ and $V$ being simply connected (s.c.). Moreover, as the sphere minus $\gamma$ is homeomorphic to the sphere minus the equator, we also get that $\overline{U}$ and $\overline{V}$ are s.c.
Let $U$ be a s.c. open subset in ${\Bbb S}^2$. We note that $\overline{U}$ need not be s.c. (take an open annulus where you make a transversal cut). On the other hand, the complement $F=S^2\setminus U$ appears to me to be s.c.
${\rm Int} F$ need not be connected, but a connected component of ${\rm Int} F$ appears to me again to be s.c.
Do the above two claims hold in general? Or have I missed some obvious counter-examples?
$F=S^2\backslash U$ need not be path connected, e.g. $F$ could be homeomorphic to the closed topologists sine curve. However, every path component of $F$ must be simply connected.
By identifying $U$ with the open unit disk (Riemann mapping), you can realize the compact set $F=S^2\backslash U$ as an intersection $\bigcap_{n\in\mathbb{N}}V_n$ where each $V_n$ is homeomorphic to the closed unit disk. This implies that $F$ has trivial shape, i.e. is cell-like.
The path-components of a cell-like continuum don't have to be contractible, e.g. if $F$ is the Knaster buckethandle continuum. However, it is known that every cell-like subset of a 2-dimensional manifold is simply connected. See Corollary 6 of:
H. Fischer, A.Zastrow, The fundamental groups of subsets of closed surfaces inject into their first shape groups, Algebraic and Geometric Topology 5 (2005) 1655-1676.
Various parts this paper could allow you to verify a positive answer to both of your questions without appealing to shape theory. For instance, Lemma 13 is a particularly handy result that would imply that once you know each path-component of $F$ is simply connected, then every component of $int(F)$ is simply connected.
Thanks a lot for your comprehensive answer. Exactly what I was hoping for. And nice twist with the sine curve being s.c. but in a somewhat pathological way. I'll look into Fischer and Zastrow.
|
2025-03-21T14:48:30.453115
| 2020-04-30T13:36:41 |
358981
|
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|
Stack Exchange
|
Stability of a continuous piecewise linear map
I am studying random perturbation of a system that is continuous and piecewise linear. More precisely: I am given a map $\Phi_1:\mathbb{R}^d\to \mathbb{R}^d$ such that
$$ \Phi_1(x) = \left\{\begin{array}{ll}Ax & \text{if $Cx\ge0$}\\
Bx & \text{otherwise,}\end{array}\right.$$
where $A$, $B$ are $d\times d$ matrices and $C$ is a vector.
I know that $\Phi_1$ is continuous (which implies that $A$ and $B$ cannot be too different for example).
I denote by $\Phi_t$ the $t$-th iterations of $\Phi$: $\Phi_t(x) = \Phi(\Phi_{t-1}(x))$. My questions concern the asymptotic properties of the iterations $\Phi_t$:
are there "easy" necessary and sufficient conditions that would guarantee that for all $t$: $\lim_{t\to\infty}\Phi_t(x)$=0.
if the above property 1. holds, does it hold uniformly for all $x$ such that $\|x\|\le1$?
if the above property 2. holds, does it imply that $\Phi_t$ is a contraction for large enough $t$?
To add some context, I am considering a noisy version of the iterations above in which $X_{t+1}=\Phi_t(X_t) + \epsilon E_{t+1}$, where $E_{t}$ is a sequence of (bounded) random variables. This implies that $X_{t} = O(\epsilon)$. The property 3. above implies that this holds uniformly in time (and I do not know how to prove it otherwise).
Have you checked results from Daniel Liberzon's book on switched dynamical system?
Thanks for pointing the link. I do not really know the literature in switching system, I will look into it.
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2025-03-21T14:48:30.453244
| 2020-04-30T13:39:02 |
358982
|
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|
Stack Exchange
|
"On the distribution of reduced residues" by Montgomery and Vaughan – missing careful argument wanted
In their paper, On the distribution of reduced residues, Montgomery and Vaughan state early on that
With a more careful argument from (2) it is easily seen that
$$\tag{*}
qhP - qhPQ + O(qhP^2) \leq M_2(q; h) \leq qhP
$$
where $Q=\prod_{\substack{{p \mid q}\\{p>h}}} (1-1/p)$.
However, the careful argument is omitted, and I haven't been able to lure out the first inequality myself. I hope asking here could help me in that direction. The introduction of Montgomery and Vaughan's paper is included below as background.
Question: How can one derive
$$
qhP - qhPQ + O(qhP^2) \leq M_2(q; h)
$$
from (2) below?
Background
Let $q$ be a natural number, let $P = \phi(q)/q$ be the "probability" that a randomly chosen integer is relatively prime to q, and let
$$
\tag{1}
M_k(q;h) = \sum_{n=1}^{q}
\left(
\sum_{\substack{{m=1}\\{(m+n,q)=1}}}^{h} 1 - h P
\right)^k.
$$
This is the $k$-th moment of the number of reduced residues modulo $q$ in an
interval of length $h$ about its mean, $hP$. Clearly $M_1(q; h) = 0$. By an elementary calculation (see Hausman and Shapiro [3]) it may be shown that
$$\tag{2}
M_2(q;h) =
qP^2
\sum_{\substack{{r \mid q }\\{r > 1}}}
\mu(r)^2
\left(
\prod_{\substack{ {p \mid q }\\{p \nmid r} }}
\frac{p(p-2)}{(p-1)^2}
\right)
r^2 \phi(r)^{-2} \left\{ \frac{h}{r}\right\}\left( 1 - \left\{ \frac{h}{r}\right\}\right).
$$
This with the simple inequality $\{\alpha\}(1 - \{\alpha\}) \leq \alpha$ gives immediately the estimate
$$\tag{3}
M_2(q;h)\leq qhP.
$$
With a more careful argument from (2) it is easily seen that
$$
qhP - qhPQ + O(qhP^2) \leq M_2(q; h) \leq qhP
$$
where $Q=\prod_{\substack{{p \mid q}\\{p>h}}} (1-1/p)$.
1. First we prove the upper bound in $(\ast)$. Using the original hint, and noting that $P=\phi(q)/q$, it suffices to show the identity
$$\sideset_{^\flat}\sum_{r\mid q}\frac{r}{\phi(r)^2}
\left(\prod_{\substack{ {p \mid q }\\{p \nmid r} }}\frac{p(p-2)}{(p-1)^2}
\right)=\frac{q}{\phi(q)},$$
where $\flat$ indicates that the summation is restricted to square-free values of $r$. The two sides are multiplicative in $q$, hence it suffices to verify the special case when $q$ is the power of a prime $p$. In that case, the identity boils down to
$$\frac{p(p-2)}{(p-1)^2}+\frac{p}{(p-1)^2}=\frac{p}{p-1},$$
which is evident.
2. Now we prove the lower bound in $(\ast)$, which can be rewritten as
$$\frac{M_2(q;h)}{qhP}\geq 1-Q+O(P).$$
Equivalently,
$$\frac{1}{\phi(q)}\sideset_{^\flat}\sum_{r\mid q}\frac{r}{h}\left\{ \frac{h}{r}\right\}\left( 1 - \left\{ \frac{h}{r}\right\}\right)\prod_{\substack{ {p \mid q }\\{p \nmid r} }}(p-2)\geq 1-Q+O(P).$$
It is clear that (cf. previous point)
$$\frac{1}{\phi(q)}\sideset_{^\flat}\sum_{r\mid q}\prod_{\substack{ {p \mid q }\\{p \nmid r} }}(p-2)=1,$$
hence the lower bound in $(\ast)$ is equivalent to
$$\frac{1}{\phi(q)}\sideset_{^\flat}\sum_{r\mid q}f(h,r)\prod_{\substack{ {p \mid q }\\{p \nmid r} }}(p-2)\leq Q+O(P),$$
where $f(h,r)$ abbreviates
$$f(h,r):=1-\frac{r}{h}\left\{ \frac{h}{r}\right\}\left( 1 - \left\{ \frac{h}{r}\right\}\right).$$
It is straightforward that
$$f(h,r)\leq\min\left(1,\frac{h}{r}\right)\leq\prod_{\substack{p\mid r\\p>h}}\frac{h}{p},$$
hence hence it suffices that
$$\frac{1}{\phi(q)}\left(\prod_{\substack{p\mid q\\p\leq h}}(p-2+1)\right)
\left(\prod_{\substack{p\mid q\\p>h}}\left(p-2+\frac{h}{p}\right)\right)
\leq Q+O(P).$$
Equivalently,
$$\prod_{\substack{p\mid q\\p>h}}\left(1-\frac{1}{p-1}+\frac{h}{p(p-1)}\right)\leq Q+O(P).$$
Now the left hand side equals
$$Q\prod_{\substack{p\mid q\\p>h}}e^{O(h/p^2)}=Q\left(1+\frac{O(1)}{\log h}\right)=Q+O\left(\frac{Q}{\log h}\right)=Q+O(P),$$
and we are done. In the last step, we used that
$$Q=P\prod_{\substack{p\mid q\\p\leq h}}\left(1-\frac{1}{p}\right)^{-1}=O(P\log h).$$
Thanks! If I haven't misunderstood anything, you prove the second inequality, $M_2(q;h)\leq qhP$. What I was interested in was the other inequality. Sorry if I caused some confusion. I have altered my question to be clearer. You got any thoughts on the question as it's now phrased?
@user45947: I included a proof of the lower bound. It was rather nontrivial, it made me think for a while!
Thanks for the effort! I've accepted the answer.
|
2025-03-21T14:48:30.453485
| 2020-04-30T13:41:29 |
358983
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"Willie Wong",
"https://mathoverflow.net/users/33741",
"https://mathoverflow.net/users/3948",
"leo monsaingeon"
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"sort": "votes",
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|
Stack Exchange
|
Gradient flows: convex potential vs. contractive flow?
Take a $\mathcal C^2$ potential $V:\mathbb R^d\to \mathbb R$, and assume that it is bounded from below (say $\min V=0$ for simplicity, so that $V\geq 0$).
Consider the autonomous gradient-flow
$$
\dot X_t=-\nabla V(X_t)
$$
and let $\Phi(t,X_0)$ be the corresponding flow.
It is well-known that if $V$ is $\lambda$-convex (i-e the Hessian $D^2V\geq \lambda Id$ in the sense of symmetric matrices) then the flow is exponentially $\lambda$-contractant,
$$
|\Phi(t,X_0)-\Phi(t,X_0')|\leq e^{-\lambda t}|X_0-X_0'|,
\qquad
\forall \, X_0,X_0'\in \mathbb R^d.
$$
In particular for $\lambda=0$ (convex potential) the flow is just nonexpanding.
Question: is this an equivalence? I-e is it true that if $V$ is smooth and $\Phi(t,.)$ is $1$-Lipschitz for all times then necessarily $V$ must be convex? I am also interested in the corresponding statement for $\lambda>0$, i-e if $\Phi(t,.)$ is $e^{-\lambda t}$-Lipschitz for all $t>0$ is it true that $D^2V\geq \lambda$?
Let me just add a few comments:
The implication "$D^2V\geq \lambda$ $\Rightarrow$ $\Phi(t,.)$ is $e^{-\lambda t}$-contractant" is classical and easy to prove. As should be clear from my question above, I am only interested in the converse implication.
I stated the problem in $\mathbb R^d$ and smooth potentials for simplicity, but in reality I'm interested in this kind of problems in infinite dimension, and in fact in the context of abstract gradient-flows in metric spaces. The point is that I want to prove that some potential $V$ is geodesically convex, assuming only that the generated flow is contractant. But I want to check that this is not a complete equivalence. For example sometimes it is easy to see that the flow map is well-behaved by pure "1st-order calculus" PDE arguments (I mean, taking just one derivative in time along solutions), but characterizing the convexity requires 2nd order calculus and is therefore more delicate to justify rigorously in infinite dimensions so the mutual implications between both concepts may not be totally clear
This is related to my previous posts:
Heat flow, decay of the Fisher information, and $\lambda$-displacement convexity
Universal decay rate of the Fisher information along the heat flow
improved regularization for $\lambda$-convex gradient flows
Doesn't this follow from dependence on initial data?
Consider the flow mapping $\Phi(t,X)$ which solves
$$ \frac{d}{dt}\Phi(t,X) = - \nabla V(\Phi(t,X)) $$
so taking the derivative in $X$ we have
$$ \frac{d}{dt} \partial_X \Phi(t,X) = - \nabla^2 V(\Phi(t,X)) \cdot \partial_X \Phi(t,X) \\= - \nabla^2 V(X) \cdot \partial_X \Phi(t,X) + O(t) \cdot \partial_X \Phi(t,X)$$
So if $-\nabla^2 V(X_0)$ has negative eigenvalue $-\lambda_0$ with eigenvector $v_0$, taking the partial in the $v_0$ direction gives
$$ \partial_{v_0} \Phi(t,X_0) = e^{\lambda_0 t} v_0 + O(t^2) $$
For $t>0$ sufficiently small you guarantee that
$$ |\partial_{v_0} \Phi(t,X_0) | \geq (1 + \frac{\lambda_0}{2}t) |v_0| $$
showing that the solution map cannot be 1 Lipschitz.
Yes indeed! I should have figured that out sooner. Thank you @Willie Wong. Actually I would be interested in a "non-differential" proof, since I want to go to infinite dimension and metric spaces.
I think the same argument can be made through differences if you are careful with your epsilons and deltas. You have that the difference between solutions $X$ and $Y$ satisfy $$ (X-Y)' = -\nabla V(X) + \nabla V(Y) $$ and if you start with $X(0) - Y(0)$ sufficiently small and only look at sufficiently small times you can approximate using Taylor series and everything goes through. The only part you may have to modify your assumption is when you work in infinite dimensions, and continuity does not guarantee local uniform continuity. This means that the term I denoted by $O(t)$ in the...
... answer may require quite a bit more work/slightly strengthened assumptions beyond $C^2$.
yes yes, I get the idea, but the problem is that in abstract metric gradient flows even the $\nabla V(X)$ object does not really exist, only its magnitude somehow (metric slope and upper gradients). So I cannot even try Taylor-expanding anything. Life is hard.
Thanks a lot anyway, at least now I'm convinced in $\mathbb R^d$, that's a start! ;-)
Ah, I understand. Sorry, cannot help more. I do still think that this is a local and not a global problem, and the key is to understanding the variation of your flow with respect to changes in initial data.
It should be noticed that already on $R^d$ equipped with a non-Euclidean norm $\|.\|$ the answer to your question is no. Ohta-Sturm [1] proved the following: let $\lambda\in R$ and consider the classes of $\lambda$-convex functions on $R^d$ on one side and the class of functions whose gradient flows are $\lambda$-contractive (they call these functions skew-convex). Then these classes coincide if and only if the norm $\|.\|$ (which is relevant for the notion of gradient flow) comes from a scalar product.
This shows that "convexity" and "contractivity" are really related only on Riemannian-like world, whereas on Finsler-like ones they are different concepts.
In particular, the answer in general metric spaces is no.
However, you may have better luck in spaces which are Hilbert-behaved on small scales (whatever this means). Judging by the other questions that you linked it seems that you are particularly interested in the Wasserstein space built over a Riemannian manifold. In this case, the paper by Otto-Westdickenberg [2] might be the reference you are looking for. They basically make the same argument Willie Wong gave above in $R^d$ but in $(P_2(M),W_2)$.
Inspired by [2], Daneri-Savare' [3] proved that if you interpret gradient flows in the $EVI_\lambda$ sense (a condition that implies $\lambda$-contractivity but that in practical situation one is often able to obtain if s/he is able to get contractivity - but beware of the results in [1]), then if a functional admits gradient flows in such sense, it must be $\lambda$-convex. This holds in arbitrary metric spaces.
[1] https://arxiv.org/abs/1009.2312
[2] http://www.instmath.rwth-aachen.de/~mwest/files/OttoWest.pdf
[3] https://arxiv.org/abs/0801.2455
EDIT
oh, small world indeed :) Let me add some comments:
one of the greatest intuitions in optimal transport is Otto’s interpretation of $(P_2(M),W_2)$ (here $M$ is a Riemannian manifold) as a sort of "infinite dimensional Riemannian manifold" (this answers your question). Quotation marks are needed because this is not really a Riemannian manifold in any reasonable sense (not even a Hilbert manifold), but still it resembles it a lot. This intuition drove the research on optimal transport (and in particular on gradient flows over the Wasserstein space) in the last 20 years, thus my advice is that if you are not familiar with this, try to build some intuition for it.
Yes, EVI is stronger than contractivity (as said above, it implies both contractivity and convexity). Now, the question "how the EVI formalism means Riemannian on small scale" is of course tricky as nobody knows what "Riemannian on small scale" means. Still, let me mention some results pointing in this direction:
[4] shows that if we have many EVIs in a Banach space then the space is Hilbert
[5] shows that gradient flows of convex functionals on CAT(k) spaces satisfy the EVI. Thus it goes the other way around by showing how to derive the EVI from some geometric property of the space, but it is interesting to see what is the key property used. They call it "commutativity of the distance" and it is a nice exercise to check that a Banach space is "commutative" in this sense if and only if it is Hilbert.
[4] https://www.sciencedirect.com/science/article/pii/S0022247X11006391
[5] https://www.researchgate.net/publication/267983274_Gradient_flows_and_a_Trotter--Kato_formula_of_semi-convex_functions_on_CAT1-spaces
Thanks Nicola, when I asked the question I was not aware of your reference [2]. I'm working on this issue with your coauthor L. Tamanini (small world!) One thing still puzzles, me, though: you said that "convexity" and "contractivity" are really related only on Riemannian-like world, or at least in situations which are Riemannian-like on small scales. But then how should I interpret the results of [2], which give precisely this type of equivalence? I really don't see how the EVI formalism means Riemannian on small scale. Is it just that EVI is much much stronger than $\lambda$-contractivity?
|
2025-03-21T14:48:30.454030
| 2020-04-30T13:48:22 |
358985
|
{
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"Alexey Ustinov",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/358985"
}
|
Stack Exchange
|
Proof of continued fraction identity of subfactorial
This question is part of a wider conjecture I have formed with someone which has its roots in Raayoni et al. (2019). The subfactorial function can be written as $$!n=\frac{n!}e+\frac{(-1)^n}{n+2-\dfrac1{n+3-\dfrac2{n+4-\cdots}}}$$ which is equation (17) of the MathWorld documentation of subfactorial.
Is there a proof of this identity that can be found in the literature or elsewhere?
It appears he has no contact details I can use.
Relevant paper https://www.emis.de/journals/JIS/VOL17/Balof/balof22.pdf
This is a particular case of a well know cfrac that can be found almost in any good textbook on continued fractions:
$$
{}_1F_1(1;c+1;z)=\sum_{k=0}^\infty\frac{z^{k}}{(c+1)_k}\\
=\cfrac{c}{c-z\,+}\,\cfrac{z}{c+1-z\,+}\,\cfrac{2z}{c+2-z\,+}\,\cfrac{3z}{c+3-z\,+}\,\ldots
$$
Just put $z=-1$, $c=n+1$.
If you don't like special functions then consider https://en.wikipedia.org/wiki/Euler%27s_continued_fraction_formula . The formula above follows by direct application of Euler's formula.
|
2025-03-21T14:48:30.454126
| 2020-04-30T15:04:05 |
358994
|
{
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"https://mathoverflow.net/users/123075",
"neverevernever"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/358994"
}
|
Stack Exchange
|
Count shortest path with different lengths in random graph
Let $G(n,p)$ be an Erdos-Renyi random graph on $n$ vertices with probability $p$, i.e. for each pair of vertices, they are connected directly by an undirected edge with probability $p$. Suppose we are interested in the regime where $p\geq C\log n/n$ where $C$ is some constant large enough. Thus the graph will be connected with high probability.
We also have that the diameter of the graph is roughly bounded by $\log n/\log(np)$, which means the longest shortest path between two vertices is bounded by this order.
There are $n(n-1)/2$ different shortest paths connecting two vertices. Let $N_l$ be the number of shortest paths with length $l$. Can we find a function $f_n(l)$ such that the following event holds with high probability:
$$\bigcap_{1\leq l\leq\frac{\log n}{\log(np)}}\{N_l\leq f_n(l)\}$$
A trivial example is $f_n(l)=n(n-1)/2$. Of course we want $f_n(l)$ to be as small as possible which can capture the dependence on $l$ and $n$ in terms of growth rate.
Maybe we should first try to calculate $\mathbb{E}(N_l)$?
@MattF. I think you meant the number of distinct pairs with shortest path length $l$? Yeah, I think this is a better definition for $N_l$.
|
2025-03-21T14:48:30.454354
| 2020-04-30T15:07:09 |
358995
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"authors": [
"Daniel Loughran",
"Jackson Morrow",
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"https://mathoverflow.net/users/56667"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/358995"
}
|
Stack Exchange
|
Criterion for generic polynomials
Generic polynomials, which are recalled below, play an important role in the constructive aspects of the inverse Galois problem.
Definition. Let $P(\mathbf{t},X)$ be a monic polynomial in $\mathbb{Q}(\textbf{t})[X]$ with $\textbf{t} = (t_1,\dots, t_n)$ and $X$ being indeterminates, and let $\mathbb{L}$ be the splitting field of $P(\textbf{t},X)$ over $\mathbb{Q}(\textbf{t})$. Suppose that:
$\mathbb{L}/\mathbb{Q}(\textbf{t})$ is Galois with Galois group $G$, and that
every $L/\mathbb{Q}$ with Galois group $G$ is the splitting field of a polynomial
$P(\mathbf{a},X)$ for some $\textbf{a} = (a_1,\dots, a_n) \in \mathbb{Q}^n$.
We say that $P(\textbf{t},X)$ parametrizes $G$-extensions of $\mathbb{Q}$ and $P(\textbf{t},X)$ is a parametric polynomial. The parametric polynomial $P(\textbf{t},X)$ is generic if $P(\textbf{t},X)$ is parametric for $G$-extensions over any field containing $\mathbb{Q}$.
Much of the literature on such polynomials concerns their existence and their constructions.
Question. Is there is a useful criterion for determining whether a monic polynomial $P(\mathbf{t},X)$ in $\mathbb{Q}(\textbf{t})[X]$ is generic or not? In particular, I have encountered the following polynomial in my research
$$
g_t(x) := x^3 + 147(t^2 + 13t + 49)x^2 + 147(t^2 + 13t + 49)(33t^2 + 637t + 2401)x + 49(t^2 + 13t + 49)(881t^4 + 38122t^3 + 525819t^2 + 3058874t + 5764801),
$$
and I would like to know whether this is a generic polynomial for $\mathbb{Z}/3\mathbb{Z}$-extensions. (For my purposes, it would be enough to know that this is a parametric polynomial for $\mathbb{Z}/3\mathbb{Z}$-extensions.) It is easy to verify condition (1) in the definition using Magma, but I do not know of a way to test condition (2).
Any references and/or suggestions are greatly appreciated!
If you have not done so already, I would recommend looking at Serre's "Topics in Galois Theory". In Section 1.1, he proves that the polynomial $x^3 - tx^2 + (t - 3)x + 1$ is generic for $\mathbb{Z}/3\mathbb{Z}$. Maybe his proof can be adapted to your case.
@DanielLoughran Thank you for the comment! Yes I have looked at Serre's notes, and the crux of his arguement seems to be that for $x^3 + tx^2 + (t-3)x + 1$, the function $t = (x^3 - 3x + 1)/(x^2 - x)$ is $\mathbb{Z}/3\mathbb{Z}$-invariant for the action of $\mathbb{Z}/3\mathbb{Z}$ on $\mathbb{P}^1$ given by $\sigma x = 1/(1-x)$. I don't see how to adapt his arguement to my setting as the coefficients in my polynomial are non-linear.
I believe there is no good way to determine in general if a polynomial $P(\mathbf{t},X)$ is generic. In fact, given a number field $K$ and a univariate polynomial $P(t,x) \in \mathbb{Q}[t,x]$, the problem of determining whether this is some $t \in \mathbb{Q}$ for which $P(t,x)$ has a root in $K$ is quite hard (and in some cases boils down to determining the $\mathbb{Q}$-points of a curve of genus greater than $1$).
The polynomial that you give is a generic $\mathbb{Z}/3\mathbb{Z}$-extension of $\mathbb{Q}(t)$
however. If you let $f(t,x) = x^{3} - tx^{2} + (t-3)x + 1$, be the example of a generic $\mathbb{Z}/3\mathbb{Z}$ extension given in the comments by Daniel Loughran, one can verify (using Magma for example) that your polynomial and $f(\frac{49}{t} + 8, x)$ define the same degree $3$ extension of $\mathbb{Q}(t)$.
This proves the claim, expect possible for the cyclic cubic extension obtained from $f(0,x) = x^{3} - 3x + 1$. But this cubic extension is also obtained from $f(-20,x)$.
EDIT: The OP asked for some info about how I found this. I did a fair amount of stumbling around to come up with $f(49/t + 8,x)$. First, I noticed that the discriminant of the maximal order of $\mathbb{Q}(t)[x]/(g_{t}(x))$ was $(t^{2} + 13t + 49)^{2}$, while the discriminant of the maximal order of $\mathbb{Q}(t)[x]/(f(t,x))$ was $(t^{2} - 3t + 9)^{2}$, and replacing $t$ with $t+8$ gives two polynomials with discriminants that are in the same square class. However, these do not define isomorphic extensions - specializing for lots of values of $t$ reveals that in most cases the discriminant of the ring of integers for $g_{t}(x)$ and that for $f(t+8,x)$ differ by a factor of $49$. This suggests that for a fixed $t$, if $K_{1}$ is the number field defined by $f(t+8,x)$ and $K_{2}$ is that defined by $g_{t}(x)$, then $K_{2} \subseteq K_{1}(e^{2 \pi i / 7} + e^{-2 \pi i /7})$ [note that $\mathbb{Q}(e^{2 \pi i /7} + e^{-2 \pi i / 7})$ is the unique cyclic extension with discriminant $49$]. One can verify that this is true. I suspected that there was a linear fractional transformation in $t$ that would take $f(t+8,x)$ to a polynomial defining one of the other cubic subfields of $K_{1}(e^{2 \pi i / 7} + e^{-2 \pi i /7})$. Looking at several specific cubic fields and searching for the values of $t$ (there are many) that result in $f(t+8,x)$ and $g_{t}(x)$ defining them leads to the suggestion that $t \to 49/t$ would work.
Thank you very much for your answer! Could you say a word about how you found that my above polynomial and $f(49/t + 8,x)$ define the same degree 3 extension of $\mathbb{Q}(t)$?
|
2025-03-21T14:48:30.454663
| 2020-04-30T15:28:45 |
358999
|
{
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"Ali Taghavi",
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|
Stack Exchange
|
For what kind of $C^*$ algebra $A$ every normal element $y\in A$ has a normal lift for every given surjective $C^*$ morphism $\phi:B\to A$
Is there a terminology for the following property of $C^*$ algebra $A$:
For every $C^*$ algebra $B$ and surjective $C^*$ morphism $\phi: B\to A$, every normal element $y\in A$ admits a normal element $b\in B$ with $\phi(b)=y$
I learned from $BDF$ theory that the Calkin algebra does not satisfy this property.
What is an example of an infinite dimensional algebra with this property? What is a simple infinite dimensional algebra with this property?Does matrix algebra satisfy that? Is this property preserved by tensor product?(any tensor product you whish to choose),
I doubt very much that any C*-algebra other than the complexes or matrix rings over them (or maybe finite products of them) satisfies this condition (assuming all algebras are unital). For example, take $B$ to be the C* algebra generated by the right (or left) shift on $l^2(N)$ ($N$ is the set of positive integers), and let $A = C(T)$ ($T$ = circle); we have an onto map $B \to A$, and the canonical generator $z$ cannot be lifted to a normal (since otherwise the shifts would be unitaries). Perhaps this type of example can be generalized.
@DavidHandelman in this example I think the index function is not identically 0(on the complement of essential spectrum). But is this property obvious for matrix algebra?
Lin showed that normal elements can be lifted from a quotient in at least one non-trivial situation. H.Lin, Approximation by normal elements with finite spectra in simple AF-algebras, J. Operator Theory, 31 (1994), 83-98.
Let $A$ be the algebra compact operators on separable Hilbert space. Then given a normal compact operator, one could map its spectrum homeomorphically into the real line, lift the resulting self-adjoint element to a self-adjoint element, and then undo the homeomorphism to get a normal lift of the original operator. So I think that the compacts give an infinite-dimensional simple example. I would suggest that $A$ must have the spectrum of every normal element homeomorphic to a subset of the reals.
@DouglasSomerset thank you very much for your comments and your attention to my question.
Every projective C*-algebra has this property: A is projective if and only if for every surjective *-homomorphism $\phi\colon B\to A$ there exists a -homomorphism (a split, or section) $\sigma\colon A\to B$ such that $\phi\circ\sigma=\mathrm{id}_A$. If $y\in A$ is normal, then $\sigma(y)$ is a normal lift. There are infinite-dimensional projective C-algebras, for example any free product $\ast_I C_0((0,1])$.
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2025-03-21T14:48:30.454862
| 2020-04-30T15:31:33 |
359000
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|
Stack Exchange
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Isomorphism of certain irreducible representations over finite fields
We are given a faithful representation of a cyclic group of order 5 $\rho: C_5=G \rightarrow End_{\mathbb{F}_3}(V) $ with $dim_{\mathbb{F}_3}V=8$ as vector space. It is also known that $V=U\oplus W$ as $\mathbb{F}_3[G]$-modules with dimension 4 as $\mathbb{F}_3$-vector spaces. Both $U$ and $W$ are irreducible representations, hence correspond to simple $\mathbb{F}_3[G]$-modules. We also know that $V$ posses a symplectic structure and that $G$ preserves the said structure, hence $G$ acts as a matrix $M\oplus (M^{-1})^T$. Furthermore, the set of their eigenvalues coincide.
Now my question is: are these two representations isomorphic? or equivalently are $\mathbb{F}_3[G]$-modules isomorphic?
Context $V$ comes as a $3$-torsion of a Jacobian of a curve of genus four and the symplectic structure comes from the Weil pairing.
The group $C_{5}$ has a (unique up to isomorphism) irreducible module over $\mathbb{F}_{3}$, so $U$ and $W$ are isomorphic, and $U \cong U^{\ast}$. This is because the isomorphism type of a module in this context is determined up to isomorphism by its Brauer character (we first do this over the algebraic closure, and use the fact that two representations over a given field which become equivalent over an extension field are equivalent over the original field). In the case of cyclic groups knowledge of the eigenvalues of a generator is equivalent to knowledge of the Brauer character.
@GeoffRobinson Thank you for your comment. What troubles me is the fact that, as vector spaces, both $U$ and $W$ are isomorphic to $\mathbb{F}{3^4}$ and are one dimensional $\mathbb{F}{3^4}$ representations (it contains the 5th roots of unity). We have four different 1-dimensional representations with different characters. Now can't $U$, $W$ correspond to different representations with different characters? All these representations have coinciding sets of eigenvalues, but this doesn't always imply the isomorphism. Or am I missing something?
No, the point is that if we extend the field to a bigger field $\mathbb{K}$ containing all $5$-th roots of unity, then $U \otimes_{\mathbb{F}} \mathbb{K}$ is a direct sum of four one dimensional $\mathbb{K}$ representations, where each of the four primitive $5$-th roots of unity represent a fixed generator of $C_{5}$ in the four respective one-dimensional representations.
I have one last issue that deals with a certain example/possibility: it is known that if $V$ is of dimension $2g$ for odd $g$ with symplectic str, $p=2g+1$ , $l$ is a prime and $l$ mod $p$ is $g$ and $\rho: C_p \rightarrow End(V)$. Then $V=U\oplus W$ with $U$ and $W$ non-isomorphic $\mathbb{F}_l$-modules both of dimension g. This result is known. Why your argument does not work in this case? I see that in this case, the eigenvalues do not coincide.
I am not sure if I understand you correctly, but here is a related example. Let $V$ be $10$-dimensional for $C_{11}$ over $\mathbb{F}{3}$ such that each generator of $C{5}$ has trace $-1$ on $V$. Then $V \cong U \otimes W$, where $U$ and $W$ are irreducible modules for $C_{11}$, but $U$ and $W$ are not isomorphic - they are dual representations. On one, the eigenvalues are $\omega^{r}$ where $\omega$ is a primitive $11$-th root of unity, and $r$ runs through (non-zero) quadratic residues (mod 11). On the other, the eigenvalues are of the form $\omega^{n}$, as $n$ runs through non-residues.
@GeoffRobinson sorry for the poorly worded comment -- and you understood my question correctly. I was not sure why in the last case of $V$ 10-dimensional $U, W$ are non-isomorphic, while in the first case they are. With the help of your last comment, I think I understand it now. Thank you so much for your comments and patience!
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2025-03-21T14:48:30.455153
| 2020-04-30T15:33:12 |
359001
|
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|
Stack Exchange
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What is the geometric description of the set of isomorphism class of $G$-torsors over a site $C$?
Let $X$ be a topological space and $G$ be a topological group. Let $\tilde{G}$ be the sheaf of groups defined by the sheaf of sections of the product $G$ bundle $ \pi_1:X\times G \rightarrow X $.
(1) Let $T\tilde{G}_X$ be the set of all isomorphism class of $\tilde{G}$-torsors over $X$.
(2) Let $H^1(X,\tilde{G})$ be the $\tilde{G}$ valued non-abelian cohomology of degree $1$ over the $X$.
(3) Let $BG_X$ be the set of all isomorphism class of Principal $G$ bundles over $X$.
Now it is well known that there exist one-one correspondence between (1) ,(2) and (3).
Now let we replace the topological space $X$ by a site $C$ then the definition of sheaf over a site is well known. Hence I am assuming that we can appropriately generalise $(1)$ and $(2)$ to the (1') and (2') as follows:(Though neither I have proved it personally nor I have seen anywhere where such generalisation is mentioned. )
(1') Let $T\tilde{G}_C$ be the set of all isomorphism class of $\tilde{G}$-torsors over $C$.
(2') Let $H^1(C,\tilde{G})$ be the $\tilde{G}$ valued non-abelian cohomology of degree $1$ over $C$.
Now my question is the following:
Is there any analogue of (3) if we replace the topological space $X$ by a site $C$?
I will also be very grateful if someone can suggest some literature in this direction.
Thank you.
What is your $G$ when you want to define notion of a $\widetilde{G}$-torsor over $\mathcal{C}$ when $\mathcal{C}$ is a site? Is it just a group object in the category $\mathcal{C}$? Do you want to give a reference from where you are borrowing your definitions 1,2 and 3?
@PraphullaKoushik I am actually not sure about what $G$ is when $C$ is a site. My reference for 1,2 and 3 is https://arxiv.org/abs/math/0212266
What distinction is being drawn between definitions (1) and (3)?
@WillSawin Sir, by (1) I mean a $\tilde{G}$-torsor i.e sheaf $S$ over a topological space $X$ with an action of a group valued sheaf $\tilde{G} $over $X$ and an existence of an open cover $\cup U_{\alpha}$ of $X$ such that each $S(U_{\alpha})$ is non empty. Now from a $\tilde{G}$ torsor we can produce an element of $H^1(X, \tilde{G})$ and from that element we can produce a principal $G$ bundle $\pi:P \rightarrow X$ where $P$ is a topological space obtained from the data given by the element of $H^1(X,G)$ and $\pi$ is a continuous function from $P$ to $X$. This $\pi:P \rightarrow X$ is my (3).
Sure, but (1) is just the sheaf of sections of (3). If you forget the topological space and pass to the site, these definitions are pretty much indistinguishable.
@WillSawin Yes I agree that (1) is the sheaf of section of (3). Actually this is my question " When we pass to site "Is indeed (1) and (3) are indistiguishable or there exist some notion of the form $\pi':P' \rightarrow X'$ where $P'$, $X'$ may be some categories and $\pi'$ is a functor. Also we may replace $G$ by Higher Groups like 2 groups or something of that sort.
@WillSawin Can you expand on what do you mean by “ forget the topological space and pass to the site”?
@WillSawin Sir, can you please give me little details about why do you think when we pass to site from topological space, (1) and (3) becomes indistinguishable? Also what do you mean by (3) when we pass to site? Unless I understand what is meant by (3) (in the context of site) I am not able to understand (3)=(1) in the context of site.
@PraphullaKoushik I just mean that there is not so much data about a map of spaces $Y \to X$ that can be expressed naturally in the language of the site of $X$ (rather than in the language of the space of $X$). The main data that would be preserved would be the sheaf of sections of $Y$, as well as the site of $Y$ and its map to $X$. But the second definition seems inappropriate for defining groups and torsors.
@WillSawin when you say site of $X$ you mean the category $\mathcal{O}(X)$ of open subsets of $X$ equipped with some Grothendieck topology or the topological stack $\underline{X}\rightarrow \text{Top}$ associated to the topological space $X$?
@PraphullaKoushik I don't know what the second one is so I guess I mean the first.
@WillSawin Sir, I have edited my question by writing some extra paragraphs at the end. It would be very helpful if you kindly go through my edits at your leisure.
"But the second definition seems inappropriate for defining groups and torsors." I do not completely understand this... Given a group $G$, one can consider the sheaf $\mathcal{G}$ on $X$, defined as $\mathcal{G}(U)={\text{smooth maps } U\rightarrow G}$. Then, there is a notion of $H^1(X,\mathcal{G})$.. The paper "Introduction to Language of stacks and gerbes" by Moerdijk says this set $H^1(X,\mathcal{G})$ is in one-one correspondence with the the set of isomorphism classes of $G$-torsors on $X$.. So, in what sense (2) is inappropriate here? @WillSawin
What do they mean when they say site? site of what? If on the site $(\mathcal{C},\mathcal{A})$ (here $\mathcal{C}$ is a category and $\mathcal{A}$ is a Grothendieck topology on $\mathcal{C}$) does not have any extra structure, why would they call a sheaf of groups on $(\mathcal{C},\mathcal{A})$ to be a "geometric" group?
I guess you are really asking if there is some "geometric" object $\mathcal C/G$ for any site $\mathcal C$? In this generality, I don't think the question even makes sense. The paper you link to are working with $\infty$-topos and there the question makes sense. In general in any geometric category, you can ask if there is some object that represents the functor in your defn (1).
@Asvin I am trying to understand your comment. But can you explain in little detail what do you mean by geometric category? Is it the same thing as mentioned in https://ncatlab.org/nlab/show/geometric+category? Also is there any precise definition of geometric object in a category?
@Asvin Also you mentioned "In general in any geometric category, you can ask if there is some object that represents the functor in your defn (1)". I did not get what did you mean by the "functor" in my defn (1).
I am just using geometric category roughly but it generalizes sets/topological spaces/schemes with zariski or etale topology /manifolds/infty spaces/whatever. The functor just assigns to a map T to X, the set of G torsors of T.
@Asvin. Ok. I have asked the later part of this question as a different question here https://mathoverflow.net/questions/359056/confusion-in-understanding-the-notion-of-g-principal-bundle-where-g-is-a-geo
@PraphullaKoushik The sheaf $\mathcal G$ is what I mean to call the first definition. I agree the first definition is very appropriate for defining torsors. The second definition is the site of $G$, which involves considering open subsets of the group $G$, or $G \times X$. I think this is not so helpful because (at least in AG) products of sites are not so well-behaved.
@WillSawin It looks like I need to read a little more.. Can you suggest some reference for "(at least in AG) products of sites are not so well-behaved."... It is expected to be not so well-behaved :D Just wanted to read some more...
@PraphullaKoushik I just mean that open sets on $X \times Y$ in the Zariski topology don't arise from the products of open sets on $X$ and open sets on $Y$, which you can find explained in any introduction to the Zariski topology.
@WillSawin Yes. That is the second exercise (I think) of Hartshorne's book :P
You should read Section 4.5 of Olsson's book Algebraic Spaces and Stacks.
The notion of a site is a piece of category theory with no intrinsic geometry, so it doesn't really make sense to ask for a geometric description of torsors for a general site.
However, in the concrete geometric contexts where site theory is typically applied, you can generalize definition 3). It will always imply definitions 1)-2) (which are always equivalent, essentially by the definition of Čech cohomology), but the converse becomes a non-trivial question about descent.
Let's assume you have some category $\mathscr{S}$ of spaces and to each object $X$ of $\mathscr{S}$, you attach a site $\mathrm{Op}(X)$ consisting of a certain full subcategory of $\mathscr{S}/X$ (e.g. the site of open subsets of a topological space, the site of étale maps into a scheme/algebraic space/DM stack, etc). Let's also assume that for a morphism $f \colon X \rightarrow Y$, the pullback map $U \mapsto f^{-1} U := U \times_Y X$ defines a continuous morphism of sites $f \colon \mathrm{Op}(X) \rightarrow \mathrm{Op}(Y)$, i.e. that if $U$ is an object of $\mathrm{Op}(X)$, then $f^{-1} U$ is an object of $\mathrm{Op}(Y)$ and that covers pull back to covers. (There are interesting contexts where this is not true, e.g. the crystalline or lisse-étale sites; in such cases, you need to be extremely careful!)
Moreover, assume that to any map of spaces $f \colon X \rightarrow Y$, the presheaf $h_X$ on $\mathrm{Op}(Y)$ defined by $U \mapsto \mathrm{Mor}_Y(U, X)$ is a sheaf, where $\mathrm{Mor}$ is the set of morphisms in $\mathscr{S}$. We say that $X$ represents the sheaf $h_X$ (note that $X$ might not be unique; the Yoneda lemma would only apply if $X$ is an object of $\mathrm{Op}(X)$).
If you want to state this abstractly, we're requiring that we have a fibered category over $\mathscr{S}$ with fiber $X \mapsto \mathrm{Op}(X)$, that this is a full subcategory of the natural fibration $X \mapsto \mathscr{S}/X$ with the same notion of pullbacks, and that this fibration satisfies the stack/descent condition for morphisms.
If $\mathcal{G}$ is a sheaf of groups on $\mathrm{Op}(X)$, the equivalent definitions 1) and 2) give a notion of when a sheaf $\mathcal{P}$ on $\mathrm{Op}(X)$ is a $\mathcal{G}$-torsor.
On the other hand, if $G$ is a group object in $\mathscr{S}/X$, we can make the following geometric notion of a $G$-torsor: a $G$-torsor is a map $P \rightarrow X$ in $\mathscr{S}$ with an action of $G$ given by a map $\rho \colon G \times_X P \rightarrow P$ (which is compatible with multiplication on $G$ in the sense that the evident diagrams commute) such that:
The map $(1, \rho) \colon G \times_X P \rightarrow P \times_X P$ is an isomorphism.
There is a covering $\{U_\alpha\}$ in $\mathrm{Op}(X)$ such that the map $P \times_X U_\alpha \rightarrow U_\alpha$ has a section (pulling back the isomorphism from point 1. along this section then gives an isomorphism $G \times_X U_\alpha \rightarrow P \times_X U_\alpha$).
Now, the sheaf $h_G$ is a sheaf of groups on $\mathrm{Op}(X)$, and the sheaf $h_P$ is a $h_G$-torsor in the sheaf-theoretic sense. Now, it makes sense to ask the following question:
If $G$ is a group object in $\mathscr{S}/X$ and $\mathscr{P}$ is an $h_G$-torsor, is there some $G$-torsor $P$ in $\mathscr{S}/X$ such that $\mathscr{P} = h_P$?
This is now a question of descent in $\mathscr{S}$.
Namely, since $\mathscr{P}$ is an $h_G$-torsor, we may find a covering $\{U_\alpha\}$ in $\mathrm{Op}(X)$ such that for each $\alpha$, we may choose a trivialization $h_G|_{U_\alpha} \simeq \mathscr{P}|_{U_\alpha}$. Therefore, $\mathscr{P}|_{U_\alpha}$ is represented by the trivial geometric $G|_{U_\alpha}$-torsor $P_\alpha = G|_{U_\alpha} \rightarrow U_\alpha$, with $G|_{U_\alpha}$-action given by left multiplication. The descent data for $\mathscr{P}$ gives us a Čech cocycle $(g_{\alpha \beta})$ with $g_{\alpha \beta} \in h_G(U_{\alpha, \beta}) = \mathrm{Mor}_X(U_{\alpha, \beta}, G)$, where $U_{\alpha, \beta} = U_\alpha \times_X U_\beta$. This is the same thing as a $G|_{U_{\alpha, \beta}}$-equivariant isomorphism $P_\alpha|_{U_{\alpha, \beta}} \rightarrow P_\beta|_{U_{\alpha, \beta}}$ in $\mathscr{S}/U_{\alpha, \beta}$. In particular, these isomorphisms satisfy the triple overlap condition because $(g_{\alpha \beta})$ is a cocycle.
If this descent datum is effective, then there is an object $P$ of $\mathscr{S}/X$ representing $\mathscr{P}$. This will always be a geometric $G$-torsor (note that this doesn't immediately follow from the Yoneda lemma, since $G$ and $P$ may not be objects of $\mathrm{Op}(X)$):
The action maps $\rho_\alpha \colon G|_{U_\alpha} \times_X P_\alpha \rightarrow P_\alpha$ glue to a map $\rho \colon G \times_X P \rightarrow P$: apply the fact that representable presheaves are sheaves to the open cover $\{G|_{U_\alpha} \times_X P_\alpha\}$ of $G \times P$. Moreover, the same argument shows that the map $(1, \rho) \colon G \times_X P \rightarrow P \times_X P$ is an isomorphism.
When you're dealing with topological spaces and open subsets, descent is always effective (in abstract terminology, the fibration $X \rightarrow (\mathscr{S}/X)$ is a stack). This is often not true in algebraic geometry!
For example, let's take $\mathscr{S}$ to be the category of schemes with the fppf topology. If $G \rightarrow X$ is affine, then we know that fppf descent is effective, and thus any sheaf-theoretic torsor $\mathscr{P}$ is represented by a geometric torsor $P \rightarrow X$. This is also true (it's a hard result of Raynaud) if $X$ is Dedekind and $G \rightarrow X$ is an abelian scheme, but it can fail in general. See this MO question and section III.4 of Milne's book Etale Cohomology.
It's a hard theorem of Artin (using the full force of his deformation-theoretic representation criteria for algebraic spaces) that fppf descent is effective for algebraic spaces, so we can in fact represent all sheaf-theoretic torsors for a group algebraic space $G$ by geometric torsors which are algebraic spaces. (See Tag 04SJ in the Stacks Project).
Edit Since you mention it in the question, I should add that this whole conversation should carry over essentially verbatim in a higher-categorical context (for example, you could replace the group $G$ by $BG$, and then talk about $G$-gerbes instead of torsors and look at cohomology in degree $2$). I'm not an expert in these things, but certainly Lurie discusses the matter comprehensively in Higher Topos Theory.
Thank you very much for the answer and references of the literature you mentioned. I will definitely go through those. I partially understood your argument. I am trying to understand it fully.
Thanks for adding the remark on higher category theory perspective of the question.
This is not a complete answer, too long for a comment.
If we start with an arbitrary site $\mathcal{C}$ and if we want to define the notion of a $G$-torsor over $\mathcal{C}$, then $G$ is not expected to be a group object in the category $\mathcal{C}$.
Observe that when $\mathcal{C}=\underline{X}$ for a topological space $X$, when you define $G$-torsor over the topological space $X$, the candidate $G$ was not a group object in the category $\underline{X}$. Instead, we have a fibered category $\underline{X}\rightarrow \text{Top}$ and $G$ is the group object in the category $\text{Top}$. The Grothendieck topology on $\text{Top}$ gives a Grothendieck topology on $\underline{X}$. This is the Grothendieck topology on $\underline{X}$ we are assuming when defining (the usual) notion of sheaf on the topolgical space $X$ or the shaef on the site $\underline{X}$.
So, if you want to imitate the notion of $G$-torsor to an arbitrary site $\mathcal{C}$, it is only reasonable to expect that there is a (fibered category ??) functor $\mathcal{C}\rightarrow \mathcal{D}$ for some $\mathcal{D}$ and $G$ is a group object in the category $\mathcal{D}$. Further, the Grothendieck topology that we have fixed on $\mathcal{C}$ is expected to come from a Grothendieck topology on $\mathcal{D}$.
For example, consider the case when $\mathcal{D}=\text{Man}$, the category of manifolds. Fix a Grothendieck topology on $\mathcal{D}$, say open cover topology. Let $\mathcal{C}$ be a differentiable stack; that is, $\mathcal{C}$ is a fibered category with the functor $\mathcal{C}\rightarrow \text{Man}$, satisfying certain special properties. Then, $\mathcal{C}$ can be made as a site. A cover $\{U_\alpha\rightarrow U\}$ is a cover for an object $U$ of $\mathcal{C}$ if, its image $\{F(U_\alpha)\rightarrow F(U)\}$ is a cover for the object $F(U)$ of $\text{Man}$. Then, fixing a group object in $\text{Man}$, that is a Lie group, we have the notion of a principal $G$-bundle over the site $\mathcal{C}$.
So, when $\mathcal{C}$ a category of special type, equipped with a functor $\mathcal{C}\rightarrow \text{Man}$ or $\mathcal{C}\rightarrow (\text{Sch}/S)$ and for a group object $G$ of $\text{Man}$ or $\text{Sch}/S$, we can define the notion of principal $G$ budnle over $\mathcal{C}$. It is not clear how one can define for sites which are not of this type.
The other two notions of $G$-torsors and $H^1(\mathcal{C},G)$ also require same assumptions as above.
References:
The notion of principal bundle over an algebraic stack; that is a special kind of fibered category $\mathcal{C}\rightarrow \text{Sch}/S$ can be found in section $1.2$ of Root stacks, principal bundles and connections.
The notion of principal bundle over an differentiable stack; that is a special kind of fibered category $\mathcal{C}\rightarrow \text{Man}$ can be found in section $4$ of Differentiable stacks and gerbes.
The notion of $G$-torsor over an algebraic space/algebraic stack can be found in Definition 04TY
Thanks. I am going through your answer(Long comment).
Can you please explain in details what do you mean by Principal $G$ bundle over a site $C$? Because in standard case from a $\tilde{G}-torsor$ over $X$i.e ( a sheaf $S$ over $X$ with (1) a free, transitive action of group valued sheaf $\tilde{G}$ over $X$ and (2) existence of an open cover $\cup U_{\alpha}$ of $X$ such that each $S(U_{\alpha})$ is non empty.) we get a Principal $G$ bundle over $X$ but not over $O(X)$. My question is what is the analogue of $X$ when $C$ is a site?
I have added the reference.. It is an abuse of notation to call a principal bundle over topological space $X$ as a principal bundle over the site $\mathcal{O}(X)$... “ My question is what is the analogue of when is a site?” I thought I answered that part.. $\mathcal{C}$ has to be the special catgeory that I have mentioned in the answer..
No no, I am not asking about the notion of Principal $G$ bundle over a stack. I am asking what do you mean by a Principal $G$ bundle over the site $C$ obtained from a $\tilde{G}-torsor$? I can understand what are you saying. But my point is when you consider the standard case [when your fibered category is $O(X) \rightarrow Top$] you get a principal $G$ bundle over $X$ but not $O(X).$ Then for a differentiable stack say $C \rightarrow Man$ by the (analogous procedure) why are you getting a Principal $G$ bundle over $C$?
All I can do is repeat what I said before :P... It is an abuse of notation to call a principal bundle over $X$ to be a principal bundle over the site $\mathcal{O}(X)$... Once you read the definition of principal bundle over a topological stack or Differentiable stack, you can realise that a principal bundle over $X$ actually gives a principal bundle over the stack $\mathcal{O}(X)$.. I have not said anything about the notion of $G$ torsor over a site $\mathcal{C}$.. So, “ what do you mean by a Principal bundle over the site C obtained from a G ~ torsor?” is not relevant..
I will try to see if I can find references for G torsor over Differentiable stack or algebraic stack and tell you..
Added reference for G torsor over an algebraic stack, that is a special kind of site.. I am seeing that definition just now, so, can’t comment more.. I will try to see the correspondence between G torsor and Principal G bundle over algebraic/Differentiable stack..
Ok. Thank you. But I think your answer and comments are in the context of $G$ torsor over a stack (considered as a special site) and Principal $G$ bundle over the stack... but I am expecting a correspondence(if possible) between a $G$ torsor over a site $C$ and Principal $G$ bundle over an object X of $D$ where $C \rightarrow D$ is the fibered category you mentioned in the answer. ( In special case for example the site $O(X)$ comes from an Object $X$ of Top and here the fibered category is $O(X) \rightarrow D$). Though I understand for an arbitrary stack such a special object may not exist.
Typographical error in the last line of my last comment . The correct version is the following :"Though I understand for an arbitrary "site" such a special object may not exist"
You did not had definition of a principal G bundle over a stack or of a G torsor over a stack. So, I gave definitions for that.. I am trying to see if I can get notion of G torsor over a stack from principal G bundle over a stack and vice versa... Till then, I request you to go through the definitions and see if there is any confusion in the definitions..
Ok. I will go through it. Thank you.
:D Stop saying thank you :P
I have edited the question by adding some extra paragraph at the end. Kindly please go through it.
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2025-03-21T14:48:30.456874
| 2020-04-30T15:49:27 |
359005
|
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|
Stack Exchange
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A non-singularity property for sets of real matrices
Let $M_N(\mathbb{R})$ be the ring of $N\times N$ real matrices. We say that a couple $(\mathcal{U},\mathcal{V})$, with $\mathcal{U},\mathcal{V}\subseteq M_N(\mathbb{R})$ is admissible if, for every $A\in\mathcal{U}$ and $B\in\mathcal{V}$, the complex matrix $A+iB$ is non-singular. Of course, if the pair $(\mathcal{U},\mathcal{V})$ is admissible then the same is $(\mathcal{V},\mathcal{U})$.
A trivial example of admissible pair is $(\{0\},GL_{N}(\mathbb{R}))$, where $GL_{N}(\mathbb{R}))$ are the non-singular real matrices. A less trivial example is $(\mathrm{Sym}_{N}(\mathbb{R}),\mathrm{Sym}^+_{N}(\mathbb{R}))$, where $\mathrm{Sym}_{N}(\mathbb{R})$ are the symmetric matrices and $\mathrm{Sym}^+_{N}(\mathbb{R})$ are the symmetric positive definite matrices.
I am particularly interested in admissible pairs which are maximal with respect to the order $$(\mathcal{U},\mathcal{V})\lesssim(\mathcal{U'},\mathcal{V'})\Leftrightarrow \mathcal{U}\subseteq\mathcal{U'},\,\mathcal{V}\subseteq\mathcal{V'}.$$
For example, one may ask if the pair $(\mathrm{Sym}_{N}(\mathbb{R}),\mathrm{Sym}^+_{N}(\mathbb{R}))$ is maximal.
Analogous notions of admissible pair and maximal admissible pair can be also formulated for more general structure, for example associative real algebras with unit.
Have been these objects already investigated?
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2025-03-21T14:48:30.456985
| 2020-04-30T17:21:59 |
359009
|
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|
Stack Exchange
|
Euler class and the real homological class of the fiber in an orientable sphere bundle
In the paper Foliations transverse to the fiber of a bundle, Plante considers the following example. Let $p:E\longrightarrow B$ a orientable fiber bundle with fiber $\mathbb{S}^k$. We have the Gysin cohomology sequence (real coefficients) associated
$$ H^{k}(B) {\longrightarrow}^{p^*} H^{k}(E) \longrightarrow H^{0}(B) \longrightarrow^\Psi H^{k+1}(B),$$
where $\Psi(1)$ is the Euler class of the bundle. Plante made the affirmation: the
fiber is null-homologous in $H_k(E)$ iff $p*$ is surjective which is true iff the Euler class is nonzero. By the exactness of the Gysin sequence, it's easy to see that $p^*$ is surjective if, and only if, the Euler class is nonzero. As the coefficients are the real numbers, we have that $p^*$ is the transpose of the map $p_*:H_k(E)\longrightarrow H_k(B)$ and, $p^*$ being surjective implies $p_*$ injective. Follows that $p^*$ being surjective implies that the class of the fiber is null in $H_k(E)$ since that is always true that $p_*[\mathbb{S}^k]=0$. I don't know how to complete the converse, to show that when the fiber is null homologous then the map $p_*$ is injective.
Ps. If the fiber is null homologous, then the map $p_*:\pi_k(E)\longrightarrow \pi_k(B)$ is an isomorphism (follows by the homotopy sequence of a fibration, page 376, Algebraic topology - Hatcher). I don't know if this says something about the map $p_*:H_k(E)\longrightarrow H_k(B)$.
|
2025-03-21T14:48:30.457132
| 2020-04-30T18:11:33 |
359013
|
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|
Stack Exchange
|
Homogeneous vector bundles on Abelian varieties
I recently encountered a result about vector bundles on Abelian varieties, which I found interesting. It characterizes homogeneous (translation invariant) vector bundles on Abelian varieties. More precisely any such vector bundle is in the form of $\bigoplus_L L\otimes U_L $, where $U_L$ is a unipotent vector bundle i.e. constructed by successive extensions of trivial line bundle and $L$ is an algebraically trivial line bundle. I wasn't able to find a stand-alone proof of this fact. All proves eventually refer to "M. Miyanishi, Some remarks on algebraic homogeneous vector bundles, in: Number theory, algebraic geometry and commutative algebra, 71–93, Kinokuniya, Tokyo,
1973.". It seems that this reference doesn't exist any more! (at least online). I'd appreciate if anyone knows where to find a proof of this fact.
What I was also interested was understanding the homogeneous sub-bundles of a homogeneous vector bundle. I was wondering whether the proof implies existence of any characterization of such sub-bundles or not? (like does it necessarily imply a homogeneous sub-bundle of $\bigoplus_{L\in A} L\otimes U_L $, is something like $\bigoplus_{L\in B \subseteq A} L\otimes U'_L $, where $U'_L$ is a unipotent sub-bundle of $U_L$ with unipotent quotient. )
I am pretty sure this result is in some papers of Indranil Biswas.
See Mukai's paper "Duality between $D(X)$ and $D(\hat{X})$ with its application to Picard sheaves".
I don't have access to Miyanishi's article either (at least during lockdown), but as Ulrich suggested, one can look at Mukai's paper "Duality between D(X) and $D(\hat X)$...", where he introduces the Fourier-Mukai transform. I'll denote this transform by $\mathcal{F}$. On page 159, Mukai gives a proof of the characterization of homogeneous bundles you mentioned. Before that in ex 2.9 and 3.2, he proves that
Theorem. $\mathcal{F}$ induces an equivalence between the category of homogenous bundles on an abelian variety $X$, and the category of coherent sheaves on the dual $\hat X$ with finite support. The unipotent bundles correspond to sheaves on $\hat X$ supported at the origin.
Given a homogenous bundle $V$, the points of the support of $\mathcal{F}(V)$ are the line bundles $L$ in your decomposition. The $U_L$ can also be recovered by taking inverse transform of the translate of $\mathcal{F}(V)_L$ back to the origin. Given all of this, it seems clear that homogenous subbundles are as you describe.
|
2025-03-21T14:48:30.457334
| 2020-04-30T18:15:54 |
359014
|
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|
Stack Exchange
|
When is the Radon-Nikodym derivative locally essentially bounded
Let $\mu\lll\nu$ be $\sigma$-finite Borel measures, which are not finite, on a topological space $X$. Under what conditions is $0<\operatorname{ess-supp}(\frac{d\mu}{d\nu}I_K)<\infty$ for every compact subset $\emptyset\subset K\subseteq X$.
In other words when is $\frac{d\mu}{d\nu} \in L^{\infty}_{\nu,\mathrm{loc}}(X)$?
My google ninja's skill got me the following: https://math.stackexchange.com/questions/3119941/what-are-sufficient-conditions-for-the-boundedness-of-a-radon-nikodym-derivative is this helpful?
@Alan It is pretty interesting, but I don't want to assume any such geometric structure. Btw, I like the ninja skills terminology I'll have to borrow it :)
I got this terminology from someone else in physicsforums, can't remember from whom though. :-D
What type of conditions do you need?
Something digestible to a non-analyst audience. This has come up as an example basically.
in fact $| \frac{d\mu}{d\nu}|{\infty,K}=\sup{H\subset K} \frac{\mu(H)}{\nu(H)}$
Let $$f:=\frac{d\mu}{d\nu}.$$ Then
$$f\in L^{\infty}_{\nu,loc}(X)\iff\text{$\forall$ compact $K\subseteq X$ $\exists$ $c_K\in(0,\infty)$ $\forall$ Borel $A$ we have $\mu(A\cap K)\le c_K\nu(A\cap K)$.}$$
Indeed, for the $\Rightarrow$ implication, take any compact $K\subseteq X$. Then $\exists$ $c_K\in(0,\infty)$ such that $f\le c_K$ $\nu$-a.e. on $K$. So, for any
Borel $A$ we have
$$\mu(A\cap K)=\int_{A\cap K}f\,d\nu\le c_K\nu(A\cap K),$$
as desired.
Vice versa, for the $\Leftarrow$ implication, take any compact $K\subseteq X$ and suppose that $\mu(A\cap K)\le c_K\nu(A\cap K)$ for some $c_K\in(0,\infty)$ and all Borel $A$. Let now $A:=f^{-1}((c_K,\infty))$, so that $f>c_K$ on $A$. Then
$$\mu(A\cap K)=\int_{A\cap K}f\,d\nu\ge c_K\nu(A\cap K),$$
and the latter inequality is strict (and hence contradicts condition $\mu(A\cap K)\le c_K\nu(A\cap K)$) if $\nu(A\cap K)>0$. So, $\nu(A\cap K)=0$, that is, $f\le c_K$ $\nu$-a.e. on $K$, as desired.
Similarly, for any compact $K\subseteq X$,
$$\operatorname{esssup}_Kf>0\iff \text{ $\exists$ $b_K\in(0,\infty)$ $\exists$ Borel $A$ such that $\mu(A\cap K)\ge b_K\nu(A\cap K)$.}$$
As a general question of interest, is this relation used/ does it have a name or is it a bit arbitrary?
@AnnieTheKatsu : I don't know of a name for this relation. It should be folklore, rather transparent from a standard proof of the Radon--Nikodym theorem; cf. the comment by Pietro Majer.
|
2025-03-21T14:48:30.457523
| 2020-04-30T18:43:48 |
359015
|
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|
Stack Exchange
|
Ergodic decomposition - how does restricting measure effect it? (Choquet Theory)
Suppose that $G$ is a discrete countable group and $\mu$ is an IRS (invariant random subgroup) of $G$: $\mu$ is conjugation invariant as a probability measure on the subgroups of $G$.
Since all the $G$-conjugation-invariant probability measures on the subgroups of $G$, IRS$(G)$,which is a convex, compact space, is a Choquet simplex, $\mu$ admits an ergodic decomposition: there is a unique measure $\upsilon$ on the extreme points of IRS$(G)$, or on the ergodic IRS's, such that for any continuous function $f: \text{Subgroups}(G) \to \mathbb{R}$ it holds $\int f d\mu = \int_{\text{ergodic IRS's}}[\int_{\text{Subgroups}(G)}fdm]d\upsilon(m)$.
Now suppose that there is a measurable $E \subset \text{Subgroups}(G)$ such that $\mu(E)=1$.
Define $E^* = \{\lambda \in \text{ergodic IRS's}: \lambda(E) = 1\}$
Can something be said about $\upsilon(E^*)$?
I'm attempting to show that $\upsilon(E^*) = 1$, in fact.
This may follow if we can show that, defining $\mu_o(A) = \mu(A\cap E)$, then the ergodic decomposition of $\mu_0$ is $\upsilon(A^* \cap E^*)$ for any measurable $A^* \subset$ Ergodic IRS's.
What you are asking about has nothing to do with the space of subgroups and is true for any measure class preserving action. One just has to write the definition of the ergodic composition in the measure category:
$$
\mu(E) = \int m(E) \,d v(m) \;.
$$
Can you please elaborate on why this equals $\upsilon(E^*)$? I guess I'm missing some background.
To reformulate: $\mu(E)=1$ iff $m(E)=1$ for $v$-a.e. $m$.
|
2025-03-21T14:48:30.457677
| 2020-04-30T19:01:41 |
359016
|
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|
Stack Exchange
|
Retracting to a bigger compact
Consider the topological spaces $X$ with the following property:
For every compact $K\subseteq X$ there is a compact set $L$ such that $K\subseteq L\subseteq X$ and $L$ is a retract of $X$.
Let us call this property RBC (Retract to a Bigger Compact). This property seems natural and useful, but I have been unable to find anything directly on it.
It is not hard to see that the following is true:
Proposition 1: Every closed convex topological subspace $C$ of a locally convex metrizable topological vector space $V$ has the RBC property.
Details on Proposition 1: Take any nonempty compact $K\subseteq C$. Then, by known results (note, in particular, Ref. [7] there), the closed convex hull $L:=\overline{\text{conv}\,K}$ of $K$ in $V$ is a retract of $V$. Since $L\subseteq C\subseteq V$, it follows that $L$ is a retract of $C$. Since $K$ is compact, $L$ is also compact, and of course $K\subseteq L\subseteq C$. Thus, $C$ has the RBC property. $\Box$
It is also clear that the RBC property is topological and thus invariant with respect to homeomorphisms. Trvially, any compact topological space has the RBC property.
Question: Can one characterize the RBC property?
That is, can one give a usable necessary and sufficient condition for it? Or a usable sufficient condition somewhat close to necessity? By "usable", I mean without the quantifier "there is" in the definition of the RBC property.
I do not know the answer even to this question: Is there an example of a topological space without the RBC property?
Thinking about the latter question, I have in mind the "non-retract" example of the $(n-1)$-sphere, which is compact but not a retract of the corresponding closed ball, whereas the ball is compact as well and of course is a retract of itself and of the corresponding $n$-space.
Update: The latter question has been answered in comments by erz, Anonymous, and Taras Banakh, who provided examples of topological spaces without the RBC property.
At this point, I would like to make the first question, to characterize the RBC property, more specific:
Specific question: Is it true that all Polish spaces have the RBC property?
My motivation for all these questions comes from probability. Indeed, any retraction $r$ from $X$ to a compact subset $K$ of $X$ naturally induces the truncation map $\xi\mapsto r\circ\xi$ of random elements $\xi$ of $X$, so that the truncated version $r\circ\xi$ of $\xi$ is a random element of the compact set $K$. Moreover, this truncation map is continuous: if $\xi_t\to\xi$ in distribution, then $r\circ\xi_t\to r\circ\xi$ in distribution.
Update 2: Following the latest comment by Anonymous, I have found a recent paper by Lipham, where on the first page one finds an example of a connected completely metrizable space which has no compact connected subset of cardinality greater than one, and therefore does not have the RBC property. By Mazurkiewicz's theorem, this space is also separable and hence Polish. Thus, it is not true that all Polish spaces have the RBC property.
Please ignore my previous comments, I was thinking of deformation retracts.
@Wojowu : I did not see your previous comments, but I think for me, with my total inexperience in this field, almost any comments could be useful.
My comments were saying that the infinite discrete space and the long line do not satisfy your stated property for deformation retracts, though they have pretty obvious retracts onto large compact sets. I'm afraid I don't really see a way to adapt these to the question at hand.
@IosifPinelis Apologies I've misread the question.
Concerning your second question: take a strongly rigid space $X$ (i.e. such that if $f:X\to X$ is continuous, then either $f$ is a constant map, or the identity. If $X$ is non-compact, and $f:X\to X$ has a relatively compact image, it must be a constant map. Hence, if you take $K$ to be a two-point set, there is no retraction on a bigger compact set containing $K$. The only remaining part is to find a non-compact strongly rigid space. So far everywhere I looked, the spaces are either compact, or it is not specified (and it is not easy to see) whether they are compact or not.
As for your first question, ANR's seem to have RBC (not sure if that is valuable)
For the second question, let X be a connected space such that no compact subset having at least two elements is connected. Then X obviously fails to have the RBC property.
@erz : Thank you for your comments, especially the first one, suggesting (to me) that non-RBC spaces are rare (if exist at all).
@Anonymous : Thank you for your comment. A trivial remark, though: If e.g. $X$ has $\le1$ element, then it obviously has your property: "$X$ is a connected space such that no compact subset having at least two elements is connected". However, then $X$, being compact, also obviously has the RBC property. So, you need your $X$ to be of cardinality $\ge2$. Then, to me, the question becomes this: Is there an example of a space of cardinality $\ge2$ having your property? (I apologize in advance if this question has an obvious answer.)
Knaster-Kuratowski fan is a connected subset of the plane such that removing a single point makes it totally disconnected. If it has a non-singleton connected compact subset, that subset would be compact, connected and such that removal of a single point makes it totally disconnected. That is impossible, and so, as @Anonymous suggests, the Knaster-Kuratowski fan is not RBC
Yes, I should have mentioned that I was considering infinite spaces. I think that spaces with the property that no compact subset with more than one point is connected are called punctiform spaces.
@Anonymous The Bernstein subset of the plane is connected but contains no infinite compact subsets, so fails to have RBC.
Yes, that is one of many known examples.
Thank you erz, Anonymous, and Taras Banakh for your very helpful comments. I have updated the post accordingly. Also, now I have the specific question: Is it true that all Polish spaces have the RBC property? Plus, I have now provided the motivation.
I don't have a reference handy, but I think that Mazurkiewicz gave an example of a non-trivial connected, completely metrizable subset of the plane such that no compact subset with more than one point is connected.
@Anonymous : Thank you very much for your latest comment as well. I have updated the post accordingly.
could you please elaborate on "Using certain known results, it is not hard to see that every closed topological subspace of a locally convex metrizable topological vector space has the RBC property."?
@erz : Sorry, I missed the convexity condition there. This is now corrected.
The proof that I thought i had that ANR's have CBR property turned out to be incorrect (i don't know if the claim is correct). Perhaps that could be a natural thing to ask? This class contains manifolds and is stable with respect to taking open subsets. Note that CBR is not stable with respect to taking neither open nor closed subsets.
@erz : Thank you for your continued interest. ANR certainly seems a more studied property than RBC, so that even a partial reduction of RBC to ANR may be helpful. However, ANR seems to be of the same logical structure as RBC: "$\exists$ a neighborhood ... such that $\exists$ a retraction ..." for ANR versus "$\exists$ a bigger compact than $K$ such that $\exists$ a retraction ..." for RBC. The two remaining $\exists$ quantifiers in the definition of ANR don't by themselves look entirely satisfying, as this leaves the job of finding the existing objects to the user.
I agree that on principle the definition of ANR is even more complicated than RBC. However, there is so much information about them available, and in particular a lot of "nice" spaces belong to that class.
I've made a question that specifically asks whether ANR implies CBR https://mathoverflow.net/questions/359384/spaces-that-are-comparable-with-their-compacts
|
2025-03-21T14:48:30.458267
| 2020-04-30T19:13:07 |
359018
|
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|
Stack Exchange
|
Holonomy groups of Hermitian, and hyper-Hermitian, manifolds
An $n$-dimensional complex manifold $M$, endowed with an Hermitian metric $g$, is Kähler if and only if the holonomy group of $g$ is contained in $U(n)$. If $g$ is Hermitian but not Kähler do we necessarily get a reduction of the holonomy group.
EDIT: To clarify, I mean holonomy group with respect to the Chern connection, i.e. the second connection specified by Robert below.
A manifold is called hypercomplex if it admits three integrable complex structures $I,J,$ and $K$ which together give a representation of the quaternions $\mathbb{H}$. Each complex structure will admit three Hermitian metrics. Can we conclude any reduction of their holonomy groups?
Any complex manifold is oriented, so a Hermitian metric must have holonomy in the special orthogonal group. I think that is the holonomy of the generic Hermitian metric on any complex manifold.
This is really a comment, but it's far too long to go into a comment window.
Your questions need to be made more precise.
First, if $(M,J)$ is complex and $g$ is a metric on $M$ that is $J$-Hermitian, there are (at least) two possible connections that you can associate with $(M,J,g)$ and the different connections can have different holonomies. There is the Levi-Civita connection $\nabla^g$ of $g$ (which may not have $\nabla^g J = 0$), and, second, there is a canonical connection $\nabla^{J,g}$ for which $J$ and $g$ are parallel but has torsion of type (1,1). These are equal iff $(M,J,g)$ is Kähler.
For which of these are you asking about the holonomy? (There is actually a whole line of connections joining the two when $(M,J,g)$ is not Kähler, as the space of connections on the tangent bundle is an affine space, so you could be asking about some connection that is a combination of these. Actually, there are even more than that associated to the triple $(M,J,g)$, but let's not go there.)
Second, I don't know what you mean, in the hypercomplex case, by the statement that "Each complex structure will admit three Hermitian metrics." Where are these metrics coming from? They can't naturally arise from any construction using the three anti-commuting complex structures alone. For example, consider $M= \mathbb{H}^n$ with its natural hypercomplex structure given by multiplication on the right by $i$, $j$, and $k$. The group $\mathrm{GL}(n,\mathbb{H})$ acts on the left (by matrix multiplication preserving the hypercomplex structure), but it does not preserve any metric.
I guess I just mean that there exists an Hermitian metric for each $I,J$, and $K$. Nothing more than this trivial observation.
@FofiKonstantopoulou: I see. Of course, there will also exist metrics that are Hermitian for all three.
|
2025-03-21T14:48:30.458529
| 2020-04-30T19:24:42 |
359019
|
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|
Stack Exchange
|
Convergence in $C_c$ but not in $C$
Let $C_c(\mathbb{R})$ be the set of compactly-supported continuous functions on $\mathbb{R}$. We can view this with a number of different topologies but I have my eye on two in particular. Let $X$ be $C_c(\mathbb{R})$ equipped with the inductive limit topology and let $Y$ be the same set with the compact-open topology.
What is an example of a convergent sequence in $Y$ which fails to converge in $X$?
Intuitively, such a sequence must exist since the map $x\to x$ is not continuous from $Y$ to $X$.
Small note: if you only work with sequences (rather than nets), it is not automatic that a discontinuous function can be detected by a sequence that converges in the source but not the target, unless you know some 'smallness' of your spaces (I don't know off hand if these spaces satisfy some condition of this form).
What you do mean?
https://en.wikipedia.org/wiki/Net_(mathematics)#Properties
Let $\phi \in C_c$ be nonzero (say for simplicity that $\phi$ is supported in $(0,1)$) and let $f_n(x) = \phi(x-n)$ be its translation to the left by $n$.
This sequence converges to 0 uniformly on compact sets (indeed, on any compact set it is eventually equal to 0). But if we think of the inductive limit topology as coming from the inclusion of the spaces $X_k = C_c((-k,k))$ into $C_c(\mathbb{R})$, then we see that $f_n$ does not converge in this topology because there is no single $X_k$ which contains all the $f_n$.
As Jochen says in the comment, it is a general fact about strict inductive limits that any convergent sequence must be contained in one of the $X_k$ (where technically to make this a strict inductive limit I ought to replace the $X_k$ with their closures). Apparently it's due to Dieudonné and Schwartz but I think the proof is a fairly simple exercise once you see how to do it.
But we can also prove it more directly in the current case. Consider the set $U = \{ f : |f(x)| \le 1/|x|\}$. Clearly $U \cap X_k$ is a neighborhood of 0 in $X_k$ (since it contains the uniform ball of radius $1/k$ centered at 0), so by definition of the inductive limit, $U$ is a neighborhood of 0 in $X$. But clearly there are infinitely many $f_n$ that are not in $U$, so $f_n$ does not converge to 0 in $X$. You can similarly show that the sequence can't converge to any other $f \in C_c$ either.
...because a theorem of Dieudonné and Schwartz says that in a strict induktive limit $X=\lim\limits_\to X_k$ of Fréchet spaces every convergent sequence is contained and converges in some "step" $X_k$.
Do you know where I can find this theorem?/ Have a reference?
|
2025-03-21T14:48:30.458774
| 2020-04-30T19:26:45 |
359021
|
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|
Stack Exchange
|
Preservation of $(C,1)$ summability of a function, when multiplied by a bounded strictly decreasing function
Let $f(x)$ be a $(C,1)$ summable function defined on $[1,∞)$:
$$ \lim_{R \rightarrow \infty} \int\limits_1^R \left(1−\frac{x}{R}\right)f(x)\,\mathrm{d}x= L, $$
where $L$ is a finite number.
Let $g(x)$ be a smooth, strictly decreasing function defined on $[1,∞)$, and satisfying the boundary conditions:
$$ g(1)=1, $$
$$ \lim_{x \rightarrow \infty} g(x) = 0. $$
QUESTION: Under the above conditions, is the function
$$f(x)g(x)$$
necessarily $(C,1)$ summable, or at least $(C,1)$ bounded?
Remark 1: G. H. Hardy proved that if $f(x)$
is as defined above, then for any $\delta > 0$ and $g(x)$ defined by
$$g(x)=x^{−\delta},$$
the function $f(x)g(x)$ is indeed $(C,1)$ summable.
Remark 2: If we replace the convergence of the $(C,1)$
integral by the convergence of the improper integral
$$ \lim_{R \rightarrow \infty} \int\limits_0^R f(x)\,\mathrm{d}x = L < \infty,$$
then it is a classical result that, under the conditions imposed above on $g(x)$, we have
$$ \lim_{R \rightarrow \infty} \int\limits_0^R f(x) g(x) \,\mathrm{d}x = M < \infty.$$
Is your $f$ nonnegative? If not, can we have $L=-\infty$?
No, $f$ is not nonnegative. I should have ruled out $L = -\infty$ too. I shall edit it accordingly.
If $f$ were nonnegative, then its $(C,1)$ summability would imply its integrability, and the question would then have an affirmative answer.
|
2025-03-21T14:48:30.459028
| 2020-04-30T19:31:45 |
359023
|
{
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"DSM",
"Jiro",
"Yemon Choi",
"https://mathoverflow.net/users/155380",
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|
Stack Exchange
|
Maximize $l_1$ norm with unitary matrix
Given an invertible matrix $A \in \mathbb{C}^{n\times n}$. How to find
$$
U^* = \max_{\text{$U$ with $U^H U = I$}} \lVert U A\rVert_1,
$$
where $\lVert\cdot\rVert_1$ is the entrywise 1-norm, i.e., $\lVert A\rVert_1 = \sum_{i,j} \lvert A_{ij}\rvert$ and $\cdot^H$ denotes the complex conjugate transpose?
For $A = I$, the solution is any complex Hadamard matrix, e.g., a scaled discrete Fourier matrix.
I'm not familiar with the notation $A^H$ -- is this what I would call "conjugate transpose"? If so, then isn't $A^H A$ always positive (semi)definite?
Also, when you say "find", are you asking for an algorithm to find the exact value, or would you still be interested in upper and lower bounds in terms of e.g. the singular values of A?
@YemonChoi Yes, conjugate transpose. And the positive definite was unnecessary, $A$ needs to be invertible. Yes, I'm mainly asking for an algorithm, but I'd be interested in bounds as well.
@LSpice Thanks for the edit
Too long to comment:
You might want to try out the following sub-optimal approaches, as the original problem is likely to be in NP-class (citations needed!). I am also going to assume that $A$ has only real entries, for simplicity.
1) Take a look at the paper "Orthogonalization of vectors with minimal adjustment" in Biometrika. The first optimality criterion might be a good surrogate for your cost function. Intuitively, it finds the an orthogonal basis which has least deviation (in terms of dot product) from the a given set of vectors (here columns of $A$).
2) You might also want to investigate the sub-optimality of $Q$ where $A=QR$ (QR decomposition). Changing the order in which Gram-Schmidt is done will yield different QR decompositions, and so you might also want to use this degree of freedom judiciously.
3) Lastly, the greedy way. Intuitively, suppose one were to find a unit vector $q_1$ such that $||q_1A||_1$ is maximized. Post this, one can find a unit vector maximizing $||q_2A||_1$, subject to the additional constraint $q_1q^\top_2=0$. This can be done till one obtains a orthonormal matrix. What remains now is:
$$
\max_{q\in R^{n\times 1}} ||qA||_1~\mbox{subject to}~qq^\top = 1 ~\&~ qC = 0.
$$
Intuitively, a unit vector has highest 1-norm if all its entries are equal in magnitude, or its alignment along on of the vectors $[\pm 1,\cdots, \pm 1]$ is maximum. This can be used in the following iterative routine. Choose any unit vector $q^{(0)}$ such that $q^{(0)}C=0$. Run the following convex routine iteratively till convergence (there is no guarantee of convergence):
$$
q^{(k+1)} = \arg \max_{qA\in R^{n\times 1}} q\left({\mbox{sign}(q^{(k)})}^\top\right) ~\mbox{subject to}~ qq^\top\leq 1 ~\&~ qC=0.
$$
(4) One could also try the following simpler idea. Note that
$\begin{bmatrix} I & X\\X^\top &I \end{bmatrix} \succeq 0 $ is a convex relaxation to orthonormal constraint (using Schur complement). And a the maximum of a linear cost functional with this relaxed constraint will always yeild an optima at its boundary, which is the set of orthonormal matrix. Using the same logic as in point (3), one can write the code below:
import numpy as np
import cvxpy as cvx
X = cvx.Variable((N,N))
Q,R = np.linalg.qr(np.random.randn(N,N))
Q = np.sign(Q@A)
count = 0
while(count<=20):
constraints = [cvx.vstack((cvx.hstack((np.eye(N),X)),cvx.hstack((X.T,np.eye(N))))) >> 0]
prob = cvx.Problem(cvx.Maximize(cvx.trace((X @ A)@(Q.T))), constraints)
prob.solve()
Q = np.sign((X.value)@A)
count = count + 1
print(np.sum(np.abs(X.value @ A)))
Z,R = np.linalg.qr(np.random.randn(N,N))
print(np.sum(np.abs(Z@A)))
Hope this helps.
Thanks. I'm going through the paper in 1). That's a side question for 2), but could $R$ of the QR decomposition to be the minimum 1-norm? Regarding 3), I'm missing the definition of $C$. It might not change the method, but restricting to real matrices might change the maximum, right (as for example Hadamard matrices do not exist of $n = 3$)?
Regarding 1) How about $A = I$? The orthogonalization would yield $U = I$, which is clearly not a good solution. The suggested surrogate only maximizes the diagonal elements of $UA$.
@SebastianSchlecht, thanks for the comment. (1) Would $R$ serve as a solution for minimum 1-norm, I have no reason to believe that. As I mentioned earlier, there are $n!$ ways of doing QR depending on the ordering for Gram-Schmidt; which one to choose needs investigation. (2) $C$ is made up by stacking row vectors obtained till that point of iteration of the greedy algorithm (not the one used to solve the inner opt-routine), (3) I agree with your observation for A=I, but not sure if that would as bad for a general A (its all sub-optimal afterall).
@SebastianSchlecht, had a few more thoughts for special cases, but just pointers. Would your idea of using DFT matrix for $A=I$ also work if $A$ were circulant? And secondly, in case you have $A=I+S$, where I is identity matrix and $S$ is skew-symmetric, would Cayley transform of $S$, given by $(I-S)(I+S)^\top$, be a decent solution?
Ok, I got now your argument with the QR. Unfortunately, I have trouble with the definition of $C$. Could you please be so kind and be more explicit there? Yes, I believe that the circulant matrix works in a similar manner.
@SebastianSchlecht, I think I am bit skeptical now about the third idea. I wrote a code to test convergence and the quality of the result. Unfortunately, neither convergence nor the quality of the result (as compared to QR) is good. I'll paste the code later, for whatever it's worth.
@SebastianSchlecht, the greedy algorithm is done iteratively and in each iteration, you find a new unit vector (or a column of Q in your notation). It is necessary for the new unit vector to be orthogonal to the vectors found till that point. This leads to the $qC=0$, where $C$ is the matrix formed using the unit vectors found till that point. Hope this clarifies the doubt.
@SebastianSchlecht, I recalled that I could have used Schur complement instead of all the jugglery of the greedy algorithm. Have included it as well.
Thank you very much for your added solutions and the attached code. I have tested the program code against a new solution I came up with. I posted it as a new answer. From some random numerical tests, it seems to perform better.
Here an alternative solution, which outperformed @DSM solution (4) in all tested cases.
Without loss of generality, we assume that $\lVert A \rVert_F = 1$. The optimal value for the $\ell_1$ norm is attained by the unitary Hadamard matrix $H$, e.g., DFT matrix.
As a proxy cost function, we use therefore
$$
\hat{U} = \min_U \| |U A| - |H| \|_F,
$$
where $|\cdot|$ is the element-wise absolute values. This is equivalent for an optimal set of phases $\hat{P}$ with $|\hat{P}_{ij}| = 1$ such that
$$
\hat{U} = \min_U \| U A - |H| \circ \hat{P} \|_F,
$$
where $\circ$ denotes the element-wise (Hadamard) product. This can be solved iteratively by unitary Procrustes solution such that
$$
U^{(i+1)} = \min_U \| U A - |H| \circ P^{(i)} \|_F \\
P^{(i+1)} = U^{(i+1)}A \oslash |U^{(i+1)}A|,
$$
where $\oslash$ is the element-wise (Hadamard) division. These iterations are guaranteed to converge:
Because of the Procrustes solution is the global minimum for the Frobenius norm, we have
$$
\| U^{(i+1)} A - |H| \circ P^{(i)} \|_F \leq \| U^{(i)} A - |H| \circ P^{(i)} \|_F
$$
Then, updating the phase also reduces the error
$$
\| U^{(i+1)} A - |H| \circ P^{(i+1)} \|_F \leq \| U^{(i+1)} A - |H| \circ P^{(i)} \|_F,
$$
which is essentially a element-wise version of
$$
\phi = \min_\theta \left(a e^{\imath \phi} - b e^{\imath \theta} \right)^2,
$$
where $a$, $b$, $\phi$, $\theta$ are real valued.
That's very nice, and thanks for posting your solution. Seems like your algorithm tends to move towards a Hadamard matrix. Why should this be the case for a general A? And, which Hadamard matrix would you choose to begin with (there are many of a given order)? I'd also like to learn about your proof of convergence/optimality for the Procrustes problem.
@DSM I added a normalization of $A$, I think, this should make the equal magnitude solution optimal. The Hadamard property is not used, it depends only on the initial phases $P^{(0)}$. I haven't found any dependency on the initial phase, thus the choice of which Hadamard matrix is not relevant. But, maybe you're right, I should replace the Hadamard just with the equal magnitude condition? I added more to the convergence, hope that helps.
|
2025-03-21T14:48:30.459622
| 2020-04-30T19:53:01 |
359024
|
{
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"Asaf Shachar",
"Connor Mooney",
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"url": "https://mathoverflow.net/questions/359024"
}
|
Stack Exchange
|
Is there a diffeomorphism of the disk with constant sum of singular values?
This question is a relaxed version of this question.
Let $D \subseteq \mathbb{R}^2$ be the closed unit disk, and let $c \ge 2$.
Does there exist a diffeomorphism $f:D \to D$ with constant sum of singular values $ \sigma_1(df)+\sigma_2(df)=c $?
The necessary condition $c \ge 2$ comes from the AM-GM inequality. For $c=2$ there is the identity map, so the question is really about $c>2$. If I am not mistaken this is an "additive" version of the Beltrami equation.
Some more context:
If we remove the origin, the answer is positive for every $c>2$; choose fixed numbers $\sigma_1,\sigma_2$ such that $\sigma_1+\sigma_2=c,\sigma_1\sigma_2=1$. Then, by the answer to this question, there exist a diffeomorphism $f:D\setminus \{0\} \to D \setminus \{0\}$ with the constant singular values $\sigma_1,\sigma_2$.
At the moment, however, I don't know how to construct such a diffeomorphism with everywhere constant singular values on the entire disk. I hope that by relaxing the requirement from fixed singular values, to a fixed sum, we will be able to use this added freedom to build a diffeomorphism of the whole disk.
The fact that I don't require $f$ to have constant Jacobian also means that we are in a different context from this previous question of mine.
A possible simplification:
Let $f$ be the map described in polar coordinates by
$$ \big(r,\theta\big )\mapsto \big(\psi(r),\theta+\phi(r)\big). \tag{1}$$
For such maps $f$, the PDE $ \sigma_1(df)+\sigma_2(df)=c $ reduces to the following ODE:
We have
$$
[df]_{\{ \frac{\partial}{\partial r},\frac{1}{r}\frac{\partial}{\partial \theta}\}}=\begin{pmatrix} \psi' & 0 \\\ \phi'\psi & \frac{\psi}{r}\end{pmatrix}.
$$
For a matrix $A=\begin{pmatrix} a & 0 \\\ b & e\end{pmatrix}$ with positive determinant, $$\sigma_1(A)+\sigma_2(A)=c \iff
c^2=\sigma_1(A)^2+\sigma_2(A)^2+2\sigma_1(A)\sigma_2(A)=|A|^2+2\det A.$$
So, $\sigma_1(A)+\sigma_2(A)=c \iff (a+e)^2+b^2=c^2.$
Thus, for $f$ given by the form $(1)$,
$ \sigma_1(df)+\sigma_2(df)=c $ if and only if
$$ (\psi'+\frac{\psi}{r})^2+(\phi'\psi)^2=c^2. \tag{2}$$
Take a map $f$ of the form that you propose, so that the equation reduces to the ODE
$$(\psi' + r^{-1}\psi)^2 + (\phi'\psi)^2 = c^2.$$
If we fix $c := 12/5 > 2$ and
$$\psi(r) := \frac{6}{5}r\left(1-\frac{1}{6}r^4\right),$$
the equation reduces further to
$$\phi'(r) = 2r(1-r^4/4)^{1/2}(1-r^4/6)^{-1}.$$
This has an analytic solution for $r < \sqrt{2}$ of the form
$$\phi(r) = r^2\left(1 + \sum_{k \geq 1} a_kr^{4k}\right).$$
The corresponding map
$$f(z) = \left(|z|^{-1}\psi(|z|) e^{i \phi(|z|)}\right) \cdot z = \frac{6}{5}z(1-|z|^4/6)e^{i |z|^2\left(1 + \sum_{k \geq 1} a_k|z|^{4k}\right)}$$
is an analytic diffeomorphism from $D$ onto itself and satisfies the desired equation.
(Alternatively, one can choose $\psi$ to be any concave, smooth increasing function with $\psi(0) = 0$ and $\psi(1) = 1$ such that $\psi$ is linear near the origin, and take $c = 2\psi'(0)$. Then $f$ is a just a dilation near the origin and we don't need to worry about singularities there).
Thank you! this is a very nice answer. Are you are implying that one can create such a diffeomorphism for any value of $c>2$? Can you please elaborate on the general scheme you have described in the last paragraph? I understand that taking $\psi$ to be linear near the origin, you get an $f$ that is a simple dilation near the origin, so there is no problem there. Now, given such $\psi$, how do you ensure that the ODE for $\phi(r)$ will be solvable up to $r=1$?...
It seems rather obvious, but I wonder whether there is something more that needs to be said here... Also, am I right that the reason for choosing $\psi$ concave is that we need to lower its derivative-since we start with derivative $\psi'(0)=\frac{c}{2}>1$ and we need to finish with $\psi(1)=1$, so we have to lower the speed. So, do you think that indeed any such $\psi$ would be OK?
Oh, after some further thought I think that I understand now. The concavity of $\psi$ is the crucial property which ensures that $\psi' + r^{-1}\psi $ remains below or equal $c^2$ at all times, so the ODE has a solution for as long as we want.
@Asaf: Yes, concavity implies that $c^2 - (\psi' + r^{-1}\psi)^2$ is nonnegative so we can solve for $\phi'$.
|
2025-03-21T14:48:30.459962
| 2020-04-30T21:16:48 |
359027
|
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"Willie Wong",
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|
Stack Exchange
|
Prove or disprove this integral of a function, defined on a countable set with infinite limit points, converges to zero
Edit: I got rid of my old definitions. Everything should be clear now
Since no one has answered my question on MSE, I’m hoping to get an answer here. I apologize if you dislike my writing. I am an undergraduate student and I don’t know whether this is a research question.
Definition
Consider a continuous $f:A\to[0,1]$ where $A\subseteq[0,1]$.
Edit: I did not use @Mathworker21's answer because I assumed it would not give the result I am looking for. Now I realize this is not the case. Here is his definition.
We can assume $A \subseteq [a,b]$, since that's where everything is happening. For ease, I'll have $[a,b] = [0,1]$. Based on the above discussion, I'll just have $P$ be defined on all of $[0,1]$ and continuous. Let $E_1 = [0,1], E_2 = [0,\frac{1}{2}], E_3 = [\frac{1}{2},1], E_4 = [0,\frac{1}{4}], E_5 = [\frac{1}{4},\frac{1}{2}], E_6 = [\frac{1}{2},\frac{3}{4}]$, $E_7 = [\frac{3}{4},1], E_8 = [0,\frac{1}{8}]$, etc.
If $A$ is finite, it's obvious how to define the average of $P$ (just do $\frac{1}{|A|}\sum_{x \in A} P(x)$). So, assume $A$ is infinite. Consider the sets $A\cap E_1, A\cap E_2, \dots$. Let $x_1$ be a point in the first nonempty one of these sets. Let $x_2$ be a point in the second nonempty of these sets, etc. Look at the measures $\delta_{x_1}, \frac{\delta_{x_1}+\delta_{x_2}}{2}, \dots, \frac{\delta_{x_1}+\dots+\delta_{x_N}}{N},\dots$. Since $[0,1]$ is a compact metric space, there is some probability measure $\mu$ on $[0,1]$ that is a weak* limit of some subsequence of these measures, i.e. there is some $(N_k)_k$ with $\frac{1}{N_k}\sum_{j=1}^{N_k} f(x_j) \to \int_0^1 f d\mu$ for each $f \in C([0,1])$.
We then define the average of $P$ over $A$ to be $\int_0^1 Pd\mu$.
For concreteness, suppose that $$f=\operatorname{id}$$ and
$$A=\left\{\frac{1}{2^x}+\frac{1}{2^y}+\frac{1}{2^z}:x,y,z\in\mathbb{Z}^{+}\right\}\cap[0,1]$$
My guess is, in this case, the average will converge to $0$. How do we prove whether I am right or wrong? Could we create a better definition that is easier to compute or gives an exact value?
Possible evidence that the average is zero
Here is a number line plot of all elements of set $A$
generateA[n_Integer] :=
Select[Union@
Flatten[Table[
1/2^x + 1/2^y + 1/2^z, {x, 1, n}, {y, x, n}, {z, y, n}]],
0 <= # <= 1 &]
NumberLinePlot[generateA[50], PlotStyle -> PointSize[0.003]]
It appears the points become "denser" near $0$. If the 'densest point' carries the most weight in the average, it is possible that the average should be $f(0)=0$.
Also by analysis, if we take the function inside $A$ and set $x$ as constant $a$ and $y$ as constant $b$, taking $z\to\infty$ we get
$$\left\{\frac{1}{2^a}+\frac{1}{2^b}:a,b\in\mathbb{Z}^{+}\right\}$$
This represents all the limit points. Since the limit points are "infinitely denser" than finite points, they should have infinitely greater weight for the average.
If we set $b\to\infty$ we get
$$\left\{\frac{1}{2^a}:a\in\mathbb{Z}^{+}\right\}$$
These represent second-order limit points "closely approximated" by first-order limit and finite points. The second-order limit points should have infinitely more weight than the lower order limit points.
As $a\to\infty$, we get $0$, the third order limit point and the densest point in the set. Third-order limit points should have infinitely more weight than second order, first order, and finite limit point.
From our analysis, it's possible, the average is $f(0)=0$.
Finally, consider the following code (from Wolfram Mathematica). This replicates $L(f,P)$ and $U(f,P)$ which I mentioned earlier:
partition[a_List, s_] := Module[{f, r}, f[{}, x_] := {x};
f[l_List, x_] := If[x - l[[1]] < s, Append[l, x], Sow[l]; {x}];
r = Reap[Fold[f, {}, a]];
Append[r[[2, 1]], r[[1]]]]
partition[{0, 1, 2, 7, 10, 11, 12}, 5]
(*{{0,1,2},{7,10,11},{12}}*)
calculate[p_, a_, s_] :=
Module[{parts = partition[a, s], n, inf, sup}, n = Length[parts];
inf = Total[Min[p /@ #] & /@ parts];
sup = Total[Max[p /@ #] & /@ parts];
{inf/n, sup/n}]
calculate[Identity, generateA[500], N[10^(-170)]] // N
I get:
{0.00598798, 0.00598798}
But it took a long time to compute. In fact, the person who gave me this answer doesn't think the sum converges to $0$ when $f=\operatorname{id}$? How do we prove otherwise?
Deleted Definition
Let $\require{enclose} \enclose{horizontalstrike}{A \subseteq [0,1]}$, and let $\require{enclose} \enclose{horizontalstrike}{P}$ be a partition of $\require{enclose} \enclose{horizontalstrike}{[0,1]}$ such that it is a finite set of sub-intervals $\require{enclose} \enclose{horizontalstrike}{X}$ with disjoint interiors and each subinterval has the same length. Define $\require{enclose} \enclose{horizontalstrike}{P' = \{ X\in P: X\cap A \neq \emptyset\}}$. Define $\require{enclose} \enclose{horizontalstrike}{n' = |P'|}$ (the cardinality of a finite set, or in this case, the number of sub-intervals whose intersection with $\require{enclose} \enclose{horizontalstrike}{A}$ is non-empty).
Calculate/define the following:
$$\require{enclose} \enclose{horizontalstrike}{L_{f,P} = \frac{1}{n^{\prime}} \sum_{X \in P^{\prime}} \bigg(\inf_{t \in X}f(t) \bigg)}$$
$$\require{enclose} \enclose{horizontalstrike}{U_{f,P} = \frac{1}{n^{\prime}} \sum_{X \in P^{\prime}} \bigg(\sup_{t \in X}f(t) \bigg)}$$
Define the limits under refinements of $P$ like so:
$$\require{enclose} \enclose{horizontalstrike}{L_f = \lim_{\|P\| \to 0}(L_{f,P})}$$
$$\require{enclose} \enclose{horizontalstrike}{U_f = \lim_{\|P\| \to 0}(U_{f,P})}$$
Where $\require{enclose} \enclose{horizontalstrike}{\|P\|=\sup_{X\in P}\|X\|}$.
$\require{enclose} \enclose{horizontalstrike}{L_f}$ is the 'lower average' of $\require{enclose} \enclose{horizontalstrike}{f}$ on $\require{enclose} \enclose{horizontalstrike}{[0,1]}$ (with respect to partition $\require{enclose} \enclose{horizontalstrike}{P}$). Likewise, $\require{enclose} \enclose{horizontalstrike}{U_f}$ is the 'upper average' of $\require{enclose} \enclose{horizontalstrike}{f}$ on $\require{enclose} \enclose{horizontalstrike}{[0,1]}$ with respect to partition $\require{enclose} \enclose{horizontalstrike}{P}$.
If these lower and upper averages limits converge to the same value (id est: are equal), we are given "my definition of average" of $f$. If they do not converge, then the average is undefined. Notice I define "upper" and "lower" averages to show when an average can not exist.
Note I describe $\require{enclose} \enclose{horizontalstrike}{L(f, P)}$ and $\require{enclose} \enclose{horizontalstrike}{U(f, P)}$ as "Riemman-like" because original Riemman-sums have upper and lower sums. However, this doesn't mean they are the same. Here we discard empty sub-interval with no points. This means my average could be anywhere between $\require{enclose} \enclose{horizontalstrike}{f(0)}$ and $\require{enclose} \enclose{horizontalstrike}{f(1)}$ depending on $\require{enclose} \enclose{horizontalstrike}{A}$, and possibly initial $\require{enclose} \enclose{horizontalstrike}{P}$.
You probably wanted to link to this question: How do prove the Darboux-like sum of a function, defined on a countable set with infinite limit points, converges to zero? You link goes to a picture instead: "Since no one has answered [my question][(https://i.sstatic.net/9oD9M.png) on MSE..." (BTW the question on [math.se] still has a bounty.)
@MartinSleziak Fixed it.
The generation procedure of the partition P is unclear to me. Suppose initially $A = { 1/6, 1/4, 3/4, 5/6}$. And you start with $n = 3$. You end up with the first and third intervals having 2 points each, and the middle interval with no point. Do you or do you not modify the middle interval? Is $n'$ here supposed to be $2$ or $3$?
If your answer is $n' = 2$, then it contradicts what you have written as your procedure. If your answer is $n' =3$, then your definition of the lower sum will involve taking the $\inf$ over an empty set.
@WillieWong You do not modify the middle interval so $n^{\prime}=3$. So we take the $\text{inf}$ over an empty set which is "undefined". According to one of the lines in my post, if $t_i$ is undefined, then $P(t_i)$ should be zero. How do I make this more clear in my post?
You should probably define what $P(t_i)$ means then. (I had no idea what that sentence was supposed to mean, until your comment.)
Can you be more precise what you mean by refinement of $P$? If you start with $m$ being a multiple of $n$ (so that the initial grid before merging is a refinement), is it obvious that the generated grid is a refinement? Specifically, suppose I start with $A = {0.0001, 1 - 0.0001}$. With $n = 3$ there is two possible $P$s, either ${[0,2/3], [2/3,1]}$ or ${[0,1/3], [1/3,1]}$. With $n = 6$, however, there are 5 possible $P$s, each containing two elements, most of which are not refinements of either of the previous partition.
Actually, here's a better example of the problem with the refinement process: let $A = {1/8, 3/8}$. Starting from $n = 2$ the partition generated with have $n' = 2$, dividing the interval at $1/2$. Starting from $n = 4$ the partition generated will also have $n' = 2$, but now the intervals are divided at $1/4$. With $n = 6$ you also get $n' = 2$ with dividing point at $1/3$.
@WillieWong You're right, the refinement $P$ does not always approach $\infty$, especially for finite $A$. What should I say instead?
Is the fact that you are doing things "Darboux-like" important? What if you just formulate it using the "Riemann" version? I believe for a fixed $A$, for each $n$, the number $n'$ is well-defined. And so it makes sense to talk about $U_{f,n}$ and $L_{f,n}$. And it also makes sense to talk about their $\limsup$ and $\liminf$ as $n\to \infty$. The Darboux formulation of integrability using Darboux sums and refinements has the benefit that under refinement, the upper sum is monotonically decreasing, and the lower sum is monotonically increasing, and so their limits are well-defined.
If there were any sort of monotonicity properties in your problem that makes the refinement language useful, you should expect that $L_{f,P}$ to be monotonically increasing. For the test case you have, however, your $f$ is non-negative on the interval, and hence $L_{f,P} \geq 0$ always. If it were to monotonically increase, and if your conjecture holds that it increases to limit 0, then you must have that $L_{f,P} \equiv 0$ for any $P$, which is obviously false. So I would advise forgetting about this whole refinement business.
@WillieWong I want cases where the average of $P$ is undefined. That is the upper and lower sum does not converge. Consider $P(x)=x^2$ on a countably dense set and $P(x)=x$ on another countable dense set. Both countable dense sets are disjoint. I want the average to be undefined. A darboux-like sum would show this, not Riemann. (Sorry my response took so long, my laptop is typing two letters each time I hit a key. I have to constatly delete a letter. sseeee wwhhaaaatt II mmeeaann. :((
It still works with Riemann: you can have an upper limit and a lower limit that do not agree with each other. (Pretty much this is what people mean when a function is not Riemann integrable.)
@WillieWong Is it possible for you to edit my post. My tutor and I will keep making careless mistakes.
@WillieWong How are the new edits?
It is not clear to me why the lower and upper limits exist. At the heart of my confusion is this process of "merging intervals". This destroys the monotonicity of the refinement process. Additionally, the "merging intervals" process is not well defined: if you have three intervals next to each other, with 1, 0, 1 point of A in the intervals respectively, do you merge the first two or the latter two?
@WillieWong The lower and upper sum exist, to define when an average can't be defined. In your example, when you "merge intervals", it could be the first two or the latter two. However, I prefer it's the first two.
For anyone who has more questions, speak in this chat.
@LSpice Why is my question unclear. I already have an answer.
For the concrete question you ask, the answer is yes.
Your set $A$ is in fact the set of all numbers between $0$ and $1$ with binary expansion that has no more than 3 bits equaling 1.
Let $\ell_+ = \lceil \log_2 n \rceil$ and $\ell_- = \lfloor \log_2 n \rfloor$. You can estimate
$$ n' \geq (\ell_- - 1) + \binom{\ell_- - 1}{2} + \binom{\ell_- - 1}{3} \approx \ell_-^3 $$
On the other hand, the sum inside either the upper/lower sum is between
$$ \sum_{i = 1}^{\ell_+} 2^{-i} + \sum_{i = 1}^{\ell_+} \sum_{j = i+1}^{\ell_+} (2^{-i} + 2^{-j}) + \sum_{i = 1}^{\ell_+} \sum_{j = i+1}^{\ell_+}\sum_{k = j+1}^{\ell_+} (2^{-i} + 2^{-j} + 2^{-k}) $$
(which is the sum of all numbers in $A$ with no more than 3 bits equaling 1 and least significant bit at least $2^{-\ell_+}$) and
$$ \sum_{i = 1}^{\ell_+} (2^{-i} + 2^{-\ell_-}) + \sum_{i = 1}^{\ell_+} \sum_{j = i+1}^{\ell_+} (2^{-i} + 2^{-j}+ 2^{-\ell_-}) + \sum_{i = 1}^{\ell_+} \sum_{j = i+1}^{\ell_+}\sum_{k = j+1}^{\ell_+} (2^{-i} + 2^{-j} + 2^{-k}+ 2^{-\ell_-}) $$
(which is the sum of the same set of numbers of the first sum, increased by the width of the interval).
The first sum is easily evaluated to be $O(\ell_+^2)$ (here we take advantage of the fact that at least one level of the summation converges using geometric series). The second sum is $O(\ell_+^2) + O(\ell_+^3 2^{-\ell_-})$.
So this means that both the upper sum and the lower sum behave like $\frac{C}{\log_2(n)} \to 0$ as $n \to +\infty$.
(This explains why your numerical computation goes very slow. At $n = 10^{-170}$, $\log_2(n)$ is something like 800. In fact, I don't even know if there is a reasonable numerical method for this, since to get convergence to the 10th decimal place, you will need a floating point arithmetic with a 2^30 bit mantissa to avoid numerical errors, or alternatively invent a new way to do exact arithmetic with rational numbers with very large numerators and denominators.)
Thank you. Do you have a better way of defining my sums (in general, not the example)? Perhaps you can edit my post. If I present my version of the definition, no one will show interest. My guess is the "highest-order limit points" will determine the average.
Can't help you there, because I have no idea why you even care about this sum. The procedure of "merging" is very strange. It would make more sense (to me) to define $n'$ from a given partition $P$ as "number of intervals in $P$ with non-empty intersection with $A$"; this makes the number a lot easier to count. Your definition on the other hand can take values between this one and twice that, and is a lot harder to count.
I can only answer this question because the expected sum is 0. If you look at $f(x) = 1 + x$, I don't know how even to prove either the upper or the lower sum has converging limits.
In my post, I state about first-order, second-order,...$n$-th order limit points. I want the $n$th order limit point to have "infinitely more weight" in the average then $n-1$th, $n-2$th...$1$st order limit points and finite points. I want the $n-1$-th order limit point to have "inifnitely more weight" then the $n-2$th, $n-3$th...first order limit points and finite points. The only way to do this is to create something similar to my definition. At the same time if we have multiple limit points of the highest order, the one where the lower-order limit points converges........
.........the slowest should have more weight depending on the rate of convergence compared to rate of convergence to the other highest order limit points.
As for your example, I expect the sums should converge to $1$, considering $A$ stays the same.
Why should it converge to 1? There may be a lot of empty intervals, where you require that the $\inf$ evaluates to 0. Up to half of the intervals can be empty in the worst case scenario. (Which is why in my comment I talked about looking at $n'$ being the number of intervals with non-zero intersection with $A$. In this case then the average of a constant function is the constant, and not something else.)
The merged sub-interval must be as long as possible without having more than one element in $A$. This means we can keep merging empty sub-intervals until the merged sub-interval until adding another sub-interval will make the merged sub-interval have more than one element in $A$, so we avoid adding it. In this case, the merged sub-intervals will always have one element.
Now consider $f(x)=x+1$ and $A=(0,1/3)\cup (2/3,1)$. As $s\to 0$, around $1/3$ of the sub-intervals, all existing in $[1/3,2/3]$, will be empty. We merge these sub-intervals. If we add another sub-interval it will have infinite more elements in $A$, which is greater than $1$. So we avoid adding another one.
There are fractal like constructions that allows very frequently half of the intervals to be empty, and not in a way you can merge. Let $B$ denote all infinite subsets of the square numbers {1, 4, 9, ...}. Let $A: { \exists b \in B: x = \sum_{i\in b} 2^{-i} }$. Then for any $s = 2^{-k^2 - 1}$, the corresponding $P$ has exactly half of its intervals containing infinitely many points, and half of its intervals exactly empty.
Incidentally, for this last example, every point in $A$ is an "infinite-order" limit point. If you define $n'$ to only range over all non-empty intervals, then the "average" process you define should be able to be expressed in terms of integrating $f$ against some singular continuous measure supported on the closure of $A$.
Here is another attempt. I am one step closer. :)
|
2025-03-21T14:48:30.461101
| 2020-04-30T22:34:01 |
359030
|
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|
Stack Exchange
|
Spherical space-form as the boundary of an Euclidean ball
Let $M^n$ be a smooth compact manifold such that the boundary $\partial M$ is diffeomorphic to a spherical space-form $S^{n-1}/\Gamma$, where $\Gamma \subset O(n)$ is a finite subgroup acting freely on $S^{n-1}$.
If the interior part $M-\partial M$ is diffeomorphic to $\mathbb R^n$, can we prove that $\Gamma$ must be trivial?
If $n=2$ of course not. If $n\ge 3$, yes. This is because $\mathbf{R}^n$ is simply connected at infinity, so whenever it's homeomorphic to the interior of a topological manifold with boundary, this boundary has a single component and is simply connected, excluding $\Gamma\neq 1$.
|
2025-03-21T14:48:30.461185
| 2020-04-30T23:03:58 |
359033
|
{
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"Gabe K",
"JHM",
"Sergey Egiev",
"Will Sawin",
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"https://mathoverflow.net/users/18060",
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"https://mathoverflow.net/users/33741",
"leo monsaingeon"
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"url": "https://mathoverflow.net/questions/359033"
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|
Stack Exchange
|
Optimal transport: find cost function given observed transport
Could you advise me please on what to read on the "inverse" problem: suppose I have a source measure, a target measure and I observe the solution to optimal transport problem -- can I "back out" the cost function e.g. find a cost function such that the observed transport is optimal conditional on this cost function?
Thanks.
Well, one probably needs further regularity requirements on the cost function, otherwise $c(x,y)=0$ for all $(x,y)\in spt(\pi)$ and $c(x,y)=+\infty$ otherwise trivially does the job ($\pi$ being the transference plan)
Thanks! Yes, I agree that the space of functions should be restricted (and may be with very stringent requirements to get uniqueness). I am looking for a systematic study of this.
Even with regularity enforced you should not expect uniqueness, at least not for a single "observation". For, you can always add a (smooth) non-negative function vanishing on $spt(\pi)$, at least when this support is smooth. So I guess it's more about requiring "structure", really (e.g. strict convexity, twist conditions, etc...) I'm sorry I can't help you more, I've never heard of such a problem.
To add another complication, in one dimension for any cost of the form $c(x,y)=h(x-y)$ for $h$ convex, the solution to optimal transport will be the identical, and is given by the so-called monotone map.
By Lagrange multipliers, such cost functions have the form $C(x,y) =a(x) +b(y)+ c(x,y)$ where $c$ is nonnegative everywhere and $0$ on the support of the transport map. If you're asking for functions that satisfy linear conditions and inequalities such as being a metric, or strict convexity as suggested by Leo, then finding a suitable cost function is a linear programming problem.
If both functions are supported on a finite space, and you demand that the cost function be a metric, it should suffice that the transport map has no loops where $A$ is sent to $B$, $B$ is sent to $C$, and $C$ is sent to $A$, or whatever. Indeed, in this case, the transport map induced a partial order, which you can complete to a total order, and then take the cost to be the number of steps in the total order
@Gabe K: good point!
It is strange question, for if the $c$-optimal coupling $\pi$ is observable, then are the dual Kantorovich potentials $-\phi=\psi^c$, $\psi^{cc}=\psi$ also observable? For we have always $-\phi(x)+\psi(y)\leq c(x,y)$ for all $x\in X$, $y\in Y$ with equality iff $x\in \partial^c \psi(y)$ iff $y\in \partial^c \phi(x)$ iff $(x,y)\in spt(\pi)$. Therefore if $\phi, \psi$ are observable, then you have pointwise lower bound for $c(x,y)$, namely $-\phi(x)+\psi(y)$ for arbitrary $x,y$. All the above answers show the problem is underdetermined, with many arbitrary parameters.
Look again at the inequality $-\phi(x)+\psi(y)\leq c(x,y)$ for all $x\in X$, $y\in Y$. If $h(x,y)$ is any function satisfying $-\phi(x)+\psi(y)\leq c(x,y)+h(x,y)$ for all $x,y$ and obtains equality iff $(x,y)\in spt(\pi)$, then the new cost $c':=c+h$ will again have $\pi$ as $c'$-optimal transport (... if i'm not mistaken).
See "Inverse Optimal Transport" By Stuart and Wolfram, SIAM J. App. Math, 80(1), 2020, and "Learning to Match via Inverse Optimal Transport" by Li et al., JMLR 2019.
|
2025-03-21T14:48:30.461417
| 2020-05-01T00:46:48 |
359038
|
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"dpistalo",
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|
Stack Exchange
|
Simple, explicit, functorial cylinder object in CDGA
In the model category of graded commutative dg-algebras CDGA over $\mathbb{Q}$ (with the projective model structure) there is a simple, functorial construction of a path object given by tensoring with the dga of polynomial differential forms on the interval. This construction comes from the simplicial model category structure on CDGA.
Is there is a similar, explicit construction of a cylinder object? If so, what is a good reference?
There is a cylinder construction for dg categories here: https://arxiv.org/abs/2109.03411. I guess one should decategorify and commutify it...
|
2025-03-21T14:48:30.461490
| 2020-05-01T01:53:21 |
359039
|
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"authors": [
"Chan Ki Fung",
"abx",
"https://mathoverflow.net/users/110093",
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/359039"
}
|
Stack Exchange
|
How big is the complement of stable locus $\operatorname{Bun}G$
Let $\Sigma$ be a smooth projective curve, and $G$ a reductive group. Let $\operatorname{Bun}G$ be the stack of principal $G$ bundles on $\Sigma$ (with a fixed topological type).
What is the codimension of the complement of the stable locus?
The complement is closed, so the codimension is $\geq 1$ if there exists a stable bundle. Is there any conditions that would guarantee the codimension is $\geq 2$?
For $G$ semi-simple simply connected, this is always the case. This is claimed (but not really proved) in Laszlo-Sorger, Ann. ENS 30, n°4 (1997), p. 523, (9.3).
@abx, it seems the quoted result says that strictly semistable bundles form a codimension $\geq 2$ subset. Do you know whether it is true the unstable bundles are also codimension $\geq 2$?
|
2025-03-21T14:48:30.461583
| 2020-05-01T03:05:26 |
359040
|
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|
Stack Exchange
|
Proving the existence of a symmetric Bayesian Nash equilibrium
I am currently faced with the following question:
Consider the public goods game. Suppose that there are $I > 2$ players and that
the public goods is supplied (with benefit of 1 for all players) only if at least $K$ players
contribute. The players' costs of contribution $\theta_1,...,\theta_I$ are private information, and
independently and uniformly distributed on $[0.5, 1.5]$.
Let $K=1$. Prove that there is a symmetric Bayesian Nash equilibrium, where
each player contributes if his cost is less than or equal to $c^*$ (same for all players) and
0.5 < $c^*$ < 1.
The probability of at least one player other than $i$ contributing = $c^∗−0.5$. If this happens, then the best response of player i is to not contribute since only one player is required to contribute for everyone to benefit
I am not sure how to proceed. I would appreciate your kind help! Thank you!
You may try the following approach: for every strategy $x$,
calculate the set of best responses of a player who faces $I-1$
players who all play $x$. Show that the set-valued function just
defined satisfies the conditions of your favorite fixed point
theorem, and show that the first point is a symmetric equilibrium.
You may want to restrict attention to some class of strategies,
like monotone strategies, and show that when $x$ is in this class,
there is always a best response in this class.
Good luck!
Got it! Thank you!
|
2025-03-21T14:48:30.461714
| 2020-05-01T03:16:02 |
359041
|
{
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"LSpice",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/359041"
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|
Stack Exchange
|
the fractional integration method of the proof of Stein-Tomas theorem?
In Schalg's Classical multilinear and Harmonic analysis, he presented two methods of the proof of Stein-Tomas theorem, one of which is called the fractional integration method. As a matter of fact, in order to prove
\begin{equation}
\lVert f * \hat\mu \rVert_{L^{p'}(\mathbb{R}^d)}\le C \lVert f \rVert_{L^p(\mathbb{R}^d)}, \quad \text{for } p=\frac{2d+2}{d+3}, \quad d\ge 3,
\end{equation}
where $\hat{\mu}\triangleq K$ is the Foureir transform of Lebesgue measure of surface with nonvanishing Gaussian curvature (and we may assume it is just the Fourier tranform of Lebesgue measure of unit sphere $\mathbb{S}^{d-1}$),
he tore the coordinates into two pieces $x=(x',t)$, where $x'=(x_1,...,x_{d-1})$, then
\begin{equation}
f*\hat{\mu} (x) =\int_{\mathbb{R}}\int_{\mathbb{R}^{d-1}} K(x'-y',t-s) f(y',s) dy'ds.
\end{equation}
Thus we may restrict our attention on the behavior of $K(x',t)$ with respect to $x'$. More precisely, if we assume that $(Ug)(x')= \int_{\mathbb{R}^{d-1}} K(x'-y',t)dt$, then Schlag claimed that $U(t)$ satifies
\begin{equation}
\lVert U(t) \rVert_{L^1(\mathbb{R}^{d-1}) \to L^\infty(\mathbb{R}^{d-1})} \le C |t|^{d-1}, \quad \lVert U(t) \rVert_{L^2(\mathbb{R}^{d-1}) \to L^2(\mathbb{R}^{d-1})} \le C <\infty,
\end{equation}
where $C>0$ is independent of $t\in \mathbb{R}$, and then we can use Riesz-Thorin interpolation theorem and then use Hardy-Littlewood-Sobolev inequality to get the desired estimates.
And my question is how to check the second estimate (i.e. the uniform bound of $L^2 \to L^2$), Schlag said it suffices to check that $K(\hat{\cdot},t) \in L_{\xi'}^\infty L_t^\infty ( \mathbb{R}^{d-1} \times \mathbb{R})$, where $K(\hat{\cdot},t)$ means the Fourier transform of $K(x',t)$ w.r.t. $x'$. For example, in $d=3$, then the Fourier transform of unit sphere can be represented by $\hat{\sigma}(x)=\frac{\sin{|x|}}{|x|}$ explicitly, but how can I check that
\begin{equation}
K(\xi',t)= \int_{\mathbb{R}^2} e^{-2\pi i x' \cdot \xi'}\frac{\sin{|(x',t)|}}{|(x',t)|} dx' \in L_{\xi'}^\infty L_t^\infty ( \mathbb{R}^{2} \times \mathbb{R}) \quad?
\end{equation}
I think 'of' doesn't belong in "the fractional integration of method".
Let's first clarify the definitions (also, there are some typos in your post, perhaps you should consider correcting them).
For $\xi\in\mathbb{R}^d$ we shall write $\xi=(\xi',\xi_d)$ with $\xi'\in\mathbb{R}^{d-1}$.
For a tempered distribution $T$ we shall denote its distributional Fourier transform by $\widehat{T}$. We will use the same symbol for distributions on $\mathbb{S}(\mathbb{R}^d)$ and $\mathbb{S}(\mathbb{R}^{d-1})$, it will be clear from the context.
We work with a surface of the form
$$
M=\{(x', \psi(x')): x'\in U\}
$$
for some bounded open set $U\subset\mathbb{R}^{d-1}$ (can think $M=\mathbb{S}^{d-1}$). The surface measure on $M$ is given for $f\in\mathbb{S}(\mathbb{R}^d)$ by
$$
\int_{\mathbb{R}^d}f(x)d\mu(x)=\int_{U}f(x', \psi(x'))\sqrt{1+|\nabla \psi(x')|^2}dx'.
$$
Note that $\sqrt{1+|\nabla \psi(x')|^2}\simeq 1$, which means that this factor is harmless.
We define
$$
K(\xi)=\widehat{\mu}(\xi),\qquad \xi\in\mathbb{R}^d.
$$
Next for a fixed $t\in\mathbb{R}$ we consider a locally integrable function $K_t$ on $\mathbb{R}^{d-1}$ given by
$$
K_t(\xi'):=K(\xi',t),\qquad \xi'\in\mathbb{R}^{d-1}.
$$
We $\textbf{shall show that}$ the distributional Fourier transform of $K_t$ coincides with an $L^\infty$ function on $\mathbb{R}^{d-1}$ which is bounded uniformly in $t\in\mathbb{R}$.
$\textbf{Solution:}$
Using the definition of a Fourier transform of a distribution and then applying Fubini's theorem, we get for $\varphi\in\mathbb{S}(\mathbb{R}^{d-1})$
\begin{align*}
\langle \widehat{K_t}, \varphi\rangle&=\langle K_t, \widehat{\varphi}\rangle=\int \widehat{\mu}(\xi',t)\widehat{\varphi}(\xi')d\xi'=\int_{\mathbb{R}^{d-1}}\int_{\mathbb{R}^{d}}e^{-2\pi i(x'\xi'+x_d t)}d\mu(x',x_d) \widehat{\varphi}(\xi')d\xi'\\
&=\int_{\mathbb{R}^{d}}e^{-2\pi i x_d t}\left(\int_{\mathbb{R}^{d-1}}e^{-2\pi i x'\xi'}\widehat{\varphi}(\xi')d\xi'\right)d\mu(x',x_d)\\
& =\int_{\mathbb{R}^{d}}e^{-2\pi i x_d t}\varphi(x')d\mu(x',x_d)=\int_U e^{-2\pi i \psi(x') t}\varphi(x')\sqrt{1+|\nabla \psi(x')|^2}dx'\\
&=:
\langle F_t, \varphi\rangle,
\end{align*}
where $F_t(x')=\chi_U(x')\sqrt{1+|\nabla \psi(x')|^2}e^{-2\pi i\psi(x') t}$. Clearly $F_t(x')\in L_{x'}^\infty L_t^\infty ( \mathbb{R}^{d-1} \times \mathbb{R})$, so the claim is proved.
thanks for you solution, but I have a question, why $\sqrt{1+|\nabla \psi (x')|}$ is harmless? For example, as for $M=\mathbb{S}^{d-1}$, then $\psi(x')=\sqrt{1-|x'|^2}$, where $x' \in B(0,1) \triangleq U$. But in this case, we can see that $\sqrt{1+|\nabla \psi(x')|^2}= \frac{1}{ \sqrt{1-|x'|^2} }$, which I think will lead to a singularity in a neighborhood of $|x'|=1$.
That's true, but notice that one can cover a sphere with finitely many charts without singularities, so it suffices to restrict attention to the part ${(x', \sqrt{1-|x'|^2}): x'\in\mathbb{R}^{d-1}$ and $|x'|<1/2 }$.
I get it, thank you very much.
You're welcome :]
|
2025-03-21T14:48:30.462287
| 2020-05-01T06:53:16 |
359051
|
{
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|
Stack Exchange
|
Local to global principle for a pair of bilinear equations?
Let $A_{i, j}, B_{i, j}, C, D \in \mathbb{Q}$, and consider the following pair of equations
$$
A_{1, 1} x_1 y_1 + A_{1, 2} x_1 y_2 + A_{2, 1} x_2 y_1 + A_{2, 2} x_2 y_2 = C
$$
$$
B_{1, 1} x_1 y_1 + B_{1, 2} x_1 y_2 + B_{2, 1} x_2 y_1 + B_{2, 2} x_2 y_2 = D.
$$
I was interested in figuring out if this system of equations satisfies the local to global principle. In other words, suppose I can find a solution $(\mathbf{x}, \mathbf{y})$ of the system in $\mathbb{R}^4$ and in $\mathbb{Q}^4_p$ for all primes $p$, then there exists a solution $(\mathbf{x}, \mathbf{y}) \in \mathbb{Q}^4$.
I would appreciate any comments or suggestions or counterexamples. Thank you very much!
Intersections of two quadrics in $\mathbb{P}^4$ are del Pezzo surfaces of degree 4. For those there are counter-examples to the Hasse principle going back to Birch and Swinnerton-Dyer. I don't know if any known example has this particular form. There are more knowledgable people active here who will give a better answer.
Actually, what the OP is looking at is the intersection of 2 planes with a smooth quadric $Q\subset \mathbb{P}^3$ -- that is, the intersection of a line with $Q$. This is an ordinary second degree equation, which certainly satisfies the local to global principle.
@abx, I had the same thought until I realised that C and D are not necessarily zero... the projective closure is a dP4 with four singular points at infinity.
May be I'm extremely stupid and missing something obvious, but jf we just pick $y$'s at random we get two linear equations in $x$. If the determinant is zero for all $y$'s then either equations are proportional in which case we have only one equation which always have solutions or both of our equations factor as $(ax_1+bx_2)(cy_1+dy_2)$ and moreover $a$ and $b$ are the same for them. But doing the same thing switching $x$ and $y$ shows that $c$ and $d$ are the same as well and we are back to the proportionality case. All in all solution either exists or not in all fields of characteristics $0$
@Martin Bright: Oops! You are right of course.
@AlekseiKulikov Could you please explain your comment in more detail please? I am trying to understand why if the determinant is $0$ for all $y$, then one obtains the mentioned conclusion. Thank you!
@JohnnyT. Determinant is $l_1(y)l_2(y) - l_3(y)l_4(y)$ for some linear forms $l_1, \ldots , l_4$. It is zero iff either $l_1 \sim l_3$ and $l_2\sim l_4$ or $l_2\sim l_3$ and $l_1\sim l_4$, where by $\sim$ I mean proportional. One of these cases give that equations are proportional as well, and the other is that in each equation we can factor something linear in $x$ (and moreover it is the same for both equations).
I think that Aleksei may have a point; The blow-up of $X$ in a rational point not on a line is a cubic surface with $2$ skew lines; these are all well-known to be rational, so it seems that $X$ is actually rational, rather than just unirational. This means that there should actually be a complete parametrisation of all the solutions, which I expect is what Aleksei's approach is leading to (writing down an explicit complete parametrisation).
Thank you for the many valuable comments. They were very helpful. And @AlekseiKulikov thank you for the clarification. It makes sense now and I am convinced that the answer to my question is affirmative.
In fact such system of equations always have a solution (at least if the coefficients are general).
Let $X$ denote the closure of your variety in $\mathbb{P}^4_{\mathbb{Q}}$. Explicitly:
\begin{align*}
&A_{1, 1} x_1 y_1 + A_{1, 2} x_1 y_2 + A_{2, 1} x_2 y_1 + A_{2, 2} x_2 y_2 = C z^2 \\
&B_{1, 1} x_1 y_1 + B_{1, 2} x_1 y_2 + B_{2, 1} x_2 y_1 + B_{2, 2} x_2 y_2 = D z^2.
\end{align*}
As remarked in comments, this defines a del Pezzo surface of degree $4$ with $4$ singular points at infinity (for general choices of coefficients).
This contains the $2$ lines at infinity
$$x_1=x_2=z=0, \quad y_1=y_2=z=0.$$
In particular $X$ contains a smooth rational point (choose a point on a line which is not one of the $4$ singular points). But it is well-known that a singular del Pezzo surface of degree $4$ with a smooth rational point is unirational (see e.g. [1, Theorem B]). Thus the rational points are Zariski dense, so there is a rational point with $z \neq 0$, which gives a solution to the original equations.
Here by "general", to apply [1, Theorem B] one needs to know that $X$ is irreducible with only finitely many singular points and is not a cone.
[1] Coray, Tsfasman - Arithmetic on singular del Pezzo surfaces.
|
2025-03-21T14:48:30.462603
| 2020-05-01T09:03:43 |
359056
|
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"Adittya Chaudhuri",
"Mike Shulman",
"https://mathoverflow.net/users/49",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/359056"
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|
Stack Exchange
|
Confusion in understanding the notion of $G$ Principal bundle where $G$ is a geometric group over a site
The first paragraph of the section Overview in the paper Principal infinity-bundles - General theory by Nikolaus, Schreiber and Stevenson https://arxiv.org/abs/1207.0248 precisely reads the following:
The concept of a G-principal bundle for a topological or Lie group G is fundamental in classical topology and differential geometry,. More generally, for G a geometric group in the sense of a sheaf of groups over some site, the notion of G- principal bundle or G-torsor is fundamental in topos theory. Its relevance rests in the fact that G-principal bundles constitute natural geometric representatives of cocycles in degree 1 nonabelian cohomology H1(−,G) and that general fiber bundles are associated to principal bundles.
My confusion is the following:
Since they are working with a sheaf of groups $G$ over a site(which they call here geometric group) so it is natural to ask what did they mean by $G$ Principal bundle or $G$-torsor in this context(Note neither they have mentioned about the "base space" of the $G$ Principal bundle nor about where $G$-torsors are defined(like over a topological space or a site?).
Though I am assuming $G$ torsors are defined over a site. But do they mean that the notion of $G$ principal bundles and $G$ torsors over a site are indistinguishable in their set up?
Or
Is there any separate meaning for $G$ Principal bundles for them?
Also they said $G$ Principal bundles constitute natural geometric representatives of cocycles in degree 1 nonabelian cohomology H1(−,G) . Note instead of writing $H^1(X,G)$ they wrote $H^1(-,G)$. So my guess is if $B$ is the base space of the $G$ Principal bundle then an isomorphism class of such $G$ principal bundle over $B$ will correspond to an element of $H^1(B,G)$.
Thank you.
Did you try reading beyond the first paragraph of the first section?
@MikeShulman Sir, I came across the notion of Principal 2 bundles from Bartels 2 bundle's (https://arxiv.org/abs/math/0410328) where a principal 2 bundle is defined within a category internal to some generalised smooth space and the structure group is a coherent 2-group as introduced in https://arxiv.org/abs/math/0307200. Baez, Schrieber in https://arxiv.org/abs/hep-th/0412325 introduced a notion of connection in it but restricted the base to dengenerate smooth 2 space and Loop spaces. Wockel in https://arxiv.org/abs/0803.3692 also mentioned about Principal 2 bundles.
@MikeShulman Though Wockel mentioned about the general Principal 2 bundles but his treatment was mostly on what they called semi strict principal 2 bundle over a manifold(considered as a degenerate smooth 2 space). Though I don't have much background in infinity category but in the paper https://arxiv.org/abs/0803.3692 from the 4th paragraph in the 1st section it seems that Principal infinity bundles is a generalisation of Principal 2 bundles when the structure group is something like infinity space(which they called geometric infinity groups).
@MikeShulman Now since I don't have have much background in infinity category I am not able to understand what geometric infinity group is!! I am also confused about the term geometric group . But it seems geometric 2 groups must be related to 2 groups (in the sense of https://arxiv.org/abs/math/0307200). Also since in the previous papers I mentioned about principal 2- bundles above they clear mentioned about their base (i.e some 2 space) but here I could not get what is the base of such Principal infinity bundles!! Is it some $(\infty,1)$ category?
@MikeShulman I am also right now not that comfortable with the language of the paper https://arxiv.org/abs/0803.3692 due to use of infinity category. But I am very curious about connecting a link of how my understanding of Principal 2 bundles (in the sense of papers by Baez, schrieber, Wockel I mentioned above) with this paper on principal infinity bundles which treats the concept of principal 2 bundles as a particular case!
@MikeShulman Sir, should I ask these things separately in a new question?
@MikeShulman In my 2nd comment there is a correction. The link I mentioned must be replaced by https://arxiv.org/abs/1207.0248 Thank you
I don't know what the point being made in of all your comments is; I was just saying that I think your original question is answered a few pages later in the paper.
@MikeShulman Thank you Sir. I will definitely look at the portion of the paper you mentioned. I am apologising for writing clumsily in the comments . I will write my doubts clearly and post it as a separate question and will mention the link here. Thank you.
|
2025-03-21T14:48:30.462895
| 2020-05-01T09:06:37 |
359057
|
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"Fedor Petrov",
"https://mathoverflow.net/users/153948",
"https://mathoverflow.net/users/4312",
"phantom"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/359057"
}
|
Stack Exchange
|
What graph's minimum vertex cover equals twice the maximum matching?
Matching: https://en.wikipedia.org/wiki/Matching_(graph_theory)
Vertex Cover: https://en.wikipedia.org/wiki/Vertex_cover
It is easy to see that
$$\texttt{minimum vertex cover} \leq 2 \texttt{ maximum matching}$$
I want to know that for what kind of graphs the equality is hold in the above inequality.
As an instance, $C_3$ is an example.
Answer. Such a graph $G$ is a disjoint union of odd complete graphs.
Obviously such graphs satisfy the equality $$\texttt{minimum vertex cover} = 2 \texttt{ maximum matching}.\quad (\star)$$
Assume that $G=(V,E)$ satisfies $(\star)$. Denote by $k$ the size of maximal independent set in $G$, then $$k=|V|-\texttt{minimum vertex cover}=|V|-2\cdot\texttt{maximum matching}=\\
\texttt{ minimum number of vertices not covered by a matching}.$$
On the other hand, by Tutte — Berge formula, if $k$ is the minimum number of vertices not covered by a matching, then there exists a subset $U\subset V$ such that $G-U$ has $|U|+k$ odd connected components. If $|U|>0$, then taking a vertex from each component we get an independent set with more than $k$ vertices. Therefore $U=\emptyset$, $G$ has $k$ odd components and if $G$ has also an even component, we again may take an independent set with more than $k$ vertices. Also if one of these connected components $C$ is not a complete graph, we may take to not-connected vertices in $C$ and a vertex from each other component, again having too large independent set. That is.
I don't understand why $G$ can't have a independent set larger than $k$.
@allfaker By definition of $k$.
I understand now, I forget the true ||= +
|
2025-03-21T14:48:30.463043
| 2020-05-01T09:40:42 |
359060
|
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"Mats Granvik",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/359060"
}
|
Stack Exchange
|
What is the asymptotic of the irregular blue curve? Is it $(8x)^{1/2}$ or is it something else?
From Terry Tao's post here there is the statement:
"Conversely, if one can somehow establish a bound of the form
$$\displaystyle \sum_{n \leq x} \Lambda(n) = x + O( x^{1/2+\epsilon} ) \tag{1}$$
for any fixed ${\epsilon}$, then the explicit formula can be used to..."
I don't know about the word "fixed", but the irregular behaviour of the blue curve below gives plenty of room for an ${\epsilon}$, if it is true that the asymptotic is $(8x)^{1/2}$, and if it is also true that it bounds the partial sums of the Möbius transform of the Harmonic numbers minus $x$. But we don't know and can't conclude any such bounds from this question. I am only asking about the asymptotics of a certain sum that is connected to / a truncated absolute value version of the numerators of the expansion of the primes.
Let:
$$\varphi^{-1}(n) = \sum_{d \mid n} \mu(d)d \tag{2}$$
Then for $n>1$:
$$\Lambda(n) = \sum\limits_{k=1}^{\infty}\frac{\varphi^{-1}(\gcd(n,k))}{k} \tag{3}$$
Form the table:
$$A(n,k)=\sum_{\substack{i=k\\\ n \geq k}}^n \varphi^{-1}(\gcd (i,k)) \tag{4}$$
From numerical evidence it appears that:
$$\sum _{k=1}^{x} \text{sgn}\left(\left(\text{sgn}\left(x+\sum _{j=2}^k -|A(x,j)|\right)+1\right)\right)+1 \sim (8x)^{1/2} \tag{5}$$
Is it true or is the asymptotic something else?
Question:
The complicated sign formula in $(5)$ comes from what we are really
doing which is to ask: What is the asymptotic of the least $k$ for which
the function $F(x)$:
$$F(x)=x+\sum _{j=2}^k -|A(x,j)| \tag{6}$$
is negative? For $k=1..x$.
Plot of the numerical evidence where the irregular blue curve is that least $k$ for which the function $F(x)$ is negative and thereby also the LHS of (5) while the smooth red curve is the conjectured asymptotic $(8x)^{1/2}$:
Efficient Mathematica program to generate the plot. Setting nn=10000 gives the plot above:
(*start*)
(*Mathematica*)
Clear[a, f, p];
nn = 1000;
p = 0;
f[n_] := n*Log[n]^p;
(*Clear[f];*)
(*f[n_] := n*Log[n]^4/(Pi*8)^2/8;*)
a[n_] := DivisorSum[n, MoebiusMu[#] # &];
Monitor[TableForm[
A = Accumulate[
Table[Table[If[n >= k, a[GCD[n, k]], 0], {k, 1, nn}], {n, 1,
nn}]]];, n]
TableForm[B = -Abs[A]];
Clear[A];
B[[All, 1]] = N[Table[f[n], {n, 1, nn}]];
TableForm[B];
TableForm[B1 = Sign[Transpose[Accumulate[Transpose[B]]]]];
Clear[B];
Quiet[Show[
ListLinePlot[
v = ReplaceAll[
Flatten[Table[First[Position[B1[[n]], -1]], {n, 1, nn}]],
First[{}] -> 1], PlotStyle -> Blue],
Plot[Sqrt[8*f[n]], {n, 1, nn}, PlotStyle -> {Red, Thick}],
ImageSize -> Large]]
ListLinePlot[v/Table[Sqrt[8*f[n]], {n, 1, nn}]]
(*end*)
Variant of the Mathematica program above: https://pastebin.com/GJ81MQez
Inefficient Mathematica program to generate the LHS in (5):
Clear[varphi];
nn = 20;
constant = 2*Sqrt[2];
varphi[n_] := Total[Divisors[n]*MoebiusMu[Divisors[n]]];
Monitor[TableForm[
A = Table[
Table[Sum[If[n >= k, varphi[GCD[i, k]], 0], {i, k, n}], {k, 1,
nn}], {n, 1, nn}]];, n]
Table[1 +
Sum[Sign[(1 + Sign[x + Sum[-Abs[A[[x, j]]], {j, 2, k}]])], {k, 1,
x}], {x, 1, nn}]
which starts:
{2, 3, 4, 5, 6, 5, 7, 7, 10, 7, 11, 10, 11, 10, 11, 11, 14, 13, 14, 13}
For my own memory to remember where to start editing tomorrow I write this Mathematica program:
Clear[varphi];
nn = 40;
varphi[n_] := Total[Divisors[n]*MoebiusMu[Divisors[n]]];
Table[1 +
Sum[Sign[(1 +
Sign[x +
Sum[-Abs[
Sum[If[x >= j, varphi[GCD[i, j]], 0], {i, j, x}]], {j, 2,
k}]])], {k, 1, x}], {x, 1, nn}]
There are previous efforts related to this question. Here is one of them.
A construction:
$$\sqrt{x} \log ^2(x)=\sqrt{x} \left(x-\left(\sqrt{x}-\log (x)\right) \left(\sqrt{x}+\log (x)\right)\right)$$
Proof of von Mangoldt function formula: https://math.stackexchange.com/a/51708/8530,
https://mathoverflow.net/a/162214/25104
Sqrt(8) in the OEIS: https://oeis.org/A010466
Could you say clearly if the "blue curve" represents $\sum_{n \leq x} \Lambda(n) -x$ or something else?
The irregular blue curve is not $\sum_{n \leq x} \Lambda(n) -x$. I will try to edit and clarify. It uses negated absolute values of the terms in the same expansion as $\sum_{n \leq x} \Lambda(n)$ though.
You don't have to give a complicated formula with signs, you can just say "Let $F(n)$ be the least $k$ such that .... is negative"
Related: https://or.stackexchange.com/a/681/671, https://math.stackexchange.com/a/3271124/8530
If one rewrites $(4)$ as:
$$A(n,k)=\sum_{i=1}^n \varphi^{-1}(\gcd (i,k))$$
Then one finds empirically that the mean of the $k$-th column in $A(n,k)$ is:
$$\lim_{n\to\infty} \frac{1}{n}\sum_{k=1}^n A(n,k) = -\frac{\varphi^{-1}(k)}{2} \tag{*}$$
and that the period length of column $k$ is $k$.
In https://oeis.org/A173557 on Jun 18 2020, Vaclav Kotesovec says that the Dirichlet generating function for: $$a(k)=|\varphi^{-1}(k)|$$
is:
$$\frac{\zeta(s)\zeta(s-1)}{\zeta(2s-2)} \prod_{p \text{ prime}} \left(1 - \frac{2}{(p + p^s)}\right)$$
and per email he explained to me that by choosing the residue at $s=2$ he finds that:
$$\sum_{k=1}^{n} a(k) \sim c \frac{n^2}{2}$$
or equivalently:
$$\sum_{k=1}^{n} |\varphi^{-1}(k)| \sim c \frac{n^2}{2} \tag{**}$$
where: $$c = A307868 = \prod_{p \text{ prime}} \left(1-\frac{2}{(p+p^2)}\right) = 0.471680613612997868...$$
In the question above we ask what is the least $k$ for which the function $F(n)$:
$$F(n)=n-\sum _{j=2}^k|A(n,j)|$$
becomes negative. Since the columns are periodic and the average of the periods in the $k$-th column is as said above: $-\frac{\varphi^{-1}(k)}{2}$, this is then approximately equal to asking what is the least $k$ for which the function $G(n)$:
$$G(n)=n-\sum _{j=2}^k\frac{|\varphi^{-1}(j)|}{2} \tag{***}$$
becomes negative.
Combining $(*)$ and $(**)$:
$$\sum _{j=1}^k\frac{|\varphi^{-1}(j)|}{2} \sim c\frac{k^2}{4}$$
Setting $j=2$ in the lower summation index:
$$\sum _{j=2}^k\frac{|\varphi^{-1}(j)|}{2} \sim c\frac{k^2}{4}-1$$
Inserting into $(***)$:
$$G(n)=n-\left(c\frac{k^2}{4}-1\right)$$
The least $k$ for which $G(n)$ becomes negative or changes sign is when $G(n)=0$. Solving:
$$c\frac{k^2}{4} - 1 = n$$ for $k$ gives the answer:
$$k(n) = \frac{2 \sqrt{n+1}}{\sqrt{c}} = \sqrt{8.4803146 (n+1)}$$
where as the conjectured asymptotic for the least $k$ when $F(n)$ becomes negative was:
$$k(n) \sim \frac{2 \sqrt{n}}{\sqrt{\frac{1}{2}}}=\sqrt{8n}$$
for comparison.
Clear[nn, t, n, k]; b = {1, -1}; nn = 1000; a[n_] := Total[MoebiusMu[Divisors[n]]*Divisors[n]]; Monitor[ g1 = ListLinePlot[ Accumulate[ Table[Sum[ If[Mod[k, 2] == 0, 1, b[[1 + Mod[n, 2]]]]*a[GCD[n, k]]/N[k], {k, 1, n}], {n, 1, nn}]]], n]; g2 = Plot[(Log[n] - Log[2])^2*Log[2] - Log[2]/2, {n, 1, nn}, PlotStyle -> Red]; Show[g1, g2]
https://mathoverflow.net/q/473639/25104
Let us denote the left hand side of $(1)$ by $\psi(x)$. It is known that $|\psi(x)-x|$ is not bounded by a constant times $x^{1/2}$. In fact Littlewood (1914) proved that
$$\psi(x)-x=\Omega_{\pm}(x^{1/2}\log\log\log x).$$
This is Theorem 15.11 in Montgomery-Vaughan: Multiplicative number theory I.
Setting:
$$F(x)=x\log(x)+\sum _{j=2}^k -|A(x,j)| \tag{1}$$
appears to give the asymptotic $\sqrt{8x\log(x)}$ for the least $k$ such that $F(x)$ is negative.
In general it appears that the least $k$ such that:
$$F(x)=f(x)+\sum _{j=2}^k -|A(x,j)| \tag{2}$$
is negative, has the asymptotic: $\sqrt{8f(x)}$.
See the Mathematica program in the question, by setting p=1.
Clear[nn, t, n, k]; nn = 1000; a[n_] := Total[MoebiusMu[Divisors[n]]*Divisors[n]]; Monitor[g1 = ListLinePlot[ Accumulate[ Table[Sum[ If[Mod[k, 2] == 0, 1, (-1)^n]*a[GCD[n, k]]/N[k], {k, 1, n}], {n, 1, nn}]]], n]; g2 = Plot[(Log[n] - Log[2])^2/(1.45), {n, 1, nn}, PlotStyle -> Red]; Show[g1, g2, PlotRange -> {0, 30}]
Monitor[g1 = ListLinePlot[ Accumulate[ Table[Sum[ If[Mod[k, 2] == 0, 1, (-1)^n]*a[GCD[n - k, k + n]]/N[k], {k, 1, n}], {n, 1, nn}]]], n]
|
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