added
stringdate 2025-03-12 15:57:16
2025-03-21 13:32:23
| created
timestamp[us]date 2008-09-06 22:17:14
2024-12-31 23:58:17
| id
stringlengths 1
7
| metadata
dict | source
stringclasses 1
value | text
stringlengths 59
10.4M
|
|---|---|---|---|---|---|
2025-03-21T14:48:31.561928
| 2020-07-20T16:18:57 |
366120
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Andreas Blass",
"Asaf Karagila",
"David Roberts",
"Rachid Atmai",
"https://mathoverflow.net/users/3859",
"https://mathoverflow.net/users/4177",
"https://mathoverflow.net/users/6794",
"https://mathoverflow.net/users/7206"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631278",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366120"
}
|
Stack Exchange
|
Specific notions of forcing from the point of view of category theory
I'm trying to learn about the topos of sheaves and the double negation topology to try to go through the independence of CH from a categorical perspective. I'm curious in general about what the standard set theoretic notions of forcing would look like in category theory.
For example: what corresponds to Prikry forcing in the topos of sheaves point of view? Is there a categorical interpretation of the Levy-Solovay theorem about large cardinals being unaffected by small forcing? Is this known or has it been investigated somewhere? Thanks in advance for any references or answers/comments.
Shouldn't the double-negation topology be called "the no-no topology"? :-)
The topos version of forcing with a poset $P$ regards $P$ as a category, forms the topos of presheaves on it, and then passes to the subtopos of double-negation sheaves. The presheaf topos amounts to strong forcing (which obeys only intuitionistic logic) and the sheaf subtopos amounts to the much more widely used weak forcing. This was, at least for the Cohen models violating CH, in the early work of Lawvere and Tierney, presented in Lawvere's paper "Quantifiers and sheaves" (Proceedings of the 1970 ICM in Nice. Big pdf (see p329), individual article). If I remember correctly, Marta Bunge worked out the topos version of the forcing that adjoins a Suslin tree (JPAA 1974, doi:10.1016/0022-4049(74)90020-6, see also the erratum).
More information about these topics is in work of John Bell (connecting it with Boolean-valued models, if I remember correctly) and in Peter Freyd's paper "All topoi are localic" (JPAA 1987, doi:10.1016/0022-4049(87)90042-9). Andre Scedrov and I included much of this material in the early, background sections of "Freyd's models for the axiom of choice" (Memoirs A.M.S. 404 (1989) --- this was before "404" became a widely known synonym for "file not found").
Thank you Andreas for your answer. Is there anything known about what Prikry's notion of forcing looks like as a category? More specifically: is there a suitable definition of measurable cardinal and singularizing sequence in category theory?
I would add the "subtopos of double-negation sheaves" can be described directly as sheaves on the poset taken as a site with the dense topology.
@DavidRoberts Right. Where topos folks say "double-negation sheaves", set folks say "every extension has an extension" (as if it were a single word because we've said it so often).
Here's Tierney's paper on his joint work with Lawvere: Tierney M. (1972) Sheaf theory and the continuum hypothesis. In: Lawvere F.W. (eds) Toposes, Algebraic Geometry and Logic. Lecture Notes in Mathematics, vol 274https://doi.org/10.1007/BFb0073963.
I think the mentioned work of Bell is all in Toposes and Local Set Theories, published by Dover, and with some lectures summarising the ideas published as Categories, toposes and sets. Synthese 51, 293–337 (1982). https://doi.org/10.1007/BF00485258, though there is a scanning error in this pdf that lost four pages.
|
2025-03-21T14:48:31.562193
| 2020-07-20T17:26:59 |
366124
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Fedor Petrov",
"Gjergji Zaimi",
"Per Alexandersson",
"https://mathoverflow.net/users/1056",
"https://mathoverflow.net/users/2384",
"https://mathoverflow.net/users/4312",
"https://mathoverflow.net/users/83671",
"thedude"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631279",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366124"
}
|
Stack Exchange
|
On the value of a skew Schur function at the identity
(Asked in MSE but got no response.)
The generating function $\frac{1}{(1-t)^N}=\sum_k {N+k-1\choose k}t^k=\sum_k h_k(1)t^k$ and the Jacobi–Trudi formula $s_{\lambda/\mu}=\det(h_{\lambda_i-i-\mu_j+j})$ tell me that the value of the skew Schur function at the identity is
$$ s_{\lambda/\mu}(1_N)=\det\left({N+\lambda_i-i-\mu_j+j-1\choose \lambda_i-i-\mu_j+j}\right).\tag{1}\label{1}$$
However, I was reading a paper by Chen and Stanley (A Formula for the Specialization of Skew Schur Functions) and they state that
$$s_{\lambda/\mu}(1,q,q^2,\dotsc)=\frac{1}{\prod_{u\in\lambda/\mu}[N+c(u)]_q}\det\left(\genfrac[]{0pt}{}{N+\lambda_i-i}{\lambda_i-i-\mu_j+j}_q\right),\tag{2}\label{2}$$
where $c(u)$ is the content of the box $u$ in the Young diagram of shape $\lambda/\mu$ and the $q$-quantities are $[x]_q=1-q^x$, $[a]_q!=[a]_q[a-1]_q\cdots$ and $\genfrac[]{0pt}{}a b_q=\frac{[a]_q!}{[b]_q![a-b]_q!}$.
I am not an expert in this $q$-business, and I am confused by this equation. I have a few closely related questions.
Since the left hand side of \eqref{2} is a polynomial in $q$, it should have a limit when $q\to 1$ and this should be the skew Schur at the identity. Is this correct? But what is the number of arguments?
The determinant at the right hand side on \eqref{2} has a limit when $q\to 1$, but the prefactor does not. How to take the limit $q\to 1$ of this equation?
How to obtain equation \eqref{1} from equation \eqref{2}?
The left hand side of (2) is not a polynomial but rather a power series. Don't confuse the full principal evaluation at infinitely many variables $1,q,q^2,\dots$ with the truncated principal evaluation $1,q,q^2,\dots,q^n,0,0,\dots $. With this in mind, it doesn't make sense to directly take a limit of (2) as $q\to 1$. If you want a framework to deal with both identities look at exercise 7.102 and its solution in Stanley's EC2. In particular the reference to the Jacobi-Trudi formula for flag Schur functions which generalizes both (1) and (2).
Usually $[x]_q=(1-q^x) /(1-q) $.
@GjergjiZaimi Is there a version of formula (2) that holds for a finite number of variables? I looked in the paper by Wachs that is mentioned in exercise 102 of Stanley, but didnt find it helpful at all.
You might want to read up on the principal specialization.
It specializes a symmetric function into a formal power series.
For example, the symmetric function $s_1(x) = x_1+x_2+ \dotsb$
has principal specialization $s_1(1,q,q^2,\dotsc) = 1+q+q^2+\dotsb = \frac{1}{1-q}$, so it does not make sense to let $q\to 1$. The expression only exist as a formal power series. However, you always compute the finite version,
$s_\lambda(1,q,q^2,\dotsc,q^{n})$, which is gonna be a polynomial in $q$.
So, the difference between your Jacob-Trudi formula and the cited q-formula,
is the number of variables (finite vs, infinite).
The principal specialization is in many instances nicer than the finite number of variables specialization.
Ok, so the limit $q\to 1$ is problematic when there are infinitely many variables. Is there a version of formula (2) that holds for a finite number of variables?
Yes you combine the Jacob trudi with the finite principal specialization of the complete homogeneous symmetric functions
|
2025-03-21T14:48:31.562423
| 2020-07-20T17:40:03 |
366126
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Andy Sanders",
"Quarto Bendir",
"https://mathoverflow.net/users/156492",
"https://mathoverflow.net/users/161399",
"https://mathoverflow.net/users/49247",
"user161399"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631280",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366126"
}
|
Stack Exchange
|
Norm of a differential form
How can we explicitly calculate the norm of a differential form?
For example let $(X, \omega) $ be a complex manifold such that locally
$$
\omega(z) =i\sum_{k,j} h_{k, j} (z) dz_k\wedge d\overline {z_j}\, ,
$$ and let $\alpha $ be a $(p, q)-$form. How we can calculate the norm of $\alpha $ with respect to the metric $\omega$?
This is a question of Hermitian linear algebra, no need for manifolds. In my experience, formally working out the linear algebra is the way to understand things. Then, you can port your computations to the tangent space of the manifold, which is a particular Hermitian vector space.
Let $h^{\overline{i}j}$ be the inverse components, so that $h^{\overline{i}j}h_{j\overline{k}}=\delta_{\overline{k}}^{\overline{i}}$. If $\alpha$ is a (2,1)-form then
$|\alpha|^2=h^{\overline{i}j}h^{\overline{k}l}h^{\overline{p}q}\alpha_{jl\overline{p}}\overline{\alpha_{ik\overline{q}}}$. Follow the same pattern for a $(p,q)$-form.
Do you have a reference on this!
I can't understand any of the textbooks, so to learn hermitian and kahler geometry I had to work it out for myself. But Griffiths & Harris "Principles of Algebraic Geometry" and chapter 2 of Besse "Einstein Manifolds" are standard references
|
2025-03-21T14:48:31.562672
| 2020-07-20T17:45:53 |
366127
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"asrxiiviii",
"https://mathoverflow.net/users/157984"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631281",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366127"
}
|
Stack Exchange
|
In a CM field, must all conjugates of an algebraic integer lying outside the unit circle lie outside the same?
This question is inspired from the post linked below:
Can an algebraic number on the unit circle have a conjugate with absolute value different from 1?
What I am curious about is the following: let $\alpha$ be an algebraic integer lying in a kroneckerian Galois extension $K/\mathbb Q$ (i.e. wherein complex conjugation commutes with the rest of the Galois group, more specifically, in a CM-field). If $|\alpha|<1$, must all the other conjugates of $\alpha$ also lie inside the unit circle?
It seems I am getting heurstic arguments both for and against the statement I am trying to investigate! (Of course, one of them is more naive than the other.)
On one hand, the answer would be positive if I can show that for any $\alpha$ lying in a CM field all conjugates of $\alpha$ must lie on the same circle as $\alpha$. I did manage to show that if $\alpha$ lies on a circle of radius $r$ for some $r>0$ such that $r^n \in \mathbb Q \text{ for some }n$ $\in \mathbb N$ (i.e. $|\alpha| \in \sqrt{\mathbb Q}$), then all conjugates of $\alpha$ must also lie on the same circle, simply because
\begin{multline*}
|\sigma\alpha|^2 = \sigma\alpha \cdot \overline{\sigma\alpha}= \sigma(\alpha)\sigma(\overline{\alpha}) = \sigma(\alpha\overline{\alpha}) = \sigma(|\alpha|^2) \\
= \sigma(r^2) = \sigma(r^{2n})^{1/n} = (r^{2n})^{1/n} = r^2 = |\alpha|^2
\end{multline*}
for every $\sigma \in \text{Gal}(K/\mathbb Q)$, hence all conjugates lie on the same circle. So things look somewhat promising, though I am stuck at this point. I am not sure if there is some continuity argument involved either. (Are Galois elements in CM fields continuous maps with respect to archimedean absolute values?). Also, it is to be noted that I haven't used the assumption that $\alpha$ is an algebraic integer, yet.
On the other hand, the Weak Approximation Theorem yields an $\alpha \in K$ for which $|\alpha|>1$ but $|\alpha|_v<1$ for every other archimedean place $v$ of $K$. This would give us an $\alpha$ which serves as a counterexample ... but $\alpha$ need not be an algebraic integer. I tried using the Strong Approximation Theorem (the only version of which I know being Bombieri-Gubler's Theorem 1.4.5.) to choose $\beta$ sufficiently close to $\alpha^{-1}$ at those nonarchimedean places $w$ for which $|\alpha|_w>1$, which gave me an algebraic integer $\beta$ but I lose control over archimedean values of $\beta$. Maybe I'm missing something really simple... In this argument, I haven't gotten any obstruction from the "CM-ness" of $K$ either.
Having said all that, it is possible that there are easy proofs/counterexamples to the statement that I am missing, although I haven't been come up with one of the latter either. I would really appreciate any suggestion or help. Thanks.
Zero is the only algebraic integer which has all its conjugates strictly inside the complex unit circle. (Look at the norm.)
For explicit examples with conjugates on either side of the unit circle, you can start with a real quadratic field with a totally positive unit that isn't already a square in this field, such as $\varepsilon = 2+\sqrt{3}$. Then take $\alpha=\sqrt{-\varepsilon}$ to obtain a CM extension and an algebraic integer (even an algebraic unit) with a pair of conjugates of absolute value less than $1$ and another pair of absolute value greater than $1$.
Thank you, I was having this misconception that the norms of algebraic integers in CM fields should be close to radicals of rational numbers.
To expand on GNiklasch's answer, and analyse what you write as well: we always have (when complex conjugation is central in the Galois group) have
$\overline{\alpha^{\sigma}} = {\bar \alpha}^{\sigma}$ when $\alpha$ is an algebraic integer in $K$ and $\sigma$ is an element of the Galois group of $K$.
In general, there is no reason why $|\alpha|^{2}$ should be rational for an algebraic integer $\alpha$ in your field $K$. It is true that all algebraic conjugates of $|\alpha|^{2}$ are positive when $\alpha \neq 0$. Then a standard argument (just by the arithmetic-geometric mean inequality) yields that the arithmetic mean of the algebraic conjugate of $|\alpha|^{2}$ is at least one (since the product of all these algebraic conjugates is a positive rational integer). Hence if some algebraic conjugate of $\alpha (\neq 0)$ has absolute value strictly less than one, there must be another algebraic conjugate of $\alpha$ with absolute value greater than one.
Thank you for your answer, I realized the error in my first thought process.
|
2025-03-21T14:48:31.562981
| 2020-07-20T18:38:06 |
366130
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Giorgio Metafune",
"https://mathoverflow.net/users/130023",
"https://mathoverflow.net/users/150653",
"user130023"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631282",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366130"
}
|
Stack Exchange
|
The proof of Tomas's restriction theorem $\|\hat{f}(\theta)\|_2\lesssim \|f\|_p$
I am trying to understand the proof of Tomas's theorem:
The proof reads
My question:
How do we get the estimates
$$\|T_k\ast f \|_{\infty}\lesssim 2^{-(n-1)k/2}\|f\|_{1},\qquad\qquad (1)$$
$$\|T_k\ast f \|_{2}\lesssim 2^{k}\|f\|_{2}.\qquad\qquad (2)$$
We can prove (1) by showing that
$$\|T_k \|_{\infty}\lesssim 2^{-(n-1)k/2}.\qquad \qquad (3)$$
And we can get (2) by proving that
$$\|T_k \|_{1}\lesssim 2^{k}\qquad\qquad (4).$$
I am stuck with (3) and (4).
By definition of $\widehat{d\theta}$ and $K$, we have that, for large enough $k$,
$$T_k (x)=\int_{2^{k-1}\leq |x|\leq 2^k} \left(K\left(\frac{|x|}{2^k}\right)-K\left(\frac{|x|}{2^{k-1}}\right) \right)\int_{\mathbb{S^{n-1}}}e^{\dot{\imath}x\cdot \theta}d\theta.$$
I am not sure of the latter claim of mine. We have
$$k(\frac{x}{2^k})=g(\frac{|x|}{2^k})$$ where $g\in \mathcal{S}(\mathbb{R})$. So, given, $\epsilon>0$, we can find $k_{\epsilon}>1$ such that $\int_{2^{k_{\epsilon}}}^{\infty} |g(r)|dr<\epsilon$.
We also have
When $|x|\leq 1$ (say 1 instead of 100 in the proof) we have $K(x)=1$.
So, for every $k\geq 1$, we have
$$D(x):=k(\frac{x}{2^k})-k(\frac{x}{2^{k-1}})=0$$ when $|x|\leq 2^{k-1}$.
That is $D$ is suppoerted in $|x|>2^{k-1}$. But, for large enough $k$, we can overlook $\int_{|x|>2^{k}}D$ because $D$ is a Schwartz function.
A second question:
The proof implicitly uses the fact that
$$\sum_{k\geq 1} \|T_k\ast f\|_{p^{\prime}}\lesssim \|f\|_{p^{\prime}}$$.
I would appreciate a hint for this too.
I agree that the original paper is not understandable because the main estimates you ask for are not explained. Actually they follow since the Fourier transform of the surface measure can be expressed through a Bessel function...but one needs some computation. The booklet by Stein: Bejing lectures on Harmonic analysis has full proofs in the chapter on oscillatory integrals but in a more general setting. A detailed proof of Tomas's result can be found in an expository paper by D. Kriventsov (available online) "The restriction problem and the Tomas Stein theorem".
Thanks a lot for the good references. I will try to understand these estimates then post their proof here hopefully.
|
2025-03-21T14:48:31.563431
| 2020-07-20T18:39:41 |
366132
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Michael Albanese",
"Oscar Randal-Williams",
"Xiao-Gang Wen",
"archipelago",
"https://mathoverflow.net/users/17787",
"https://mathoverflow.net/users/21564",
"https://mathoverflow.net/users/318",
"https://mathoverflow.net/users/32022"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631283",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366132"
}
|
Stack Exchange
|
Intersection form of surface bundle over surface
Let $\Sigma_g$ be a Riemannian surface of genus $g$. Let $M^4$ be a surface bundle over surface: $\Sigma_g \to M^4 \to \Sigma_h$. $\Sigma_g$ is the fiber and $\Sigma_h$ is the base space.
My question: is there a surface bundle over surface $M^4$ such that it has a 2-cocycle $c \in H^2(M^4;Z)$ satisfying
(1) $\int_{M^4} c^2 =\pm 1$, and
(2) $\int_{\Sigma_g} c =0$
Also: How to construct surface bundles with known odd intersection form? This may help to answer the first question.
See a related question:
Oddness of intersection form of surface bundle
A further question: is there a surface bundle over surface $M^4$ such that it has a 2-cocycle $c \in H^2(M^4;Z)$ satisfying
(1) $\int_{M^4} c^2 =\pm 1$,
(2) $\int_{\Sigma_g} c =0$, and (3) $c = w_2$ mod 2.
What is the signature and $g$ for such surface bundle?
Note that the condition (1) implies that $M^4$ is not spin and $w_2$ is non-trivial.
If $c \in H^2(M; \mathbb{Z})$, what do you mean by $\int_{\Sigma_g}c$?
Integration along fibres.
Yes, such a thing exists, but I don't know an explicit example.
To see that it exists, it is clearest to me to consider the universal situation. For any $k \in \mathbb{Z}$ there is a space $\mathcal{S}_g(k)$ which classifies oriented surface bundles
$$\Sigma_g \to E \overset{\pi}\to B$$
equipped with a class $c \in H^2(E; \mathbb{Z})$ such that $\int_{\Sigma_g} c = k$. Associated to such a family there are characteristic classes
$$\kappa_{i,j} = \int_\pi e(T_\pi E)^{i+1} \cdot c^j \in H^{2(i+j)}(B;\mathbb{Z}),$$
where $T_\pi E$ denotes the tangent bundle of $E$ along the fibres of $\pi$, and $e(T_\pi E)$ denotes its Euler class. (The classes $\kappa_{i,0}$ are the usual Miller--Morita--Mumford classes $\kappa_i$.)
In
J. Ebert and O. Randal-Williams, Stable cohomology of the universal Picard varieties and the extended mapping class group. Doc. Math. 17 (2012), 417–450.
Johannes Ebert and I studied, among other things, the low-dimensional integral cohomology of $\mathcal{S}_g(k)$, and showed that as long as $g$ is large enough (I think $g \geq 6$ will do) one has
$$H^1(\mathcal{S}_g(k);\mathbb{Z})=0 \quad\quad H^2(\mathcal{S}_g(k);\mathbb{Z})\cong\mathbb{Z}^3$$
where the isomorphism in the second case is given by a basis of cohomology classes $\lambda, \kappa_{0,1}, \zeta$, where the outer two are related to the $\kappa_{i,j}$ by the identities
$$12 \lambda = \kappa_{1,0} \quad\quad 2\zeta = \kappa_{0,1} - \kappa_{-1,2}.$$
In particular, applying this with $k=0$ and using that every second homology class is represented by a map from an oriented surface, it follows that there is a surface bundle
$$\Sigma_g \to E \overset{\pi}\to \Sigma_h$$
for some $h$ (which is uncontrollable using this method) with a class $c \in H^2(E;\mathbb{Z})$ satisfying $\int_{\Sigma_g}c = 0$, and having
$$\int_{\Sigma_h}\lambda=\text{whatever you like} \quad\quad \int_{\Sigma_h}\kappa_{0,1}=1 \quad\quad \int_{\Sigma_h}\zeta = 0$$
and hence having
$$\int_E c^2 = \int_{\Sigma_h} \int_\pi c^2 = \int_{\Sigma_h} \kappa_{-1,2} = \int_{\Sigma_h} \kappa_{0,1} = 1.$$
Thank you very much for the answer, so soon for such a specific question. The surface bundle you described has a fiber $\Sigma_g$ with $g\geq 6$. I wonder if a stronger result with $g\geq 2$ exits. I know $g\leq 1$ does not work.
It is also possible for $g=0$: you can take the Hirzebruch surface $S^2 \to H_1^4 \to S^2$. In the basis of homology given by the fibre and the section at infinity, its intersection form is $\begin{bmatrix}
0 & 1 \
1 & 1
\end{bmatrix}$.
Sorry, the bottom-right entry should be $-1$.
For $S^2_f\to H_1^4\to S^2_b$, are two conditions (1) $\int_{H_1^4} c^2 =\pm 1$, and
(2) $\int_{S^2_f} c =0$ satisfied? I think (2) is not satisfied.
Also, I like to know if the $c$ satisfying the above two conditions exists for each $g\geq 6$?
Sorry, you're right: I forgot about condition (2). I agree it is not possible for $g=0$.
|
2025-03-21T14:48:31.563690
| 2020-07-20T18:59:37 |
366135
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Max Alekseyev",
"https://mathoverflow.net/users/7076"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631284",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366135"
}
|
Stack Exchange
|
What is the order of binary partition function when number of partitions is fixed?
Given a number $N$, the number of ways to write it as a sum of powers of $2$ is called the binary partition function of $N$ and is well studied. But if the number partitions are fixed, then how do you write the number $N$ as the sum of $k$ powers of $2$, that is, are there are any known upper and lower bounds for number of partitions?
See OEIS A089052.
|
2025-03-21T14:48:31.563751
| 2020-07-20T19:27:25 |
366136
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Peter Humphries",
"https://mathoverflow.net/users/3803"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631285",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366136"
}
|
Stack Exchange
|
Why don't Dirichlet series with higher order zeros at $s=1$ have faster converging partial sums at $s=1$?
When examining the Dirichlet series generated by $\mu(n)\gcd(n,k)$ for some fixed value $k$, I found something counterintuitive. To set this up, we can expand as an Euler product to see that
\begin{align*}
\sum_{n=1}^{\infty}\frac{\mu(n)\gcd(n,k)}{n^s}&=\prod_{p}\left(1-\frac{\gcd(p,k)}{p^s}\right)\\
&=\prod_{p|k}\left(1-\frac{\gcd(p,k)}{p^s}\right)\cdot\prod_{p\nmid k}\left(1-\frac{\gcd(p,k)}{p^s}\right)\\
&=\prod_{p|k}\left(1-\frac{p}{p^s}\right)\cdot\prod_{p\nmid k}\left(1-\frac{1}{p^s}\right)\\
&=\prod_{p|k}\left(\frac{1-\frac{p}{p^s}}{1-\frac{1}{p^s}}\right)\cdot\prod_{p}\left(1-\frac{1}{p^s}\right)\\
\end{align*}
We can now define $g(s)=\prod_{p|k}\frac{1-p^{1-s}}{1-p^{-s}}$ we get that
$$\sum_{n=1}^{\infty}\frac{\mu(n)\gcd(n,k)}{n^s}=\frac{g(s)}{\zeta(s)}$$
The thing I find strange is that, about $s=1$, $1-p^{s}$ is bounded but $1-p^{1-s}\sim \log(p)(s-1)$ and so if we let $\omega(k)$ be the number of distinct prime factors of $k$ we have that
$$\sum_{n=1}^{\infty}\frac{\mu(n)\gcd(n,k)}{n^s}\sim C_0 (s-1)^{\omega(k)+1}$$
where
$$C_0=\prod_{p|k}\frac{\log(p)}{1-p^{-1}}$$
Since $\sum_{n=1}^{\infty}\frac{\mu(n)\gcd(n,k)}{n^s}$ has a higher order zero at $s=1$ and vanishes very quickly, it would be my intuition that around $s=1$ the sum $\sum_{n=1}^{\infty}\frac{\mu(n)\gcd(n,k)}{n}$ would converge to $0$ very fast, but numerical evidence would suggest that it converges much slower than $\sum_{n=1}^{\infty}\frac{\mu(n)}{n}$. What is the intuition behind why the speed of Dirchlet series vanishing at $s=1$ and convergence $s=1$ not being related?
As an aside, it is obvious by Newman's Tauberian theorem that this sum does indeed converge at $s=1$, it is just the speed that bothers me.
The order of the zero of $\sum_{n = 1}^{\infty} \frac{\mu(n) (n,k)}{n^s}$ at $s = 1$ determines the rate at which this Dirichlet series tends to zero as $s \searrow 1$. However, this says nothing about the size of the partial sums $\sum_{n \leq x} \frac{\mu(n) (n,k)}{n}$ as $x \to \infty$ other than this is $o(1)$ (by a Tauberian theorem or Perron's formula). Instead, the decay rate of $\sum_{n \leq x} \frac{\mu(n) (n,k)}{n}$ is dictated by the zero free region of the analytic continuation of the Dirichlet series left of the line $\Re(s) = 1$.
|
2025-03-21T14:48:31.563908
| 2020-07-20T20:06:41 |
366137
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Martin Rubey",
"Richard Stanley",
"Ville Salo",
"https://mathoverflow.net/users/123634",
"https://mathoverflow.net/users/2807",
"https://mathoverflow.net/users/3032"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631286",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366137"
}
|
Stack Exchange
|
Permutations with bounded displacement on a circle
I am interested in references to results on permutations $\sigma$ of $\{0,\ldots,n-1\}$ satisfying $\text{min}\{ \sigma(i)-i\text{ (mod } n),i-\sigma(i) \text{ (mod } n) \} \leq k$ for some $k$ for all $i$. I don't know if there is a common term for these, so I settled for the "bounded displacement" in the title.
I was only able to find two related topics, but both with the linear order, without the "mod $n$" part. These are the band permutations and probabilistic results about them, concerned with random walks (and relaxing the "all $i$" assumption to just "most $i$"), discussed in this video, and the enumeration problem for permutations with bounded drop (bounding the value of $i-\sigma(i)$ instead) inspired by sorting problems, and solved in this paper. It also seems like such permutations might arise with Schreier graphs, but I was unable to find any references, nor analogues of the previous two on the circle.
I am not primarily concerned with counting these permutations (but will be happy to see that too). Ideally, I would like to see dynamical results about such a permutation of a discrete set, or about quantities invariant or bounded under such permutation acting on $\mathbb{R}^n$, since this is the context we encountered them in.
Given the last paragraph, https://arxiv.org/abs/1301.4736 might be related.
Let $f_k(n)$ be the number of such permutations. By the technique used to prove Proposition 4.7.10 of Enumerative Combinatorics, vol. 1, second ed., if we interpret $f_k(n)$ suitably for small $n$ then $\sum_{n\geq 0}f_k(n)x^n=-xQ_k'(x)/Q_k(x)$ for some polynomial $Q_k(x)$.
It would be great if you could add this to findstat.org. (I'm on a cellphone right now) here is code:
def statistic(pi):
n=len(pi)
return max(min((e-1-i)%n,(i-e+1)%n) for i,e in enumeate(pi))
|
2025-03-21T14:48:31.564053
| 2020-07-20T20:25:56 |
366139
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Mark L. Stone",
"https://mathoverflow.net/users/75420"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631287",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366139"
}
|
Stack Exchange
|
SDP relaxation vs. Monte Carlo for MaxCut: which one performs better?
the Goemans Williamson SDP relaxation of the MAXCUT problem famously gives a polynomial approximation ratio of .87856 for the MAXCUT on regular graphs.
Another popular approach to obtain efficient approximations to NP-complete combinatorial optimization problems is that of sampling from a high enough temperature distribution using Markov chains. For the case of MAXCUT, this reduces to sampling from the Gibbs state on an Ising model. And there is a lot of literature which gives bounds on the temperatures that can be sampled from efficiently.
However, I cannot find any comment on how SDP relaxations compare to polynomial-time Monte Carlo methods. What is the approximation ratio we obtain from Monte Carlo methods for say, 3 regular graphs?
Any help is appreciated!
if you don't get an answer here, https://or.stackexchange.com/ might be a better bet, although it's a rather big "ask" there.
|
2025-03-21T14:48:31.564159
| 2020-07-20T20:29:49 |
366141
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Gerry Myerson",
"Jeremy Rouse",
"Stanley Yao Xiao",
"https://mathoverflow.net/users/100231",
"https://mathoverflow.net/users/10898",
"https://mathoverflow.net/users/3684",
"https://mathoverflow.net/users/48142",
"vidyarthi"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631288",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366141"
}
|
Stack Exchange
|
Solutions to $(a^c-b^c)+m(r^c-s^c)=0$ in integers
Let $c\ge2$ be a fixed positive integer. How many nontrivial solutions in the integers does the equation $(a^c-b^c)+3(r^c-s^c)=0$ have? If $c=2$, I think it has infinitely many solutions as it seems to be similar to Pell's equation. If $c\ge 3$, I think the solutions are finite, by Falting's theorem. Is it that there are no nontrivial solutions to the equation in integers if $c\ge3$ and odd. By nontrivial I mean not all of the integers are zero and neither $a=b$ nor $r=s$. Any hints? Thanks beforehand.
For a fixed $c$, your polynomial defines a surface, not a curve, and hence Faltings's theorem does not apply to it. If $c = 3$, it is a rational surface and has infinitely many solutions (most of which have $a \ne b$ and $r \ne s$). If $c = 4$, it is a K3 surface and probably has infinitely many points with $a \ne b$ and $r \ne s$ (a quick search turned up at least $48$).
@JeremyRouse so you mean it has infinitely many solutions always?
@vidyarthi for $c \geq 5$ the equation you wrote down defines a surface of general type, so it is unlikely to have infinitely many integral points unless they accumulate on some subvariety.
Rewriting your equation as, for fixed $m$ and $k$,
$$\displaystyle x^k + my^k = u^k + mv^k, x,y,u,v \in \mathbb{Z},$$
we see that this is of the form $F(x,y) = F(u,v)$ for a binary form of degree $k$ and defines a surface $X_F \subset \mathbb{P}^3$. Heath-Brown showed in this paper that if one deletes the rational lines on this surface, necessarily formed by rational automorphisms of the binary form $F$, then on the remaining open subset $U_1$ of the surface $X_F$ contains at most $O_\epsilon \left(B^{\frac{12k + 16}{9k^2 - 6k + 16} + \epsilon} \right)$ primitive integral points of height at most $B$. This kind of result is the best kind we have in general, as it is not easy to access the geometry of higher degree curves on $X_F$ for arbitrary $F$. The best exponent $\beta_k$ known, which depends only on the degree $k$, is contained in the following paper by myself and Cam Stewart: On the representation of integers by binary forms .
In the special case you are interested in, because the surface is geometrically a Fermat surface which is very special, more can be said about what kind of curves can lie on the surface (which we expect to contribute the bulk of the points). I am not entirely familiar with this but this paper of Browning and Heath-Brown may help.
thanks, I think the solution space is not Zariski dense in the surface, is it true?
@vidyarthi it is conjectured to be true by the Bombieri-Lang conjecture, but we are far from being able to prove something like this. The best we can do is that after deleting the "obvious" subvarieties which we expect to contain the bulk of points, the remaining rational points are sparse (in the sense that there are few of them of bounded height)
In a joint paper with Andrew Bremner (https://www.sciencedirect.com/science/article/pii/S0022314X13002527#se0070) we obtained several result concerning exitence of rational and integral points on surfaces of the form
$$
a(x^p-y^q)=b(z^r-w^s),
$$
where $1/p+1/q+1/r+1/s=1$.
If $p=q=r=s=4$ and $h=b/a$, we were able to treat the case of curves that have arithmetic genus 0, hence geometric genus 0, and hence parameterizable, lying on the surface under consideration. If $h$ is not a perfect rational square then there are no curves of arithmetic genus 0.
$ax^k + by^k = au^k + bv^k\tag{1}$
$a,b,x,y,u,v$ are integer.
Case $k=3$:
If equation $(1)$ has a known solution, then equation $(1)$ has infinitely many integer solutions below.
Let $(x0,y0,u0,v0)$ is a known solution.
p,q are arbitrary.
Substitute $x=pt+x0, y=qt+y0, u=pt+u0, v=qt+v0$ to equation $(1)$, then we get
$$t=\frac{-ax0^2p+by0^2q-au0^2p-bv0^2q}{ax0p^2+by0q^2-au0p^2-bv0q^2}$$
Hence equation $(1)$ has a parametric solution.
Example: $x^3 + 3y^3 = u^3 + 3v^3$
$(x0,y0,u0,v0)=(3,4,6,1)$.
$(x,y,u,v)=(-3(p-q)(2p-3q), -9pq+3q^2+4p^2, -3(p-2q)(p-3q), -9pq+12q^2+p^2)$
p,q are arbitrary.
Without assumption of a known solution , we have another solution below.
$(x,y,u,v)=(3pq-3p^2+3p+8q^2-19q+8, 2q^2+3pq-7q+8-9p+3p^2, -15pq-3p^2+21p-10q^2+35q-28, -25q-15p+20+8q^2+9pq+3p^2)$
p,q are arbitrary.
Case $k=4$:
According to Richmond's theorem for $ax^4 + by^4 + cz^4 + dw^4 = 0$, if equation $(1)$ has a known solution, then equation (1) has infinitely many integer solutions.
Example: $x^4 + 3y^4 = u^4 + 3v^4$
$(x0,y0,u0,v0)=(4,1,2,3)$.
$(x,y,u,v)=(1068p-1424,-489p+652,-114p+152,837p-1116)$
There's a simpler way to show one solution of $x^4+3y^4=u^4+3v^4$ implies infinitely many; if $(x,y,u,v)$ works, and $k$ is any integer, then $(kx,ky,ku,kv)$ works.
|
2025-03-21T14:48:31.564449
| 2020-07-20T20:57:18 |
366144
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Simon Parker",
"https://mathoverflow.net/users/123207"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631289",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366144"
}
|
Stack Exchange
|
Lifting property for proper morphism
Let $X \subseteq \mathbb{C}^n$ be a complex affine variety and $\tilde{X} \to X$ a surjective proper morphism where $\tilde{X}$ is smooth. Is it true that every morphism $\mathbb{C} \to X$ can be lifted to $\mathbb{C} \to \tilde{X}$?
I'm particularly interested in the case where $X$ has an isolated singularity and $\tilde{X} \to X$ is a resolution.
Also on MSE: https://math.stackexchange.com/q/3742478/202132
Regarding your general question, the answer is no.
Take any hyperbolic projective variety $Y$ (for instance, a ball quotient) of dimension $n$, and project it generically onto $\mathbb{P}^n$. Removing a hyperplane from $\mathbb{P}^n$ and its preimage from $Y$, we get a surjective finite morphism $f \colon Y^{\circ} \to \mathbb{C}^n$. There are plenty of morphisms $\mathbb{C} \to \mathbb{C}^n$, but none of them lifts to $Y^{\circ}$, since by the hyperbolicity assumption there are no non-constant entire curves in $Y^{\circ}$.
As a toy model, you can consider the case where $Y$ is a genus $2$ curve mapping $2:1$ to $\mathbb{P}^1$, and then remove a point from $\mathbb{P}^1$ and the two points over it from $Y$.
Regarding your question about the isolated singularity, the answer is yes. In fact, the resolution is an isomorphism away from the singular point $p \in X$. Thus, you can take the image of your entire curve in $X-{p}$, lift it to the resolution via the resolution map and then consider the Zariski closure in $\tilde{X}$.
|
2025-03-21T14:48:31.564575
| 2020-07-20T21:03:16 |
366145
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Anthony Quas",
"R W",
"https://mathoverflow.net/users/11054",
"https://mathoverflow.net/users/8588"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631290",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366145"
}
|
Stack Exchange
|
Friedland metric entropy
I was asking if it is possible to extend the definition of topological Friedland entropy for $\mathbb{Z}^d$ continuos actions to measure preserving actions.
The topologica Friedland entropy is constructed as follows. Let $T$ be a $\mathbb{Z}^d$ action on a compact topological space $X$. Consider then teh sequence space
$$
\mathcal{X}=\mathcal{X}_T=\left\{(x_n)_{n\in \mathbb{N}} \in \prod_{n \in \mathbb{N}}X: T_i(x_n)=x_{n+1} \text{ for some } i =1,\dots,d\right\}.
$$
This is a compact space. There is a natural shift on $\mathcal{X}$, given by $\sigma((x_n)_{n\in \mathbb{N}})=(x_{n+1})_{n \in \mathbb{N}}$. We can see that $\sigma$ is continuos and then define the entropy of the action $T$, $ent_{top}(T)=h_{top}(\sigma)$.
This definition is motivated from the fact that when $d=1$, $h_{top}(\sigma)=h_{top}(T)$. This comes from the fact that the map $\phi:X \ni x \mapsto orb(x)=(T^{i}(x))_{i \in \mathbb{N}}$ is a topological conjugacy map.
This definition can be given also for metric non compact spaces, considering Bowen topological entropy.
Is it possible to define an extension of the metric entropy of Kolmogorov using this method? Consider the case of measure preserving actions on probability spaces, the problem I am facing are the following
How can define a probability measure on $\mathcal{X}$? I have the product probability on $\prod_{n \in \mathbb{N}}X$, but i cannot see a way to restric it $\mathcal{X}$.
How can I see that $\sigma$ is measure preserving?
How can I see that those definition are the same in the case $d=1$, using a conjugacy argument?
Thank you for you suggestions!
@Anthony Quas - This does not have much to do with the natural extension as OP's construction results in a one-sided shift. Of course, one can make it two-sided by taking its natural extension, but this is not what OP is asking about.
Thanks @RW: I see now I misread the question. I will delete my comment.
The space $\mathcal X$ can be identified with the product of the base space $X$ and the space $\mathcal I$ of (one-sided) sequences $(i_1,i_2,\dots)$ of symbols $i_k\in I$ (where $I=\{1,2,\dots,d\}$) by the map
$$
(x;i_1,i_2,\dots) \mapsto (x, T_{i_1}x, T_{i_1}T_{i_2} x,\dots \;.
$$
Therefore, a measure on $\mathcal X$ is uniquely determined by its projection onto $X$ (for which one can naturally take an invariant measure $m$) and a family of measures on $\mathcal I$ parameterized by the points from $X$. The choice here is enormous. The most "natural" is to take just the Bernoulli measure on $\mathcal I$ corresponding to the uniform distribution on $I$. The related notions of entropy (in a somewhat more general situation) are discussed, for instance, in the paper by Tim Austin "Entropy of probability kernels from the backwards tail boundary".
|
2025-03-21T14:48:31.564888
| 2020-07-20T21:07:45 |
366147
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Jason Starr",
"Mohan",
"Nikolas Kuhn",
"RandomMathUser",
"https://mathoverflow.net/users/112142",
"https://mathoverflow.net/users/13265",
"https://mathoverflow.net/users/161405",
"https://mathoverflow.net/users/9502"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631291",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366147"
}
|
Stack Exchange
|
Proper morphisms with geometrically reduced and connected fibers
Let $f: X \to S$ be a proper morphism ($S$ locally noetherian), and $X \to S' \to S$ its Stein factorisation. By Zariski's Main Theorem the number of geometric connected components of the fibers of $f$ can be read from the cardinal of the fibers of the finite $S' \to S$. In particular if all fibers of $f$ are geometrically connected, then $S' \to S$ is radicial.
I expect that if furthermore the fibers of $f$ are geometrically reduced (and $f$ is surjective and $S$ reduced in order to remove trivial counterexamples), then $S'=S$ that is $f$ is an $\mathcal{O}$-morphism (viz. $f_*\mathcal{O}_X = \mathcal{O}_S$). Strangely I only find this fact when $f$ is furthermore assumed to be flat, for instance: https://stacks.math.columbia.edu/tag/0E0L.
Here is an outline of a demonstration (suggested by a friend): we want to show that $S' \to S$ is an isomorphism. Since it is a surjection by assumption on $f$, it suffices to show that it is an immersion. By our assumptions on $f$, $S' \to S$ has geometrically connected and reduced fibers. We way assume that $S=\textrm{Spec} A$ and $S'=\textrm{Spec} B$, with $A \to B$ finite. Let $C$ be the cokernel of $A \to B$ (seen as a $A$-module). If $p$ is a prime ideal in $A$, $B \otimes_A \overline{k}(p) = \overline{k}(p)$ (since it is connected and reduced over $\overline{k}(p)$), so $C \otimes_A \overline{k}(p)=0$, so $C=0$.
Is the above proof indeed correct? Does the hypotheses already imply that $f$ is flat? Is there a reference to this result somewhere in the literature, presumably in EGA?
What is your argument for showing $S'\to S$ has reduced fibers without flatness of $f$?
That is false when $S$ is a cuspidal plane cubic, when $X$ is the union of a twisted cubic and a tangent line in $3$-space, and the morphism is linear projection from a general point on the line.
I miscomputed the pushforward of the structure sheaf. The example in my previous comment is wrong.
@Mohan: you are right, unlike for geometric connectedness which is a topological condition so obvious, it is not obvious that $S' \to S$ has geometrically reduced fibers. Still I would expect that if it were not the case then $X \to S$ would have non geometrically reduced fibers too, but my intuition is probably wrong.
Here is a standard example. Take $\mathbb{P}^1\subset\mathbb{P}^3$ of large degree and let $S$ be the cone, with the vertex $p$, only singular point. Let $f:X\to S$ the blow up of $p$. One can check that $X$ is smooth and thus the Stein factorization $S'$ is the normalization of $S$. The fiber over $p$ in $X$ is smooth irreducible (scheme-theoretically), but the fiber in $S'$ is not reduced.
Thanks for the counterexample. Now this makes me wonder if there is a nice condition, less strong than flatness, that ensures that $S'$ has reduced geometric fibers.
How does one see that the fiber in S' is non-reduced? (or that S is not normal)
@NikolasKuhn One has a natural map $\oplus H^0(O_{\mathbb{P}^3}(n))\to \oplus H^0(O_{\mathbb{P}^1}(nd))=B$, where $d$ is the degree and let $A$ be the image of the first ring in $B$. Then, $A$ is a proper subring of $B$ since $d$ is large and $B$ is an integral birational extension of $A$. One checks that spectrum of $B$ is $S'$ and that of $A$ is $S$. Also, $B$ is normal. Hope you can fix the rest.
|
2025-03-21T14:48:31.565144
| 2020-07-20T22:23:28 |
366150
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"https://mathoverflow.net/users/41650",
"user41650"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631292",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366150"
}
|
Stack Exchange
|
The locus of lines intersecting with another fixed line on a Fano threefold
Let $Y$ be an index $2$, degree $5$, Picard number $1$ Fano threefold, i.e $Y$ is a linear section of Grassmannian $\operatorname{Gr}(2,5)$. Let $\Sigma(Y)$ be the Hilbert scheme of lines on $Y$, it is isomorphic to $\mathbb{P}^2$. Let $\mathcal{B}\in \lvert\mathcal{O}_Y(2)\rvert$ be a smooth quadric hyersurface, it is a degree $10$ K3 surface. Now, I consider the following two situations:
I fix a line $L_1\in Y$, consider all lines $L_t$ intersects with $L_1$. Since the intersection with a fixed line is a codimension $1$ condition, I think such a family of lines is parametrized by $\mathbb{P}^1$? Or at least, can I choose a pencil of lines intersecting with the fixed $L_1$?
I consider a family of lines $L_t$ tangent to $\mathcal{B}$, is this family also a $\mathbb{P}^1$ or just a smooth curve?
Maybe the general question is how to describe those families rigorously?
Question 1. Let $I(Y) \subset \Sigma(Y) \times \Sigma(Y) \cong \mathbb{P}^2 \times \mathbb{P}^2$ be the incidence scheme (parameterizing pairs of intersecting lines). Then $I(Y) \cong \mathrm{Fl}(1,2;3) \subset \mathbb{P}^2 \times \mathbb{P}^2$; I think you can find this in Sanna, Giangiacomo. Small charge instantons and jumping lines on the quintic del Pezzo threefold. Int. Math. Res. Not. IMRN 2017, no. 21, 6523-6583. In particular, lines intersecting a given line $L$ are parameterized by $p_1(p_2^{-1}([L])) \subset \Sigma(Y)$ which is indeed a line on $\mathbb{P}^2$ (here $p_i$ denote the projections of $I(Y)$ to the factors).
Question 2. Recall that $Y \subset \mathbb{P}^6 = \mathbb{P}(V)$. In particular, every line on $Y$ is a line in $\mathbb{P}(V)$. This defines an embedding
$$
\Sigma(Y) \to \mathrm{Gr}(2,V).
$$
It is defined by a rank-2 vector bundle $\mathcal{U}$ on $\Sigma(Y)$. A description of this bundle can be found in the same reference, for now it is important that $\det(\mathcal{U}) \cong \mathcal{O}(-3)$. A quadric in $Y$ is cut out by a quadric in $\mathbb{P}(V)$; its equation is in $S^2V^\vee$, and it induces a global section of $S^2\mathcal{U}^\vee$. The tangency locus is the degeneracy locus of the corresponding section of $S^2\mathcal{U}^\vee$, or equivalently of the induced morphism
$$
q \colon \mathcal{U} \to \mathcal{U}^\vee.
$$
Its equation is $\det(q) \colon \mathcal{O}(-3) \cong \det(\mathcal{U}) \to \det(\mathcal{U}^\vee) \cong \mathcal{O}(3)$; thus the the tangency locus is a sextic curve in $\Sigma(Y) \cong \mathbb{P}^2$. For general $q$ it is smooth, but it is not true that it is smooth for any smooth quadric divisor --- if, for instance, a divisor contains a line, this line is contained in the tangency locus and gives a singular point on it.
Thanks so much! In my case, the real thing I am considering is the curve $\Gamma$ as a double cover of this sextic curve on X_{10}, and the ruled surface $S$ over this $\Gamma$ and the fibers are exactly coming from all the lines intersecting $\Gamma$. And your comment on if a line is in this branch divisor , then it will give a singular point on $\Gamma$ also clears my confusion
Hi Sasha, I have a confusion. The vector space V is dimension 7 right? So I think Gr(2,V) and $\mathbb{P}^2=\Sigma(Y)$ is embedding to $\mathbb{P}^{20}$, so this is Veronese degree 5 embedding, so deg($\mathcal{U}$) should be $O(-5)$? Otherwise, I think the dimension of $S^2V$ and $H^0(Y,S^2\mathcal{U}^{\vee})$ does not match, the previous one is 28 and the latter seems is 10
Sorry, I think the dimension of $H^0(Y,S^2\mathcal{U}^{\vee})$ is of dimension 24. By the way, the computation of $ch(S^2\mathcal{U}^{\vee})$ in the paper you mentioned should be wrong, the correct number should be $3+3H+\frac{27}{2}L-\frac{1}{2}P$, thus $\chi(S^2\mathcal{U}^{\vee})=3+\frac{8}{3}\times 3+\frac{27}{2}-\frac{1}{2}=24$. I think higher cohomology would vanish. Of course, I think the bundle $\mathcal{U}^{\vee}$ should be on $\Sigma(Y)$ instead of $Y$, it seems that the vector bundle $\mathcal{U}$ in that paper is the tautological bundle on $Y$.
|
2025-03-21T14:48:31.565429
| 2020-07-20T22:31:27 |
366151
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Peter Humphries",
"https://mathoverflow.net/users/140336",
"https://mathoverflow.net/users/3803",
"user15243"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631293",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366151"
}
|
Stack Exchange
|
A question related to Hilbert modular form
This is a question related to Hilbert modular forms.
Let $\mathbb{K}=\mathbb{Q}(\sqrt D)$ be an imaginary quadratic field with discriminant $D<0$ and $\zeta (\text{mod } m)$ a Hecke character such that $$\zeta((a))= \left( \frac{a}{|a|} \right)^u \text{ if } a \equiv 1 \pmod{m}$$
where $u$ is any non-negative integer. Then
$$ f(z)= \sum_a \zeta(a) N_{\mathbb{K}/\mathbb{Q}}(a)^\frac{u}{2} e(zN_{\mathbb{K}/\mathbb{Q}}(a))\in M_k(\Gamma_0(N),\chi) $$
where $k=u+1, N= |D|N_{\mathbb{K}/\mathbb{Q}}(m)$ and $\chi (\text{mod } m)$ is the Dirichlet character given by
$$ \chi(n)= \chi_D(n) \text{ if } n\in \mathbb{Z} .$$
Moreover $f$ is a cusp form if $u>0$.
This is a theorem from the book "Topics in Classical Automorphic Forms" by Henryk Iwaniec (page 213). This theorem actually provides a connection between the conductor of a Hecke character and the level and weight of a modular form by automorphic induction.
I would like to know if there is a similar thereom in the case of Hilbert modular forms too; that is, if a theorem precisely shows how to construct a hilbert modular form over a totally real field $\mathbb{F}$ from a Hecke character of an imaginary quadratic extension $\mathbb{E}$ of $\mathbb{F}$.
Thank you in advance.
Yes. This is automorphic induction.
Please provide some references.
Did you try googling automorphic induction? For example, see section 13 of https://doi.org/10.2307/2944321
If you are unsatisfied with my answer, please do let me know what you are instead looking for. (Also, you should accept answers to your questions when they are satisfactory - for some reason, you have only done this once out of the ten questions that you have asked.)
Hi @PeterHumphries ,thanks a lot for the answer. Last few weeks I was busy with some other works and couldn't manage to give it enough time. I will surely accept the answer once I completely understand it.
Results of this form are best stated adèlically. Perhaps the canonical reference is this paper of Shalika and Tanaka:
https://doi.org/10.2307/2373316
Sadly the paper was written pre-Jacquet-Langlands and is rather hard to read. Jacquet-Langlands do treat automorphic induction themselves in Section 12 of their seminal book:
http://doi.org/10.1007/BFb0058988
Alternatively, one can try reading this later paper of Labesse and Langlands, which discusses converses to automorphic induction:
https://doi.org/10.4153/CJM-1979-070-3
(See also my answer here: Reference for: CM Hilbert Modular forms arise from Hecke characters)
All of these deal with automorphic induction for Hecke characters; automorphic induction in more general settings is known due to the work of Arthur and Clozel:
https://www.jstor.org/stable/j.ctt1bd6kj6
In what follows, I summarise the correspondence between Hecke characters and automorphically induced automorphic representations.
Let $E/F$ be a quadratic extension of number fields, and let $\Omega$ be a unitary Hecke character of $\mathbb{A}_E^{\times}$, so that $\Omega$ is the idèlic lift of a classical (primitive) Größencharakter $\psi$ of $E$. This has a completed $L$-function $\Lambda(s,\Omega)$ whose finite part $L(s,\Omega)$ has an Euler product of the form
$$\prod_{\mathfrak{P}} \frac{1}{1 - \psi(\mathfrak{P}) \mathrm{N}_{E/\mathbb{Q}}(\mathfrak{P})^{-s}},$$
where the product is over the prime ideals $\mathfrak{P}$ of $\mathcal{O}_E$. Note that $\psi(\mathfrak{P}) = 0$ whenever $\mathfrak{P}$ divides the conductor $\mathfrak{Q}$ of $\Omega$.
Automorphic induction associates to $\Omega$ an automorphic representation $\pi = \pi(\Omega)$ of $\mathrm{GL}_2(\mathbb{A}_F)$ whose completed $L$-function $\Lambda(s,\pi)$ is equal to $\Lambda(s,\Omega)$. (One can prove this via the converse theorem.)
Let $\omega_{\pi}$ denote the central character of $\pi$, so that this is a Hecke character of $\mathbb{A}_F^{\times}$ that is the idèlic lift of a classical (primitive) Größencharakter $\chi_{\pi}$ of $F$; when $F = \mathbb{Q}$, $\chi_{\pi}$ is just a Dirichlet character (it is the nebentypus of the newform associated to $\pi$). One can check that $\omega_{\pi} = \omega_{E/F} \Omega|_{\mathbb{A}_F^{\times}}$, where $\omega_{E/F}$ denotes the quadratic Hecke character associated to the quadratic extension $E/F$. Let $\lambda_{\pi}(\mathfrak{n})$ denote the $\mathfrak{n}$-th Hecke eigenvalue of $\pi$, where $\mathfrak{n}$ is an integral ideal of $\mathcal{O}_F$. (Here I am normalising the Hecke eigenvalues as an analytic number theorist would, namely that $\lambda_{\pi}(\mathfrak{p})$ is the sum of two complex numbers of absolute value $1$ when $\mathfrak{p}$ does not divide the conductor of $\pi$.) Then the finite part $L(s,\pi)$ has an Euler product of the form
$$\prod_{\mathfrak{p}} \frac{1}{1 - \lambda_{\pi}(\mathfrak{p}) \mathrm{N}_{F/\mathbb{Q}}(\mathfrak{p})^{-s} + \chi_{\pi}(\mathfrak{p}) \mathrm{N}_{F/\mathbb{Q}}(\mathfrak{p})^{-2s}},$$
where the product is over prime ideals $\mathfrak{p}$ of $\mathcal{O}_F$. Note that the conductor $\mathfrak{q}$ of $\pi$ satisfies $\mathfrak{q} = \mathrm{N}_{E/F}(\mathfrak{Q}) \mathfrak{d}_{E/F}$, where $\mathfrak{d}_{E/F}$ denotes the relative discriminant.
Now for each prime ideal $\mathfrak{p}$, write $\lambda_{\pi}(\mathfrak{p}) = \alpha_{\pi,1}(\mathfrak{p}) + \alpha_{\pi,2}(\mathfrak{p})$, where $\alpha_{\pi,1}(\mathfrak{p}), \alpha_{\pi,2}(\mathfrak{p})$ denote the Satake parameters. Note that $\alpha_{\pi,1}(\mathfrak{p}) \alpha_{\pi,2}(\mathfrak{p}) = \chi_{\pi}(\mathfrak{p})$. Then by comparing Euler products, we have the following:
If $\mathfrak{p}$ splits in $E$, so that $\mathfrak{p} \mathcal{O}_E = \mathfrak{P} \sigma(\mathfrak{P})$ for some prime ideal $\mathfrak{P}$ of $\mathcal{O}_E$ with $\mathrm{N}_{E/F}(\mathfrak{P}) = \mathrm{N}_{E/F}(\sigma(\mathfrak{P})) = \mathfrak{p}$, where $\sigma$ denotes the nontrivial Galois automorphism of $E/F$, then $\alpha_{\pi,1}(\mathfrak{p}) = \psi(\mathfrak{P})$ and $\alpha_{\pi,2}(\mathfrak{p}) = \psi(\sigma(\mathfrak{P}))$.
If $\mathfrak{p}$ is inert in $E$, so that $\mathfrak{p} \mathcal{O}_E = \mathfrak{P}$ for some prime ideal $\mathfrak{P}$ of $\mathcal{O}_E$ with $\mathrm{N}_{E/F}(\mathfrak{P}) = \mathfrak{p}^2$, then $\alpha_{\pi,1}(\mathfrak{p}) = -\alpha_{\pi,2}(\mathfrak{p}) = \psi(\mathfrak{P})^{1/2}$.
If $\mathfrak{p}$ is ramified in $E$, so that $\mathfrak{p} \mid \mathfrak{d}_{E/F}$ and $\mathfrak{p} \mathcal{O}_E = \mathfrak{P}^2$ for some prime ideal $\mathfrak{P}$ of $\mathcal{O}_E$ with $\mathrm{N}_{E/F}(\mathfrak{P}) = \mathfrak{p}$, then $\alpha_{\pi,1}(\mathfrak{p}) = \psi(\mathfrak{P})$ and $\alpha_{\pi,2}(\mathfrak{p}) = 0$.
From this and multiplicativity, one can deduce that
$$\lambda_{\pi}(\mathfrak{n}) = \sum_{\substack{\mathfrak{N} \subset \mathcal{O}_E \\ \mathrm{N}_{E/F}(\mathfrak{N}) = \mathfrak{n}}} \psi(\mathfrak{N}).$$
I haven't yet described what happens at the archimedean places. At each archimedean place $w$ of $E$, the local component of $\Omega$ is a unitary character $\Omega_w : E_w^{\times} \to \mathbb{C}^{\times}$ with image in the unit circle.
If $E_w \cong \mathbb{R}$, then $\Omega_w(x_w) = \mathrm{sgn}(x_w)^{\kappa_w} |x_w|_w^{it_w}$ for some $\kappa_w \in \{0,1\}$ and $t_w \in \mathbb{R}$. The local component of the completed $L$-function is $\Gamma_{\mathbb{R}}(s + \kappa_w + it_w)$, where $\Gamma_{\mathbb{R}}(s) = \pi^{-s/2} \Gamma(s/2)$.
If $E_w \cong \mathbb{C}$, then $\Omega_w(x_w) = e^{i\kappa_w \arg(x_w)} |x_w|_w^{it_w}$ for some $\kappa_w \in \mathbb{Z}$ and $t_w \in \mathbb{R}$. The local component of the completed $L$-function is $\Gamma_{\mathbb{C}}(s + \frac{|\kappa_w|}{2} + it_w)$, where $\Gamma_{\mathbb{C}}(s) = 2(2\pi)^{-s} \Gamma(s)$.
From this, we can describe the local components of $\pi$ at each archimedean place $v$ of $F$.
If $F_v \cong \mathbb{R}$ and $v$ splits in $E$ into two real places $w_1$ and $w_2$, then $\pi_v$ is a principal series representation of the form $\mathrm{sgn}^{\kappa_{w_1}} |\cdot|_v^{it_{w_1}} \boxplus \mathrm{sgn}^{\kappa_{w_2}} |\cdot|_v^{it_{w_2}}$.
If $F_v \cong \mathbb{R}$ and $v$ ramifies in $E$, so there exists a single complex place lying over $v$, then $\pi_v$ is a discrete series representation of the form $D_{|\kappa_w| + 1} \otimes \left|\det\right|_v^{it_w}$; in particular, the weight is $|\kappa_w| + 1$.
If $F_v \cong \mathbb{C}$ then $v$ splits in $E$ into two complex places $w_1$ and $w_2$, and $\pi_v$ is a principal series representation of the form $e^{i\kappa_{w_1} \arg} |\cdot|_v^{it_{w_1}} \boxplus e^{i\kappa_{w_2} \arg} |\cdot|_v^{it_{w_2}}$.
Note that there are restrictions on the parameters $t_w$, since $\Omega$ is trivial on $E^{\times}$ and in particular on $\mathcal{O}_E^{\times}$.
(I write much of this down in section 4 of this paper of mine: https://doi.org/10.1093/imrn/rnx283)
At this point, you know the Hecke eigenvalues of $\pi$ and also all of its archimedean data. From here, you can write down explicitly the Fourier expansion of the newform of $\pi$ (adèlically, this is its Whittaker expansion). Note that you need to be a little careful, since the constant term in the Fourier expansion does not necessarily vanish: $\pi$ is cuspidal if and only if $\Omega$ does not factor through the norm map; otherwise, the newform associated to $\pi$ is an Eisenstein series.
Belatedly noticed this question... Yes, by this year, as @PeterHumphries said, the fact is an instance (perhaps the simplest) of "automorphic induction". That's the larger picture.
The specific case in the question can be described (and proven) in a much more direct way: binary (pluriharmonic) theta series attached to everywhere locally positive-definite quadratic forms (probably fooling around with generalized idea classes, if done classically). The proof that such things are modular forms is not entirely trivial, but can be (and was) done in a fashion that does not require the full-blown set-up of Segal-Shale-Weil/oscillator repn. (R. Gunning's little orange book on modular forms proves that theta series attached to even-dimensional quadratic forms are elliptic modular forms, for example. Closer examination of the argument leads a person to the theta-correspondence idea... My old book on Hilbert modular forms does a mildly adelic, but not overtly representation-theoretic, proof that Hilbert modular theta series are Hilbert modular forms.)
And, yes, the archimedean theta correspondence sends the trivial repn of real-anisotropic orthogonal groups to holomorphic discrete series. Also, non-trivial repns, attached to spherical harmonics, go to holomorphic discrete series... That's just a local question, answerable by looking for "lowest" K-types.
At unramified finite primes, some computations with Jacquet modules tell which unramified principal series the (local) theta correspondence maps to. (Thinking in terms of the Borel-Casselman-Matsumoto theorem about imbeddings to unramified principal series.)
A big part of the point of "adelizing" is to realize that the same arguments that work over $\mathbb Q$ really do work generally, because most of it is local... As with Iwasawa-Tate for $GL(1)$: once one gets a grip on the slight fanciness, the point is that it's the same shape as Riemann's argument, simply allowing the symbols/terminology to refer to somewhat subtler things. As in $\mathbb A/\mathbb Q$ in place of $\mathbb R/\mathbb Z$.
|
2025-03-21T14:48:31.566102
| 2020-07-20T23:27:09 |
366152
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631294",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366152"
}
|
Stack Exchange
|
Continued fractions and class groups
Let $d$ be a positive integer. It is well-known (due to Lagrange) that the continued fraction of $\sqrt{d}$ is eventually periodic. Moreover, it is known that the equation
$$\displaystyle x^2 - dy^2 = -1$$
is soluble in integers $x,y$ if and only if the continued fraction of $\sqrt{d}$ has odd period.
An alternative criterion is given in terms of the class group $\text{CL}(K_d)$ of $K_d = \mathbb{Q}(\sqrt{d})$ and the narrow class group $\text{CL}^\sharp(K_d)$. The difference is that the former mods out all principal ideals, while the latter only mods out principal ideals of positive norm. Indeed, we see that the above equation, which exactly identifies when $\mathcal{O}_{K_d}$ has a unit of norm $-1$, implies that the two groups are equal. The converse is also true.
More precisely, we don't need to look at the entire class group/narrow class group, but only the $2^k$-torsion subgroups for $k = 1, 2, \cdots$. Indeed, we have the following alternative criterion: the equation $x^2 - dy^2 = -1$ is soluble in integers $x,y$ if and only if $\text{CL}(K_d)[2^k] \cong \text{CL}^\sharp(K_d)[2^k]$ for all $k \geq 1$ (note by the finiteness of the class group, both sides eventually stabize).
Is there deeper connection between continued fractions of $\sqrt{d}$ and the $2^k$-torsion criterion above?
For instance, we know that for all primes $p \equiv 1 \pmod{4}$ the equation $x^2 - py^2 = -1$ is soluble, which means that for any such prime $\sqrt{p}$ always has odd period. Is there any reason why one should believe this looking only from the continued fraction side?
|
2025-03-21T14:48:31.566240
| 2020-07-20T23:36:30 |
366153
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Dragon lala lalo",
"https://mathoverflow.net/users/161411"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631295",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366153"
}
|
Stack Exchange
|
Literature on the polynomials and equations, in structures with zero-divisors
I need literature about zeroes of polynomials and equation resolution in associative algebraic structures with zero-divisors, but I am having difficulties to find it.
For example, there is literature in the resolution of quadratic(and remarks on other degrees) in quaternions and split-quaternions.
It seems that search engines do something with the keyword zero, that has two meanings in this context, this is, zero-divisor and zeroes of the polynomials.
So far I do not find ways to more concrete examples.
It would be fine also, algebraic literature ( as opposed to specific examples )
Thanks in advance
For associative algebras, as your required, see Plotkin, Algebras with the same (algebraic) geometry, Israel J. Math., 96 (2) (1996), 511–522.
This is, being more precise, part of this nice relatively new field of Universal Algebraic Geometry which discuss such things.
For a survey I recommend A. Shevlyakov, Lectures notes in universal algebraic geometry, https://arxiv.org/abs/1601.02743.
EDIT depending on the kind of question you are interested in, pherhaps the theory of (associative) PI-algebras might be also interesting. For this I recommend V. Drensky's Free Algebras and PI-algebras.
I forget to add other details, specifically, the property of multiplicative inverse for nonzero elements. In anycase, I will examine the literature that you had provided ;)
|
2025-03-21T14:48:31.566371
| 2020-07-21T00:53:01 |
366155
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Boris Bilich",
"Marco Farinati",
"https://mathoverflow.net/users/140292",
"https://mathoverflow.net/users/98863"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631296",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366155"
}
|
Stack Exchange
|
Examples of "non equivalent" algebras that are derived equivalent?
One can define different equivalence relations between algebras depending on what one want to study, but also these definitions may have their own life and not result as one expected at first.
My interest is in derived equivalence. Which is your favorite example of two algebras A and B such that D(A) is triangulated equivalent to D(B) but one feel that they are really ''different''? Maybe another way to formulate the question is: what is your favorite invariant that ''should be'' derived invariant but it is not?
My personal example is an algebra A with finite global dimension, and B another algebra derived equivalent to A but with different global dimension. Do someone has any other ''example'' of ''non-equivalent'' algebras A and B that are still derived equivalent?
Can you give a reference for your example?
I hope someone else give a concrete minimal example. But what I can tell you is that derived equivalences given by non projective tilting objects do not preserve global dimension.
|
2025-03-21T14:48:31.566478
| 2020-07-21T03:37:58 |
366157
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Narutaka OZAWA",
"Rodrigo Dias",
"https://mathoverflow.net/users/101270",
"https://mathoverflow.net/users/7591"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631297",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366157"
}
|
Stack Exchange
|
Equivalence of families indexes of Fredholm operators
Let $F=F(H,H)$ be the space of bounded Fredholm operators in a Hilbert space $H$ with topology inherited from the norm operator topology, and let $X$ be a compact topological space.
For a continuous map $T\colon X\to F$, there exists a closed subspace $W\subseteq H$ with $\dim H/W<\infty$ such that $W\cap\ker T_x=0$ for all $x\in X$ and $H/T(W) =\bigcup_{x\in X} H/T_x(W)$ is a vector bundle over $X$ (See appendix of K-Theory, Anderson & Atiyah).
Then one can show that
$$\mbox{Ind}_1(T) = [X\times H/W] - [H/T(W)] \in K(X)$$
does not depend on $W$.
On the other hand, there exists a finite dimensional subspace $V\subseteq H$ such that $V+T_x(H) = H$ for all $x\in X$, so we can define $T^V\colon X\to F(H\oplus V, H)$ by $T^V_x(u,v) = T_x u + v$. Then $T^V_x$ is surjective and $\dim\ker T^V_x$ is constant on $x$. Thus $\ker T^V = \bigcup_{x\in X} \ker T_x$ is also a vector bundle over $X$. One can show that
$$ \mbox{Ind}_2(T) = [\ker T^V] - [X\times V] \in K(X)$$
does not depend on $V$.
These index maps are called the family index of families of Fredholm operators in $H$, and it made me suspect that they are equal.
Question: Is it true that $$[X\times H/W] - [H/T(W)] = [\ker T^V] - [X\times V] \qquad (1)$$ in $K(X)$ ?
Is there any reference that proves the equivalence of these indexes?
Edit: We can shrink $W$ or augment $V$ in order to have $\dim H/W = \dim V$. Say $H/W \cong V \cong \mathbb{C}^N$, so that $X\times H/W \cong X\times V \cong X\times\mathbb{C}^N$, and therefore
$$\mbox{Ind}_1(T) = [X\times\mathbb{C}^N] - [H/T(W)]\ ,$$
$$\mbox{Ind}_2(T) = [\ker T^V] - [X\times\mathbb{C}^N]\ .$$
Equation $(1)$ becomes
$$[X\times\mathbb{C}^N] - [H/T(W)] = [\ker T^V] - [X\times\mathbb{C}^N]$$
and it holds iff there exists $k\geq0$ such that
$$\ker T^V \oplus H/T(W) \oplus (X\times\mathbb{C}^k) \cong X\times\mathbb{C}^{2N+k}$$
But why does there exists such $k$?
Notice that $V^\perp\cap\ker T_x^*=0$ for every $x$, because for every $w\in V^\perp\cap\ker T_x^*$, $u\in H$, and $v\in V$, one has
$$\langle w,T_x(u)+v \rangle = \langle w,T_x(u) \rangle = 0.$$
By composing two isomorphisms $\ker P_{V^\perp}T \ni u \mapsto u\oplus T(-u) \in \ker T^V$ and $\ker P_{V^\perp}T\cong H/T^*(V^\perp)$, one sees $\mathrm{Ind}_2(T)=-\mathrm{Ind}_1(T^*)$.
Thus the quality of two indices follows from the fact that self-adjoint Fredholm operator
$\left[\begin{smallmatrix} & T^*\\ T & \end{smallmatrix}\right]$ has index zero in either definition.
Thank for your answer, Narutaka! I've already understood the isomorphisms, but I am struggling to prove that $\mbox{Ind}_2(T\oplus T^*)=0$ in $K(X)$. Do you have any hint?
@Rodrigo Dias: $T\oplus T^*$ is homotopic to $T^*T\oplus I$ and then to $(I+T^*T)\oplus I$, which has index zero because it is invertible.
|
2025-03-21T14:48:31.566682
| 2020-07-21T04:53:28 |
366159
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Pavel Čoupek",
"https://mathoverflow.net/users/60903"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631298",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366159"
}
|
Stack Exchange
|
An example of a special $1$-dimensional non-Noetherian valuation domain
I am looking for a $1$-dimensional non-Noetherian valuation domain $R$ such that there exists a sequence $\{a_i\}_{i=1}^\infty$ of elements of $R$ such that $\langle a_1\rangle \subsetneqq\langle a_2\rangle \subsetneqq\langle a_3\rangle \subsetneqq\cdots$ and for each $n\in \mathbb{N}$, $\langle a_i\rangle \subsetneqq\langle a_{i+1}^n\rangle$ for all $i$.
This does not exist. The fact that a valuation ring has dimension $1$ is equivalent to the value group being "rank $1$", i.e. archimedean, which in turns translates to the property of the valuation ring: For any two nonzero elements $a, b$, there is $n$ such that $(b^n) \subseteq (a)$. The condition on ideals you impose would in fact mean that the valuation is of "infinite rank", which translates to $R$ having infinite dimension.
This isn't possible in any $1$-dimensional quasi-local domain $D$. If $a,b\in D$ and $b$ is not a unit then consider the multiplicative set $S$ generated by $b$. Clearly $S$ is not disjoint from any nonzero prime of $D$, so $S = D \setminus 0$. Thus $a \in S$ which means some power of $b$ divides $a$.
If you had an ascending chain $(a_i)$ which satisfied, for any $i$, the condition $(a_i) \subsetneq (a_{i+1}^n)$ for all $n \in \mathbb{N}$, then that chain would have to stabilize at $(a_{i+1}) = D$.
|
2025-03-21T14:48:31.566931
| 2020-07-21T05:26:47 |
366160
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"David Roberts",
"https://mathoverflow.net/users/4177"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631299",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366160"
}
|
Stack Exchange
|
Size of the category of cohomology theories
I'd like to understand the structure of the functor category Coh whose objects are cohomology functors from a category of Spaces to the category of graded commutative rings GCR. Spaces could be any of the familiar geometric categories: topological spaces, manifolds, algebraic varieties, schemes, etc.
My first question is about just the size of Coh. For any choice of Spaces there are several well known cohomology functors (singular, de Rham, etale, ...) and new ones keep cropping up (syntomic, prismatic) - and they all agree too, on suitable subcategories of Spaces after suitable extensions of scalars, hinting at a fundamental core to it all (e.g., motives) - but I don't know how many more there can be. Is there a systematic way to enumerate or even construct them all? The latter is extremely unlikely because construction of any one we have has been a highly creative and painstaking task, but can we at least know how many in some sense are still out there? Is it even a discrete set, or can we in fact "deform" cohomology theories in families in certain settings?
Same question of size applies to the sets of natural transformations among cohomology theories, i.e., the Hom sets in Coh. Apart from the standard comparison isomorphisms what do we know about other natural transformations, even just for two well known cohomologies, for example, Betti and de Rham?
Sorry if the scope of the question is too broad and I should have made it more manageable by fixing a particular category of spaces and coefficient system for their cohomology. Please feel free to pick a setting that makes for a satisfactory answer.
When you say "size" of a category. what do you mean? Large vs small? Countably-many (vs $\kappa$-many) isomorphism classes of objects? Do you allow things like differential cohomology? Twisted cohomology? This is an interesting question, but an answer will have to explain the things you haven't asked.
Welcome to MathOverflow nms! I guess that there are at least two possible answers to your question:
in algebraic topology, Brown's representability theorem says that cohomology theories are represented by spectra, so you can construct a new cohomology theory by constructing a new spectrum. The category of spectra can also be proved (see Remark <IP_ADDRESS> in Lurie's "Higher Algebra" for instance) to be presentable, which is a suitable "smallness" condition (see Chapter Five of Lurie's "Higher Topos Theory");
in algebraic geometry, there are also suitable analogues of Brown's representability, which assert that cohomology theories with suitable properties are representable as motivic spectra, by which one means objects in the categories $\mathrm{DA}_\tau(S;\Lambda)$ and $\mathrm{SH}_\tau(S;\Lambda)$ built out of a scheme $S$, a ring $\Lambda$ and a Grothendieck topology $\tau$ using $\mathbb{A}^1$-homotopy theory (see Ayoub's thesis or the book of Cisinski and Déglise for a detailed analysis of the construction of these categories). Such representability results appear for instance in the paper "Mixed Weil cohomologies" by Cisinski and Déglise, or in §1 of the paper "The rigid syntomic ring spectrum" by Déglise and Mazzari, or in Drew's thesis. I suggest also taking a look at the appendix of the book "Many variations of Mahler Measures" by Brunault and Zudilin, for an application of this representability theorems to Deligne-Beilinson cohomology. Finally, questions about the sizes of the sets of morphisms in the categories $\mathrm{DA}_\tau(S;\Lambda)$ and $\mathrm{SH}_\tau(S;\Lambda)$ are very difficult to answer: for instance, it seems completely out of reach to prove that the motivic cohomology groups $\mathrm{Hom}(\mathbf{1}_X,\mathbf{1}_X(j)[i])$ are finitely generated, under suitable assumptions on $X$ (e.g. $X$ smooth and proper over $\mathbb{Q}$).
It is perhaps more natural to study homology theories (or cohomology theories) up to `Bousfield equivalence', where two theories are equivalent if they send the same maps to isomorphisms. (So, for example, classical cohomology with coefficients in a field $k$ are sorted by the characteristic of $k$.) This has been studied, with much sophistication, beginning with papers by Bousfield around 1970, who wrote about the lattice of such localization functors, and gave examples of both orderly and unusual behavior.
I would suggest looking up his papers, and also those of Mark Hovey (e.g. Cohomological Bousfield classes. J. Pure Appl. Algebra 103 (1995), no. 1, 45–59, or Hovey, Mark; Palmieri, John H. The structure of the Bousfield lattice. Homotopy invariant algebraic structures (Baltimore, MD, 1998), 175–196, Contemp. Math., 239, Amer. Math. Soc., Providence, RI, 1999.) A paper that shows that versions of your question can be proved to be undecidable is Casacuberta, Carles; Scevenels, Dirk; Smith, Jeffrey H. Implications of large-cardinal principles in homotopical localization. Adv. Math. 197 (2005), no. 1, 120–139. Have fun exploring this!
[I should add that, when restricted to the category of finite CW complexes, the Bousfield classes are known: they are detected by the sequence of Morava K-theories associated to each ordinary prime.]
|
2025-03-21T14:48:31.567292
| 2020-07-21T05:35:01 |
366161
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631300",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366161"
}
|
Stack Exchange
|
What are some interesting applications of the Archimedean Property?
So a wile back I managed to prove the The Remainder Theorem starting from the Archimidean property and since then I've thought what could be other results which can be proved using it. But I haven't found any other applications for the Archimidean property, except some very small, almost trivial, results in Analysis. So are there any important results that use it?
|
2025-03-21T14:48:31.567351
| 2020-07-21T05:52:37 |
366162
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"George Shakan",
"Seva",
"https://mathoverflow.net/users/50426",
"https://mathoverflow.net/users/9924"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631301",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366162"
}
|
Stack Exchange
|
Origins of the ``baby Freiman'' theorem
It is a basic folklore fact from the area of additive combinatorics that a subset $A$ of an abelian group satisfies $|2A|<\frac32\,|A|$ if and only if $A$ is contained in a coset of a (finite) subgroup $H$ with the density $|A|/|H|>\frac23$. What are the historical origins of this fact? It was certainly known to Freiman around years 1961-1965, but I suspect that it can appear somewhere in the earlier papers by Kemperman / Scherk / Mann / Olson.
https://terrytao.files.wordpress.com/2013/11/expander-book.pdf In this book, Theorem 1.5.2, Tao attributes it to a 73 paper of Freiman.
@GeorgeShakan: It is true that Theorem 1.5.2 from Tao's book is presented in Freiman's [1973b], but I am by far not sure that the abelian version of this theorem had not appeared elsewhere before.
|
2025-03-21T14:48:31.567447
| 2020-07-21T06:17:53 |
366164
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"ABIM",
"Jochen Wengenroth",
"Martin Väth",
"https://mathoverflow.net/users/165275",
"https://mathoverflow.net/users/21051",
"https://mathoverflow.net/users/36886"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631302",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366164"
}
|
Stack Exchange
|
Conditions for continuity of an integral functional
Let $X$ be a metric space, $\nu,\mu$ be Borel measures on $X$, $f:X\times \mathbb{R}\rightarrow [0,\infty)$ be a measurable function. Under what conditions is the integral functional $F_f$, defined by:
$$
\begin{aligned}
F_f: L^1(X) & \rightarrow [0,\infty]
\\
g&\mapsto \int_{x \in X} f(x,g(x))d\mu(x)
\end{aligned}
$$
continuous?
I expect that $f$ should at-least be Carath'{e}odory; i.e.: measurable in its first argument and continuous in its second, and probably some sort of growth condition such as $\int_{x\in X} \sup_{y \in \mathbb{R}}f(x,y)d\mu(x)<\infty$ to ensure that $F_f$ is finite-valued...
I'm particularly interested in the case where $X$ is $\mathbb{R}^n$ or is a topological manifold modelled thereon.
(Does the situation simplify when $F_f$ is instead considered on $C(X)$ with the uniform topology and $X$ were a compact space?)
A simple sufficient condition for the continuity is that $f$ is Lipschitz with respect to the second argument, i.e., $|f(x,y)-f(x,z)|\le c|y-z|$ for some constant independent of $x,y,z$.
What about when $L^1$ is replaced by $C(\mathbb{R})$? I don't this works then... we probably need some integrability contstraint also?
Besides some condition ensuring that the map is well defined a simple and natural condition for the continuity is a Lipschitz condition with respect to the second variable, i.e., $$|f(x,y)-f(x,z)|\le c|y-z|$$ for some constant $c$ independent of $x,y,z$. This is a standard assumption for (a version of) the Picard-Lindelöf theorem for ODEs.
I repeated my comment as an answer to make the following remark more visible: Please do not try to minimize space but try to maximize the readability.
I hate to read somthing like
Using [xx, lemma 2.3] in combination with [yy, theorem 4.17] the reader can easily check that $F_f$ is continuous whenever $f$ satisfies condition ($\ast$) in [zz, proposition 4].
Meanwhile the OP deleted his comment to which my answer was a reaction.
Thank you for posting this and especially for the writing tip. I appreciate it.
Since $\nu$ does not appear anywhere else in the question, I suppose that $L^1(X)=L^1(\nu)$.
In order that the functional be defined, one should then assume (probably without loss of generality) that $\mu$ is absolutely continuous w.r.t. $\nu$.
In this case, the two conditions in the question are more than sufficient, and far from being necessary.
To prove that they are sufficient, it suffices to show that for each sequence $x_n\to x$ in $L_1(\nu)$ there is a subsequence with $F_f(x_{n_k})\to F_f(x)$. Since $x_n\to x$ in $L_1(\nu)$, there is a subsequence with $x_{n_k}\to x$ $\nu$-a.e., hence by hypothesis $\mu$-a.e. Since $f$ is Carathéodory, it follows that $g_{n_k}(t)=f(t,x_{n_k}(t))\to g(t)=f(t,x(t))$ for $\mu$-a.e. $t$. It remains to apply Lebesgue's dominated convergence theorem with the dominating function being $\sup_y f(\cdot,y)$.
Using Vitali's instead of Lebesgue's dominated convergence theorem, one can replace the strong integral hypotheses by various sorts of growth conditions, depending on $\mu/\nu$. For instance, in case $\mu=\nu$ the growth condition $f(t,y)\le a(t)+C|y|$ with a constant $C$ and some $\mu$-integrable $a$ is sufficient.
Edited the last condition so that the result becomes more general, and the growth condition is of Krasnosel'skij type.
|
2025-03-21T14:48:31.567699
| 2020-07-21T07:15:00 |
366166
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"0xbadf00d",
"Zach Teitler",
"https://mathoverflow.net/users/88133",
"https://mathoverflow.net/users/91890"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631303",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366166"
}
|
Stack Exchange
|
Best approximation with tensors of rank $\ge2$
Let $k\in\mathbb N$, $H_i$ be a (finite-dimensional, if necessary) $\mathbb R$-Hilbert space for $i\in I:=\{1,\ldots,k\}$, $H:=\bigotimes_{i\in I}H_i$ denote the tensor product$^1$ of $(H_i)_{i\in I}$ and $$H^{(r)}:=\left\{u\in H:\operatorname{rank}u=r\right\}\;\;\;\text{for }r\in\mathbb N_0.$$
I'm struggling to understand the importance and implication of the following result: Let $v\in H$.
There is a $u\in H$ with $\operatorname{rank}u=1$ and $$\left\|u-v\right\|_H=\inf_{u\in H^{(1)}}\left\|u-v\right\|_H\tag1.$$
There is a $u\in H^{(3)}$ and a $(u_n)_{n\in\mathbb N}\subseteq H^{(2)}$ with $$\left\|u_n-u\right\|_H\xrightarrow{n\to\infty}0\tag2.$$
Okay, by 1., there is a (not necessarily unique) minimizer of $$H^{(1)}\to[0,\infty)\;,\;\;\;u\mapsto\left\|u-v\right\|_H\tag3.$$
Question 1: But why can we infer from 2. that the analogous problem of minimizing $$H^{(2)}\to[0,\infty)\;,\;\;\;u\mapsto\left\|u-v\right\|_H\tag4$$ may have no solution? I guess we need to take $v=u$ (with $u$ as in 2.), but why does the existence of $(u_n)_{n\in\mathbb N}$ imply that there is no solution?
Question 2: That we can only guarantee the existence of a minimizer of $$H^{(r)}\to[0,\infty)\;,\;\;\;u\mapsto\left\|u-v\right\|_H\tag5$$ for $r=1$ is unsatisfactory only if it would actually be beneficial to take $r$ as large as possible. I could imagine that the error $\left\|u^{(r)}-v\right\|_H$ of a hypothetical minimizer $u^{(r)}$ of $(5)$ is nonincreasing in $r\in\mathbb N_0$. Is this the case? If so, how can we show this?
Question 3: Can we infer from 2. that $(5)$ may have no minimizer for all $r\ge2$?
$^1$ If $E_i$ is a $\mathbb R$-vector space, I'm defining $$(x_1\otimes x_2)(B):=B(x_1,x_2)\;\;\;\text{for }B\in\mathcal B(E_1\times E_2)\text{ and }x_i\in E_i,$$ where $\mathcal B(E_1\times E_2)$ is the space of bilinear forms on $E_1\times E_2$, and $$E_1\otimes E_2:=\operatorname{span}\{x_1\otimes x_2:E_i\in E_i\}\subseteq{\mathcal B(E_1\times E_2)}^\ast.$$
The rank strata are generally not closed so it should not be surprising that minimzers may not exist. It’s more natural to take the union of rank strata for ranks up to $r$ (sadly these unions are still generally not closed, except in the matrix case or when $r=1$), and being nested it’s then clear that errors are nonincreasing in $r$.
I don't know about infinite-dimensional Hilbert spaces and arbitrary $v \in H$. For any $I, J, K \in \mathbb{Z}_{\geq 1}$ and for $H_1 = \mathbb{R}^I, H_2 = \mathbb{R}^J, H_3 = \mathbb{R}^K$, there is an example of a rank-3 $v \in H_1 \otimes H_2 \otimes H_3$, which can be approximated arbitrarily well by a rank-2 $u \in H_1 \otimes H_2 \otimes H_3$. You can find the example in http://www.kolda.net/publication/koba09/ page 469.
Thank you for your answer. Yes, I know this particular example. That's the point of (2.). The question is, why the desired conclusions are possible. The existence of $u$ as in this example shows that $H^{(r)}$ is not closed though.
|
2025-03-21T14:48:31.567889
| 2020-07-21T07:49:52 |
366168
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Paul",
"https://mathoverflow.net/users/126827"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631304",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366168"
}
|
Stack Exchange
|
$x '(t) + g (x (t)) = f (t),\quad \forall t\in \mathbb R$ have periodic solution $\iff\; \frac 1T \int_0 ^ T f (t) dt \in g (\mathbb R) $
I have a research work concerning the equation: $$x '(t) + g (x (t)) = f (t),\quad \forall t\in \mathbb R$$
f and g are defined and continuous in $\mathbb R$ and with values in $\mathbb R$.
Furthermore f is assumed to be T periodic (There is no initial condition)
First question:
Assume that there is a periodic solution. Using the mean formula, to show that $\frac 1T \int_0 ^ T f (t) dt \in g (\mathbb R) $
Second question
Show that this condition : $\frac 1T \int_0 ^ T f (t) dt \in g (\mathbb R) $ is sufficient for the existence of a periodic solution
For first question:
$ g \circ x $ being continuous, there is a $ c $
between $ 0 $ and $ T $ such as $ \frac 1T\int_0 ^ T
g (x (s)) ds = g (x (c)) $. but $ x $ is a solution
$ T $ periodic, then $ x (T) = x (0) $, and we
will have:
$$\frac1T \int_0 ^ T f (s) ds = \frac 1T \int_0 ^ T
(g (x (s)) + x '(s)) ds = \frac1T \int_0 ^ T
g (x (s)) ds = g (x (c)) \in g (\mathbb R). $$
For second question:
I need help how to use some one of fixed point theorem or any other way to prove that the condition $\frac 1T \int_0 ^ T f (t) dt \in g (\mathbb R) $ is sufficient
The condition is not sufficient. Take for example the equation
$$x'=x^2+\sin t$$ with $g(x)=-x^2$ in your notation. Then
$$\int_0^{2\pi} \sin t\, dt=0=g(0).$$ If $x$ were a $2\pi$-periodic solution, then
$$
0=x(2\pi)-x(0)=\int_0^{2\pi} x'(t)\, dt=\int_0^{2\pi}(x^2(t)+\sin t)\, dt= \int_0^{2\pi}x^2(t)\, dt$$
would imply $x\equiv 0$ in $[0, 2\pi]$.
indeed, I will discuss the hypotheses with the supervisor. A big thank-you
|
2025-03-21T14:48:31.568010
| 2020-07-21T09:32:32 |
366175
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631305",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366175"
}
|
Stack Exchange
|
On the difference between Malliavin derivative and Gross-Sobolev derivative
Let $W=C_0([0,1],\mathbb R^d)$ be the classical Wiener space equipped with $\mu$ the Wiener measure.
If $F:W\to\mathbb R$ is a cylindrical function of the form
\begin{align*}
F(w)=f(W_{t_1}(w),\cdots,W_{t_n}(w)), f\in\mathcal S(\mathbb R^n)
\end{align*}
where $W_t$ is the $t$-th component projection, we define for $h\in H$ (the Cameron-Martin space),
\begin{align*}
\nabla_h F(w)=\frac{d}{d\epsilon}F(w+\epsilon h)\rvert_{\epsilon=0}
\end{align*}
Notice that $W_t(w+h)=W_t(w)+h(t)$ and then
\begin{align}
\nabla_h F(w)&=\frac{d}{d\epsilon}f(W_{t_1}(w+\epsilon h),\cdots,W_{t_n}(w+\epsilon h))\rvert_{\epsilon=0}\\
&=\sum_{i=1}^n \partial_i f(W_{t_1}(w+\epsilon h),\cdots,W_{t_n}(w+\epsilon h))) h(t_i) \rvert_{\epsilon=0}\\
&=\sum_{i=1}^n \partial_i f(W_{t_1}(w),\cdots,W_{t_n}(w)) h(t_i).
\end{align}
At this point we can see that the map $h\mapsto \nabla_h F(w)$ defines a random variable taking values in the dual space $H^*$, hence if we identify $H$ with its dual we have that $\nabla F$ is an $H$-valued random variable.
If we extend this operator to all Wiener functionals (using the density of the cylindrical ones) we obtain the so called "Gross-Sobolev" derivative.
This definition looks pretty similar to the definition of the Malliavin derivative, but I don't believe that the Mallivin derivative is a random variable taking values in the Cameron Martin space.
My idea on how things go (I am not completely sure and that's why I am asking you this) is the following:
The last expression above can be written as
\begin{align}
&=\sum_{i=1}^n \partial_i f(W_{t_1}(w),\cdots,W_{t_n}(w)) \int_0^1 1_{[0,t_i]}(s)\dot{h}(s)ds\\
&=\sum_{i=1}^n \partial_i f(W_{t_1}(w),\cdots,W_{t_n}(w)) \langle 1_{[0,t_i]},\dot{h}\rangle_{L^2(\mu)}\\
&=\langle DF,\dot{h}\rangle_{L^2(\mu)}\\
&=\bigg\langle \int_0^{\cdot}D_rF dr,h\bigg\rangle_{H}=\langle \nabla F,h\rangle_H.
\end{align}
Where $DF$ is the classical Malliavin derivative of $F$. Then if what I've done above is correct the Gross-Sobolev derivative equals the time integral of the Malliavin derivative.
Can we thing of them as being the same since the space $L^2(\mu)$ is isometric and isomorphic to $H$?
Is this correct, or is it just a coincident?
Thanks in advance for any feedback.
|
2025-03-21T14:48:31.568148
| 2020-07-21T09:48:56 |
366177
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Wojowu",
"https://mathoverflow.net/users/30186"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631306",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366177"
}
|
Stack Exchange
|
What subsystem of second-order arithmetic is needed for the recursion theorem?
In its simplest version, the recursion theorem states that for any $m\in\mathbb{N}$ and any function $g:\mathbb{N}\rightarrow\mathbb{N}$, there exists a function $f:\mathbb{N}\rightarrow\mathbb{N}$ such that $f(0)=m$ and $f(n+1) = g(f(n))$. There are many more complicated versions, with multiple variables and parameters and course-of-values recursion and so on. But that’s the gist of it.
Now the recursion theorem, no matter which version of it you take, is a statement in the language of second-order arithmetic. And I'm pretty sure that it’s provable in $Z_2$, i.e. full second-order arithmetic. But my question is, what is the weakest subsystem of second-order arithmetic capable of proving it?
Do different versions of the recursion theorem require different subsystems to prove it?
It looks to me like any such version should be provable in RCA_0.
As Wojowu already pointed out $\mathsf{RCA}_0$ proves recursion theorem. You could find a proof in Simpson's book [1], Section II.3.
In fact primitive recursion theorem is equivalent to $\Sigma^0_1\textsf{-Ind}$ over $\mathsf{RCA}_0^{\star}$. Here $\mathsf{RCA}_0^{\star}$ is $\mathsf{EA}+\Delta^0_1\text{-}\mathsf{CA}+\Delta^0_0\text{-}\mathsf{Ind}$ and $\mathsf{RCA}_0=\mathsf{RCA}_0^{\star}+\Sigma^0_1\text{-}\mathsf{Ind}$. So we need to prove in $\mathsf{RCA}_0^{\star}+\mathsf{PrimRec}$ any given instance
$$\exists y\;\varphi(0,y)\land \forall x\;(\exists y\;\varphi(x,y)\to \exists y\;\varphi(x+1,y))\to \forall x\;\exists y\varphi(x,y)$$
of $\Sigma^0_1\textsf{-Ind}$, where $\varphi$ is $\Delta^0_0$.
Indeed let us reason in $\mathsf{RCA}_0^{\star}+\mathsf{PrimRec}$. We assume $\exists y\;\varphi(0,y)\land \forall x\;(\exists y\;\varphi(x,y)\to \exists y\;\varphi(x+1,y))$ and claim $\forall x\;\exists y\varphi(x,y)$ and claim that $\forall x\exists y\varphi(x,y)$. Using $\Delta^0_1\textsf{-CA}$ and premise of induction we form the following function $g(x)$:
$$g(x)=\begin{cases}\langle y_0,\ldots,y_{n-1},\min\{y_{n}\mid \varphi(n,y_{n})\}\rangle &\text{, if $x=\langle y_0,\ldots,y_{n-1}\rangle$ and}\\ & \text{$\;\;\;$ $\varphi(i,y_i)$, for all $i< n$}\\ 0&\text{, otherwise}\end{cases}$$
We applying primitive recursion to $g$ and put $f(0)=\langle \rangle$. The resulting $f$ maps $n$ to a sequence $\langle y_0,\ldots,y_{n-1}\rangle$ such that $\varphi(0,y_0),\ldots,\varphi(n-1,y_{n-1})$. Note that the latter fact could be verified by $\Delta^0_0\textsf{-Ind}$. Thus we prove the instance of $\Sigma^0_1\textsf{-Ind}$.
[1] Simpson, S. G. (2009). Subsystems of second order arithmetic (Vol. 1). Cambridge University Press.
|
2025-03-21T14:48:31.568316
| 2020-07-21T11:15:12 |
366186
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Gerry Myerson",
"gmath",
"https://mathoverflow.net/users/158000",
"https://mathoverflow.net/users/23689"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631307",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366186"
}
|
Stack Exchange
|
How to calculate possible arrangements of hexagons?
I was wondering someone could help. I've developed a board game which is made up of six, large hexagonal board pieces, which can be arranged in any order, and with any rotation/arrangement of sides within that overall layout.
I was playing with a friend who made a laughable attempt to calculate how many possible layouts there could be, and got me thinking I literally have no idea where to start on the maths of it. Hence I'm here.
So for example the six could be laid out in a single row, or could be bunched together like in the picture... and then within that, each tile can be rotated.
The number must be astronomical, but how would I (or you!) figure it out?
Many thanks in advance and sorry to trouble you all.
Astronomical numbers are quite small, by mathematical standards.
http://oeis.org/A000228 may give you a start.
Just a thought: if you think of the dual, I think you are looking at number of ways of connecting the edges of 6 vertices of degree 6, so that the final graph is connected. Since you have numbered the sides, you would also need to number the edges. Maybe looking at the configuration model (https://en.wikipedia.org/wiki/Configuration_model) is useful.
|
2025-03-21T14:48:31.568435
| 2020-07-21T11:56:28 |
366188
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"LSpice",
"Rodrigo de Azevedo",
"heng",
"https://mathoverflow.net/users/112113",
"https://mathoverflow.net/users/161434",
"https://mathoverflow.net/users/2383",
"https://mathoverflow.net/users/91764",
"lambda"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631308",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366188"
}
|
Stack Exchange
|
How to solve a system of quadratic equations?
Suppose we have a system of $p$ quadratic equations about $\mathbf{x} \in \mathbb{R}^3$ and $\mathbf{x} > 0$
$$ \left\{
\begin{array}{lr}
\mathbf{x}^\top \mathbf{C}_1 \mathbf{x} = 1, \\
\mathbf{x}^\top \mathbf{C}_2 \mathbf{x} = 1, \\
\quad\quad \vdots\\
\mathbf{x}^\top \mathbf{C}_p \mathbf{x} = 1, \\
\end{array}
\right.$$
where matrices $\mathbf{C}_1, \dotsc, \mathbf{C}_p \in\mathbb{R}^{3 \times 3}$ are symmetric and positive definite.
Suppose $\mathbf{x} = [a,b,c]^\top$, and $\mathbf{y} = [a^2,b^2,c^2, ab, ac, bc]^\top$. It is known that the above quadratic equations can be written as
\begin{equation}
\mathbf{Ay} = \mathbf{1},
\label{eq:linearsolver}
\end{equation}
where $\mathbf{A} \in \mathbb{R}^{p \times 6}$. If $p \geq 6$, we can obtain $\mathbf{y}$ and further estimate $\mathbf{x}$.
Questions:
Is there any methods besides $\mathbf{Ay}=\mathbf{1}$ to solve the above quadratic equations?
What is the minimal value of $p$ to guarantee a unique solution of $\mathbf{x}$ of the above quadratic system?
What do you mean by "What is the minimal value of $p$ to guarantee a unique solution"? Do you mean the minimal value for which a solution can be unique? No matter how big $p$ is, $\mathbf A$ could be, say, the $0$ matrix, in which case there are no solutions, or some low-rank matrix, in which case there could be multiple solutions.
If $\mathbf x$ is a solution then so is $-\mathbf x$, so you can't ever have uniqueness. Up to sign you can certainly recover $\mathbf x$ from $\mathbf y$ so the condition for having only two solutions is indeed just that the matrix has full rank.
Have you tried in $\mathbb{R}^2$ first?
Related: https://mathoverflow.net/q/308163
@LSpice Let's say $\mathbf{A}$ is always full rank in this case. I want to know how many quadratic equations are required to obtain $\mathbf{x} \in \mathbb{R}^3$, is p=3 enough?
@lambda I agree, from $\mathbf{y}$ to $\mathbf{x}$ there's a sign ambiguity even if $\mathbf{A}$ is full rank. Actually in my case $\mathbf{x} > 0$. I want to know whether there are other methods to solve quadratic equations?
X-posted: https://math.stackexchange.com/q/3764350/339790
|
2025-03-21T14:48:31.568598
| 2020-07-21T12:23:24 |
366190
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631309",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366190"
}
|
Stack Exchange
|
There is no general method to construct n-regular polygon such that the given n-polygon inscribed the n-regular polygon
Conjecture 1: With $n\ge 5$, given general n-polygon, there is no general method to construct n-regular polygon such that the given n-polygon inscribed the n-regular polygon (with one and only one vertex belong to one sideline)
$n=3, n=4 $ the method without word in the Figure as follows, and how can show that $MNPQ$ is the right Figure is square?
Conjecture 2: With $n\ge 5$, given general n-polygon, there is no general method to construct n-regular polygon such that n-regular polygon inscribed the n-polygon (with one and only one vertex belong to one sideline).
Question: I am looking for a proof of the conjectures above? Or please give a reference to me.
|
2025-03-21T14:48:31.568670
| 2020-07-21T12:40:32 |
366192
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Iosif Pinelis",
"gsa",
"https://mathoverflow.net/users/36721",
"https://mathoverflow.net/users/50551",
"https://mathoverflow.net/users/83682",
"user123124"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631310",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366192"
}
|
Stack Exchange
|
English translation of "Une inégalité pour martingales à indices multiples et ses applications"
Does anyone know of a English translation of "Une inégalité pour martingales à indices multiples
et ses applications" by Renzo Cairoli. Or could translate the statement of the martingale convergence theoerm and his definition of multiindex martingale.
Link; http://archive.numdam.org/article/SPS_1970__4__1_0.pdf
References where theorem is stated clearly also works!
I found the notes French for Mathematicians: A linguistic approach by Joël Bellaïche very helpful. He argues that someone who knows English can learn how to read a mathematical text in French without too much effort.
@gsa thanks for the tip
$\newcommand\Om\Omega$ $\newcommand\F{\mathcal F}$ $\newcommand\M{\mathcal M}$
With the help from Google Translate:
Throughout the work $m$ is a fixed integer $\ge2$ and $j$ runs through the
integers from $1$ to $m$. For each $j$, $(\Om_j,\F_j,P_j)$ is a probability space. Set $\Om=\prod\limits_j\Om_j$, $\F=\bigotimes\limits_j\F_j$, $P=\bigotimes\limits_j P_j$. The expectation with respect to $P$ will be denoted by $E$.
The processes that we are going to consider are (unless otherwise stated)
real, defined on $(\Om,\F,P)$ and admitting as a set of indices
a set of points with $m$ coordinates of which each coordinate traverses
a countable subset of $\mathbb R$. This set will be endowed with the relation $(r_1,\dots,r_m)\le(r'_1,\dots,r'_m)$ if $r_1\le r'_1,\dots,r_m\le r'_m$.
We will denote by $\M$ the class of martingales
$$(X_{r_1,\dots,r_m}, \F_{r_1}\otimes\cdots\otimes\F_{r_m})$$
relative to an increasing family of product [$\sigma$-]fields contained in $\F$.
Apparently, here the general definition of a martingale over a directed partially ordered index set is assumed; see e.g. Section Filtrations and martingales.
Theorem 2. If for the martingale $(X_{n_1,\dots,n_m})\in\M$ we have
$$\sup_{n_1,\dots,n_m}E\{|X_{n_1,\dots,n_m}|(\log^+|X_{n_1,\dots,n_m}|)^{m-1}\}<\infty$$
(therefore, in particular, if $\sup\limits_{n_1,\dots,n_m}E|X_{n_1,\dots,n_m}|^p<\infty$ for some $p> 1$), then the limit
$$\lim_{n_1\to\infty,\dots,n_m\to\infty}X_{n_1,\dots,n_m}$$
exists (and is finite) a.s.
The notes contained what I wanted, isnt theorem 1.8 more general?
@user1 : I don't understand your comment. What notes? What theorem 1.8? Where?
http://www.gatsby.ucl.ac.uk/~porbanz/teaching/G6106S15/Chapter1_27Jan15.pdf page 6
@user1 : Yes, Theorem 1.8 in the notes linked to in my answer is more general than Theorem 2 translated in my answer, in two ways: (i) the index set in Theorem 1.8 can be any directed set and (ii) the integrability condition in Theorem 2 is of course narrower than the general uniform integrability condition in Theorem 1.8 in the notes linked to in my answer.
|
2025-03-21T14:48:31.569343
| 2020-07-21T13:22:50 |
366196
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Brendan McKay",
"Brian Hopkins",
"Béart",
"https://mathoverflow.net/users/14807",
"https://mathoverflow.net/users/161287",
"https://mathoverflow.net/users/9025"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631311",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366196"
}
|
Stack Exchange
|
There is a 3-connected 5-regular simple $n$-vertex planar graph iff $n$ satisfies....?
Is there any characterization on the set of integers $n$ such that there is a 3-connected 5-regular simple $n$-vertex planar graph?
There is a 3-connected 5-regular simple $n$-vertex planar graph if and only if $n=12$ or $n \ge 16$ is even. See Recursive generation of 5-regular graphs by Mahdieh Hasheminezhad, Brendan D. McKay, Tristan Reeves in WALCOM: Algorithms and Computation, eds. Das and Uehara, Lecture Notes in Computer Science, vol 5431, Springer 2009. The number of such graphs is given in OEIS A308489.
They use a set of 7 graphs that are irreducible under a system of expansions & reductions and, as is common for contemporary graph theory, computer assistance. E.g., "The program completed execution in 21 seconds. In total, 39621 induced subgraphs were found..."
Those up to 36 vertices are at http://users.cecs.anu.edu.au/~bdm/data/planegraphs.html .
There are no such graphs when $n$ is odd, by the handshaking lemma.
Conversely, for all even $n \geq 224$, we claim such a graph exists.
In particular, given two planar 5-regular graphs $G$, $H$ each drawn on the surface of a sphere, we can define the 'connected sum' of the graphs as follows:
remove a small disk (containing one vertex) from the sphere on which $G$ is drawn;
remove a small disk (containing one vertex) from the sphere on which $H$ is drawn;
combine the two resulting hemispheres at their equator.
The resulting graph (which may depend on the chosen vertices) has $|G| + |H| - 2$ vertices, and inherits the planarity, 5-regularity, and 3-connectedness of $G$ and $H$.
Now, given an even integer $n \geq 224$, we can find integers $i, j \geq 0$ such that $n = 2 + 10i + 58j$. Then we can construct an $n$-vertex graph with the desired properties by taking the connected sum of $i$ copies of the icosahedron and $j$ copies of the snub dodecahedron.
This leaves finitely many values of $n$ to check, namely the even numbers between 14 and 222.
great answer! Great that you present this simple construction
Fun application of the Frobenius problem going from $2+10i+58j$ to 224. Using your construction with the icosahedron and the 16-vertex graph Brendan & collaborators found covers graphs with $n = 2+10i+14j$ vertices and brings the cases to check down to 14, 18, 20, 24, 28, 34, 38, 48.
|
2025-03-21T14:48:31.569529
| 2020-07-21T14:16:46 |
366202
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"D. Thomine",
"IsingX",
"R W",
"alesia",
"https://mathoverflow.net/users/112954",
"https://mathoverflow.net/users/144686",
"https://mathoverflow.net/users/75670",
"https://mathoverflow.net/users/8588"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631312",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366202"
}
|
Stack Exchange
|
Random products of $SL(2,R)$ matrices and Furstenberg's theorem
I was recently learning Furstenberg's theorem on random products of $SL(2,R)$ matrices, and came across with a simple example that confused me:
Considering random products of two matrices $A=\begin{pmatrix}
2 &0\\
0 &1/2
\end{pmatrix}$ and $B=\begin{pmatrix}
0 &1\\
-1 &0
\end{pmatrix}$ with probability $1/2$ and $1/2$, it is mentioned in literature that the Lyapunov exponent for this random product is $0$. See, e.g., page 29 in Damanik's survey article: https://arxiv.org/abs/1410.2445
I understand that if we focus on a specific matrix element of the product, it has no well-defined limit. However, if we study its norm, say the Frobenius norm $||A||=\sqrt{\sum_{i,j}|a_{ij}|^2}$. I found that by taking average, this norm indeed grows exponentially, and the Lyapunov exponent is positive.
So my question is, why it is said "the lyapunov exponent for this example is $0$" in literature? Thanks!
Because its Lyapunov exponent is zero, but that is not what you are computing.
Instead of looking at random walks on $SL(2, \mathbb{R})$, let me focus on random walks on $\mathbb{R}_+^*$, as there is the same issue. Let $(X_n)$ be i.i.d. in $\mathbb{R}_+^*$, and to make things simple, assume that there are only finitely many values. Let $P_n := X_n \ldots X_1$.
The Lyapunov exponent of this random walk is the real $\Lambda$ such that
$$\lim_{n \to + \infty} \frac{\ln (P_n)}{n} = \Lambda.$$
By the law of large numbers, $\Lambda = \mathbb{E} (\ln(X_1))$. For instance, if $X_1$ takes values $2$ and $1/2$ each with probability $1/2$, the Lyapunov exponent is $0$: the Markov chain $(P_n)$ will oscillate between very large and very low values.
However, if you compute the expectation of the norm, a short computation gets you $\mathbb{E} (P_n) = (5/4)^n$, which grows exponentially fast. But that does not mean that the Lyapunov exponent is $\ln (5/4)$. The issue is merely that the exponential does not commute with the expectation:
$$1 = e^{\mathbb{E}(\ln(P_n))} \leq \mathbb{E} (e^{\ln(P_n)}) = (5/4)^n.$$
To get back to a general random walk, and very roughyl, we have $\ln(P_n) \simeq \mathcal{N} (n\mu, n\sigma^2)$. The Lyapunov exponent is the constant $\mu$. However,
$$\mathbb{E} (P_n) \simeq \mathbb{E} (e^{\mathcal{N} (n\mu, n\sigma^2)}) = \mathbb{E} (e^{n(\mu+\frac{\sigma^2}{2})}),$$
so taking the norm as you get gives you an error coming from the diffusion of the random walk (well, in practice, the exact value of $\sigma^2/2$ for this error is wrong, but I don't think the heuristics is too bad at this level).
I seem to struggle finding a good reference for the multiplicative ergodic theorem in the context of markov processes, does that exist?
Dear @D.Thomine, thanks! You answered my question! Based on your answer, I realized that what I calculated is the expectation value of the norm. You comment that " exponential does not commute with the expectation" solved my puzzle.
You are not alone to be confused by this fact - an economist was recently so much impressed by this fact ("exponential does not commute with the expectation") that made it a whole new economical discipline "ergodicity economics".
@alesia: unfortunately, I don't know of one. I learnt it via study groups in ergodic theory (using Oseledets theorem, etc.), so without the vocabulary of Markov processes.
|
2025-03-21T14:48:31.569772
| 2020-07-21T14:33:25 |
366203
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631313",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366203"
}
|
Stack Exchange
|
Motivic cohomology and resolution of singularities
Suppose $X$ is a singular variety and we somehow know its motivic cohomology groups. Can the knowledge of motivic cohomology help in understanding the singularities of $X$ and how to resolve them?
|
2025-03-21T14:48:31.569825
| 2020-07-21T14:35:06 |
366205
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"KConrad",
"asrxiiviii",
"https://mathoverflow.net/users/157984",
"https://mathoverflow.net/users/3272"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631314",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366205"
}
|
Stack Exchange
|
Do roots of polynomial with coefficients in a CM field lie in a CM field?
This is something that I have been thinking about for a while now, not sure if it is standard (or even true at all) or not:
Let $K/ \mathbb Q$ be a CM number field, that is, it is closed under complex conjugation, which induces an automorphism that commutes with the other $\mathbb Q$-embeddings of $K$ into the field of complex numbers. Let $P(z) \in K[z]$ be irreducible and monic. Will the splitting field of $P$ over $K$ be CM? If not, does there exist a CM field containing the roots of $P$?
Can we say something if $K/\mathbb Q$ is Galois and CM (so complex conjugation commutes with the rest of the Galois group)?
Partial progress(?):
If $\alpha_1, \cdots , \alpha_d$ denote the roots of $P(z):=\sum_{j=0}^d a_jz^j \in K[z]$ lying in some splitting field $E/K$, then Vieta's relaions tell us that $a_j = (-1)^{n-j}e_j(\alpha_1, \cdots , \alpha_d)$ where $e_i$ denotes the $i$-th symmetric function. Thus for any $\sigma \in \text{Gal}(E/\mathbb Q)$, we have
$$(-1)^{n-j} e_j(\overline{\sigma\alpha_1}, \cdots ,\overline{\sigma\alpha_d}) = \overline{\sigma a_j} = \sigma({\overline{a_j}}) = (-1)^{n-j} e_j(\sigma(\overline\alpha_1), \cdots , \sigma(\overline\alpha_d))$$
Since the tuples $\{\overline{\sigma\alpha_1}, \cdots ,\overline{\sigma\alpha_d}\}$ and $\{\sigma(\overline\alpha_1), \cdots , \sigma(\overline\alpha_d)\}$ have equal values of elementary symmetric functions, they must be roots of the same polynomial and hence must be equal upto permutation. What I want however is that $(\overline{\sigma\alpha_1}, \cdots ,\overline{\sigma\alpha_d}) = (\sigma(\overline\alpha_1), \cdots , \sigma(\overline\alpha_d))$, the equality holding as ordered tuples.
Maybe I am missing something really simple but the observation that I just obtained also makes me fear that a counterexample might also be constructible (no pun intended) by just enforcing these conditions, although I haven't been able to construct the same. If none of these hold, can we say something along these lines under some (hopefully minor) additional conditions? I would really appreciate any help, thanks.
Edit: So I've already asked this in a comment, but I feel like it is more appropriate a an edit. As Prof. KConrad has pointed out, there is a very simple counterexample. That leads me to ask the following:
Is there anything we can say concerning the relationship between $\sigma\overline\alpha$ and $\overline{\sigma\alpha}$ for some $\alpha$ which is a root of a monic irreducible polynomial with coefficients in a CM field $K$ (where $\sigma$ is some $\mathbb Q$-automorphism of a finite Galois extension over $K$ containing the conjugates of $\alpha$)? What about when we impose conditions on $\sigma$? Or on $\alpha$ (like say, $\alpha$ has to be an algebraic integer)? I might be naive to expect this, in which case I would appreciate some examples suggesting very erratic behaviour.
Edit 2: I suppose a more precise reformulation of the question asked in the above edit might be the following: what we can say about the size (absolute value) of the element $|(\sigma\tau - \tau\sigma)\alpha|$ where $\alpha$ and $\sigma$ are as above and $\tau$ is complex conjugation.
Let $K = \mathbf Q(i)$, which is Galois over $\mathbf Q$, and set $P(z) = z^2-(1+2i)$. The splitting field of $P(z)$ over $K$ is $K(\sqrt{1+2i})$. The field $K(\sqrt{1+2i})$ is not CM since its degree over $\mathbf Q$ is 4 and the only quadratic field inside it is $K$, which is not totally real.
Very nice and simple counterexample, thank you sir! Is there anything we can say concerning the relationship between $\sigma\overline\alpha$ and $\overline{\sigma\alpha}$ for some $\alpha$ which is a root of a monic irreducible polynomial with coefficients in a CM field $K$ (where $\sigma$ is some automorphism of a finite Galois extension over $K$ containing the conjugates of $\alpha$)? What about when we impose some condition on $\sigma$? Or on $\alpha$ (like say, $\alpha$ has to be an algebraic integer)?
|
2025-03-21T14:48:31.570083
| 2020-07-21T15:41:22 |
366208
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631315",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366208"
}
|
Stack Exchange
|
Exit time for Brownian motion with stochastic barriers
I am interested in the expected exit time of a one-dimensional Brownian particle from a stochastically evolving interval as follows.
Context:
If $L_t$ and $R_t$ denote the distance to the left and right boundaries at time $t$ respectively, then the process $(L,R)$ has generator
\begin{multline}
Af(l,r) = \frac{1}{2} ( \partial^2_{ll} - 2 \partial^2_{lr} + \partial^2_{rr})f(l,r) + \lambda \bigg\{ \int_0^l (f(u,r)-f(l,r))du + \int_0^\infty (f(l+s,r)-f(l,r))e^{-s} ds \bigg\} \\
+ \lambda \bigg\{ \int_0^r (f(l,u)-f(l,r))du + \int_0^\infty (f(l,r+s)-f(l,r))e^{-s} ds \bigg\}.
\end{multline}
for some $\lambda > 0$. This process has the following interpretation. If $(L_0,R_0) = (l,r)$, then
a particle, initially at the origin, diffuses until it first exits $(-l,r)$; instead here we are evolving the boundaries and waiting until the first time either $L_t=0$ or $R_t = 0$;
the boundaries are allowed to jump:
at rate $\lambda l$, the process $L$ jumps to a uniformly chosen point in $(0,l)$;
at rate $\lambda$, the process $L$ jumps to $l + E$, with $E \sim \mathtt{Exponential}(1)$ (independent of everything else);
similarly for the right boundary $R$.
It is straightforward to check that $\phi(l,r) = e^{-(l+r)}$ is a stationary density for the process, ie if $L_0$ and $R_0$ are independent exponential random variables with unit mean, then so are $L_t$ and $R_t$ for all $t$. That is,
\begin{equation}
\int_0^\infty \int_0^\infty Af(l,r) e^{-(l+r)} dl dr = 0.
\end{equation}
Aim:
I wish to compute
\begin{equation}
\mathbf{E}_\phi \tau = \int_0^\infty \int_0^\infty \mathbf{E}[\tau \, |\, L_0 = l, R_0 = r] e^{-l} e^{-r} dl dr,
\end{equation}
where
\begin{equation}
\tau = \inf\{t\geq0: L_t R_t = 0\},
\end{equation}
that is, the expected time to hit either $L_t = 0$ or $R_t =0$ starting from the stationary distribution.
Thoughts thus far:
In the case that $\lambda = 0$, then the problem reduces to the mean first exit time of $(-l,r)$ of a one-dimensional Brownian motion. It is well known that this is given by $lr$.
Although in this setting the boundaries are jumping, since we start from the stationary (exponential distribution), then on average the boundary is a distance 1 to the left and a distance 1 to the right, leading us to guess that $\mathbf{E}_\phi \tau = 1$.
A standard, maybe naive, approach would be to find a solution $g$ the boundary value problem
\begin{equation}
\begin{cases}
Ag(l,r)=-1, & \text{for all } (l,r) \in (0,\infty)^2, \\
g(0,r) = g(l,0) = 0, & l,r\in[0,\infty);
\end{cases}
\end{equation}
given which,
\begin{equation}
\mathbf{E}_\phi \tau = \int_0^\infty \int_0^\infty g(l,r) e^{-(l+r)} dl dr.
\end{equation}
However, it is not easy to identify a solution. Taking cues from the classical case, with $\lambda = 0$, it is straightforward to compute that with
\begin{align}
&h(l,r) = lr,\\
&Ah(l,r) = -1 + \lambda \big\{ -\frac{1}{2} l^2 r + r - \frac{1}{2} l r^2 + l \big\},
\end{align}
so that
\begin{equation}
L_t R_t -L_0R_0 + t - \lambda \int_0^t \big\{ -\frac{1}{2} L_s^2 R_s + R_s - \frac{1}{2} L_s R_s^2 + L_s \big\} ds
\end{equation}
is a martingale. Taking $t = \tau$ and taking expectations with respect to $\mathbf{E}_\phi$, a naive application of Dynkin's formula or the optional stopping theorem would then 'yield'
\begin{equation}
\mathbf{E}_\phi \tau = 1 + \lambda \mathbf{E}_\phi \bigg[ \int_0^\tau \big\{ -\frac{1}{2} L_s^2 R_s + R_s - \frac{1}{2} L_s R_s^2 + L_s \big\} ds \bigg].
\end{equation}
Specific questions:
Given stationarity, can I conclude that the expectation on the right hand side here is 0? This isn't trivial because I am integrating up to the stopping time, but very naively passing the expectation under the integral would give zero, as hoped.
Broadly, is attempting to solve to the boundary value problem the right approach? What other methods may I try?
My approach has been to decompose $A = A_1 + A_2$, where $A_1$ corresponds to diffusion of the boundaries and $A_2$ corresponds to jumping; I have solved the boundary value problem for $A_1$. Is there something like a Feynman--Kac correction which would then give me the solution for the whole problem? My PDE knowledge is limited so I would be grateful of even general pointers here.
|
2025-03-21T14:48:31.570365
| 2020-07-21T15:51:26 |
366209
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Igor Belegradek",
"asv",
"https://mathoverflow.net/users/1573",
"https://mathoverflow.net/users/16183"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631316",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366209"
}
|
Stack Exchange
|
Concavity of distance to the boundary of Riemannian manifold
Let $(M^n,g)$ be a smooth Riemannian manifold with non-empty boundary $\partial M$. Assume (for simplicity) that $M$ is compact. Let $M$ be locally geodesically convex, i.e. any shortest path in $M$ connecting any two sufficiently close points is a geodesic in $M$ (namely satisfies the standard second order ODE).
Assume in addition that $M$ has non-negative sectional curvature. Is it true that the function on $M$ given by $$x\mapsto dist(x,\partial M)$$
is concave, i.e. its restriction on any shortest path in concave?
A reference would be helpful.
When $M$ is convex this is theorem 8.10 in Cheeger-Ebin book "Comparison theorems in Riemannain geometry".
@IgorBelegradek : Many thanks. It practically answers my question. I have to read the proof. Probably it gives everything.
|
2025-03-21T14:48:31.570458
| 2020-07-21T15:58:49 |
366210
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Andreas Blass",
"https://mathoverflow.net/users/6794"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631317",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366210"
}
|
Stack Exchange
|
About functions primitive recursively definable using a $ \Sigma _ 1 $ oracle
In a discussion with one of my friends about degrees of computability, I was informed about something that was somehow new to me. As I'm not that much familiar with computability theory, I've partially forgotten what exactly the statement was. I'm trying to recover the exact statement, using the part I can recall, and I'll try to formulate the question in the context of first order arithmetic, with which I'm more familiar. The statement was something like this:
Every function primitive recursively definable using a $ \Sigma _ 1 $ oracle is $ \Sigma _ 2 $.
The "is $ \Sigma _ 2 $" part is what I'm the least sure about, and I can't recall whether it was "is $ \Sigma _ 2 $-definable in the language of arithmetic", "provably total using $ \Sigma _ 2 $ induction", a combination of them, or even something else. In the other part, if I recall correctly, "primitive recursively definable using a $ \Sigma _ 1 $ oracle" means a function in the class defined as follows: add the characteristic function of a recursively enumerable subset of $ \mathbb N $ to the primitive recursive initial functions (zero, successor and projections), and recursively define new functions by composition an primitive recursion, using the previously defined functions. If I again recall correctly, this class is denoted by $ \mathrm { PR } ^ { \Sigma _ 1 } $.
To reformulate the question, let $ \mathcal L $ be the language of primitive recursive arithmetic $ \mathrm { PRA } $, in which there is a constant symbol for $ 0 $, and a function symbol for every (primitive recursive definition of a) primitive recursive function. Let $ \mathcal L ' $ extend $ \mathcal L $, adding a new unary function symbol $ f $ (as the initial function) and recursively adding new function symbols, each corresponding to a definition by either composition or primitive recursion. I'll denote everything related to the theories in the language $ \mathcal L ' $ with a prime symbol and those related to $ \mathcal L $ without it. For example $ \Sigma _ n $ will denote the class of $ \Sigma _ n $ formulas in the language $ \mathcal L $, and $ \Sigma _ n ' $ will denote the class of $ \Sigma _ n $ formulas in the language $ \mathcal L ' $. $ \mathrm { PRA } $ is the theory in the language $ \mathcal L $ with the usual axioms of first order classical logic with equality, primitive recursive function defining axioms, and $ \Delta _ 0 $ induction axiom schema. $ \mathrm I \Sigma _ n $ is the theory in the language $ \mathcal L $ extending $ \mathrm { PRA } $ by adding $ \Sigma _ n $ induction axiom schema. $ \mathrm { PRA } ' $ and $ \mathrm I \Sigma _ n ' $ are the corresponding theories in the language $ \mathcal L ' $, with the following additional axioms:
$$ \forall x \big( f ( x ) = 0 \lor f ( x ) = 1 \big) \text , $$
$$ \forall x \big( f ( x ) = 1 \leftrightarrow A ( x ) \big) \text , $$
where $ A ( x ) $ is in $ \Sigma _ 1 $.
I can now make sense of the above struggle asking the following questions:
Is there, for every function symbol $ g $ in $ \mathcal L ' $, a natural number $ n $ such that $ g $ is $ \Sigma _ n $-definable (i.e. is there a $ \Sigma _ n $ formula $ B ( x , y , \dots , z ) $ such that $ \mathbb N \models \forall x , y , \dots , z \big( g ( x , y , \dots ) = z \leftrightarrow B ( x , y , \dots , z ) \big) $)? If yes, how low can this $ n $ be (can it be, say, equal to $ 2 $)?
For a $ g $ in $ \mathcal L ' $ which is $ \Sigma _ n $-definable by the formula $ B ( x , y , \dots , z ) $, is there a natural number $ m $ such that $ \mathrm I \Sigma _ m ' \vdash \forall x , y , \dots , z \big( g ( x , y , \dots ) = z \leftrightarrow B ( x , y , \dots , z ) \big) $? If yes, how low can this $ m $ be?
For a $ g $ in $ \mathcal L ' $ which is $ \Sigma _ n $-definable by the formula $ B ( x , y , \dots , z ) $, is there a natural number $ m $ such that $ g $ is provably total in $ \mathrm I \Sigma _ m $? More accurately, can we have
$$ \mathrm I \Sigma _ m \vdash \forall x , y , \dots , \exists z \ B ( x , y , \dots , z ) $$
and
$$ \mathrm I \Sigma _ m \vdash \forall x , y , \dots , z _ 0 , z _ 1 \big( B ( x , y , \dots , z _ 0 ) \land B ( x , y , \dots , z _ 1 ) \rightarrow z _ 0 = z _ 1 \big) \text ? $$
If yes, how low can this $ m $ be?
I realize that these are separate questions and maybe not suitable to be asked in a single post, but they seem very much related, and I thought maybe they could be answered all together. This thought is mostly driven by the fact that something similar was claimed to be true in the mentioned discussion. I would appreciate any insights about why the above questions can be answered positively/negatively, as well as any source where I can read about them and find proofs. I would also appreciate it if someone could inform me about what the original statement probably was, in case the above questions do not formulate any well-known fact in computability theory, while there is some known fact in line with the vague statement that I started with.
Question 1 is answered by an old theorem due (if I remember correctly) to Post: $\Delta^0_2$ is equivalent to "recursive in a $\Sigma^0_1$ set" (and thus also to "recursive in $0'$, the halting problem").
|
2025-03-21T14:48:31.570756
| 2020-07-21T16:11:45 |
366212
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"AlexIvanov",
"LSpice",
"https://mathoverflow.net/users/148992",
"https://mathoverflow.net/users/2383"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631318",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366212"
}
|
Stack Exchange
|
Infinite sum in power series ring
Let $R$ be a commutative ring with $1$, $R[[x]]$ be the power series ring over $R$ and $A$ be an (prime) ideal of $R[[x]]$ with $x\not\in A$ and $\{f_i\}_{i=1}^\infty$ be a sequence of element of $A$. Now I have two questions:
Is $f\mathrel{:=}f_1+xf_2+x^2f_3+\dotsb+x^nf_{n+1}+\dotsb$ a well defined element of $R[[x]]$?
(Since we can find the coefficient of $x^n$ in $f$ for each $n$, it seems that $f$ is well defined.)
If (1) is true is $f\in A$? (If (1) is true and (2) is not true, under what conditions is (2) true?)
The usual definition of the topology on $R[[x]]$ sets its ideals $x^n R[[x]]$ as a neighbourhood base at the identity. According to that definition, the answer to (1) is obviously yes. For (2), of course the answer is that $A$ should be closed, but I don't know if there's a nice characterisation of closed ideals in $R[[x]]$.
Here is at least some characterization (not sure it can be called "nice"): An ideal $I$ is in $R[[x]]$ is closed in the $x$-adic topology if and only if $\bigcap_{n \geq 1} I + (x)^n = I$.
|
2025-03-21T14:48:31.570878
| 2020-07-21T16:14:43 |
366213
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Fernando Muro",
"Ian Agol",
"YCor",
"https://mathoverflow.net/users/12166",
"https://mathoverflow.net/users/1345",
"https://mathoverflow.net/users/14094",
"https://mathoverflow.net/users/161437",
"sadok kallel"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631319",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366213"
}
|
Stack Exchange
|
Euler characteristic with compact support of spaces of Euclidean lattices
Has the Euler characteristic with compact support of $\mathrm{SL}_n(\mathbb R)/\mathrm{SL}_n(\mathbb Z)$ been computed ? References? Thanks.
For $n$ even, this will be odd dimensional, and hence have euler characteristic 0. In general, this will be a bundle over the symmetric space $X$ of symmetric positive definite matrices of determinant 1 (unimodular positive definite quadratic forms or metrics on $R^n$), with fiber $SO(n)$. Hence if the base or fiber has odd dimension, then the Euler characteristic will be 0. For the other dimensions, the volume has been computed and should be proportional to the euler characteristic by applying Chern-Gauss Bonnet to the base. https://arxiv.org/abs/math/0402085
Many thanks Ian. My question is about $\chi_c$ (compact support). So odd dimensional (non-compact) it is non zero. However your answer and reference are useful. My more general interest is actually finding a stratification of this space into a finite number of -interesting-strata, possibly open cells. In case you also have a thought on this, thank you in advance for sharing it.
Ah, I see. Somehow I was thinking you wanted the euler characteristic of the compactification (which I think was shown to exist by Borel-Serre).
Do you know the value for small $n$? at least $n=2$?
Yes. For $n=2$, this space is homeomorphic to the complement in $S^3$ of a trefoil knot, and so $\chi_c=0$.
Hi, Sadok. By Poincaré duality, this would be the same as the ordinary Euler characteristic, computed from homology, right?
Hi Fernando. Yes it is (up to sign depending on the dimension of the manifold). I guess I meant to say in my response that $\chi$ is not necessarily zero in odd dimensions because the manifold is non-compact.
I think that the euler characteristic is 0 for the following reasons.
Firstly, the space $SL_N(\mathbb{R})$ is a bundle over the symmetric space $SO(N,\mathbb{R})\backslash SL_N(\mathbb{R}) = SP(n,\mathbb{R})=X$, the space of symmetric positive-definite real matrices of determinant 1. For a discussion of this symmetric space, see e.g. Bridson-Haefliger II.10. Then $SL_N(\mathbb{R})/SL_N(\mathbb{Z})$ is a bundle over $X/SL_N(\mathbb{Z})$ with fiber $SO(N,\mathbb{R})$. Note that this is an orbifold bundle, but that by passing to a torsion-free subgroup, one can assume that it is a manifold (and since you're interested in euler characteristic, this just multiplies by the index).
Now the space $X/SL_N(\mathbb{Z})$ admits a bordification by Borel-Serre. Hence $SL_N(\mathbb{R})/SL_N(\mathbb{Z})$ has a bordification by an $SO(N,\mathbb{R})$-bundle over the Borel-Serre bordification. Hence it is the interior of a manifold with boundary $M$. In this case, $H^*_c(SL_N(\mathbb{R})/SL_N(\mathbb{Z}))\cong H^*(M,\partial M)$. Then by Lefschetz duality, $\chi(H^*_c(M,\partial M))=\chi(M)$.
But since $M$ is a bundle with fiber $SO(N,\mathbb{R})$, and $\chi(SO(N,\mathbb{R}))=0$ (any Lie group has a nowhere vanishing vector field), we have $\chi(M)=\chi(SO(N,\mathbb{R}))\times \chi(X/SL_N(\mathbb{Z})) =0$, since the euler characteristic of bundles is the product of the euler characteristic of the base and the fiber.
This seems right. I think we can avoid the compactification by restricting the orbi-bundle over strata (locally closed). Over some strata of the base, the fiber is SO(n) and over other strata it is a quotient of SO(n) by a finite group I believe. In all cases, the fiber is compact with zero Euler characteristic. We then get a stratification of $SL_n(\mathbb R)/SL_n(\mathbb Z))$ by bundles whose $\chi_c=0$ since $\chi_c$ is multiplicative on bundles. Since $\chi_c$ is additive over strata, this gives the answer $0$ as well.
Well, you need to know also that the cohomology is finite so that the Euler characteristic is well-defined. This follows from the compactification, or from a finite spine. But as I said, thinking about strata can be avoided by passing to a manifold cover (a level 3 principal congruence subgroup should suffice).
|
2025-03-21T14:48:31.571297
| 2020-07-21T16:37:45 |
366215
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Carlo Beenakker",
"Francois Ziegler",
"Jochen Glueck",
"KConrad",
"Mary Sp.",
"Michael Greinecker",
"YCor",
"https://mathoverflow.net/users/102946",
"https://mathoverflow.net/users/11260",
"https://mathoverflow.net/users/14094",
"https://mathoverflow.net/users/161374",
"https://mathoverflow.net/users/19276",
"https://mathoverflow.net/users/3272",
"https://mathoverflow.net/users/35357"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631320",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366215"
}
|
Stack Exchange
|
Oldest abstract algebra book with exercises?
Per the title, what are some of the oldest abstract algebra books out there with (unsolved) exercises? Maybe there are some hidden gems from before the 20th century out there. I am already aware of the books of Dickson and van der Waerden.
In older books (or most non-English language books), the ridiculous "abstract" adjective was not present.
@YCor Do you prefer the timeless "modern" instead?
Isn't it just the meaning of "algebra"?
It is also the art of solving equations involving real or complex numbers.
A year ago you posted a large number of separate questions of this kind, asking for oldest books in various areas of math with exercises (always hoping for "some hidden gems from before the 20th century"). Please consider posting just once a big-list request for the oldest books in different areas of math with exercises, rather than all of these separate questions marching (potentially) through all areas of mathematics that are all the same question.
I agree wirh @KConrad. By the way, it might be just my ignorance, but I have difficulties to see the point of the requirement "with exercises". When you are looking for old books on various subjects, is there a specific reason why you are only interested in books that contain exercises?
@JochenGlueck see the rationale explained in the post of the earliest question https://mathoverflow.net/questions/313967/reference-request-oldest-linear-algebra-books-with-exercises. It does not explain why there have to be repeated posts about all these different areas of math instead of one single post on the general topic. See also https://mathoverflow.net/questions/327697/reference-request-oldest-number-theory-books-with-unsolved-exercises.
For a pre-20th century textbook: Modern Higher Algebra by George Salmon (1876) has exercises (with solutions).
If I may broaden the query from "abstract algebra" to more general "algebra", I note that Elements of Algebra by Euler (1770) has more than one hundred exercises. The exercises are discussed in The origin of the problems in Euler’s algebra.
Here is an example, from the chapter on cubic equations
[source].
And another example (find an integer $x$ such that $2x^2-5$ is a cube) [source]
Problem 10 has a typo :-(
yep, the square should be a cube, well spotted!
It seems that the unsolved problems may not have been in Euler’s original, but rather added by a later translator.
Two important early 20th-century abstract algebra textbooks that were superseded by van der Waerden are:
Hasse, H., 1926. Höhre Algebra.
Haupt, O., 1929, Einführung In Die Algebra, Zweiter Band - Mit Einem Anhang Von W. Krull, Akademische Verlagsgesellschaft M. B. H., Leipzig.
For an in-depth historical account of the transition from algebra to abstract algebra (including discussions of the various textbooks) see:
Leo Corry, 1996, Modern Algebra and the Rise of Mathematical Structures, Birkhäuser Verlag.
I think the following references might be useful:
H. Weber (1895/1896): Lehrbuch der Algebra, 2 volumes. Vieweg, Braunschweig. It includes examples, but it does not have any separate exercises.
E. Artin (1938): Foundations of Galois Theory. New York University Lecture Notes, New York.
N. Bourbaki (1947): Éléments de Mathématique, Algèbre. Hermann, Paris.
I am not totally sure whether the last two references include exercises.
Lehrbuch der Algebra is here -- it doesn't seem to have exercises...
@CarloBeenakker Thank you very much for the link! You are right. I will edit my answer.
|
2025-03-21T14:48:31.571564
| 2020-07-21T16:43:36 |
366216
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"A.Skutin",
"Mohan",
"https://mathoverflow.net/users/100359",
"https://mathoverflow.net/users/9502"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631321",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366216"
}
|
Stack Exchange
|
Question about Jacobian subalgebra
Assume that the algebraically independent polynomials $f, g\in\mathbb{C}[x, y]$ are such that the Jacobian matrix $\text{Jac}_{x, y}^{f, g}\in\mathbb{C}\setminus\{0\}$.
Is it true that $\mathbb{C}[x, y] = \mathbb{C}[f, g]+g\cdot\mathbb{C}[x, y]$?
Do you have some reason to suspect this? This is equivalent to the Jacobian conjecture.
Do you have proof of its equivalence to JC?
If JC is true, the above equality is obvious. If the above equality is true in general, consider $g_i\in\mathbb{C}[x_1,\ldots, x_{n+1}]$ with $g_i=f_i, i\leq n$ and $g_{n+1}=x_{n+1}$ and apply the equality for the $g_i$s.
So You know that if $Q(n)$ is my question and $JC(n)$ is the Jacobian conjecture, then $JC(n)=>Q(n)=>JC(n-1)$. Assume we have proven $JC(i)$ for $i<n$ then can $Q(n)$ be proved?
As I know Jacobian conjecture is open even for $n=2$, so I edited the question to the case $n=2$.
This is still is equivalent to JC.
Your equality says, $\mathbb{C}[x,y]=\mathbb{C}[f,g]+g\mathbb{C}[x,y]$, the last term is equal to $\mathbb{C}[f]+g\mathbb{C}[f,g]+g\mathbb{C}[x,y]=\mathbb{C}[f]+g\mathbb{C}[x,y]$, since $g\mathbb{C}[f,g]\subset g\mathbb{C}[x,y]$. This says, the map $\mathbb{C}[f]\to \mathbb{C}[x,y]/g\mathbb{C}[x,y]$ is onto and then it is clear that this is an isomorphism. Then, $g=0$ is an embedded line in $\mathbb{C}^2$ and by Abhyankar-Moh, is a co-ordinate line after an automorphism. Then, it is easy to verify that $\mathbb{C}[f,g]=\mathbb{C}[x,y]$.
|
2025-03-21T14:48:31.571699
| 2020-07-21T17:16:53 |
366222
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Jules Lamers",
"JustWannaKnow",
"https://mathoverflow.net/users/150264",
"https://mathoverflow.net/users/45956"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631322",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366222"
}
|
Stack Exchange
|
What is the definition of the thermodynamic limit of a thermodynamic quantity?
Statistical mechanics is all about taking thermodynamic limits and, as far as I know, there are more than one way to define such limits. Consider the following theorem:
Theorem: In the thermodynamic limit, the pressure:
$$\psi(\beta,h) := \lim_{\Lambda \uparrow \mathbb{Z}^{d}}\psi_{\Lambda}^{\#}(\beta, h) $$
is well-defined and independent of the sequence $\Lambda \uparrow \mathbb{Z}^{d}$ and of the type of the boundary condition $\#$.
Here, I'm using the same notation and conventions from chapter 3 of Velenik and Friedli's book. The notation $\Lambda \uparrow \mathbb{Z}^{d}$ stands for the convergence in the sense of Van Hove.
Definition [Convergence in the sense of Van Hove] A sequence $\{\Lambda_{n}\}_{n\in \mathbb{N}}$ of (finite) subsets of $\mathbb{Z}^{d}$ is said to converge to $\mathbb{Z}^{d}$ in the sense of Van Hove if all three properties listed below are satisfied:
(1) $\{\Lambda_{n}\}_{n\in \mathbb{N}}$ is an increasing sequence of subsets.
(2) $\bigcup_{n\in \mathbb{N}}\Lambda_{n} = \mathbb{Z}^{d}$
(3) $\lim_{n\to \infty}\frac{|\partial^{in}\Lambda_{n}|}{|\Lambda_{n}|} = 0$, where $|X|$ denotes the cardinality of the set $X$ and $\partial^{in}\Lambda:=\{i\in \Lambda: \hspace{0.1cm} \exists j \in\Lambda^{c} \hspace{0.1cm} \mbox{with} \hspace{0.1cm} |i-j|=1 \}$
My point here is the following. Convergence in the sense of Van Hove is a notion of convergence of sets, not functions of sets. But what does $\lim_{\Lambda\uparrow \mathbb{Z}^{d}}\psi^{\#}_{\Lambda}(\beta, h)$ mean?
Just an aside: for certain stat-mech models thermodynamic quantities are known to depend on the choice of boundary conditions, e.g the free energy of the six-vertex model (which has d=2) with periodic vs domain-wall boundaries. I believe that the conditions of the definition are met; the point is that the local (divergencelessness) constraint of the model causes the choice of boundary conditions to influence a macroscopic portion of the lattice.
It means that if you consider any sequence of sets $(\Lambda_n)_{n\in\mathbb{N}}$ converging to $\mathbb{Z}^d$ in the sense of van Hove, then the sequence of numbers $(\psi_{\Lambda_n}^\#(\beta,h))_{n\in\mathbb{N}}$ is convergent. (Moreover, the theorem claims that the limit is independent of the sequence chosen and that the limiting function has nice properties.)
(It may be more natural to consider nets rather than sequences, but we refrained from that generality in the book since this was not necessary in the cases we discuss.)
perfectly clear! Thank you so much!
|
2025-03-21T14:48:31.571915
| 2020-07-21T17:48:38 |
366224
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Grisha Papayanov",
"Joshua Mundinger",
"Tobias Fritz",
"Yemon Choi",
"https://mathoverflow.net/users/125523",
"https://mathoverflow.net/users/160378",
"https://mathoverflow.net/users/27013",
"https://mathoverflow.net/users/43309",
"https://mathoverflow.net/users/763",
"jg1896"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631323",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366224"
}
|
Stack Exchange
|
Sufficient conditions for $\mathrm{Der}_k(A)$ to be f.g. projective
Let $k$ be a field and $A$ a commutative $k$-algebra. What are sufficient conditions for the module of derivations $\mathrm{Der}_k(A)$ to be finitely generated projective?
I'm looking for conditions which apply in particular when $A = C^\infty(M)$ for a manifold $M$. In this case, the derivations are the vector fields and the module of derivations is finitely generated projective by Swan's theorem. Note that the module of Kähler differentials is not finitely generated unless $M$ consists only of isolated points.
The question is confusing because it implies that you would like conditions for the case you already know about (smooth functions on a manifold)... are you saying that you would like to consider when $A/k$ is not necessarily of finite type?
@JoshuaMundinger: I'd like to have conditions which apply to some reasonably general class of algebras, which in particular includes those of the form $C^\infty(M)$ in case that the ground field is $\mathbb{R}$. Indeed, since I understand this case, a sufficient condition of the form "$A$ is isomorphic to some $C^\infty(M)$" would not be interesting.
@JoshuaMundinger: if it's still confusing, then just ignore my statements about $C^\infty(M)$. I'm just looking for sufficient conditions to guarantee that the module of derivations is fg projective. Thus being regular Noetherian would be one possible answer, since then (if I understood correctly) the module of Kähler differentials is already fg projective and the module of derivations is its dual.
Tobias: regarding your final statement, the dual of the Kahler module is ${\rm Der}(A,A^*)$, which is not going to be naturally isomorphic to ${\rm Der}(A)$ although perhaps they coincide for some reason when $A$ is regular Noetherian. Indeed, while there are natural instances where the Kahler module is projective, why would that make its dual projective? Surely you'd expect the dual of something projective to be injective ?
Also, as soon as you want to take $C^\infty(M)$ you probably want to be imposing topological constraints: there is an old MO question https://mathoverflow.net/questions/6074/kahler-differentials-and-ordinary-differentials/6138#6138
@YemonChoi: thanks! $\mathrm{Der}(A)$ consists of the derivations $A \to A$, which by definition of Kähler differentials $\Omega^1_K$ coincides with $A$-module maps $\Omega^1_K \to A$, i.e. with the dual module $(\Omega^1_K)^*$; this is from the nLab. What am I missing? And I don't want topological constraints, although I'm well aware of the difference between Kähler differentials $\Omega^1_K$ and the usual module of 1-forms $\Omega^1$.
Already purely algebraically, derivations on $C^\infty(M)$ form a fg projective module, and I'd like to know whether there are any more general statements of which this would be a special case.
Tobias, I am sufficiently fusty that I would recommend consulting books like Weibel when it comes to homological algebra rather than the nLab. Perhaps we are using different terminology from each other; if A is a unital K-algebra then K-linear derivations from A to a symmetric A-module correspond to elements of ${\rm Hom}A( \Omega{A/K} , M)$. For me the dual of an $A$-module $M$ is ${\rm Hom}_K (M,K)$ but perhaps for you the dual of $M$ is ${\rm Hom}_A(M,A)$ ?
BTW, why is it obvious that derivations from $C^\infty(M)$ to itself are automatically continuous and hence given by vector fields? I am not a diff geometer so maybe I am overlooking something. Certainly automatic continuity problems for Banach algebras can be much tougher, but the Frechet category is niceer I admit
A remark here: for commutative rings $A$, if $M$ is a finitely generated projective module, then so is $M^*= Hom_A(M,A)$ by the dual basis lemma: see McConnell, Robinson, Noncommutative Noetherian Rings, revised edition, 3.5.2
@YemonChoi: yes, for me the dual is $\mathrm{Hom}_A(M,A)$. In differential geometry, being a derivation is one of several equivalent standard definitions of a vector field, and no additional continuity condition is needed. Everything happens locally, and manifolds are finite-dimensional and hence locally compact, which I think is what makes this work.
@jg1896: great, I was actually looking for a reference for that! BTW I had stupidly forgotten to mention that $A$ can be assumed commutative, and have just edited the question accordingly.
Thanks Tobias and @jg1896 - I withdraw my earlier caveats
a silly example is that this is true if A is an algebra of continious functions on a topological space (with no restictions on the space whatsoever), because these algebras do not have any derivations at all (except for the zero one)
@YemonChoi the existence of cut-off functions shows that all derivations are local and then Hadamard lemma (Taylor series up to order two) shows that a derivation is determined by its values on coordinate functions
For finitely generated domains over a base field $k$ of characteristic 0, we have that if $A$ is regular, then both $Der_k \, A$ and the module of Kähler differentials are finitely generated projective (McConnell, Robson, Noncommutative Noetherian Rings, revised edition, 15.2.11).
Zariski-Lipman's Conjecture says that if $Der_k \, A$ is finitely generated projective (or, in a more modest version, free), then $A$ is regular.
So for this class of algebras (roughly, regular functions on smooth affine varieties), it is expected that $A$ is regular if and only if $Der_k \, A$ is a finitey generated projective module.
Your example does not fit here as smooth functions on a real manifold are not even domains, but I think that the literature on this subject (i.e. Zariski-Lipman's Conjecture) might be a good direction to look.
|
2025-03-21T14:48:31.572283
| 2020-07-21T17:50:42 |
366226
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Deane Yang",
"Spoilt Milk",
"Willie Wong",
"https://mathoverflow.net/users/3948",
"https://mathoverflow.net/users/613",
"https://mathoverflow.net/users/99716"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631324",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366226"
}
|
Stack Exchange
|
Curvature collineation and the Killing identity
The Lie derivative of a general covariant $4$-tensor is given by
$$\mathcal{L}_{K}R_{abcd} = X^{e}\nabla_{e}R_{abcd} + R_{ebcd}\nabla_{a}X^{e} + R_{aecd}\nabla_{b}X^{e} + R_{abed}\nabla_{c}X^{e} + R_{abce}\nabla_{d}X^{e},$$
where $X^{a}$ is a y smooth vector field. If the $(0,4)$ covariant tensor $R$ is the Riemann tensor (and admits all the symmetries of the same) and $X^{a}$ is a Killing vector, then by the theorem on the inheritance of symmetries, a Killing symmetry would imply an affine symmetry which in-turn would imply a curvature symmetry (assuming the no non-metricity and no torsion). Let $K$ be a vector field that preserves the Riemann tensor $R_{abcd}$ such that we have the condition of curvature collineation to hold, i.e.,
$$\mathcal{L}_{K}R_{abcd} = 0.$$
I now want to use this property of curvature collineation and derive the Killing identity: $\nabla_{a}\nabla_{b}K_{c} = R_{dabc}K^{d}$ but haven't been able to do so. Any help is appreciated.
My attempt:
Replacing with the total derivatives, we have
$$\nabla_{a}\left(R_{ebcd}K^{e} \right) + \nabla_{b}\left(R_{aecd}K^{e} \right) + \nabla_{c}\left(R_{abed}K^{e} \right) + \nabla_{d}\left(R_{abce}K^{e} \right) + \left(R_{abcd;e} - R_{ebcd;a} - R_{aecd;b} - R_{abed;c} - R_{abce;d} \right)K^{e} = 0,$$
Now, using the symmetries of the Riemann tensor, $R_{abcd;e} = -R_{abde;c} - R_{abec;d}$ and the definition of the Riemann tensor $R_{ebcd}K^{e} = \nabla_{c}\nabla_{d}K_{b}$, we obtain
$$\left[\nabla_{c},\nabla_{a}\right]\nabla_{b}K_{d} + \left[\nabla_{d},\nabla_{b}\right]\nabla_{a}K_{c} - \nabla_{b}\left(\left[\nabla_{c}, \nabla_{d}\right]K_{a} \right) - \nabla_{d}\left(\left[\nabla_{a},\nabla_{b}\right]K_{c}\right) = \left(R_{ebcd;a} - R_{eacd;b}\right)K^{e}.$$
I am stuck here, I don't exactly know how to handle the $\left[\nabla_{c},\nabla_{d}\right]\nabla_{b}K_{d}$ type terms.
In your formulas, is $K = X$?
Notice that this can't work in general. In particular, if the metric is flat, then the equation you've written holds for any vector field and not just Killing fields.
@DeaneYang Yeah $K=X$ in this case, and yes this does work for any smooth vector field. But if $X$ is a Killing field and the $(0,4)$ covariant tensor is the Riemann tensor, curvature collineation holds and this should simplify to the Killing identity (right?)
If the metric is flat, the vector field need not satisfy the identity.
No, that's not right. The Killing identity does not follow from the collineation equation.
Using the identity in here, we have
$$ [\nabla_{[a}, \mathcal{L}_X] R_{bc]de} = \pi_{d[a}{}^f R_{bc]fe} + \pi_{e[a}{}^f R_{bc]df} \tag{1}$$
where we used that the first order deformation tensor is symmetric in its first two indices.
The first order deformation tensor is, explicitly
$$ \pi_{lm}{}^n = \frac12 g^{np} \left( \nabla_l (\nabla_m X_p + \nabla_p X_m) + \nabla_m(\nabla_l X_p + \nabla_p X_l) - \nabla_p (\nabla_m X_l + \nabla_l X_m) \right) $$
Rearranging we get
$$ \pi_{lm}{}^n = \frac12 g^{np}( 2 \nabla_l \nabla_m X_p + [\nabla_m ,\nabla_l ] X_p + [\nabla_l, \nabla_p]X_m + [\nabla_m, \nabla_p] X_l ) $$
or
$$ \pi_{lm}{}^n = \frac12 g^{np}( 2 \nabla_l \nabla_m X_p + (R_{spml} + R_{smlp} + R_{slmp})X^s ) $$
From the first Bianchi identity we finally get
$$ \pi_{lm}{}^n = g^{np} ( \nabla_l \nabla_m X_p + R_{slmp} X^s) $$
Notice that the terms in the brackets are exactly the "Killing identity" terms. (I think we use different sign convention for Riemann.)
An immediate consequences of the above computation:
Killing vectors satisfy the Killing identity
For Killing vectors, the 0th order deformation tensor vanishes (Killing's equation), and since the 1st order deformation tensor is formed through the covariant derivative of the 0th order ones, it must also vanish. And thus Killing's identity must hold.
Now let's see what this allows us to say for curvature collineation. By the second Bianchi identity, $\nabla_{[a}R_{bc]de} = 0$. Assuming curvature collineation, the left hand side of equation (1) vanishes. This requires
$$ \pi_{d[a}{}^fR_{bc]fe} + \pi_{e[a}{}^fR_{bc]df} = 0 $$
Let $\Gamma$ be the vector space of type $(1,2)$-tensors symmetric in the covariant indices, and define the linear mapping $\Upsilon$ from $\Gamma$ to type $(0,5)$ tensors by
$$ \gamma_{ab}{}^c \mapsto \gamma_{d[a}{}^fR_{bc]fe} + \gamma_{e[a}{}^fR_{bc]df} $$
For your assertion to hold (that curvature collineation implies Killing's identity), it is sufficient that the Riemann curvature tensor is such that the mapping $\Upsilon$ is injective. Obviously, as Deane noted, this fails when your manifold is flat.
In low dimensions injectivity however can fail in general. In two dimensions the antisymmetry in the $a, b, c$ indices means that $\Upsilon$ is by definition the trivial map. In three dimensions the domain $\Gamma$ is 18 dimensional, while the target is 3 dimensional (isomorphic to the space of antisymmetric matrices) and so must have a non-trivial Kernel. Even in four dimensions, the dimensional counting argument is not optimistic: let $R^*_{abcd} = \frac12 R_{abef} \epsilon^{ef}{}_{cd}$ the right Hodge dual of the Riemann curvature tensor. Taking the Hodge dual w.r.t. to the antisymmetric $a,b,c$ indices in $\Upsilon(\gamma)$ we can study the equivalent linear map
$$ \Upsilon^*: \gamma \mapsto \gamma_{d[a]}{}^f R^*_{b]fc}{}^d $$
The image is a $(0,3)$ tensor that is antisymmetric in its first two indies, so $\Upsilon^*$ is a mapping from a 40 dimensional space to a 24 dimensional one, and still must have a non-trivial kernel in general.
(Remark: in the two dimensional case, however, using that Riemann curvature has special structure, you see that curvature collineation implies that the vector field $X$ is conformal Killing, so there's still some hope of getting the desired inequality through a different route.)
In conclusion, I doubt that there is a general statement that curvature collineation implies Killing identity, because there are too many special cases to worry about. At best (for this line of argument) you may have a statement that states that for generic metrics in sufficiently high dimension that curvature collineation implies Killing identity.
Shouldn't the Riemann tensor terms be $R_{splm} + R_{smpl} + R_{slpm}$ so that using $R_{s[plm]} = 0$ would give the result mentioned? You have mentioned that your sign convention is different, but I don't get it. Could you please clarify this. Thanks for the answer.
Some people take $R_{ijkl} X^l = [\nabla_i, \nabla_j ]X_k$, some people define it to be the negative. I used in this computation the convention indicated in this comment, you are using the opposite one. So my version of the "Killing identity" is off from yours by a minus sign.
Note that in dimension 5, things start looking better for $\Upsilon$: the domain is 75 dimensional, while the image is $\binom{5}{2} \binom{5}{3} = 100$ dimensional (we note the image is also antisymmetric in the $d$ and $e$ indices), so we have hope.
|
2025-03-21T14:48:31.572704
| 2020-07-21T19:07:55 |
366228
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Dieter Kadelka",
"Mateusz Kwaśnicki",
"as1",
"https://mathoverflow.net/users/100904",
"https://mathoverflow.net/users/108637",
"https://mathoverflow.net/users/157313"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631325",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366228"
}
|
Stack Exchange
|
Random walk in a switching scenery
For each $x \in \mathbf{Z}$ let $(\eta_t(x))_{t\geq0}$ denote independent copies of a process $(\eta_t(0))_{t\geq0}$ defined as follows. The process $\eta_t(0)$ takes values in $\{-1,1\}$, where $-1$ and $1$ denote the values of a scenery with two possible states, and
$\eta_0(0) = -1$ with probability $1/2$ and $1$ otherwise;
at the arrival times of a Poisson process $N_t(0)$ with rate $\lambda > 0$, $\eta(0)$ switches state, ie
\begin{equation}
\eta_t(0) = (-1)^{N_t(0)}\eta_0(0).
\end{equation}
Let $X = (X_t)_{t\geq0}$ denote an independent continuous-time nearest neighbour random walk on $\mathbf{Z}$, ie at rate $1$, $X$ jumps to one of its nearest neighbours chosen with equal probability.
What is the mean first time that $X$ encounters a scenery different from the one in which it started?
That is, denoting
\begin{equation}
T := \inf\{ t > 0 : \eta_t(X_t) = - \eta_0(X_0) \},
\end{equation}
what is the value of $\mathbf{E}T$?
There are potentially loads of applications but the ones I have in mind are biological. Imagine the diffusing particle represents the motion of an organism through a one-dimensional habitat and $\eta_t(x)$ denotes the presence of a virus at $x$ at time $t$. How long until the organism encounters the virus?
Can you tell where this scenery comes from? I think the question would be more interesting.
There are potentially loads of applications but the ones I have in mind are biological. Imagine the diffusing particle represents the motion of an organism through a one-dimensional habitat and $\eta_t(x)$ denotes the presence of a virus at $x$ at time $t$. How long until the organism encounters the virus?
Interesting. I think you should edit your question and add this.
Some trivial comments: Clearly something between $\lambda^{-1}$ (the first change of state along the path of $X_t$ occurs at rate $\lambda$) and $(\lambda+1)^{-1}$ (nothing will change until either $X_t$ jumps or the state at $0$ changes). For $\lambda = 1$, something close to $0.681$ (based on a sample of one million trials).
|
2025-03-21T14:48:31.572862
| 2020-07-21T19:23:53 |
366229
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"KConrad",
"Kimball",
"Rachid Atmai",
"https://mathoverflow.net/users/3272",
"https://mathoverflow.net/users/3859",
"https://mathoverflow.net/users/6518"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631326",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366229"
}
|
Stack Exchange
|
Modern example of a reciprocity law and intuition behind it
I'm very new to the Langlands program and I was going through the Gauss reciprocity law, Hilbert's 9th problem, Artin's reciprocity law which allowed him to identify the Artin's L-functions with the Hecke L-functions, all the way to the Shimura-Tayinama conjecture and the corresponding reciprocity law.
Could someone provide a modern example of a reciprocity law and how it fits in the Langlands picture? These laws all seem to hinge on the identification of certain eigenvalues with cardinalities of solution sets of equations, which is quite suprising. References are also very welcome. Thank you.
Thank you crispr, I'm gonna look up the Grothendieck trace formula.
Gelbart and Weinstein have survey articles about reciprocity/Langlands program, both in the Bulletin of the AMS if I recall. Have you tried reading either of those?
Thanks for the references Kimball, I wasn't aware of these survey articles
Artin did not "identity the Artin's $L$-functions with the Hecke $L$-functions": neither collection of $L$-functions is a subset of the other. Artin was able to identity the $L$-functions of finite-order Hecke characters with the 1-dimensional Artin $L$-functions.
Very cool KConrad, thanks for the info.
|
2025-03-21T14:48:31.572997
| 2020-07-21T21:01:38 |
366234
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Alex",
"Alexandre Eremenko",
"Conrad",
"Iosif Pinelis",
"https://mathoverflow.net/users/133811",
"https://mathoverflow.net/users/25510",
"https://mathoverflow.net/users/33278",
"https://mathoverflow.net/users/36721"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631327",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366234"
}
|
Stack Exchange
|
Analyzing the decay rate of Taylor series coefficients when high-order derivatives are intractable
This could be a soft question. I am trying to show that the $n$-th Taylor series coefficient of a function is $O(n^{-5/2})$. However, because the function is a function composition of another function with itself, it seems intractable to compute high-order derivatives. I was wondering if there are methods that can bound the asymptotic decay rate of Taylor series coefficients without obtaining the exact coefficients. For example, can complex analysis help here?
Thank you so much!
The function that I am trying to analyze is $f(x)=g(g(x))$, where $g(x) = \frac{1}{\pi}\left( x\cdot (\pi-\arccos(x)) + \sqrt{1-x^2} \right)$. I conjecture that its $n$-th Taylor coefficient about $x=0$ is $O(n^{-5/2})$. I have shown that the $n$-th Taylor coefficient of $$g(x)= \frac{1}{\pi} + \frac{x}{2} + \sum_{n=1}^\infty \frac{(2n-3)!!}{(2n-1)n!2^n \pi} x^{2n}$$ is $O(n^{-5/2})$.
Complex analysis can help. The rate of Taylor coefficients is determined by:
a) the radius of convergence, which is equal to the radius of the largest disk $|z|<r$
where your function is analytic. This radius is responsible for the exponential asymptotics, and
b) the nature of singularities on the circle $|z|=r$.
Your function $f$ can have only square root singularities, since $g$ has only square root singularities. Since the singularities of $g$ are $\pm1$, to determine the radius of convergence, you have to show that equation $g(z)=\pm1$ has no solutions
in $|z|<1$. This will justify your asymptotics.
I have not done the calculation, perhaps you can do it yourself.
Ref: P. Flajolet and R. Sedgewick, Analytic combinatorics, Chap. VI.
Edit: Conrad gave a simple argument in the comment below below which shows that $f$ has no other
singularities in the closed unit disk, except at $z=\pm1$, so your conjecture on the asymptotics is correct.
Thank you Professor Eremenko! Can complex analysis give a fine-grained rate, like n^{-5/2}?
Yes, of course. Even finer than that, like $\sim cn^{-5/2}$ where $c$ can be also computer, and higher order terms as well. The difficulty in justifying this is showing that there are no singularities of $f$ in the unit disk, except those of $g$. But I believe this fact can be even checked with a computer, if no simple argument is found.
Thank you so much Professor Eremenko! I am reading Chap VI of Analytic combinatorics and see if I can figure it out myself. I learned a really helpful tool from you.
@Cechco: The main part which is somewhat doubtful is solving the equation $g(x)=1$, not the Chap VI of Flajolet.
$g$ has positive coefficients so $|g(x)| \le g(|x|) <g(1)=1, |x| < 1$
@Conrad: Thanks! I just tried to verify this on Maple:-)
I thought you were initially interested in solutions in $|z|\le1$ rather than only in $|z|<1$.
Yes, but since the coefficients are positive, and $g(1)=1$, I conclude that $1$ is the ONLY solution of $g(z)=1$ in $|z|\leq 1$. Moreover, since $g'(1)\neq 0$, the nature of all singularities of $f$ in $|z|\leq 1$ is clear.
Now that I have looked at Ch. VI you referred to (especially Theorem VI.5 there), it seems to me that we have to have the analyticity of $f$, not just in the unit disk, but also in a certain kind of domain containing points not in the unit disk.
Our function has isolated singularities on the boundary. In fact only one singularity. Subtracting the elementary function with the same singularruty we obtain that the difference is analytic in the larger disk, therefore the contribution from this difference to the Taylor series is negligibly small.
@AlexandreEremenko : I am sorry for all these continuing questions, as I have absolutely no real experience in this area and am trying to learn. However, (i) I think here we have two points of singularity on the unit circle, $\pm1$. (ii) Finding elementary functions mimicking the singularities does not seem quite obvious to me. (iii) Here we also have branch cuts going from $\pm1$ outside the unit disk, and so, it does not seem possible to extend $g$ analytically to an entire larger disk. (iv) we also seem to have to ensure that $g$ does not map any "good" point to a cut.
Yes, two points, $-1,1$. The contributions from them are added. Branch cut is irrelevant here. Both singularities are of the form (some power)+analytic function. When we subtract those powers the remainder has radius of convergence $>1$, so it can be neglected and only powers contribute. These powers have explicit expansions at 0. The computation in your answer seems correct to me.
I think the subtraction would work if we only had $g$, but we actually have $f=g\circ g$. So, I think we do have to take care of point (iv) in my previous comment.
@Iosif Pinelis: by "subtraction" I mean "subtraction of the main term with non-integer exponent" and "neglecting all other terms with non-integer exponent". In other words, all we need for asymptotics of coefficients is the principal terms of expansion of $g\circ g$ near $-1$ and $1$.
@AlexandreEremenko : I still think that we can restrict our worries about $g\circ g$ to just the points $\pm1$ only after taking care of the previously listed point (iv): to ensure that $g$ does not map any "good" point to a cut.
@Iosif Pinelis: I agree with the first part of your sentence, and this is the same that I said (there are only two points $\pm$ to consider), but the second part of the sentence I do not understand: what do you refer to as (iv)? And the "cuts" are irrelevant here.
@AlexandreEremenko : Point (iv) was first stated in my big four-point comment and restated in my last comment: that we seem to have to ensure that $g$ does not map any "good" point to a cut. What I mean here is the following. Let $\mathbf D$ be a domain as in Theorem VI.5 you referred to, with $\rho=1$ and $\zeta_{1,2}=\pm1$, so that $\mathbf D$ contains some points outside the unit disc.
Previous comment continued: It would be bad if $g$ mapped some point of $\mathbf D$ to a branch cut for $g$, because then $g\circ g$ would not be analytic in $\mathbf D$. I did not see this bad situation taken care of in your answer. So, I thought this could be something quite trivial, but so far I don't see such a trivial way to deal with this.
We know that $g$ maps $\overline{D}$ into $D\cup{1}$. So there is always a cut from $1$ outside $f(D)$..
@AlexandreEremenko : Alas, I don't understand your latest comment. In particular, what do you mean by "a cut" -- a cut for $g$ or for $f$? Also, I don't see how the 1st sentence in that comment implies the 2nd one (except that $g$ obviously has a cut, even two cuts). Most importantly, I don't see at all how that comment addresses the concern expressed in my most recent two-part comment: to ensure that $g$ does not map any "good" point to a cut.
@Iosif Pinelis: I was talking about the cut of g. Since $g$ maps the unit disk into itself $g=g\circ g$ also does. So there are no problems with slit. But I do not understand this long discussion: your argument in your answer seems correct to me, what else are you trying to find out? –
@AlexandreEremenko : As I wrote in a previous comment: "It would be bad if $g$ mapped some point of $\mathbf D$ to a branch cut for $g$, because then $g\circ g$ would not be analytic in $\mathbf D$. I did not see this bad situation taken care of in your answer. So, I thought this could be something quite trivial, but so far I don't see such a trivial way to deal with this." This has been the main point I have been trying to clarify.
@Iosif Pinelis: I think I clarified this point in my previous comment: if $g$ is analytic in the unit disk and maps the unit disk into itself, then evidently $g\circ g$ is analytic in the unit disk and maps the unit disk into itself as well. Do you disagree with this statement?
@Iosif Pinelis: And I do not claim that my answer contains a compete solution: read the sentence: "I have not done the calculation...". I only explained how one can do it. And you did it. So what is the point of further discussion?
Note that $g(1)=g'(1)=1$ and for real $x\in(-1,1)$
\begin{equation*}
g''(x)=\frac1{\pi\sqrt{1-x^2}}.
\end{equation*}
The map $z\mapsto1-z^2$ maps the set
\begin{equation*}
R:=\mathbb C\setminus(-\infty,-1]\setminus[1,\infty)
\end{equation*}
onto $\mathbb C\setminus(-\infty,0]$. So, the map
\begin{equation*}
R\ni z\mapsto h(z):=\frac1{\pi\sqrt{1-z^2}}
\end{equation*}
is analytic, and hence $g$ can be continued analytically to $R$ by the Taylor formula
\begin{equation*}
R\ni z\mapsto g(z):=g(1)+g'(1)(z-1)+\int_1^z(z-u)h(u)\,du \\
=z+\int_1^z(z-u)h(u)\,du. \tag{0}
\end{equation*}
Take a real $r>1$ and let
\begin{equation*}
D_r:=\{z\in R\colon|z|<r,|\arg(z-1)|>\pi/4\}.
\end{equation*}
The main difficulty is to show that $g(D_r)\subseteq R$ for some $r>1$.
First here, take any real $t>0$. Then there is some real $u_t>0$ such that for all complex $z$ with $|z|\le1$ and $|\arg z|\ge t$ we have $|\frac1\pi+\frac z2|\le\frac1\pi+\frac12-u_t$, whence
\begin{equation*}
|g(z)|\le\Big|\frac1\pi+\frac z2\Big|+ \sum_{n=1}^\infty \frac{(2n-3)!!}{(2n-1)n!2^n \pi}|z|^{2n}
\le\frac1\pi+\frac12-u_t+ \sum_{n=1}^\infty \frac{(2n-3)!!}{(2n-1)n!2^n \pi}
=g(1)-u_t=1-u_t.
\end{equation*}
Since $g$ is analytic on $R$, it is uniformly continuous on any compact subset of $R$. So, there is some real $r_t>1$ such that $|g(z)|<1$ for all complex $z\ne1$ with $|z|\le r_t$ and $|\arg z|\ge t$.
Also, it follows that $|g(z)|<1$ for all complex $z$ with $|z|<1$.
Thus, to prove that $g(D_r)\subseteq R$ for some $r>1$, it suffices to show that
$\Im g(z)\ne0$ for all $z\ne1$ with $|z|\ge1$ and $|\arg(z-1)|>\pi/4$ that are close enough to $1$.
To see this, note that $h(u)\sim\frac1{\pi\sqrt2\,\sqrt{1-u}}$ as $u\to1$. Then (0) yields
\begin{equation*}
g(z)=z+c_1\cdot(z-1)^{3/2}, \tag{1}
\end{equation*}
where $c_1=c_1(z)$ converges to a nonzero complex number as $z\to1$. So, the conclusion that
$\Im g(z)\ne0$ for all $z\ne1$ with $|z|\ge1$ and $|\arg(z-1)|>\pi/4$ that are close enough to $1$ follows, which does show that $g(D_r)\subseteq R$ for some $r>1$.
So, $f=g\circ g$ is analytic on $D_r$ for such an $r$.
Moreover, (1) implies
\begin{equation*}
f(z)=g(g(z))=g(z+c_1\cdot(z-1)^{3/2})\\
=z+c_1\cdot(z-1)^{3/2}+\hat c_1\cdot(z-1+c_1\cdot(z-1)^{3/2})^{3/2} \\
=z+\tilde c_1\cdot(z-1)^{3/2}, \tag{2}
\end{equation*}
where $\hat c_1=\hat c_1(z)\sim c_1(z)$ and $\tilde c_1=\tilde c_1(z)\sim2c_1(z)$ as $z\to1$.
Similarly, since $h(u)\sim\frac1{\pi\sqrt2\,\sqrt{1+u}}$ as $u\to-1$, the Taylor formula for $g(z)$ at $z=-1$ (together with the observation $g(-1)=g'(-1)=0$) yields $g(z)=c_{-1}\cdot(z+1)^{3/2}$, where $c_{-1}=c_{-1}(z)$ converges to a complex number as $z\to-1$. Therefore and because $g$ is analytic at $0$, we have
\begin{equation*}
f(z)=g(g(z))=c_0+\hat c_0\cdot g(z)
=c_0+\tilde c_0\cdot(z+1)^{3/2}, \tag{3}
\end{equation*}
where $c_0:=g(0)$, $\hat c_0=\hat c_0(z)=O(1)$, and $\tilde c_0=\tilde c_0(z)=O(1)$ as $z\to-1$.
Now we are finally in a position to use (with thanks to Alexandre Eremenko) Theorem VI.5 (with $\alpha=-3/2$, $\beta=0$, $\rho=1$, $r=2$, $\zeta_1=1$, $\zeta_2=-1$, $\mathbf D=D_r$, $\sigma_1(z)=z$, $\sigma_2(z)=c_0$), which yields the $n$th coefficient for $f$:
\begin{equation*}
[z^n]f(z)=O(n^{\alpha-1})=O(n^{-5/2}),
\end{equation*}
as you conjectured.
For an illustration, here is the set $\{g(z)\colon z\in R,|z|<2\}$:
while the idea is right, not sure your analysis is quite correct since $g(1-r)$~$1-r-cr^{3/2}, r>0, r \to 0$ which gives $f(1-r)$ a $3/2$ singularity too, so the coefficients are $O(n^{-5/2})$; it may be easier seen taking derivatives since $g'(1-r)$~$1-c_1r^{1/2}$ so $f'=(g'\circ g)g'$ is also $1-c_2r^{1/2}$ so $f'$ has coefficients $O(n^{-3/2})$ etc
@Conrad : Thank you for your comment. More recently, I too have realized that $-5/4$ cannot be right. I had, rather blindly, relied on Mathematica in computing the composition $f=g\circ g$ and its asymptotics at $z=1$. Apparently, there was a mistake either in my or Mathematica's coding. I have now redone this part of the proof (more) manually, which hopefully yields the correct result.
I think now all is good - since $g$ "contracts" the unit disc it didn't make sense to get a worse estimate for the coefficients of $f=g \circ g$ (while we cannot get a better one because of positivity of the coefficients and the fact that $f(x)=g(x)/2+\sum {b_nx^n}, b_n \ge 0$)
Thank you so much Professor Pinelis for your detailed proof! I have a quick question: Theorem VI.5 assumes that $\sigma_1,\sigma_2$ are a linear combination of elements from ${(1-z)^{-\alpha} \lambda(z)^\beta| \alpha,\beta\in \mathbb{C}}$. To see this, notice that $z$ is in $\operatorname{span}{(1-z)^{-\alpha} \lambda(z)^\beta| \alpha,\beta\in \mathbb{C}}$ (because $z=-(1-z)+1$) and a constant $c_0$ is also there. Am I right?
@Cechco : Yes, any polynomial in $z$ is such a linear combination, and it clearly does not affect the asymptotics of the coefficients as $n\to\infty$.
@IosifPinelis I am sorry there is one thing I tried two days but couldn't understand. I was wondering why checking the function values of the boundary points of $R$ implies $g$ maps $R$ into $R$. It seems to me that $\partial g(R)\subseteq g(\partial R) = {0,1,\infty}\cup {g(s\pm 0i)| |s|>1}$. However, I couldn't prove that $g(\partial R) = \partial g(R)$ if it is true. If it happens to be true, I have a perhaps feasible argument: assume there is a point $z$ that gets mapped to, say, $[1,\infty)$ (similar for $(-\infty,-1]$. Note that $g(0)=1/\pi$.
2/2: Consider a path that connects $0$ and $z$. Then the image of the path (it connects $1/\pi$ and $g(z)\ge 1$) will cross the curve ${ g(s\pm 0i| s\ge 1 }$ by an argument like Jordan curve theorem. But I couldn't prove $g(\partial R)=\partial g(R)$. My question may be stupid. I am sorry for it.
@IosifPinelis I got some numerical results from Mathematica tonight, which made me a little confused. There seems to be two rays starting from $-1$ and going to $\infty$ that get mapped to $(-\infty,-1]$. The real part of the two rays is $\le -1$. I used the following code: g[u_] := (Sqrt[1 - u^2] + u (Pi - ArcCos[u]))/Pi; ContourPlot[Im[g[a + b I]] == 0, {a, -100, 0}, {b, -1, 100},
AxesLabel -> Automatic]
@Cechco : You cannot get $\Re g\le-1$ on the entire two rays emanating from $-1$, because $g(-1)=0$.
@Cechco : As for the containment $g(D)\subseteq R$ for an adequate domain $D$ -- which is the crucial part of the proof, I did that previously in some haste, having in mind principles of boundary correspondence such as ones in https://www.impan.pl/en/publishing-house/journals-and-series/colloquium-mathematicum/all/139/1/87556/a-topological-dichotomy-with-applications-to-complex-analysis or https://arxiv.org/pdf/0907.2960.pdf . That did not work quite well, though. I have now redone that part of the proof. Hopefully, this is correct now.
|
2025-03-21T14:48:31.574027
| 2020-07-21T22:28:28 |
366237
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Agno",
"Carlo Beenakker",
"Kurisuto Asutora",
"Stopple",
"Sylvain JULIEN",
"https://mathoverflow.net/users/11260",
"https://mathoverflow.net/users/12489",
"https://mathoverflow.net/users/13625",
"https://mathoverflow.net/users/46852",
"https://mathoverflow.net/users/6756"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631328",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366237"
}
|
Stack Exchange
|
Are the ordinates of the non-trivial zeros of $\zeta(s)$ uniformly distributed around the mid points of Gram point intervals they can be mapped to?
Let $\rho_n$ be the $n$-th non-trivial zero of $\zeta(s)$ and $z_n = \Im(\rho_n)$ with $z_n > 0$ and $z_{n+1} \ge z_n$.
A well known method to establish that all $\rho$s reside on the critical line ($\Re(\rho)=\frac12)$ up to a certain height $T$, is to make use of the regularly distributed and easy to compute Gram points ($g_n$).
The process boils down to assuring that each Gram point interval $[g_n,g_{n+1})$ contains only a single $z$. Heuristics show this is already the case for most intervals, however it is also known to fail infinitely often resulting in empty intervals or intervals containing multiple $z$ (i.e. violations of Gram's law and Rosser's rule).
Failures will typically be induced when larger swings in the $Z(t)$-function occur, where $Z(t) = e^{i\Theta(t)} \zeta\left(\frac12 + it\right)$ and $\Theta(t)$ is the Riemann-Siegel Theta function. Some accounting is then required to ensure there are as many $z$ as intervals in a Gram block with 'bad' Gram points. Morally this process is equivalent to mapping each 'wrongly' located $z$ to its unique interval.
So, when $Z(t)$ oscillates rather 'calmly', a $z$ will typically reside not too far from the middle of its Gram point interval. However, when oscillations of $Z(t)$ become more 'fierce' in amplitude, the $z$ will be pushed more towards the edges of their intervals or even hop into the neighbouring intervals.
With this in mind, we could compute how far ($=d_n$) each $z_n$ has been 'pushed' away from the mid point of the interval it 'belongs to'. This could be done as follows (note: the index of a Gram point $g_n$ starts at $-1$ and for $z$ at $1$):
$$d_n = z_n - \frac{g_{n-2}+g_{n-1}}{2}$$
The picture below aims to visualise the process and how $d_n$ is computed for real data around the first Lehmer pair. (note: the legend of the red line should be: $\Im\left( e^{(i\sin(\Theta(t))}\right)$)
The next graphs show the normalised histograms of the differences $d_n$ for increasing ranges of $n$.
Question:
The shape of the distribution seems to become increasingly 'normal' for higher $n$. Could this indeed be the case and if so, is there a logical explanation for this phenomenon?
Here is the $d_n$-data for $n=1..999999$.
ADDED 1:
Since, on average, the densities of $z_n$ and $g_n$ increase when we go up the critical line, the variances of the distributions shown above will become smaller and their peaks higher. This can be normalised as follows:
$$d_n^* = z_n \log(z_n) - \frac{g_{n-2}+g_{n-1}}{2}\log\left(\frac{g_{n-2}+g_{n-1}}{2}\right)$$
Here is the $d_n^*$-data for $n=1..999999$.
A Normal distribution with $\mu=0$ and $\sigma=2.65$ then already provides a pretty decent visual fit (red line).
ADDED 2:
Here is the distribution of the first 100 mln $d_n^*$ (blue dots). The red line is a Normal distribution with $\mu=0$ and $\sigma=2.71$.
The real data have a $\mu=-5.301\cdot 10^{-8}$ and $\sigma=2.642$. The fit with the Normal distribution clearly gets weaker in the tails. I do expect this to improve for larger $n$, since the "turmoil" in $Z(t)$ is expected to increase thereby inducing a relatively higher proportion of larger deviations $d_n^*$.
A Gram point is defined as the $t$ where $\Theta(t)=k\pi$ (i.e. where $\Im\left(\zeta(\frac12+it)\right)=0, t\ne z$). I realised that the mid point of the Gram interval (calculated as a simple average above), quickly converges to the $t$ where $\Theta(t)=\frac{(2k+1)\pi}{2}$ (i.e. where $\Re\left(\zeta(\frac12+it)\right)=0, t\ne z$). Let's label these points as $\hat{g}_n$.
My question is then equivalent to: "Is $\hat{d}_n=z_n -\hat{g}_{n-2}$ uniformly distributed?" or in other words: "Is the difference between 'paired' zeros of $\Re\zeta(\frac12+it)$ uniformly distributed?".
Maybe you can find relevant results about some kind of "normal distribution of critical zeros related quantities" in Selberg's collected papers.
Not what you ask for, but somewhat related and maybe interesting for you (if you don't know it already): Arias de Reyna, J.: On the distribution (mod 1) of the normalized zeros of the Riemann zeta-function. J. Number Theory 153 (2015), 37–53. That paper is not about the distance between the zero and the Gram point it "belongs to", but about the relative location of the zero with respect to the Gram points between which it is actually located.
@Kurisuto Asutora Thanks for the useful link! Arias de Reyna indeed follows a related but different approach. The uniformity of the resulting distributions (for $n=1..2mln$) also appears to be a bit less strong, but this is based on inspecting the visuals only. An intriguing sentence is his paper I find: "In a non published report, Odlyzko [14, p. 60] conjectured that the ordinates of the zeros $z_n$ are not related to the Gram points for $z_n$ large". This conjecture seems to contradict my observations above.
Odlyzko's unpublished paper is at http://www.dtc.umn.edu/~odlyzko/unpublished/zeta.10to20.1992.pdf
Is there a rationale for $\sigma=2.65$? I observe that the $10^7$-th zero occurs at height $T=4.99238\cdot 10^6$, and $\log \log T=2.73589$.
@Stopple. Unfortunately I have no rationale for the $\sigma = 2.65$, it is just the value I picked that gave the best visual fit with the relative distribution of the first $10^6$-th $d_n^*$. For higher $n$ this could get closer to your value (I had even hoped that $\sigma = e$ for a moment ;-) ).
Concerning "Could this indeed be the case and if so, is there a logical explanation for this phenomenon?": For comparison, you could study a similar statistic for randomly generated points instead of the zeta zeros. For example, take a Poisson process with intensity one, which gives you (random) points $x_1,x_2,...$, and then look at the distribution of difference $x_n - n,~1 \leq n \leq N$ (suitably normalized).
@Kurisuto Asutora Apologies for my slow response, just back from holidays. I have checked the distributions for the edges of the intervals i.e. $z_n-g_{n-1}$ and $z_n-g_{n-2}$ and as expected their means shift and they become skewed. Have also tried the Poisson process with $\lambda=1$ and produced sample paths for the first 1 mln steps $x_n$. Then computed the differences $x_n-n$ and the associated Histograms. The resulting distributions seem quite chaotic and are far from normal.
Can you post a picture of the distribution that you obtain for the points coming from the Poisson process? (I am traveling and don't have a computer with me to try to experiment myself).
PS: I don't understand the last two sentences in your post. What do they mean? (You mean "normally distributed instead of "uniformly distributed"?)
@Kurisuto Asutora The Maple code used is: J := PoissonProcess(1.0); A := SamplePath(J(t), t = 1 .. 10000, timesteps = 10000, replications = 1); L := [seq(A(x) - x, x = 1 .. 10000)]; Histogram(L, color = "blue", bincount = 200, frequencyscale = relative, style = polygon, symbolsize = 1, range = -100 .. 100); Running this, results (for instance) in the following plot: https://ibb.co/W32L0Bj. Not sure wether I have done this correctly, but the first column of this picture is the distribution we're after https://ibb.co/6DW60NS . To your PS, I indeed meant 'normally distributed'.
|
2025-03-21T14:48:31.574836
| 2020-07-21T22:30:14 |
366239
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Connor Malin",
"Liding Yao",
"Robert Bruner",
"https://mathoverflow.net/users/118469",
"https://mathoverflow.net/users/134512",
"https://mathoverflow.net/users/6872"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631329",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366239"
}
|
Stack Exchange
|
Can a restriction of a null-homotopic spherical map be null-homopotic?
Let $n,q$ be positive integers. We are interested to the cases where $n>q$.
Let $F:\mathbb B^n\to\mathbb S^{q-1}$ be a continuous (differentiable, if needed) map, such that $F(1,0^{n-1})=(1,0^{q-1})$, $F(-1,0^{n-1})=(-1,0^{q-1})$ and $F(\{0\}^1\times\mathbb S^{n-2})\subset\{0\}^1\times\mathbb S^{q-2}\simeq\mathbb S^{q-2}$.
Can we find a map $G:\mathbb B^{n-1}\to \mathbb S^{q-2}$ such that $G(\theta)=F(0,\theta)\in \mathbb S^{q-2}$ for all $\theta\in\mathbb S^{n-2}$?
Perhaps add a paragraph explaining what you are looking for without formulas. I think this would make it more readable.
@ConnorMalin Sure. But I don't know how to start with. I am trying to show a null-homotopy of a total map implies a null-homotopy of a component. If $f(s,y)=(s,f''(s,y))$ is a null-homotopic map between spheres, I want to show $y\mapsto f''(0,y)$ is null-homotopic. Do you have any suggestions on what to edit?
So you are asking if a map of spheres sends a codimension 1 equator to a codimension one equator, and the original map is nullhomotopic (with no conditions on the null homotopy), does this imply the restriction of the map to the two equators is nullhomotopic?
@ConnorMalin Yes!
There exists nullhomotopic maps preserving codimension one equators such that the restriction to the equators is not nullhomotopic. Very easy examples come from taking non nullhomotopic maps of spheres and then extending them to a sphere of dimension higher by a nonsurjective map.
However, one can in fact come up with much stranger examples. I claim that for any even n, one can construct a map $S^{2n} \rightarrow S^{n+1}$ so that the restriction of the domain to every (2n-1)-sphere $x_1^2 + \dots x_{2n} ^2=t$ and codomain to the n-sphere $x_1^2 + \dots x_n ^2=t$ is nonnullhomotopic, and this map is nullhomotopic.
This fact follows immediately from the statement, "There is a nonnullhomotopic map $S^{2n-1} \rightarrow S^n$ such that the suspension of this map is trivial." This is because the suspension of the map is given by sending meridians to meridians via the map we are suspending.
Such a nullhomotopic map (for $n=2$) exists for the following reason: the Hopf map $S^3 \rightarrow S^2$ is a map that has infinite order in the homotopy groups of $S^2$. By a cute argument involving complex conjugation, the suspension of the Hopf map has order 2. Hence, twice the Hopf map is in the kernel of the suspension homomorphism, so we have proven a specific case.
For general n, this follows by computing the rational homotopy groups of $S^n$ and finding only one nontorsion group and applying the above reasoning.
Would you mind explaining the "cute argument"? Or is there any reference talk about the concrete construction? Sorry I am an analysis student and I am on the way of understanding.
@yaoliding So we define multiplication by n in $\pi_k(X)$ by precomposing with a degree n map from the sphere to itself. If $X$ is itself a sphere, we can ask whether precomposing with a degree n map and postcomposing with a degree n map are the same. By an unenlightening argument, one can conclude that at least after applying the suspension homomorphism they agree. To see they don't agree in general, we consider the Hopf map $\mathbb{C}^2 \supset S^3 \rightarrow \mathbb{C}P^1 = S^2$. We have conjugation actions on both sides $(x,y) \rightarrow (\bar{x},\bar{y})$ and
$[x,y] \rightarrow [\bar{x},\bar{y}]$. Which obviously commute with the Hopf map. This conjugation in $S^3$ is given by two reflections so is degree 1. In $S^2$ this is given by 1 reflection, so is degree -1. Since the Hopf map has infinite order, post composition by a degree -1 map is an example where the resulting map is not negative the original map. But we now deduce that its suspension has order two (or is trivial) because the above argument implies $\Sigma \eta = -\Sigma \eta$. I saw this argument in Schwede's book on symmetric spectra. You can find the fact about suspension in there.
I also like the following argument that the suspension of the Hopf map has order 2. We use S^3 = S(C^2) ---> P^1(C) = S^2 by sending (z1,z2) to [z1,z2] as our model of the Hopf map. The map from S^3 to S^3 sending (z1,z2) to (z1^k,z2^k) covers the map on P^1 sending [z1,z2] to [z1^k,z2^k]. It is easy to check that these maps have degree k^2 and k, respectively. Therefore, $\eta \circ k^2 = k \circ \eta$. Suspending, composition becomes commutative, so we get that (k^2-k) annihilates $\eta$ for any integer k. Since the gcd over all k of k^2-k is 2, $\eta$ has order 2.
|
2025-03-21T14:48:31.575202
| 2020-07-21T23:10:41 |
366241
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Jochen Wengenroth",
"https://mathoverflow.net/users/21051"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631330",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366241"
}
|
Stack Exchange
|
Continuity of eigenvectors
Let $\mathbb{C} \ni z \mapsto M(z)$ be a square matrix depending holomorphically on a parameter $z$ with the property that $\operatorname{dim}\ker(M(z)))=1$ for $z $ away from a discrete set $D \subset \mathbb{C}$ and $\operatorname{dim}\ker(M(z)))\ge 1$ for $z \in D.$
I ask: Is it always possible to choose a continuous vector $\mathbb{C} \ni z \mapsto v(z)$ such that $M(z)v(z)=0?$
The second requirement dim ker$(M(z))\ge 1$ for $z\in D$ follows from the first: Otherwise $M(z)$ would be invertible in $z$ and hence in a neighbourhood of $z$.
Yes. Let the size of your matrix be $n$.
Your condition implies that there is an $n-1\times n-1$ submatrix whose determinant is not identically equal to $0$.
Assume without loss of generality that this is the submatrix formed by the first $n-1$ rows and columns. Then we can set $u_n=1$ and find a vector $u(z)$ such that
$M(z)u(z)=0$ by solving the system of $n-1$ linear equations in $n-1$ variables. This requires only arithmetic operations
on the entries of $M$, so $u(z)$ will be meromorphic on $\mathbb{C}$. Let $D$ be the divisor of poles of $u$. According to a theorem of Weierstrass there is an entire function $f$ having zeros at $D$. Then $v(z)=f(z)u(z)$
is a solution to your problem, which is not only continuous but holomorphic.
|
2025-03-21T14:48:31.575332
| 2020-07-22T00:00:14 |
366245
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Eric Naslund",
"https://mathoverflow.net/users/12176"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631331",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366245"
}
|
Stack Exchange
|
Is there any relationship between Szemerédi's theorem and Sunflower conjecture?
I have observed some similar things between a reformulation of the Sunflower conjecture (see also conjecture 1.3 in Improved bounds for the sunflower lemma) and Szemerédi's theorem such that for Szemerédi's theorem we have an often-used equivalent finitary version which states that:
Theorem: for every positive integer $k$ and real number $\delta\in (0,1]$ there exists a positive integer $N=N(\delta,k)$ such that every subset of $\{1,2,\cdots,N\}$ of size at least $δN$ contains an arithmetic progression of length $k$ say $k\text{-arithmetic progression}$
The Sunflower conjecture states that :
Conjecture: Let $r ≥ 3$. There exists $c = c(r)$ such that any $w$-set system $F$ of size $|F| ≥ c^w$ contains an $r$-sunflower
We see that both of them investigate about existence of such constant with such size bounds for which we have $k\text{-arithmetic progression}$ or $r$-sunflower. Also, another thing that attracted my attention is that for Erdős conjecture on arithmetic progressions, Erdős and Turán made in 1936 the weaker conjecture that any set of integers with positive natural density contains infinitely many 3-term arithmetic progressions. This was proven by Klaus Roth in 1952, and generalized to arbitrarily long arithmetic progressions by Szemerédi in 1975 in what is now known as Szemerédi's theorem. In the opposite direction, Kostochka proved that any $w$-set system of size
$$
|F| \geq
cw! \cdot {(\log \log \log w/ \log \log w)}^{w}
$$
must contain a $3$-sunflower for some absolute constant $c$. Now for quantitative bounds of $r_k(N)$(the size of the largest subset of $\{1, 2, \ldots, N \}$ without an arithmetic progression of length $k$) it is an open problem to determine its exact growth rate. In the same time it is an open problem to determine the exact rate growth or exact bounds for The sunflower (the sunflower lemma via Shannon entropy). My question here is:
Question: According to similar things which I have cited above between the Sunflower conjecture and Szemerédi's theorem, is there any non-trivial relationship between them? And in which context we can consider Sunflower coincide with arithmetic progression?
I do not know of a direct connection to Roth or Szemerédi over the integers. However, the paper
N. Alon, A. Shpilka and C. Umans, On Sunflowers and Matrix Multiplication, 2012 IEEE 27th Conference on Computational Complexity, Porto, 2012, pp. 214-223, doi:10.1109/CCC.2012.26 (author pdf)
shows that a proof of the Sunflower Conjecture would imply a proof of the Erdos-Szemeredi sunflower theorem, which also follows from a bound of $(3-\delta)^n$ for the capset problem, a strong form of Roth over $\mathbb{F}_3^n$ (which is already known due to Croot-Lev-Pach). See this 2016 blog post by Gil Kalai for more discussions along this line.
It is also noteworthy that the Erdős-Szemerédi sunflower conjecture (which has been proved and follows from the capset problem) also implies that if $|S|=C\log(n)$ is a subset of $[n]$ for a large constant $C$, then there are three disjoints $X, Y, Z$ whose subset sums are identical, and thus the sums of the subsets $X, X \cup Y, X \cup Y \cup Z$ are in arithmetic progression; see
P. Erdős, A. Sárközy, Arithmetic progressions in subset sums, Discrete Mathematics 102 Issue 3 (1992) pp 249–264, doi:10.1016/0012-365X(92)90119-Z
(Core pdf).
Minor comment: "... and is equivalent to the capset problem" I don't believe it is equivalent (other than trivially since they are both true), as I have never seen a proof of the reverse direction, but exponential bounds for the capset problem imply exponential bounds for the Erdos-Szemeredi sunflower conjecture (An explicit bound is Theorem 8 here https://arxiv.org/abs/1606.09575)
|
2025-03-21T14:48:31.575730
| 2020-07-22T00:09:14 |
366246
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"David Roberts",
"Niels",
"curious math guy",
"https://mathoverflow.net/users/11682",
"https://mathoverflow.net/users/152554",
"https://mathoverflow.net/users/4177"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631332",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366246"
}
|
Stack Exchange
|
Commutative group stacks and Galois cohomology
"Classically", if we consider an abelian variety $A$ over some number field $k$, we get a $Gal(\bar{k}/k)$-module $A(\bar{k})$, or equivalently a sheaf of abelian groups on the étale site $\text{Spec}(k)_{ét}$. In particular, we can apply the machinery of Galois cohomology.
My question is if there is a similar construction for abelian stacks. If we have commutative group stacks over number field $k$ with a short exact sequence
$$0\rightarrow A\rightarrow B\rightarrow C\rightarrow 0$$
is there a universal way of extending the left exact sequence
$$0\rightarrow A(k)\rightarrow B(k)\rightarrow C(k)$$
i.e. something akin to $H^i(G_k,A)$? My thinking is that we have an exact triangle
$$A^\flat\rightarrow B^\flat\rightarrow C^\flat \rightarrow A^\flat[1]$$
in $D(\text{Spec}(k)$ on which we can apply the global sections functor. Would that give us something relevant?
This paper by Noohi might be relevant: Group cohomology with coefficients in a crossed module Journal of the Institute of Mathematics of Jussieu, 10(2) (2011) 359–404. https://doi.org/10.1017/S1474748010000186 https://arxiv.org/abs/0902.0161
including the definition of $A^\flat$ looks like a good idea
@Niels, I appolozige, I assumed this is standard notation. Denote by $C(k)$ the $2$-category of complexes of abelian sheaves $G^\bullet$ over $\text{Spec}(k)$ such that $^G^i=0$ for $i\neq 0,-1$ and such that $G^{-1}$ is injective. Then there exists an equivalence of $2$-categories between commutative group schemes and $C(\text{Spec}(k))$, which we denote by $\flat$. My reference for this is Section 2 of https://arxiv.org/pdf/1404.0285.pdf.
|
2025-03-21T14:48:31.575862
| 2020-07-22T00:59:38 |
366251
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"KConrad",
"https://mathoverflow.net/users/3272"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631333",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366251"
}
|
Stack Exchange
|
Maximal order of $x^n-d$ and its dependence on $d$
It's well known that the structure of the maximal order of $\mathbb{Q}[\sqrt{d}]$ depends on $d$ modulo $4$: (assuming $d$ is squarefree), the maximal order is $\mathbb{Z}\left[\frac{1+\sqrt{d}}{2}\right]$ if $d \equiv 1 \mod{4}$ and $\mathbb{Z}[\sqrt{d}]$ otherwise. The structure of the biquadratic extension $\mathbb{Q}[\sqrt{d},\sqrt{e}]$ has a similar but more complicated description.
Now let's consider $K=\mathbb{Q}[\sqrt[n]{d}]$. For simplicity, let's consider only $d$ prime. Based on some SAGE computation, it seems that there is a degree $n$ extension $M$ of $\mathbb{Q}[\mu_{n^2}]$ such that $\mathcal{O}_K$ contains more than just $\mathbb{Z}[\sqrt[n]{d}]$ iff $d$ splits in $M$. (Note that if $n$ is prime, then $M$ is unique.) For $n=2$, one sees that $M=\mathbb{Q}[i]$.
Is this known? It doesn't seem too hard to prove that if $d$ is split, then there are more algebraic integers in $K$. But does this appear anywhere?
You are asking when the ring of integers of $\mathbf Q(\sqrt[n]{d})$ is $\mathbf Z[\sqrt[n]{d}]$. Presumably you intend for $x^n - d$ to be irreducible (before you take $d$ prime for simplicity). See the theorems in https://kconrad.math.uconn.edu/blurbs/gradnumthy/integersradical.pdf. You might also be interested in https://arxiv.org/abs/1909.07184, although it is about a setting over a base ring larger than $\mathbf Z$.
|
2025-03-21T14:48:31.575981
| 2020-07-22T02:46:23 |
366254
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631334",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366254"
}
|
Stack Exchange
|
Comparison of two Fourier transforms
I am looking for $\delta>0$, such that
$$
\delta \int_{-\infty}^{\infty} \exp(its)
{ \Gamma\{2(it+1)/3\}\over \Gamma\{(it+1)/2\} }dt \le \\
\int_{-\infty}^{\infty} \exp(its)
{ \Gamma (it+1)\over \Gamma\{(it+1)/2\} }dt
$$
for any $s>0$.
Ideally I want to find $\delta(\beta)$ such that
$$
\delta(\beta_1) \int_{-\infty}^{\infty} \exp(its)
{ \Gamma\{(it+1)/\beta_1\}\over \Gamma\{(it+1)/2\} }dt \le \\
\delta(\beta_2) \int_{-\infty}^{\infty} \exp(its)
{ \Gamma \{(it+1)/\beta_2\}\over \Gamma\{(it+1)/2\} }dt
$$
for any $s>0$, $1<\beta_2<\beta_1<2$. However it is far more complicated problem. Any hints will be greatly appreciated
|
2025-03-21T14:48:31.576047
| 2020-07-22T03:17:53 |
366255
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Andy Putman",
"Arun Debray",
"Mark Grant",
"https://mathoverflow.net/users/161496",
"https://mathoverflow.net/users/317",
"https://mathoverflow.net/users/8103",
"https://mathoverflow.net/users/97265",
"user_501"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631335",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366255"
}
|
Stack Exchange
|
Simple examples of equivariant cobordism
Let $Y$ be an oriented 3-manifold with a free action by a finite group $G$. If I understand correctly, there exists a multiple of $NY$ of $Y$ and an oriented manifold $X$ such that $\partial X = NY$ and $G$ extends to a free action on $X$. (That is, the equivariant oriented cobordism group is finite. Here, I believe $NY$ should be interpreted as $N$ disjoint copies of $Y$ - note that $N$ is nonzero.) I am trying to understand some very simple examples of this. For instance, if $Y = S^3$ and $G$ is a cyclic group (so that the quotient is a lens space), what is the manifold $X$?
EDIT: For a concrete mention of this claim, see the bottom of the first page of https://www.maths.ed.ac.uk/~v1ranick/papers/aps002.pdf
I realize that the claim is from equivariant bordism theory (as mentioned in one of the comments) but I am not very familiar with this, so I just gave the place where I first saw it.
I'm confused -- applying your claim to the trivial group $G$, it would imply that the ordinary oriented cobordism group is finite, which is of course false.
@AndyPutman I think this bordism theory is equivalent to the bordism theory of oriented manifolds with a principal $G$-bundle (depending on whether $G$ must act by orientation-preserving diffeomorphisms -- if not, this gets more complicated). Since $G$ is finite, this contains infinite-order elements when $n\equiv 0\bmod 4$, and is torsion otherwise (e.g. by an Atiyah-Hirzebruch argument). So the second part of the question, asking about $n = 3$, is still valid.
@AndyPutman My apologies - I think indeed I only care about $n = 3$. The question has been edited accordingly.
Following @ArunDebray's comment, the AHSS shows that the relevant bordism group is $\Omega_3(BG)\cong H_3(BG;\mathbb{Z})$, which is annihilated by $|G|$, so we may take $N=|G|$. In fact, $G\times Y$ (where $G$ acts on $Y$) is equivariantly cobordant to $G\times Y$ (where $G$ acts on $G$). Is there a geometic way to see this?
For the concrete case of the cyclic group $C_p$ acting linearly on $S^3$, there's a very explicit construction. Call $\omega = e^{2\pi i/p}$. Fix an integer $q$ coprime with $p$, and let us look at the action $\lambda_q$ on $\mathbb{C}^2$ generated by the diagonal matrix with entries $\omega, \omega^q$. $\lambda_q$ restricts to an action (which I will still call $\lambda_q$) on $S^3$ (as the unit sphere in $\mathbb C^2$ whose quotient is $L(p,q)$.
Take $M = V(x_0^p + \dots + x_3^p)$, the Fermat hypersurface of degree $d$ in $\mathbb{CP}^3$. On $\mathbb{CP}^3$ we define two actions $\psi$ and $\phi$ of $C_p$. Calling $g$ the generator of $C_p$, the two actions are given by:
$\phi_g(x_0:x_1:x_2:x_3) = (\omega x_0 : \omega^{-q}x_1 : x_2 : x_3)$, and $\psi_g(x_0:x_1:x_2:x_3) = (x_0 : x_1 : \omega x_2 : x_3)$.
I claim that:
Both actions preserve $M$.
The fixed point set of the action $\phi$ on $M$ is the set of points $(0:0:1:\omega^a)$ as $a$ varies, and the action is semi-free (i.e. there are no new fixed points appearing when you take powers of $g$).
$\psi_g$ cyclically permutes the fixed points of $\phi_g$ on $M$.
$\phi_g$ and $\psi_g$ commute.
The linearised action of $\phi_g$ on $M$ at $(0:0:1:1)$ is $\lambda_{-q}$.
I will not justify these points (the only one that maybe requires some care is the last one), and I will just take them for granted. Now take a small $\phi_g$-invariant ball $B$ in $M$ centred at $(0:0:1:1)$ and remove all its $\psi_g$-orbit in $M$, to get $M_0$. The action of $\phi$ on $M_0$ is free (because $\phi_{g^k}$ has the same fixed points as $\phi_g$ for each $0 < k< p$) and it extends the linear action $\lambda_q$ on $pS^3 = \sqcup_k \psi_{g^k} \partial B$. (Note that there is a $-q$ in an exponent of $\omega$ when defining $\phi$: this is because the boundary of $M_0$ is the boundary of $B$ with its orientation reversed.)
|
2025-03-21T14:48:31.576335
| 2020-07-22T03:22:35 |
366256
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631336",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366256"
}
|
Stack Exchange
|
Has the following generalization of monotropic programming been studied in the literature?
I am interested in problems of the form
$$\min_{x \in C} \sum_{i=1}^n\sum_{j=1}^n f(x_i,x_j)$$
where $C$ is a convex subset of $\mathbb{R}^{n}$, and $f \colon \mathbb{R}^{2} \to \mathbb{R}$ is convex.
Question: Has this class of optimization problems being studied in some detail?
Thank you in advance for your help.
The Extended Monotropic Programming problem deals with a more general extension to all n variables at a time than the two variables at a time extension you are interested in, except that the literature I have seen deals only with the constraint set $C$ being a closed linear subspace, not a general convex set.
Specifically:
$\min_{x \in \text{closed linear subsapce}} \Sigma_{i=1}^nf_i(x_i)$ with respect to $x$, where $x = (x_1,...,x_n)$, $x_i \in \mathbb{R}^{i}$
See "Extended Monotropic Programming and Duality", Dimitri P. Bertsekas,
Journal of Optimization Theory and Applications volume 139, pages209–225(2008)
2010 postprint available at https://www.mit.edu/~dimitrib/Extended_Mono.pdf
Additional reference:
On a zero duality gap result in extended monotropic programming,Radu Ioan Bot, Ern ̈o Robert Csetnek"
Older published version: https://link.springer.com/article/10.1007/s10957-010-9733-y
|
2025-03-21T14:48:31.576460
| 2020-07-22T03:43:29 |
366257
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Cheng-Chiang Tsai",
"LSpice",
"https://mathoverflow.net/users/2383",
"https://mathoverflow.net/users/31327"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631337",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366257"
}
|
Stack Exchange
|
Slodowy slice intersecting a given orbit "minimally"?
Let $\mathfrak{g}$ be a complex semisimple Lie algebra. Is it true that for any $X\in\mathfrak{g}$, there exists an $\mathfrak{sl}_2$-triple $(e,h,f)$ in $\mathfrak{g}$ such that
We have $X\in e+Z_{\mathfrak{g}}(f)$, and
$\dim Z_{\mathfrak{g}}(X)=\dim Z_{\mathfrak{g}}(e)$?
($Z_{\mathfrak{g}}(-)$ always the centralizer in $\mathfrak{g}$. One can deduce from the above conditions that the conjugacy orbit of $X$ must then meet the slice transversally, like in classical Kostant section situation.)
When $X$ is regular this is the well-known result about Kostant section. When $\mathfrak{g}=\mathfrak{gl}_n$ this is also true and can be deduced from the rational form of $X$. It might be that this question can be answered by a simple reference, but thanks a lot in advance in all cases!
Although it's a different question, you may be interested in @AlexanderPremet's answer to my question Commutativity and Kostant sections.
Thank you! Yeah I had been attracted to your question when I looked for an answer to this 1.5 years ago :)
Btw I think you will be amused by the following proof: choose $X=X_s+X_n$ in a split $G$ over $\mathbb{Q}_p$ with $X_s$ split. Consider the Shalika germ expansion (do we need $p\gg\operatorname{rank}(G)$ to have germs for arbitrary $X$?) for $X$. We claim that for any nilpotent $e$ with non-zero germ $\Gamma_e(X)$ we have the asserted property.
I then turn the above into AG following inspiration from $\S5$ of Langlands-Shelstad '87 (... transfer factor) which is based on Langlands '83 (Orbital Integrals on Forms of SL(3), I).
(Edited) A natural question then, is that suppose $X_s$ is possibly non-split. Is the same assertion true, say for any $e$ in the "stable wave-front set" of $X$ (maximal e with stable Shalika germ $\Gamma_e^{st}(X)\not=0$), or just some $e$ with $\Gamma_e^{st}(X)\not=0$?
Explicit asymptotic expansions in $p$-adic harmonic analysis II gives fairly explicit conditions for existence of germs, but not much weaker than you'd expect from DeBacker+Kim–Murnaghan-type results. I don't think I follow the second step of your proof, but I like the first step!
Nice, and thanks! The 2nd step not quite a step but rather an intuition. I guess one can read it as "following the AG tool of Langlands and Shelsted in (cited sections of papers) they used to treat Shalika germs, ..."
Woooo, I think I finally find a proof!$
\newcommand{\Lie}{\operatorname{Lie}}
\newcommand{\Lg}{\mathfrak{g}}
\newcommand{\Ll}{\mathfrak{l}}
\newcommand{\Ad}{\operatorname{Ad}}$
Let $G=\operatorname{Aut}(\Lg)^\circ$ be a corresponding complex Lie group. Write $X=X_s+X_n$ the Jordan decomposition into semisimple and nilpotent part. The centralizer $L:=Z_G(X_s)$ is a Levi subgroup. Denote by $\Ll=\Lie L$ and $Z(\Ll)=\Lie Z(L)$ the center of $\Ll$. Write $\mathcal{O}:=Z(\Ll)\cdot\Ad(L)X_n$, a locally closed subvariety in $\Ll$. Note that $\mathcal{O}\cong Z(\Ll)\times\Ad(L)X_n$ has an obvious map to $Z(\Ll)$.
Pick a parabolic subgroup $P$ containing $L$ as its Levi subgroup. Let $U$ be the unipotent radical of $P$ so that $P=LU$. There exists a nilpotent $G$-orbit in $\Lg$ that intersects $X_n+\Lie U$ at a dense open subset of $X_n+\Lie U$. Let $e$ be any element in the intersection. We have to prove the two asserted property in the question, i.e.
$\Ad(G)X$ meets the Slodowy slice $e+Z_{\Lg}(f)$ and
$\dim Z_G(X)=\dim Z_G(e)$.
We have $Z_G(X)=Z_L(X_n)$ and also $\dim Z_G(e)=\dim Z_L(X_n)$ [LS79, Theorem 1.3]. This proves (2).
Consider the “generalized Grothendieck-Springer alternation” given by
$$
\tilde{\mathfrak{g}}_X:=\{(g,\gamma)\in (P\backslash G)\times\mathfrak{g}\;|\;\Ad(g)\gamma\in\Lie P\text{ is such that its image under}\Lie P\twoheadrightarrow\Lie L\text{ lies in }\mathcal{O}\}.
$$
It is the usual Grothendieck–Springer alternation if $X$ is regular semisimple, hence the name. There is a natural smooth map $\mu:\tilde{\mathfrak{g}}_X\rightarrow Z(\Ll)$ via $\mathcal{O}\twoheadrightarrow Z(\Ll)$ and another natural $G$-equivariant map $\pi:\tilde{\Lg}_X\rightarrow\Lg$ sending $(g,\gamma)\mapsto\gamma$. For $\zeta\in Z(\Ll)$ that are $(G/L)$-regular (this means $Z_G(\zeta)=L$), we have $\pi(\mu^{-1}(\zeta))=\Ad(G)(\zeta+X_n)$. For the central fiber we have instead $\pi(\mu^{-1}(0))\supset\Ad(G)e$.
Since $e+Z_{\Lg}(f)$ meets $\Ad(G)e$ transversally, there exists a neighborhood (analytically or algebraically, whichever) $U$ of $e$ in $\Lg$ such that for any orbit $\Ad(G)Y$ we have $\Ad(G)Y\cap U\not=\emptyset\implies\Ad(G)Y\cap (e+Z_{\Lg}(f))\ne\emptyset$. The preimage $\pi^{-1}(U)$ intersects $\mu^{-1}(0)$. Thanks to smoothness of $\mu$, there exists a neighborhood $U_{Z(\Ll)}$ of $0\in Z(\Ll)$ such that $\pi^{-1}(U)$ intersects $\mu^{-1}(\zeta)$ for any $\zeta\in U_{Z(\Ll)}$. For $\zeta\in U_{Z(\Ll)}\cap Z(\Ll)^{\text{$(G/L)-reg$}}$ this means $U$ intersects $\Ad(G)(\zeta+X_n)$, and thus the asserted property (1) is true for ($X$ replaced by) $\zeta+X_n$.
Since nilpotent orbits are stable under scalar scaling, that (1) is true for $\zeta+X_n$ implies that it is true for $c(\zeta+X_n)$ for all $c\in\mathbb{C}^{\times}$ and $\zeta\in U_{Z(\Ll)}\cap Z(\Ll)^{\text{$(G/L)-reg$}}$, and thus also for $c\zeta+X_n$ since $c(\zeta+X_n)$ and $c\zeta+X_n$ are also conjugate. Since such $c\zeta$ covers all of $Z(\Ll)^{\text{$(G/L)-reg$}}$, the assertion for $X=X_s+X_n$ is proved.
[LS79] Lusztig, G.; Spaltenstein, N., Induced unipotent classes, J. Lond. Math. Soc., II. Ser. 19, 41-52 (1979). ZBL0407.20035.
Sadly, I can't recall why I needed this cute property ….
p.s. I now remember my motivation: I wanted to study the affine Springer fiber over $X$, and wanted a definition of a "regular locus" (and have it non-empty!) like in the situation when $X$ is regular semisimple. As I commented under the question, this answer will lead to a definition at least when $X_s$ is split. Analogous construction might or might not work in the non-split case ...
Emile Okada had kindly taught me that my question is indeed answered affirmatively (that there exists a canonical choice of such $\mathfrak{sl}_2$-triple up to conjugacy, with some nice properties) by the well-established theory of $G$-sheets of Borho [Bor81]. See also [PS18] for positive characteristic situation.
[Bor81] Borho, Walter, Über Schichten halbeinfacher Lie-Algebren, Invent. Math. 65, 283-317 (1981). ZBL0484.17004.
[PS18] Premet, Alexander; Stewart, David I., Rigid orbits and sheets in reductive Lie algebras over fields of prime characteristic, J. Inst. Math. Jussieu 17, No. 3, 583-613 (2018). ZBL1429.17020.
In fact, with Dan Ciubotaru they use it to obtain some nice results (https://arxiv.org/abs/2307.06780v1) that is highly related to what I hoped for (regarding Shalika germs and their wave-front set). They work over essentially over a finite field (and obtain results regarding unramified elements). The general case including ramified elements still seem rather difficult to me, though ...
Also, something that Emile taught me but I did not understand four months ago: as a corollary of their Lemma 4.2, the nilpotent orbit (i.e. the $\mathfrak{sl}_2$-triple) is unique and is given by the Lusztig-Spaltenstein induction as in the other answer, in which I seemed to re-produce a tiny subset of the theory of $G$-sheets.
|
2025-03-21T14:48:31.576910
| 2020-07-22T05:20:12 |
366259
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Andrew",
"David Roberts",
"Robert Bryant",
"https://mathoverflow.net/users/13972",
"https://mathoverflow.net/users/14551",
"https://mathoverflow.net/users/147073",
"https://mathoverflow.net/users/4177",
"liding"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631338",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366259"
}
|
Stack Exchange
|
Is there a solution of a first order nonlinear PDE?
Let $a^{ij}, b_{i}, c$ and $f>0$ are smooth function. Suppose $\Lambda I\geq (a^{ij})\geq \lambda I$, where $I$ is identity matrix, $\lambda, \Lambda$ is positive constant. Is there solution of the following equation
$$ \sum_{i,j}^{n}a^{ij}u_{i}u_{j}+\sum_{i}b_{i}u_{i}+cu=f$$
on a domain $\Omega\subset\mathbb{R}^{n}$ or a closed $n-$dimension manifold, where $u_{i}=\frac{\partial u}{\partial x_{i}}$? If not, are there known condidtions ensuring that this equation admits a soulution or locally solution?
Is $u_i$ the partial derivative with respect to $x_i$, or something else?
Yes ,it is the partial derivative with respet to $x_{i}$. Thank you and D. Tampieri for pointing out it.
There's not always a solution. Just look at $(u')^2 = -1$ when $n=1$. Are you asking for sufficient conditions that a solution exist? Would existence of local solutions suffice for what you want? Perhaps you can make your question more precise.
I have a PDE similar with this problem on closed manifold. I just want to find a solution. Both the sufficient conditions and the local solution will help me.
Could you give me some references
@liding The eikonal equation $|\nabla u|=f$ is a special case. Lot is known about it.
This isn't a general first order scalar PDE, but of a particular sort: If $M$ is a Riemannian manifold with metric $a$, $b$ is a vector field on $M$, and $c$ and $f$ are functions on $M$, the equation becomes $$|\nabla u|_a^2 + 2b,\cdot\nabla u+c,u - f =0.$$ (In the flat case, the inequalities on $a$ ensure that the metric is complete.) Writing it as $$|\nabla u+b|_a^2 + c,u - (f+|b|^2_a) = 0,$$ makes it clearer when solutions exist in a neighborhood of a point. For example $c(x)\not=0$ or $f(x)+ |b(x)|^2_a>0$ would guarantee local solutions near $x$, by the method of characteristics.
Thank you very much!
|
2025-03-21T14:48:31.577061
| 2020-07-22T05:53:12 |
366260
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"https://mathoverflow.net/users/33128",
"joaopa"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631339",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366260"
}
|
Stack Exchange
|
Logarithmic Weil height
Let $a_0,\cdots,a_n$ be algebraic integers. Is $h(a_0,\cdots,a_n)\le\max_{0\le i\le n}\log(\max(1,|a_i|))$ where $h(a_0,\cdots,a_n)$ denotes the logarithmic Weil height?
Thanks in advance.
It depends on what you mean by $|\cdot|$, but probably no.
If by $|\cdot|$ you mean the absolute value on $\mathbb C$, and your algebraic integers are elements of $\mathbb C$, then the answer is no. The logarithmic Weil height of $1-\sqrt3\in\mathbb A^1(\overline{\mathbb Q})$ is $\frac12\log(2)$, which is strictly bigger than $\log(\max(1,|1-\sqrt3|))=0$. (To show that the logarithmic height is $\frac12\log(2)$, use the formula for the logarithmic height on $\mathbb A^1(\mathbb Z[\sqrt3])$ below.)
However, a similar-looking inequality is true if we take account of all possible archimedean norms on the algebraic integers. Here is a precise statement. Let $a_1,\dots,a_n$ be algebraic integers in $\mathbb C$. Then
$$
h(a_1,\dots,a_n)\leq\max_{1\leq i\leq n}\left(\max_{\sigma\in G_{\mathbb Q}}\log(|\sigma(a_i)|)\right)\,,
$$
where $h$ denotes the logarithmic Weil height on $\mathbb A^n(K)$ and the second $\max$ is taken over all field automorphisms of $\overline{\mathbb Q}$.
This follows relatively straightforwardly from the definition of the Weil height. To cut a long story short, if $K$ is a number field of degree $d$ over $\mathbb Q$, then the Weil height on $\mathbb A^n(\mathcal O_K)$ is given by
$$
h(a_1,\ldots,a_n)=\frac1d\sum_{\sigma\colon K\hookrightarrow\mathbb C}\log(\max_{1\leq i\leq n}(|\sigma(a_i)|)) \,,
$$
where the sum is taken over all complex embeddings $\sigma\colon K\hookrightarrow\mathbb C$. This is just what you get by specialising the usual formula for the Weil height on $\mathbb P^n(K)$. It follows from this that
$$
h(a_1,\dots,a_n) \leq \max_{\sigma\colon K\hookrightarrow\mathbb C}\max_{1\leq i\leq n}\log|\sigma(a_i)| \,,
$$
which implies the claimed inequality.
Remark: This all depends, of course, on the chosen normalisation of the Weil height. Since the OP was talking about "algebraic integers", I presumed they were considering Weil heights with the normalisation which leads to a Weil height on $\mathbb P^n(\overline{\mathbb Q})$.
Wahoo!! Thanks for your answer. And you are right. I took the Weil height on $\mathbb P^n(\overline{\mathbb Q})$.
|
2025-03-21T14:48:31.577215
| 2020-07-22T06:21:44 |
366262
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Alexandre Eremenko",
"Emil Jeřábek",
"Gerry Myerson",
"https://mathoverflow.net/users/12705",
"https://mathoverflow.net/users/158000",
"https://mathoverflow.net/users/25510"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631340",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366262"
}
|
Stack Exchange
|
Solutions of equation $\sin \pi x_1\sin \pi x_2=\sin \pi x_3\sin \pi x_4$
I am interested in finding all the solution $(x_1,x_2,x_3,x_4)\in \mathbb{R}^4$ of equations:
$$\sin \pi x_1\sin\pi x_2=\sin \pi x_3\sin\pi x_4.$$
I have found out a paper: Rational products of sines of rational angles --- GERALD MYERSON, which answers my question partially.
Here is my interpretation:
Suppose $\alpha,\beta,\gamma,\delta\in \left(0,\dfrac{\pi}{2}\right]\cap \mathbb{Q}\pi$ and $\sin \alpha\sin \beta=\sin\gamma\sin\delta$.
Then $(\alpha,\beta,\gamma,\delta)=$
$(x,y,x,y)$
or $(x,y,y,x)$
or $\left(\dfrac{\pi}{6},\phi,\dfrac{\phi}{2},\dfrac{\pi}{2}-\dfrac{\phi}{2}\right)$
or $\left(\dfrac{\pi}{6},\phi,\dfrac{\pi}{2}-\dfrac{\phi}{2},\dfrac{\phi}{2}\right)$
or $\left(\phi,\dfrac{\pi}{6},\dfrac{\phi}{2},\dfrac{\pi}{2}-\dfrac{\phi}{2}\right)$
or $\left(\phi,\dfrac{\pi}{6},\dfrac{\pi}{2}-\dfrac{\phi}{2},\dfrac{\phi}{2}\right)$
or $\left(\dfrac{\pi}{2}-\dfrac{\phi}{2},\dfrac{\phi}{2},\phi,\dfrac{\pi}{6}\right)$
or $\left(\dfrac{\phi}{2},\dfrac{\pi}{2}-\dfrac{\phi}{2},\phi,\dfrac{\pi}{6}\right)$
or $\left(\dfrac{\pi}{2}-\dfrac{\phi}{2},\dfrac{\phi}{2},\dfrac{\pi}{6},\phi\right)$
or $\left(\dfrac{\phi}{2},\dfrac{\pi}{2}-\dfrac{\phi}{2},\dfrac{\pi}{6},\phi\right)$
or $\left(x_1^0,x_2^0,x_3^0,x_4^0\right)$
or $\left(x_1^0,x_2^0,x_4^0,x_3^0\right)$
or $\left(x_2^0,x_1^0,x_3^0,x_4^0\right)$
or $\left(x_2^0,x_1^0,x_4^0,x_3^0\right)$
or $\left(x_4^0,x_3^0,x_2^0,x_1^0\right)$
or $\left(x_3^0,x_4^0,x_2^0,x_1^0\right)$
or $\left(x_4^0,x_3^0,x_1^0,x_2^0\right)$
or $\left(x_3^0,x_4^0,x_1^0,x_2^0\right)$,
where $x,y,\phi\in \left(0,\dfrac{\pi}{2}\right]$ and $(x_1^0,x_2^0,x_3^0,x_4^0)\in
\left\{
\left(
\dfrac{\pi}{21}, \dfrac{8\pi}{21}, \dfrac{\pi}{14}, \dfrac{3\pi}{14}
\right),
\left(
\dfrac{\pi}{14}, \dfrac{5\pi}{14}, \dfrac{2\pi}{21}, \dfrac{5\pi}{21}
\right),
\left(
\dfrac{4\pi}{21}, \dfrac{10\pi}{21}, \dfrac{3\pi}{14}, \dfrac{5\pi}{14}
\right),
\left(
\dfrac{\pi}{20}, \dfrac{9\pi}{20}, \dfrac{\pi}{15}, \dfrac{4\pi}{15}
\right),
\left(
\dfrac{2\pi}{15}, \dfrac{7\pi}{15}, \dfrac{3\pi}{20}, \dfrac{7\pi}{20}
\right),
\left(
\dfrac{\pi}{30}, \dfrac{3\pi}{10}, \dfrac{\pi}{15}, \dfrac{2\pi}{15}
\right),
\left(
\dfrac{\pi}{15}, \dfrac{7\pi}{15}, \dfrac{\pi}{10}, \dfrac{7\pi}{30}
\right),
\left(
\dfrac{\pi}{10}, \dfrac{13\pi}{30}, \dfrac{2\pi}{15}, \dfrac{4\pi}{15}
\right),
\left(
\dfrac{4\pi}{15}, \dfrac{7\pi}{15}, \dfrac{3\pi}{10}, \dfrac{11\pi}{30}
\right),
\left(
\dfrac{\pi}{30}, \dfrac{11\pi}{30}, \dfrac{\pi}{10}, \dfrac{\pi}{10}
\right),
\left(
\dfrac{7\pi}{30}, \dfrac{13\pi}{30}, \dfrac{3\pi}{10}, \dfrac{3\pi}{10}
\right),
\left(
\dfrac{\pi}{15}, \dfrac{4\pi}{15}, \dfrac{\pi}{10}, \dfrac{\pi}{6}
\right),
\left(
\dfrac{2\pi}{15}, \dfrac{8\pi}{15}, \dfrac{\pi}{6}, \dfrac{3\pi}{10}
\right),
\left(
\dfrac{\pi}{12}, \dfrac{5\pi}{12}, \dfrac{\pi}{10}, \dfrac{3\pi}{10}
\right),
\left(
\dfrac{\pi}{10}, \dfrac{3\pi}{10}, \dfrac{\pi}{6}, \dfrac{\pi}{6}
\right)
\right\}.$
Is my interpretation correct? Do I miss any possibilities?
What is about the non-rational solutions? Is there any way to solve the equation over $\mathbb{R}$?
The question does not make much sense over $\mathbb R$, as for most choices of $x_1,x_2,x_3$, there is a solution $(x_1,x_2,x_3,\pi^{-1}\arcsin(\sin(\pi x_1)\sin(\pi x_2)/\sin(\pi x_3)))$. There is nothing to classify here.
What do you mean by "solving over R"? This equation defines a hypersurface in $R^4$ which evidently contains infinitely many points. What does it mean to solve? Parametrize this hypersurface?
I'm glad you found my paper.
|
2025-03-21T14:48:31.577509
| 2020-07-22T08:20:15 |
366267
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Dave L Renfro",
"Mary Sp.",
"Todd Eisworth",
"Wojowu",
"https://mathoverflow.net/users/15780",
"https://mathoverflow.net/users/161374",
"https://mathoverflow.net/users/18128",
"https://mathoverflow.net/users/30186"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631341",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366267"
}
|
Stack Exchange
|
Reference for graduate-level text or monograph with focus on "the continuum"
I always had the dream to design a course for my graduate students like "mathematical models of the continuum". This course should cover history of real numbers, the Measure Problem, the Banach-Tarski Paradox, but also Baire/Cantor/Polish spaces, and it should look at infinite games and determinacy as well.
Obviously I am conscious about the text books and articles covering all these concepts. However, I seem to be unable to find a rigorours, graduate-level monograph that captures all of this in one, which I could potentially use as a text.
Would someone be able to point me to a good reference for that?
You might be interested in the topic of descriptive set theory, which treats many of the topics you seem to have in mind.
I like your idea of such a course a lot! If it is appropriate to recommend a book in German language, I think this one could be the perfect match:
Oliver Deiser (2007): Reelle Zahlen: Das klassische Kontinuum und die natürlichen Folgen
I own this book and can say it covers all the topics that you mentioned, and it is certainly a graduate-level text. I find the quality of the exposition outstanding, and the range of topics quite unique. It covers the historical development quite extensively and gives many references, with a focus on the original sources.
I think there is also a 2nd edition from 2008.
A great textbook for your course would be "The Structure of the Real Line" by Lev Bukovský. It covers all of the topics you mentioned, except for the Banach-Tarski Paradox, and provides all necessary topological and measure-theoretic background.
I was just going to suggest this one, Santi.
I finally got around to getting a copy of this book (arrived a few days ago). This is probably the most useful book (in English, known to me at this time) to have on hand for what the OP has in mind, although it may be a bit too encyclopedic to be useful as an actual text, at least as the primary text. If someone wanted a short list of books in English to pursue after Dasgupta's book, I would definitely include Bukovský's and Krechis's books in my list.
For english references:
An history of mathematics book, but extremely well written and mathematically sophisticated, with tons of references (that might be useuful) that adress all such things is
G. H. Moore's Zermelo's Axiom of Choice: Its Origins, Development, and Influence.
This should be my top pic. Other sources I know of:
D. L. Cohn's Measure Theory.
has a very nice introduction to Polish spaces and analytic sets, and
Folland's Real Analysis: Modern Techniques and Their Applications.
discusses the measure problem and Banach-Tarksi's Theorem, and has plenty references.
However, the level is more of upper-undergraduate then graduate, I think.
For descriptive set theory we have Krechis Classical descriptive set theory and Moschovakis Descriptive Set theory, but I guess that, by what you've said you know their content already.
Maybe Set Theory. With an Introduction to Real Point Sets by Abhijit Dasgupta. See the chapter and section titles at amazon.com, especially the section titles, which indicate much better than just the chapter titles in showing the topics included. However, the book is probably better described as an "advanced undergraduate / beginning graduate" level text (U.S. standards) than a true graduate-level monograph.
I can recommend two books to you which I think give a rather good coverage of the foundations for determinacy and infinite games
Hugh Woodin: The Axiom of Determinacy, Forcing Axioms, and the Nonstationary Ideal. Walter de Gruyter, Berlin 1999.
Gerald Sacks (ed.): Mathematical Logic in the 20th Century, SUP and World Scientific Publishing Singapore and London, 2003 (note especially the article from Donald Martin on Analytic Games) https://www.worldscientific.com/doi/pdf/10.1142/9789812564894_fmatter
I know Woodin's monograph and can say it is excellent. I would say clearly upper graduate / research level. Maybe an important consideration for OP, one of the main results is Woodin gives a canonical model in which the Continuum Hypothesis is FALSE.
|
2025-03-21T14:48:31.577803
| 2020-07-22T09:19:18 |
366273
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Alkan",
"Gerry Myerson",
"Robert Israel",
"https://mathoverflow.net/users/13650",
"https://mathoverflow.net/users/158000",
"https://mathoverflow.net/users/161528"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631342",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366273"
}
|
Stack Exchange
|
Numbers of the form $x^2(x-1) + y^2(y-1) + z^2(z-1)$ with $x,y,z\in\mathbb Z$
Roger Heath-Brown conjectured that any integer $k\not\equiv\pm4\pmod9$ can be expressed as $x^3+y^3+z^3$ with $x,y,z\in\mathbb Z$ in infinitely many different ways. Also it is well-known that some solutions are extremely difficult to construct and currently there are interesting and notable discoveries in that direction such as representations of 33 and 42.
I am not sure that my question is relatively easy. It is related to https://oeis.org/A336205. I conjecture that any even integer can be represented as $x^2(x-1) + y^2(y-1) + z^2(z-1)$ with $x, y, z\in\mathbb Z$. I also conjecture that this is possible in infinitely many different ways for any even integer. In particular, there can be some related results on that direction since I easily formulated below parametric one based on comment of Robert Israel at https://oeis.org/A336240. $$x^2(x-1) + y^2(y-1) + z^2(z-1) = 2a^2(a-1)$$ with $x = 1 - (6a-2)m^2$, $y = a - m(1-(6a-2)m^2)$, $z = a + m(1-(6a-2)m^2)$. I don't know that there are some references on that equation or further similar results. If someone can see additional nontrivial ones, comments on parametric solutions are also very welcome.
It is obvious that first conjecture can be rewritten based on definition of generalized pentagonal pyramidal numbers, i.e., any integer can be expressed as $xT_x + yT_y + zT_z$ with $x,y,z\in\mathbb Z$ where $T_n$ is nth triangular number.
It is relatively easy to find solutions to this spesific representation, at least for any even integer $t$ with $-1000 \leq t \leq 1000$. Is it well-known fact or conjecture ? If this is not known, can someone prove this conjecture ? (Also more experimental evidences would be very welcome if proof is probably very hard.)
Finally, I would like to share this note kindly. Although I am not sure that it is sufficiently interesting or meaningful, I also believe that it is probable to construct further questions and related generalized conjectures to some sort of similar representations such as $f(x)T_x + f(y)T_y + f(z)T_z$ or $g(x)x^2 + g(y)y^2 + g(z)z^2$ based on several natural selections of $f$ and $g$. For example, $g(w)=w-2$ could be fruitful or investigation of sum of three generalized $r$-gonal pyramidal numbers can provide interesting experimental results for certain values of $r$ : https://math.stackexchange.com/questions/2472205/is-every-number-a-sum-of-3-tetrahedral-numbers and Does the set $\{\binom x3+\binom y3+\binom z3:\ x,y,z\in\mathbb Z\}$ contain all integers? ($r = 3$).
How far up have you verified the conjecture?
If my program is correct, I found representations for even integers from $-164$ to $324$.
For 326; x = -1752, y = 698, z = 1715 for example. Best regards.
|
2025-03-21T14:48:31.578003
| 2020-07-22T09:48:06 |
366275
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Carlo Beenakker",
"https://mathoverflow.net/users/11260",
"https://mathoverflow.net/users/37545",
"user1611107"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631343",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366275"
}
|
Stack Exchange
|
How to solve a differential equation in the form $\frac{\partial}{\partial t}g(x,t)=g(x-\Delta,t)+\frac{\partial^2}{\partial x^2} g(x,t)$?
How to find the general solution of a differential equation with a shift, in the following form?
$$\frac{\partial}{\partial t}g(x,t)=g(x-\Delta,t)+\frac{\partial^2}{\partial x^2} g(x,t)$$
where $\Delta > 0$. And what about the following?
$$\frac{\partial}{\partial t}g(x,t)=g(x,t-\Delta)+\frac{\partial^2}{\partial x^2} g(x,t)$$
Edit1: Here are few follow-up details about my question. Is there a ``nice" way to represent the solution in $x$-space, as opposed to e.g., Fourier? Is the solution real + positive + normalizable? Does it have the correct properties of a probability density function?
as you can see from the general solution I wrote down, the normalization $N(t)=\int_{-\infty}^\infty g(x,t)dt=G(0,t)=e^{t}G(0,0)=e^{t}N(0)$ increases exponentially with time; so this is not a probability density function (why did you expect that?)
Fourier transform $G(k,t)=\int_{-\infty}^\infty e^{ikx} g(x,t)dx$ with respect to $x$, then
$$\frac{\partial}{\partial t}G(k,t)=e^{ik\Delta}G(k,t)-k^2 G(k,t),$$
hence
$$G(k,t)=\exp\left(te^{ik\Delta}-tk^2\right)G(k,0).$$
For the second differential equation you would similarly Fourier transform with respect to $t$.
Thanks. Can you please continue the solution for few more steps? Is it real, or have an imaginary part? How do you than transform back from k to x-space?
to transform back to $x$-space you calculate $g(x,t)=(2\pi)^{-1}\int_{-\infty}^\infty e^{-ikx}G(k,t)dk$. This is the general solution, it will be real if $g(x,0)$ is real; it cannot be worked out further without further information on $g(x,0)$.
Ok, thank you. About the normalization: I was thinking that it is simpply a diffusion equation, with a drift term (even if the drift dependson x-\Delta). But you are right, it's a source. My mistake.
The thing is, the idea of Fourier I am familiar with, but the backward transformation of the exp[t exp(i k Delta)] is something that I can't work out. Is there a simple form of g(x,t) that you can demonstrate the inverse transform on? E.g., if g(x,t)=delta(x) or a narrow Gaussian? Thanks again
the inverse Fourier transform does not have a closed form expression for the delta function initial condition; you would need to calculate integrals of the form $\int_0^{\infty } e^{\cos k-k^2} \cos( k x) \cos (\sin k) , dk$, which can only be done numerically.
|
2025-03-21T14:48:31.578174
| 2020-07-22T09:51:55 |
366276
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Alapan Das",
"Alec Rhea",
"Andrea T",
"Iosif Pinelis",
"Sylvain JULIEN",
"https://mathoverflow.net/users/115044",
"https://mathoverflow.net/users/13625",
"https://mathoverflow.net/users/156029",
"https://mathoverflow.net/users/161542",
"https://mathoverflow.net/users/36721",
"https://mathoverflow.net/users/92164",
"pregunton"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631344",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366276"
}
|
Stack Exchange
|
Multidimensional series: an application of quantum field theory
While computing the quantum vacuum energy of a real scalar field defined on $\mathbb{R}\times \mathbb{T}^3$, I encountered the following sum:
$$ \sum_{n_1^2+n_2^2+n_3^2\geq 1}^{\infty} \frac{1}{(n_1^2+n_2^2+n_3^2)^2} $$
Does it have a known exact result?
Thank you.
EDIT: What is the value of Epstein $\zeta_3(2)$? Where the Epstein zeta function is defined as
$$\zeta_k(s)=\sum_{\mathbb{Z}^k/\{0\}} \frac{1}{(n_1^2+....n_k^2)^s}$$
Maybe you can get relevant information typing "multiple zeta values" in a search engine.
According to Legendre's theorem, all numbers not of the form $4^a(8k-1), k=1,2,...$ can be written as sum of three squares. So, this is $$\zeta(2)-\sum_{a=0}^{\infty} \sum_{k=1}^{\infty} \frac{1}{16^a(8k-1)^2}=\zeta(2)-\frac{16}{15}\left(\sum_{k=1}^{\infty} \frac{1}{(8k-1)^2}\right)$$.
@AlapanDas : I don't think this will work, because the representation as the sum of three squares is not unique in general, even if $n_1,n_2,n_3$ are assumed to be natural numbers -- see e.g. https://math.stackexchange.com/questions/2068160/determining-the-number-of-ways-a-number-can-be-written-as-sum-of-three-squares
@Losif Pinelis Yes. As nothing was mentioned clearly about $n_1,n_2,n_3$ I just assumed it as the sum of all squares of inverses of all those number which are sum of three squares.
What do you mean by $\sum\limits_{n_1^2+n_2^2+n_3^2\ge1}^\infty$?
@IosifPinelis I think they mean $\sum^\infty_{(n_1,n_2,n_3)\in{(n_1,n_2,n_3)\in\mathbb{R}^3:n_1^2+n_2^2+n_3^2\geq1}}$.
I mean that $n_{1,2,3}$ can take any integer value, except clearly $n_1=n_2=n_3=0$.
Possibly related question: https://mathoverflow.net/questions/228760/is-special-value-of-epstein-zeta-function-in-3-variables-a-period
|
2025-03-21T14:48:31.578312
| 2020-07-22T09:55:27 |
366278
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Nathanael Skrepek",
"Stefan Waldmann",
"https://mathoverflow.net/users/114751",
"https://mathoverflow.net/users/12482"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631345",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366278"
}
|
Stack Exchange
|
Uniform distance from a discontinuous function is continuous
Define the metric $d(f,g)\triangleq \sup_{x \in [0,1]} \|f(x)-g(x)\|$ on the set $\operatorname{B}$ of uniformly bounded functions from the interval $[0,1]$ to $\mathbb{R}$, fix $g \in \operatorname{B}$, and define the map $F:\operatorname{B}\rightarrow [0,\infty)$ by $F(f):=d(g,f)$. Is the map $F$ continuous? It certainly is on the subset $C([0,1],\mathbb{R})$ but what about on the rest of the space?
If $d$ is a metric on $B$ then the mapping $F(f) := d(g,f)$ is certainly continuous with respect to the topology induced by the metric $d$:
Let $(f_n)_{n\in\mathbb{N}}$ a sequence in $B$ that converges to $f \in B$ w.r.t. $d$. This is equivalent to
$$
d(f,f_n) \to 0.
$$
Hence, by the triangular inequality
$$
F(f_n) = d(g,f_n) \leq d(g,f) + d(f,f_n) \to d(g,f) + 0 = F(f).
$$
On the other hand by the reversed triangular inequality
$$
F(f_n) = d(g,f_n) \geq d(g,f) - d(f,f_n) \to d(g,f) + 0 = F(f).
$$
This means
$$
F(f) \leq \lim_{n\in\mathbb{N}} F(f_n) \leq F(f)
$$
which implies the continuity of $F$.
In fact, the argument works for any metric space.
yes, i didn't specify the space and metric on purpose.
|
2025-03-21T14:48:31.578442
| 2020-07-22T10:46:24 |
366280
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Abdelmalek Abdesselam",
"David Roberts",
"Jochen Wengenroth",
"Peter",
"Yemon Choi",
"https://mathoverflow.net/users/15629",
"https://mathoverflow.net/users/21051",
"https://mathoverflow.net/users/4177",
"https://mathoverflow.net/users/7410",
"https://mathoverflow.net/users/763",
"https://mathoverflow.net/users/99745",
"paul garrett"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631346",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366280"
}
|
Stack Exchange
|
What is known about the "unitary group" of a rigged Hilbert space?
Suppose that $(E,H)$ is a rigged (infinite dimensional, separable) Hilbert space, i.e. $H$ is a Hilbert space, and $E$ is a Fréchet space, equipped with a continuous linear injection $E \rightarrow H$ with dense image.
Define now the group $U(E,H)$ to consist of those invertible, continuous, linear transformations on $E$, which extend to unitary transformations of $H$.
What is known about the group $U(E,H)$?
For example:
Does the map $U(E,H) \rightarrow U(H)$ have dense image?
Is it a Fréchet Lie group? If so:
What is its Lie algebra? Is it just the continuous linear skew-adjoint operators on $E$?
Is the map $U(E,H) \times E \rightarrow E$ smooth?
Do the smooth vectors of the representation $U(E,H) \times H \rightarrow H$ consist exactly of $E$?
Is $U(E,H)$ contractible?
I am most interested in the case that $E$ is a nuclear Fréchet space.
Perhaps a good first step towards a Fréchet Lie group structure on $U(E,H)$ would be to equip the space of skew operators on $E$ with the structure of Fréchet (Lie) algebra.
It's not so obvious to me how this should go. Is there anything known in this direction?
Presumably you mean for H to be infinite-dimensional... An interesting question!
I am not very familiar with definition of rigged Hilbert space, so does it allow $E$ to be a Banach space, such as $\ell_1$ sitting inside $\ell_2$?
Perhaps @Peter Michor can help?
@YemonChoi Yes, in general this would be allowed. However, the only Banach spaces that are nuclear are finite dimensional. So in this case that I'm mostly interested in, $E$ will not be Banach. A good example of what I'm thinking about is $C^{\infty}(S^{1})$ sitting inside $L^{2}(S^{1})$.
@YemonChoi, and, adding a few details to Peter's comment, because of a Sobolev imbedding theorem + Rellich lemma, the space of smooth functions on the circle is also the projective limit of Sobolev spaces $H^s$, and the transition maps are Hilbert-Schmidt, so even by the strictest definition the smooth functions on the circle are nuclear Frechet. I think this is the simplest example...
Thanks - the point of my comment is that I suspect I could cook up a counterexample to the first question if I was allowed to take $\ell_1$ inside $\ell_2$. BTW, when you say "dense image" inside $U(H)$, which topology are you taking on $U(H)$? The weak operator topology, or the norm topology?
@YemonChoi I would be most interested in the norm topology, but really either result would be interesting. Though maybe I buried the lead somewhat, because I really am most interested in equipping $U(E,H)$ with a topology and Lie group structure.
The simplest interesting example is $E=s(\mathbb{N})$ the space of rapidly decaying sequences sitting inside $\ell^2(\mathbb{N})$. I would have to double check but linear continuous maps $E\rightarrow E$ are represented by infinite matrices $m_{i,j}$ with $|m_{i,j}|\le a_i b_j$ where $a_i$ is of rapid decay and $b_j$ is of at most polynomial growth. This is a very concrete starting point from which I think a complete understanding of $U(E,H)$ can be derived.
I spoke too fast in my previous comment. The condition on the matrix $m_{i,j}$ is not that $\exists a_i,b_j$, with $a_i$ of rapid decay and $b_j$ of polynomial growth such that $|m_{i,j}|\le a_i b_j$ but rather $\forall c,d$ with $c_i$ of polynomial growth and $d_j$ of rapid decay $\sum_{i,j}c_id_j|m_{i,j}|<\infty$. One can also use $\sup_{i,j}$ instead of the sum. I counterexample to my previous erroneous statement is $m_{i,j}=\delta_{ij}$ for the identity operator.
Have you found out anything more about the contractibility since this question? Eg of the group $GL_\infty(E)$ in your answer, or an analogous $U_\infty(E)$?
I've found a construction of a Fréchet Lie group structure on a subgroup of $\mathrm{GL}(E,H)$ in the literature that works under the additional assumption that $E$ is countably normed. Here is an overview:
First, let me recall the definition of a Sobolev Chain from page 1 of Omori's book.
Definition: A Sobolev chain is a system $\{ E, E_{k}\}_{k \in \mathbb{N}}$ where each $E_{k}$ is a Banach space, and for each $k \in \mathbb{N}$, there is a linear and dense embedding $E_{k+1} \rightarrow E_{k}$. And $E$ is the intersection of $E_{k}$ equipped with the inverse limit topology.
Let $L(E)$ then be the collection of continuous linear operators on $E$.
Following Chapter IX in Omori we let $L_{0}(E)$ consist of those $T \in L(E)$ that extend to bounded linear operators $E_{k} \rightarrow E_{k}$ for all $k$.
We let $L_{\infty}(E)$ consist of those $T \in L_{0}(E)$ such that there exist positive constants $C$ and $D_{k}$ such that
\begin{equation}
\| Tu \|_{k} \leqslant C\|u\|_{k} + D_{k}\|u\|_{k-1}.
\end{equation}
We then define $|T|$ to be the infimum over all such $C$.
The strong PLN-topology is then defined by the semi-norm $| \cdot |$ and the projective limit topology of the Banach spaces of bounded operators on $E_{k}$ equipped with the norm topologies.
We then define $\mathrm{GL}_{\infty}(E)$ to consist of those $T \in L_{\infty}(E)$ such that $T^{-1}$ exists and is an element of $L_{\infty}(E)$ itself.
Omori then proves that $\mathrm{GL}_{\infty}(E)$ is open in $L_{\infty}(E)$, and is, moreover, a strong ILB-Lie group.
In section 6 of this paper, it is proved that each strong ILB-Lie group is a regular Fréchet Lie group.
Now, to go back to the setting I was originally interested in, if $(E,H)$ is a rigged Hilbert space, then it admits a fundamental system of seminorms $\| \cdot \|_{1} \leqslant \| \cdot \|_{2} \leqslant \| \cdot \|_{3} \leqslant ...$
For each $k$ we let $E_{k}$ be the completion of $E$ with respect to $\| \cdot \|_{k}$.
The assumption that $E$ is countably normed then means that all the linking maps $E_{k+1} \rightarrow E_{k}$ are linear and dense embeddings, and thus that $\{E, E_{k}\}_{k \in \mathbb{N}}$ is a Sobolev chain.
We thus obtain a regular Fréchet Lie group $\mathrm{GL}_{\infty}(E)$.
This is (to some extent) an affirmative answer to the second question.
What remains are its subquestions.
Furthermore, we have not used nuclearity of $E$, I would be interested to see if this allows us to simplify anything or say anything more.
Passing to the completion may destroy injectivity. Frechet spaces weich are inverse limits of Banach spaces with injective connecting maps are called countably normed.
@JochenWengenroth Thanks for pointing that out! I changed my answer accordingly.
These are just some thoughts on you question, so not an answer but too long for a comment. I waited a few days before posting this since, rather than addressing your post directly, I am going to look at a variant. This is usually not the done thing but my excuse is that I hope that it will contain material which could be of interest to you.
First of all, as has already been pointed out, your questions only make sense if you specify the relevant topologies. Experience suggests that the norm is not suitable for such questions. However, there are available a range of well-behaved and well-studied weaker ones which might do very well.
My main thesis is that most of the rigged Hilbert spaces which arise in mathematical physics have a special form and it is worth while starting there. The setting is an unbounded self adjoint operator $T$ on a Hilbert space. $E$ is then the intersection of the domains of definition of its powers. This is a Fréchet space with its natural structure, even Fréchet nuclear (Pietsch) under conditions on the spectral behaviour of $T$ which are satisfied by many of the classical operators (Laplace, Laplace-Beltrami, Schrödinger).
My main suggestion is that one study the space of unitary operators on the Hilbert space which commute with $T$. Such operators then map $E$ continuously into itself.
This space can be explicitly calculated for most of the situations mentioned above and this may provide some insight into your original question.
As an example, for the case of the standard one-dimensional Schrödinger operator the Fréchet space is that of the rapidly decreasing smooth functions and the corresponding operator space is the countable product of circle groups, regarded as multipliers on the coefficients of the Hermite expansion. As a compact group it won’t be dense anywhere in a non-trivial way but it will, presumably, be an infinite dimensional Lie group.
Similar remarks apply to the other classical cases where the spectral behaviour of the operator is known.
This sounds like an interesting approach. Do you have any references for this? Maybe in particular explaining the details of your example of the one-dimensional Schrödinger operator? Do you think you could relax the condition that the operators need to commute with T? Perhaps we could consider the unitary operators $U$ such that $[T,U]$ is a compact operator for example?
|
2025-03-21T14:48:31.579008
| 2020-07-22T11:09:09 |
366281
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Scott Armstrong",
"https://mathoverflow.net/users/5678",
"https://mathoverflow.net/users/69642",
"user69642"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631347",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366281"
}
|
Stack Exchange
|
Error rate implying regularity
My question is a bit general/vague.
It is well known that the regularity of certain functions can be measured through the rate of decay of certain error quantity based on an approximation procedure (see, e.g., the famous result of DeVore, Jawerth and Popov for Besov regularity in terms of the rate of decay for the error of the best n-term approximation).
I would like to know if there is a standard well-developed theory which allows to obtain regularity results thanks to rates of decay of approximation.
Are there any general references in this direction ?
Many thanks in advance.
I am not sure what you mean exactly, can you elaborate? For instance, many regularity spaces are defined in these terms, e.g., Holder (C^\alpha) functions are simply those that can be approximated in L^\infty by constant functions in every ball of radius r up to an error of O(r^\alpha). If you relax L^\infty to L^2 (in an appropriate sense) you get the space H^\alpha. The space C^{k,\alpha} is the space of functions which can, in every ball, be approximated in L^\infty by a polynomial of degree k up to an error of O(r^{k+\alpha}).
Since these are the very definitions of the regularity spaces, many arguments in elliptic regularity work like this. Google "Campanato iterations" or "epsilon regularity lemma" for example. Read any paper with "partial regularity" in the title. Read basically any paper of Caffarelli. Read the recent papers on the regularity of the Monge-Ampere equation/ optimal transportation by De Philippis & Figalli or Otto & collaborators.
If you want to see the Schauder estimates proved like this, see the book of Han-Lin. If you want to see the Calderon-Zygmund estimates proved like this, see Lemma 7.2 of my book with Kuusi-Mourrat. This is all very elliptic/parabolic, but perhaps only because it is what I know.
Thanks for the comments! Indeed, several functional spaces are termed directly in this way but what I am looking for is more in the spirit of approximation spaces and more specifically in the systematic identification/characterization of such spaces as known functional spaces. The way I am approximating my function is not necessarily classical (not a Taylor expansion) and therefore it might be not completely clear what is the underlying functional space associated with such an error rate.
|
2025-03-21T14:48:31.579207
| 2020-07-22T11:13:44 |
366282
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631348",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366282"
}
|
Stack Exchange
|
Pairing up vertices in a graph
Given a connected (undirected) graph with an even number of vertices, consider how many ways are there to pair up vertices so that each pair is connected by an edge. Is there a known classification of all such graphs into those with $0$ ways, $1$ way, and more than $1$ way?
For example, a star belongs to the first category, a path to the second, and a cycle to the third (with $2$ ways).
In general, finding the number of perfect matchings is #P-complete. But using the Edmonds "blossom" algorithm, one can decide effectively if the number is zero or not. And it's also easy using the same algorithm to figure out if the number is exactly 1 (in principle we can try removing the edges one by one and check in each case whether there is still a perfect matching). So if we just want to classify into 0, 1 or many, it's doable in polynomial time.
|
2025-03-21T14:48:31.579295
| 2020-07-22T11:49:32 |
366285
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631349",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366285"
}
|
Stack Exchange
|
Minimizing an f-divergence and Jeffrey's Rule
My question is about f-divergences and Richard Jeffrey's (1965) rule for updating probabilities in the light of partial information.
The set-up:
Let $p: \mathcal{F} \rightarrow [0,1]$ be a probability function on a finite algebra of propositions.
Suppose that the probability of $E$ in $\mathcal{F}$ shifts from its prior value $p(E)$ to its posterior value $p'(E) = k$.
Jeffrey's Rule then says, for all $X$ in $\mathcal{F}$,
$p'(X) = \sum_{E}p(X|E)p'(E)$.
In other words, it offers a fairly straightforward generalization of Bayesian conditioning for partial information.
The concept of an f-divergence seems to be a fairly natural generalization of the Kullback-Leiber divergence. And minimizing the Kullback-Leibler divergence between prior and posterior probability functions is known to agree with Jeffrey's Rule (Williams, 1980).
Here is where I get stuck.
I have seen it written that "minimizing an arbitrary f-divergence subject to the constraint $p(E_{i}) = k$ is equivalent to updating by Jeffrey's Rule". However, I can only find proofs going in one direction, namely, all f-divergences agree with Jeffrey's Rule (e.g., Diaconis and Zabell, 1982, Theorem 6.1).
Q: Is it also true that only f-divergences agree with Jeffrey's Rule? Or might there be some non-f-divergence $\mathcal{D}$ such that minimizing it subject to the same constraint also agrees with Jeffrey's Rule?
Any pointers would be awesome.
Refs:
Jeffrey, R. (1965) The Logic of Decision
Williams, P. (1980) "Bayesian Conditionalisation and the Principle of Minimum Information"
https://philpapers.org/rec/WILBCA
Diaconis, P. and Zabell, S. (1982) "Updating Subjective Probability"
https://www.semanticscholar.org/paper/Updating-Subjective-Probability-Diaconis-Zabell/6079c34933e5b95558fa343d294c4b734b8682e5
|
2025-03-21T14:48:31.579538
| 2020-07-22T12:12:44 |
366286
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631350",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366286"
}
|
Stack Exchange
|
Rational functions with trivial Weil symbols at every point
Let $f, g$ be a pair of nonzero rational functions in $\mathbb{C}(t).$ For $\lambda\in \mathbb{C}$ let $a$ be multiplicity of $g(t)$ at $\lambda$ and $b$ - multiplicity of $f(t)$ at $\lambda.$ Weil symbol of $f$ and $g$ at $\lambda$ is defined by the following formula:
$$
(f,g)_{\lambda}=(-1)^{ab}\frac{f^a}{g^b}(\lambda).
$$
Question: For which pairs of rational functions $f, g$ Weil symbol $
(f,g)_{\lambda}$ equals to $\pm1$ at every point $\lambda\in \mathbb{C}?$
It is easy to find some pairs of such functions. For every $r(t)\in \mathbb{C}(t)$ we can take $f=r(t)^a(1-r(t))^b$ and $g=r(t)^c(1-r(t))^d.$
This problem can be generalized to arbitrary Riemann surfaces, but it is probably very hard, because nontrivial examples of such pairs of functions come from $A-$polynomials of some knots.
|
2025-03-21T14:48:31.579615
| 2020-07-22T13:19:55 |
366288
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Igor Khavkine",
"Loïc Teyssier",
"Simon Parker",
"https://mathoverflow.net/users/123207",
"https://mathoverflow.net/users/24309",
"https://mathoverflow.net/users/2622"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631351",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366288"
}
|
Stack Exchange
|
Regular singular point of non-linear ODE: $\dot{x}(t) + t^{-1}Ax(t) = Q(x(t))$
Consider a system of ordinary differential equations of the form
$$
\dot{x}(t) + \frac{1}{t}Ax(t) = Q(x(t))
$$
where $x(t) \in \mathbb{C}^n$, $A \in \mathrm{Mat}_{n\times n}(\mathbb{C})$ is a constant matrix, and $Q: \mathbb{C}^n \to \mathbb{C}^n$ is homogeneous of degree $2$, i.e. $Q(\lambda x) = \lambda^2 Q(x)$ for $\lambda \in \mathbb{C}$.
What is known about existence of solutions near $t = 0$?
If it were not for the quadratic term $Q$, the point $t = 0$ would be a regular singular point of the ODE and then we could use the Frobenius method. But in all the references I know, regular singular points are only discussed for linear systems.
mse: https://math.stackexchange.com/q/3763730/202132
By "solutions near $t=0$", do you actually mean solutions analytic/holomorphic at $t=0$ as well ? Do you have any assumptions on the eigenvalues of $A$ ?
@LoïcTeyssier I'm trying to gather as much information as possible about those systems. So if you can say anything by adding more assumptions (distinct eigenvalues, analyticity, etc), please let me know!
What I meant is: do you want the solution to extend in any way at $t=0$, or could it be meromorphic or even with an essential singularity?
@LoïcTeyssier Whatever helps you get an answer.
There's nothing inherently linear about constructing power series solutions à la Frobenius. The existence and uniqueness theory for a class of singular non-linear ODEs, of which yours is a special case, is treated for instance in Ch.IX of
Wasow, W., Asymptotic expansions for ordinary differential equations, (Dover, 1987) reprint from 1965. ZBL0169.10903
@DanielHatton This sounds like it should be its own question. Please formulate clearly your situation and post it as a question.
|
2025-03-21T14:48:31.579966
| 2020-07-22T14:02:52 |
366294
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Fedor Pakhomov",
"François G. Dorais",
"Sam Sanders",
"https://mathoverflow.net/users/2000",
"https://mathoverflow.net/users/33505",
"https://mathoverflow.net/users/36385"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631352",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366294"
}
|
Stack Exchange
|
What subsystem of third order arithmetic proves the real numbers are Dedekind complete?
Reverse mathematics is mainly about subsystems of second-order arithmetic, but in recent years it’s expanded to cover subsystems of third-order arithmetic as well. Now the fact that the real numbers are Dedekind complete is (encodable as) a statement in the language of third order arithmetic. And I think it’s probably provable using full third order arithmetic.
But my question is, what is the weakest subsystem of third-order arithmetic capable of proving that the real numbers are Dedekind complete?
By the way, the fact that the real numbers form a real closed field is provable even in $RCA_0$, so my question is really about the interpretability of the second-order theory of real numbers.
$\Pi^1_1$-CA (allowing third order parameters) is surely enough. Given a nonempty $X \subseteq \mathbb{R}$ which is bounded above, we can form the set $U = { q \in \mathbb{Q} : (\forall x \in \mathbb{R})(r \in X \to r \leq q)}$ of all rational upper bounds of $X$ using $\Pi^1_1$-comprehension with parameter $X$. Note that $U$ is really just a second-order object, so $\inf U$ exists (by arithmetic comprehension) and $\sup X = \inf U$. So there is no need for any third-order axioms per se.
It is fairly easy to show that over $\mathsf{ACA}0$ Dedekind completeness for $\Pi^0\infty$-sets of reals is equivalent to $\Pi^1_1\textsf{-CA}_0$. It is possible to derive $\Pi^1_1\textsf{-CA}_0$ from this form of Dedekind completeness using the fact that $\Pi^1_1\textsf{-CA}_0$ equivalent over $\mathsf{ACA}_0$ to the principle "every ill-founded tree $T\subseteq \omega^{<\omega}$ has a leftmost path." The latter equivalence could be proved using Kleene's normal forms for $\Pi^1_1$-sets.
It seems to me that in a similar way it is possible to show that over some some reasonable base system of higher-order reverse math (I guess something like $\mathsf{RCA}_0^{\omega}+\exists^2$ would work here) the third-order Dedekind completeness principle will be equivalent to the variant of $\Pi^1_1\textsf{-CA}_0$ that allows higher order parameters. Unfortunately, I am not comfortable enough with higher-order reverse math to write an answer.
@Pakhomov: one has to be very careful with the combination of $\exists^2$ and principles that allow for third-order parameters. See my "Plato and the foundations of mathematics" paper on arxiv.
The answer to your question (unsurprisingly) depends on the formalisation of "being a subset of $\mathbb{R}$". Alex Kreuzer [1] has used characteristic functions to represent subsets of Cantor space $2^\mathbb{N}$. Dag Normann and I have adopted this formalism in e.g. [2, 3] for subsets of $\mathbb{R}$, as it yields nice results that generalise the notion of open/closed set from Reverse Mathematics.
Using the "sets as characteristic functions" formalism, Kohlenbach'ssystem RCA$_0^\omega$ from [0] plus Every bounded subset of $\mathbb{R}$ has a surpremum
is a conservative extension of WKL$_0$. One uses the intuitionistic fan functional from [0] to establish this.
References
[0] Kohlenbach, U., Higher order reverse mathematics, Reverse mathematics 2001, Lect. Notes Log., vol. 21, ASL, 2005, pp. 281–295.
[1] Kreuzer, A., Measure theory and higher order arithmetic. Proc. Amer. Math. Soc. 143 (2015), no. 12, 5411–5425.
[2] Normann D. and Sanders S., Open sets in Reverse Mathematics and Computability Theory, Journal of Logic and Computability 30 (2020), no. 8, pp. 40.
[3]____, On the uncountability of R, Submitted, arxiv: https://arxiv.org/abs/2007.07560 (2020), pp. 29.
|
2025-03-21T14:48:31.580226
| 2020-07-22T17:07:50 |
366302
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Carlo Beenakker",
"Johannes Trost",
"Michael Renardy",
"Tanya Vladi",
"https://mathoverflow.net/users/11260",
"https://mathoverflow.net/users/12120",
"https://mathoverflow.net/users/158421",
"https://mathoverflow.net/users/37436",
"https://mathoverflow.net/users/90189",
"user90189"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631353",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366302"
}
|
Stack Exchange
|
Prove $\int_{0}^{\infty} \cos(\omega x) \exp(-x^{\alpha}) \, {\rm d} x \ge {\alpha^2 \sqrt{\pi} \over 8} \exp \left( -\frac{\omega^2}{4} \right)$
I would like to prove that
$$\int_{0}^{\infty} \cos(\omega x) \exp(-x^{\alpha}) \, {\rm d} x \ge
{\alpha^2 \sqrt{\pi} \over 8} \exp \left( -\frac{\omega^2}{4} \right)$$
for any $\omega > 0$ and $1 < \alpha < 2$.
Here is some research effort. By using the representation of the generalized Gaussian density ($c_\alpha\exp(-|x|^{\alpha})$, $c_{\alpha}>0$ normalizing constant) as Gaussian mixture, I can prove that
$$\int_{0}^{\infty} \cos(\omega x) \exp(-x^{\alpha}) \, {\rm d} x \ge
{\Gamma(1/\alpha) \over \alpha} \exp \left( -\frac{\omega^2}{2}
{\Gamma(3/\alpha)\over \Gamma(1/\alpha)} \right).$$
When $\alpha=2$, ${\Gamma(3/\alpha)\over 2 \Gamma(1/\alpha)}={1\over 4}$ and $ {\Gamma(1/\alpha) \over \alpha}= {\alpha^2 \sqrt{\pi} \over 8} $ and $\mbox{Var}(\xi)= {\Gamma(3/\alpha)\over\Gamma(1/\alpha)}$ when $ \xi \sim c_\alpha\exp(-|x|^{\alpha})$.
Both ${\Gamma(3/\alpha)\over 2 \Gamma(1/\alpha)}$ and $ {\Gamma(1/\alpha) \over \alpha}$ are decreasing in $\alpha \in (1,2)$. I need the result as originally stated as the coefficient in front of $\omega$ in the exponent has to be free from $\alpha$. All of my attempts are tied to probablisitic arguments and hence I fail to get rid of ${\Gamma(3/\alpha)\over 2 \Gamma(1/\alpha)}$ coefficient.
Related question
How to prove that $ \sum_{m=0}^{\infty} { \Gamma\{(1+2m)/\alpha\}\over \Gamma(1/2+m)} { (-t^2/4)^{m}\over m !} \ge (\alpha/2)^{3}\exp(-t^{2}/4) $
It is false for $\omega=0$, $\alpha=2$.
I presume your $dc$ should be $dx$; the inequality is badly violated; say for $\omega=1$ the l.h.s. and r.h.s. cross at $\alpha=1.4$.
this inequality does hold ($\omega>0$, $1<\alpha<2$): $$\int_{0}^{\infty}\cos(\omega x)\exp(-x^{\alpha})dx \ge
{\alpha^2 \sqrt{\pi} \over 8} \exp(-\omega^2/4) .$$ Equality is reached at $\alpha=2$.
@CarloBeenakker, yes, thank you, it should be $\sqrt{\pi}$- I have edited the question
More precise bounds are known; see e.g. Chen, Z-Q., Kim, P. and Song, R., Heat kernel estimates for the
Dirichlet fractional Laplacian, 2010.
@user90189, thank you so much for the reference, all the bounds are are given without specific constants, here, I need that specific constant . I know as $i\to\infty$ $f(\omega) \sim \omega^{(-\alpha-1)}$, the next step is to get the constants for that tail.
Maybe this is helping for the large $\omega$ expansion: https://arxiv.org/pdf/0911.4796.pdf
... in the paper arxiv.org/pdf/0911.4796.pdf that I recommended above, the asymptotics for your integral is $Q_{\beta}$, Eq. (13), The author's $\beta$ is your $\alpha$.
@JohannesTrost, thank you for the reference. I do not see how to use it though. It has a different representation, yes, very useful for asymptotics. I still do not see how I can get my lower bound. The truncation error cam provide some upper bound only. What am I missing?
@TanyaVladi I was referring to your answer to user90189's comment.The constants for the tail $\sim \omega^{-\alpha-1}$ are given in the paper.
@JohannesTrost, I understand, thank you for the clarification, however I am not interested in the tail, I need inequality to hold, you can think about it as for medium $\omega$, neither small nor large
|
2025-03-21T14:48:31.580707
| 2020-07-22T17:48:57 |
366303
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"0xbadf00d",
"DCM",
"https://mathoverflow.net/users/61771",
"https://mathoverflow.net/users/91890"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631354",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366303"
}
|
Stack Exchange
|
Prove of the shape-derivative identity relating the shape and material derivative of a shape-dependent function
I've started reading about shape optimization. Most of the concepts I've encountered so far (such as the shape derivatives of domain and boundary integrals and the corresponding) seem to be complex, but turned out to be quite simple. However, I really struggle to wrap my head around the different notions of "derivatives" for a "shape-dependent" function $y$.
The setting is as follows: Let
$d\in\mathbb N$;
$D\subseteq\mathbb R^d$ be open and $\mathcal A\subseteq 2^D$ with $D\in\mathcal A$;
$E_\Omega\subseteq\mathbb R^{\Omega}$ be a $\mathbb R$-Banach space for $\Omega\in\mathcal A$ and $E:=\bigcup_{\Omega\in\mathcal A}E_\Omega$;
$y:\mathcal A\to E$ with $$y(\Omega)\in E_\Omega\;\;\;\text{for all }\Omega\in\mathcal A;\tag1$$
$\tau>0$, $T_t$ be a $C^1$-diffeomorphism from $U$ onto an open subset of $\mathbb R^d$ for $t\in[0,\infty)$ and $$V:=\bigcup_{t\in[0,\:\tau)}T_t(D);$$
$v:[0,\tau)\times V\to\mathbb R^d$ be differentiable in the second argument with $$v\left(t,T_t(x)\right)=\frac{\partial T}{\partial t}(t,x)\;\;\;\text{for all }(t,x)\in[0,\tau)\times D;\tag2$$
$\Omega\in\mathcal A$ and $\Omega_t:=T_t(\Omega)$ for $t\in[0,\tau)$.
Now the "shape derivative* is defined as follows:
Definition 1 (shape derivative) Let $Y:[0,\tau)\to E_d$ with $$\left.Y(t)\right|_{\Omega_t}=y(\Omega_t)\;\;\;\text{for all }t\in[0,\tau).$$ Then $y$ is called shape differentiable at $\Omega$ in direction $v$ if $Y$ is Fréchet differentiable at $0$. In that case, $$y'(\Omega;v):=\left.Y'(0)\right|_{\Omega}\tag4.$$
(Please note that we most probably need to assume a certain regularity (at least continuity) of the time-dependence of $Y$ (and most probably of $T$ as well). I've omitted them, cause it's part of my question what we need to assume precisely.)
The second definition is given by the "material derivative*:
Definition 2 (material derivative) $\dot y(\Omega;v)\in E_\Omega$ is called **material derivative of $y$ at $\Omega$ in direciont$ v$ if $$y(\Omega_t)\circ\left.T_t\right|_{\Omega}\in E_\Omega\;\;\;\text{for all }t\in[0,\tau)\tag5$$ and $$[0,\tau)\to E_\Omega\;,\;\;\;t\mapsto y(\Omega_t)\circ\left.T_t\right|_{\Omega}\tag6$$ is Fréchet differentiable at $0$ with derivative equal to $\dot y(\Omega;v)\in E_\Omega$, i.e. $$\frac{y(\Omega_t)\circ\left.T_t\right|_{\Omega}-y(\Omega)}t\xrightarrow{t\to0+}\dot y(\Omega;v)\tag7.$$
Question 1: What do we need to assume in order to show that $(3)$ is well-defined, i.e. independent of the choice of $Y$?
Question 2: How can we relate the shape and material derivative?
I guess we need to assume that there is a continuous linear $$\iota_A:E_A\to E_D$$ for all $A\in\mathcal A$. Let $$y_t:=y(\Omega_t)\circ\left.T_t\right|_{\Omega}\;\;\;\text{for }t\in[0,\tau).$$ Then we could write $$\frac{Y(t)-Y(0)}t=\frac{Y(t)-\iota_\Omega y_t}t+\frac{\iota_\Omega y_t-Y(0)}t\;\;\;\text{for all }t\in(0,\tau)\tag8.$$ If the answer to question 1 is positive, then we could assume $$Y(t)=\iota_{\Omega_t}y(\Omega_t)\;\;\;\text{for all }t\in[0,\tau)\tag9$$ and, assuming $y$ has a material derivative at $\Omega$ in direction $v$, we could conclude $$\frac{\iota_\Omega y_t-Y(0)}t=\iota_\Omega\frac{y_t-y(\Omega)}t\xrightarrow{t\to0+}\iota_\Omega\dot y(\Omega;v)\tag{10}.$$
Note that there is a proof of the "shape-derivative identity" in this paper, but I think their proof is missing rigor and they seem to assume $E_\Omega$ is a closed subspace of $L^1(\Omega)$:
This is probably me being silly, but isn't the second definition just the first one with $E_\Omega=L^1(\Omega)$ for all $\Omega\in \mathcal{A}$? Is there a difference beyond the space in which you take the limit?
@DCM No, it's not you. It's me being confused about these concepts. I've completely updated the question. Hopefully it makes more sense now.
@DCM Why did you delete your answer?
I feared it would put other people off answering. I can reinstate it if you like :)
The classical material derivative $D\varphi/Dt$ of a test function $\varphi \in C_c^\infty(\mathbb{R}_+\times D)$ is obtained by setting
$$
\dfrac{D\varphi}{Dt}(x) := \dfrac{\partial\tilde\varphi}{\partial t}(0,x)\;\;\mbox{with}\;\; \tilde\varphi(t,x) = \varphi(t,T_t(x))
$$
for $x\in D$. Expanding out using the chain rule, we have
$$
\dfrac{D\varphi}{Dt}(x) = \dfrac{\partial \phi}{\partial t}(0,x)+ \sum_{i=1}^d v^i(x)\dfrac{\partial \phi}{\partial x^i}(0,x)\;\;(x\in D).
$$
with $v(x) = \lim_{t\to 0}t^{-1}(T_t(x)-x)$. I'm imagining here that it's $(T_t)_{t>0}$ that has been prescribed, but one can go in the other way too (i.e. go from a vector field to a flow rather than a flow to a vector field).
The 'shape identity' is the natural generalisation of the formula above to distributions $y(\Omega_t)$ of the form
$$
\langle y(\Omega_t),\varphi\rangle=\int_{\Omega_t} y_{\Omega_t}(x)\varphi(x)\mathrm{d}x\;\;(\varphi\in C^\infty_c(D))
$$
with $y_{\Omega_t}\in L^1_\mathrm{loc}(\Omega_t)$ and $\Omega_t = T_t(\Omega)$. In this formulation $y_{\Omega_t}(x)$ is trying to be $\varphi(t,x)$ from the smooth formulation while the 'shape derivative' is trying to be $x\mapsto (t\mapsto y_{\Omega_t}(x))'(0)$.
Regarding your first question, it seems like the easiest ways to make everything work are:
Regard everything as a distibution on $D$; or
Make sure that $E_\Omega = \{f_{|\Omega}: f\in E_D\}$ for all $\Omega\in \mathcal{A}$, do what you need in $E_D$, then restrict back to $\Omega$.
These two approaches both let you form linear combinations and take limits 'normally', so remove the problems associated with everything living in different spaces. It's important to check with the second approach that the behavior of the limit in $\Omega$ doesn't depend on the extensions chosen, but there's a result in the reference which shows how to do that (i.e. by testing against a smooth bump supported in $\Omega$).
Regarding your comment about the reference assuming that $E_\Omega$ is closed in $L^1(\Omega)$; I'm not sure I agree - isn't the fact that quotients converge to something in $L^1$ just part of their definition?
As for minimum requirements, I think you at least want the quotients
$$
\dfrac{\langle y(\Omega_t)\circ T_t, \varphi\rangle - \langle y(\Omega),\varphi\rangle}{t}\;\;\mbox{and}\;\;\dfrac{
\langle y(\Omega_t),\varphi\rangle - \langle y(\Omega),\varphi\rangle}{t}
$$
to converge as $t\to 0$ for all test functions $\varphi$, since these are what give you the distributional 'material' and 'shape' derivatives.
Thank you for your answer. I think what you're doing is considering the "weak" versions of the material and shape derivative. They are defined in the same way as in the question, but the limits in the definition of the Fréchet derivatives have to be understood with respect to the weak topology. So, for example, $$\left\langle\frac{y(\Omega_t)\circ\left.T_t\right|{\Omega}-y(\Omega)}t,\varphi\right\rangle\xrightarrow{t\to0+}\left\langle\dot y(\Omega;v),\varphi\right\rangle\tag{11}$$ for all $\varphi\in E\Omega'$.
Now you've chosen $E_\Omega=L^1_{\text{loc}}(\Omega)$ and I guess that weak convergence of a sequence $(f_n){n\in\mathbb N}\subseteq L^1{\text{loc}}(\Omega)$ to $f\in L^1_{\text{loc}}(\Omega)$ holds if and only if $$\int f_n\varphi:{\rm d}\lambda^{\otimes d}\xrightarrow{n\to\infty}\int f\varphi:{\rm d}\lambda^{\otimes d}\tag{12}$$ for all $\varphi\in C_c^\infty(\Omega)$ (I don't remember whether we need additional assumptions to show that). So, while useful for its own, the question for the "strong" derivatives in my post is still open.
Side note: You've used $\varphi\in C_c^\infty(D)$ instead of $\varphi\in C_c^\infty(\Omega)$. Could you elaborate on why this is important or at least useful?
Taking $C^\infty_c(D)$ as the space of test functions seems sensible because it gives you a way to take linear combinations and limits of the $y(\Omega_t)$ in $\mathscr{D}'(D)$. The other way natural way to do this is to choose your $E$ functor $\Omega\mapsto E_\Omega$ to be one for which $E_\Omega = {f_{|\Omega}: f\in E_D}$ for all $\Omega\in \mathcal{A}$, at which point you can do the same in $E_D$ (although this latter approach requires you to check that your limits are independent of which extensions you choose - your reference does it like this).
It took me a while before I could come back to this question. I think it's worth considering the pointwise analogue of the material/shape derivative which is considered here. I've asked a separate question for that: https://mathoverflow.net/q/371076/91890. It would be great if you could take a look.
|
2025-03-21T14:48:31.581184
| 2020-07-22T18:52:53 |
366306
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Anonymous",
"David Hansen",
"Donu Arapura",
"Tabes Bridges",
"https://mathoverflow.net/users/14044",
"https://mathoverflow.net/users/1464",
"https://mathoverflow.net/users/27219",
"https://mathoverflow.net/users/4144"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631355",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366306"
}
|
Stack Exchange
|
Example of an intersection complex not concentrated in a single degree
I'm having trouble finding references for in-depth examples of perverse sheaves, so answers in the form of such a reference would be most helpful.
I want to construct an example of an intersection complex not concentrated in a single (natural) cohomology degree. Reading BBD, it seems the definition of intermediate extension needs to be made in the derived category, even to discuss intermediate extension of constant sheaves. So I think I can find an example of an open inclusion $j: U_0 \hookrightarrow X_0$ such that $j_{!*} \bar{\mathbb{Q}}_{\ell} [d]$ is not concentrated in degree $d$. I'm looking for a simplest example, but I'm having trouble verifying my work so far. So I would also appreciate if someone could point out any glaring errors in my reasoning (and lack thereof).
The first few attempts I made all seem to have $R^1 j_! \bar{\mathbb{Q}}_{\ell} = 0$, and so $Rj_! \bar{\mathbb{Q}}_{\ell} [d] = {}^p j_!\bar{\mathbb{Q}}_{\ell}[d]$; so ${}^pj_! \bar{\mathbb{Q}}_{\ell} [d] \hookrightarrow {}^p j_* \bar{\mathbb{Q}}_{\ell} [d]$; hence $j_{!*} (\bar{\mathbb{Q}}_{\ell} [d]) = j_!(\bar{\mathbb{Q}}_{\ell}) [d]$.
In particular, the above seems that to hold whenever $X_0$ is smooth and $j: U_0 \hookrightarrow X_0$ is the inclusion of dense open. So this is not the right direction.
Looking now at singular varieties, the first two examples that come to mind are $C_0 = \mathrm{Proj} (\mathbb{F}_q[S,T,U]/(T^2U-S^3))$ (projective cubic curve with a cusp) and $C'_0 = \mathrm{Proj} (\mathbb{F}_q[S,T,U]/(T^2U - S^3 - S^2U))$ (projective cubic curve with a node). Note the nonsingular loci $C_{ns, 0} \cong \mathbb{A}^1_0$ and $C'_{ns, 0} \cong \mathbb{G}_{m, 0}$. (Assume $\mathrm{char}(\mathbb{F}_q) > 2$ for $C'_0$.)
But in the case of $C_0$, taking $j: C_{ns,0} \hookrightarrow C_0$ to be the inclusion of the nonsingular locus, it appears to me that $Rj_!$ is exact. In particular, the stalk at a geometric point ${\bar{x}}$ lying over the node $x \in C_0$
$$
(R^1 j_! \bar{\mathbb{Q}}_{\ell})_{\bar{x}} =
\lim_{\to} H^1 (U, j_! \bar{\mathbb{Q}}_{\ell}) \overset{(a)}{=}
\lim_{\to} H^1_c (U \times_{C_{0}} C_{ns, 0}, \bar{\mathbb{Q}}_{\ell}) \overset{(b)}{\cong}
H^1_c (\mathbb{A}^1, \bar{\mathbb{Q}}_{\ell}),
$$
where the limit is taken over étale $U \to C_0$ over $\bar{x}$. Then we have $H^1_c(\mathbb{A}^1, \bar{\mathbb{Q}}_{\ell})$ vanishes by Poincaré dualtiy as $H^1 (\mathbb{A}^1, \bar{\mathbb{Q}}_{\ell}) = 0$. (I think $(a)$ holds by definition of $H^*_c$, and $(b)$ I can't justify.) So, assuming every link in this chain holds, we have $j_! = j_{!*}$, and I have not found my example.
But I believe—if my reasoning is at all accurate for $C_0$—that I have found an example in $j': C'_{ns, 0} \hookrightarrow C'_0$. Repeating the argument above, with $x' \in C'_0$ the self-intersection point,
$$
(R^1 j'_! \bar{\mathbb{Q}}_{\ell})_{\bar{x}'} =
\lim_{\to} H^1 (U, j'_! \bar{\mathbb{Q}}_{\ell}) =
\lim_{\to} H^1_c (U \times_{C'_0} C'_{ns, 0}, \bar{\mathbb{Q}}_{\ell}) \cong
H^1_c (\mathbb{G}_{m}, \bar{\mathbb{Q}}_{\ell}).
$$
In this case, we have $H^1 (\mathbb{G}_{m}, \bar{\mathbb{Q}}_{\ell}) = \bar{\mathbb{Q}}_{\ell}(-1)$ (this is my understanding after reading Milne's and de Jong's notes on étale cohomology), and so $(R^1 j'_! \bar{\mathbb{Q}}_{\ell})_{\bar{x}'} = \bar{\mathbb{Q}}_{\ell}(1) \ne 0$. Since we have determined now that $j'_!$ is not exact, we need to calculate ${}^p j'_!$, ${}^p j'_*$, and finally calculate $j'_{!*}$. Should I keep going? Am I on the right track? Have I made glaring errors? Is there a reason $(b)$ should hold? What can I read to speed up my progress on these questions? I've read BBD and Kiehl-Weissauer, and a couple of less formal notes on perverse sheaves, and I've seen precious few examples in any detail. I recognize I haven't read the entire literature, so does anyone know where I should look next?
Tiny comment that isn't really that helpful: the singularity on $C_0$ is a cusp, not a node.
The most beautiful exposition of perverse sheaves is (https://arxiv.org/abs/0712.0349). There are examples there of intersection cohomology complexes supported in multiple degrees.
Sorry I haven't read your entire question, which is a bit long. This is really just an extended comment to address the "where I should look next?" part. Suppose $X$ has an isolated singularity $x$, and $j:U\to X$ is the smooth complement. Then the formula on top of page 60 of BBD would simplify to
$$j_{!*}\overline{\mathbb{Q}}_\ell[n]= (\tau_{\le n-1}\mathbb{R} j_* \overline{\mathbb{Q}}_\ell)[n]$$
where $n=\dim X$ and I'm using middle perversity. Now let $X$ be a sufficiently complicated singularity, a cone over an elliptic curve will do. Then this won't be a translate of a sheaf. Look at the stalk at $x$, it will have cohomology in 2 degrees.
Should the index of the truncation be $n-1$ rather than $n$?
Thanks. Fixed..
|
2025-03-21T14:48:31.581594
| 2020-07-22T19:44:38 |
366307
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Erel Segal-Halevi",
"Jan Kyncl",
"https://mathoverflow.net/users/24076",
"https://mathoverflow.net/users/34461"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631356",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366307"
}
|
Stack Exchange
|
Is the following generalization of the Caratheodory theorem true?
The colorful Carathéodory theorem (Bárány, 1982) considers $d+1$ "colors" $X_1,\ldots,X_{d+1}\subseteq \mathbb{R}^d$, and a point $x$ in the convex hull of each color ($x\in \text{conv}(X_i)$ for each $i\in[d+1]$). It says that there exists a set of $d+1$ points of distinct colors, such that $x$ is in their convex hull.
Suppose that, instead of a single point $x$, there are $d+1$ points $x_1\in \text{conv}(X_1),\ldots,x_{d+1}\in \text{conv}(X_{d+1})$. Let $\bar{x} := (x_1+\cdots+x_{d+1})/(d+1) = $ the average of these points. Is it true that there exists a set of $d+1$ points of distinct colors, such that $\bar{x}$ is in their convex hull?
Note that the Bárány's theorem is implied by this claim.
The motivation comes from this paper: https://dx.doi.org/10.1007/s00493-019-4019-y it says that a special case of Theorems 1.6 and 1.7 can be proved the the colorful Carathéodory theorem, and I wondered if the general case could be proved by a generalization of it.
The answer seems trivially true, since $\bar{x}$ is in the convex hull of the points $x_1, \dots, x_{n+1}$ and they have distinct colors. The question might be more interesting if the assumption was weakened to $x_i \in \mathrm{conv}(X_i)$.
In such a case there would be a counterexample in the plane, for example: $X_1=(0,0), X_2=(0,1), X_3={(10,0),(10,10)}$.
@JanKyncl Indeed this is what I meant. Thanks for the counter-example
An even simpler example is $X_1 = X_2 = {(0,0)}$ and $X_3 = {(-3,3),(3,3)}$. Taking $x_1=x_2 =(0,0)$ and $x_3 = (0,3)$ gives $\bar{x} = (0,1)$, but for any selection of 3 points of different colors, their convex hull is an $45^{\circ}$ line that does not cross $(0,1)$.
I tried to avoid degeneracies and find an example in general position. I got an inspiration from the following paper, which may be relevant: https://doi.org/10.1016/j.comgeo.2012.01.006
There is a counterexample in the plane, for example the following four points in convex position: $X_1=(0,0), X_2=(0,1), X_3=\{(10,0),(10,10)\}$.
(This was originally posted as a comment to a previous version of the question).
|
2025-03-21T14:48:31.581756
| 2020-07-22T19:48:47 |
366308
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Leo Herr",
"https://mathoverflow.net/users/86614"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631357",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366308"
}
|
Stack Exchange
|
Extending rational maps of nodal curves
Let $R$ be a discrete valuation ring with fraction field $K$ and $C, D$ two nodal (=prestable) curves over $\operatorname{Spec} R$. If I have a map $C_K \to D_K$ between the restriction of the curves to the generic point $\operatorname{Spec} K$,
can I blow up $C$ to extend this to a map $C \to D$ over $\operatorname{Spec} R$?
If I blow up $C$ to do this, can I ensure that the result remains a nodal curve and the generic fiber $C_K$ is unchanged?
If it helps, the map $C_K \to D_K$ is a "partial stabilization" -- it contracts $\mathbb{P}^1$'s.
Remark: This is pretty clear for smooth curves, and well documented in the stacks project.
(My plan of attack so far:
A) Stacks 0BX7 should let me extend the map away from a finite set of closed points in the special fiber of $C$.
B) I believe these closed points should be nodes that get smoothed out in the generic fiber -- if the node persists, then maybe I can define the map at the node via specialization of the two generic points without a problem.
C) If the node gets smoothed out, this is etale-locally pulled back from $\mathbb{A}^2 \to \mathbb{A}^1$ and one can see directly the blowup at this node remains nodal.
D) By taking the closure $\Gamma$ of the graph of the rational map $C \dashrightarrow D$, I can argue that the projection $\Gamma \to C$ is a blowup at nodes in the special fiber that get smoothed out, since the map can be extended already away from those points.
I don't completely believe any of these steps -- this is more of a sketch than a proof. I'd very much like citations and references to any argument, esp. to the stacks project. )
I should mention that I'm over the complex numbers and the map $C \to D$, if it exists, would also be a partial stabilization.
After lots of help from others, I realized the question boils down to some sort of stable reduction, which is a properness statement for the moduli of stable or relative stable maps $\overline{M}(X/V)$.
The isomorphism $C_K \simeq D_K$ provides a point $C_K \to C_K \times D_K$ over $K$ of $\overline{M}(C \times D/\operatorname{Spec} R)$. Properness (=stable reduction) extends this to a map $C' \to C \times D$ over $\operatorname{Spec}R$. The maps $C' \to C, C' \to D$ must be partial stabilizations because that's a closed condition in the base, and they're blowups because they're proper birational.
The blowups are at closed subschemes supported at a finite number of points in the special fiber, and I wish I could say the centers are smooth or something (or iterated blowups at smooth centers), but I don't need it.
Can anyone see why that would be, say by the fact that the exceptional divisors are chains of $\mathbb{P}^1$'s or something?
I apologize if this question and answer were obvious to the broader community.
|
2025-03-21T14:48:31.581963
| 2020-07-22T20:03:28 |
366309
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631358",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366309"
}
|
Stack Exchange
|
Dense embeddings into Euclidean space
The question is a follow-up on this old post. Fix a positive integer $d$ and consider $\mathbb{R}^d$ with its usual Euclidean topology. Given a metric space $(X,\delta_X)$, what conditions are needed for $(X,\delta_X)$ to admit a continuous (preferably bi-Lipschitz) embedding into a dense subset of $\mathbb{R}^d$?
|
2025-03-21T14:48:31.582025
| 2020-07-22T20:09:04 |
366310
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Adam P. Goucher",
"Claus",
"David Roberts",
"Nik Weaver",
"Vervious",
"https://mathoverflow.net/users/152231",
"https://mathoverflow.net/users/156936",
"https://mathoverflow.net/users/23141",
"https://mathoverflow.net/users/39521",
"https://mathoverflow.net/users/4177"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631359",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366310"
}
|
Stack Exchange
|
Why did Robertson and Seymour call their breakthrough result a "red herring"?
One of the major results in graph theory is the graph structure theorem from Robertson and Seymour
https://en.wikipedia.org/wiki/Graph_structure_theorem. It gives a deep and fundamental connection between the theory of graph minors and topological embeddings, and is frequently applied for algorithms.
I was working with this results for years, and now heard someone saying that Robertson and Seymour called this result a "red herring". Is this true, and why would they possibly call such a breakthrough result a red herring?
(Edit: my question refers to the 2003 graph structure theorem, not to the graph minor theorem which they established later)
Seymour and Robertson have indeed said that, and in fact they wrote that in their 2003 article in which they published the graph structure theorem.
Here is the quote from Robertson and Seymour „Graph Minors. XVI. Excluding a non-planar graph“ (Journal of Combinatorial Theory, Series B, Vol. 89, Issue 1, Sept. 2003, pages 43–76, doi:10.1016/S0095-8956(03)00042-X).
Their theorem 1.3 is the earliest version of what we now call the graph structure theorem.
While 1.3 has been one of the main goals of this series of papers, it turns out to have been a red herring. There is another result (theorem 3.1) which is proved in this paper, and from which 1.3 is then derived; and in all the future applications in this series of papers, it is not 1.3 but 3.1 that will be needed.
My reading is, they were calling it a red herring because, at this point, they realized the importance of the concept of a tangle. (I think Reinhard Diestel said that the notion of a tangle is perhaps the deepest single innovation for graph theory stemming from this proof.)
Continuing the quote of Robertson and Seymour (the bold font is from me, not from the article):
Let us explain how theorem 3.1 is used to prove 1.3. Evidently we would like to eliminate the ‘‘tree-structure’’ part of 1.3 and concentrate on the internal structure of one of the ‘‘nodes’’ of the tree. How can we do so? An inductive argument looks plausible at first sight; if there is no low order cutset of G dividing it into two substantial pieces then G itself must be almost a ‘‘node’’ if the theorem is to be true, while if there is such a cutset we may express G as a clique-sum of two smaller graphs, and hope to apply our inductive hypothesis to these graphs. But there is a difficulty here; it is possible that these smaller graphs have an L-minor while G does not. Fortunately there is a way to focus in on a ‘‘node’’ which does not involve any decomposing, as follows. We can assume that the tree is as refined as possible in the sense that no node can be split into two smaller nodes, and so for every low order cutset of G; most of any node will lie on one side or the other of the cutset (except for nodes of bounded cardinality, which we can ignore.) Therefore if we fix some node, every small cutset has a ‘‘big’’ side (containing most of the node) and a ‘‘small’’ side—and it turns out that no three small sides have union G: Thus a node defines a ‘‘tangle’’, which is such an assignment of big and small sides to the low order cutsets; and conversely, it can be shown that any tangle in G of sufficiently high ‘‘order’’ will be associated with some node of the tree-structure. Hence a convenient way to analyze the internal structure of the nodes is to analyze the local structure of G with respect to some high order tangle, and this is the content of theorem 3.1.
The graph minor theorem is of interest in reverse mathematics because its proof requires a surprisingly strong subsystem of second order arithmetic. I wonder what the proof-theoretic strength of this Theorem 3.1 is?
This is about a different theorem of Robertson and Seymour (namely the 'graph structure theorem', not the 'Robertson-Seymour theorem').
Oh, thank you for the clarification.
@Claus is the 2013 result this one: Joret and Wood, Complete graph minors and the graph minor structure theorem, https://doi.org/10.1016/j.jctb.2012.09.001 ?
@DavidRoberts David thank you for your edit, and thank you for asking this clarifying question. I realized it was the year 2004 I was refering to: the year when Robertson and Seymour finally proved their graph minor theorem, which is now often called the Robertson-Seymour Theorem. But I have removed this reference as it had created confusion.
@ClausDollinger ok, thanks for clearing that up. I guess you meant Graph Minors XX, with the culmination of the proof of Wagner's conjecture (now the Robertson–Seymour theorem).
Do any of you have a good resource / introduction to tangles? A brief search yields a dearth of lecture notes, wikipedia articles, and the like
@Vervious the following is a great introduction, putting everything in context: the text from Reinhard Diestel (2017): Graph Theory 5th Edition.
@DavidRoberts Yes exactly. Thanks again David.
If you are interested in tangles in the sense of Robertson and Seymour, this is just to provide some perspective on it. I am working on this for my Ph.D. project and I thought maybe it is considered helpful if I share this high-level, intuitive perspective here (it is not a detailed definition):
The best and shortest description, I think, is given in the following quote. It mentions the proof of the graph minor theorem, but in fact it is the same concept of tangles as in the proof of the graph structure theorem:
Quote: „Originally devised by Robertson and Seymour as a technical device for their proof of the graph minor theorem, tangles have turned out to be much more fundamental than this: they define a new paradigm for identifying highly connected parts in a graph.
Unlike earlier attempts at defining such substructures—in terms of, say, highly connected subgraphs, minors, or topological minors—tangles do not attempt to pin down this substructure in terms of vertices, edges, or connecting paths, but seek to capture it indirectly by orienting all the low-order separations of the graph towards it.
In short, we no longer ask what exactly the highly connected region is, but only where it is. For many applications, this is exactly what matters.
Moreover, this more abstract notion of high local connectivity can easily be transported to contexts outside graph theory. This, in turn, makes graph minor theory applicable beyond graph theory itself in a new way, via tangles.“ End of Quote.
This quote is from Reinhard Diestel, in the preface to the 5th edition of his Graph Theory book.
|
2025-03-21T14:48:31.582495
| 2020-07-22T20:13:26 |
366312
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Florian Lehner",
"Moritz Firsching",
"https://mathoverflow.net/users/100231",
"https://mathoverflow.net/users/39495",
"https://mathoverflow.net/users/97426",
"vidyarthi"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631360",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366312"
}
|
Stack Exchange
|
The edge precoloring extension problem for complete graphs
Consider coloring the edges of a complete graph on even order. This can be seen as the completion of an order $n$ symmetric Latin square except the leading diagonal. My question pertains to whether we can always complete the edge coloring in $n-1$ colors given a certain set of colors? The number of colors I fix is exactly equal to $\frac{(k)(k+2)}{2}$, where $k=\frac{n}{2}$ and form $4$ distinct consecutive last four subdiagonals (and, by symmetry, superdiagonals) in the partial Latin square.
For example, in the case of $K_8$, I fix the following colors:
\begin{bmatrix}X&&&&1&3&7&4\\&X&&&&2&4&1\\&&X&&&&3&5\\&&&X&&&&6\\1&&&&X&&&\\3&2&&&&X&&\\7&4&3&&&&X&\\4&1&5&6&&&&X\end{bmatrix}
A completion to a proper edge coloring in this case would be:
\begin{bmatrix}X&5&6&2&1&3&7&4\\5&X&7&3&6&2&4&1\\6&7&X&4&2&1&3&5\\2&3&4&X&7&5&1&6\\1&6&2&7&X&4&5&3\\3&2&1&5&4&X&6&7\\7&4&3&1&5&6&X&2\\4&1&5&6&3&7&2&X\end{bmatrix}
Can the above be always done if the colors I fix follow the same pattern for all even order complete graphs? Note that the pattern followed in the precoloring consists of two portions-
i) the last $k-1$ subdiagonals are actually taken from a canonical $n$-edge coloring of the complete graph on $n-1$ vertices, where $n$ is even. By canonical, I mean the commutative idempotent 'anti-circulant' latin square. Like in the example above, the canonical coloring of the complete graph on $7$ vertices is
\begin{bmatrix}1&5&2&6&3&7&4\\5&2&6&3&7&4&1\\2&6&3&7&4&1&5\\6&3&7&4&1&5&2\\3&7&4&1&5&2&6\\7&4&1&5&2&6&3\\4&1&5&2&6&3&7\end{bmatrix}
ii)The $k$-th subdiagonal just consists of entries in the pattern $1-2-3-$ so on and takes into account the previous entries to create an appropriate entry. Like in the example above the last diagonal I took was $1-2-3-6$. It could also have been $1-2-3-7$.
And, if the completion exists, would the completion be unique? Any hints? Thanks beforehand.
I don't quite understand your description of the pre-coloring. Are there any restrictions on how you colour? For general $n$, shouldn't it be the last $n/2$ subdiagonals?Transversal also has a concise meaning in latin squares, but I don't think this is what you mean by transversal.
@FlorianLehner edited the post, thanks
At this point, it is still unclear to me, what "follow the same pattern" is. It might help if you clarify what exactly is the precouloring that you are interested in.
@MoritzFirsching edited the post. See now for the pattern I follow
What pattern exactly for the k-th subdiagonal? Any valid completion for the k-th row? One that starts with 1-2-3-...-k-1? They lexicographically smallest completion?
@MoritzFirsching yes, sort of
I'm unable to grasp the precise definition from "yes, sort of"
@MoritzFirsching I meant that yes, my preference is for the lexicographically smallest order, but, if there are clashes, the the color would be the nearest to that. Suppose, in the coloring I give, I took the string $1-2-3-6$ and not $1-2-3-4$, which was not possible on account of $4$ color in the previous row
ok, so in the example above, $1-2-3-7$ is not the coloring you are looking for? Maybe it would be more clear to remove the sentence "It could also have been $1-2-3-7$?
I updated my question to show non-uniqueness for the case $n=10$, using the clarified definition of the precoloring, I hope I understood correctly now.
For the case $n=8$, with the precoloring you describe the completion you give is indeed unique. I checked by writing the corresponding boolean program and let a solver enumerate all solutions: there is only one.
For the case $n=10$, consider the pre-colored $K_{10}$
$$\left(\begin{array}{rrrrrrrrrr}
X & & & & & 1 & 8 & 4 & 9 & 5 \\
& X & & & & & 2 & 9 & 5 & 1 \\
& & X & & & & & 3 & 1 & 6 \\
& & & X & & & & & 4 & 2 \\
& & & & X & & & & & 7 \\
1 & & & & & X & & & & \\
8 & 2 & & & & & X & & & \\
4 & 9 & 3 & & & & & X & & \\
9 & 5 & 1 & 4 & & & & & X & \\
5 & 1 & 6 & 2 & 7 & & & & & X
\end{array}\right)$$
This can be completed in $77$ ways, for example
$$\left(\begin{array}{rrrrrrrrrr}
X & 6 & 7 & 3 & 2 & 1 & 8 & 4 & 9 & 5 \\
6 & X & 8 & 7 & 3 & 4 & 2 & 9 & 5 & 1 \\
7 & 8 & X & 5 & 4 & 2 & 9 & 3 & 1 & 6 \\
3 & 7 & 5 & X & 9 & 8 & 6 & 1 & 4 & 2 \\
2 & 3 & 4 & 9 & X & 5 & 1 & 6 & 8 & 7 \\
1 & 4 & 2 & 8 & 5 & X & 3 & 7 & 6 & 9 \\
8 & 2 & 9 & 6 & 1 & 3 & X & 5 & 7 & 4 \\
4 & 9 & 3 & 1 & 6 & 7 & 5 & X & 2 & 8 \\
9 & 5 & 1 & 4 & 8 & 6 & 7 & 2 & X & 3 \\
5 & 1 & 6 & 2 & 7 & 9 & 4 & 8 & 3 & X
\end{array}\right)
$$ or
$$\left(\begin{array}{rrrrrrrrrr}
X & 7 & 2 & 6 & 3 & 1 & 8 & 4 & 9 & 5 \\
7 & X & 8 & 3 & 4 & 6 & 2 & 9 & 5 & 1 \\
2 & 8 & X & 5 & 9 & 4 & 7 & 3 & 1 & 6 \\
6 & 3 & 5 & X & 8 & 7 & 9 & 1 & 4 & 2 \\
3 & 4 & 9 & 8 & X & 5 & 1 & 6 & 2 & 7 \\
1 & 6 & 4 & 7 & 5 & X & 3 & 2 & 8 & 9 \\
8 & 2 & 7 & 9 & 1 & 3 & X & 5 & 6 & 4 \\
4 & 9 & 3 & 1 & 6 & 2 & 5 & X & 7 & 8 \\
9 & 5 & 1 & 4 & 2 & 8 & 6 & 7 & X & 3 \\
5 & 1 & 6 & 2 & 7 & 9 & 4 & 8 & 3 & X
\end{array}\right)$$
This answers your question about uniqueness.
So it looks very plausible to me, that the completion can always be done for $n\geq 8$ and it is not unique for $n\geq 10$.
thanks, edited the question slightly. My question is, can we do this unique extension for every complete graph of even order?That is given the last $k$ subdiagonals, can we always extend the coloring to the whole graph?
yes, thanks for the case $n=10$, but is there a formal proof for the possibility in every case, because the general problem of precoloring extension is NP-complete. So, I think in this case it is in P. Is it true? Would the uniqueness depend on the choice of the fixed colors?
So many questions, but I think they can all be answered. For existence of a solution for every n, just take a coloring for all edges (which can always be done with n-1 colors) and forgot all but the $k$ outer diagonals as a precoloring. This can then clearly be extended. to a full coloring again. Uniqueness fails for n=10 already and it might not be too hard to construct more than one coloring for every larger n. I would expect that the number of extensions to a precoloring depends on the choice of precoloring. Perhaps asked more things as a actual questions and not in a comment..
But, whether our precoloring sequence of fixed colors would appear in the normal edge coloring is the main question we must ask. See the answer by Florian. Yes, uniqueness would not be true, but even completion is also not true, as shown by Florian. But, given the colors having some intersection of colors between any two rows would warrant a completion is my view.
Assuming that you mean to precolour $k$ subdiagonals and have no further constraints on the precolouring, the answer to both of your questions is no.
For every $n$ there is a precolouring which cannot be extended: choose colours $1, \dots n/2$ in the first row and colours $n/2+1, \dots, n-1$ in the second row (and thus the second column). Then there is no valid colour for the entry in the first row/second column, so we cannot complete the colouring.
If we can complete the colouring, then the completion is not necessarily unique: note that we can always give a valid precolouring only using colours $1 \dots k$. Thus in any completion of this precolouring we can permute the colours $k+1, \dots, n-1$ to obtain a different completion.
I didnt quite get your last paragraph. You meant we can have at most $k$ in a single row? And then what is the permutation you are talking about? In the case $n=8$, as shown in the answer by Moritz, it is unique it seems
Since you have updated the question and are now asking about one specific precolouring, this answer does not fit the question anymore. The pre-colouring I had in mind for the second part was $1,2,\dots,k$ in the first row, $1,2,\dots, k-1$ in the second row, $1,2,\dots, k-2$ in the third row etc. If this can be extended to a full colouring (which I think should be possible), then there is more than one way to do it (e.g. swap the roles of $k+1$ and $k+2$).
why do you think the extension is possible for the kind of coloring I give?
I have no idea whether it is or isn't possible for your specific precolouring, all I'm saying is that my examples don't fit your pattern.
|
2025-03-21T14:48:31.583061
| 2020-07-22T20:55:13 |
366316
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Jan Kyncl",
"Joseph O'Rourke",
"https://mathoverflow.net/users/159965",
"https://mathoverflow.net/users/24076",
"https://mathoverflow.net/users/6094",
"zjs"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631361",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366316"
}
|
Stack Exchange
|
Polytope with most faces
Fix $m,n \in \mathbb{N}$ with $m \ge n+1$. Take $m$ points in general position in $\mathbb{R}^n$ and let $P$ be their convex hull. What is the maximal number of (external, codimension-one) faces that $P$ can have, in terms of $m$ and $n$?
(Apologies if this is a well-known quantity.)
dual question: https://mathoverflow.net/questions/127423/how-many-vertices-can-a-convex-polytope-have/
See the answers at Jan's link. "Cyclic polytopes maximize the number of facets for a fixed number of vertices." For you, $m$ is the number of vertices.
Thank you both! This is extremely helpful, particularly the closed form.
The upper bound conjecture of Motzkin, made a theorem by McMullen in 1970, states that the highest number of facets among all polytopes with $m$ vertices in $\mathbb R^n$ is the number of facets of the cyclic polytope $\Delta(m,n)$.
|
2025-03-21T14:48:31.583158
| 2020-07-22T21:05:43 |
366317
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Frode Alfson Bjørdal",
"https://mathoverflow.net/users/37385"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631362",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366317"
}
|
Stack Exchange
|
Algebraization of arithmetic and stronger theories?
Intuitionistic and classical propositional logic, and even classical first-order logic with identity, have algebraic counterparts. Algebraizable logics, 1989, by Willem J. Blok and Don Pigozzi, is a classical reference.
Is more now known about whether stronger systems are algebraizable?
At least that was discovered after 1989. :)
Indeed. Anyway, I would like to know some about what such algebras look like if they exist.
|
2025-03-21T14:48:31.583220
| 2020-07-22T21:13:23 |
366318
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"KReiser",
"Longma",
"R. van Dobben de Bruyn",
"https://mathoverflow.net/users/112114",
"https://mathoverflow.net/users/122681",
"https://mathoverflow.net/users/82179"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631363",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366318"
}
|
Stack Exchange
|
Is the restriction of a projection map open?
Let $X$ be a locally closed algebraic sub variety of $\mathbb C^{m+n}$. Let $p: \mathbb C^{m+n}\to \mathbb C^{m}$ be the projection map. Suppose that the restriction of $p$ to $X$ is injective. Is the map $p: X \to p(X)$ open in Euclidean topology? Note that $p(X)$ is Zariski constructible in $\mathbb C^{m}$.
What if $\tilde Y \to Y$ is the resolution of a node $y$ with preimages $p,q$, and $X = \tilde Y \setminus {q}$? Then $X \to Y$ is a bijection, but not a homeomorphism: a small neighbourhood of $p$ will map to a locally closed subset of $Y$ contained in only one of the sheets through $y$. (Just stick these inside affine spaces if that's what you want.)
Crossposted from MSE. In general, you should wait much longer than 43 minutes before cross-posting, and you ought to link all versions of your post together.
Okay. Thanks for the example and the notice.
|
2025-03-21T14:48:31.583317
| 2020-07-22T21:43:07 |
366321
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Quarto Bendir",
"Sebastian",
"https://mathoverflow.net/users/156492",
"https://mathoverflow.net/users/4572"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631364",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366321"
}
|
Stack Exchange
|
Vector field along an immersion whose covariant derivative is the differential
Let $(M,g)$ be a Riemannnian manifold and let $f:\Sigma\to M$ be a smooth immersion. Then the vector bundle $f^\ast TM\to\Sigma$ has a natural bundle metric and metric-compatible connection. Can one characterize the situations in which there must exist a section $V$ such that $\nabla V=df$? This is trivially possible if $(M,g)$ is Euclidean space. It feels like it should not be possible in general.
This seems to be equivalent to the existence of a closed 1-form $\omega$ on $\Sigma$ and a normal vector field $w$ along $\Sigma$ such that
\begin{align}\nabla \omega-\langle h,w\rangle&=f^\ast g\\ h(\cdot,\omega^\sharp)+\nabla^\perp w&=0\end{align}
where $h$ is the second fundamental form. I can't see any immediate conclusions to make.
Your differential $df\in\Omega^1(\Sigma,f^*TM)$ satisfies the integrability condition $$d^\nabla df=0$$ where $d^\nabla$ is the induced exterior derivative from the (pull-back of) the Levi-Civita connection on $M.$ If $df=\nabla V$ the integrability condition is that the curvature tensor $R$ applied to the vector field $V$ does vanish. As you have guessed this is clearly not possible in general.
Thank you, that was essentially my intuition but I failed to realize it explicitly. Could one say anything in the situation that $(M,g)$ is a flat manifold?
Yes, one can say the following: it is always possible locally, but not always possible globally. As a counterexample you can consider a cone $C$ in 3-space with the induced flat metric, and $f$ to be the identity $\Sigma=C\to C.$ If the cone angle is not $2\pi$, i.e., you do not have a degenerate cone (plane), $V$ does not exist globally.
|
2025-03-21T14:48:31.583462
| 2020-07-22T23:54:03 |
366327
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Michael Engelhardt",
"Norregaard",
"https://mathoverflow.net/users/134299",
"https://mathoverflow.net/users/161594"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631365",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366327"
}
|
Stack Exchange
|
On proving the absence of limit cycles in a dynamical system
I'm studying this classical paper in nonlinear dynamics in biophysics and I want to understand it properly. I'm stuck at this two-variable problem which I've struggled with for too long now.
$$ \dot M = \frac{1}{1+E^m} - a M $$
$$ \dot E = M - b E $$
Where $ \dot z = \frac{dz}{dt} $. It turns out later that $ m \geq 8 $. Note also that, because it's concentrations of molecules, $ M,E \geq 0 $.
This coupled system of equations has a fixed point $ (M_0,E_0) $ where the nullclines $ \dot M = 0 $ and $ \dot E = 0 $ intersect. This occurs when:
$$ M_0 = b E_0 $$
$$ a b E_0 (1+E_0^m) = 1 $$
So far, so good. Now the article says: "We expand near this point by writing $ M = M_0 + X $, $ E = E_0 + Y $"
$$ \dot X = - m a^2 b^2 E_0^{m+1} Y - a X + O(Y^2) $$
$$ \dot Y = X - b Y $$
I get why they expand near the fixed point and I get that $ O(Y^2) $ means neglecting higher-order terms. And I think I can derive the equation for $ \dot Y $:
$ \dot Y = \dot {(E-E_0)} = M_0 + X - b (E_0 + Y) = b E_0 + X - b E_0 - b Y = X - b Y $
But I need your help to understand how to get $ \dot X $ just from taylor expansion and the implicit equation for $ E_0 $.
The paper can be found here: http://www.math.us.edu.pl/mtyran/dydaktyka/biomatematyka/griffith_1968_I.pdf
J. S. Griffith "Mathematics of Cellular Control Processes" J. Theoret. Biol. (1968)
Thanks in advance.
$$
\dot{X} = \dot{M} = \frac{1}{1+(E_0 + Y)^m } -a(M_0 +X) =
$$
$$
\frac{1}{1+E_0^m + mE_0^{m-1} Y + O(Y^2) } -aM_0 -aX =
$$
$$
\frac{1}{1+E_0^m} \left( 1-\frac{mE_0^{m-1} }{1+E_0^m } Y + O(Y^2) \right) -aM_0 -aX
$$
Now use $1/(1+E_0^m) =abE_0 $ as well as $aM_0 = abE_0 $, yielding
$$
\dot{X} =-a^2 b^2 m E_0^{m+1} Y +O(Y^2 ) -aX
$$
as desired.
Thanks! It looks correct. Can I just ask how you split the fraction and got the Y-terms from the denominator to the numerator?
From the second to the third line, I factored out $1/(1+E_0^m )$ and then expanded in a geometric series.
|
2025-03-21T14:48:31.583732
| 2020-07-23T00:46:16 |
366329
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Abdelmalek Abdesselam",
"LSpice",
"Rivers McForge",
"https://mathoverflow.net/users/155425",
"https://mathoverflow.net/users/2383",
"https://mathoverflow.net/users/24078",
"https://mathoverflow.net/users/7410",
"truebaran"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631366",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366329"
}
|
Stack Exchange
|
Computing the infinite dimensional Lebesgue measure of "cubes"
There is no Lebesgue measure in infinite dimensions—this slogan is familiar to every student interested in analysis. One possible, precise statement of this result may be as follows: if $X$ is an infinite dimensional Banach space there is no Borel, translation invariant measure with the property that each point has a neighborhood with finite measure. However people tried some constructions weakening some of the assumptions. One approach is described in Gill, Pantsulaia, and Zachary - Constructive analysis in infinitely many variables. The authors claim that they were able to construct an analogue of Lebesgue measure on the space of all sequences $\mathbb{R}^{\infty}$ (of course this space is not Banach). I wonder whether there is someone familiar with this paper: unfortunately I got lost (and I'm not quite sure whether the authors put everything correctly).
Sketch of the strategy: Denote $I_n=\prod_{k=n+1}^{\infty}[-\frac{1}{2},\frac{1}{2}]$: they use the symmetric closed interval of length $1$ to "stabilize" the construction: they split the space $\mathbb{R}^{\infty}$ into two disjoint pieces $X_1=\bigcup_{k=1}\mathbb{R}^{k}_I$ where $\mathbb{R}^k_I=\mathbb{R}^k \times I_{k}$ and $X_2=\mathbb{R}^{\infty} \setminus X_1$.
On $X_1$ they consider the product topology, on $X_2$ they consider discrete topology (sic!) and finally they consider the topology of disjoint union on $\mathbb{R}^{\infty}$. Then they define an analogue of Lebesgue measure, denoted by $\lambda_{\infty}$ in several steps.
First they define (on page 113) this measure for sets of the form $A \times I_1$ (where $A$ is Borel in $\mathbb{R}$ with finite measure) as $\lambda(A)$. I'm guessing that they proceed similarly for $A \times I_n$ where $A \subset \mathbb{R}^n$ (they didn't state it explicitly but seem to be using this definition in the proof of theorem 2.5). Note that it is possible that $A$ is of measure $0$.
Then they define this measure on the class $\Delta_0$ consisting of sets of the form $K \times I_n$ where $K$ is compact with nonzero, finite (which is automatic) measure, then for the class $\Delta$ of finite, (almost) disjoint sums of sets from $\Delta_0$.
After that they define $\lambda_{\infty}$ for open sets as $\lim_{N \to \infty} \sup \lambda_{\infty}(P_N)$ where $P_N \subset G$, $P_N \in \Delta$ (I guess that they meant just supremum together with convention that it is $0$ if there is no $P_N \subset G$: which occurs quite often since $X_2$ was discrete).
Then they define $\lambda_{\infty}$ for compact sets (as I understand this was not necessarily covered before since for example $[0,\frac{1}{2}]^{\infty}$ is compact but does not belong to $\Delta$).
Furthermore having defined this measure for compact and for open sets they define outer and inner measures as infimum over open sets/supremum over compact sets. The inner measure is defined only for sets with finite outer measure. They then call a set bounded measurable if these two measures coincide. Finally the set is called measurable it its intersection with any bounded measurable set is again bounded measurable-then $\lambda_{\infty}(A)$ is defined as the supremum of $\lambda_{\infty}(A \cap M)$ over all bounded measurable sets $M$.
Question How to understand this measure well enough to be able to compute measures of simple sets? For example I would be happy if I could compute $\lambda_{\infty}(J_1 \times J_2 \times J_3 \times \dotsb)$ where each $J_k$ is either an interval (could be closed, could be open) or the real line or a point or a half line.
There are plenty of possibilities: some of them seem understandable for me, some are not. Here are some examples:
$\lambda_{\infty}([a_1,b_1] \times \dots \times [a_n,b_n] \times I_n)=(b_1-a_1)\cdot ... \cdot (b_n-a_n)$—this set belongs to $\Delta_0$,
$\lambda_{\infty}([-\frac{1}{2},\frac{1}{2}]^{\infty})=1$—this set is not open but is compact and $(-\frac12-\delta,\frac12+\delta) \times I_1$ is open and its measure is $1+2\delta$ which could be arbitrarly close to $1$
$\lambda_{\infty}((-\frac{1}{2},\frac{1}{2})^{\infty})=1$—this set is not open but one can take $\delta_n \to 0^+$ which converges quickly enough to ensure that the open set $\prod_n (-\frac12-\delta_n,\frac12+\delta_n)$ has measure arbitrarily close to $1$: similarly one can take compact $\prod_n [-\frac12+\delta_n,\frac12-\delta_n]$ with measure arbitrary close to 1 (which by the way requires considering again open sets, this time containing this compact set)
$\lambda_{\infty} (\prod_n(-\frac12-a_n,\frac12+a_n))=\prod_n(1-2a_n)$ (here $a_n$ are positive)—this set is open, as a union of open sets $\prod_k^n(-\frac12-a_k,\frac12+a_k) \times I_n$.
Now let me list some sets for which I'm not sure how to proceed:
$A=[1,x]^{\infty}$ (where $x>1$)—this set is contained in $X_2$ therefore it is open but there is no $P_N \in \Delta$ such that $P_N \subset A$ so its measure should be $0$.
$A=\mathbb{R} \times \{0\} \times I_3$: on the one hand if $\lambda_{\infty}$ has to be consistent with finite dimensional Lebesgue measure then it should have measure $0$: on the other hand any open set containing $A$ would have infinite measure while any compact set contained in $A$ would have measure $0$—so we get into trouble with unbounded measurable sets.
$A=\mathbb{R} \times \{0\} \times \{0\} \times \dotsb$: intuition tells me that this set should have measure $0$. However tracking all the definition leads to the same trouble as in the example above.
I'm not sure whether this question is appropriate for this site for it asks about computing measure of very simple sets. However even if this measure should serve as an analogue of Lebesgue measure its definition is quite involved—so I'm looking for simple formulas.
EDIT: I've adressed the comment below. Let me also mention that the authors state that they are using the following convention: $0 \times \infty=0$ but $0 \times \infty^{\infty}=\infty$.
I did some proofreading, I think without changing any meaning; but, in the third example of an understandable set, at some point I got confused whether we were dealing with open sets of width $2a_n$, compact sets of width $2\delta_n$, or some mix and match of these. I didn't change what you had except to format it as a list, but I think there may be a typo there.
It seems like you could get a "measure" if you're willing to weaken $\sigma$-additivity to finite additivity, since the proof that there's no Lebesgue measure in infinite dimensions relies on $\sigma$-additivity.
Just in case the motivation is to understand path integrals in QFT, this type of construction is not useful for this particular purpose.
@Abdelmalek Abdesselam thank you for this remark-in fact this is what I've expected since I don't see much interest in those constructions
I didn't generalize and say the construction is not interesting. There could be other areas where this is interesting. I was just talking about QFT.
|
2025-03-21T14:48:31.584171
| 2020-07-23T01:39:19 |
366334
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Aleksandar Milivojević",
"Arun Debray",
"Denis Nardin",
"Fernando Muro",
"Joshua Mundinger",
"https://mathoverflow.net/users/104342",
"https://mathoverflow.net/users/12166",
"https://mathoverflow.net/users/125523",
"https://mathoverflow.net/users/43054",
"https://mathoverflow.net/users/97265"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631367",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366334"
}
|
Stack Exchange
|
Construction of $K(\operatorname{Gal}(\bar{k}/k), 1)$
Take any field $k$. Is there a construction of the Eilenberg-MacLane space $K(\operatorname{Gal}(\bar{k}/k), 1)$ as a CW complex in terms of $k$?
For any group G, there is a construction of $K(G, 1)$ as a CW complex: choose a presentation. Then, begin with a single 0-cell, attach a 1-cell for each generator, and glue 2-cells in along the relations.
You might be interested more in $BGal(\bar{k}/k)$, since $Gal(\bar{k}/k)$ is topological.
Can you explain a little better what kind of construction are you looking for? $K(\mathrm{Gal}(\bar k/k),1)$ is the étale homotopy type of $k$, but I'm not sure if that would be an answer to this question.
@ArunDebray you will generally have to add more cells to kill the higher homotopy groups
@AleksandarMilivojevic ah that's right; thank you!
You can construct the algebraic closure from $k$. Then the Galois group and then the classifying space as the Milnor construction. All this is master level at most.
|
2025-03-21T14:48:31.584298
| 2020-07-23T05:01:20 |
366339
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Mateusz Wasilewski",
"Weiqiang Yang",
"https://mathoverflow.net/users/161607",
"https://mathoverflow.net/users/24953"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631368",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366339"
}
|
Stack Exchange
|
Maximum eigenvalue of a covariance matrix of Brownian motion
$$ A := \begin{pmatrix}
1 & \frac{1}{2} & \frac{1}{3} & \cdots & \frac{1}{n}\\
\frac{1}{2} & \frac{1}{2} & \frac{1}{3} & \cdots & \frac{1}{n}\\
\frac{1}{3} & \frac{1}{3} & \frac{1}{3} & \cdots & \frac{1}{n}\\
\vdots & \vdots & \vdots & \ddots & \frac{1}{n}\\
\frac{1}{n} & \frac{1}{n} & \frac{1}{n} & \frac{1}{n} & \frac{1}{n}
\end{pmatrix}$$
How to prove that all the eigenvalues of $A$ are less than $3 + 2 \sqrt{2}$?
This question is similar to this one.
I have tried the Cholesky decomposition $A = L^{T} L$, where
$$L^{T} = \left(\begin{array}{ccccc}
1 & 0 & 0 & \cdots & 0\\
\frac{1}{2} & \frac{1}{2} & 0 & \cdots & 0\\
\frac{1}{3} & \frac{1}{3} & \frac{1}{3} & \cdots & 0\\
\vdots & \vdots & \vdots & \ddots & \vdots\\
\frac{1}{n} & \frac{1}{n} & \frac{1}{n} & \frac{1}{n} & \frac{1}{n}
\end{array}\right)$$
then
$$(L^{T})^{-1}=\left(\begin{array}{ccccc}
1 & & & \cdots\\
-1 & 2 & & \cdots\\
& -2 & 3 & \cdots\\
\vdots & \vdots & \vdots & \ddots\\
& & & -(n-1) & n
\end{array}\right)$$
$$A^{-1}=L^{-1}(L^{T})^{-1}$$
How to prove the eigenvalues of $A^{-1}$
$$\lambda_{i}\geq\frac{1}{3+2\sqrt{2}}$$
Further, I find that $A$ is the covariance matrix of Brownian motion at time $1, 1/2, 1/3, \ldots, 1/n$
It's positive definite, so all the eigenvalues are positive.
In this answer I show that the largest eigenvalue is bounded by $5< 3 + 2\sqrt{2}$. I will first use the interpretation of this matrix as the covariance matrix of the Brownian motion at times $(\frac{1}{n},\dots, 1)$ (I reversed the order so that the sequence of times is increasing, which is more natural for me).
We have $A_{ij} = \mathbb{E} (B_{t_{i}} B_{t_j})$. The largest eigenvalue will be the supremum over the unit ball of the expression $\langle x, A x\rangle$, which is equal to $\sum_{i,j} A_{ij} x_{i} x_{j}$. This is equal to $\mathbb{E} (\sum_{i=1}^{n} x_{i} B_{t_{i}})^2$. In order to exploit the independence of increments of the Brownian motion, we rewrite the sum $\sum_{i=1}^{n} x_i B_{t_{i}}$ as $\sum_{i=1}^{n} y_{i} (B_{t_{i}} - B_{t_{i-1}})$, where $y_{i}:= \sum_{k=i}^{n} x_{k}$ and $t_0:=0$. Thus we have
$
\mathbb{E} (\sum_{i=1}^{n} x_{i} B_{t_{i}})^2 = \sum_{i=1}^{n} y_{i}^2 (t_{i}-t_{i-1}).
$
The case $i=1$ is somewhat special and its contribution is $\frac{y_1^2}{n} \leqslant \sum_{k=1}^{n} x_{k}^2 = 1$. For the other ones we have $t_{i} - t_{i-1} = \frac{1}{(n-i+1)(n-i+2)}\leqslant \frac{1}{(n-i+1)^2}$. At this point, to get a nicer expression, I will reverse the order again by defining $z_{i}:= y_{n-i+1}$. So we want to estimate the expression
$
\sum_{i=1}^{n} \left(\frac{z_i}{i}\right)^2.
$
We can now use use Hardy's inequality to bound it by $4 \sum_{i=1}^{n} x_{i}^2 =4$. So in total we get 5 as an upper bound, if I haven't made any mistakes.
Brilliant. Thank you very much. But why $y_1$ is special? Can't we get 4 as upper bound?
The $i=1$ case is special because $t_1 - t_0 = \frac{1}{n}$, which is not quadratic like the other differences.
Thank you so much. I have spent much time trying to prove it through linear algebra, but failed. We get L through Cholesky decomposition, and (Lx)'(Lx) is hard to deal with.
Inspired by @Mateusz Wasilewski I find another method.
\begin{eqnarray*}
\langle x,Ax\rangle & = & \langle Lx,Lx\rangle\\
& = & \sum_{i=1}^{n}u_{i}^{2}
\end{eqnarray*}
where $u_{i}=\sum_{j=i}^{n}\frac{1}{j}x_{j}$.
\begin{eqnarray*}
\sum_{i=1}^{n}u_{i}^{2} & = & \sum_{i=1}^{n}(\sum_{k=i}^{n}b_{k})^{2}\quad(\text{where} \ b_{k}=\frac{1}{k}x_{k})\\
& = & \sum_{i=1}^{n}(\sum_{k=i}^{n}b_{k}^{2}+2\sum_{k>j\geq i}b_{k}b_{j})\\
& = & \sum_{k=1}^{n}\sum_{i=1}^{k}b_{k}^{2}+2\sum_{j=1}^{n-1}\sum_{k=j+1}^{n}\sum_{i=1}^{j}b_{k}b_{j}\\
& = & \sum_{k=1}^{n}\frac{x_{k}^{2}}{k}+2\sum_{j=1}^{n-1}\sum_{k=j+1}^{n}b_{k}x_{j}\\
& = & \sum_{k=1}^{n}\frac{x_{k}^{2}}{k}+2\sum_{k=2}^{n}\sum_{j=1}^{k-1}b_{k}x_{j}\\
& = & \sum_{k=1}^{n}\frac{x_{k}^{2}}{k}+2\sum_{k=2}^{n}b_{k}z_{k-1}\\
& = & x_1^2 +\sum_{k=2}^{n}\frac{(z_{k}-z_{k-1})^{2}}{k}+2\sum_{k=2}^{n}\frac{(z_{k}-z_{k-1})}{k}z_{k-1}\\
& = & x_{1}^{2}+\sum_{k=2}^{n}\frac{z_{k}^{2}-z_{k-1}^{2}}{k}\\
& = & \sum_{k=1}^{n}\frac{z_{k}^{2}}{k}-\sum_{k=1}^{n-1}\frac{z_{k}^{2}}{k+1}\\
& = & \sum_{k=1}^{n-1}z_{k}^{2}(\frac{1}{k}-\frac{1}{k+1})+\frac{z_{n}^{2}}{n}
\end{eqnarray*}
|
2025-03-21T14:48:31.584549
| 2020-07-23T06:29:10 |
366340
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Jérôme Poineau",
"curious math guy",
"efs",
"https://mathoverflow.net/users/109085",
"https://mathoverflow.net/users/152554",
"https://mathoverflow.net/users/4069"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631369",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366340"
}
|
Stack Exchange
|
Identity theorem in $p$-adic geometry/analysis
If one wants to do $p$-adic analysis and geometry, it is often bad so adapt "naively" complex analytic ideas, basically because $\mathbb{Q}_p$ is disconnected. The modern approach to this is, to my knowledge, the theory of rigid analytic spaces and Berkovich spaces. For instance in the theory of Berkovich spaces, the maximum modulus principle holds. I'm curious what we can say about the Identity theorem. Does it hold in any of the $p$-adic geometries?
I've been told to think of rigid analytic spaces as varieties and (Huber's-)adic spaces as schemes with Berkovich being closer to an "analytic" object, so probably I'd expect it (if at all) to hold there.
I am not sure what you mean about the identity theorem. Could you write down the statement you would like to see? Anyway, I am pretty sure it should be true, and if it holds for one of the theories, it certainly does for the others. The spaces may be presented a bit differently, but they are mostly equivalent.
I'm happy to do so. I choose the name based on what it is called on wikipedia (https://en.wikipedia.org/wiki/Identity_theorem). Let $D$ be some domain in $\mathbb{C}^n$ (i.e. open an connected). If two holomorphic functions $f$,$g$ on $D$ agree on a subset $S$, which has an accumulation point, then in fact they agree on $D$.
Thanks for the clarification. You probably mean $\mathbb{C}$ and not $\mathbb{C}^n$.
This also holds at the "more elementary" level of Krasner's theory of analytic elements.
Do you have a nice reference for this?
@curiousmathguy I suppose you are asking me for the reference (you can "call me" by writing my username after an @). The original references are the papers by Krasner in the Comptes Rendus (which you can find online at gallica). Also, you may look at Alain Robert's book on p-adic analysis.
Especially this one: "Prolongement analytique dans les corps valués complets: démonstration de la loi d'unicité: fonctions analytiques uniformes",
C. R. Acad. Sci. Paris 239 (1954), 745–747.
Let me assume that you have an analytic function $f$ defined on a closed one-dimensional unit disc, which is to say the spectrum of the Tate algebra $k\{T\}$. Then, Weierstrass preparation theorem tells you that $f$ may be written as a product of a polynomial $P$ and a nowhere vanishing function. If $P$ is non-zero, it has only finitely many zeroes, and so does $f$.
As you see, it can really be turned into a statement about Tate algebras, so the precise theory you want to work with does not really matter.
Please, let me know if you had a more general context in mind.
Thanks you Jérome! This is exactly what I hoped for, thanks for showing this to me!
|
2025-03-21T14:48:31.584756
| 2020-07-23T07:20:42 |
366342
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Aleksei Kulikov",
"Carlo Beenakker",
"https://mathoverflow.net/users/104330",
"https://mathoverflow.net/users/11260"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631370",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366342"
}
|
Stack Exchange
|
Small-$r$ asymptotics of an integral of $1/\log ^\alpha r$
Let $f(r)=\frac{1}{(-log r)^{\alpha}}$ and let $\xi_r$ be the unique value in $(0,r)$ such that$f(\xi_r)=\frac{1}{r}\int_{0}^rf(t)dt$, where $\alpha\in (0,1)$ and $r>0$. My question is about the order of $\xi_r$?
Can we have $\xi_r=o(r)$ as $r$ tends to $0$? moreover, can we get $\xi_r=o(\frac{r}{log^2(1/r)})$?
Is there any reference about this question?
Starting from the integral (in terms of an incomplete Gamma function)
$$g(r)=\frac{1}{r}\int_0^r (-\ln t)^{-\alpha}\,dt=\frac{1}{r}\Gamma(1-\alpha,-\ln r),$$
we expand for $r\rightarrow 0$, or $y\equiv -\ln r\rightarrow\infty$,
$$g(r)=y^{-1-\alpha}\bigl(y-\alpha+{\cal O}(1/y)\bigr).$$
We then wish to solve $g(r)=(-\ln \xi)^{-\alpha}$ for $\xi$, or for $z\equiv-\ln\xi$,
$$\Rightarrow z=y^{1/\alpha+1}\bigl(y-\alpha+{\cal O}(1/y)\bigr)^{-1/\alpha}=y+1+{\cal O}(1/y).$$
The resulting small-$r$ asymptotics is
$$\xi=\frac{r}{e}\bigl(1+{\cal O}(1/\ln r)\bigr).$$
In the end I believe it should be $y^{1/a+1}$ which gives $\xi \sim r$, disproving OP’s conjecture. Also $\xi \to 1$ doesn’t make sense from the qualitative considerations — $g(r)$ tends to zero while $f(1)=\infty$.
right, minus sign corrected, thanks @AlekseiKulikov
|
2025-03-21T14:48:31.584859
| 2020-07-23T07:32:01 |
366343
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Joshua Mundinger",
"YCor",
"https://mathoverflow.net/users/100140",
"https://mathoverflow.net/users/125523",
"https://mathoverflow.net/users/14094",
"https://mathoverflow.net/users/41291",
"solver6",
"მამუკა ჯიბლაძე"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631371",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366343"
}
|
Stack Exchange
|
Does $\mathcal{A}\otimes\mathbb{C}(t)\cong\mathcal{D}\otimes\mathbb{C}(t)$ imply an isomorphism of Lie algebras?
Assume that $\mathcal{A}$ is a Lie $\mathbb{C}$-algebra. Also denote the Lie $\mathbb{C}$-algebra $\mathcal{D}=\mathbb{C}[x_1,\ldots, x_n]\partial_{x_1}\oplus\ldots\oplus\mathbb{C}[x_1,\ldots, x_n]\partial_{x_n}$. My questions are as follows
Let $\mathcal{A}\otimes_{\mathbb{C}}\mathbb{C}(t)$, $\mathcal{D}\otimes_{\mathbb{C}}\mathbb{C}(t)$ are isomorphic $\mathbb{C}(t)$-Lie algebras.
Is it true that $\mathcal{A}, \mathcal{D}$ are isomorphic $\mathbb{C}$-Lie algebras?
Is it true that $\mathcal{A}\otimes_{\mathbb{C}}\mathbb{C}[t], \mathcal{D}\otimes_{\mathbb{C}}\mathbb{C}[t]$ are isomorphic $\mathbb{C}[t]$-Lie algebras?
It's not tensor product of Lie algebras, it's extension of scalars of Lie algebras. Also I assume that the isomorphism in the assumption is an isomorphism of Lie algebras over $C(t)$?
Yes, you are right
Put $\mathcal{A}'=\mathcal{A}\otimes_{\mathbb{C}}\mathbb{C}(t)$ and similarly for $\mathcal{D}'$. Choose an isomorphism $f\colon\mathcal{D}'\to\mathcal{A'}$.
Choose a (countable) basis $\mathcal{D}_0=\{d_i:i\in\mathbb{N}\}$ for $\mathcal{D}$ over $\mathbb{C}$. Then $f(\mathcal{D}_0)$ is a basis for $\mathcal{A}'$ over $\mathbb{C}(t)$ but $\dim_{\mathbb{C}}(\mathcal{A}')=\dim_{\mathbb{C}}(\mathcal{A})$ so we can also choose a countable basis $\mathcal{A}_0=\{a_i:i\in\mathbb{N}\}$ for $\mathcal{A}$ over $\mathcal{C}$. We must have $[d_i,d_j]=\sum_kp_{ijk}d_k$ and $[a_i,a_j]=\sum_kq_{ijk}a_k$ for some structure constants $p_{ijk},q_{ijk}\in\mathbb{C}$. Now let $K$ be a countable subfield of $\mathbb{C}$ containing all these structure constants, and also all the constants needed to ensure that $f(\mathcal{A}_0)\subseteq K(t).\mathcal{D}_0$ and $f^{-1}(\mathcal{D}_0)\subseteq K(t).\mathcal{A}_0$. Put $\mathcal{A}_1=K.\mathcal{A}_0$ and $\mathcal{D}_1=K.\mathcal{D}_0$, so these are Lie algebras over $K$ that become isomorphic over $K(t)$. If $\mathbb{C}$ were algebraic over $K$ then it would be countable, which is false. Thus, we can choose an embedding $i\colon K(t)\to\mathbb{C}$ extending the identity on $K$. By applying this to the coefficients of $f$, we obtain an isomorphism $\mathcal{D}\simeq\mathcal{A}$.
Thank you. I don't have enough knowledge about structure constants, so right now I cannot check your proof completely. From the answer we have that 1), 2) are true, but what about 3)?
If by structure constants you mean coordinates of commutators of elements from a countable basis then it looks as a correct proof for me. I need to think more about it to mark as accepted
So you proved the more general fact that if $\mathcal{A}\otimes_{\mathbb{C}}\mathbb{C}(t)$ is isomorphic as $\mathbb{C}(t)$-Lie algebra to $\mathcal{B}\otimes_{\mathbb{C}}\mathbb{C}(t)$ and $\dim_{\mathbb{C}}\mathcal{A}=\dim_{\mathbb{C}}\mathcal{B}$-countable then $\mathcal{A}$ is isomorphic to $\mathcal{B}$ as $\mathbb{C}$-Lie algebra. Is there any known reference to this fact existed in literature?
I wonder why a similar argument would not work with $\mathbb C((t))$...
@მამუკაჯიბლაძე the $t$-adic topology on $K((t))$ doesn't play well with embedding into $\mathbb C$.
|
2025-03-21T14:48:31.585064
| 2020-07-23T07:51:06 |
366345
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Carlo Beenakker",
"François Brunault",
"Geoff Robinson",
"Johnny T.",
"https://mathoverflow.net/users/11260",
"https://mathoverflow.net/users/14450",
"https://mathoverflow.net/users/6506",
"https://mathoverflow.net/users/84272"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631372",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366345"
}
|
Stack Exchange
|
When does $\det \begin{pmatrix} A & X \\ X^T & A \end{pmatrix} = (\det A)^2 + (\det X)^2$?
Let $A$ be an $n \times n$ real symmetric matrix.
Let
$$
M = \begin{pmatrix} A & X \\ X^T & A \end{pmatrix}
$$
where $X$ is a real invertible $n \times n$ matrix. I am interested in finding $X$ such that
$$
\det M = (\det A)^2 + (\det X)^2.
$$
Given $A$, is there a systematic way of finding such $X$ under some hypotheses?
Any suggestions, comments or reference for a related topic is very much appreciated. Thank you!
This case should be amenable to a brute-force search.
for $A=0$ no nonzero $X$ exists that satisfies the equation, while $X=0$ satisfies it for any $A$; are you sure you don't want a minus sign instead of a plus sign in your equation?
Just a small remark: It suffices to consider the case that $A$ is diagonal, since $A$ may be daigonalized via an orthogonal matrix $U$, and then we can replace $M$ by $VMV^{T}$, where $V$ is the "direct sum" of two copies of $U$.
$$B=\left[\matrix{1&1&1&1\\1&1&-1&-1\\-1&-1&-1&1\\-1&1&-1&1}\right]$$
and probably many other solutions. I'm also voting to close because you didn't pose a research-level problem. If you have an interesting general case, pose that.
Thank you for the example, let me change the problem.
|
2025-03-21T14:48:31.585180
| 2020-07-23T08:45:47 |
366350
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Patrick Elliott",
"https://mathoverflow.net/users/121425"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631373",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366350"
}
|
Stack Exchange
|
When does cohomology of a pro-algebraic group commute with filtered colimits of coefficients?
Let $G$ be a pro-algebraic group, that is, a projective limit of algebraic groups $G_i$. Let $V_i$ be an inductive system of finite dimensional rational $G_i$-representations, so that the colimit $V$ is a rational $G$ representation.
What conditions are required on $G$ to ensure that
$$
H^n(G,V)\simeq \mathrm{colim}_iH^n(G, V_i)?
$$
I am specifically interested in the case where $G$ is the pro-algebraic completion of a discrete group $\Gamma$. In this case, what conditions are required of $\Gamma$ for the above isomorphism to hold?
For maybe a vacuous comment, I know groups of type (F) have this property for their pro-algebraic completions, but (and maybe this is the only definition) the definition I know has this property built in. I guess I really want to dig into what this phenomenon MEANS.
This isomorphism always holds, and no conditions are needed. Here I presume that the pro-algebraic group $G$ is pro-affine (as the context of the question seems to suggest).
Let $G$ be a pro-affine pro-algebraic group over a field $k$. Denote by $C=O(G)$ the ring of regular functions on $G$. What is important for us is not the ring but the coalgebra structure on $C$, which is induced by the group structure on $G$. So $C$ is a coassociative, counital, not necessarily cocommutative coalgebra over $k$.
A (rational) representation of $G$ over $k$ is the same thing as a $C$-comodule. (It does not matter whether to consider the left or right $C$-comodules, as the antipode map induced by the inverse element map $g\mapsto g^{-1}\colon G\longrightarrow G$ provides an isomorphism between $C$ and its opposite coalgebra.)
Generally, let $C$ be a coassociative, counital coalgebra over aa field $k$, and let $M$ be a left $C$-comodule. The $k$-vector spaces of cohomology of $C$ with the coefficients in $M$, defined as the $\operatorname{Ext}$ spaces $H^*(C,M)=\operatorname{Ext}_C^*(k,M)$ taken in the category of left $C$-comodules ($=$ discrete/rational modules over the $k$-algebra $C^*$ dual to $C$), are computable as the cohomology spaces of the cobar-complex
$$
M\longrightarrow C\otimes_kM\longrightarrow C\otimes_kC\otimes_kM\longrightarrow C\otimes_kC\otimes_kC\otimes_kM\longrightarrow\dotsb
$$
As it is clear from the form of this complex, its cohomology, viewed as a functor of $M$, commutes with the (filtered) direct limits. So for any (directed) inductive system of left $C$-comodules $M_i$, we have $H^*(C,\varinjlim_i M_i)=\varinjlim_iH^*(C,M)$.
Returning to the case of a pro-affine proalgebraic group $G$ and its coalgebra $C=O(G)$, the cohomology of $G$ with the coefficients in a (rational) $G$-module $V$ is the same thing as the cohomology of the coalgebra $C$ with the coefficients in the $C$-comodule $V$, that is $H^*(G,V)=H^*(C,V)$. Thus the functor $H^*(G,{-})$ preserves filtered direct limits.
I'm a little confused now. In this paper (https://arxiv.org/pdf/math/0503418.pdf) of Toen, Pantev, and Katzarkov it is implied on p.50 that in the case of a pro-algebraic completion of a discrete group G, we need added hypotheses to get this result. Are they just being overly cautious?
|
2025-03-21T14:48:31.585384
| 2020-07-23T09:04:37 |
366353
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Jorge E. Cardona",
"Robert Furber",
"https://mathoverflow.net/users/14870",
"https://mathoverflow.net/users/61785"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631374",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366353"
}
|
Stack Exchange
|
Bounded weak and weak-$\star$ topologies and metrics
Let $X$ be a reflexive and separable Banach space. Let $(h_n)$ be a sequence dense in $\overline{B^*_1}$ (the closed ball in $X^*$ of radius $1$). Set
$$
d(x,y) := \sum_{n=1}^\infty 2^{-n} |(x-y, h_n)|,
$$
where $(x-y, h_n)=(x-y, h_n)_{X,X^*}$ is the dual pairing on $X$.
Then $d$ is symmetric, satisfies the triangle inequality and $d(x,y) = 0 \implies x = y$, hence $d$ is a metric.
On closed bounded (by the norm) balls of $X$, the metric $d$ induces the weak topology.
Can the topology induced by $d$ be described in a functional way in relation to the weak topology or the bounded weak-topology? Would any answer change if $d$ is defined using the ratio $\frac{|(x-y,h_n)|}{1 + |(x-y,h_n)|}$ ?
This seems to be something that should be in Megginson's book on Banach spaces, I can only see something related in exercise 2.86-88 of section 2.7.
For $E$ a Banach space, the bounded weak-$\ast$ topology on $E^$ is not metrizable unless $E$ is finite-dimensional. The reason is that a base of neighbourhoods of $0$ in $E^$ is defined by the polars of the norm-compact absolutely convex subsets of $E$. If the bounded weak-$\ast$ topology is first-countable, then this implies that there are countably many norm-compact absolutely convex sets covering $E$. By the Baire category theorem, this means one of them has nonempty interior, so $E$ is locally compact and therefore finite-dimensional.
I've not read the book, but I bet Megginson is using "complete" in the sense of uniform spaces, not metric spaces (because that makes it a true statement). If $E$ is a separable infinite-dimensional Banach space, then both the weak-$\ast$ and the bounded weak-$\ast$ topology restrict to the same, metrizable, topology on any bounded subset of $E^*$, but are not metrizable themselves.
@RobertFurber Megginson has in Definition 2.3.1 the following: "A topological space is topologically complete if some complete metric induces its topology."
@RobertFurber what exactly do you mean by "the polars of the norm-compact absolutely convex subsets of $E^\star$"?
I commented because I don't have time to write a proper answer at the moment, unfortunately. I will see if I can look at what the Megginson book says (I take it that it is An Introduction to Banach Space Theory?). The polar of a subset $S \subseteq E$ is the set of elements of $E^*$ that are bounded above by $1$ on $S$ (some authors define it slightly differently).
Proving that the bounded weak-$\ast$ topology is not completely metrizable is easier, actually. By Banach-Alaoğlu the closed ball of radius $n$ in $E^$ is compact, $E^$ is the union of these closed balls, so by the Baire category theorem if $E^$ is completely metrizable in the bounded weak-$\ast$ topology then one of these balls has nonempty interior, so $E^$ is locally compact and therefore finite-dimensional.
Megginson does not say that the bounded weak-$\ast$ topology on $E^$ is topologically complete, he says it is complete (meaning in the sense of uniform spaces, see Definition 2.1.48). This is true (so I win the bet I made earlier). As I have mentioned, the bounded weak-$\ast$ topology it is not metrizable unless $E^$, and therefore $E$, is finite-dimensional.
@RobertFurber Thank you for helping to clarify this! You are right, "complete" is different to "topologically complete", it now makes sense. I removed the reference to completeness from the question. I still have my original question though, how does the topology induced by $d$ looks in relation to the weak, bounded weak and norm topologies?
|
2025-03-21T14:48:31.585733
| 2020-07-23T09:51:11 |
366358
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"David Hansen",
"François Brunault",
"curious math guy",
"https://mathoverflow.net/users/1464",
"https://mathoverflow.net/users/152554",
"https://mathoverflow.net/users/6506"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631375",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366358"
}
|
Stack Exchange
|
Is semi-simplicity of Galois representations local?
Let $\rho:G_{\mathbb{Q}}\rightarrow \text{Gl}(V)$ be a finite dimensional $\ell$-adic Galois representation. Then for each prime, by pre-composing $\rho$ with the natural inclusion $G_{\mathbb{Q}_p}\rightarrow G_{\mathbb{Q}}$, we get a representation of the local Galois group. I'm curious about how the property of being semisimple behaves under localization, that is to say if $\rho$ is semisimple, can we conclude that $\rho\vert_{G_{\mathbb{Q}_p}}$ is semisimple?
Conversely, if for all $p\neq \ell$ (or maybe all primes) $\rho_{G_{\mathbb{Q}_p}}$ is semisimple, does it follow that $\rho$ is semisimple? If I'm not mistaken, all the $\rho_{G_{\mathbb{Q}_p}}$ being semisimple implies that $\rho\vert_{\bigcup_p G_{\mathbb{Q}_p}}$ is semistable, and since ${\bigcup_p G_{\mathbb{Q}_p}}$ is of finite index in $G_\mathbb{Q}$ (I think), this would mean $\rho$ is semisimple.
If $E/\mathbf{Q}$ is a non-CM elliptic curve and $\ell>3$ is a prime where $E$ has good ordinary reduction, then $V_\ell E$ is a semisimple $G_\mathbf{Q}$-representation but it is not semisimple as a $G_{\mathbf{Q}_\ell}$-representation.
If a newform $f$ is ordinary at $p$ then the local Galois representation at $p$ is upper-triangular (Deligne, Mazur-Wiles) but usually it does not split. See the papers of Eknath Ghate, e.g. On the local behaviour of ordinary modular Galois representations
@DavidHansen Out of curiosity, if $\ell$ is a prime of bad reduction, is it clear what to expect?
|
2025-03-21T14:48:31.585857
| 2020-07-23T10:11:26 |
366362
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Gerald Edgar",
"Nate Eldredge",
"Titouan Vayer",
"https://mathoverflow.net/users/151188",
"https://mathoverflow.net/users/33741",
"https://mathoverflow.net/users/454",
"https://mathoverflow.net/users/4832",
"leo monsaingeon"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631376",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366362"
}
|
Stack Exchange
|
Regularity with respect to the Lebesgue measure through dimensions
Let us consider two probability measures $\mu \in \mathcal{P}(\mathbb{R}^{p})$ and $\nu \in \mathcal{P}(\mathbb{R}^{q})$ with $p,q \in \mathbb{N}^{*}$. We note $\#$ the push forward operator i.e for $f: \mathbb{R}^{p} \rightarrow \mathbb{R}^{q}$ measurable $f\#\mu \in \mathcal{P}(\mathbb{R}^{q})$ defined by
$f\#\mu(A)=\mu(f^{-1}(A))=\mu(\{x \in \mathbb{R}^{p} | f(x) \in A\})$ for $A$ measurable set of $\mathbb{R}^{q}$.
Let $l: \mathbb{R}^{p} \rightarrow \mathbb{R}^{q}$ be a linear application and assume that $\mu$ has a density $g$ with respect to the Lebesgue measure on $\mathbb{R}^{p}$.
My question is:
If $q <p$ (we go from higher to lower dimension) does $l\#\mu$ has a density with respect to the Lebesgue measure on $\mathbb{R}^{q}$ ?
I think this is true using the coarea formula. More precisely if $L\in \mathbb{R}^{q\times p}$ the matrix associated to $l$ and $J_L=\sqrt{det(LL^{T})}$ the density of $l\#\mu$ looks like:
\begin{equation}
h(y)=\int_{l^{-1}(y)} \frac{g(x)}{J_L} dV_{l^{-1}(y)}(x)
\end{equation}
where denotes $dV_{l^{-1}(y)}(x)$ the volume element. Is it correct ?
What could go wrong if $q>p$ (we go from lower to higher dimension) ?
Your argument is fine but implicitly requires that $l$ is surjective, otherwise $J_L = 0$. If $l$ is not surjective then it is clear that $l # \mu$ does not have a density because it gives measure 1 to the image of $l$, which is a proper linear subspace of $\mathbb{R}^q$ and therefore has Lebesgue measure zero. In particular this is the case for every $l$ when $q>p$.
Building up on Nate Eldredge's comment: in the worst case scenario that your map $l$ is actually trivially zero, you see that $l|\mu=\delta_{0_{\mathbb R^q}}$ is as far as possible from being absolutely continuous! For the positive answer, one way to put it is that if $\mu\ll dx_p$ then automatically $l#\mu\ll l#(dx_p)$. Now if $l$ is surjective it is an easy exercise to check that $l#(dx_p)\ll dx_q$ (and this is actually an if and only if), hence in particular $l#\mu\ll l#(dx_p)\ll dx_q$.
$\LaTeX$ comment... $$f#\mu\text{ or }f^\sharp\mu\text{ ??}$$
Thank you very much !
|
2025-03-21T14:48:31.586283
| 2020-07-23T12:34:15 |
366371
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Fedor Pakhomov",
"Mauro ALLEGRANZA",
"https://mathoverflow.net/users/16302",
"https://mathoverflow.net/users/36385",
"https://mathoverflow.net/users/42676",
"jeq"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631377",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/366371"
}
|
Stack Exchange
|
Reference request on Gentzen's proof of the consistency of PA
I've always been interested in having a good understanding of Gentzen's proof of the consistency of arithmetic.
Being more precise, he showed that $PRA + WF(\epsilon_0) \vdash Con(PA)$.
I am interested in an exposition of his work that
1) Is transparent on which parts of the consistency proof uses the well-foundness of $\epsilon_0$.
2) Discuss (in mathematical and philosphical way) to what extent this proof deviate from the original Hilbert's Program, and in what extent it fits in a nice way to a more flexible formulation of this programme.
I should also add:
0) Is technically as simple as possible.
Item 0) is because I think this question might be of interest to the curious general mathematician.
Not mainstream, but Chapter 3 of my 1980 thesis https://arxiv.org/abs/1904.10540 gives an 8-page exposition of Ackermann, Dreben, Denton, and Scanlon's proof[s], recast using Kripke's notion of fulfillability.
@jeq Nice to learn that you thesis is now available on-line. Just recently I have mentioned it in my answer to this question https://mathoverflow.net/questions/365425/the-inconsistency-of-graham-arithmetics-plus-forall-n-n-g-64 .
Maybe useful: Anna Horská, Where is the Gödel-point hiding: Gentzen's Consistency Proof of 1936 (Springer, 2014)
Pohlers's 1989 book 'Proof Theory, An Introduction' gives a very clean, streamlined approach (based on work by Tait.)
Takeuti's presentation in his 'Proof Theory' is closer to Gentzen's original proof, but is much less readable than Pohlers.
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.