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
|
|---|---|---|---|---|---|
2025-03-21T14:48:31.999014
| 2020-09-10T12:26:26 |
371333
|
{
"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:632879",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371333"
}
|
Stack Exchange
|
Scaling behavior of Wasserstein distances
Let $p>1$ and $\mu\neq \nu$ be two probability measures on $\Omega\subset \mathbb{R}^d$ a bounded set. For $\alpha \geq 0$, we let $$C_\alpha(\mu,\nu) = \inf_\sigma \frac{W_p(\mu+\sigma,\nu+\sigma)}{W_p(\mu,\nu)},$$
where $W_p$ is the $p$-Wasserstein distance and the infimum is taken over all nonnegative measures on $\Omega$ of mass $\alpha$. By using the compactness of such a set, it can easily be shown that $C_\alpha(\mu,\nu)>0$. Is it true that $C_\alpha=\inf_{\mu\neq \nu} C_\alpha(\mu,\nu) >0$ ? (here the infimum is taken over $\mu,\nu$ probability measures on $\Omega$). This result looks intuitive, but I could not find a proof of it in the litterature, nor could I prove it.
If $(\mu_n),(\nu_n)$ are sequences attaining the infimum, and if $C_\alpha=0$, then we may without loss of generality assume that $\mu_n$ and $\nu_n$ converge to the same limit measure $\mu$, and therefore I suspect that whether $C_\alpha=0$ or not is related to some fine properties of $W_p(\mu,\nu)$ for $\mu$, $\nu$ very close.
Note that this problem is related to Remark 2.4 in A description of transport cost for signed measures - Edoardo Mainini.
Actually, we have $C_\alpha =0$ for any $\alpha>0$. Indeed, let $\mu_t = (1-t)\delta_{x_0} + t\delta_0$ and $\nu_t = (1-t)\delta_{x_0} + t\delta_1$ for some $t\in (0,1)$ and $x_0$ far away enough from $0$ and $1$. Then $W_p^p(\mu_t,\nu_t)= t$, whereas if $\sigma = t\sum_{k=1}^{n} \delta_{k/(n+1)}$, then one can check that $W_p^p(\mu_t + \sigma,\nu_t + \sigma) \leq \frac{t }{(1+n)^{p-1}}$. By letting $t_n = \alpha/(n+1)$, we obtain that $C_\alpha(\mu_{t_n},\nu_{t_n})^p \leq (1+n)^{1-p}$. By letting $n$ goes to $\infty$, we see that $C_\alpha= \inf C_\alpha(\mu,\nu)=0$.
|
2025-03-21T14:48:31.999144
| 2020-09-10T12:45:24 |
371335
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"LSpice",
"https://mathoverflow.net/users/14443",
"https://mathoverflow.net/users/2383",
"kneidell"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632880",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371335"
}
|
Stack Exchange
|
Centralizers of $\mathbb{F}_q$-rational semisimple elements of a finite group of Lie type
Let $\mathbb{G}$ be a connected reductive $\mathbb{F}_q$ algebraic group over its algebraic closure $\bar{\mathbb{F}_q}$, and $\mathbb{T}$ be an $\mathbb{F}_q$-defined maximal torus. Let $\Phi$ be the root system of $\mathbb{G}$ wrt $\mathbb{T}$, and, given $g\in\mathbb{T}$, put $$\Phi(g)=\lbrace\alpha\in \Phi:\alpha(g)=1\rbrace.$$
There is a very clean criterion by Deriziotis for which closed subsystems $\Sigma\le \Phi$ can occur as $\Phi(g)$ for some $g\in \mathbb{T}$; namely, these are precisely the subsystems $\Sigma\le \Phi$ which admit a basis which is a subset of the set of affine simple roots of $\Phi$, and all such subsystems occur for some $g\in \mathbb{T}$. These are often called pseudo-Levi subsystems. Furthermore, if $g\in\mathbb{T}(\mathbb{F}_q)$, then $\Sigma$ is stable under the action of the Frobenius map associated with the $\mathbb{F}_q$-structure on $\mathbb{G}$.
Question Given a subsystem $\Sigma\le \Phi$ as in the last paragraph, stable under the Frobenius map, does there exist $g\in \mathbb{T}(\mathbb{F}_q)$ for which $\Sigma=\Phi(g)$? Do there exist counterexamples for this?
What I know by now: If $\Sigma=\Phi(g)$ is a Levi subsystem, meaning that it has a basis of simple elements of $\Phi$ then one can always take $g$ to be $\mathbb{F}_q$-rational. To show this, one can compute the dimension of the subgroups $$\mathbb{T}_{\Sigma'}=\bigcap_{\alpha\in \Sigma'}\ker(\alpha)\le \mathbb{T}$$
for all $\Sigma\le \Sigma'\le \Phi$, and verify that
$$\dim\mathbb{T}_{\Sigma'}\le \mathrm{rk}(\Phi)-\mathrm{rk}(\Sigma),$$
with equality iff $\Sigma'=\Sigma$, and, consequently, deduce that $\mathbb{T}_{\Sigma}^\circ\setminus(\bigcup_{\Sigma<\Sigma'}\mathbb{T}_{\Sigma'})$ is irreducible of dimension $\mathrm{rk}(\Phi)-\mathrm{rk}(\Sigma)$, and therefore admits an $\mathbb{F}_q$-rational point.
In the more general case, where $\Sigma$ is merely pseudo-Levi, this argument fails more-or-less completely. However, in all cases I have computed thus far it seems that one can find elements $g\in\mathbb{T}$ with $\Phi(g)=\Sigma$ whose representing matrices only have the entries $0,1$ and $-1$... I wonder if maybe there is a simpler argument that my dimension computation above overshoots.
The quantifiers aren't entirely clear for your question—you want to fix $\mathbb T$, right? Also, do you really mean to switch from $\mathbb F_q$ to $\mathbb F_p$? (It's unclear to me, because you refer to "a $\mathbb F_q$-group defined over $\bar{\mathbb F}_q$"—I'm not sure what to make of that—so I'm not sure how one even would speak of $\mathbb F_p$-rational points; and then your last paragraph starts off talking about $\mathbb F_p$-rationality, and then ends with $\mathbb F_q$-rationality.)
Oh sorry, there are several mistakes here.. I'll edit
However, I do want to fix $\mathbb{T}$ in this question
I still think there are some mistakes: what does "a connected reductive $\mathbb F_q$ algebraic group over its algebraic closure $\bar{\mathbb F}_q$" mean? I think you just mean "a connected reductive algebraic group over $\mathbb F_q$". Also, I think you mean to say that Deriziotis (not Deriztiois) characterised the groups that can occur as $\Phi(g)$ for some $g \in \mathbb T(\bar{\mathbb F}_q)$ (not for some $g \in \mathbb G$).
With no conditions on $q$, you will run into problems. For example, for $\mathbb G = \operatorname{SL}_2$ with $q = 2$, the only rational element of the diagonal torus is the identity element, so you cannot realise $\Sigma = \emptyset$. Would you be interested in a "large-$q$-only" condition? (I don't know if it holds—it might even be a "large-$p$-only" condition—but such results are more likely when dealing with rationality than uniform ones.)
Yes, I'm definitely okay with large p only conditions.
As @LSpice already pointed out, you need $q$ to be sufficiently large even in the case of a Levi subgroup. Just take $G = \operatorname{GL}_n(\overline{\mathbb{F}}_q)$ and $G^F = \operatorname{GL}_n(\mathbb{F}_q)$ under the usual Frobenius endomorphism. If $T \leqslant G$ is the maximal torus of diagonal matrices then $(\mathsf{C}_{q-1})^n \cong T^F = C_G(s)$ for some semisimple element $s \in G^F$ if and only if $q-1 \geqslant n$. That is because $s$ needs $n$ distinct eigenvalues to be regular.
Deciding the exact conditions for your subgroup to be realisable as the centraliser of a rational semisimple element involves a detailed case by case analysis. For the exceptional groups Frank Lübeck's GAP calculations Centralizers and numbers of semisimple classes in exceptional groups of Lie type are invaluable here. But if you're just interested in a "$q$ sufficiently large" statement then this was obtained by Deriziotis's advisor R. W. Carter in Corollary 20 of the following paper:
"Centralizers of semisimple elements in finite groups of Lie type", Proc. London Math. Soc. (3), vol. 37, (1978), 491–507.
It is essentially a counting argument. It's pointed out explicitly in Theorem 21 that a Levi subgroup is always the centraliser of a rational semisimple element assuming $q$ is sufficiently large.
Another paper that streamlines things here is the following paper of Bonnafé:
"Quasi-isolated elements in reductive groups", Comm. Algebra (7), vol. 33, (2005), 2315–2337.
This paper makes parts of Carter's and Deriziotis' work much clearer. In particular, Bonnafé gives a clean construction for elements whose centraliser is not contained in any proper parabolic subgroup.
|
2025-03-21T14:48:31.999779
| 2020-09-10T13:07:10 |
371339
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Alex M.",
"Hans-Peter Stricker",
"JimN",
"https://mathoverflow.net/users/2672",
"https://mathoverflow.net/users/54780",
"https://mathoverflow.net/users/62043"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632881",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371339"
}
|
Stack Exchange
|
The distance distribution of graphs
The degree distribution of a graph is of main importance, especially for large graphs, and namely random graphs. Its expected value and its higher moments tell a lot about a graph – but of course not everything.
The distance distribution $D(k)$ of a graph is possibly of equal importance. First determine for each vertex $i$ of the graph its distance distribution $d_i(k)$ giving the numbers of $k$-order neighbours ($1$-order neighbours being the immediate neighbours). Then calculate the normalized sums
$$D(k) = \frac{1}{N}\sum_{i=1}^N d_i(k)$$
The expected distance between two nodes is the expected value of this distribution. But the distribution has higher moments which may tell something interesting about the graph.
My impression is that the so-defined distance distribution (and no other definition comes to my mind) is roughly as informative as the degree distribution (even though in other respects), but in the literature it is significantly underrepresented:
Googling for "degree distribution" "graph theory" yields 180,000 results.
Googling for "distance distribution" "graph theory" yields only 10,000 results.
There is an extended article on degree distributions on Wikipedia.
There is no article on distance distributions on Wikipedia.
How can I understand this?
In the following examples one can see that for regular graphs like Euclidean tilings (examples 1 to 4) the distance distributions look very much the same (even more than the degree distributions), but for the hyperbolic tiling $(3.4)^3$ and an Erdös-Renyi graph, the distance distribution looks qualitatively different.
The shades of green indicate node degrees.
For some random graph distributions generated by the configuration model (625 nodes, mean degree 4). A power-law distribution of node degree is to come.
can you generate those diagrams from the random graph distributions which have been studied specifically for their degree distribution: random k-trees and Barabasi-Albert models, maybe also the small-world model of Watts-Strogatz?
@JimN: I added diagrams for a random $4$-regular graph. A Watts-Strogatz graph and a power-law graph (Barabasi-Albert) is to come.
random k-trees share the expected/desirable power law distribution in node degrees, like Barabasi-Albert, but B-A lacks higher-order statistics like clique/clustering distributions that k-trees also exhibit. And there are other higher-order distributions that k-trees have which match real-world distributions (like edge embeddedness and overlapping clusters) which other models have.
To make a random partial k-tree:start with a large clique,store all cliques of size k, choose one uniformly at random and make a new node adj to that chosen k-clique.This creates a clique of size (k+1), so you store more new k-cliques. Now delete some small number of edges(this destroys a number of k-cliques). Again, add a new vertex adj to a uniformly-chosen existing k-clique and repeat. I have a paper on higher-order structures in random models where we tried to argue the importance of going beyond degree distribution: https://tinyurl.com/yxna36n7 , and (partial) k-trees had the desirables
I'll do my best.
good luck! To address your question, clearly the distance distribution provides more information than a degree distribution, but can you find examples of when the distance distribution explains any phenomenon in real networks that the degree-distribution is insufficient to detect or predict? The distance distribution may provide more info, but at the cost of a more complex measure. There needs to be incentive to use a complex measure when a degree distribution would do.
My first and main example: The spreading of contagious diseases depends of course on the degree distribution of the underlying contact network, but maybe even more on the distance distribution (which in turn correlates with the degree distribution).
Are you referring to your first picture? Yes, that shows there is a difference in distance distribution and degree distribution. But can you show two 6-regular graphs with different distance distributions where one distribution (correctly) predicts something like contagion-spreading better than the other? If not, the branching factor of 6 is still going to be the main measure an epidemiologist will use to estimate a disease's r-naught value.
Let us continue this discussion in chat.
Q: Why is the degree distribution of vertices more widely studied than the distance distribution of links?
A: Vertices are more fundamental than links, in the following sense:
A vertex is associated with a scale, its degree (= number of nodes connected to it), which is determined only by the topology of the network. A link, on the other hand, has no intrinsic scale. One would need to attribute a weight to each link, such as its length, to define a scale, but there is no unique way to do that for a given network topology.
The lack of an intrinsic scale diminishes the fundamental interest of the link distance distribution. For that reason, Zhou, Meng, and Stanley [1] have recently studied an alternative distance distribution, the degree distance distribution, which does have an intrinsic scale. The degree distance $d$ of a link connecting two vertices of degree $k$ and $k'$ is defined as $d=\log|k-k'|$.
The authors argue that a power law degree distance distribution better represents the scale-free property of a network than a power law degree distribution.
[1] Bin Zhou, Xiangyi Meng, and H. Eugene Stanley, Power-law distribution of degree–degree distance: A better representation of the scale-free property of complex networks (2020)
With respect to your claim "vertices are more fundamental than links" please allow me to point you to this other question: Primacy of arcs/arrows over vertices/objects. But of course I noted that you wrote "in the following sense".
I see several reasons why degree distributions may be preferred to distance distributions.
First, degrees are more robust than distances, in the following sense: adding or removing one or a few links have a small impact on the degree distribution, whereas it may have a dramatic impact on distance distributions. In particular, adding a few random links (which may reflect erroneous measurements in practice) makes the average distance drop very quickly. This is one of the key principles underlying the Watts and Strogatz model, by the way.
Second, in practice, computing degree distributions is trivial, whereas computing distance distributions is costly; it typically requires $O(n\cdot m)$ time on graphs with $n$ nodes and $m$ links. This is often prohibitive. Approximations may help, but the problem remains much harder than degree distributions, and some values missed by the approximation may be important for the distribution.
Third, as one may see in your examples, distance distributions are often quite centered around their mean, with few exceptions. Actually, as I said, adding just a few random links is often sufficient to obtain this effect. Instead, observed degree distributions often span several orders of magnitude, which may be seen as more informative: degrees somehow make more difference between nodes than distances.
Finally, notice that distances make sense only between pairs of nodes in the same connected components. If the considered graphs are not connected, then it may be unclear how to interpret distance distributions.
One may also argue that sampling random graphs with prescribed degree distributions makes more sense (or is easier?) than sampling random graphs with prescribed distance distribution. This makes degree distributions more appealing both in practice, when one indeed generates random graphs, and in theory, as proving property of these random objects may be tractable.
I only see now that you already answered a question of mine in December last year! That's nice - and I'm still after this issue. Anything new from your side?
I think that distance distributions are actually much more interesting and informative than degree distributions of graphs. The reason is that graphs with the same degree distributions can have very different properties, including very different distance distributions. Examples are the Node Duplication model of random graphs and the Barabasi-Albert (preferential attachment) that exhibit a power-law degree distribution but have a very different structure.
The reason there is much more interest in the degree distribution is because it is easier to analyze. However, much higher quality information about nodes is obtained via distances.
There are some results available for distances in random graphs, notably
Distances in Erdos-Renyi graphs:
E. Katzav, M. Nitzan, D. ben-Avraham, P.L. Krapivsky, R. Kuhn, N. Ross and O. Biham, Analytical results for the distribution of shortest path lengths in random networks,
EPL 111, 26006 (2015).
as well as in:
I. Tishby, O. Biham, E. Katzav and R. Kuhn, Revealing the Micro-Structure of the Giant Clusters in Random Graph Ensembles, Phys. Rev. E 97, 042318 (2018).
and
E. Katzav, O. Biham and A. Hartmann, The distribution of shortest path lengths in subcritical Erd{\H o}s-R'enyi networks, Phys. Rev. E 98, 012301 (2018).
Distances in configuration model networks:
M. Nitzan, E. Katzav, R. Kuhn and O. Biham, Distance distribution in configuration model networks, Phys. Rev. E 93, 062309 (2016).
In particular, there is an exact result for random regular graphs
Distances in undirected node-duplication networks:
C. Steinbock, O. Biham and E. Katzav, The distribution of shortest path lengths in a class of node duplication network models, Phys. Rev. E 96, 032301 (2017).
Distances in directed node-duplication networks:
C. Steinbock, O. Biham and E. Katzav, The distribution of shortest path lengths in directed random networks that grow by node duplication, Eur. Phys. J. B 92, 130 (2019).
What is frustrating about all the three answers that you have published so far is that you only promote your own articles. It is a very subtle form of spam. In fact, I'm tempted to flag your posts as spam, really.
The point is that the nature of these two distributions is completely different. The degree distribution is local: in order to find it one one just has to know how the 1-neighbourhoods of vertices look like. The distance distribution is global. Essentially, it measures the dependence of the size of the graph distance spheres on radius, i.e., the growth of the graph. Needless to say that, for instance, two regular graphs with the same (constant) vertex degrees may look very very different.
Several reasons for the "preference" given to degree distributions have already been exposed here. I think that the main one is, actually, a manifestation of the "streetlight effect". As Matthieu Latapy put it in his answer, "computing degree distributions is trivial, whereas computing distance distributions is costly".
|
2025-03-21T14:48:32.000439
| 2020-09-10T13:28:17 |
371343
|
{
"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:632882",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371343"
}
|
Stack Exchange
|
Definite integral of modified Bessel function of second kind
How do I integrate a modified Bessel function of the second kind as shown below? A good approximation of the definite integral is also ok, I do not need an exact solution.
$\int_\frac{1}{\lambda}^{\infty}K_{5/3}(x) dx$
$$\int_\frac{1}{\lambda}^{\infty}K_{5/3}(x) dx=2^{2/3} {\lambda}^{2/3} \Gamma \left(\tfrac{2}{3}\right) \, _1F_2\left(-\tfrac{1}{3};-\tfrac{2}{3},\tfrac{2}{3};\frac{1}{4 {\lambda}^2}\right)+\frac{27 \pi \sqrt{3} \, _1F_2\left(\frac{4}{3};\frac{7}{3},\frac{8}{3};\frac{1}{4 {\lambda}^2}\right)}{160\ 2^{2/3} {\lambda}^{8/3} \Gamma \left(-\frac{1}{3}\right)}-\frac{\pi }{\sqrt{3}}$$
The integral grows as $2^{-1/3}\,3 \Gamma \left(5/3\right)\, \lambda^{2/3}$ for large $\lambda$, see plot (blue is integral, gold is large-$\lambda$ asymptotics).
|
2025-03-21T14:48:32.000516
| 2020-09-10T14:02:03 |
371347
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Eris",
"Willie Wong",
"YCor",
"https://mathoverflow.net/users/14094",
"https://mathoverflow.net/users/165116",
"https://mathoverflow.net/users/3948"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632883",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371347"
}
|
Stack Exchange
|
What is the physical meaning of torsion
The torsion tensor in 4 dimensions $S_{ab}^{\hphantom0\hphantom0 c}$ has 24 components and it can be split into a vector part $\hphantom0^{V}S_{ab}^{\hphantom0\hphantom0 c}=\frac{1}{3}(S_a\delta^c_b-S_b\delta^c_a)$ (4 components) and an axial-vector part $\hphantom0^AS_{ab}^{\hphantom0\hphantom0 c}=g^{cd}S_{[abd]}$ (4 components) and a traceless part $\hphantom0^TS_{ab}^{\hphantom0\hphantom0c}=S_{ab}^{\hphantom0\hphantom0 c}-\hphantom0^AS_{ab}^{\hphantom0\hphantom0 c}-\hphantom0^VS_{ab}^{\hphantom0\hphantom0 c}$ (16 components).
Torsion is usually interpreted physically as a twist in space because it causes a parallelogram to fail to close. How does the vector components particularly contribute to that twist geometrically? and what happen if it is a complex quantity i.e. $S_a=S_b=i\phi(t)$?
Related: What is torsion in differential geometry intuitively?
(I fixed "axialvector part" to "axial vector part" but maybe it should be "axial part", please check.)
24 components only in certain spatial dimensions, you should probably specify the dimension (I think you mean 4 dimensions, since the "axial vector part" is determined by a three form.) Also; please explain what you mean by "complex". Are you working with an almost complex manifold?
Thanks YCor. Both Axial-vector and Axial are used in literature.
@WillieWong sure, I edited the question. I'm not specifically on a complex manifold but I'm trying to understand what kind of manifold would result from such a choice of a complex torsion.
How exactly are you defining torsion then? On a real differential manifold with an affine connection, the torsion tensor is manifestly real by definition. You can't just slot in complex values and hope it would work out.
|
2025-03-21T14:48:32.000645
| 2020-09-10T14:33:46 |
371353
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Tim Campion",
"https://mathoverflow.net/users/2362"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632884",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371353"
}
|
Stack Exchange
|
Arrangement of subspaces over finite fields
I'm trying to find out what is already known about the following setup.
Let $V$ be an $n$-dimensional vector space over a finite field $F_q$ (I'm mostly interested in the case where $q$ is prime), and let $V_1, V_2, \ldots, V_k$ be proper linear subspaces of $V$ such that the following hold:
$\bigcup\limits_{i=1}^k V_i = V$ (the $V_i$ cover $V$),
$V_i\not\subset \bigcup\limits_{j\neq i} V_j$ (the cover doesn't have a subcover), and
$\bigcap\limits_{i=1}^k V_i = \{0\}$ (the cover isn't lifting a cover from a quotient of $V$).
For example, one can always take $\{V_i\}$ to be the collection of all 1-dimensional subspaces of $V$, in which case $k= |\mathbb{P}(V)| = \frac{q^n - 1}{q-1}$. If $n=2$, this is the only choice of $k$ and of $\{V_i\}$.
For a different example, if $q=2$ and $k=4$, then $n= 4$ is impossible, as one can check. However, $n= 3$ is possible, and there is one cover with $\dim V_1 = \dim V_2 = \dim V_3 = 2$, $\dim V_4 = 1$ and another with $\dim V_1 = \dim V_2 = 2$, $\dim V_3 = \dim V_4 = 1$. If $V$ is spanned by $x, y, z$, then these correspond to $V_1 = <x, y>$, $V_2 = <x, z>$, $V_4 = <x+y+z>$, and $V_3=<y,z>$ or $V_3 = <y+z>$. These are "essentially" the only two choices of $\{V_i\}$, up to relabellings, for $(n,k,q) = (3, 4, 2)$.
Here are my questions:
A) For fixed $n$ and $q$, what are the possible values of $k$? What can the arrangements $\{V_i\}$ look like for these $k$?
B) For fixed $k$ and $q$, what are the possible values of $n$? What can the arrangements $\{V_i\}$ look like for these $n$?
What happens in characteristic 0? Are we then forced to use all 1-dimensional subspaces for any $n$?
|
2025-03-21T14:48:32.000773
| 2020-09-10T14:45:08 |
371355
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Ian Agol",
"Noah Snyder",
"Terry Tao",
"https://mathoverflow.net/users/1345",
"https://mathoverflow.net/users/22",
"https://mathoverflow.net/users/51189",
"https://mathoverflow.net/users/766",
"zeraoulia rafik"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632885",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371355"
}
|
Stack Exchange
|
What are applications of Jones polynomial on von Neumann algebras?
I have read according list of below papers a basic connection between Jones polynomial and statistical mechanics is that the Kauffman bracket or Kauffman polynomial a polynomial invariant of knots is in different special cases the Jones polynomial for knots and the partition function for the Potts model in statistical mechanics. The Jones polynomial and its relations to the Yang-Baxter equations in Statistical mechanics, has been generalized to other invariants of knot theory by Kauffman via the Kauffman bracket .Witten has shown that one can use knot theory in the context of quantum field theory to produce invariants of 3- dimensional manifolds. Michael Atiyah also is using the Jones-Witten theory to explore functional integration in gauge theories and quantization. Now my question here is :
Question
What are applications of Jones polynomial on von von Neumann algebras ? or what the Jones polynomials has to do with von Neumann algebras?
Reference list
[1]:The book "Exactly Solved Models in Statistical Mechanics" by Baxter is a really good source if you are interested in the connection between statistical physics and the work of Jones http://physics.anu.edu.au/theophys/_files/Exactly.pdf
[2]:"Statistical Mechanics and the Jones Polynomial" by Louis Kauffman http://www.maths.ed.ac.uk/~aar/papers/kauffmanjones.pdf
[3]:A good source of information on the connection between QFT and the Jones polynominal is Witten's paper "Quantum field theory and the Jones polynomial" http://projecteuclid.org/download/pdf_1/euclid.cmp/1104178138
[4]:A brief version: certain algebras arising in Jones' work also occur in the study of exactly solvable models in statistical mechanics. See here for details:
J.S. Birman, The Work of Vaughan F. R. Jones, in ICM'1990 proceedings:
http://www.mathunion.org/ICM/ICM1990.1/Main/icm1990.1.0009.0018.ocr.pdf
The second question is already covered in https://mathoverflow.net/questions/97788/on-connection-between-knot-theory-and-operator-algebra
@TerryTao, Thanks for the link , I din't get attention to it , am going to check it
I don't think it's quite right to think of knot polynomials as having applications to von Neumann algebras. Instead I think it's more accurate to say that the Temperley-Lieb-Jones algebras (and more generally "towers of algebras with Markov traces" or equivalently quantum groups or tensor categories) have applications both to von Neumann algebras (via the theory of standard invariants) and to low-dimensional topology (via their connection to the braid group).
A great place to start on reading about applications from TLJ to subfactor theory is Vaughan's paper "Index for Subfactors" where he rediscovered the Temperley-Lieb algebras in the context of subfactors. (I believe it was David Evans who pointed out that they'd appeared previously in the context of statistical mechanics in Temperley-Lieb's work.)
In this paper, Vaughan attributes the observation of the similarity of the Temperley-Lieb relations and the braid group relations to D. Hatt, P. de la Harpe and N. Stoltzfus:
Jones, Vaughan, Groupes de tresses, algèbres de Hecke et facteurs de type (II_ 1). (Braid groups, Hecke algebras and type (II_ 1) factors), C. R. Acad. Sci., Paris, Sér. I 298, 505-508 (1984). ZBL0597.20034.
(Hatt and de la Harpe are also mentioned in the paper "A POLYNOMIAL INVARIANT FOR KNOTS
VIA VON NEUMANN ALGEBRAS"). He attributes the observation of the similarity between the Temperley-Lieb algebra presentation and the Hecke algebra to R. Steinberg. Since the Hecke algebra is a quotient of the group ring of the braid group, one obtains finite dimensional representations of the braid group into Hecke algebras, and similarly into the Temperley-Lieb algebra, satisfying the skein relation. These braid group representations were discovered by Jones in 1983 (see "Braid groups, Hecke algebras and type II1 factors").
He announced the knot polynomial in 1985 in the paper cited above. In this paper, he acknowledges Joan Birman's help for identifying the trace that is invariant under the Markov move (this is essentially fixing the dependence of the trace on the writhe).
Clearly Vaughan discovered the Temperley-Lieb algebra from his study of subfactors. But I speculate that his derivation of representations of the braid group and the knot polynomials was more fortuitous stemming from the suggestions of people that he acknowledges.
I think one of Vaughan's strengths as a mathematician was that he was very persistent in wringing out all that he could from a technique. There's obviously something fortuitous in that the Temperley-Lieb algebra leads to interesting knot invariants (when it could just have well have given boring ones), but I think there's also a highly non-fortuitous part which is that once he had the Temperley-Lieb algebras in hand he really stuck with them to see everything you could do with them including knot polynomials.
That is, both Jones and Temperley-Lieb independently discovered the Temperley-Lieb algebras in contexts completely unrelated to knot theory. In principal Temperley-Lieb's work could have lead to the discovery of the Jones polynomial if someone had kept at thinking about what else you could with those algebras the same way Vaughan did a few years later. (That said, Vaughan did have a critical advantage, the subfactor approach gives not only the algebras themselves but also a natural trace on them, which plays a critical role.)
Yes, I was merely postulating that he did not, as far as I know, intend to derive braid representations and knot invariants from his investigations of von Neumann algebras. But as you indicate, luck favors the prepared mind (and he certainly mastered these and related subjects quickly after discovering these connections). As far as I know, a direct connection between the Jones polynomial and von Neumann algebras / subfactors is still mysterious. Henriques and collaborators have derived conformal field theories from von Neumann algebras, but I don't know if this gives any relation.
|
2025-03-21T14:48:32.001289
| 2020-09-10T16:01:17 |
371360
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Connor Malin",
"Igor Belegradek",
"archipelago",
"https://mathoverflow.net/users/134512",
"https://mathoverflow.net/users/1573",
"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:632886",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371360"
}
|
Stack Exchange
|
Normal bundle information in the surgery structure set
Recall that the surgery structure set of a smooth, compact n-manifold $M$ with normal bundle $\eta$ is the set of maps $f:N \rightarrow M$ that are homotopy equivalences up to the relation $f:N \rightarrow M $ is equivalent to $f':N' \rightarrow M$ if there is an h-cobordism $(W,N,N')$ with a map $(W,N,N') \rightarrow (M \times I, M \times \{0\}, M \times \{1\})$, restricting to the maps involved. This definition is from Ranicki's "Algebraic and Geometric Surgery".
We denote the structure set $\mathscr{S}(M)$.
This set of equivalence classes is of interest because it sits in the exact sequence of pointed sets $\mathscr{S}(M) \rightarrow \mathscr{N}(M) \rightarrow L_n(\pi_1(M))$, where $\mathscr{N}(M)$ is the set of normal invariants of $M$, i.e. the degree 1 maps from manifolds with lifts to fiberwise isomorphisms of the normal bundles, up to cobordisms which extend the map on normal bundles, so-called normal cobordisms.
The $L$ groups are the surgery obstruction groups, but these are not relevant to my question. My question is about how we can define the map $\mathscr{S}(M) \rightarrow \mathscr{N}(M)$ with no normal bundle information.
It turns out that between compact manifolds, a homotopy equivalence can always be covered by fiberwise isomorphism of normal bundles. Hence, we can always take our homotopy equivalence and make a $choice$ to turn it into a normal map. However, these choices are in general not fiberwise homotopic (e.g. $S^1$ has multiple different framings of the normal bundle).
So for this map to be well defined, it needs to be the case that we can find a normal cobordism between any two different lifts of the homotopy equivalence to the normal bundle. Why is this true?
Your definition of $N(M)$ is off: a class in $N(X)$ for a Poincaré complex $X$ is represented by a degree 1 map $N\rightarrow X$ from a manifold covered by a bundle map $TN\oplus \varepsilon^k\rightarrow \xi$ for some $k\ge0$ and \emph{some} bundle $\xi$ over $X$. If $X$ happens to be a manifold itself, $\xi$ need not agree with $TX$.
@archipelago Ah, so then is the map just if we have a homotopy equivalence we can push the normal bundle of the domain manifold forward and have a canonical normal map?
You pull back the tangent bundle of $N$ along a homotopy inverse. All this can be reformulated in terms of normal bundles, but I prefer the tangential perspective.
@archipelago Thanks this answers my questions. So I’m not sure about something I said which is that all tangent/normal bundles are stably equivalent no matter which smooth structure we pick, at least if we are compact. I mostly believed this because I thought it was necessary for the maps to make sense in the exact sequence, but now you’ve explained it it needn’t be true. Do you know a counter example?
I suggest to look for examples of homotopy equivalent closed smooth manifolds with different first Pontryagin class by considering linear $S^3$ bundles over $S^4$.
I do not quite follow your terminology but the claim "homotopy equivalence can always be covered by fiberwise isomorphism of [stable] normal bundle" in incorrect. Do a web search "non-tangential homotopy equivalences". What is true is that homotopy equivalence (of closed manifolds) pulls back the stable spherical fibrations.
@IgorBelegradek Thank you, I was not sure if I could upgrade the equivalence of spherical fibrations to one of vector spaces.
Ranicki's book is nice but I would start studying surgery with the original papers by Milnor-Kervaire, Browder, and Novikov, who take the time to explain things. In particular, Browder's textbook on simply-connected surgery is a good start.
|
2025-03-21T14:48:32.001549
| 2020-09-10T16:03:47 |
371361
|
{
"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:632887",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371361"
}
|
Stack Exchange
|
Sufficient conditions for the continuity of an improper integral concerning the finite-time stability of a dynamical system
Consider the initial value problem
\begin{equation}\label{fainait ve}
\dot{\boldsymbol{x}}(t) = \boldsymbol{f}(\boldsymbol{x}(t)), \;\; t \geq 0, \; \;\boldsymbol{f}(\boldsymbol{0}_n) = \boldsymbol{0}_n, \;\; \boldsymbol{x}(0) = \boldsymbol{\xi} \in \mathbb{R}^n
\end{equation}
where $\boldsymbol{f}(\cdot) \in \mathcal{C} \left(\mathcal{D} ; \mathbb{R}^n \right) $ and $\mathcal{D} \subseteq \mathbb{R}^n$ and $\mathcal{D}$ is open with $\boldsymbol{0}_n \in \mathcal{D}$, and $\boldsymbol{f}(\cdot)$ is locally Lipschitz apart from the origin $\boldsymbol{0}_n$. Let $\boldsymbol{\psi}\left(t, \boldsymbol{\xi} \right) $ be the unique solution of the above initial value problem. Suppose there exists $\mathsf{v}(\cdot) \in \mathcal{C}^{\infty}\left( \mathcal{D} ; \mathbb{R} \right) $ such that $\mathsf{v} (\boldsymbol{0}_n) = 0$ and
\begin{align}
& \forall \boldsymbol{\xi} \in \mathcal{D} \setminus \{ \boldsymbol{0}_n \} , \; \mathsf{v} (\boldsymbol{\xi}) > 0 \quad \text{and} \;\; \; \left. \dfrac{\partial\,\mathsf{v}\left(\boldsymbol{\psi}\left(t, \boldsymbol{\xi}\right) \right)}{\partial t} \right|_{t = 0}
= \dfrac{\partial \mathsf{v} (\boldsymbol{\xi})}{\partial \boldsymbol{\xi}} \boldsymbol{f}(\boldsymbol{\xi}) < 0,
\\[1mm]
& \forall \boldsymbol{\xi} \in \mathcal{D},\;\; \sigma (\boldsymbol{\xi}) : = \int_{\mathsf{v}(\boldsymbol{\xi})}^{0}
1 \left / \left. \dfrac{\partial\,\mathsf{v}\left(\boldsymbol{\psi}\left(t, \boldsymbol{\xi}\right) \right)}{\partial t} \right. \right|_{t = \vartheta(\tau, \boldsymbol{\xi})} \mathsf{d} \tau =
\int_{\mathsf{v}(\boldsymbol{\xi})}^{0}
\dfrac{1}{\left. \dfrac{\partial\,\mathsf{v}\left(\mathbf{x} \right)}{\partial \mathbf{x} }
\boldsymbol{f}\left( \mathbf{x} \right) \right|_{ \mathbf{x} = \boldsymbol{\psi} \left( \vartheta(\tau, \boldsymbol{\xi}), \boldsymbol{\xi} \right)} } \mathsf{d} \tau < + \infty
\end{align}
where $\vartheta(\cdot, \boldsymbol{\xi})$ with a given $\boldsymbol{\xi}$ is the inverse function of $[0, \sigma(\boldsymbol{\xi})) \ni t \mapsto \mathsf{v}\left( \boldsymbol{\psi}\left(t, \boldsymbol{\xi} \right) \right) $. Note that the inverse function of $[0, \sigma(\boldsymbol{\xi})) \ni t \mapsto \mathsf{v}\left( \boldsymbol{\psi}\left(t, \boldsymbol{\xi} \right) \right) $ is well defined since $\mathsf{v}\left( \boldsymbol{\psi}\left(t, \boldsymbol{\xi} \right) \right)$ is strictly decreasing and differentiable in $t$.
Question: Can we prove that $\sigma(\boldsymbol{\xi})$ is continuous at $\boldsymbol{\xi} = \boldsymbol{0}_n$?
Background: The aforementioned problem is encountered in the characterization of finite-time stability of differential equations. The function $\mathsf{v}(\cdot)$ is actually a Lyapunov function of the dynamical system, and $\sigma(\boldsymbol{\xi})$ is the setting-time of the system. ($\sigma(\boldsymbol{\xi})$ always has finite value which is why we say the system is finite-time stable)
For the scalar case $n = 1$, the continuity of $\sigma(\boldsymbol{\xi})$ at $\boldsymbol{0}_n$ can be proved via the application of Dominated convergence theorem without using $\mathsf{v}(\cdot)$. However, the multi-dimensional case is much more difficult to be deal with due to the presence of $\mathsf{v}(\cdot)$ and $\boldsymbol{\psi}(t, \boldsymbol{\xi})$.
Thank you so much!
|
2025-03-21T14:48:32.001726
| 2020-09-10T16:25:00 |
371362
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Mark Schultz-Wu",
"https://mathoverflow.net/users/101207",
"https://mathoverflow.net/users/12481",
"joro"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632888",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371362"
}
|
Stack Exchange
|
Can we make cryptography signature algorithm based on hardness of isomorphism?
In public key cryptography, Alice knows functions $f$ and its inverse
$f^{-1}$. $f$ is public and $f^{-1}$ is secret. To sign a message
$m$, she gives $(m,a=f^{-1}(m))$. To verify a signature, the verifier
checks if $f(a)=f(f^{-1}(m)) ?= m$.
$f$ and $f^{-1}$ are related by some secret $S$ and it is computationally
infeasible an adversary to compute $f^{-1}$ or the secret $S$.
Examples of secrets $S$ are integer factorization and discrete logarithms.
One possible approach is to take the secret $S$ to be obfuscation permutation
and $f$,$f^{-1}$ to be some "objects" for which isomorphism (and the permutation $S$) is hard to compute, while still have "some properties" for signature.
The question from 2016 Finite objects for which isomorphism is NP-hard or harder? appears to
give hard isomorphism.
Q Can we make cryptography signature algorithm based on hardness of isomorphism
with security at least NP-hard?
There is related work based Isomorphism of Polynomials, but we
believe its security is only Graph Isomorphism (could be wrong on this).
NP-hardness of Hamiltonian Cycle is used in Zero-knowledge proof
Added 2020-09-15
From answer the linked question
two candidates for NP-hardness are CIRCUIT ISOMORPHISM
and FORMULA ISOMORPHISM.
On p.1: The formula isomorphism problem is in $\Sigma_2^p$ , NP-hard, and unlikely to be $\Sigma_2^p$.-complete
Crossposted at cstheory: https://cstheory.stackexchange.com/questions/47551/can-we-make-cryptography-signature-algorithm-based-on-hardness-of-isomorphism
This question seems to be asking if we can construct a trapdoor one-way function based on the hardness of some isomorphism problem (which is what $f$ appears to be), that we can use in a digital signature scheme. But it is well-known that digital signature schemes can be constructed from the weaker notion of a one-way function (note that if you were interested in public-key encryption, known techniques require trapdoor OWFs). Is your primary interest a trapdoor OWF, or a digital signature scheme?
@Mark I am mainly interested as trapdoor, which can also be used in encryption like RSA.
One candidate might be Circuit Isomorphism.
Then you should know that there are no known constructions of public-key encryption from an NP-complete problem (despite ones that get "close but barely miss", say LWE-based protocols), so if you suggest building such a thing this would be a major result independently of this specific question.
As for basing cryptography on "circuit isomorphism", I've heard some practitioners express hope on basing cryptography on the "Minimum Circuit Size Problem". This is a problem asking if a certain circuit has a "small representation" (generally in terms of some size bound). The hardness of MCSP is a rather intricate story though (and there have been many developments in the last 5 years), but MCSP can be thought of better as finding a "compact normal form" for circuits than of asking if two circuits are isomorphic.
@Mark Thanks. I edited with NP-hardness of FORMULA ISOMORPHISM and CIRCUIT ISOMORPHISM.
crossposted on crypto: https://crypto.stackexchange.com/questions/84023/can-we-make-cryptography-signature-or-encryption-algorithm-based-on-hardness-of
@orgesleka Interesting, thanks.
|
2025-03-21T14:48:32.001948
| 2020-09-10T16:26:55 |
371363
|
{
"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:632889",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371363"
}
|
Stack Exchange
|
Trade-off between covering number, ball radius and diameter of $d$-dimensional shapes
Given any $d$-dimensional shape $X$ in the Euclidean space, let $\ell(X)$ be the length of the longest line segment connecting two points of $X$. How can we prove the following statement?
There exists a positive constant $c$ such that, for all integers $n \ge 1$, we can completely cover any $d$-dimensional shape $X$ with $n$-many $d$-dimensional balls having radius $r\le c\frac{\ell(X)}{n^{1/d}}~.$
Note: I can prove the above inequality by combining the bound of the question
Trade-off between hypervolume and diameter of $d$-dimensional shapes having a hypercubic smallest bounding box with the well known bound $n\le\left(\frac{3}{r}\right)^d\cdot\frac{V(X)}{V(B)}$, where $V(\cdot)$ is the volume of a shape, and $B$ is the unit norm ball. I am wondering whether it is possible to avoid the use of the inequality of my previous question proving the above statement in a simpler and more direct way.
|
2025-03-21T14:48:32.002033
| 2020-09-10T16:39:18 |
371365
|
{
"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:632890",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371365"
}
|
Stack Exchange
|
Kernel of the adjoint of the infinitesimal generator of Levy SDE
Consider S.D.Es driven by a combination of Brownian and non-Brownian Levy noise (like say Gamma). Then we know that the flow of the density of the S.D.E variable is given by the adjoint of the infinitesimal generator (say ${\cal L}^*$) of the S.D.E .
How often is it true that the S.D.E variable's density is converging to the kernel of ${\cal L}^*$?
Are there examples of (a) being able to show when the kernel of this ${\cal L}^*$ is unique and non-trivial and (b) being actually be able to compute this density in the kernel in such a case?
For comparison consider this paper (https://epubs.siam.org/doi/pdf/10.1137/0325042) where it was shown that for the basic Brownian motion driven SDE, the iterates converge to the Gibbs' measure which happens to be the kernel of the corresponding ${\cal L}^*$. (Is the asymptotic measure being the kernel of ${\cal L}^*$ a coincidence?)
My question is essentially about asking what is the analogue of such a result for non-Brownian Levy noise.
For the Brownian case more general results of this type can be seen in,
https://www.sciencedirect.com/science/article/pii/S0304414902001503
|
2025-03-21T14:48:32.002133
| 2020-09-10T19:26:07 |
371370
|
{
"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:632891",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371370"
}
|
Stack Exchange
|
Expansion of random subgraphs of a bi-regular bipartite graph
Let $G = (L, R, E)$ be a bi-regular bipartite graph, with $|L|=n$ and $|R| = C \cdot n$, where $C$ is a large constant. Let $d$ be its (constant) right-degree.
We know $G$ is a good spectral expander. Say, for concreteness, that its normalized second eigenvalue is roughly $\frac{1}{\sqrt{d}}$, so in particular we have good vertex expansion from both sides.
I wish to pick a random subset $S \subseteq R$ of cardinality, say, $\frac{n}{4}$. Let's think of $d > 4$, so we are in a situation in which with high probability, $\Gamma(S)$ covers nearly all of $L$. Let $G' = (\Gamma(S), S, E')$ be the induced subgraph, where we keep only the edges that come out of $S$. Note that $G'$ is still $d$ right-regular, but is no longer left-regular.
The question is: What can we say about the expansion properties of a typical $S$? Specifically, what is the expected vertex expansion from left to right? I don't expect it to be as good, but is it only mildly good, say that each small enough set $X \subseteq \Gamma(S)$ is mapped to a set of size $\frac{|X|}{2}$? (note that it's actually a good expansion, as the "imbalanceness" is roughly $\frac{1}{4}$). A simple union-bound does not seem to give anything meaningful.
|
2025-03-21T14:48:32.002239
| 2020-09-10T20:10:00 |
371371
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Ben C",
"David E Speyer",
"Jef",
"https://mathoverflow.net/users/110362",
"https://mathoverflow.net/users/154157",
"https://mathoverflow.net/users/297",
"https://mathoverflow.net/users/519",
"naf"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632892",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371371"
}
|
Stack Exchange
|
Curve with no embedding in a toric surface
I am looking for a smooth proper curve $C$ such that there does not exist any closed embedding $C \to S$ where $S$ is a (normal projective) toric surface.
Since $C$ is smooth I believe it suffices to consider smooth projective toric surfaces $S$ since we may always perform a toric resolution of singularities and the strict transform of $C$ will be isomorphic to $C$ since $C$ is smooth.
Using the result on p.25 of Harris Mumford, On the kodaira dimension of the moduli space of curves, I can conclude that a very general curve cannot have any such embedding.
However, I am not able to write down an explicit example. Does anyone know such an example or what sort of obstruction might work to check this in particular examples.
You might already know this, but there is the notion of a nondegenerate curve which is related but (I think) stronger than your condition, for a definition see Castryck and Voight's article. In that paper they prove that every curve over an algebraically closed field is nondegenerate if its genus is $\leq 4$, so your example will necessarily have genus at least $5$.
@Jef thanks for the comment. I am familiar with Castryck's work and I think you are correct nondegenerate is a stronger condition it corresponds to an embedding that intersects the toric divisor transversally. There is a related notion of weakly nondegenerate which I do believe corresponds to what I am looking for. However, it appears less is know about when a curve admits a weakly nondegenerate equation.
Some observations: The divisor $C$ on $S$ corresponds to a lattice polygon, call it $P$. The number of interior lattice points of this polygon is $g$. If the polygon has width $k$ in any lattice direction, then the curve has gonality $\leq k$. Thus, if we are to get a generic curve of genus $g$, the width must be $\geq g/2+1$ in every direction. Also, if $N$ is the number of lattice points and we want to get a generic curve, we must have $N-3 \geq 3g-3$, since the moduli space of curves is $3g-3$ dimensional, and we lose $3$ dimensions rescaling coordinates.
To be clear, by the width of a lattice polytope $P$, I mean the following: the minimum, over nonzero lattice vectors $v$, of $\max_{p \in P} \langle p,v \rangle - \min_{q \in P} \langle q,v \rangle$.
A generic curve of genus $5$ is not a hypersurface in a toric surface. This argument is going to use conceptual ideas from Haase and Schicho's paper "Lattice polygons and the number $2i+7$", plus a bunch of case analysis.
Let's start with generalities about a curve $C$ in a toric surface $S$, other than one of the boundary divisors. The divisor $C$ gives a nef line bundle $\mathcal{O}(C)$ on $S$, which gives a lattice polygon $P$ in $\mathbb{Z}^2$. (See, for example, Lecture 4 here.) The number of interior lattice points of $P$ will be the genus of $C$.
Let $Q$ be the convex hull of the interior lattice points. So, in our genus $5$ case, $Q$ will be a lattice polytope with $5$ lattice points. Such a polytope either (Case 0) has no interior lattice points (Case 1) has one interior lattice point (Case 2) is a triangle with two interior lattice points or (Case 3) is a line segment of length $4$.
Case 0 Polytopes with no interior lattice points are trapezoids with vertices of the form $(0,0)$, $(a,0)$, $(0,1)$, $(b,1)$. If we want $5$ lattice points, our options are
$$\mbox{polygon 0a} = \begin{matrix}
\bullet&&& \\ \bullet&\bullet&\bullet&\bullet \\
\end{matrix} \qquad
\mbox{polygon 0b} = \begin{matrix}
\bullet&\bullet& \\ \bullet&\bullet&\bullet \\
\end{matrix}$$
Case 1 There are $12$ polytopes with $1$ interior lattice point, which you can see in Figure 6 of Haase and Schicho. There are $3$ of these with $5$ lattice points:
$$
\mbox{polygon 1a} =\begin{matrix} &\bullet& \\ \bullet&\bullet&\bullet \\ &\bullet& \\ \end{matrix} \qquad
\mbox{polygon 1b} =\begin{matrix} \bullet&& \\ \bullet&\bullet&\bullet \\ &\bullet& \\ \end{matrix} \qquad
\mbox{polygon 1c} =\begin{matrix} \bullet&& \\ \bullet&\bullet&\bullet \\ \bullet&& \\ \end{matrix}.$$
Case 2 Wei and Ding, "Lattice polygons with two interior lattice points", list all lattice polytopes with $2$ interior lattice points. Only one of them has $5$ lattice points:
$$
\mbox{polygon 2} = \begin{matrix} &&\bullet& \\ \bullet&\bullet&\bullet& \\ &&&\bullet \\ \end{matrix}
$$
Case 3 Finally, here is the line segment:
$$\mbox{polygon 3}=\begin{matrix} \bullet&\bullet&\bullet&\bullet&\bullet \end{matrix}$$
So $Q$ must be one of the $7$ polygons above.
Haase and Schicho show that not every polytope can occur as $Q$. Hasse and Schicho make the following definition: Let $v$ be a vertex of a lattice polytope $Q$, and let $x$ and $y$ be the minimal lattice vectors along the edges incident to $v$. Hasse and Schicho say that $v$ is a "good corner" if $(\vec{x}, \vec{y})$ is $SL_2(\mathbb{Z})$ equivalent to $((1,0), (-1,k))$ for some $k$. Haase and Schicho, Lemma 9, show that $Q$ is always either the empty set, a line segment, or a polytope with good corners. Polygons 0a and 2 do not have good corners, so we can eliminate them now.
We now look at Haase and Schicho, Lemma 8. This lemma says: Let $a_1$, $a_2$ and $b$ be integers with $GCD(a_1, a_2)=1$. Suppose that $a_1 x_1 + a_2 x_2 \leq b$ on $Q$, and that equality occurs at at least two lattice points. Then $a_1 x_1 + a_2 x_2 \leq b+1$ on $P$.
We will apply this result to eliminate polygons 0b and 3. Take $(a_1, a_2) = (0,1)$, so $a_1 x_1 + a_2 x_2$ is the second coordinate. Then Hasse and Schicho's lemma show that the polytope P is contained in a horizontal strip of width $3$ when $Q$ is polygon 0b, and width $2$ in polygon 3. This means that the corresponding curve is trigonal or hyperelliptic, respectively. A generic genus $5$ curve is not trigonal or hyperelliptic.
This leaves cases 1a, 1b and 1c. Haase and Schicho define a polygon they call $Q^{(-1)}$, whose definition I will let you read in their paper, and show that $P \subseteq Q^{(-1)}$. In the figure below, I have redrawn polygons 1a, 1b and 1c with solid dots $\bullet$, and drawn the additional vertices of $Q^{(-1)}$ as hollow circles $\circ$.
$$
\begin{matrix} &&\circ&& \\ &\circ&\bullet&\circ& \\ \circ&\bullet&\bullet&\bullet&\circ \\ &\circ&\bullet&\circ& \\ &&\circ&& \\ \end{matrix} \qquad
\begin{matrix} \circ&&&& \\ \circ&\bullet&\circ&&& \\ \circ&\bullet&\bullet&\bullet&\circ \\ &\circ&\bullet&\circ& \\ &&\circ&& \end{matrix} \qquad
\begin{matrix} \circ&&&& \\ \circ&\bullet&\circ&&& \\ \circ&\bullet&\bullet&\bullet&\circ \\ \circ&\bullet&\circ&&& \\ \circ&&&& \\ \end{matrix}.$$
In each case, there are $13$ lattice points in $Q^{(-1)}$, and thus $\leq 13$ lattice points in $P$. The curve $C$ is the zero locus of a polynomial with this Newton polytope, so it depends on $13$ parameters. But rescaling either of the variables, or rescaling the whole polynomial, does not change the isomorphism class of the curve, so we really only have $10$ parameters. The moduli space of curves of genus $5$ has dimension $12$, so the generic curve of genus $5$ is not a zero locus of a polynomial of any of the above forms. This concludes our case analysis.
Writing down an explicit genus $5$ curve which works still seems interesting. We can avoid cases 0b and 3 by just not making our curve hyperelliptic or trigonal. It isn't clear to me what explicit criterion avoids cases 1a, 1b or 1c.
Hasse should be Haase.
@ulrich Thanks!
|
2025-03-21T14:48:32.002666
| 2020-09-10T22:27:05 |
371375
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"FKranhold",
"Phil Tosteson",
"https://mathoverflow.net/users/124042",
"https://mathoverflow.net/users/52918"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632893",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371375"
}
|
Stack Exchange
|
Different ways to “deloop” a (topological) $A_\infty$-algebra
Let $\varphi:A\to \mathrm{Ass}$ be an $A_\infty$-operad in topological spaces, and let $X$ be an $A$-algebra. I see three possibilities to construct a delooping out of $X$:
Rectify $X$ by taking the pushforward $\varphi_!X$, which is now a topological monoid. Now take the classical bar construction $|N(\varphi_!X)|$. This should be the notion of rectification taken from [Berger/Moerdijk, 2003].
Rectify $X$ by forming the 2-sided bar construction $B(\mathrm{Ass},A,X)$ which is again a topological monoid, and then take the classical bar construction $|N(B(\mathrm{Ass},A,X))|$. This is the description given in [Basterra et. al., 2016].
We see that the reduced suspension is a right $A$-module functor by the splitting
$$\Sigma A X \to \Sigma\left(\bigvee_{k\ge 1}X^{\wedge k}\right)\to \Sigma X,$$
so we can directly form the 2-sided bar construction $B(\Sigma,A,X)$. For $A=\mathcal{C}_1$, this should be the classical description from [May, 1972].
Is there any reference comparing these three descriptions? I would expect that (at least for well-pointed $X$), all descriptions agree up to (weak) equivalence?
Just a quick clarification: you need to take a cofibrant replacement before applying $\phi_!$, right? If so, then to compare 1. and 2. you can compare the topological monoids.
According to Berger/Moerdijk, $\varphi_!$ and $\varphi^*$ forms a Quillen adjunction, so in some sense yes, the induced homotopy functor would at first cofibrantly replace $X$. I guess this is the more canonical functor to consider? Is it easy to say what cofibrant $A$-algebras are? I’d expect that at least those with a “cellular decomposition” (start with the trivial $A$-algebra and inductively attach free $A$-algebras along cofibrations of based spaces) are cofibrant?
It's not so much about being "canonical" as it is about making sure all your constructions preserve homotopy equivalences. It looks as though BM work in a very general setting. According to nlab, in the topological case the right induced model structure will be cofibrantly generated by celluar attachments like the ones you descibe.
https://ncatlab.org/nlab/show/cofibrantly+generated+model+category
E.g. suppose your $A_\infty$ algebra has a non-trivial associativity paths, $(ab)c \to a(bc)$. Then $\phi_!$ will naively collapse them in order to make the algebra associative. This will probably change the homotopy type of your space-- which is not what you want to do when you rectify.
Okay, I agree! So assume $X$ to be cofibrant, and now you claim that on the level of topological monoids, there is a morphism between $\varphi_! X$ and $B(\mathrm{Ass},A,X)$? Maybe this is obvious but I don’t see it. In which direction does it go?
I think there is a natural transformation $B({\rm Ass},A,-) \to \phi_! -$. The bar construction is the realization of a simplicial space $B_n(-)$ and you should be able to identify $coeq(B_1 \to B_0)$ with $\phi_!-$. This proof is analogous to the homological algebra fact that you can compute $Tor_A(M,N)$ via the bar construction or by resolving $N$. (Or the equivalent fact for hocolims).
There are two old papers that address this topic in some detail: R. W. Thomason. Uniqueness of delooping machines. \url{https://projecteuclid.org/euclid.dmj/1077313403}
Z. Fiedorowicz. Classifying spaces of topological monoids and categories. \url{https://www.jstor.org/stable/2374307?seq=1#metadata_info_tab_contents}. Thomason's paper compares 2 and my original version of 3 and others. I haven't looked up Berger and Moerdijk.
|
2025-03-21T14:48:32.003027
| 2020-09-10T22:29:25 |
371377
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Andrea Marino",
"Dmitri Panov",
"Ryan Budney",
"Strongly Negative Amphicheiral",
"YCor",
"https://mathoverflow.net/users/140013",
"https://mathoverflow.net/users/14094",
"https://mathoverflow.net/users/1465",
"https://mathoverflow.net/users/173304",
"https://mathoverflow.net/users/51189",
"https://mathoverflow.net/users/943",
"zeraoulia rafik"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632894",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371377"
}
|
Stack Exchange
|
Invariant knot for finite group actions on $S^3$
Inspired by the Smith conjecture, is there a finite group action on $S^3$ (by smooth or analytic diffeomorphisms) which possesses an invariant knotted circle?
Certainly. To get a simple example, take any toric knot, i.e. something given by equation $z_1^p=z_2^q$ in the unit sphere $|z_1|^2+|z_2|^2=1$. For any such knot there is a (linear) circle action on $S^3$ that preserves the knot, and you can take any finite subgroup of this circle.
Since invariant knot include knot Group, Other conventions consider knots to be embedded in the 3-sphere, in which case the knot group is the fundamental group of its complement in $S^3$
Probably you want a nontrivial action, and then the question reduces to faithful actions of cyclic groups of prime order. Any knot in Euclidean $\mathbf{R}^3$ or in the round sphere, with nontrivial isometry group, yields an example. For example, the usual representation of the trefoil knot has a isometry group of order $6$.
Turning your question around a little, you are indirectly asking about the symmetry groups of knots in $S^3$. These are readily computed, by the techniques of Sakuma as mentioned. But you can also compute them for hyperbolic knots fairly rapidly with SnapPea. This was originally done by Jeff Weeks, at essentially the same time as Sakuma. I believe Sakuma's table of symmetry groups for "small" knots is in the back of Kawauchi's reference book on knot theory. A nice feature of the Snappea approach is you can algorithmically describe the fixed point sets.
I think the comments answered the question, but it seems like some pictures would be helpful as well. There are many examples, but one of my favorites is this paper by Sakuma which concludes with a table of knots preserved by orientation-preserving involutions on $S^3$ which reverse the orientation on $K$.
I also want to point out that in general studying prime order group actions is not sufficient. For example, you may ask if the figure eight knot $4_1$ is preserved by a symmetry which reverses the orientation on $S^3$, but preserves the orientation on $4_1$. This is the case, but the symmetry has order 4 (and there is no such order 2 symmetry). Visually, this can be seen in the following image as a $\pi/2$ rotation within the plane of the diagram followed by a reflection across an $S^2$ intersecting the diagram in the shown dotted green circle.
+1, but i suggest you to use inkscape for drawings. I discovered it this year and changed my drawing quality (in my thesis I had very poor graphics XD)
Fair enough - edited to include a better diagram.
|
2025-03-21T14:48:32.003233
| 2020-09-10T23:11:52 |
371380
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Andrei Coman",
"Qiaochu Yuan",
"https://mathoverflow.net/users/165142",
"https://mathoverflow.net/users/290"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632895",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371380"
}
|
Stack Exchange
|
Algorithm for the nilpotence of matrix polynomials
Let $P$ be a multivariate polynomial of real-valued $N \times N$ matrices. Given $X_1, X_2, ..., X_M \in \mathcal{M}_N\{\mathbb{R}\}$, is there any optimal algorithm to determine whether the result of $P(X_1, X_2, ..., X_M)$ is nilpotent?
How about binary exponentiation? Are you expecting an answer that doesn’t just use an algorithm that checks whether a matrix is nilpotent?
I'm looking for anything that behaves better than the straight-forward computation of the polynomial. Preferably something involving as few matrix multiplications as possible (even fewer than what binary exponentiation uses). I am unsure if such an algorithm exists.
|
2025-03-21T14:48:32.003327
| 2020-09-11T00:09:27 |
371383
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"David Handelman",
"David Roberts",
"MaoWao",
"Ulrich Pennig",
"Victor V Albert",
"Yemon Choi",
"https://mathoverflow.net/users/101376",
"https://mathoverflow.net/users/3995",
"https://mathoverflow.net/users/4177",
"https://mathoverflow.net/users/42278",
"https://mathoverflow.net/users/763",
"https://mathoverflow.net/users/95776"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632896",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371383"
}
|
Stack Exchange
|
What is a type $\text{II}_1$ factor von Neumann algebra?
After finding formal definitions in various texts (see, eg, Witten, Notes On Some Entanglement Properties Of Quantum Field Theory, Rev. Mod. Phys. 90, 45003 (2018), doi:10.1103/RevModPhys.90.045003 arXiv:1803.04993), I have ($\text I$) not been able to obtain a good intuition for them and ($\text{II}_1$) have not seen anyone relate such factors to more commonly understood Hilbert spaces such as $L^2(\mathbb R)$. This is very likely for good reason, but it would be nice to have these ideas come down from the clouds so that one can present them to non-mathematicians and (maybe this is too ambitious) laypeople.
Some more technical sources which helped me answer this are here and here.
You say that you "have not seen anyone relate such factors to more commonly understood Hilbert spaces". This phrasing suggests you think that factors are examples of Hilbert spaces? If so, that is already a misconception
Moreover, it is not clear what "intuition" is supposed to mean here. What algebraic objects of a similar nature have you encountered? Obviously there is no point trying to say that two-one factors are analogues of simple groups, if you are not already familiar with results about simple groups. But if you are familiar with examples of simple and non-simple groups then one can start to explain what von Neumann factors are, and then one can try to say something about what makes two-one factors special
@YemonChoi one could be charitable and guess the OP meant how one could see $II_1$ factors as having canonical representations on certain Hilbert spaces arising from other known structures. But I agree it's not worded that way.
@DavidRoberts Fair enough. For what it's worth, I am not sure why the OP is linking to a paper of Witten as their example of a formal definition of a two-one factor. The Wikipedia page on von Neumann algebras is actually fairly good at leading up to the definition of a factor and then the classification result of Murray and von Neumann https://en.wikipedia.org/wiki/Von_Neumann_algebra
@Yemon I agree. One can at minimum chase the WP definition chain to at least get an idea of what objects are involved, and what examples to look at. I was going to suggest von Neumann's original work, which of course was motivated by physics, but I imagine it's low on intuition based on decades of experience, as a modern survey would be!
Sorry all for posting such a vague question. I've looked into it a little, and wrote up something that I think could help the people I had in mind. Feel free to let me know any more thoughts.
It is still not clear to me what the question here hopes to achieve: the updated version has a link to a short set of notes which gives the basic definitions, and to my eyes as a mathematician those notes are more accurate than the answer you've written for your own question. It seems like you want someone to write a version of the standard definition which uses the language/perspective of quantum information theory; but what you've written below has many imprecisions/errors from a mathematician's point of view
I’m voting to close this question because not enough effort has been put in by the proposer.
Consider first the usual Fock space (countably $\infty$-dim vector space) whose basis states $|n\rangle$ are labeled by nonnegative integers $n\geq 0$. This space admits what are called minimal projections: matrices $|n\rangle\langle n|$ which serve as the fundamental building blocks of all matrices that are diagonal in this basis. These building blocks are trace one, and cannot be further divided into sums of two other projections which have lower trace. The presence of such minimal projections implies that any von Neumann algebra on this space is of Type I.
Now consider a tensor product space of $n$ qubits (the vector space $\mathbb{C}^2$) with $n$ being arbitrarily large, and with allowed operators acting nontrivially only on a finite subset of the qubits. Such a space does not admit a minimal projection. Consider a projection
$$
\Pi = |0\rangle\langle0|^{\otimes k}\otimes I^{\otimes (n-k)}
$$
onto the basis state $|0\rangle\langle 0|$ for the first $k$ qubits and identity $I$ on the remaining $n-k$. The trace of this projection is $2^{n-k}$. It is not minimal because we can resolve the identity on the $k+1^{\text{st}}$ qubit and construct projections (with $\ell\in\{0,1\}$)
$$
\Pi_{\ell} = |0\rangle\langle0|^{\otimes k}\otimes |\ell\rangle\langle\ell|\otimes|0\rangle\langle0|^{\otimes (n-k-1)}
$$
that have smaller trace and sum up to $\Pi$. The lack of a minimal projection makes this space a candidate for housing Type II factors.
This ability to always resolve identities of other qubits allows us to keep cutting up projections into finer and finer pieces. This difference between the two spaces reminds me of the fact that any positive real number can always be expressed as a sum of two other positive reals, while there exists an element of the integers (namely, $1$) that cannot be expressed as a sum of two positive integers.
Many of the mathematical definitions/concepts mentioned here are being used in ways that seem to differ from how an "orthodox" functional analyst would use them. For a start, any Hilbert space admits "minimal projections", they just might not be of the form that you are thinking of
To maybe make more clear what Yemon Choi said already: There seems to be a crucial misunderstanding of the role of the Hilbert space. Whether or not there are minimal projections really depends on which projection you "choose" to belong to the von Neumann algebra, not which projections exist on the underlying Hilbert space. Of course there are $\text{II}_1$ factors that can be faithfully represented on the Fock space, in fact, every separable $\text{II}_1$ factor can.
@MaoWao thanks, this helps. Let me see if I understand. One can embed the space of $n$-qubits into a $2^n$-dim Fock space, meaning that the above projections can also be constructed there. So one should not focus on the Hilbert space when constructing these various factors.
Maybe this is helpful: https://arxiv.org/pdf/2302.01958 ? (In particular Section 4.)
|
2025-03-21T14:48:32.003739
| 2020-09-11T02:41:51 |
371391
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Gerry Myerson",
"https://mathoverflow.net/users/158000"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632897",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371391"
}
|
Stack Exchange
|
Smooth function approximating pi(x)
We can define the prime number function as $$\pi(x) = \int_{-\infty}^x \sum_{p}\delta(p-x).dx$$ That is, we include each prime p as a delta function $\delta_p(x) = \delta(p-x)$, set $P(x) = \sum_{p}\delta_p(x)$, and then integrate $P(x)$ from $-\infty$ to $x$ to calculate $\pi(x)$.
In this form, $\pi(x)$ is a step-wise function, discontinuous at each prime p, and constant between consecutive primes.
I wondered whether it might be possible to get a smooth approximation to $\pi(x)$ by smoothly approximating $P(x)$. For example, consider $$q_p(x) = (1/\sqrt(2\pi))\exp(-(x-p)^2/2),$$ and then let $Q(x) = \sum_{p}q_p(x)$. So $Q(x)$ is a function on the real line that has normal distribution "bumps" at each prime, and is close to zero between primes.
I will admit to not having gone through the details, but it feels intuitively clear that $Q(x)$ is a well-defined function, that is, $\sum_{p}q_p(x)$ converges for all $x$ because $\exp(-(x-p)^2/2) < (x-p)^{-2}$.
Similarly, I am pretty sure that $q(x) = \int_{-\infty}^x Q(x).dx$ is well-defined for all $x$. But for clarity, I'd like to understand the following questions:
Is $Q(x)$ well-defined?
Is $Q(x)$ continuous?
Is $Q(x)$ smooth?
Is $q(x) = \int_{-\infty}^x Q(x).dx$ well-defined?
Is $\pi(x) \sim q(x)$?
Apologies if these are dumb or naive questions, but I wonder if functions like $q(x)$ provide a handle on $\pi(x)$ that allow for a different angle of attack on $\pi(x)$.
You may be interested in the smooth approximation discussed at http://empslocal.ex.ac.uk/people/staff/mrwatkin/zeta/encoding1.htm where there's a nice visual showing the convergence to $\pi(x)$ as you take more terms.
|
2025-03-21T14:48:32.003869
| 2020-09-11T03:09:29 |
371392
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Willie Wong",
"https://mathoverflow.net/users/3948"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632898",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371392"
}
|
Stack Exchange
|
About Dirac function
In Vladimirov's book "A Collection of Problems on the Equations of Mathematical Physics", p129, 11.16, there is a equality about Dirac function, which is the fundamental solution of three dimensional wave equation,
$$\frac{\theta(t)}{4\pi a^2t}\delta_{S_{at}}(x)=\frac{\theta(t)}{2\pi a}\delta(a^2t^2-|x|^2),$$ where $(t,x)\in\mathbb{R}\times\mathbb{R}^3$, $\theta$ is Heaviside function, $S_{at}:\{x\in\mathbb{R}^3;|x|=at\}.$
It confuses me, since I always regard $\delta_{S_{at}}(x)$ the same as $\delta(at-|x|)$, $\delta(a^2t^2-|x|^2)$.
This is a consequence of the standard change of variables formula for distributions. See e.g. Section 4 starting at page 69 of https://web.math.princeton.edu/~seri/homepage/courses/Analysis2011.pdf
Your notion and the equality in the book are consistent, since
$$
\delta (a^2 t^2 - |x|^2 ) = \frac{1}{2|at|} (\delta (at-|x|) + \delta (at+|x|) )
$$
(I suppose an assumption is being made that $at>0$).
|
2025-03-21T14:48:32.003966
| 2020-09-11T03:34:58 |
371394
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"AgnostMystic",
"David Roberts",
"https://mathoverflow.net/users/158175",
"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:632899",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371394"
}
|
Stack Exchange
|
Request for books/articles on random polynomials
Can somebody kindly recommend me a couple of introductory books/articles on random polynomials with clear expositions of fundamental results (like the distribution of roots, expected number of real zeroes etc) and some indications of connections and probable applications? Thanking you in advance for any help.
I posted this question on mathestackexchage as well but received no responce. Here is the link https://math.stackexchange.com/questions/3821219/request-for-recommendations-for-random-polynomials
Half a day is far too soon. People live in all the timezones, you know. Maybe the person who could answer your question hasn't been on the internet since you posted it...
@ David Roberts ,my apologies ,i will bear this thing in mind next time i cross post
The two old but interesting books for random polynomials are
A. T. Bharucha-Reid, M. Sambandham. Random polynomials. Probability and Mathematical Statistics, Academic Press, Inc., Orlando, Fla., 1986.
K. Farahmand. Topics in random polynomials. Pitman research notes in mathematics series 393, Longman, Harlow, 1998.
I highly recommend the article by Edelman and Kostlan: https://www.ams.org/journals/bull/1995-32-01/S0273-0979-1995-00571-9/
|
2025-03-21T14:48:32.004086
| 2020-09-11T04:13:35 |
371397
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Sebastian Goette",
"YCor",
"https://mathoverflow.net/users/14094",
"https://mathoverflow.net/users/70808"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632900",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371397"
}
|
Stack Exchange
|
Genericity of equivariant embeddings
I'd like to ask an equivariant version of this question.
Let $M$ be a closed manifold equipped with the action of a compact Lie group $G$. By the Mostow-Palais embedding theorem, $M$ can be embedded equivariantly into $\mathbb{R}^n$ for some $n$, where $G$ is realized as a subgroup of $GL(n,\mathbb{R})$.
Vague question: Suppose $f:M\to\mathbb{R}^n$ is a continuous $G$-equivariant map. I'd like to know how close $f$ is to an embedding, if we are allowed to make $n$ arbitrarily large.
One attempt at a more precise phrasing:
Question 1: Let $f:M\to\mathbb{R}^n$ be a continuous $G$-equivariant map. Is it true that (assuming $n$ is sufficiently large) for any $\epsilon>0$, there exists a continuous $G$-equivariant embedding $f':M\to\mathbb{R}^n$ such that the image of $f'$ is contained in the $\epsilon$-ball around the image of $f$?
More generally:
Question 2: What if $G$ is a non-compact linear group but the quotient $G\backslash M$ is compact?
In the title you wrote "genericity" but you're asking about "density"
It seems from the two answers below, the condition you want is not "$n$ sufficiently large" but "all relevant $G$-representations occur with sufficiently large multiplicity". That means if you fix $G$ compact but not finite and some dimension $k$ , then there might not be a $G$ action on some fixed $\mathbb R^n$ that works for all $G$-manifolds of dimension $\le k$.
I erased my answer as it was inaccurate. Still Q1 is unclear, as it is not said in how $\mathbb{R}^n$ comes with a $G$-action, which in particular makes "for large $n$" very unclear.
Wasserman (Equivariant Differential Topology, Topology, 8(1969), pp. 127-150) proved a generalized Whitney embedding theorem under the assumption that $G$ is compact.
Below you can find the statement of one of his theorems (Corollary 1.10 in the above-mentioned paper).
Def. Let $V$ be a finite dimensional representation of $G$. A $G$-manifold $M$ is subordinate to $V$ if $\forall x \in M \exists U$ invariant neighbourhood of $x$ and equivariant smooth embedding $U\to V^k $ for some $k>0$.
Thm(Wasserman) If $M^n$ is subordinate to $V$, then any equivariant smooth map $f:M\to V^k$ can be approximated $C^r$ and uniformly by an equivariant immersion if $k\geq 2n$ and equivariant 1-1 immersion if $k\geq 2n+1$. If $C$ is a closed subset of $M$, and $f|C$ is an immersion (or embedding)the approximation may be chosen to agree with $f$ on $C$.
|
2025-03-21T14:48:32.004259
| 2020-09-11T04:28:53 |
371398
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Jason Starr",
"abx",
"https://mathoverflow.net/users/127776",
"https://mathoverflow.net/users/13265",
"https://mathoverflow.net/users/40297",
"user127776"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632901",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371398"
}
|
Stack Exchange
|
On the locality of the Lefschetz conditions
Given a smooth projective variety $X$ of dimension $\geq 3$ and an ample divisor $Y$. The pair satisfy the effective Lefschetz condition $Leff(X,Y)$. This implies that the category of vector bundles on a neighborhood of $Y$ is equivalent to the category of vector bundles on the formal completion $\hat{X}_Y$. Now given an open $U$ that does NOT necessarily contain $Y$ and a vector bundle $E$ on $X$. Can we extend any section of $E|_{\hat{U}_{U\cap Y}}$ to $E|_U$? If not are there any obstructions?
"the vector bundle can be uniquely extended to the whole $X$": that's not true.
You are right,extension exists as a unique coherent sheaf not necessarily a vector bundle.
You need that $X$ has dimension at least 3, not 2.
|
2025-03-21T14:48:32.004355
| 2020-09-11T04:39:35 |
371400
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Yvan Velenik",
"https://mathoverflow.net/users/165085",
"https://mathoverflow.net/users/5709",
"probsg"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632902",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371400"
}
|
Stack Exchange
|
GKS inequality with boundary condition
I want to know whether the following version of GKS inequality with boundary condition for Ising model hold or not. Consider Ising model on $\mathbb{Z}^d$ and $\varnothing \neq A\subset \Lambda_1 \subset \Lambda_2 \subset \mathbb{Z}^d$. Let $x \in \Lambda_1 \setminus A$ and $\eta_A$ be a fixed configuration on $A$. Is the following true?
\begin{equation}
\langle \sigma_x \rangle_{\Lambda_1}^{\eta_A} \leq \langle \sigma_x \rangle_{\Lambda_2}^{\eta_A}, \qquad (\star)
\end{equation}
where $\langle \cdot \rangle_{\Lambda}^{\eta}$ is the expectation w.r.t to Gibbs measure on $\Lambda$ with boundary condition $\eta$.
I know that the original GKS inequality works with free boundary condition, i.e. $A=\varnothing$, and implies that for all $B \subset \Lambda_1$ we have $\langle \sigma_B \rangle_{\Lambda_1} \leq \langle \sigma_B \rangle_{\Lambda_2}$, with $\sigma_B=\prod_{y\in B} \sigma_y$.
So, can we prove or disprove the inequality $(\star)$?
Thank you.
Update after the comment by Prof. Yvan Velenik.
As shown by Yvan, the inequality $(\star)$ is not correct in general setting.
I am working with a particular setting as follows. Consider a connected graph $G$ containing several components: $G=A\cup B \cup C$, where $V(A)\cap V(B)=\varnothing$, there is only one edge, say $\{x,y\}$ between $A$ and $B$ with $y \in A$ and $x \in B$; and $C=D \cup E$ with $D\cap A = \varnothing$, $E\cap B=\varnothing$ and $B$ is connected to $D$ and $E$ is connected to both $A$ and $D$. The question is still to check
$$\langle \sigma_x \rangle^{\eta_A}_{B} \leq \langle \sigma_x \rangle^{\eta_A}_G ? \qquad (1)$$
I'am sorry for the complicated construction of $G$. Actually, I'am considering the case $G$ is a random graph locally tree like. I want to truncate the Gibbs measure on $G$ to the measure on a ball (say $B$) around the vertex $x$. So we can expect $A$ and $B$ are very near to trees (there are only a few cycles in $A\cup B$). Assume $|G|=n$ and $|A\cup B|$ is more ore less $(\log n)^3$.
If the inequality is not true, can we expect an approximation like
$$\langle \sigma_x \rangle^{\eta_A}_{B} \leq \langle \sigma_x \rangle^{\eta_A}_G + o_n(1)? \qquad (2)$$
where $o_n(1)$ depends only on $n$.
Thank you.
You can consider the measures in $\Lambda_1\setminus A$ and $\Lambda_2\setminus A$ with free boundary condition, replacing the effect of $\eta_A$ by a suitable magnetic field acting on the vertices on the exterior boundary of $A$.
If this induced magnetic field is nonnegative, then your inequality $(\star)$ is just GKS. This is in particular the case if $\eta_A$ has only $+$ spins along the inner boundary of $A$.
In general, however, the inequality $(\star)$ does not hold. If it did hold for general boundary conditions $\eta$, then you'd get absurd results. Indeed, consider the Ising model without magnetic field. Then,
$$
\langle\sigma_x\rangle_{\Lambda_1}^{\eta_A} = -\langle\sigma_x\rangle_{\Lambda_1}^{-\eta_A} \geq -\langle\sigma_x\rangle_{\Lambda_2}^{-\eta_A} =
\langle\sigma_x\rangle_{\Lambda_2}^{\eta_A},
$$
where the two equalities follow the spin-flip symmetry, and the inequality from $(\star)$. Since, by $(\star)$, you'd also have
$$
\langle\sigma_x\rangle_{\Lambda_1}^{\eta_A} \leq
\langle\sigma_x\rangle_{\Lambda_2}^{\eta_A},
$$
one would deduce that $\langle\sigma_x\rangle_{\Lambda_1}^{\eta_A} =
\langle\sigma_x\rangle_{\Lambda_2}^{\eta_A}$, which is easily seen to be false in general.
What I mean by "replacing the effect of $\eta_A$ by a suitable magnetic field acting on the vertices on the exterior boundary of $A$" is the following: the energy in $\Lambda_1\setminus A$, given that there is a boundary condition $\eta_A$ on $A$, is
$$
- \sum_{\{i,j\}\subset\Lambda_1\setminus A} J_{i,j}\sigma_i\sigma_j - \sum_{i\in \Lambda_1\setminus A}\sum_{j\in A} J_{i,j}\sigma_i\eta_j
=
- \sum_{\{i,j\}\subset\Lambda_1\setminus A} J_{i,j}\sigma_i\sigma_j - \sum_{i\in \Lambda_1\setminus A} \Bigl(\sum_{j\in A} J_{i,j}\eta_j \Bigr) \sigma_i
,
$$
so that the effect of the boundary condition can be seen as effective magnetic fields $h_i$ acting on vertices $i\in\Lambda_1\setminus A$ and given by
$$
h_i = \sum_{j\in A} J_{i,j}\eta_j.
$$
thank you so much Prof. Yvan Velenik for your answer. I have some further questions, please help me if you have suggestions.
Yes, thank you. This is indeed helpful in the case that the external filed $h$ is large enough, say $h \geq -h_i$ for all $i$.
@probsg I do not have an answer to your second, more specific question. If I had to solve it, then I'd try to use the high-temperature representation. It's a bit of a mess when you can have arbitrary boundary conditions, because of signs popping up, but they allow you to represent the expected value of $\sigma_x$ as a sum over paths from $x$ to $A$. This might allow you to exploit the fact that you have a tree structure (and thus only few available paths), which might lead to an inequality with an $o_n(1)$ term as you'd like. Not saying it would be trivial though.
Yes, thanks a lot. I'll try it out.
|
2025-03-21T14:48:32.004646
| 2020-09-11T07:07:12 |
371402
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Harry Gindi",
"https://mathoverflow.net/users/1353"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632903",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371402"
}
|
Stack Exchange
|
Every Spectral Deligne-Mumford stack satsifies fpqc descent?
In SAG Remark <IP_ADDRESS>, Lurie asserts that if we have a representable (by Spectral Deligne-Mumford stacks) natural transformation $X\to Y$ where $Y$ is a functor satisfying fpqc descent, then so too does $X$.
In particular, if we let $Y$ be the representable functor $\operatorname{Spec}(\mathbb{S})$ the sphere spectrum, this functor of points satisfies fpqc descent, so this statement would imply that the functor of points of any spectral Deligne-Mumford stack also satisfies fpqc descent.
However, later on, in the paragraph immediately following Remark <IP_ADDRESS>, he says that the functor of points $h_X$ of a spectral Deligne-Mumford stack need not satisfy fpqc descent.
Is this a mistake? Is it true, at any rate, that such a functor at least always satisfies fppf descent (where fppf means 'faithfully flat and locally almost of finite presentation')? Do we need to require that the functor is representable by a relative Deligne-Mumford n-stack for n finite?
(It seems like at least classically, this also isn't true.)
I proved something weaker today. I don't know if it's the strongest thing you can say, but I showed that if $\mathcal{X}$ is a spectral Deligne-Mumford stack whose underlying $\infty$-topos is a bounded $\infty$-topos in the sense of SAG Appendix A, then $\mathcal{X}$ satisfies fppf descent. The proof goes by showing that the functor sending a spectral DM stack to its category of truncated étale sheaves is a stack for the fppf topology (and this goes through Toën's theorem for derived algebraic stacks, exploiting the nil-invariance of this functor to reduce to the case of ordinary rings).
|
2025-03-21T14:48:32.005033
| 2020-09-11T07:45:10 |
371405
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Christian Gaetz",
"Max Alekseyev",
"Sil",
"https://mathoverflow.net/users/136794",
"https://mathoverflow.net/users/17773",
"https://mathoverflow.net/users/33089",
"https://mathoverflow.net/users/7076",
"kodlu"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632904",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371405"
}
|
Stack Exchange
|
Irreducibility of a palindromic polynomial
I have strong reasons to believe that the palindromic polynomial $p_n(x)$ defined by
$$p_n(x) = x^{2n}+2x^{2n-1}+3x^{2n-2}+ \cdots+ nx^{n+1}+(n+3)x^{n}+nx^{n-1}+\cdots+2x+1$$
is irreducible in $\mathbb{Z}[x]$. I have manually (using Mathematica) checked this for $n$ up to 100.
The Gauss Lemma does not seem to help and I have also considered the irreducibility of $p_n(x) \pmod{p}$, for some primes $p$, but these approaches provide proofs only in specific cases. How do I go about proving this in general?
I don't have any insight for your specific polynomial, but note that it looks similar to the polynomials in https://mathoverflow.net/questions/332958/irreducibility-of-root-height-generating-polynomial for which, according to the comment by Gjergji Zaimi, it is a well-studied and still open problem to prove irreducibility. So there might be some technique which works for your polynomial, but you should not expect such a thing in general.
The polynomial $p_n(x)$ can also be written as
$$p_n(x) = \frac{x^{2n+2}+ 2x^{n+2} - 6x^{n+1}+2x^n+1}{(x-1)^2}.$$
for whatever it's worth, I have tested this up to $n=660,$ using Magma and the polynomial turned out to be irreducible
Also,
$$p_n(x) = \left(\frac{x^{n+1}-1}{x-1}\right)^2 + 2x^n.$$
Based on criteria (Theorem 11) in Irreducibility Criteria for Reciprocal Polynomials and Applications by Cafure and Cesaratto it is sufficient to show that 1) $|p_n(-1)|$ is not a perfect square (in this case simple as $|p_n(-1)| \in {2,3}$) and 2) $q(y)$ is irreducible in $\mathbb{Q}[y]$ where $p_n(x)=x^nq(x+\frac{1}{x})$. Not sure how to prove the second part though, only I have observed that constant coefficients of $q$ are $2,3$ or $4$.
|
2025-03-21T14:48:32.005284
| 2020-09-11T08:16:37 |
371407
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Ben McKay",
"Igor Belegradek",
"YCor",
"coudy",
"https://mathoverflow.net/users/13268",
"https://mathoverflow.net/users/14094",
"https://mathoverflow.net/users/1573",
"https://mathoverflow.net/users/6129"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632905",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371407"
}
|
Stack Exchange
|
Homogeneous metric surfaces
I am looking for a reference for this result.
Let $S$ be a metric space such that
it is homeomorphic to a two-dimensional manifold,
it is 2-homogeneous: given two pairs of points
$(x,y)$ and $(x',y')$ such that $d(x,y) = d(x',y')$ there exists an isometry of $S$ that sends $x$ to $x'$ and $y$ to $y'$.
it is a geodesic space: given two points $x$ and $y$, for $l=d(x,y)$ there exists a curve $\gamma:[0,l]\rightarrow S$ such that $d(\gamma(s), \gamma(t)) =|s-t|$, $\gamma(0) = x$, $\gamma(l) = y$.
Then $S$ is isometric to either the Euclidean plane, some hyperbolic plane, a sphere or its quotient by $\{\mathrm{id}, -\mathrm{id}\}$.
The result appears in a survey in French of Étienne Ghys in a volume entitled "L'héritage scientifique de Poincaré". There is no attribution, only a vague sketch of proof.
I browsed Ghys' papers in search of this result without success.
I think it is in Busemann's work, either Geometry of Geodesics, or Metric methods in Finsler spaces and in the foundations of geometry
It's not true as stated: you should replace "isometric" with "homothetic".
@YCor Indeed there are several hyperbolic and spherical spaces, classified up to isometry by the area of their unit disc. And only one euclidean space. So isometric to one of these.
See theorem 7 in the survey https://arxiv.org/abs/1412.7893, and also https://arxiv.org/abs/2007.11917.
@McKay thanks, indeed the two books of buseman contain results
pretty close to Ghys'statement, e.g. 55.3 in "geometry of geodesics".
I am not sure how Ghys' hypotheses imply axiom IV of Busemann's G-space but the reference to Berestovskii by Igor seems to sort it out.
Busemann is citing Tits (1952) for the two-dimensional case
but I can't find the article at the moment.
|
2025-03-21T14:48:32.005429
| 2020-09-11T08:55:36 |
371412
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Jeremy Rickard",
"Mare",
"https://mathoverflow.net/users/22989",
"https://mathoverflow.net/users/61949"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632906",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371412"
}
|
Stack Exchange
|
Permanent of matrices of finite order
Assume $M$ is a $n \times n$-matrix with entries in $\mathbb{Z}$ such that $M^k$ is the identity matrix for some $k \geq 1$.
Question 1: Is the permanent of $M$ non-zero?
This is tested for many random matrices and for example for $3 \times 3$-matrices with entries in $\{-1,0,1 \}$.
Question 2: What is the largest absolute value the permanent can have for such $M$ with entries in $\{-1,0,1 \}$ for a fixed $n$?
If $M^k$ is the identity, then $\text{det}M=\pm1$, and the permanent is congruent to the determinant mod $2$, so is nonzero.
@JeremyRickard Nice solution. So even being invertible is enough. If you want you can post this as an answer to question 1.
|
2025-03-21T14:48:32.005513
| 2020-09-11T09:11:43 |
371413
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"ABIM",
"Christian Remling",
"Matthew Daws",
"Nate Eldredge",
"https://mathoverflow.net/users/36886",
"https://mathoverflow.net/users/406",
"https://mathoverflow.net/users/4832",
"https://mathoverflow.net/users/48839"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632907",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371413"
}
|
Stack Exchange
|
Multivarate "RKHS" Examples
I've been reading about RKHSs and Hilbert spaces of functions these days a bit these days and I haven't yet come across an example of a hilbert space $H$ whose elements are all functions $f:\mathbb{R}^n\rightarrow \mathbb{R}^m$ for $n,m>1$ and for which the evaluation functions $E_x:f\mapsto f(x)$ are bounded.
Do such objects exist and if so what are some well-known examples?
The only thing I have at the moment is the space of $n\times m$ matrices with Frobenius norm...which is a bit underwhelming...
What is your definition of a RKHS? Because wikipedia thinks a RKHS is a Hilbert space of real/complex-valued functions on a set. So I don't know how one could consider functions to $\mathbb R^m$; nor do I see how matrices fit into this framework, unless we think of matrices as functions from $[n^2]$ to $\mathbb C$?
Oh for me its a hilbert space whose elements are functions (not to the base-field but to a topological vector space) and whose evaluation map is bounded linear operator
Do you have a reference for this sort of generalisation?
Unfortunately not, but I guess any such space should be of the form $H\cong H' \otimes \mathbb{R}^m$ for some (classical) RKHS H'.
What about Sobolev spaces $H^s(\mathbb{R}^n; \mathbb{R}^m)$ for $s > n/2$? I thought this was kind of the canonical example.
Oh really? I didn't know this worked when $m\neq 1$. Do you happen to have a reference? (I've only seen the recent paper of E. Novak ~ 2017)
Maybe I'm missing something, but why would it not work? Each coordinate function $f^{(i)}$ is $H^s(\mathbb{R}^n)$ and so it's continuous and its evaluation functions are bounded, and thus the evaluation functions for $f$ should likewise be bounded with a norm $\sqrt{m}$ times larger.
The standard spaces of entire functions (de Branges space, or just Paley-Wiener space) give you $m=n=2$, and for higher values, you can just take orthogonal sums of these and interpret them as functions on $\mathbb C^N$.
|
2025-03-21T14:48:32.005671
| 2020-09-11T09:47:10 |
371414
|
{
"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:632908",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371414"
}
|
Stack Exchange
|
Existence of smooth function that characterizes boundary and interior of set
It is well known that every closed set $A \subset \mathbb{R}^{n}$ is the zero level set of some smooth function. It follows that every closed set is also the zero sublevel set of some smooth function, i.e.
\begin{align*}
A &= \{x \in \mathbb{R}^{n} : f(x) \le 0 \}.
\end{align*}
I am wondering if one can easily characterize closed sets for which there exist a smooth function $f$ such that the following stronger conditions hold
\begin{align*}
\partial A &= \{x \in \mathbb{R}^{n} : f(x) = 0 \}, \text{ and}\\
A^{\circ} &= \{x \in \mathbb{R}^{n} : f(x) < 0 \}?
\end{align*}
I think every closed set $A \subset \mathbb{R}^{n}$ has this property. Let $\{\phi_k\}_{k\in\mathbb{N}}\subset C^\infty_c(\mathbb{R}^{n})$ a countable collection of non-negative smooth functions with compact support such that $A^\circ=\bigcup_{k\in\mathbb{N}}\{\phi_k>0\}$ (for instance, $\{\phi_k>0\}$ may be balls of some countable covering of $A^\circ$). Then the series
$$ \sum_{k=0}^\infty\, {2^{-k}\| \phi_k\|_{C^k}^{-1}}\,\, \phi_k$$
normally converges with derivatives of any order to a smooth function $f_-$ with $\{f_->0\}=A^\circ$. Here $\displaystyle\| \phi \|_{C^k}=\max_{ \alpha\in\mathbb{N}^n\atop |\alpha|\le k }\|\partial^\alpha\phi \|_{\infty,\mathbb{R}^{n}}$ is the standard ${C^k}$-norm. The same construction for $A^c$ produces a smooth $f_+$ with $\{f_+>0\}=\mathbb{R}^{n}\setminus \overline{A}$; then $f:=f_+-f_-$ has the required properties, that is $\{f<0\}=A^\circ$, $\{f>0\}=\mathbb{R}^{n}\setminus \overline{A}$, and $\{f=0\}=\partial A$.
|
2025-03-21T14:48:32.005789
| 2020-09-11T10:30:58 |
371415
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Liviu Nicolaescu",
"dohmatob",
"https://mathoverflow.net/users/20302",
"https://mathoverflow.net/users/494410",
"https://mathoverflow.net/users/78539",
"tony"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632909",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371415"
}
|
Stack Exchange
|
Concentration inequalities for gradient flows induced by random fields
Let $G=(G(x))_{x \in \mathbb R^m}$ be a conservative random field with values in $\mathbb R^m$, for large positive integer $m$. That is, there exists a scalar random field $g=(g(x))_{x \in \mathbb R^m}$, which depends continuously-differentiably on $x$, such that $G(x) = \nabla g(x)$ for all $x \in \mathbb R^m$. Further ssume that:
$G$ looks thesame every where. $G(x)$, $-G(x)$, and $G(x')$ are equal in distribution for all $x,x' \in \mathbb R^m$. In particular, this implies $G$ is centered., i.e $E[G(x)] = 0$.
Moment bounds. There exists $\alpha,\beta,\gamma > 0$ such that for all $x \in \mathbb R^m$,
$ E[\|G(x)\|^2] \le \alpha$,
$E[\|G(x)\|^4] \le \beta$ (note that this automatically implies the correlation bound $E[\|G(x)\|^2\|G(x')\|^2] \le \beta^2$),
$\gamma_1 \le Var[\|G(x)\|^2] \le \gamma_2$.
Fix $x_0 \in \mathbb R^m$, set $g_0:=\mbox{sign}(g(x_0))$, and consider the differential equation
$$
\begin{split}
\dot{x}(t) &= -g_0G(x(t)), \;t \in \mathbb R\\
x(0) &= x_0.
\end{split}
\tag{1}
$$
For each $t \ge 0$, define $r_t := \|x(t)-x_0\|$ and $c_t := \int_0^t \|G(x(s))\|^2ds$
Goal. My ultimate goal to prove that there is a time $t^\star \ge 0$ (which is deterministic, but may depend on dimension parameter $m \to \infty$) such w.p $1-o(1)$, $g_0g(x(t^\star))<0$ (i.e $g(x(t))$ flips sign from $t=0$ to $t=t^\star$) and $r_{t^\star}$ is as small as possible.
I figured out I could use concentration of measure techniques. This is because, for all $t \ge 0$, $g(x(t)) = g(x_0) - \int_0^t \|G(x(s))\|^2ds$.
Question. How to go about large and small deviation inequalities for $r_t$ and $c_t$ as a function of time $t$ ? More specically, I'd like to how (anti-)concentration inequalities for both $r_t$ and $c_t$.
Softer question. I don't know under what topic such problems fall "concentration of random fields" ?, "oncentration of gradient flows" ?, etc. High-level help and references on this point, would already be very useful.
Observations
Some parts of my problem can be crudely solved. For example, if $u > 0$, then for all time $0 < t < u/\alpha$ we have
$$
P(c_t > u) \le \frac{E[\int_0^t \|G(x(s))\|^2 ds]}{u} = \frac{\int_0^t E[\|G(x(s))\|^2] ds}{u} \le \frac{t\alpha}{u}
$$
Similarly, a concentration bound for $r_t$ can be obtained using Hoelder's inequality combined with Markov like so
$$
\begin{split}
P(r_t > r) &= P\left(\|\int_0^t G(x(s))ds\|^2 > r^2\right) \le \ldots \le t^2 P\left(\int_0^t \|G(x(s))\|^2ds > r^2\right)\\
& \le \frac{t^2\alpha}{r^2},
\end{split}
$$
valid for all $r > 0$ and time $0 < t < r/\sqrt{\alpha}$.
Completely worked-out example
Consider a static random field given by $G(x) = \nabla g(x)$ with $g(x) := w^Tx$, where $w \sim N(0,(1/m)I_m)$. Then integrating (1) gives $x(t) = x_0 - g_0tw$. Thus, $r_t = t\|w\|$ and $c_t = t\|w\|^2$, and there is no shortage of anti-/concentration inequalities in this case. We can take $t^\star = \|x_0\|^2/m \cdot e_m$ for any deterministic $e_m \to \infty$ and get the following w.p $1-o(1)$
$r_{t^\star} = \mathcal O(t^\star)$.
$c_{t^\star} = \Theta(t^\star)$.
$g_0g(x(t^\star)) < 0$.
I think something is missing in the statement of your Goal. What is the goal concerning $r_{t^}$? Also, is $t^$ to be considered deterministic?
Fixed the wording of the Goal (there was a missing piece of text). Concerning, $t^\star$, it should be deterministic (as in the worked example at the end of my post).
Hi. Did you find what would be the relevant literature regarding "concentration of gradient flow"?
|
2025-03-21T14:48:32.006012
| 2020-09-11T11:22:41 |
371419
|
{
"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:632910",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371419"
}
|
Stack Exchange
|
What is a random eigenfunction on the hyperbolic plane?
Is there an (invariant under isometries) notion of a random eigenfunction on the hyperbolic plane, for a given eigenvalue?
It is a reference request because the answer is probably positive and I even have some ideas about how such a thing can be constructed. (This question was inspired by the well known quantum unique ergodicity conjecture but is not really related to it.)
|
2025-03-21T14:48:32.006068
| 2020-09-11T11:54:58 |
371421
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Moishe Kohan",
"https://mathoverflow.net/users/39654",
"https://mathoverflow.net/users/40804",
"mme"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632911",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371421"
}
|
Stack Exchange
|
Homotopy groups of ball complement
Let $X$ be a connected closed topological manifold. Let $n$ be an integer such that $\pi_i(X)=\{0\}$ for $1\leq i \leq n$.
Let $f:B^m\to X$ be a topological embedding, where $B^m$ is the $m$-dimensional ball. Is $\pi_i(X\setminus f(B^m))=\{0\}$ for $1\leq i \leq \mathrm{min}(m, n)$?
The Alexander horned ball, whose complement has nontrivial $\pi_1$, gives a counterexample. Here $\text{min}(m,n) = 2$.
You even can get an example with $m=1, n=3$.
|
2025-03-21T14:48:32.006128
| 2020-09-11T12:23:58 |
371423
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Max Lonysa Muller",
"https://mathoverflow.net/users/93724"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632912",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371423"
}
|
Stack Exchange
|
Are there variations of Ramaswami's formula for the analytic continuation of the Riemann zeta function?
On p. 286 of Borwein's paper entitled "Computational Strategies for the Riemann zeta function", the author mentions a formula due to Ramaswami: $$(1-2^{1-s})\zeta(s) = \sum_{n=1}^{\infty} \binom{s+n-1}{n}\zeta(s+n). $$
I wonder whether variations of this identity also exist. For instance, are there similar binomial sums for $$(1-a^{1-s})\zeta(s) $$ for $a \in \mathbb{Z}\setminus\{2\}$, or is there something special about $a=2$ that makes it work?
And what about products like $$\zeta(s) \prod_{k=1}^{p} (1-a_{k}^{1-s})$$ for some sequence $a_{1}, \dots, a_{p} \in \mathbb{Z}$, does that expression equal any binomial sum(s) in terms of values of the Riemann zeta function?
N.B. I've also asked this question on MSE.
I've corrected some typos. The $2^{-s}$, $a^{-s}$, and $a_{k}^{-s}$ factors should have been $2^{1-s}$, $a^{1-s}$, and $a_{k}^{1-s}$, respectively.
First note that there is a typo in the formula you cite: it should be
$$ (1-2^{1-s})\zeta(s) = \sum_{n=1}^{\infty} \binom{s+n-1}{n}\zeta(s+n) $$
($1-s$, not $-s$). Something "special" in the number $2$ can be found, since $(1-2^{1-s})\zeta(s) = \eta(s)$ (Dirichlet eta function). However, the above formula can be generalised for a general integer $a \geq 2$. I will refer to the paper Lee, H.; Ok, B. M.; Choi, J. Notes on some identities involving the Riemann Zeta function (2002). Communications of the Korean Mathematical Society 17(1):165-173 for such identities.
First of all, we have the following identities obtained by Ramaswami:
$$ (1-3^{1-s})\zeta(s) = 1+2 \sum_{n=1}^{\infty} \frac{(s)_{2n}}{(2n)!} \zeta(2n+s) 3^{-2n-s} $$
$$ (1-2^{-s}-3^{-s}-6^{-s})\zeta(s) = 1 +2 \sum_{n=1}^{\infty} \frac{(s)_{2n}}{(2n)!} \zeta(2n+s) 6^{-2n-s}$$
The first one is the case $a=3$ of your first question, while the second one is a special case of the analogous of your second question, but with a sum instead of a product.
The general case for any integer $a \geq 2$ has been proven by Apostol. Actually, he obtained many interesting identities of this kind:
$$ (1-a^{1-s})\zeta(s) = \sum_{n=1}^{\infty} \frac{(s)_{n} \zeta(n+s)}{n! a^{n+s}} \frac{B_{n+1}(a) - B_{n+1}}{n+1} $$
$$ (1-a^{1-s})\zeta(s) = \sum_{h=1}^{a-1} h^{-s} + \sum_{n=1}^{\infty} (-1)^n \frac{(s)_{n} \zeta(n+s)}{n! a^{n+s}} \frac{B_{n+1}(a) - B_{n+1}}{n+1} $$
$$ (1-a^{1-s})\zeta(s) = \frac{1}{2} \sum_{h=1}^{a-1} h^{-s} + \sum_{n=1}^{\infty} \frac{(s)_{2n} \zeta(2n+s)}{(2n)! a^{2n+s}} \frac{B_{2n+1}(a)}{2n+1} $$
I'm not aware of any formula of this kind for a product as the one of your second question.
Thank you! This is what I was looking for. I'll also correct the typo.
|
2025-03-21T14:48:32.006297
| 2020-09-11T12:27:56 |
371424
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Nulhomologous",
"https://mathoverflow.net/users/128235",
"https://mathoverflow.net/users/158462",
"k.j."
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632913",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371424"
}
|
Stack Exchange
|
The Weil pairing on a generalized elliptic curve
Now I'm trying the section 6 (and 3.20) of chapter IV of Deligne-Rapoport's "Les schemas de module de courbes elliptiques".
I can't understand what $e_n$ (of 6.5.(d)) is.
It seems to be the Weil pairing.
So my question is:
Let $n$ a natural number, $S$ a scheme on which $n$ is invertible, and $C/S$ a generalized elliptic curve, whose each geometric fibre is smooth or the Neron $m$-gon for $m | n$.
Then how can I define the Weil pairing on $C^\text{sm}[n]$?
Here is what I have tried.
First, if $C/S$ is smooth, then trivially we have the Weil pairing.
And so it defines $\mathscr{M}_n^\circ[1/n] \to \mathbb{Z}[\zeta_n]$.
($\mathscr{M}_n$ is the stack classifying the generalized elliptic curves with the level $n$-structures.
And $\mathscr{M}_n^\circ$ is the smooth part.
And $\zeta_n$ is the primitive $n$-th root of unity.)
The authors say that by the definition of normailziation, this induces $\mathscr{M}_n \to \mathbb{Z}[\zeta_n]$.
(I can't understand it, because $\mathscr{M}_n$ is the normalization of $\mathscr{M}_n^\circ[1/n] \to \mathscr{M}_1$.)
And even if we get $\mathscr{M}_n \to \mathbb{Z}[\zeta_n]$, this defines the Weil pairing only for generalized elliptic curves whose each fibre is smooth or $n$-gon.
$e_n$ is explained in IV.3.21, isn't it? Moreover, at least if $n>5$, ${\mathcal M}_n[1/n]$ is the normalization of ${\mathcal M}_n^0[1/n]$ (and this is what you need, since $n$ is invertible in $S$ by hypothesis in 6.5).
@Nulhomologous In 3.21, they define the Weil pairing only for $\mathscr{M}_m$, for $n|m$.
I want to define it for a generalized elliptic curves whose fibres are smooth or $m$-gon, for various $m|n$.
But they show this parings for various $m\mid n$ are compatible with each other with respect to the natural morphisms, isn't it?
@Nulhomologous Where is it? I can't find it.
|
2025-03-21T14:48:32.006434
| 2020-09-11T12:29:19 |
371425
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Anton Petrunin",
"C.F.G",
"https://mathoverflow.net/users/1441",
"https://mathoverflow.net/users/83398",
"https://mathoverflow.net/users/90655",
"mrprottolo"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632914",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371425"
}
|
Stack Exchange
|
Diameter comparison for manifolds with $\text{Ric}\ge 0$
Let $M$ be a Riemannian manifold with $\text{Sec}\ge 0$. From Topogonov Theorem follows that for every $p \in M$ the quantity
$$
\frac{\text{Diam}(B_r(p))}{r}
$$
is non-increasing in $(0,\infty)$. Does this hold also if we only assume $\text{Ric}\ge 0$? I suspect that this is no longer true, however I am not able to build a counterexaple.
Clarification: $\text{Diam}(B_r(p)):=\sup_{x,y\in B_r(p)}d(x,y).$
It is a theorem due to Abresch-Gromoll that diameter growth is linear. Any complete Riemannian manifold
with $Ric \geq 0$ has diameter growth of order $O(r)$ with respect to any point.
The notion of diameter growth in that paper is not the one I am interested in, here by $\text{Diam}(B_r)$ I mean the supremum of $d(x,y)$ among all $x,y \in B_r$, which clearly implies $\text{Diam}(B_r)\le 2r.$
I would try to modify the Perelman's example http://library.msri.org/books/Book30/files/perex.pdf
|
2025-03-21T14:48:32.006534
| 2020-09-11T13:40:17 |
371427
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Richard Stanley",
"Sam Hopkins",
"Zach H",
"darij grinberg",
"https://mathoverflow.net/users/112113",
"https://mathoverflow.net/users/25028",
"https://mathoverflow.net/users/2530",
"https://mathoverflow.net/users/2807",
"https://mathoverflow.net/users/7717",
"lambda"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632915",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371427"
}
|
Stack Exchange
|
Derivations for symmetric functions
A symmetric function is a formal power series in infinitely many variables $x_1,x_2,\dots$ invariant under the permutation of variables (as opposed to a polynomial). Let $\Lambda$ denote the algebra of symmetric functions with coefficents in $\mathbb{Q}$ and $\Lambda_n$ be the subspace of homogeneous degree $n$ symmetric functions. Recall, a derivation is a a linear map $D: \Lambda \to \Lambda$ satisfying the product rule:
$$
D(f \cdot g) = (Df) \cdot g + f \cdot (Dg).
$$
I'm aware of several examples of derivations for $\Lambda$:
Differentiating with respect to a power-sum $\partial/\partial p_k$ where $p_k = x_1^k + x_2^k + \dots$.
In this paper, Nenashev introduces(?) the operator $\nabla$, which acts on the Schur basis by
$$
\nabla s_\lambda = \sum_{(i,j) \ \text{an outer corner of $\lambda$}} (j-i) s_{\lambda \setminus (i,j)}
$$
(here we view $\lambda$ as a subset of $\mathbb{N} \times \mathbb{N}$), and shows it is a derivation.
Additionally, Nenashev proves $\partial/\partial p_k = (\partial/\partial p_{k-1}\nabla - \nabla \partial/\partial p_k)/(k-1)$.
Question: Are there other derivations of $\Lambda$? If yes, have the derivations of $\Lambda$ been classified?
The derivations $\partial/\partial p_1$ and $\nabla$ are both of degree 1, in that they map from $\Lambda_n \to \Lambda_{n-1}$. It'd be especially interesting to identify other degree 1 derivations.
Note $\partial/\partial p_k$ has a nice combinatorial formulation when acting on the Schur basis closely related to the Murnaghan-Nakayama rule. If there are other derivations, do they also have nice combinatorial formulations when acting on the Schur basis? Can derivations be characterized by their actions on the Schur basis?
As a commutative $\mathbb{k}$-algebra, $\Lambda$ is just a polynomial ring in (for example) $h_1, h_2, h_3, \ldots$. Thus, its derivations form a free $\Lambda$-module with basis $\partial / \partial h_1, \partial / \partial h_2, \partial / \partial h_3, \ldots$. That's a lot of derivations :) The derivations that decrease degree form a free $\mathbb{k}$-module with basis $\partial / \partial h_1, \partial / \partial h_2, \partial / \partial h_3, \ldots$; that's still a lot. The question is which of them are combinatorially interesting.
I think "free $\Lambda$-module" is not quite right, because you can also have infinite sums.
Maybe to restrict attention, focus on derivations that interact with $\omega$ in a simple way: e.g., I think we have $\Delta \circ \omega = -\omega \circ \Delta$, right?
If you differentiate the Jacobi-Trudi matrix with respect to $h_i$, then one expresses $\frac{\partial}{\partial h_i}s_\lambda$ (or more generally $s_{\lambda/\mu}$) as a "canonical" linear combination with $\pm 1$-coefficients of skew Schur functions $s_{\nu/1^k}$. E.g., $\frac{\partial}{\partial h_2}s_{2211}=s_{211}+s_{311/1}-s_{22}$. However, this does not seem so interesting to me.
@darijgrinberg I suppose you can just differentiate with respect to the $h_i$'s (or the $e_i$'s for that matter). One thing that's interesting to me is that all of the $\partial / \partial p_k$'s come from $\partial /\partial p_1$ and $\nabla$. Maybe the other families can be expressed similarly?
@lambda: You're right! It's a direct product, not a direct sum.
@ZacharyHamaker You can always just conjugate everything by the algebra automorphism that sends the $p$'s to some other generating set.
|
2025-03-21T14:48:32.006760
| 2020-09-11T14:35:25 |
371429
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Jeff Strom",
"Simon Henry",
"https://mathoverflow.net/users/22131",
"https://mathoverflow.net/users/3634"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632916",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371429"
}
|
Stack Exchange
|
Pushouts and products in categories
This has to do with the "pushout-product" construction.
In a category $\mathcal{C}$, suppose we have $C\gets A\to B$ with pushout $D$
and $Y\gets W\to X$ with pushout $Z$. Then we can form
$$
(C\times Z) \cup_{C\times Y} (D\times Y)
\gets
(A\times X) \cup_{A\times W} (B\times W)
\to
B\times X .
$$
This diagram cones to $D\times Z$.
Questions:
Is $D\times Z$ the pushout?
It is the pushout in set and in various topological contexts, so is $D\times Z$ the pushout for certain good-enough categories?
Is there a good reference for these answers?
The simplest standard assumption for this to work is that the two pushouts you assume at the begining should be preserved by products (at least by products by all the object appearing in the question). So this for example holds in any cartesian closed category (as in this case, product are left adjoint functors and hence preserve all colimits). I'm affraid writting a complete proof involves a bit to much diagrams for MO. The rough idea is as follows:
Using this preservation of the pushout involved by products, you can expand $D \times Z$ as the colimit of some diagram with the shape of $3 \times 3$ array whose node are each products of the form $C \times Y, A \times Y, A \times X, \dots...$. You can then do some formal manipulation on the diagram to regroup the terms in the pushout and write the resulting colimits as this pushout of pushout... Let me know if you need more details.
As Simon says in the comments, it is sufficient that the product preserves pushouts in each variable, which is the case in Set and in any cartesian closed category of spaces. (Indeed, the product can be replaced by any two-variable functor that preserves pushouts in each variable.)
Unfortunately at the moment I can't find a good reference that says exactly this, although I would be more surprised than not if it doesn't exist somewhere in the literature. Simon sketched one proof method in the comments; here's a slightly more abstract one that is at least closer to some things in the literature.
First notice that the goal is equivalent to saying that the pushout product functor $\hat{\times} : \mathcal{C}^{\mathbf{2}} \times\mathcal{C}^{\mathbf{2}} \to \mathcal{C}^{\mathbf{2}} $ takes a pair of pushout squares (regarded as morphisms in the arrow category $\mathcal{C}^{\mathbf{2}} $) to a pushout square. Since pushout squares are closed under composition (again, as morphisms in $\mathcal{C}^{\mathbf{2}} $), it suffices to show that $\hat{\times}$ preserves pushout squares in each variable separately. Thus, we can reduce to the case where we have a pushout $D$ of $C\leftarrow A \to B$ and a morphism $W\to X$, and we want to show that $D\times X$ is the pushout of
$$ (C\times X) \cup_{C\times W} (D\times W) \leftarrow (A\times X) \cup_{A\times W} B\times W \to B\times X $$
Now there is a commutative cube in which the top and bottom faces are the images of our given pushout square $D = C\cup_A B$ under the functors $(-)\times W$ and $(-)\times X$, while the vertical arrows are induced by the map $W\to X$. I trust you can draw this cube; let's orient it so that $A$ and $B$ appear on the back face and $C$ and $D$ appear on the front face.
The back and front faces of this cube are not pushouts. But if we take the pushouts of their underlying spans, the induced maps from these pushouts to the lower-right corners are the two pushout-product maps in question, and the induced square between them is the one we're interested in. A diagram of this sort of "pushouts in two faces of a cube" can be found, for instance, at the top of page 9 of this paper; it's not there in the situation of a pushout product, but the immediate goal is the same, namely to show that the relevant square is a pushout. This follows by repeated application of the pushout pasting lemma (in both directions) to the four squares that are known to be pushouts.
This is great! Thanks
|
2025-03-21T14:48:32.007140
| 2020-09-11T14:54:10 |
371432
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Boris Kunyavskii",
"Mikhail Borovoi",
"Pol van Hoften",
"https://mathoverflow.net/users/4149",
"https://mathoverflow.net/users/56856",
"https://mathoverflow.net/users/84626"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632917",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371432"
}
|
Stack Exchange
|
Globalising tori and weak approximation
Let $G$ be a semi-simple and simply connected reductive group over $\mathbb{Q}$ and let $T \subset G_{\mathbb{Q}_p}$ be a maximal torus. A classical result of Harder tells us that we can find a maximal torus $S \subset G$ such that $S_{\mathbb{Q}_p}$ is $G(\mathbb{Q}_p)$ conjugate to $T$. In fact we can find infinitely many of them; we are allowed to specificy the $G(\mathbb{Q}_v)$-conjugacy class of $S_{\mathbb{Q}_v}$ for finitely many places $v$ of $\mathbb{Q}$.
Question 1: Can we choose $S$ such that $S$ satisfies weak approximation at $p$, i.e., such that $S(\mathbb{Q})$ is dense in $S(\mathbb{Q}_p)$? If $T$ splits over an unramified extension of $\mathbb{Q}_p$, then any $S$ works (c.f. Proposition 7.8 of Platonov-Rapinchuk). One could hope that by changing $S$ at places away from $p$, one might be able to get rid of the obstruction to weak approximation. Perhaps this is too much to ask for, and in fact the following weaker statement would suffice for the applications that I have in mind.
Question 2: Let $x \in T(\mathbb{Q}_p)$, can we always choose $S$ such that $x$ is contained in the closure of $S(\mathbb{Q})$?
I think that Question 1 can be answered in the affirmative using
(the proof of) Theorem 1 in the paper by Gopal Prasad and Andrei Rapinchuk
"Irreducible Tori in Semisimple Groups", IMRN, 2001, No. 23, 1129-1242. The idea is to specialise a generic torus in G and use a result due
to Alexander Klyachko (Borovoi provided an independent proof) which
implies weak approximation for such tori.
Do you mean the paper of Klyachko referred to by Prasad and Rapinchuk? And what text of Borovoi do you mean?
Yes, I meant the paper of Klyachko mentioned by Prasad and Rapinchuk. They also mention your proof, which only exists as a handwritten letter you sent me many years ago.
The question follows from Theorem 1.1.(i) of your link, which says that we make sure that $S$ is "without affect" by changing $S$ at finitely many places away from $p$ (and tori "without affect" have weak approximation). Of course the cited theorem is about absolutely almost simple groups over number fields, but our $G$ will be a product of restrictions of scalars of those.
|
2025-03-21T14:48:32.007321
| 2020-09-11T14:57:29 |
371433
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Moishe Kohan",
"Noel",
"Pierre PC",
"https://mathoverflow.net/users/129074",
"https://mathoverflow.net/users/165187",
"https://mathoverflow.net/users/39654"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632918",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371433"
}
|
Stack Exchange
|
Complement of contractible locally Euclidean subspace
Let $X$ be a connected closed topological manifold. Let $S\subset X$ be a contractible locally Euclidean subspace. Is $X\setminus S$ connected?
This question is more suitable for math stack exchange. How many contractible compact manifolds do you know?
@MoisheKohan what do contractible compact manifolds have to do with the question?
A remark: I don't think it will always be path connected. If $X$ is the torus, we can find, for a given meridian $\ell$, a curve $C$ that loops around the torus an infinite amount of times, converging to $\ell$ (in $\mathbb S^1\times\mathbb S^1$, take $\ell=\lbrace x=0 \rbrace$ and $C=\lbrace (t,\exp(i/t)),t\leq\varepsilon \rbrace$). Take two such curves $C_1$ and $C_2$, and take $S=C_1\cup C_2$. You can even have $S$ connected. I think in this example $X\setminus S$ is connected, though.
Are you interested in partial results? It is not too complicated to show that $X\setminus S$ is connected when $S$ is of codimension at least 2 (in which case $S$ connected is enough, instead of contractible). I think I can also show that if the connectedness of $X\setminus S$ holds for all $S$ of codimension 0 ($S$ is a contractible open set), then it is true for arbitrary contractible $S'$, regardless of the codimension.
|
2025-03-21T14:48:32.007439
| 2020-09-11T16:54:48 |
371440
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Aaron Bergman",
"Igor Khavkine",
"Tim Campion",
"https://mathoverflow.net/users/2362",
"https://mathoverflow.net/users/2622",
"https://mathoverflow.net/users/35508",
"https://mathoverflow.net/users/947",
"user1504"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632919",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371440"
}
|
Stack Exchange
|
What is the relationship between spinors and supermanifolds and fermions?
I have the following two impressions about fermions in physics. I'm confused about their accuracy, and their compatibility:
To consider the behavior of a fermion, whose intrinsic spin is described by a representation $V$ of the group $Spin(p,q)$, on a pseudo-Riemannian manifold $M$ of signature $(p,q)$, you first introduce a spin structure on $M$. Then the fermion field is a section of the bundle associated to $V$.
To consider the behavior of a fermion on a pseudo-Riemannian manifold $M$ of signature $(p,q)$, you first turn $M$ into a supermanifold. Then the fermion field is a superfunction on $M$ with some constraints coming from its intrinsic spin.
Question: Is either of (1) or (2) close to accurate? What major points or subtleties have I missed? If both are close to accurate, then how does one "translate" between the formalism of (1) and the formalism of (2)?
You've got the right concepts, but they're presented in a way that makes me think some context could be helpful.
In #1, you're really talking about the special case where $V$ is one of the spinor representations. In this case, yes, the fermionic fields are sections of the associated spinor bundle $\mathcal{V}$. Sometimes, people write $\Pi \mathcal{V}$ to emphasize that we take the spinor fields to be anti-commuting when we use them to generate an algebra over the structure sheaf of $M$.
The spin-statistics theorem forces this choice on us in 4+ dimensional relativistic QFT. We get causality violations if we don't choose the spinor fields to be anti-commutative. In lower dimensions, the relation between spin & statistics is more complicated. You can have 2d scalars which anti-commute and spinors which commute.
The connection between 1 & 2 is given by Batchelor's Theorem: The structure sheaf of any supermanifold is (non-canonically) isomorphic to the sections of the exterior algebra of some vector bundle on the underlying manifold.
Locally, the idea is elementary: A function on a supermanifold looks like
$$f(x,\theta) = f_0(x) + \sum_i f_1^i(x) \theta_i + \sum_{ij} f_2^{ij}(x) \theta_i\theta_j + ....$$
If the $\theta_i$ form a basis for a spinor representation, then the coefficient functions $f_1$ are the components of a section of the (dual) spinor bundle. The $\theta_i$ anti-commute, so the $f^i_1$ must also anti-commute.
Supermanifolds don't do much for you when you're only thinking about fermions. They're helpful when you want to start packaging the spinors and other fields into representations of a supersymmetry group.
Thanks -- this is exactly the sort of thing I was hoping for, right from the very first sentence! I guess one aspect that puzzles me is that when things are presented as in (2), it's not clear how extract a spin structure from the supermanifold data. And in this respect, the appeal to Batchelor's theorem confuses me because Batchelor relates relates the "super" structure to a plain vector bundle -- with structure group $GL(n)$ -- so where does the spin come from?
But now I think maybe I see -- when you expand a superfunction in local coordinates, it looks a lot like a section of a bundle of Clifford algebras. So maybe the spin structure is obtained as a subbundle of this associated Clifford bundle (via the multiplicative embedding of $Spin(p,q)$ into $Cl(p,q)$)? Part of what makes it confusing is that ordinary manifolds are special cases of supermanifolds, and not every ordinary manifold admits a spin structure...
Yes, the structure sheaf is locally isomorphic to $C^\infty \otimes \wedge E$, and the transition functions for E give you the spin structure. But E has rank 0 in the special case of real manifolds, so there's no transition data.
@TimCampion For the record, your (2) need not presume that the fermion field is a spinor. The point is to capture the anti-commutativity, not the half-integer spin property. If you want your fermion to also be a spinor, a spin structure has to be provided just as in (1) and you have to parity shift $\Pi\mathcal{V}$ the appropriate spinor bundle, as indicated in this answer. Only when $\mathcal{V}$ is locally trivialized to such spinor fermion fields look like simple superfunctions on $M$.
This is a bit of a repackaging of the same info in the other answer, but maybe it will be more clear.
The short answer is (almost) both: A fermion is a section of the parity shifted spinor bundle on a manifold. As such, you can't have a fermion without a spin structure.
Each aspect of this can be considered separately: there is no classical reason that an anti-commutative field has to be a section of the spinor bundle, and there is no reason that a section of the spinor bundle has to be anti-commutative. However, in physics the Spin-Statistics Theorem says that to have a consistent, Lorentz-invariant theory in >2 spatial dimensions, all anti-commutative fields must be spinors (have half-integral spin).
However, you only need the parity shifted bundle here. The full formalism of supermanifolds is for when you have supersymmetry, which is an odd (ie, anti-commuting) symmetry that relates bosons and fermions.
You can look at this in two ways. The first is as supersymmetric quantum mechanics, where you have maps from, say, the super manifold $\mathbb{R}^{1|1}$ to a Riemannian manifold. Here, the need for a spin-structure arises when you try to quantize theory in order to patch together the Clifford algebras that arise on each local chart.
The second way to look at this is to have your fields be functions on a supermanifold. Here, the supermanifold is modeled on super-Minkowski space, which is acted on by the super-Poincare group. In super-Minkowski space, the odd part is (some number of copies of) the parity shifted the spinor bundle, so the need for the spin-structure is part of the definition.
Dan Freed's notes Classical field theory and supersymmetry on this stuff are very good.
Thanks! So it seems the conclusion I took from the other answer was wrong -- (1) and (2) are not equivalent at all! There are many things I still don't understand, but I think the most pressing one is the phrase "anticommutative field" -- because I don't understand this operation of "multiplying" field configurations. Am I correct in understanding that the phrase "anticommutative field" is meaningless classically, and only has meaning when you are doing quantum physics, where observables are supposed to form an algebra?
@TimCampion Yes and no. You can say that fermions are fundamentally quantum mechanical and stop there, but physicists like to do path integrals, which are based on 'classical' actions and involve multiplying (and integrating!) these classical fields. Thus, the formalism of anti-commuting fields/Grassman variables and Berezin integration was used (developed?). Basically, you develop the formalism so that the path integral gives the right answer for fermions, and you can think of the anti commuting fields/Grassman variables as being the (semi-)classical limit of the quantum fields.
|
2025-03-21T14:48:32.007917
| 2020-09-11T17:34:21 |
371442
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Dmitri Pavlov",
"Phil Tosteson",
"Tim Campion",
"https://mathoverflow.net/users/2362",
"https://mathoverflow.net/users/402",
"https://mathoverflow.net/users/52918"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632920",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371442"
}
|
Stack Exchange
|
Higher homotopy groups of Joyal fibrant replacements of 2-coskeletal simplicial sets
Suppose $X$ is a 2-coskeletal simplicial set (meaning $X^{Δ^k}→X^{∂Δ^k}$ is an isomorphism for all $k≥3$).
What is the easiest example of $X$ such that the Joyal fibrant replacement $Y$ of $X$
is not Joyal weakly equivalent to a 2-coskeletal quasicategory?
(Equivalently, mapping simplicial sets between objects of $Y$ have nontrivial homotopy groups in degree 1 or higher.)
If $X$ satisfies the Segal conditions, then $X$ is the nerve of a 1-category, hence is Joyal fibrant,
so such $X$ cannot be an example.
In the Kan model structure on simplicial sets, examples are easy to construct:
the Kan fibrant replacement of the nerve of the delooping of a monoid $M$
is the homotopy group completion of $M$, which can have nontrivial higher homotopy groups.
Up to $\infty$-categorical equivalence, the $n$-coskeletal quasicategories are precisely the quasicategories with $(n-1)$-truncated mapping spaces. So a good place to start would be to ask to slightly weaker question "What are some 2-coskeletal simplicial sets which are not the nerves of 1-categories?"
Maybe you can promote your Kan example into a Joyal example by taking the simplicial category with two objects $a,b$ and $Hom(a,b) = N(M)$ (self maps are just the identity), and then applying the simplicial nerve to get a simplicial set. Haven't checked whether this works.
Emily Riehl has shown that Dugger-Spivak mapping spaces are always 3-coskeletal, and that the homspaces of $\mathfrak C X$ are 2-coskeletal when $X$ is a 1-category.
Regarding the Kan model structure: since every homotopy type is the classifying space of a category, every simplicial is in particular weakly equivalent to a 2-coskeletal simplicial set. Better yet, every homotopy type is the classifying space of a poset, so every simplicial set is weakly equivalent to a 1-coskeletal simplicial set. But of course, a Kan complex which is $n$-coskeletal is $n$-truncated.
@TimCampion: Yes, I forgot to include “fibrant” in the original statement, but you already added it.
Let $P$ be the poset $(\partial \Delta[1]) \star (\partial \Delta[1])$ (where $\star$ means "join"). Note that the classifying space of $P$ is $S^1$. Moreover, as a poset, (the nerve of) $P$ is 1-coskeletal.
There is a "suspension" $\Sigma P$ of $P$, like Phil Tosteson suggests, but constructed in a more hands-on way: $\Sigma P$ has
two objects $\{-,+\}$,
4 nondegenerate 1-cells, all going from $-$ to $+$, corresponding to the 4 elements of $P$, and
4 nondegenerate 2-cells corresponding to the 4 1-cells of (the nerve of) $P$. (in each of these one of the 1-faces is degenerate; there's a choice to make of which one -- let's say that the $\partial_0$ face is degenerate)
An exhaustive (but not too bad) search reveals that $\Sigma P$ is 2-coskeletal -- this is essentially because $P$ is 1-coskeletal and has no nontrivial "composable pairs". But clearly the Joyal fibrant replacement of $\Sigma P$ is not 2-coskeletal -- we have $Hom_{\Sigma P}(-,+) \simeq S^1$ which is not essentially discrete.
To be a bit more careful about that last claim, think about it this way. If we apply $\mathfrak C$ to $\Sigma P$, then I think it's pretty clear that we get the simplicial category which I'd also denote $\Sigma P$, with two objects $\{-,+\}$, and with the homspace $Hom(-,+)$ given by (the nerve of) $P$. Since every simplicial set is Joyal-cofibrant and $\mathfrak C$ is left Quillen, we haven't messed up the $\infty$-categorical equivalence class of $\Sigma P$.
Then, a Bergner-fibrant replacement of this simplicial category can be found by simply Kan-fibrantly replacing the homspaces levelwise, and we find that indeed we have an $\infty$-category with two objects $-,+$ and the only nontrivial homspace being $Hom(-,+) \simeq S^1$. This is a model-independent statement, so the Joyal-fibrant replacement of $\Sigma P$ likewise has this property, which shows it's not equivalent to an ordinary 1-category, and hence not equivalent to anything Joyal-fibrant and 2-coskeletal.
After a bit more thought, I'm pretty sure that the above construction works in greater generality: let $P$ be an arbitrary 1-coskeletal simplicial set (note that every homotopy type is modeled by a poset, which is in particular 1-coskeletal). Construct $\Sigma P$ as indicated above. Then there is still a close relationship between $n$-simplices of $P$ and $(n+1)$-simplices of $\Sigma P$, with the result that $\Sigma P$ is 2-coskeletal. Moreover, the same argument as above shows that $Hom_{\Sigma P}(-,+) \simeq P$ (which can be an arbitrary homotopy type).
|
2025-03-21T14:48:32.008223
| 2020-09-11T17:40:41 |
371443
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Andrés Felipe",
"Giorgio Metafune",
"Nik Weaver",
"https://mathoverflow.net/users/150653",
"https://mathoverflow.net/users/152735",
"https://mathoverflow.net/users/23141"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632921",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371443"
}
|
Stack Exchange
|
On the dimension of the range of the resolution of the identity
I want to prove the following: Let $A,B$ be bounded self-adjoint operators in a complex-Hilbert space and $E_A(\lambda)$, $E_B(\lambda)$ its corresponding spectral resolutions, i.e.,
$$A=\int_{[m_A,M_A)}t\;dE_A(t)\qquad\text{and}\qquad B=\int_{[m_B,M_B)}t\;dE_B(t).$$
If $A\geq B$ (in the sense of positive operators) then $\mathrm{dim}(\mathrm{rg}E_A(\lambda))\leq \mathrm{dim}(\mathrm{rg}E_B(\lambda))$ for all $\lambda\in\mathbb{R}$.
I think for each $\lambda$ we can define the linear operator $T_\lambda:\mathrm{rg}(E_B(\lambda))\to\mathrm{rg}(E_A(\lambda)),\;x\mapsto E_A(\lambda)x$ and prove that this opeator is surjective but I do not know how to do it.
Can someone give me an idea? I will be grateful.
Suppose the range of $E_A(\lambda)$ has strictly larger dimension than the range of $E_B(\lambda)$, for some $\lambda$. Then we can find a vector $v$ in the first range which is orthogonal to the second range, i.e., is in the range of $I-E_B(\lambda)$. Let $P$ be the orthogonal projection onto the (one dimensional) span of $v$. Since $A\geq B$, also $PAP \geq PBP$. But $PAP \leq \lambda P < PBP$, contradiction.
(I am taking $E_B(\lambda)$ to be the spectral projection of $B$ for the interval $(-\infty, \lambda]$; if you want it to be $(-\infty, \lambda)$ then you would have $PAP < \lambda P \leq PBP$.)
You mean $v\in\mathrm{rg}(I-E_B(\lambda))$?
There is also a second part in the question which follows similarly. If $X$ is the Hilbert space and $E_AE_B(X) \neq E_A (X)$ (I omit $\lambda$), there is a norm-one $x \in E_A(X)$ orthogonal to $E_AE_B(X)$. Then $0=(x, E_AE_By)=(x,E_By)=(E_Bx,y)$ for all $y$ gives $E_Bx=0$. Then $x \in (I-E_B)(X)$ and then $(Bx,x)>\lambda$ and $(Ax,x) \geq (Bx,x) >\lambda$, in contrast with $x \in E_A(X)$.
@AndrésFelipe you are right, corrected.
It is not clear to me why $PAP\leq \lambda{P}$. Do I have to use the integral representation of A?
@AndrésFelipe the cheap way to do this is to assume, without loss of generality, that $A$ is a multiplication operator, $A = M_f$ on some $L^2(X)$, and the range of $E_A(\lambda)$ is the set of vectors supported on ${x: f(x) \leq \lambda}$.Then $\lambda P - PAP = (\lambda I - A)P = M_{\lambda - f)P$ is multiplication by a positive function on that range, so it is positive.
I think it also works: If $x\in\mathrm{rg}(E_A(\lambda))$ then $$(Ax,x)=(AE_A(\lambda)x,x)=\left(\int_{\mathbb{R}} t\chi_{(-\infty,\lambda]}(t)dE_A(t)x,x\right)=\int_{\mathbb{R}} t\chi_{(-\infty,\lambda]}(t)d\left(E_A(t)x,x\right)\leq \lambda(x,x)$$
|
2025-03-21T14:48:32.008427
| 2020-09-11T19:04:25 |
371448
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Chris Gerig",
"Henry Adams",
"Igor Belegradek",
"https://mathoverflow.net/users/12310",
"https://mathoverflow.net/users/1573",
"https://mathoverflow.net/users/45451"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632922",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371448"
}
|
Stack Exchange
|
Existence of equivariant triangulations
I'm looking for a compatibility result which links two types of structures that could be imposed on a topological space $X$:
Call $X$ triangulable if there exists a finite simplicial complex $K$ whose geometric realization $|K|$ admits a homeomorphism to $X$.
Call $X$ involutive if it admits a nontrivial $\mathbb{Z}/2$-action, in the sense that there is a continuous $\iota:X \to X$ satisfying $\iota \circ \iota = \text{id} \neq \iota$.
Here is the question: let's say a given space $X$ admits a triangulation $K$ and an involution $\iota$ in the senses described above. Is it true that we can find another triangulation $K'$, ideally a subdivision of $K$, with respect to which $\iota$ is a simplicial map?
There are more general versions of this question about which I'm curious, obtained by replacing $\mathbb{Z}/2$ by another finite group $G$ acting on a triangulable $X$. If the desired statement is true, I would love to be pointed to a reference; and if it is false, I wonder if there is a nice obstruction to the existence of $K'$.
I imagined Bredon's book on transformation groups discusses this, but I just skimmed it without luck. What if we pass to the quotient $X/G$, try to put a simplicial complex on that, and then lift it? (A related result is that a CW-complex of the base of a regular covering map lifts to a G-CW-complex on the cover, where G is the deck group.)
Try https://shareok.org/handle/11244/319144. The page has a pdf link to J.Dover's thesis "Equivariant Piecewise-Linear Topology and Combinatorial Applications".
In the very particular case when $X$ is a smooth compact manifold, see the "equivariant triangulation theorem" at https://ncatlab.org/nlab/show/equivariant+triangulation+theorem and https://link.springer.com/article/10.1007/BF01405351. There appears to also be a version where $X$ is allowed to have boundary.
There are involutions $\sigma$ of the 3-sphere, whose fixed-point sets are wild 2-spheres: The fixed-point set cannot be a subcomplex of any triangulation, hence, $\sigma$ cannot be PL in any triangulation.
Bing, R. H., A homeomorphism between the 3-sphere and the sum of two solid horned spheres, Ann. Math. (2) 56, 354-362 (1952). ZBL0049.40401.
See also here for Calegari's take on Bing's proof.
Edit. There is even (unique in some sense) free involution of the 4-sphere which cannot preserve a triangulation (this is due to Ruberman). Thus, a bad fixed point set is not the only obstruction.
|
2025-03-21T14:48:32.008643
| 2020-09-11T20:02:15 |
371456
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Balazs",
"Phil Tosteson",
"R. van Dobben de Bruyn",
"Tabes Bridges",
"Yellow Pig",
"https://mathoverflow.net/users/104669",
"https://mathoverflow.net/users/12395",
"https://mathoverflow.net/users/27219",
"https://mathoverflow.net/users/52918",
"https://mathoverflow.net/users/6107",
"https://mathoverflow.net/users/82179",
"ssx"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632923",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371456"
}
|
Stack Exchange
|
Philosophical underpinnings of Grothendieck's construction of the Hilbert scheme
Long ago when I was in grad school I was told that Grothendieck's construction of the Hilbert scheme is rooted in two main technical points: Castelnuovo-Mumford regularity and Mumford flattening stratification. In grad school I made sure to learn these two constructions very carefully, and I have been dealing with the Hilbert scheme (mostly of points in the plane) closely in my work since then, but I still don't understand at all on a philosophical level why the Hilbert scheme exists and why these two technical constructions play the main role in its existence. Could someone enlighten me regarding these questions? In general, why is the subject of Hilbert schemes so insanely technical if they are so incredibly useful in mathematics, and what could be a 'correct' point of view on Hilbert schemes which would make them easier to understand from the philosophical perspective? What would be good suggestions for literature which could help to understand the technicalities of the subject of Hilbert schemes on a more philosophical level?
I have no idea if this is historical, but here is one way to look at it. If you want to represent a functor, you need to find some space you can embed everything in. For example, for moduli of smooth curves of genus $\geq 2$, you can use the tricanonical embedding to put them all in the same projective space. For subschemes of a projectivised scheme $(X,H)$, the natural thing to look at is embeddings defined by $nH$. Just like in the curve case, you need some a priori bound on what multiple of $H$ you need, and that's where Castelnuovo–Mumford regularity comes in.
Thanks a lot! Well, I know that the construction of the Hilbert scheme is based on embedding into a bigger space, I guess this is what it starts with. Thanks a lot for the information though!!
Right, I figured none of this was new information, which is part of the reason I put it in the comments. But to me it does offer most of the explanation of why CM regularity exists and is used in the construction, and even how one would come up with it.
Sure, thanks a lot, I'll take this into account.
As for the flattening stratification, I have never viewed this as something incredibly deep. We know that Hilbert polynomials are constant in flat families, and in general vary in a constructible way. The flattening stratification is just a neat way to package this geometrically. If you want to be philosophical about it, it's some incarnation of the (model-theoretic) idea that finite type things (like coherent sheaves) behave in a constructible way. For example, for coherent sheaves on the base itself, there is an open where it is locally free, and then proceed by Noetherian induction.
Hmm, okay, thanks, maybe you are right and there indeed isn't much philosophy behind the technicalities of Hilbert schemes.
The could be, and I would be very happy to hear if someone has something more useful to say! But at least I am somewhat convinced that I might be able to come up with them myself if I already know the EGA language well.
Cool! What about Hilbert, did he have something to do with how Grothedieck came up with the idea of Hilbert schemes?
Not an expert, but the use of CM regularity is somewhat intuitive to me. The idea is that for $\mathbb P^n$ you want to parameterize graded ideals in $k[x_1, \dots, x_n]$ which a priori is a closed subset of an infinite product of Grassmannians. CM regularity lets you cut this down to the Grassmannian of monomials of degree equal to the regularity. This is just a restatement of what R. van Dobben De Bruyn said.
@YellowPig Hilbert invented the Hilbert polynomial :P
Sorry, I suspected from the beginning that my comment about Hilbert didn't have much depth (and I am certainly aware of the fact that Hilbert invented the Hilbert polynomial :) ), but I guess I'm still curious, did the idea of Hilbert schemes first occur to Grothendieck, or were there precursors to it?
Historically the Chow variety came first, and before that, the Grassmanian. All these are attempts to parametrise classes of sub-objects of projective space. I would say the new "philosophical" underpinning of Grothendieck is the functor of points point of view, and the related notion of a flat family. The results you mention are tools to make this work.
A couple of random thoughts: the Chow variety, which parametrizes cycles, seems to go hand in hand with the primitive language of varieties, with its destructive operations like taking the radical of an ideal. There is a great deal of information lost, and if I recall correctly, the Chow variety is not functorial. The cleaner notion of a scheme made a cleaner construction possible. As was often the case though, I have a feeling Grothendieck was the only person both bold and blind enough (in terms of his minimal background in classical AG) to think that such a thing was possible.
|
2025-03-21T14:48:32.009029
| 2020-09-11T20:13:41 |
371457
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Fedor Petrov",
"Mare",
"darij grinberg",
"https://mathoverflow.net/users/2530",
"https://mathoverflow.net/users/4312",
"https://mathoverflow.net/users/61949"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632924",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371457"
}
|
Stack Exchange
|
Trace of a matrix associated to posets
Let $P$ be a finite connected poset with $n$ elements. Let $C=(c_{x,y})$ be the $n \times n$ matrix with entry 1 in case $x \leq y$ and 0 else.
The Coxeter matrix of $P$ is defined as the matrix $M_P=-C^{-1}C^T$.
Let $u_P$ be defined as the trace of $M_P^2$.
Question 1: Is it true that $u_P$ is an odd integer in case $P$ is a lattice? Does it have a nice interpretation in this case?
I can prove question 1 for distributive lattices.
Question 2: Is $u_P$ an odd integer in case $P$ is just a bounded poset?
Questions 1 and 2 have a positive answer for posets with at most 9 points. In case this is true for general bounded posets, there surely is a nice reason.
For a general connected poset $P$ $u_p$ can be zero.
Here is how to get a poset (not bounded) with $u_p=0$ using Sage:
n=6
posets=[P for P in Posets(n) if P.is_connected()]
U=[P for P in posets if ((P.coxeter_transformation())^2).trace()==0]
P=U[0]
display(P)
Note that every integer $n\times n$-matrix $A$ satisfies $\operatorname{tr}\left(A^2\right) \equiv \operatorname{tr}A \mod 2$. Thus you don't need to square the matrix when defining $u_P$.
And yes: Whenever $P$ has a global maximum, we have $\operatorname{tr}\left(M_P\right) = 1$, so $u_P$ is odd. The proof is really nice! I'm wondering what can be said about the parity of $u_P$ in general.
@darijgrinberg The hint with the trace is nice, in fact we have in general $Tr(A^2)=(Tr(A))^2-2 o_2 (A)$ when $o_2$ is the coefficient of the degree $n-2$ monomial of the characterisitc polynomial of $A$ when $A$ is an $n \times n$-matrix. See https://math.stackexchange.com/questions/506962/expressing-the-trace-of-a2-by-trace-of-a/508536 .
@darijgrinberg Im sorry, I just realised that the example has no global maximum (it just looks like that in the picture). It would be interesting to see your proof.
$u_P$ has the same parity as the number of non-empty chains in $P$ (this immediately proves the case mentioned by @darijgrinberg when the global maximum $M$ exists: you may add or remove it to any chain, so partition all non-empty chains except ${M}$ onto pairs). But what if $P={a_1,a_2,b_1,b_2}$ and inequalities $a_i<b_j$? I see 8 non-empty chains.
@FedorPetrov Thanks, for this nice solution. I turned Darji's and your comments into a (commutatiy) answer. If you want you can also turn your comments into an answer and I delete my answer.
@darijgrinberg Thanks for your comment with $tr(A^2)=tr(A)$ mod 2. This made the problem much easier. This problem has by the way a nice application to homological algebra. I might write you an email soon.
Here is a nice proof for posets $P$ with a global maximum $M$ (it works dually for posets with a global minimum, but not for general posets as the example in the question shows) suggested by comments of Darij Grinberg and Fedor Petrov.
Note first that the entries of the coxeter matrix $Co=Co_P$ (formerly known as $M_P$) are given by $Co_{x,y}=- \sum\limits_{z \in P: z \geq x}^{}{\mu(y,z)}$, where $\mu$ is the Moebius function of $P$.
Now for any square matrix $A$ over the ring of integers, we have $\operatorname{tr}(A^2) \equiv \operatorname{tr}(A)^2 \equiv \operatorname{tr}(A) \mod 2$.
Thus when we show that $\operatorname{tr}(A) \equiv 1 \mod 2$, we are done.
Now by Hall's theorem on chains we have $\mu(x,y)=-c_1+c_2-c_3+c_4-....$ when $c_i$ denotes the number of length $i$ chains starting at $x$ and ending at $y$.
Mod 2 this is equal to the number of chains from $x$ to $y$ (since there is difference between - and + in mod 2).
Thus $\operatorname{tr}(Co)=- \sum\limits_{x \in P} \sum\limits_{z \in P: z \geq x}^{}{\mu(x,z)}$ is equal mod 2 to the total number of non-empty chains in the poset.
Now for any chain $K$ we can add or remove the maximum $M$ of the poset $P$, so we can partition all non-empty chains except $\{ M \}$ into pairs.
Thus $\operatorname{tr}(Co) \equiv 1 \mod 2$.
|
2025-03-21T14:48:32.009428
| 2020-09-11T22:40:50 |
371463
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Ben McKay",
"Robert Bryant",
"Robin Goodfellow",
"https://mathoverflow.net/users/13268",
"https://mathoverflow.net/users/13972",
"https://mathoverflow.net/users/58187"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632925",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371463"
}
|
Stack Exchange
|
When are principal bundles supporting Cartan connections isomorphic?
Suppose I have two Cartan geometries $(\mathscr{G}_1,\omega_1)$ and $(\mathscr{G}_2,\omega_2)$ of type $(G,H)$ over the same manifold $M$. What conditions on $G$ and $H$ allow us to conclude that $\mathscr{G}_1$ and $\mathscr{G}_2$ are isomorphic as principal $H$-bundles?
It seems to be a common implicit assumption in the literature that $\mathscr{G}_1$ is always isomorphic to $\mathscr{G}_2$ in the cases we usually look at. In particular, for parabolic geometries, it seems to be folklore that this is true.
Previously, I had implicitly assumed that such an isomorphism always exists for Cartan geometries of all types, but I recently thought of the following example. If I have a Hermitian holomorphic line bundle, then I can construct a Cartan geometry of type $(\mathbb{C}^m\rtimes\mathrm{U}(1),\mathrm{U}(1))$ corresponding to the Chern connection. However, in general, there are too many line bundles over a given complex manifold for them to all be associated (in the sense that $L\cong\mathscr{G}\times_{\mathrm{U}(1)}\mathbb{C}$) to the same principal $\mathrm{U}(1)$-bundle, so there must be nonisomorphic principal $\mathrm{U}(1)$-bundles admitting Cartan connections of this type over the same manifold.
I’ve thought about this for a few days now, and I imagine there’s probably a nice general condition on $(G,H)$, but I’m not seeing what that condition might be.
I had examples in one of my papers of non-isomorphic holomorphic bundles arising as total spaces of holomorphic Cartan geometries. I think I proved that this doesn't happen for holomorphic parabolic geometries. But I don't know about real geometries.
In fact, I have a simple example of non-isomorphic holomorphic bundles arising as total spaces of holomorphic Cartan geometries, with the same model, on complex surfaces: https://arxiv.org/abs/1105.4732, but they are isomorphic as real bundles.
This is not a complete answer, but I think that it might help to clear up some misunderstandings. It is not, in general, true that all the principal $H$-bundles over $M$ supporting a Cartan connection of type $(G,H)$ are isomorphic, though the OP's proposed example does not actually show this. I think that the following discussion may help.
To fix notation, let's recall what we mean by a "Cartan connection of type $(G,H)$": Here $G$ is a Lie group with Lie algebra $\frak{g}$ and $H$ is a Lie subgroup with Lie algebra ${\frak{h}}\subset{\frak{g}}$. The representation $\mathrm{Ad}:H\to\mathrm{Aut}({\frak{g}})$ preserves the subalgebra ${\frak{h}}$ and so induces a representation $\rho:H\to \mathrm{Aut}({\frak{g/h}})$. If $\pi:B\to M$ is a principal right $H$-bundle, let $X_v$ for $v\in\frak{h}$ be the vertical vector field on $B$ whose flow is right action by $\mathrm{exp}(tv)$. Then a Cartan connection of type $(G,H)$ on $\pi:B\to M$ is a $\frak{g}$-valued $1$-form $\gamma:TB\to \frak{g}$ with the following properties:
$\gamma_u:T_uB\to{\frak{g}}$ is an isomorphism for all $u\in B$.
$\gamma\bigl(X_v(u)\bigr) = v$ for all $u\in B$ and all $v\in\frak{h}$.
$R^*_h(\gamma) = \mathrm{Ad}(h^{-1})(\gamma)$ for all $h\in H$.
It is important to note that not every principal right $H$-bundle over $M$ supports a Cartan connection of type $(G,H)$. This is because such a Cartan connection $\gamma$ defines an isomorphism $\iota_\gamma:TM\to B\times_\rho {\frak{g/h}}$. To see this, let $\omega = \gamma\,\mathrm{mod}\,{\frak{h}}:TB\to {\frak{g/h}}$. The above axioms imply that $\omega_u:T_uB/V_uB\to {\frak{g/h}}$ is an isomorphism for all $u\in B$, where $V_uB\subset T_uB$ is tangent to the fiber of $\pi:B\to M$. Since we have a canonical isomorphism $T_uB/V_uB\to T_{\pi(u)}M$, it follows that we can regard $\omega$ as defining an isomorphism $\omega_u:T_{\pi(u)}M\to {\frak{g/h}}$ for all $u\in B$ that satisfies $\omega_{u\cdot h} = \rho(h^{-1})(\omega_u)$ for all $u\in B$ and all $h\in H$. By the very definition of $B\times_\rho{\frak g/h}$, this establishes the claimed isomorphism $\iota_\gamma:TM\to B\times_\rho{\frak g/h}$.
Conversely, if an isomorphism $\iota:TM\to B\times_\rho{\frak g/h}$ is given, then one can construct a Cartan connection of type $(G,H)$ on $B$.
Thus, one can see why the OPs construction starting with a line bundle $L$ endowed with a $\mathrm{U}(1)$-connection does not automatically imply that there is a Cartan connection of the desired type on $M$. For example, in this case, if a Cartan connection existed, then $TM$ would have to be isomorphic to $L\otimes \mathbb{C}^n = B\times_\rho {\frak g/h}$, and this is generally not the case.
However, there is a simpler example to demonstrate that not all $H$-bundles that admit Cartan connections of type $(G,H)$ are isomorphic: Here, let $n=3$, let $H=\mathrm{SO}(2)$ and let $G = \mathbb{R}^3\rtimes H$, where $H=\mathrm{SO}(2)$ acts on $\mathbb{R}^3$ by rotation in the second and third coordinates. An $H$-bundle $\pi:B\to M^3$ is just an $\mathrm{SO}(2)$-bundle, so it has an Euler class (which could be nonzero) and the associated bundle $B\times_\rho \mathbb{R}^3$ is a sum of a trivial bundle and a $2$-plane bundle. If there is a Cartan connection on $B$, then we get an isomorphism of $TM$ with the sum of a trivial bundle and a $2$-plane bundle. In particular, this means that $M$ is oriented and we have a nonvanishing vector field on $M$ together with a $2$-plane subbundle that has a well-defined Euler class.
Now, every oriented $3$-manifold has a trivial tangent bundle, but once one chooses a nonvanishing vector field, the Euler class of the complementary $2$-plane bundle is determined and may very well be nonzero. For example, let $M = S^1\times S^2$. If we choose the vector field tangent to the $S^1$-fibers, then the complementary $2$-plane field is nontrivial on each $S^2$-fiber. Meanwhile, if we choose a trivialization of the tangent bundle of $M$, then letting the vector field be one of the three trivializing vector fields, the complementary $2$-plane bundle will be trivial.
Thus, we can have two $H$-bundles over $M$ that are not isomorphic even though they both admit Cartan connections of type $(\mathbb{R}^3\rtimes H,\ H)$.
It follows that the very first criterion one needs to have in order for all the Cartan connections of type $(G,H)$ to have isomorphic underlying $H$-bundles is that all of the structure reductions of the canonical $\mathrm{GL}(n,\mathbb{R})$-structure on $TM$ to a $\rho(H)$-structure be isomorphic. This is a very strong condition on $\rho(H)$ and $M$, and whether it is met depends on both $\rho(H)$ and $M$.
Meanwhile, for most of the familiar examples in parabolic geometry, $\rho(H)$ is some large group such as $\mathrm{GL}(n,\mathbb{R})$, $\mathrm{SL}(n,\mathbb{R})$, $\mathrm{CO}(n)$, or $\mathrm{SO}(n)$, and it happens that this uniqueness is met trivially. This may account for the common (false) belief that prompted this question in the first place.
Thank you so much. I’m kind of embarrassed that I convinced myself Chern connections induce Cartan connections, but your post was helpful in clearing up why I was wrong and (partially) answering the main question.
@RobinGoodfellow: Well, I now think that, probably, the answer to the main question is going to be that "$\rho(H)$ contains a maximal compact subgroup of $\mathrm{Aut}(\frak{g/h})$" is the necessary and sufficient condition that all the $H$-bundles over any $M$ that support a Cartan connection of type $(G,H)$ be isomorphic.
As a small follow-up: you say that most familiar examples in parabolic geometry satisfy your criterion trivially. While I agree for the usual examples, for more general parabolic geometries this seems like it might be questionable. For example, if I have a parabolic geometry from the (adjoint) split-real form of the exceptional simple Lie group of rank 2 coming from crossing out the smaller node of the Satake diagram, then $G_0$ will be isomorphic to $\mathrm{GL}_2\mathbb{R}$, which feels too small to trivially satisfy the criterion...
...Is it known whether it holds for all parabolic geometries?
Automorphisms of $\mathfrak{g/h}$ with what additional structure? Or do you just mean a copy of $\mathrm{O}(\dim(\mathfrak{g/h}))<\mathrm{GL}(\mathfrak{g/h})$?
@RobinGoodfellow: I didn't say for all parabolic geometries, just the most familiar. There is no reason that it would hold for the one you are talking about, with $\rho(H)$ being a $5$-dimensional representation of $\mathrm{GL}(2,\mathbb{R})$, and I wouldn't expect it to be so. I think that, for most cases when $\rho(H)$ is 'small', uniqueness will fail. For your second question, yes, I just meant $\mathrm{GL}({\frak{g/h}})$ when I wrote $\mathrm{Aut}({\frak{g/h}})$.
Fascinating! Do you think, then, that there is an example of two Lorentzian geometries (torsion-free Cartan geometries of type $(\mathbb{R}^m\rtimes\mathrm{O}(1,m-1),\mathrm{O}(1,m-1))$) on the same manifold whose “orthonormal” frame bundles are non-isomorphic? I’m not particularly familiar with indefinite signature metrics, but this sounds like a very interesting (and counterintuitive) result.
@RobinGoodfellow: Well, yes. There will be such examples for $m=3$: Just take the two non-isomorphic examples for $\bigl(\mathbb{R}^3\rtimes \mathrm{SO}(2), \mathrm{SO}(2)\bigr)$ on $3$-manifolds that I mentioned and notice that $\rho\bigl(\mathrm{SO}(2)\bigr)$ sits as a maximal compact in $\mathrm{SO}^\circ(1,2)\subset\mathrm{GL}(3,\mathbb{R})$. In other words, there will be $3$-manifolds (e.g. $M=S^1\times S^2$) that admit two oriented, time-oriented conformal Lorentzian structures whose underlying Cartan conformal connection bundles are not isomorphic.
|
2025-03-21T14:48:32.010024
| 2020-09-11T23:26:04 |
371466
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Andrés E. Caicedo",
"https://mathoverflow.net/users/164757",
"https://mathoverflow.net/users/6085",
"thatguythatroamsforums"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632926",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371466"
}
|
Stack Exchange
|
Can someone explain to a "newbie" of number theory how Matijasevič demonstrated the impossibility of hilbert’s tenth problem?
As the title stated, I'm an amateur in the number theory that has just approached hilbert's tenth problem and the demonstration given by Matijasevic, but I couldn't find much on it, and what I could find was overly complicated for my self taught knowledge or just poorly traslated from russian up to the point of becoming impossible to read. Can someone please give me an explanation of the demonstration without taking for granted my knowledge of PhD mathematics?
This is a very accessible book. It explains the details and provides motivation.
There is also this book.
@AndrésE.Caicedo I was looking for something more synthetic, or is the explanation just to big to fit somewhere else than in a book?
I don't know your background to know whether a one line description would be meaningful. The books cover all prerequisites as well.
For instance, one can state the result as saying that (PA proves that) $\Sigma^0_1=\Sigma_1$. I don't know whether this makes sense to you or, even if it does, whether you see why it is indeed a solution to the tenth problem.
My suggestion is that you state in the body of the problem what your background is. For instance, are you familiar with Turing machines? With first-order logic? In terms of number-theoretic prerequisites, not much is needed. Some familiarity with Pell equations is probably enough.
|
2025-03-21T14:48:32.010170
| 2020-09-11T23:47:33 |
371470
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Harry Gindi",
"curious math guy",
"https://mathoverflow.net/users/1353",
"https://mathoverflow.net/users/152554"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632927",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371470"
}
|
Stack Exchange
|
Deligne's theorem for $n$-topos
Deligne's theorem states that a coherent topos has enough points, i.e. that we can prove that a morphism of sheaves on a "nice" site is an isomorphism by showing that the induced morphism on stalks are isomorphisms.
I'm looking for a higher categorical analogue. Specifically, if I have a morphism of $n$-sheaves on a "nice" site, can I test if it is an isomorphism by testing it on points, i.e. do the fibre functors form a conservative family? I would be very grateful if someone could point me to a nice reference, thanks!
PS: I assume that we need to assume $n$-coherent, but I might be wrong.
There are two cases:
1.) If your ∞-topos is locally coherent and hypercomplete, then you have Lurie's ∞-categorical version of Deligne's completeness theorem (SAG A.4.0.5).
2.) If your ∞-topos is bounded and coherent, Lurie shows another version of this theorem, similar to Makkai's conceptual completeness theorem (SAG A.9.0.6). Part of the proof of this theorem shows that you can test for equivalence of truncated objects on points.
Note: The ∞-categorical version of coherence is rather different from the 1-categorical version. In particular, coherence no longer implies compact generation (they are quite different, in fact).
I apologize, but does that also imply similar statements for $n=2$ and not $\infty$?
@curiousmathguy You can deduce this from the bounded coherent version, since every $n$-localic coherent ∞-topos is bounded. This says that you can test for equivalence of all truncated objects on the points in question.
Thanks, I'm sorry, I still feel very uncertain with everything $\infty$-theoretc.
|
2025-03-21T14:48:32.010302
| 2020-09-11T23:53:12 |
371471
|
{
"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:632928",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371471"
}
|
Stack Exchange
|
Moduli of continuity and Wasserstein differentiability of functions between measures
Let $X=\mathbb{R}^n$; I am also interested in the general case $X$ is a metric space but for simplicity let's focus on Euclidean space. Let $\mathcal{P}(X)$ denote the space of Borel probability measures on $X$. I am interested in maps $f:(\mathcal{P}(X), W_q) \to (\mathcal{P}(X), TV)$ and $f:(\mathcal{P}(X), TV) \to (\mathcal{P}(X), W_q)$, where $W_q$ is the $q$th-order Wasserstein distance and $TV$ is the strong topology induced by the total variation norm on measures. In other words, the domain is given the weak (probabilist's) topology and the codomain is given the strong topology (or vice versa).
I want to study the smoothness properties of such functions $f$. I am particularly interested in computing moduli of continuity and/or differentiability of $f$. I am being deliberately vague here: Since we are dealing with a map between two metric spaces, differentiability isn't even well-defined, although one can imagine reasonable notions such as metric derivatives along paths or even
$$
f'(P)
:= \lim_{Q\to P} \frac{TV(f(P),f(Q))}{W_q(P,Q)}.
$$
I am curious if smoothness/regularity of functions in this setting (or something closely related) has been studied before, and if so, what references/techniques there are. For example, what are reasonable notions of "differentiability" for such functions, and if so, can they be related to Lipschitzness/moduli of continuity?
|
2025-03-21T14:48:32.010417
| 2020-09-12T00:01:20 |
371472
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Ernest Davis",
"LSpice",
"https://mathoverflow.net/users/22344",
"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:632929",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371472"
}
|
Stack Exchange
|
Is the space of metric topologies over a given set dense (in the order sense)?
Suppose that $S$ is an infinite set and that $\alpha$ and $\beta$ are metrics over $S$ such that the topology induced by $\alpha$ is everywhere strictly finer than the metric induced by $\beta$, meaning that every open set $U$ in $\beta$ contains a set $V$ that is open in $\alpha$ but not in $\beta$. Suppose further that $S$ is dense (in the metric sense) with respect to both metrics, in the sense that, for any $x \in S$, for any $\epsilon > 0$ there exists $y \neq x$ within distance $\epsilon$ of $x$. Does there exist a metric $\gamma$ over $S$ that is everywhere strictly finer than $\beta$ but everywhere strictly coarser than $\alpha$? Or, contrarily, are there cases where it is known that no such $\gamma$ exists?
I think you also want the "in the metric sense" 'dense', in addition to the "in the order sense" 'dense', in your title. (And, of course, infinitude of $S$ is redundant.)
$\def\cl{\operatorname{cl}}$
A large family of counterexamples can be constructed using the following
proposition:
Let $(S, T_1)$ be a topological space with two complementary dense
subspaces $A, B$.
Define $T_3 = \{ (A \cap U) \cup (B \cap V) \mid U, V \in T_1 \}$,
in other words $(S, T_3)$ is the topological sum of $A$ and $B$.
Let $T_2$ be a topology on $S$, finer than $T_1$, such that $T_3$ is
everywhere strictly finer than $T_2$.
If $T_2$ is semi-regular, then $T_2 = T_1$.
Sketch of a proof:
Observe that $T_1|A = T_2|A = T_3|A$ and $T_1|B = T_2|B = T_3|B$. Moreover,
$\cl_3 V = \cl_A (V \cap A) \cup \cl_B (V \cap B)$ for every $V \subset S$.
From the assumption that $T_3$ is everywhere strictly finer than $T_2$
it follows that both $A$ and $B$ are dense in $(S, T_2)$. Hence for
every $U \in T_2$ we have $\cl_2 (U \cap A) = \cl_2 (U \cap B) = \cl_2 U$.
Show that for any $U \in T_2$ we have $\cl_2 U = \cl_3 U = \cl_1 U$.
Deduce that if $U$ is a regular open set in $(S, T_2)$, then $U \in T_1$.
Therefore, if $T_2$ is semi-regular, $T_2 \subset T_1$.
The application to your problem is easy. If $(S, T_1)$ is nonempty,
metrizable and dense in itself, there are many choices of $A,B$ and
$T_3$ will also be metrizable and dense in itself. Of course for any
$U \in T_1 \setminus \{\emptyset\}$ we have $U \cap A \in T_3 \setminus T_1$
and if $T_2$ is to be metrizable it must certainly be semi-regular.
Great! Thanks very much.
|
2025-03-21T14:48:32.010591
| 2020-09-12T01:46:24 |
371476
|
{
"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:632930",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371476"
}
|
Stack Exchange
|
Hilbert transform on a Besov space
Consider the usual Hilbert transform of periodic functions
$$H(f) = \frac{1}{2\pi}P.V.\int_{-\pi}^{\pi}\cot(\frac{x-y}{2})f(y)dy.$$
We know $H$ does not map $L^\infty$ continuously to $L^\infty$. Now we consider $H$ on the Besov space $B^0_{\infty, \infty}$, which is a little larger than $L^\infty$. Can we say $H$ maps $B^0_{\infty, \infty}$ continuously to $B^0_{\infty, \infty}$? If no, are there any counter-examples?
|
2025-03-21T14:48:32.010650
| 2020-09-12T02:00:00 |
371478
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Benjamin Dickman",
"C.F.G",
"https://mathoverflow.net/users/22971",
"https://mathoverflow.net/users/90655"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632931",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371478"
}
|
Stack Exchange
|
Source of infection on chessboard
I am looking for the original source of the following well known problem.
Seven unit cells of a 8×8-chessboard are infected. In one time unit, the cells with at least two infected neighbors (having a common side) become infected. Can the infection spread to the whole chessboard?
(It follows that the answer is no since the perimeter of infected part cannot increase.)
This problem appears in "connoisseur's collection" of Peter Winkler, with the following note:
This lovely problem appeared in the Soviet magazine KVANT around 1986, then migrated to Hungary.
I am also interested about Hungary.
P.S. It is found: Moscow mathematical olimpiad 1986, (8-4). Indeed, it appeared in Квант 1986, № 8, с. 57.
I think Laszlo Babai can help about the origin of this problem, because he is from Hungary and this problem is in his list of puzzles. (posted around 2008).
Not helpful for the main wonder, but the Babai list from @C.F.G above begins with a problem whose history I investigated in another question!: MO 105400
The Hungarian connection is in
Gábor Pete: Hogyan gyepesítsünk kockát? [How to make the cube weedy?] Polygon (Szeged) VII:1 (1997), 69-80.
József Balogh and Gábor Pete, Random Disease on the Square Grid (1997)
|
2025-03-21T14:48:32.011125
| 2020-09-12T04:20:53 |
371484
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Max Alekseyev",
"Shivin Srivastava",
"https://mathoverflow.net/users/120939",
"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:632932",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371484"
}
|
Stack Exchange
|
Is this recurrent sequence decreasing?
Let $S_n$ be defined as $\frac{1}{n}\sum_{t=1}^{t=n} [px_t^2 - (p+q)x_t]$ where $x_t = 1-(1-p-q)^t$. We want to find the conditions on $p$ and $q$ such that $S_n$ is monotonically decreasing for all $n$. $0 < p,q < 1$ and $-1 < 1-p-q < 1$.
Note:
I tried to prove this by taking the difference between consecutive terms of the summand i.e. $[px_t^2 - (p+q)x_t] - [px_{t+1}^2 - (p+q)x_{t+1}]]$ and trying to show that it is greater than 0 but I am not able to make progress on this.
Since $x_0=0$, it will be convenient to do summation starting from $t=0$.
Denoting $r:=1-p-q$, we have
\begin{split}
S_n &= \frac1n\sum_{t=0}^n \left(pr^{2t} + (q-p)r^{t} - q\right) \\
&=\frac{p}{1-r^2} \frac{1-r^{2(n+1)}}{n} + \frac{q-p}{1-r} \frac{1-r^{n+1}}{n} - \frac{n+1}{n}q.
\end{split}
Given that $|r|<1$, it can be easily verified that
$$\frac{\partial S_n}{\partial n} = \frac{(p - (r + 1) q)r}{n^2(1-r^2)} + O(r^n)$$
as $n$ grows.
Hence, $S_n$ decreases for large enough $n$ iff $p-(r+1)q\lessgtr 0$ or $$p\lessgtr\frac{(2-q)q}{1+q},$$
where the inequality sign is "$<$" if $r>0$, and "$>$" if $r<0$. In other words, either of the following two conditions works:
$$p < \min\left\{\frac{(2-q)q}{1+q},1-q\right\}$$
or
$$p > \max\left\{\frac{(2-q)q}{1+q},1-q\right\}.$$
Thanks for the answer! So $S_n$ will decrease in the long run but we cannot say anything about it when $n$ is small?
You'd need to take a close look at $\frac{\partial S_n}{\partial n}$ to make conclusion about the behavior of $S_n$ for small $n$. At very least, the two conditions I derived are necessary for $S_n$ to decrease.
|
2025-03-21T14:48:32.011264
| 2020-09-12T05:12:29 |
371485
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Thomas Kojar",
"https://mathoverflow.net/users/40793"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632933",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371485"
}
|
Stack Exchange
|
Poisson-like random walk expressed as Bernoulli-like random walks (splitting scheme)
In our problem we have the transition density for $x,y\in \mathbb{Z}$ and $t\in \mathbb{N}$
$$R_{t}(x,y):=e^{-t}\frac{t^{x-y}}{(x-y)!}1_{x\geq y},$$
which is the Poisson distribution pdf. (This is also in semigroup form of the operator $I+\nabla^{-}$:
$$R_{t}(x,y)=e^{-t(I+\nabla^{-})}(x,y),$$
where $If(x)=f(x)$ and $\nabla^{-}f(x)=f(x)-f(x-1)$).
Q: I am curious if we can write this as a convolution of unit step operators (i.e. $P_{t,k}(x,y):=c(x,y,t,k)1_{1\geq x-y\geq 0}$).
By convolution we mean
$$A\ast B(x,y):=\sum_{z\in \mathbb{Z}}A(x,z)B(z,y).$$
So I am curious if we can decompose a Poisson jump into Bernoulli jumps
$$R_{t}(x,y)=P_{t,1}(x,y)\ast \cdots \ast P_{t,n}(x,y)$$
for some n, possibly infinite. The $R_{t}(x,y)$ is a probability a random walk of jumping to the left from $x$ to $y$ with Poisson jumps. So heuristically I am wondering if I can "split" this Poisson-random walk into Bernoulli-random walks. In a way I am trying to run over all possible unit step paths from from $x$ to $y$.
Same operator
Let $P_{t}(x,y)=c(x,y,t)1_{1\geq x-y\geq 0}$ for some function $c$ ,ideally of the form $c(x,y,t)=c(x-y,t)$. These are unit steps transition and so we ask to find some n (possibly infinite) to have the n-convolution equal $R_{t}$:
$$R_{t}(x,y)=P_{t}\ast\cdots \ast P_{t}(x,y),$$
where $A\ast B(x,y):=\sum_{z\in \mathbb{Z}}A(x,z)B(z,y)$. That doesn't seem to work because by a simple induction argument we get
$$P_{t}\ast\cdots \ast P_{t}(x,y)=\sum_{k=0}^{n}\binom{n}{k}c(1,t)^{n-k}c(0,t)^{k}c(k,t) 1_{1\geq x-y-n+k\geq 0} $$
$$= \binom{n}{n-(x-y)+1}c(1,t)^{x-y-1}c(0,t)^{n-(x-y)+1}c(n-(x-y)+1,t)+\binom{n}{n-(x-y)}c(1,t)^{x-y}c(0,t)^{n-(x-y)}c(n-(x-y),t)$$
$$= (c(1,t)/c(0,t))^{x-y}c(0,t)^{n}[\binom{n}{n-(x-y)+1}\frac{c(1,t)}{c(0,t)}c(n-(x-y)+1,t)+\binom{n}{n-(x-y)}c(n-(x-y),t)].$$
Here are some cases that show it is not true if we use the same operator.
$\bullet$ If $\frac{c(1,t)}{c(0,t)}c(n-(x-y)+1,t)=c(n-(x-y),t)$, then using the summation identity for the binomial we get
$$= (c(1,t)/c(0,t))^{x-y}c(0,t)^{n}\binom{n+1}{n-(x-y)+1}c(n-(x-y),t).$$
So to get $\frac{1}{(x-y)!}$ requires $\binom{n+1}{n-(x-y)+1}=\frac{1}{(x-y)!}$, which implies $(n-(x-y)+1)!$ and so $x-y=0$, which is not necessarily true for general $x\geq y$.
$\bullet$ If we let $n=x-y$, we get
$$=\binom{x-y}{1}c(1,t)^{x-y}c(0,t)+\binom{x-y}{0}c(1,t)^{x-y}c(0,t)=c(1,t)^{x-y}c(0,t)(x-y+1),$$
which is not equal to $(x-y)!$.
Different operators
Using the property
$$e^{-t(I+\nabla^{-})}(x,y)=e^{-\sum_{k=1}\frac{t}{2^{k}}(I+\nabla^{-})}(x,y)=\ast_{k=1}^{\infty}R_{t/2^{k}}(x,y)=\lim_{n\to +\infty}R_{t/2}\ast\dots\ast R_{t/2^{n}}(x,y),$$
we could use $P_{t,k}(x,y):=R_{t/2^{k}}(x,y)=e^{-t}\frac{(t/2^{k})^{x-y}}{(x-y)!}1_{x\geq y}$. But these are not unit steps.
Splitting scheme point of view
One point of view is from the splitting scheme for semigroups literature (eg. see here ). Here they decompose semigroup operators into smaller steps.
If you insist on $P_t(x,y) = c(t,x-y) \mathbb{1}_{1\geqslant x-y \geqslant 0}$, then it is not possible to factorise $R_t$ as $P_t Q_t$ for whatever "reasonable" $Q_t$. Indeed, this would mean that the corresponding Z-transforms satisfy
$$ \sum_{x=0}^\infty R_t(x,0) z^n = \sum_{x=0}^\infty P_t(x,0) z^n \times \sum_{x=0}^\infty Q_t(x,0) z^n $$
for all $z$. However
$$ \sum_{x=0}^\infty P_t(x,0) z^n = c(t,0) + c(t,1) z $$
has a zero at some negative $z$, while
$$ \sum_{x=0}^\infty R_t(x,0) z^n = e^z $$
is non-zero everywhere.
(Here "reasonable" means, for example, non-negative. Indeed, in this case $Q_t(x,0)$ necessarily vanishes exponentially fast as $x \to \infty$, and all Z-transforms converge well for all $z$.)
Thank you, I should have thought of that. I might have to drop the translation invariance $x-y$. My deeper heuristic question is whether there is a way to split this Poisson random walk into Bernoulli random walks. So I have to find a way to formalize this, maybe with some Feyman Kac representation summing over all unit step paths.
Actually I don't need it to be in kernel form. That was only what I was trying but the question is for general kernel.
|
2025-03-21T14:48:32.011645
| 2020-09-12T06:27:02 |
371486
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Alexander Schmeding",
"Anderson Brasil",
"Ben McKay",
"Daniele Tampieri",
"Danu",
"Džuris",
"Gabe K",
"Geoff Robinson",
"Harry Gindi",
"J W",
"JimN",
"Lubin",
"Manfred Weis",
"Noah Snyder",
"Noam D. Elkies",
"Pietro Majer",
"R.P.",
"Sam Hopkins",
"Timothy Chow",
"Will Orrick",
"Will Sawin",
"YCor",
"bof",
"gmvh",
"https://mathoverflow.net/users/113756",
"https://mathoverflow.net/users/11417",
"https://mathoverflow.net/users/125275",
"https://mathoverflow.net/users/13268",
"https://mathoverflow.net/users/1353",
"https://mathoverflow.net/users/140385",
"https://mathoverflow.net/users/14094",
"https://mathoverflow.net/users/14450",
"https://mathoverflow.net/users/14830",
"https://mathoverflow.net/users/152049",
"https://mathoverflow.net/users/17907",
"https://mathoverflow.net/users/18060",
"https://mathoverflow.net/users/22",
"https://mathoverflow.net/users/2312",
"https://mathoverflow.net/users/25028",
"https://mathoverflow.net/users/3106",
"https://mathoverflow.net/users/31310",
"https://mathoverflow.net/users/43266",
"https://mathoverflow.net/users/45250",
"https://mathoverflow.net/users/46510",
"https://mathoverflow.net/users/484",
"https://mathoverflow.net/users/52890",
"https://mathoverflow.net/users/6101",
"https://mathoverflow.net/users/62043",
"https://mathoverflow.net/users/88583",
"user676464327"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632934",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371486"
}
|
Stack Exchange
|
Why is it still common to not motivate results in publications?
This is a question about practice and publication of research mathematics.
On the Wikipedia Page for Experimental Mathematics, I found the following quote:
Mathematicians have always practised experimental mathematics. Existing records of early mathematics, such as Babylonian mathematics, typically consist of lists of numerical examples illustrating algebraic identities. However, modern mathematics, beginning in the 17th century, developed a tradition of publishing results in a final, formal and abstract presentation. The numerical examples that may have led a mathematician to originally formulate a general theorem were not published, and were generally forgotten.
My question concerns the last two sentences. I've heard several of my professors complaining about precisely this: in many mathematics papers, one can read the whole paper without ever understanding how the authors came up with the arguments in the first place.
Questions:
1.) (Historical:) Why did mathematicians stop publishing motivational steps when publishing mathematics? (For example, were there particular schools of mathematicians who actively advocated for this?)
2.) Is there any movement amongst mathematicians today to change this tradition? (By this, I do not mean movement on a personal level; I know that many mathematicians motivate results in their publications with preliminary calculations they have performed when they initially thought about the problem. Instead, what I am looking for is a movement on the community level to initiate a change)
3.) There seems to be a lot of disadvantages to the above practice; by communicating how one thinks about a problem, others would be able to copy the author's ways of thinking which will add to the knowledge of the greater mathematical community. Is there any practical benefit for not motivating results in papers?
Clarification. By "publication", I mean everything that one makes available to the greater mathematical community (so anything that one makes available on one's webpage is included (e.g. preprints) as well as anything uploaded to the ArXiv.
4.) Is the assertion made in the last two sentences in the quote accurate. (Thanks to YCor for pointing this out.)
This is a quote, which you put in the title as an unquestionable universal assertion. What about asking about accuracy of that assertion?
I would interpret "motivating a result" as explaining how it relates to other Mathematics (or maybe, sometimes, other sciences) and what potential applications it has there (and maybe what new insight it might give there). This is very field dependent. Personally, I don't really care what steps an an author had to go through to arrive at their result, as long as the proof is is correct and clearly written.
“Clearly written” is the crucial element here.
A small comment: often the first proof of a result is not the "right" one, and the "correct" way of thinking about the problem (that is, the one which properly "motivates" it and makes clear the proof) is only understood much later.
Mathematical research is an activity which involves as much competition as cooperation. I wonder if some mathematicians don't keep for themselves some of the ideas behind their solutions also because they feel their way of thinking as an advantage which might let them get into other new results first than their competitors.
Others have given excellent answers. In general I do think that the pendulum has swung too far in the rabbit-out-of-a-hat direction, and it would help the mathematical community if there were more sharing of the process of discovery and not just the final result. But let me try to answer question 3 and suggest that it would not necessarily be an unqualified good for the pendulum to swing the other way and for everyone to be expected to give an account of how they arrived at their results.
For many of my results, the honest answer to the question of how I came up with them is, I don't know. I mean, I can usually say how the problem came to my attention, and if I were forced by some bureaucrat to report how many hours I spent working on it and what I did during those hours, I could come up with something. But if you're interested in the answers to questions such as, "How did you come up with that argument?" then often I will have no idea how to answer that question. In a few cases I might be able to quote chapter and verse for where I saw a similar argument in the literature. More often, the argument, or at least the line of attack, seems obvious to me, and I wouldn't know how to explain (to someone who doesn't find it obvious) why it seems obvious. Other times, if the idea isn't obvious, I still won't have any explanation for where it came from. This may be especially true if it's a collaborative effort—was it something my colleague said that triggered a thought in my brain? Again, usually, I don't know.
Even in the cases where I am able to answer such questions, being able to give an account of the process of discovery that is helpful to other people is as much of an art as any other kind of writing. What do you write down and what do you leave out? How do you organize your narrative? Writing down something like, "Well, I spent three weeks chasing down this dumb idea before finally realizing that I had misremembered a certain theorem and what I was hoping for couldn't possibly work" might provide emotional comfort and reassurance to insecure junior mathematicians, but do we really want to read detailed accounts of every single dead end in every single mathematical paper? What might help one person might not help another, and it's not a simple business to craft an account that will help a significant number of people.
Writing a paper is already a difficult task, and it would be burdensome to impose an additional requirement, on top of existing expectations, to provide a "account of discovery process" section. I know you weren't asking for such a requirement to be imposed, but I just want to sound a note of caution to be careful what you wish for. Before calling for the community to move in a certain direction, do you have a clear idea of what the destination is, and would it necessarily be better than where we are now? Maybe the way things are now is fine—people can share the process of discovery if they think it would be illuminating, but don't have to if they don't.
Of course research reports should be justified in some sense, maybe because the result was a breakthrough and the proof and its key ideas surprising enough to qualify as a historical event in mathematical research.
@ManfredWeis : See my comment about Aristotle's distinction between efficient cause and final cause.
There is a small ambiguity in the expression motivate a result.
You seem to use it for (A): "explain why the authors came out with certain arguments, definitions, methods etc, in order to prove the result".
But it can also refer to (B): "explain why they think the result is interesting/important; what is it aimed for; what is the reason to do such a research, why we should buy it".
It is true that the issue (A) is sometimes neglected in written papers (maybe it's more present in seminars, in anecdotical form); possible good reasons are:
mental paths that lead to the truth in some cases may help, yes, but in some other may be convoluted and distorted and of no help; the final point of view may be far shorter, clearer and simpler for understanding the logical structure.
the editorial issue of saving room in an article.
another (maybe less good) reason, a bit of vanity: remove all scaffolding and leave an aesthetic, shining and unintelligible object -- and let you think the author is genius.
I'd say issue (A) becomes important, both from the historical and pedagogical side, later, once (B) is agreed and the result accepted in the mathematical community.
Artists have always been protective about their tools and methods; it's their bread. Here are some historical examples.
I think the habit of vanity mentioned above was quite common in the European mathematical style of a century ago or more, and hopefully has been reduced in favor of a more pedagogical American style.
In Tartaglia's time, mathematicians would not even give you a proof, just the plain result.
Everybody would like to know how Gauss reached his neat conclusions, for instance. But the introduction of a paper is not necessarily the best moment and place.
Archimedes wrote the computation of the surface area of the sphere in perfect style and rigor, by means of the exhaustion method. But the explanation of how he arrived to "four times the largest inscribed circle" is not there. He first computed the volume of the sphere, by means of his favorite tool, the lever. He explained this later, in a letter to Erathostenes, The Method, a magnificent piece of scientific communication.
Good point about the distinction between what Aristotle might have called (A) the efficient cause and (B) the final cause.
I'd so much prefer (A) to the current (B). I feel like (B) is an explanation to the editor/reviewer why this article should be published, while I as the reader already know I am interested in the topic if I am reading the article at all. I've always seen (B) as waste of reader's time.
good point, the current (B) is often just advertising
While I agree that (B) can degenerate into mere advertising, I don't agree that that's its only function. Perhaps it's because I'm more of a generalist than a specialist, but I frequently find myself skimming papers in areas that I don't know too much about. A priori, I might not know how interested I am in the paper. If the paper makes no effort to connect itself with something I already know something about, I'm likely to ignore it. And yet maybe the paper actually contains something that would interest me if only someone took the time to point out the connection with my interests.
Note that the Discrete Analysis journal includes editorial introductions. I suppose that is more (B) than (A).
C.F. Gauß may have been one of the earliest proponents of that "ideal"; if I remember correctly, E. Galois once said that Gauß is like tricky fox that wipes its trail with its tail. Also the motto "pauca sed matura" supports the assumption that Gauß played an important role in that development.
Another reason may be rooted in publishing mathematical results in journals; for one part there are often limitations on article length and for the other part articles that are published in such journals tend to be influential, which in turn may leave on young mathematicians the impression that being able to strip down publications to the absolutely necessary is a precondition to fame.
Or maybe it is also owed to mathematician's passion in abstracting (from latin "abstrahere", pull away, strip off) so that leaving only the bare bones of a mathematical result to the generations to come may be a special joy to certain mathematicians.
There are however signs of hope that mathematicians that are more in the vein of L. Euler provide insight into the stories behind their results via accompanying online publications as indicated in the answers to my question Examples of Mathematical Papers that Contain a Kind of Research Report
(Regarding limitations to journal pages) Page limits should not be a constraint if one uploads to the arXiv, correct? Or if one publishes papers on their webpages or leaves them as preprints?
@user676464327 the possibilities you mention are only available since a comparably very short time, but they may leverage more informative publications
Euler is wonderful. Most of us are not Eulers. We are lousy writers, even less competent to write about what you call the "motivation" than we are to write up the "bare bones", and the result of our attempt would be boring and unilluminating. Moreover, for some of us, the process of solving a problem consists of forgetting the various false steps and bad ideas, so that by the time we are ready to write the paper, there is nothing left but the "bare bones", and we have no better idea than anyone else how they were found.
I think there is value in having published papers being written somewhat "efficiently," so that a reader can quickly find the important results. On the other hand, sharing the motivation is also important and makes the math more human. One way to do both is to have a blog and write a less formal summary with your papers which gives a more complete story.
The arXiv is very recent compared with the history of mathematics.
Even TeX is only a few decades old.
Mathematical typesetting used to be a specialized and expensive skill,
making journal pages a limited resource.
Mathematical style takes a long time to evolve.
(Heck, people still swear by illogical relics of physical typesetting like
"," or "." when the punctuation isn't part of the quote.)
@NoamD.Elkies This is a bit off-topic, but why did that come from physical typesetting? I didn't know that.
@WillSawin https://style.mla.org/punctuation-and-quotation-marks/
Euler was such a wonderful writer. It's remarkable how much variation there is among even the greatest mathematicians in quality of writing. Euler and Gauss were both all time great mathematicians, but the difference in quality of writing is dramatic.
1.) I think I remember hearing that Gauss and Cauchy both liked to present results 'from nowhere'. Rumors, not history, but maybe it's right. I wouldn't bet my life on it.
2.) I think a lot of this goes on at seminars and in personal interactions. As Markus Land repeatedly stressed to us, mathematics is a social activity. That said, I don't know that I agree that this practice is extremely widespread. Maybe it depends on your subfield?
3.) Some referees and editors don't like to keep too much of that stuff in, or so I hear. I do have a cool example, though: Justin Noel, who recently left academia, used to include explanatory notes, motivation, and more detailed computations as comments in his TeX files, which you could find on the arXiv!
Why do referees and editors not like to keep too much motivational stuff in papers?
From a recent peer review on one of my papers: "The proofs are as expected and boring." So I guess the referee would like a shorter presentation because he/she feels that the details are unneccessary
The arXiv versions of most of my papers are longer than the published versions. Referees ask to cut things for the published versions, most often examples, but I keep them in the arXiv version. I suspect many people do that.
I apologize for my perhaps silly curiosity, but in what sense Justin Noel "left mathematics"?
@DanieleTampieri: perhaps he is working at Booking.com: http://www.nullplug.org/
@BenMcKay so he is no more working as a professional mathematician, though it seems that he still likes the topic. I apologize again for my perhaps pedantic comment (and I do not intended to be rude): it is only that in my opinion, Mathematics is first of all passion, therefore it is very difficult if not impossible to "leave it".
@DanieleTampieri I just meant he left academia. He was for a very short amount of time my Master's advisor before he decided to go into industry =].
@DanieleTampieri : Indeed, according to some, I "left mathematics" in 1998, even though I still usually list my occupation as "mathematician" and occasionally publish research papers in mathematics.
I don't have any data or evidence of the following ... the following is just my opinions and perspectives on the matter.
(The 'why') Mathematics is a field of rigour, sometimes to excruciating details. If a reviewer needs to verify facts, they often just want the statements and the argument and they can verify it - reading a story of how the facts came around are secondary or even a waste of time to some. Even at conferences, I feel that mathematicians spend a lot of time presenting a proof rather than telling a story of how they came across the results or what the implications of the result are. Also, publishing mathematics is rarely meant to be an activity for non-specialists to participate in. Since a lot of modern math publications are in sub-sub-areas of a particular field of math, the only people reading and/or reviewing that work are specialists in that area and so there is not usually effort put it to telling the story of how it came to be, or the importance of it in the current research landscape because you are preaching to the choir.
I agree at seminars / conferences, I've seen little attempt at trying to appeal to a wider audience with stories of how the work arrived. It's not uncommon to see many titles at a conference simply be a re-statement of the main theorem of the result. I try to stray from that.. once, I was presenting at the Fields Institute and before I even started speaking, an audience member in the front row said my slide had "a soft title" and once I started and talked about motivations behind the work, he walked out for a smoke before I even got to my first theorem statement. (And this guy is a very notable individual of the research area.) There are a few people who will tell a good story at a conference and leave the proof details to the paper, but there is a common attitude that this is some kind of sign of weakness. I have sat through plenty of talks that were not exactly my area and 30 seconds was given to the slide with all the required definitions and the presenter essentially went through them with a "we all know that blah is defined to be this expression" and moves on faster than a person could read/absorb the definitions. Then proceeded to talk about proof details, which was a waste of time for anyone who didn't get the definitions. I don't know if there is a movement to change these attitudes. I've just decided that to go less math conferences and more computer science ones (or other fields).
I agree - I've been asked to remove storytelling in math submissions yet in other venues, I was told that I needed to increase the background, motivation, implications, etc (much to my surprise at the time). I essentially began in a math field and migrated to computer science.
I think you're being a little hard on people in #2. Some of that is really just a lack of skill in knowing your audience and in organizing your talk. I think I can say that I have never given a really good talk, but it's not because I try to tailor my talks toward experts. I'm just not that good…
@HarryGindi I don't know which one of is being more hard on people in #2... me saying that they present in a smug manner or you saying that they lack a skill in presenting :D
Disclaimer: I am myself a junior mathematician who has done no real research (yet), but have read many publications.
At the beginning of my journey, I would have agreed with you. But now, especially as I learn more advanced topics, I find that my opinions have changed.
You say
...by communicating how one thinks about a problem, others would be able to copy the author's ways of thinking which will add to the knowledge of the greater mathematical community
Again, I once would have agreed with you, but now I think this claim is false. I think what you are saying here is that you wish for the author to "implant" their intuition on the subject into your brain. But I argue that this is impossible. It seems that there is always an intuitive way to think about a subject for any individual mathematician, but there is no single explanation that is intuitive for all mathematicians.
If you allow me to use an analogy, you are saying that you wish for the author to write down some algorithm which, if you hadn't been presented the answer, would have allowed you to come up with the proof on your own. But this is exactly what writing the proof without any motivation is, it is an algorithm (albeit a trivial one) which says "you could have read the references and written <insert proof here> and you would have produced the proof".
If we treat your brain and the author's like a computer, and the proof as a program running on it, the problem you are trying to solve is that, in your brain, there are a lot of functions and variables and types used in the program which are undefined, thus you are getting a "compiler error". However in the author's brain, the program compiles because all of these functions and variables and types are defined. That is a way to give more meaning to the phrase "the author's way of thinking".
The task you are asking of the author is to predict which functions and variables in your brain are defined and only use those to produce the proof. This is clearly an impossible task for all mathematicians, as what one mathematician knows is never the same as another. It is also the purpose of references. By giving references, the author is saying "we have tested that this program runs if you have these dependencies installed (i.e. that you read or at least understand the references), but we give no guarantee that it will run without them". If you don't install the dependencies then you can't guarantee that the program will run. When you think about it, the references (and recursively their references) are probably close to the path the author took to come up with the proof (maybe over the course of their entire life).
If you ask for an algorithm which is non trivial and produces the proof (i.e. not one which simply says "read these references, then write this"), what you are really asking is for the mathematician to produce a proof of a more general theorem, or at the very least, a more amazing insight than the one which has already presented, which is an unreasonable expectation (since the author would have to again extend that result, and then extend that one, etc.).
Thus, it is up to the individual to find such a motivation that works for them, to fix the "compilation errors" in their own brain, by reading references, reading books on the subject, and asking questions on websites like mathoverflow.
It is not the researcher's responsibility to completely fill in your understanding to allow you to think about the problem in the same way that they do, just like it is not the responsibility of a software developer to manually install every dependency on the computer of every person that uses their program.
The researcher's only responsibility is to present factual information in a way that is not overly complicated, and to fix any "bugs" in their proof which are found.
As to Question 1: Euclid's Elements is written in the style you deplore.
I think it does partially answer Question 1. It points out that omitting "motivation" didn't start with Gauss but goes back to Euclid.
But the style of the Elements was generally seen as something to emulate if one strove for rigor (cf. Spinoza's Ethics).
@Matt F. Parts of Elements are known to be based on work of earlier thinkers, e.g. Eudoxus and Theatetus, but it isn't possible to be confident that this was true of the entire work. The number theory, for example, doesn't seem to have known antecedents. The lives of Euclid and Eudoxus probably overlapped, if only slightly. Most present-day scholars of Pythagoras believe he did not do anything that could be called "research mathematics". It's true that, where there are predecessors, Euclid does not mention them, much less their motivations.
@Matt F. My "probably" was too strong. "Possibly" would have been better, but having looked at a variety of sources, later dates now seem to me more likely. Really, we know nothing about Euclid's life. Proclus, writing over 700 years later made some surmises and concluded Euclid lived between the time of Plato's students and the time of Appolonius and that he worked during the era of Ptolemy I Soter (c. 367 - 282). But Pappus says that Appolonius (birth often estimated c. 240) studied with the students of Euclid, which is hard to reconcile with earlier dates.
@WillOrrick Or you could say "arguably", you can use that to prefix practically any statement whatsoever...
@RP Not every statement is arguable, or, at the very least, not every argument is a reasonable one. The point is that "c. 325 BC" gives a false sense of precision. Britannica says Euclid "flourished c. 300 BC", and says nothing about birth or death dates, which I think better reflects the actual state of knowledge.
Community Wiki answer, too long for a comment, addressing things one may do about it rather than the practice itself.
A personal story. I once met T. Y. Lam's daughter, Fumei Lam, during a lunch. She works in graph theory. She was upset about taking months to find a counterexample to one of her own conjectures, plus an annoyingly simple example at that. I said (I must paraphrase, years ago) "But you know what happens next. You thought you had everything in one big lump. Now you have a little bit outside the lump. You continue to work on it; the big lump will shrink a little more, the little bit grow somewhat. If things go unusually well, you may produce a classification." I think I added something about others working on it. She did seem relieved at that.
This is not in any way a comprehensive answer to your question, but it may be worth pointing out that while there are certainly several disadvantages to "erasing the trail" up to a result, there is also (at least) one major advantage: it lessens the "path dependence" and encourages others to think about possible applications of the result that would not have occured to the original discoverer.
For example, many people discover a result by thinking very carefully about one specific example. But the beauty of abstract math is that the same general result often applies to a huge range of concrete examples that differ hugely in their details, but share the minimal mathematical structure necessary for the result. The original discoverer may be so used to thinking about the result in the context of one type of example that they miss applications to other classes of examples. Presenting the detailed example that they worked through to motivate the result risks similarly locking other readers into the same mode of thought. Whereas if a reader approaches a new result with a completely different example/application in mind, then they may be able to more easily extend the result to new corollaries that the original discoverer didn't think of.
|
2025-03-21T14:48:32.013628
| 2020-09-12T09:02:28 |
371493
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Max Alekseyev",
"domotorp",
"https://mathoverflow.net/users/7076",
"https://mathoverflow.net/users/955"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632935",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371493"
}
|
Stack Exchange
|
Computation of cyclic van der Waerden numbers
Van der Waerden's theorem gives us a finite number $W(k,r)$ defined as the smallest positive integer $N$ such that for any $n\geq N$, any $r$-coloring of $[n]=\{1,\dots,n\}$ admits a monochromatic $k$-AP. We can ask the same question except with $[n]$ replaced by $\mathbb{Z}/n\mathbb{Z}$, calling the answer the "cyclic van der Waerden number" and denoting it by $W_c(k,r)$ (seems to be first mentioned in Burkert and Johnson, 2011). An immediate bound is that $W_c(k,r)\leq W(k,r)$, so we know that $W_c$ is finite.
Is there any progress on determining the values of $W_c(k,r)$ that is not just "check every number not greater than $W(k,r)$"? Even if the exact values are not known for larger $k$ and $r$, are there any improved asymptotics on $W_c$ that are better than the Gowers bound for $W$? My quick literature search seems to only produce the Burkert and Johnson paper and a single other one (Grier, 2012) which computes $W_c(3,2)$ but nothing else.
$W(k,r)\le W_c(k,2r)$ follows from taking an $r$-coloring of $[W(k,r)]$ that has no $k$-AP, and adding 'r' to the colors of the first $W(k,r)/2$ numbers to obtain a $2r$-coloring that has no circular $k$-AP.
Since the best bound for $W(k,r)$ is about $2^{2^{r^{2^{2^k}}}}$, any better bound for $W_c(k,r)$ would improve this as well.
Why the constructed $2r$-coloring has no circular $k$-AP?
To give an example: If $k=3$ and $r=2$, then 12122121 becomes 34342121. A circular AP would need to make a small ($<W(k,r)$) jump and a big ($>W(k,r)$) jump.
I'm still not convinced. Can you add a proof for the general case?
Case 1: the circular AP's difference is a positive number $<W(k,r)/2$.
Case 2: the circular AP's difference is a negative number $>W(k,r)/2$.
In both cases the circular AP is just a regular (non-circular) AP, as otherwise it would change colors.
Thanks, I see now. The only question remains - what if $W(k,r)$ is odd?
We didn't use in the argument that it is even.
You said "first $W(k,r)/2$ numbers".
|
2025-03-21T14:48:32.013830
| 2020-09-12T09:43:27 |
371497
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"David Loeffler",
"https://mathoverflow.net/users/143589",
"https://mathoverflow.net/users/2481",
"xlord"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632936",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371497"
}
|
Stack Exchange
|
Roadmap for studying Galois deformation theory/modularity theorems from a modern perspective
I am a graduate student with some background in Galois deformation theory. I am familiar with the basics (the existence of a universal deformation space with prescribed conditions) and with some examples in Galois deformation theory, as well as with the some of the conjectural relations with $p$-adic systems of Hecke eigenvalues (i.e. $R=T$ theorems). I have seen applications and arguments in the case of $\operatorname{GL}_2$.
I'd like to get myself to a spot where I am somewhat familiar with the more modern advancements in the theory, by which I mean, Taylor–Wiles–Kisin patching, Kisin's deformation rings, potential modularity theorems, the Calegari–Geraghty method, the 6-author "Patching and the $p$-adic local Langlands correspondence" paper, etc...
It seems not so easy to start reading such papers without an idea where you're going because frequently they approach 100 pages with many logical dependencies. On the other hand, the more introductory accounts like the Darmon-Diamond-Taylor papers appears to be slightly outdated (as far as I understand).
So, what would be a good roadmap for studying Galois def theory/modularity theorems from a modern perspective? Where should I begin?
Thanks!
A fantastic place to start would be Toby Gee's notes from the 2013 Arizona Winter School. This gives a nice overview of the theory as it then existed -- things have of course moved on further since then, but it's significantly more "modern" than Darmon--Diamond--Taylor, for instance. The course concludes with a proof of a GL(2) modularity lifting theorem over totally real fields, a result which goes back to Fujiwara in 2000 or so, but rather than presenting Fujiwara's proof, the notes give a "modernized" proof using more up-to-date machinery.
Dear David, thank you for your answer. I've read most of these notes and did indeed find them useful. Do you have an opinion on what could be a second (more advanced) source to look at after these notes? For example, what could one read in order to understand better the geometry of the various deformation spaces/Hecke algebras?
Once you've got your head around Toby's notes, you should be in a good position to start reading the research literature. The notes include some suggestions for further reading, which might give you some good pointers. (Your request for a "second thing to read" suggests that you believe the subject to be more "linear" in its development than it actually is -- like any subject with >1 person working on it, there isn't a unique correct order to read things. There are many roads to modularity lifting, none of which is the royal road.)
|
2025-03-21T14:48:32.014190
| 2020-09-12T09:54:34 |
371498
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Joao Vitor",
"https://mathoverflow.net/users/165195"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632937",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371498"
}
|
Stack Exchange
|
Connecting homomorphism in Cech cohomology
Let $M$ be a smooth manifold and $\mathcal{U}$ be a good open cover of $M$. If I have an exact sequence of sheaves
$$0 \longrightarrow A \stackrel{f}\longrightarrow B \stackrel{g}\longrightarrow C \longrightarrow 0,$$
then there is an exact long sequence from Cech's cohomology under what chances?
$$...\rightarrow \check{H}^{q}(\mathcal{U}, A) \rightarrow \check{H}^{q}(\mathcal{U}, B) \rightarrow \check{H}^{q}(\mathcal{U}, C) \stackrel{\delta^q} \rightarrow \check{H}^{q+1}(\mathcal{U}, A) \rightarrow ...$$
How would connecting homomorphism $\delta^q$ be? Can you recommend any literature that deals with this?
Appreciate.
A short exact sequence of sheaves will give you a sequence of Cech complexes $0\to \mathcal{\check{C}}^\bullet(\mathcal{U}, A)\to \mathcal{\check{C}}^\bullet(\mathcal{U}, B)\to \mathcal{\check{C}}^\bullet(\mathcal{U}, C)\to 0$, which is in general not exact on the right and the connecting homomorphism has to be defined by going to a refinement (see the proof in https://stacks.math.columbia.edu/tag/09V2). However if $H^1(U_{i_0,\ldots, i_n}, A) =0$ for all $n\ge 0$ and all $i_0,\ldots, i_n$ (since you assume that $\mathcal{U}$ is a good cover then this is true if for example $A$ is a locally constant sheaf) then the above sequence of Cech complexes is exact on the right and the connecting homomorphism is the usual one defined by diagram chasing (https://stacks.math.columbia.edu/tag/0111).
Thank you, Chris.
|
2025-03-21T14:48:32.014309
| 2020-09-12T10:20:34 |
371502
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Taras Banakh",
"bof",
"https://mathoverflow.net/users/43266",
"https://mathoverflow.net/users/61536"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632938",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371502"
}
|
Stack Exchange
|
A selection principle in measure theory
A Borel subset $B$ of the unit interval $\mathbb I=(0,1)$ is defined to be a density neighborhood of a set $A\subseteq\mathbb I$ if for every $a\in A$ we have $$\lim_{\varepsilon\to0}\frac{\lambda(B\cap[a-\varepsilon,a+\varepsilon])}{2\varepsilon}=1$$where $\lambda$ denotes the Lebesgue measure on $\mathbb I$.
Problem. Let $A\subseteq\mathbb I$ be a set of Lebesgue measure zero and $(B_n)_{n\in\omega}$ be a sequence of Borel density neighborhoods of $A$. Is there a sequence of compact sets $(K_n)_{n\in\omega}$ such that $K_n\subseteq B_n$ for all $n\in\omega$ and the set $K=\bigcup_{n\in\omega}K_n$ is a density neighborhood of $A$?
Unlike topological neighborhoods, $A\subseteq B$ is not required in order for $B$ to be a density neighborhood of $A$, right?
@bof You are right: this is not required. I thought how to call such a set. But nothing better than "density neighborhood" could not find.
What are the trivial cases? I think it's trivial if $(B_n)$ is a constant sequence, am I right? What if $A$ is a countable set, is it true in that case, and is it trivial?
@bof Very good question. If $A$ is $\sigma$-compact (or maybe even Menger), then the answer is affirmative.
Professor Wladyslaw Wilczynski kindly informed me that the answer to this problem is negative.
Take any Lebesgue null dense $G_\delta$-set $A$ in the real line $\mathbb R$. Choose a countable dense subset $\{x_n\}_{n\in\omega}$ in $\mathbb R\setminus A$. Since the density topology $\tau_d$ on the real line is Tychonoff, for every $n\in\omega$ there exist disjoint $\tau_d$-open Borel sets $D_n,E_n\subseteq\mathbb R$ such that $x_n\in D_n$ and $A\subseteq E_n\subseteq\mathbb R\setminus\{x_k\}_{k\in\omega}$.
For every $n\in\omega$ consider the Borel set $B_n=\bigcap_{k\le n}E_k$. It is clear that $B_n$ is a density neighborhood of $A$. It can be shown that the sequence $(B_n)_{n\in\omega}$ has the required property: for any sequence of compact sets $K_n\subseteq B_n$, the union $\bigcup_{n\in\omega}K_n$ is not a density neighborhood of $A$.
|
2025-03-21T14:48:32.014485
| 2020-09-12T10:56:36 |
371504
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Liam Keenan",
"Maxime Ramzi",
"Yonatan Harpaz",
"https://mathoverflow.net/users/102343",
"https://mathoverflow.net/users/113828",
"https://mathoverflow.net/users/51164"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632939",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371504"
}
|
Stack Exchange
|
References about "monoidal fibrations" in $\infty$-category theory
$\newcommand{\cat}{\mathsf} \newcommand{\fun}{\mathrm{Fun}} \newcommand{\calg}{\mathrm{CAlg}}$
Let $\cat C^\otimes,\cat D^\otimes, \cat E^\otimes$ be symmetric monoidal $\infty$-categories, and $p^\otimes: \cat D^\otimes \to \cat E^\otimes$ a map of $\infty$-operads (aka a lax symmetric monoidal functor).
Assume $p: \cat D\to \cat E$ is a cartesian fibration.
I'd like to know under what natural conditions $\fun^{lax}(\cat C^\otimes,\cat D^\otimes)\to \fun^{lax}(\cat C^\otimes,\cat E^\otimes)$ is still a cartesian fibration, and more specifically if there are references regarding this type of situation.
A few remarks:
In the cases I'm interested in, $p^\otimes$ is a map of symmetric monoidal $\infty$-categories, so a strict symmetric monoidal functor. If that's necessary to give an interesting statement, I'm willing to assume it.
Still in those cases, $\cat{D,E}$ are presentable with a compatible tensor product, and I can reduce to the case where $\cat C$ is small, so using Day-convolution, one can reduce to the question of whether $\calg(\cat D)\to \calg(\cat E)$ is still a cartesian fibration, i.e. to the case $\cat C = N(Fin_*)$.
Finally, I'm also in a situation where $p$ is a "monoidal fibration", by which I mean that if $x\to y$ is a $p$-cartesian edge, then so is $x\otimes z\to y\otimes z$; I think this can be relevant - and hopefully this, together with the first bullet point, should be enough.
Has something like this been written up anywhere ?
EDIT : I wrote down what I think is a complete proof, and indeed items 1 and 3 are the ones that make it work. But the proof is long for this sort of technical result and I'm still interested in references so as not to lengthen what I'm writing too much.
You might want to check out Nikolaus’ paper on stable oo-operads. The content around Corollary 3.7 might give you what you want, but I’m not sure off the top of my head.
@LiamKeenan : Hi Liam; thanks for the reference ! Unfortunately I just had a look at it and it seems like there isn't anything like what I'm after in that paper - Nikolaus looks at how the functorialities of $Fun(A,B)$ are compatible with the operadic structure, but not (as far as I can tell) with questions of (co)cartesian-ness
I don't know a reference but here is a not-too-long proof. The condition that $\mathsf{D} \to \mathsf{E}$ is a cartesian fibration implies that for every $\langle n \rangle \in \mathrm{Fin}_*$ the map $\mathsf{D}^{\otimes}_{\langle n\rangle} \to \mathsf{E}^{\otimes}_{\langle n\rangle}$ is a cartesian fibration and that for every inert map $\alpha : \langle n\rangle \to \langle m\rangle$ the transition functor $\alpha_*:\mathsf{D}^{\otimes}_{\langle n\rangle} \to \mathsf{D}^{\otimes}_{\langle m\rangle}$ sends $p^{\otimes}_{\langle n\rangle}$-cartesian edges to $p^{\otimes}_{\langle m\rangle}$-cartesian edges.
Now apply (the dual of) [HTT, Corollary <IP_ADDRESS>] to deduce that every $p^{\otimes}_{\langle n\rangle}$-cartesian edge in the fiber $\mathsf{E}^{\otimes}_{\langle n\rangle}$ is also $p^{\otimes}$-cartesian as an edge in $\mathsf{E}$ (note that being a cartesian edge is a form of a relative limit). This means that the map $p^{\otimes}: \mathsf{D} \to \mathsf{E}$, though possibly not a cartesian fibration itself, still admits cartesian lifts for a certain collection of edges in $\mathsf{E}$: all the edges which are contained in a fiber $\mathsf{E}^{\otimes}_{\langle n\rangle}$ for some $\langle n\rangle$. Otherwise put: all the arrows which map to an equivalence in $\mathrm{Fin}_*$. It then follows that the functor
$$ \mathrm{Fun}_{\mathrm{Fin}_*}(\mathsf{C}^{\otimes},\mathsf{D}^{\otimes}) \to \mathrm{Fun}_{\mathrm{Fin}_*}(\mathsf{C}^{\otimes},\mathsf{E}^{\otimes}) $$
is a cartesian fibration, where $\mathrm{Fun}_{\mathrm{Fin}_*}$ denotes functors preserving the projection to $\mathrm{Fin}_*$ (but not necessarily preserving inert edges). Indeed, any natural transformation of functors $\mathsf{C}^{\otimes} \to \mathsf{E}^{\otimes}$ whose projection to $\mathrm{Fin}_*$ is constant consists object-wise of arrows in $\mathsf{E}^{\otimes}$ which admit cartesian lifts in $\mathsf{D}^{\otimes}$ by the above, and hence itself admits cartesian lifts as a natural transformation. By base change we then conclude that the functor
$$ \mathrm{Fun}^{\mathrm{lax}/\mathsf{E}}(\mathsf{C}^{\otimes},\mathsf{D}^{\otimes}) \to \mathrm{Fun}^{\mathrm{lax}}(\mathsf{C}^{\otimes},\mathsf{E}^{\otimes}) $$
is a cartesian fibration, where $\mathrm{Fun}^{\mathrm{lax}/\mathsf{E}}$ here stands for the those functors $\mathsf{C}^{\otimes} \to \mathsf{D}^{\otimes}$ over $\mathrm{Fin}_*$ whose projection to $\mathsf{E}^{\otimes}$ preserves inert edges. To finish the proof it will suffice to show that for cartesian edge in $\mathrm{Fun}^{\mathrm{lax}/\mathsf{E}}(\mathsf{C}^{\otimes},\mathsf{D}^{\otimes})$, if its target preserves inert edges then its domain preserves inert edges. Given that inert edges in $\mathsf{D}^{\otimes}$ are exactly the cocartesian lifts of the inert edges in $\mathrm{Fin}_*$, this follows from the fact that inert transition functors $\alpha_*:\mathsf{D}^{\otimes}_{\langle n\rangle} \to \mathsf{D}^{\otimes}_{\langle m\rangle}$ sends $p^{\otimes}_{\langle n\rangle}$-cartesian edges to $p^{\otimes}_{\langle m\rangle}$-cartesian edges.
It seems like you're claiming that condition 3 is not necessary, is that correct ? or am I missing the point in your argument where you use that ? (in the argument I had in mind, I used it to prove that $p^\otimes_{\langle n \rangle}$-cartesian edges were $p^\otimes$-cartesian; which it seems you're proving more generally)
So I used the HTT reference for this, and looking at it again I see that for this one needs condition 1, i.e., that $p^{\otimes}$ preserves cocartesian edges, but not condition 3. That HTT reference (and especially the following particular case of it) is actually quite useful in practice. Applying its dual version with $K=\mathcal{E}_0 = \Delta^0$ one obtains the statement that for a map of cocartesian fibrations (which preserves cocartesian edges), an edge contained in a fiber is cartesian with respect to the induced map on fibers if and only if it is cartesian on the total spaces.
That does sound very useful ! I might change the way I wrote things up then (I ended up just writing it because there were no references - but the way I wrote it, I use condition 3, and it's longer than what you wrote)
Condition 3 will be needed if you wanted to prove that the induced map on categories of strong symmetric monoidal functors is also a cartesian fibration, see the last part of the proof above.
Yes, that's actually what I used (using the enveloppe of an $\infty$-operad, I reduced to the case of strong symmetric monoidal functors). But I only need the statement about categories of lax symmetric monoidal functors (although in any case, I had the additional condition at my disposal)
I have accepted Yonatan's answer because ultimately his proof is what's appearing in the reference, but for future reference, we included his proof in this preprint, as Appendix B (rather than the one I had in mind at first, which was longer and used more hypotheses). So now, there is a reference.
|
2025-03-21T14:48:32.014947
| 2020-09-12T11:03:49 |
371505
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Gerry Myerson",
"https://mathoverflow.net/users/158000"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632940",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371505"
}
|
Stack Exchange
|
On the average density of non-zero digits of NAFs of fixed length
An NAF is a non-adjacent form of a positive integer $k$.
One of the five properties of NAFs is "The average density of non-zero digits among all NAFs of length $l$ is approximately $1/3$."
How to prove it?
Thanks.
If you don't know how to prove it, how do you know it's true?
It seems this is Theorem 3.29 in D. Hankerson, A. Menezes, S. Vanstone, Guide to Elliptic Curve Cryptography, Springer, New York (2004). Perhaps the authors give a proof.
OK, it seems you are quoting from that book – it would have been nice of you to cite your source. The Theorem is on page 98, but no proof is given there.
The number of NAFs of length $n$ and $k$ nonzero digits is equal to $2^{k-1}\binom{n-k}{k}$. Therefore the total number of NAFs of length $n$ is
$$A_n=\sum_{k\geq 0} 2^{k-1}\binom{n-k}{k}$$
and the total number of nonzero digits that appear in the set of all NAFs of length $n$ is
$$B_n=\sum_{k\geq 0} k\cdot2^{k-1}\binom{n-k}{k}.$$
Now, the density you seek is given by $\frac{B_n}{nA_n}$ and perhaps there is a slick way to get the approximation quickly, but a boring way is to proceed by determining this value exactly.
You can routinely check the recurrence relations satisfied by both $A_n$ and $B_n$, or their generating functions. We have
$$\sum_{n\geq 0}A_{n+1}x^{n}=\frac{1}{1-x-2x^2}$$
$$\sum_{n\geq 0} B_{n+1}x^n=\frac{1-x}{(1-x-2x^2)^2}$$
from which you can deduce explicit formulas for each sequence. You'll find
$$A_n=\frac{2^n-(-1)^n}{3} \, , \, B_n=\frac{n2^{n}}{9}+O(2^n)$$
which means that the average density of nonzero digits is $B_n/nA_n=\frac{1}{3}+O(1/n)$.
|
2025-03-21T14:48:32.015083
| 2020-09-12T11:16:11 |
371506
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Gottfried Helms",
"Max Lonysa Muller",
"Piotr Hajlasz",
"https://mathoverflow.net/users/121665",
"https://mathoverflow.net/users/7710",
"https://mathoverflow.net/users/93724"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632941",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371506"
}
|
Stack Exchange
|
Does the functional square root of the cosine admit a vector-based interpretation?
In linear algebra, the cosine of the angle between two vectors $a$ and $b$ is defined as $$\cos(a,b) = \frac{\langle a, b \rangle}{||a||\cdot||b||} .$$
The functional square root of the cosine has at various times been studied by mathematicians. It is the function $f(\cdot)$ such that $$f(f(x)) = \cos(x). $$ See for instance this MO question. I wonder whether it is possible to have a vector-based interpretation of this functional square root of the cosine, similar to the cosine itself as defined above.
I have also asked this question on MSE.
Could the person(s) who voted down and/or to close please explain his/her motivations?
@PiotrHajlasz I think this question is interesting because it may cast a renewed understanding of the functional square root of the cosine. It seems to me most effort to find this root has so far been directed at finding its Taylor series coefficients. While progress has been made in this direction, it has so far not enabled us to retrieve a closed-form analytic expression for the root. In contrast to many other functions, the cosine of two vectors has an elegant and simple representation in vectorial terms... (cont'd)
... It therefore lends itself well to be interpreted from a geometric and linear-algebraic perspective. Such a perspective may be lacking in the current literature on functional equations
I am removing my vote to close.
Hmm, for me the idea of iteration of a function means, that the input and the output have the same structure. So they both may be a scalar, maybe a two-component vector or $2 \times 2$-matrix (for instance for $2$-D rotation matrices) , or infinite vectors of coefficients (for instance in the context of Carleman-matrices, which give a "vandermonde-vector" as input as well as output - here I use "vand. vector" meaning an infinite vector of an argument $x$ as $V(x) = [1,x,x^2,x^3,...]$). Could we define some structure first which is stable between input & output for your problem?
|
2025-03-21T14:48:32.015254
| 2020-09-12T11:21:34 |
371509
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Carlo Beenakker",
"Fedor Petrov",
"Focus",
"https://mathoverflow.net/users/11260",
"https://mathoverflow.net/users/119037",
"https://mathoverflow.net/users/4312"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632942",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371509"
}
|
Stack Exchange
|
Convergent of improper integral
Let $f \in C^1[0,\infty)$ be an increasing function with $f(0)>0$, suppose $\int_0^\infty \frac{1}{f(x)+f'(x)} < \infty$, prove that $\int_0^\infty \frac{1}{f(x)} < \infty$.
I find it weird since the behaviour of $f'$ is random, so I don't know how to control $f$ in terms of $f+f'$.
You ask to prove it, what is the evidence that this is true?
isn't it just a consequence of $f(x)+f'(x)>f(x)>0$ for all $x>0$?
@CarloBeenakker no, it seems that it is not
@FedorPetrov It is an exercise
In general exercises are not welcome here (although I like this one)
For $a\geqslant 0,b>0$ we have $$\frac1b\leqslant \frac2{a+b}+\frac{a}{b^2}$$
(if $a\leqslant b$, then $\frac1b\leqslant \frac2{a+b}$; if $a\geqslant b$, then $\frac1b\leqslant \frac{a}{b^2}$). Apply this for $b=f$, $a=f'$ and use that $\int f'/f^2=\int (-1/f)'$ converges.
|
2025-03-21T14:48:32.015348
| 2020-09-12T13:37:06 |
371516
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"https://mathoverflow.net/users/12481",
"joro"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632943",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371516"
}
|
Stack Exchange
|
Zero entries in matrix powers over finite rings
Let $R$ be the finite ring of the integers modulo $q$ or $GF(2^k)$.
Let $M$ be $n \times n$ matrix with entries from $R$.
Assume $N,I,J$ are integers and for $ 1 \le i \le N-1$ we have $M^i[I,J]=0$
and $M^N[I,J] \ne 0$.
Q1 How large can $N$ be in terms of $n$, can it be $\exp(Cn)$?
Second question:
For $A \in R, A \ne 0$ assume for $ 1 \le i \le N-1$ we have $M^i[I,J]=A$
and $M^N[I,J] = 0$
Q2 How large can $N$ be in terms of $n$, can it be $\exp(Cn)$?
The Cayley-Hamilton theorem tells us that for each fixed pair $I,J$ the matrix entries $M^i[I,J]$ satisfy a length $n$ linear recurrence $$M^i[I,J] = a_1M^{i-1}[I,J] + a_2 M^{i-2}[I,J] + \cdots + a_n M^{i-n}[I,J]$$
for some fixed constants $a_1, a_2, ..., a_n$. In particular this means if $M^i[I,J] = 0$ for the first $n$ values of $i$ then $M^i[I,J] = 0$ for all $i$, and if $M^i[I,J] = A$ for the first $n+1$ values of $i$ then $M^i[I,J] = A$ for all $i$.
Thanks for the educational answer.
|
2025-03-21T14:48:32.015437
| 2020-09-12T13:44:02 |
371517
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Angelo",
"Anonymous",
"Jason Starr",
"Mellon",
"Nitin Nitsure",
"Piotr Achinger",
"https://mathoverflow.net/users/13265",
"https://mathoverflow.net/users/14044",
"https://mathoverflow.net/users/148928",
"https://mathoverflow.net/users/3847",
"https://mathoverflow.net/users/4790",
"https://mathoverflow.net/users/58651"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632944",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371517"
}
|
Stack Exchange
|
Algebraic spaces as functors on complete local rings
Let $X$ be an algebraic space locally of finite presentation, and let $\tilde{X}$ denote the restriction of $X$ (as a functor on schemes) to the category of complete local rings. Is it true that the mapping $X \mapsto \tilde{X}$ (of algebraic spaces to functors on complete local rings) is a fully faithful functor?
I.e. can we uniquely determine a morphism $f : X \to Y$ of algebraic spaces locally of finite presentation simply by specifying its values on complete local rings?
How do you define the category of complete local rings? What are the arrows?
Ordinary homomorphisms of rings, not necessarily local homomorphisms.
This cannot be true without any finiteness conditions. For example, take a non-reduced ring $R$ with a unique prime $m$ satisfying $m=m^2$ (e.g, a quotient of a rank $1$ nondiscrete valuation ring by a nonzero nonunit). Then any map $R \to S$ to a local ring $(S,n)$ is necessarily a local map, and thus kills $m$ if $S$ is also complete (as $m$ maps into $n^k$ for all $k$ since $m=m^k$). So the spectrum of both $R$ and $R/m$ represent the same functor on complete local rings.
I suppose the Noetherian is assumption will eliminate these types of counter-examples? I will modify the question, there should reasonably be at least some finiteness assumptions.
There is an exponential map from complex line to itself. This is not algebraic, but it is defined on complete local ring valued points. So the functor does not appear to be full.
What happens at the generic point of $\mathbb{A}^1$? The fraction field is a complete local ring. Or did I misunderstand the question?
I agree with Achinger. My answer works only for residue field being finite over the base.
You can make an example where $X$ is a nodal plane cubic and where $Y$ is the quotient of the nontrivial double cover by the involution acting only on the smooth locus. Thus, the algebraic space $Y$ is nonseparated with two closed points over the node of $X$. There is a “section on complete local rings” of the natural morphism from $Y$ to $X$.
|
2025-03-21T14:48:32.015621
| 2020-09-12T13:53:10 |
371519
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"LSpice",
"Qiaochu Yuan",
"Sam Hopkins",
"Troshkin Michael",
"Vít Tuček",
"https://mathoverflow.net/users/157863",
"https://mathoverflow.net/users/2383",
"https://mathoverflow.net/users/25028",
"https://mathoverflow.net/users/290",
"https://mathoverflow.net/users/6818"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632945",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371519"
}
|
Stack Exchange
|
Bilinear forms in compact/semisimple Lie group theory
If you look up the list of compact or semisimple Lie groups, you will see that three out of four infinite families (B, C and D) are defined in terms of a bilinear form on a vector space, either symmetric or skew-symmetric.
Are there any underlying reasons for this prominence of bilinear/quadratic forms in Lie group theory? Why do they, and not any other geometric objects, play such a fundamental role?
If you know that simple Lie groups and simple algebraic groups are mostly the same things, then you know that these have to be defined by algebraic equations. These kind of forms certainly give algebraic equations. Other constructions in geometry can be more of the flavor, your object satisfies a differential equation, which are not going to be algebraic in the same way.
(Edit: I rewrote this answer. In the first draft I tried to take some shortcuts and found that they didn't work.)
Let $G$ be a compact Lie group acting faithfully on a f.d. vector space $V$ over $\mathbb{C}$. It's a nice exercise to show that every f.d. irreducible representation of $G$ appears in some tensor product $V^{\otimes n} \otimes (V^*)^{\otimes m}$ (see, for example, this old MO question). What this implies is that the entire structure of the category $\operatorname{Rep}_\text f(G)$ of f.d. representations of $G$ is contained in the data of the invariant tensors $\operatorname{Hom}(V^{\otimes m}, V^{\otimes n})^G$: more formally, these invariant tensors describe the subcategory of $\operatorname{Rep}_\text f(G)$ generated by $V$ under tensor product and dual and direct sum, and the nice exercise implies that $\operatorname{Rep}_\text f(G)$ is the idempotent completion of this subcategory.
Furthermore, the Tannaka half of Tannaka–Krein duality tells us that $G$ is determined by $\operatorname{Rep}_\text f(G)$ in a suitable sense, although depending on how you take "suitable sense" means you may instead recover the complexification $G_{\mathbb{C}}$. From here on I will blithely ignore the difference between $G$ and its complexification. (Really I should say something here about averaging over a compact group and $\operatorname U(n)$ being the maximal compact subgroup of $\operatorname{GL}_n(\mathbb{C})$.)
Taken together, these two results tell us that $G$ or maybe its complexification is determined as a subgroup of $\operatorname{GL}(V)$ by its $G$-invariant tensors $\operatorname{Hom}(V^{\otimes m}, V^{\otimes n})^G$. What this means is that we ought to be able to define various $G$ of interest by saying "the $G$ preserving such-and-such tensors," and we do.
Moreover, if we decompose a given space of tensors $\operatorname{Hom}(V^{\otimes m}, V^{\otimes n})$ into its irreducible components under the action of $GL(V)$, then $G$ preserves some tensor iff it preserves the projection of the tensor to each irreducible component, so we can restrict our attention to collections of "irreducible tensors."
The tensors of rank $1$ are not so interesting; the stabilizer of a nonzero vector $v \in V$ is a general affine group, so we don't get anything new. Next are the tensors of rank $2$. The tensors in $\operatorname{Hom}(V, V)$ are again not so interesting; generically their stabilizers look like products of $\operatorname{GL}(V_i)$ where $V_i$ are the eigenspaces of a diagonalizable $T \in \operatorname{Hom}(V, V)$, so we again don't get anything new. So the next candidate is bilinear forms, and since $V^{\otimes 2} \cong \operatorname S^2(V) \oplus \bigwedge^2(V)$ is the irreducible decomposition here, we are naturally led to considering the stabilizers of symmetric resp. skew-symmetric forms, hence to the orthogonal and symplectic groups.
It's not just bilinear stuff out there though; to get the special linear groups we have to go all the way to a tensor in $V^{\otimes \dim V}$, namely any choice of a nonzero element of $\bigwedge^{\dim V}(V)$, and we can get, for example, $G_2$ using trilinear forms. But bilinear stuff is the simplest stuff after linear stuff.
Part of the mystery here concerns the relationship between compact groups and algebraic groups, for which see BCnrd's answer here: https://mathoverflow.net/a/16269/290
It should be possible to avoid the use of Tannaka's theorem above, which is what I tried to do in the first draft. What I want to show is that every Zariski closed subgroup $G \subset GL(V)$ is precisely the subgroup preserving the $G$-invariant tensors. I bet this has a simple proof but I'm not seeing it.
I did some TeX tidying. In one place, you had $\operatorname{Hom}(V^{\otimes m}, V^{\otimes n})^)^G$, which I assumed was supposed to be the fixed points in $\operatorname{Hom}(V^{\otimes m}, V^{\otimes n})$ as elsewhere, so I deleted the spare $$.
Whoops, thanks, thought I fixed that already.
How hard is it to see that you actually need just finitely many $G$-invariant tensors?
Is $\mathrm{Hom}_G(V, W) = \mathrm{Hom}(V, W)^G$ the reason behind that invariant tensors determine the structure of $\mathrm{Rep}(G)$?
@Vit: I believe this but have no idea how to prove it directly (for the classical groups in their defining representations one can just check all of them, I guess). It is some kind of noncommutative generalization of finite generation of the ring of invariants...
@Troshkin: yes, that’s right. I thought it would be more concrete to talk about invariant tensors rather than carrying around the entire category.
According to a theorem of Serre, all semi-simple Lie groups are linear algebraic groups. See https://en.wikipedia.org/wiki/Complex_Lie_group for the precise statement and a reference.
That shows why we should look at "algebraic functions" when looking for complex semi-simple Lie groups. Natural place to start is subgroups of $GL(n, \mathbb{C})$ that preserve some linear forms. But then we get something isomorphic to $GL(n, \mathbb{C})$ as such subgroup has to preserve the kernel. Bilinear forms are the next best things and it turns out they provide plenty of examples. But we don't have to stop there! The complex Lie group $F_4$ can be defined as the subgroup of $GL(26, \mathbb{C})$ fixing a symmetric trilinear form. And the complex simple Lie group $G_2$ can be defined as the stabilizer of a generci $3$-form on $\mathbb{C}^7.$ There are similar descriptions for $E$-series. See e.g. this answer by Robert Bryant https://mathoverflow.net/a/99795/6818 (I think he wrote about this more explicitly somewhere else on MO, but I have trouble finding it. Anyway... this description of $E$-series goes back to Elie Cartan.)
|
2025-03-21T14:48:32.016164
| 2020-09-12T15:50:52 |
371526
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Igor Khavkine",
"JustWannaKnow",
"gmvh",
"https://mathoverflow.net/users/150264",
"https://mathoverflow.net/users/2622",
"https://mathoverflow.net/users/45250"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632946",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371526"
}
|
Stack Exchange
|
Creation and Annihilation operators in QFT - Part II
Following some suggestions on my previous posts, I'm trying to reformulate my question in a more specific way. This is a continuation of my original post. Since the mentioned post, I think I've learned some things, so let me elaborate a little more on the issue.
Let $\mathcal{H}$ be a Hilbert space and $\mathcal{F}_{f}(\mathcal{H}) = \bigoplus_{n=0}^{\infty}\mathcal{H}^{-}_{n}$ the fermionic Fock space associated to $\mathcal{H}$. Here, $\mathcal{H}^{-}_{n}$ is the space of all anti-symmetric tensors $\mathcal{H}_{n} := \overbrace{\mathcal{H}\otimes\cdots\otimes\mathcal{H}}^{\text{$n$ times}}$. Let $\varphi \in \mathcal{H}$ be fixed and $a(\varphi): \mathcal{H}_{n}^{-}\to \mathcal{H}_{n-1}^{-}$ and $a^{\dagger}(\varphi):\mathcal{H}_{n}^{-}\to \mathcal{H}_{n+1}^{-}$ be the usual annihilation and creation operators, respectivelly. These operators are bounded operators and we can extend them to $\mathcal{F}_{f}(\mathcal{H})$ (actually to a dense subset of it).
Now, let $\mathcal{H} = L^{2}(\mathbb{R}^{3})$. Suppose we identify $\psi \in \mathcal{H}_{n}^{-}$ with $\psi \in L^{2}_{-}(\mathbb{R}^{3n})$ the space of all anti-symmetric $f\in L^{2}(\mathbb{R}^{3n})$. In other words:
$$f \in L^{2}_{-}(\mathbb{R}^{3n}) \iff f(x_{\sigma(1)},...,x_{\sigma(n)}) = \epsilon_{\sigma}f(x_{1},...,x_{n})$$
for every permutation $\sigma$ of $\{1,...,n\}$. Then $a(\varphi)\psi \in H_{n-1}^{-}$ is identified with:
\begin{eqnarray}
(a(\varphi)\psi)(x_{1},...,x_{n-1}) = \sqrt{n}\int \overline{h(x)}\psi(x,x_{1},...x_{n-1})dx \tag{1}\label{1}
\end{eqnarray}
If $\psi \in L^{2}_{-}(\mathbb{R}^{3n})$ is a continuous function, then we can take $\varphi$ to be a delta distribution, so we can define an operator:
\begin{eqnarray}
(a(x)\psi)(x_{1},...,x_{n-1}) = \sqrt{n}\psi(x,x_{1},...,x_{n-1}) \tag{2}\label{2}
\end{eqnarray}
This operator can be extended to a suitable dense domain on the Fock space.
My original question was to understand the creation and annihilation operators $\varphi(x,\sigma)$ and $\varphi^{\dagger}(x,\sigma)$ introduced on page 18 of Feldman, Trubowitz and Knörrer's notes. I believe the mentioned extension of my $a(x)$'s and $a^{\dagger}(x)$'s is precisely what FTK's notes are talking about if the spins $\sigma$ were not to be considered. However, I'd like to consider spins in my above description, so to define $\varphi(x,\sigma)$ and $\varphi^{\dagger}(x,\sigma)$.
Question: How to introduce spin variables within the above theory?
My guess is as follows: a particle now is described not only by not only its wave function $\psi \in L^{2}(\mathbb{R}^{3})$ but also by its spin $\sigma$, so I should probably consider, say, $L^{2}(\mathbb{R}^{3})\otimes \mathbb{C}^{2}$, or something like this. Then, maybe use $L^{2}(\mathbb{R}^{3})\otimes \mathbb{C}^{2} \cong L^{2}(\mathbb{R}^{3}; \mathbb{C}^{2})$? But then, if $\mathcal{H} = L^{2}(\mathbb{R}^{3};\mathbb{C}^{2})$, an element $\psi \in H_{n}^{-}$ should be identified to what? An element $\psi \in L^{2}_{-}(\mathbb{R}^{3n};\mathbb{C}^{2})$?
Your guess is correct.
Your very last question is essentially this: what is $\mathcal{H}_n^-$? Not unexpectantly, it is always $\mathcal{H}_n^- = \bigwedge^n \mathcal{H}$, denoting the fully anti-symmetrized Hilbert tensor product. The further isomorphism $L^2(\mathbb{R}^3; \mathbb{C}^2) \cong L^2(\mathbb{R}^3\times{+1,-1})$ should make it clear what $\mathcal{H}_n^-$ is. Exercise: fill in the blank in $\mathcal{H}_n^- \subset L^2(\mathbb{R}^{3n}; ?)$.
@IgorKhavkine Maybe I could use the following: if $\mathcal{H} = L^{2}(\mathbb{R}^{3};\mathbb{C}^{2}) \cong L^{2}(\mathbb{R}^{3}\times {+1,-1})$ it seems that $\mathcal{H}_{n} \cong L^{2}(\mathbb{R}^{3}\times {+1,1})\otimes \cdots \otimes L^{2}(\mathbb{R}^{3}\times {+1,-1}) \cong L^{2}(\mathbb{R}^{3n}\times {+1,-1}^{n})$? (Don't know for sure if the last isomorphism holds tho).The latter seems more useful to define $\varphi(x,\sigma)$ and $\varphi^{\dagger}(x,\sigma)$ since $x \in \mathbb{R}^{3}$ and $\sigma$ is a spin variable, probably taking values ${+1,-1}$.
@IamWill All correct. Don't see any reason for you to doubt yourself there.
|
2025-03-21T14:48:32.016419
| 2020-09-12T16:17:09 |
371530
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"ABIM",
"Logan Fox",
"Nik Weaver",
"https://mathoverflow.net/users/160011",
"https://mathoverflow.net/users/23141",
"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:632947",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371530"
}
|
Stack Exchange
|
Terminology: Co-completion of Met?
In main-stream mathematical literature, the term metric space is reserved for $(X,d)$ where $X$ is a set and $d:X\times X\rightarrow [0,\infty)$ satisfies the usual properties of a metric. However, at times it is convenient to allow $d$ to take infinite values (for example if we would like to give meaning to a "co-product" of metric spaces). In that case, what is the most standard terminology for such an "extended-real-valued" metric (i.e.: $codom(d)=[0,\infty]$)?
I'm not sure if there is a standard terminology, but I have seen "infinite-valued metric" used (in Gerald Beer's Topologies on Closed and Closed Convex Sets).
Interesting, I have seen the same author (in "The Structure of Extended Real-valued Metric Spaces") calling these objects extended real-valued metric spaces... now I'm even more confused.
I think "extended metric" is pretty standard. That's the term I used in my book Lipschitz Algebras.
True, yes I have seen that there. Thanks Nik.
I do not think that there is a standard name for such spaces (and hence for such generalised metrics). It is quite common to see the terms '$\infty$-metric space' and 'extended metric space' (or some slight modifications). However, the latter name is also used in a more general sense, where the metric $d$ is allowed to take values in any ordered set. Consequently, the former term can be seen as a special case of the latter, where the ordered set is the one of extended positive real numbers.
|
2025-03-21T14:48:32.016551
| 2020-09-12T08:59:36 |
371531
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Yemon Choi",
"https://mathoverflow.net/users/100231",
"https://mathoverflow.net/users/763",
"vidyarthi"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632948",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371531"
}
|
Stack Exchange
|
Where can I find journal contents of Ars Combinatoria
In the journal website, there are table of contents available only from 1995-2019. Where can I find the table of contents before that? And, is the journal only offline through subscription? Thanks beforehand.
Small correction of English: "up to 1995" suggests "years before 1995" whereas I think you mean that there are only TOC going back as far as 1995
MathSciNet does show the issues going back to the start of the journal in 1976, but I admit this is a slow and inconvenient way to skim
The tables of contents are not available online, but you can reconstruct them using Web of Science or MathSciNet (if you have access), or Google Scholar (query source:"Ars Combinatoria").
It should not be too much work to recreate the listings for all missing years; for starters, here I have listed the 764 papers from the years 1985-1994 (volumes 19 through 38).
Thanks for your effort! It seems you are more interested in Mathematics than physics!
|
2025-03-21T14:48:32.016660
| 2020-09-12T16:33:12 |
371532
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Tim Campion",
"https://mathoverflow.net/users/2362"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632949",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371532"
}
|
Stack Exchange
|
Varieties where any subalgebra is a regular sub-object
Following [LPAC] Chap.3, p.132 let $S$ a set of sorts and $\Sigma$ a $S$-sorted signature. From the [LPAC] treatment we have the category $Alg(\Sigma)$ with a $(\mathit{regular.Epi,Mono})$-factorization, and where all epimorphism are all regular, and are just the surjective morphisms.
Consider a equational class $\operatorname{Alg}(\Sigma, E)$ defined by a set of equation $E$ as in [LPAC] p.138, from point $(1),\ (5)$ of remark 3.6 of [LPAC] p.140, follow that the full inclusion $\operatorname{Alg}(\Sigma, E)\subset \operatorname{Alg}(\Sigma)$ create the $(\mathit{regular.Epi,Mono})$-factorization.
By [LPAC] corollary 3.7 $\operatorname{Alg}(\Sigma, E)$ is locally presented, then by [LPAC] 1.61 pag. 50 has a $(\mathit{regular.Epi,Mono})$-factorization and a $(\mathit{Epi, regular.Mono})$-factorization, then we have that:
*) all epimorphisms are regular iff all monomorphisms are regular:
let $f: X\to Y$ a epimorphism, and $f=m\circ e: X\to I\to Y$ with $m$ regular monomorphism, for $f=m\circ e$ and $f$ epimorphism, follow that $m$ is a epimorphism, then it is an isomorphism. Dually for the rest.
For $E=\emptyset$ the above result $(*)$ is true for $\operatorname{Alg}(\Sigma)$, but in $\operatorname{Alg}(\Sigma)$ we have that "Epi=regular Epi" i.e. all epimorphism are regular (and then all monomorphism are regular). Also for the (algebraic) theory of groups "Epi=regular Epi" is valid (Barry Mitchell, "Theory of Categories" pag.38, ex. 13). Of course "Epi=regular Epi" isn't even valid, consider "rings with unity" theory, and the inclusion $\mathbb{Z}\subset \mathbb{Q}$ is monomorphism, and is a epimorphism but no a regular epimorphism (bacause isnt a isomorphism and: (reg.Epi +Mono) $\to$ Iso).
My question is:
"for what kind of (finitary) algebraic theory $\Sigma$ and equation class $E$, is valid that "Epi=regular Epi" for the category $Alg(\Sigma, E)$" ?
[LPAC] Locally Presentable and Accessible Categories – J. Adamek, J.Rosicky.
I believe an alternate formulation of the same question would be: Which (possibly multi-sorted) varieties are balanced (i.e. have the property that every monic epic is an isomorphism).
|
2025-03-21T14:48:32.016820
| 2020-09-12T16:54:28 |
371534
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"LSpice",
"Sara",
"Yemon Choi",
"https://mathoverflow.net/users/165240",
"https://mathoverflow.net/users/2383",
"https://mathoverflow.net/users/763"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632950",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371534"
}
|
Stack Exchange
|
Understanding a step in a proof on diagnosability of graphs
I am currently writing a paper about diagnosability of graphs. For this, I am looking into a paper published in 1967 by Preparata, Metze and Chien called "On The Connection Assignment Problem of Diagnosable Systems".
I am currently working through the proof of Theorem 5 within this paper and there is one step that I just can't understand. In the link above, this step happens right at the top of page 22 with the sentence:
The above stated condition (1) can be met if the number $n$ of units cannot be partitioned in less than $s$ sequences of the given maximum length.
Why is this true?
It would be helpful if you could include some of the relevant definitions and statements in your question, to save people having to look through the whole of the article that you are referring to
You're definitely right about that, but the definitions come from an algorithm further up in the paper, so it is very hard to summarize all of those in just one post.
Hmm, is diagnosability of graphs related to the graph isomorphism disease? (I thought for sure it was a typo for 'diagonalisable', but it appears not!)
I think I just solved it myself, so I'm just gonna put the answer here:
I ended up reshuffling the proof and using two properties. One is equation (3) from the same proof. The other property is
$$s \ r_{max} \geq n,$$
which is true because we have split $n$ into $s$ segments of maximum length $r_{max}$. Then we can combine the two properties to get (1).
|
2025-03-21T14:48:32.017213
| 2020-09-12T17:06:04 |
371536
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Andrej Bauer",
"Mare",
"Mirco A. Mannucci",
"Sam Hopkins",
"YCor",
"https://mathoverflow.net/users/1176",
"https://mathoverflow.net/users/14094",
"https://mathoverflow.net/users/15293",
"https://mathoverflow.net/users/25028",
"https://mathoverflow.net/users/61949"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632951",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371536"
}
|
Stack Exchange
|
On free lattices
Free distributive lattices on a finite set exist and are finite, while free modular lattices on a finite set exist but are not finite when the set has at least 4 elements.
Question: Is there a class (presumably, a variety in the sense of universal algebra) $C$ of lattices larger than the class of distributive lattices such that free $C$-lattices on a finite set exist (for all finite sets) and are finite?
If yes, is there a "largest" such class $C$?
Comment: free modular lattices on $> 3$ elements are infinite.
@SamHopkins Thanks, I added that.
An interesting class of lattices a bit bigger than distributive lattices, and different than modular, would be semidistributive lattices. But I don't know enough about universal algebra to say whether those qualify as a variety, so that it makes sense to talk about free objects.
Ah, pg. 82 of https://www1.chapman.edu/~jipsen/Jipsen%20Rose%201992%20Varieties%20of%20Lattices.pdf: "The fourth statement shows that semidistributivity cannot be characterized by a set of identities, and so the class of all semidistributive lattices does not form a variety. "
An outsider's question: is it common to cay "classs" instead of "variety" in universal algebra?
@AndrejBauer: also an outsider; but: strictly speaking, I don't think it is required that the class of objects be a variety for free objects to exist. Just that certainly being a variety guarantees the existence of free objects.
Wikipedia suggests free objects might exist in more general contexts: https://en.wikipedia.org/wiki/Free_object.
Here is what you need: http://math.hawaii.edu/~jb/math618/os8uh.pdf Look up theorem 8.1 . Then use it to create a "mini-lattice" which is not distributive and close with respect to HSP
PS to address Andrej's question and Sam's comment: even if one looked at a class which is not a variety, by closing with respect to HSP operations you could extend it to a variety. So I think we can safely assume that the OP uses variety in the question
@MircoA.Mannucci: Those are interesting notes, but can you say explicitly what the conclusion is, i.e., what's the answer to Mare's question?
answer: right now I have just eaten penne with heavy sour cream and pesto, I could not even recite the multiplication table up to 9 :). Mine is simply a comment, maybe tomorrow, if nobody else has written an answer, I can try. But here is what I can tell you now: I would use theorem 8.1 and a judiciously chosen little lattice to generate a variety of lattices which includes distributives but not modular. As for the second part,
namely whether there is a maximal variety with the finite property, I would perhaps consider ascending list of mini-lattices and their corresponding varieties, and take, as it were, the sup. Is this a strategy? I would not bet my house on that, but perhaps worth trying....
Ah, I see, distributive lattices are HSP(2), and as explained towards the end of http://www.math.hawaii.edu/~jb/math618/os9uh.pdf, this variety is covered by both HSP($N_5$) and HSP($M_3$), and by the theorem 8.1 you linked to since these are generated by a single lattice they will have finite free objects.
Right! Theorem 8.1 and subsequent consequences give you control on the generated variety. How about this my newly found MO friend: you write the full answer. If you do, I tell you my secret recipe to make penne al pesto + sour cream
What about the variety generated by modular lattices and a fixed finite non-modular lattice?
(I should have written "generated by distributive lattices and a fixed non-distributive modular lattice)
A variety is called locally finite if it has the property that its finitely generated algebras are finite. This is equivalent to the property that its finitely generated free algebras are finite. So, the questions may be rewritten as:
(1) Is there a locally finite variety of lattices larger than the variety of distributive lattices?
(2) Is there a largest locally finite variety of lattices?
Every variety generated by a single finite algebra is locally finite, so there are lots of locally finite varieties of lattices. Any variety generated by a finite, nondistributive lattice will properly contain the variety of distributive lattices, hence will be an answer to Question (1).
[There do exist locally finite varieties of lattices that are not generated by a single finite lattice, like the variety generated by all lattices of height 2.]
If there were a largest locally finite variety of lattices, then by the answer to Question (1) it would have to contain every finite lattice. But the variety generated by all finite lattices is the variety of all lattices, and the variety of all lattices is not locally finite. Hence the answer to Question (2) is No.
A comment on the comments:
The class of semidistributive lattices forms a quasivariety, and therefore
has free objects over any set. But this quasivariety is not locally finite. The free semidistributive lattice on 3 generators is infinite.
|
2025-03-21T14:48:32.017538
| 2020-09-12T17:16:03 |
371538
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Ira Gessel",
"Mark Wildon",
"Max Alekseyev",
"Richard Stanley",
"Sam Hopkins",
"https://mathoverflow.net/users/10744",
"https://mathoverflow.net/users/25028",
"https://mathoverflow.net/users/2807",
"https://mathoverflow.net/users/7076",
"https://mathoverflow.net/users/7709"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632952",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371538"
}
|
Stack Exchange
|
How do I determine the number of "second degree" derangements?
The number of derangements for a given set is saying, the number of total possible ways of shuffling the members such that no member sits in its original place. This is given by the closed form:
$$!n = n! \cdot \sum_{k=0}^{n}\frac{(-1)^k}{k!}$$
Now, given a set S and one of its derangements D1, how many ways are there to find a new (let's say second degree) derangement D2 with members that are not in the same place as S and D1 at the same time...
Does "not in the same place as $S$ and $D1$ at the same time" mean that $S,D2$ and $D1,D2$ are both derangements? In that case, the number of ways depends on the choice of $D1$. You are extending a $2\times n$ Latin rectangle to a $3\times n$ Latin rectangle. There are four ways to extend $1234,2143$ but only two ways to extend $1234,2341$. I think that Sam Hopkins wants two permutations $\pi,\sigma$ such that $i,\pi(i)$, and $\sigma(i)$ are distinct for all $i$.
Yes, sorry, what I wrote earlier was wrong. This problem can be rephrased as counting/understanding the structure of the set of pairs of permutations u and v such that u, v, and $u^{-1}v$ are all derangements. If the question is specifically about how to extend a fixed derangement u to such a pair, then as Richard said it depends on the specific u.
This problem is discussed in John Riordan's Combinatorial Analysis but I don't have access to my copy right now.
The Riordan reference is Chapter 8, Section 3, page 201. Please excuse a bit of self-publicity: Riordan's method for solving this problem is the hardest part in Section 4.3 of my draft combinatorics textbook. In particular, Exercise 4.19 gives the formula $\sum_k \binom{n}{k} !k !(n-k) v_{n-2k}$ for the number of $ 3 \times n$ Latin rectangles, where $v_m$ is a Ménage number. See http://www.ma.rhul.ac.uk/~uvah099/Maths/CombinatoricsWeb.pdf.
I just discovered that this question is essentially a duplicate of https://mathoverflow.net/questions/144899/number-of-permutations.
(Corrected and expanded, again!)
As mentioned in the comments, the number of third permutations depends on the relationship between the first two. Asymptotically, the number of third permutations is $\sim e^{-2} n!\, (1-1/n - 1/(2n^2)+O(n^{-3}))$ regardless of the first two permutations. For the exact number, follow Ira's hint: Riordan, Introduction to Combinatorial Analysis, chapter 8, part 3.
I can give a hint of how much the number of third permutations varies according to the first two. Let $s$ be the number of intercalates in the first two permutations. (An intercalate is two positions where the two permutations have the same two entries in opposite order: where one has $ab$ the other has $ba$.) Since intercalates cannot overlap, the number of them can't exceed $n/2$. Asymptotically, the number of third permutations which are a derangement of the first two is
$$ e^{-2} n!\, \Bigl( 1 - \frac 1n - \frac 1{2n^2} + \frac 1{3n^3} + \frac {s}{n^4} + O(n^{-4})\Bigr).$$
This is from C. D. Godsil and B. D. McKay, Asymptotic enumeration of Latin rectangles, J. Combinatorial Theory, Ser. B, 48 (1990) 19-44. Corrected version.
A triple of permutations, each two of which are derangements of each other, is a 3-row Latin rectangle. There is a simple summation for the number of them due to Yamamoto, see page 18 in Stone's survey.
For Ira's paper: http://people.brandeis.edu/~gessel/homepage/papers/3latin.pdf
For another approach to 3-row Latin rectangles, with generalizations and further references, see my paper Counting three-line Latin rectangles, Combinatoire énumérative (Montreal, Que., 1985/Quebec, Que., 1985), 106–111, Lecture Notes in Math., 1234, Springer, Berlin, 1986. (I had a bad link in the original version of this comment; use the one in Sam's comment.)
The answer depends on the cycle structure of $D_1$. Let $n:=|S|$ and $c_i$ be the number of cycles of length $i$ in $D_1$ (with $\sum_i ic_i=n$). Since $D_1$ is a derangement, we have $c_1=0$, but what follows well applies to any permutation (not necessarily derangement) $D_1$ of $S$.
The number of permutations $D_2$ that are derangement w.r.t. the identity permutation as well as w.r.t. $D_1$ equals
$$\sum_{j=0}^n (-1)^j\cdot (n-j)!\cdot [z^j]\ F(z),$$
where $[z^j]\ F(z)$ is the coefficient of $z^j$ in
$$F(z) := (1+z)^{c_1}\cdot \prod_{i=2}^n \left( \left(\frac{1+\sqrt{1+4z}}2\right)^{2i} + \left(\frac{1-\sqrt{1+4z}}2\right)^{2i} \right)^{c_i}.$$
This can be obtained with the method described in my paper (in particular, see formula (4) and Lemma 1).
Particular cases:
when $c_1=n$ (i.e., $D_1$ is the identity permutation), we get just the number of derangements;
when $c_n=1$ (i.e., $D_1$ is a cyclic permutation), we get the menage number A000179(n);
when $n=2m$ and $c_2=m$ (i.e., $D_1$ is a derangement and an involution), we get A000316(m) = A000459(m)$\cdot 2^m$.
Can you say what you mean by "possible derangements $D_2$" in the case that $D_1$ is not a derangement of $S$ (e.g., if $D_1$ is the identity)?
@SamHopkins: $D_2$ should be a derangement w.r.t. the identity permutation as well as a derangement with respect to $D_1$. I've clarified this in my answer.
Thanks, makes sense!
|
2025-03-21T14:48:32.017881
| 2020-09-12T17:39:35 |
371541
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632953",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371541"
}
|
Stack Exchange
|
When are all average trajectories of $w_{k+1}=Aw_k+b$ bounded?
Below is an open-problem in my field, and I'm wondering if someone has insights I'm missing. (cross-posted on math.se)
Suppose observation $x$ is drawn from some distribution $\mathcal{D}$, $w_0\in \mathbb{R}^\mathrm{d}$, and my update has the following form
$$w_{k+1}=(I-xx')w_k+b$$
When are average trajectories $u_k=E[w_k]$ bounded? Expectation is taken over all sequences of IID observations $x_1,\ldots,x_k$. Motivation for this recurrence is here.
I have worked out the case of $b=0$ and I'm wondering if global asymptotic stability for $b=0$ also implies boundedness for some other value of $b$. An interesting special case is when $x$ comes from Gaussian centered at 0.
Cases of increasing difficulty are
$b=0$
$b=c$ is some vector $c$ from $\mathbb{R}^\mathrm{d}$
$b=Bx$ for some $\mathrm{d}\times \mathrm{d}$ matrix $B$
$b=d$ where $d$ is drawn from some distribution $\mathcal{D}_2$ independent of $x$
$b=c+Bx+d$
$b,x$ are drawn jointly from some distribution $\mathcal{D}_3$
For the case of $b=0$, condition below seems to be a necessary and sufficient condition for all trajectories $u_k$ to converge to 0. The following must hold for all symmetric matrices $A$
$$E[(x'Ax)^2]<2 E[x'A^2 x]$$
For the case of $x$ coming from zero-centered Gaussian with covariance $\Sigma$, this becomes the following equation, which can be checked in practice using generalized eigenvalue solver (using identities 20.18 and 20.25c of Seber's Matrix Handbook book)
$$(\text{Tr}(A\Sigma))^2+2\text{Tr}(A\Sigma)^2<2\text{Tr}A^2\Sigma$$
|
2025-03-21T14:48:32.018019
| 2020-09-12T17:58:23 |
371543
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"David Roberts",
"Zhen Lin",
"https://mathoverflow.net/users/11640",
"https://mathoverflow.net/users/152679",
"https://mathoverflow.net/users/4177",
"varkor"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632954",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371543"
}
|
Stack Exchange
|
When is a finitary functor induced by Ind (co)continuous
Let $\mathbf C$ and $\mathbf D$ be small categories. $\mathrm{Ind}(\mathbf C)$ is an accessible category (by definition), and is locally finitely presentable (i.e. cocomplete, or equivalently complete) iff $\mathbf C$ has finite colimits. Let $\mathbf C$ and $\mathbf D$ have finite colimits, and consider a functor $F : \mathbf C \to \mathrm{Ind}(\mathbf D)$. By the universal property of $\mathrm{Ind}$, this extends to a finitary functor $\tilde F : \mathrm{Ind}(\mathbf C) \to \mathrm{Ind}(\mathbf D)$.
In terms of $F$, assuming such a characterisation exists:
When is $\tilde F$ continuous? (Equivalently, when does $\tilde F$ have a left adjoint?)
When is $\tilde F$ cocontinuous? (Equivalently, when does $\tilde F$ have a right adjoint?)
I imagine (2) should hold when $F$ preserves finite colimits, though I wasn't able to find a reference in Locally presentable and accessible categories.
Your "iff" is not quite correct, I think. You should assume idempotents split.
@ZhenLin: just to clarify, what would be the corrected statement?
Assuming idempotents in $\mathcal{C}$ split, $\textbf{Ind}_\kappa (\mathcal{C})$ has colimits if and only if $\mathcal{C}$ has $\kappa$-small colimits.
If I understand correctly: if $\mathcal C$ has $\kappa$-small colimits, then idempotents in $\mathcal C$ split. So is this to say that $\mathbf{Ind}_\kappa(\mathcal C)$ may be cocomplete even if $\mathcal C$ does not have $\kappa$-small colimits, but idempotents in $\mathcal C$ do split?
No, if we don't know that idempotents in $\mathcal{C}$ split then we cannot conclude that $\mathcal{C}$ has $\kappa$-small colimits even if $\textbf{Ind}_\kappa (\mathcal{C})$ is cocomplete.
Sorry, I realise I omitted a critical negation in my previous comment. Thank you for your patience! Suppose $\mathbf{Ind}\kappa(\mathcal C)$ is cocomplete. If idempotents in $\mathcal C$ split, then we know that $\mathcal C$ has $\kappa$-small colimits. However, if idempotents in $\mathcal C$ do not split, then $\mathcal C$ cannot have $\kappa$-small colimits, or otherwise its idempotents would split. Is there a specific example of a small category in which idempotents do not split, but whose $\mathbf{Ind}\kappa$-completion is cocomplete?
To clarify where my confusion is coming from, I was basing my original statement of Theorem 5.5(ii) of A classification of accessible categories (which I see I must have misunderstood). That paper does not mention idempotents, so I had not appreciated their importance here.
Ah, perhaps the answer is that, if $\mathbf{Ind}\kappa(\mathcal C)$ is cocomplete, then there is a category $\mathcal C'$ with $\kappa$-small colimits such that $\mathbf{Ind}\kappa(\mathcal C) ≃ \mathbf{Ind}_\kappa(\mathcal C')$, but $\mathcal C$ itself may not be.
That’s right. Every locally $\kappa$-presentable category is equivalent to the category of $\kappa$-ind objects of some $\kappa$-cocomplete category, but that may be a different category from what you start with. (Not too different, in reality: it is the Cauchy completion.)
Allow me to generalise to $\kappa$-accessible categories for infinite regular cardinals $\kappa$. Your guess for (2) is correct: if $F$ preserves $\kappa$-small colimits then $\tilde{F}$ preserves colimits. The proof is a little bit indirect.
Proposition. Let $\mathcal{I}$ be a category and let $\mathcal{C}$ be a small category with $\kappa$-small colimits. If $\mathcal{I}$ is $\kappa$-small, then the comparison functor
$$\textbf{Ind}_\kappa ([\mathcal{I}, \mathcal{C}]) \to [\mathcal{I}, \textbf{Ind}_\kappa (\mathcal{C})]$$
is fully faithful and essentially surjective on objects.
(The key point is to show that every diagram of shape $\mathcal{I}$ can be written as a $\kappa$-filtered colimit of diagrams of $\kappa$-presentable objects of the same shape $\mathcal{I}$. Actually, the statement of the proposition is equivalent to this fact, and this is what we need for the next step.)
Proposition. Let $\mathcal{C}$ be a category with $\kappa$-small colimits, let $\mathcal{E}$ be a category with $\kappa$-filtered colimits, let $F : \mathcal{C} \to \mathcal{E}$ be a functor, and let $\tilde{F} : \textbf{Ind}_\kappa (\mathcal{C}) \to \mathcal{E}$ be the extension. Then $\tilde{F}$ preserves colimits if and only if $F$ preserves $\kappa$-small colimits.
(We already know $\tilde{F}$ preserves $\kappa$-filtered colimits, so it is enough to check whether $\tilde{F}$ preserves $\kappa$-small colimits. The "only if" direction is easy. The "if" direction is proved using the previously mentioned decomposition of $\kappa$-small diagrams.)
I don't have a good answer for (1), but perhaps this will be enough for your purposes:
Proposition. Let $\mathcal{C}$ and $\mathcal{D}$ be small categories with $\kappa$-small colimits. A functor $F : \mathcal{C} \to \mathcal{D}$ has a left adjoint if and only if $\textbf{Ind}_\kappa (F) : \textbf{Ind}_\kappa (\mathcal{C}) \to \textbf{Ind}_\kappa (\mathcal{D})$ has a left adjoint.
(The "only if" direction is easy: after all, $\textbf{Ind}_\kappa$ is a pseudofunctor so it preserves adjunctions. The "if" direction amounts to saying that the left adjoint of a $\kappa$-accessible functor between locally $\kappa$-presentable categories preserves $\kappa$-presentable objects, which is straightforward to check.)
Off-topic, but the url listed on your user page is broken. It's good to see you here again, btw!
Thanks! I updated the link. (Nothing new there though.)
Thank you, this is helpful! Is there reason to believe there may not be a nice characterisation for (1)?
Well, I wouldn't expect anything as nice as (2). For example you might consider $\textbf{CRing}^X$ where $X$ is an infinite set. This is a locally finitely presentable category for general reasons, but the terminal object – the limit of the empty diagram! – is not finitely presentable. So one has to do more work to figure out how a finitely accessible functor defined on finitely presentable objects is going to act on the terminal object.
Thanks, this is a nice example. I think simply knowing that there isn't such an elegant characterisation for (1) is enough to answer that part of my question. I had hoped there would be a condition not much more complex than that for (2), but suspected it might not be the case.
|
2025-03-21T14:48:32.018488
| 2020-09-12T19:14:20 |
371548
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Ben Ciotti",
"Daniele Tampieri",
"Pietro Majer",
"https://mathoverflow.net/users/113756",
"https://mathoverflow.net/users/137640",
"https://mathoverflow.net/users/6101"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632955",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371548"
}
|
Stack Exchange
|
Showing integrability of a locally integrable function on a bounded domain under some additional assumptions
Suppose $\Omega\subset \mathbb{R}^3$ is a smooth and bounded domain, and $f:\Omega\to[0,\infty]$ is a given function which is finite almost everywhere and satisfies
Assumption A: For all $g\in C_0^1(\Omega)$ we have the product $fg\in L^1(\Omega)$. (Here $C_0^1(\Omega)$ refers to functions which are continuously differentiable in $\Omega$ and extend continuously to $0$ on $\partial\Omega$).
Question 1: Can we show that $f\in L^1(\Omega)$?
Question 2: Does the answer to Question 1 change if we include some or all of the following assumptions:
Assumption B: $f$ possesses a weak derivative which is finite almost everywhere in $\Omega$;
Assumption C: There exists a nonnegative function $f_0 \in H^2(\Omega)\cap C(\bar{\Omega})$ such that $f-f_0=0$ (in the sense of trace) on $\partial\Omega$;
Assumption D: There exists a nonnegative function $h\in H^1(\Omega)$ such that $h$ is nonzero almost everywhere in $\Omega$ and $f=-\ln h$ in $\Omega$.
Note: Assumption D more or less implies Assumption B. I wrote them separately in the hopes of formulating the problem as simply as possible.
Notation: Here $H^k$ is the standard Sobolev space notation for $W^{k,2}$.
9/14/20 Edit:
Question 1 has been answered in the affirmative. I additionally pose the following
Question 3: Answer Questions 1 and 2 in the case that Assumption A is replaced by
Assumption A': $f\in L^1_{\text{loc}}(\Omega)$.
In the definition of $C^1_0$, do you mean: (i) continuously differentiable on $\Omega$, continuous on $\overline \Omega$, vanishing on $\partial \Omega$, or (ii) : continuously differentiable on $\overline \Omega$, and vanishing on $\partial \Omega$. For instance : if $\Omega$ is the unit ball, and $f(x)=\sqrt{1-|x|^2}$, does $f\in C^1_0$ ?
@PietroMajer Your first definition. Thank you for pointing that out. I have edited the post to make it more clear.
Let $(g_k)_{k\ge0}$ be a sequence of smooth functions such that $g_k(x)=1$ if $\text{dist}(x,\partial\Omega)\ge 2^{-k}$,
$g_k (x)=0$ if $\text{dist}(x,\partial\Omega)\le 2^{-k-1}$ and $0\le g_k\le 1$ everywhere.
To prove the affirmative answer to Question 1 by contrapositive, let $f\not\in L^1(\Omega)$ be given: we want to find $g\in C^1_0(\Omega)$ such that $fg\not\in L^1(\Omega)$. We can assume $fg_k\in L^1(\Omega)$ for all $k$, otherwise we are done with $g=g_k$ for some $k$. Then $ \int_\Omega fg_k$ is an increasing sequence of positive real numbers, that diverges to $+\infty$, for if it were bounded, $f\in L^1(\Omega)$ by Beppo Levi's theorem. So for some subsequence $(g_{k_j})_j$ we have $ \int_\Omega fg_{k_{j+1}}\ge \int_\Omega fg_{k_j}+1 $, that is $ \int_\Omega f(g_{k_{j+1}}-g_{k_j})\ge 1$. For all $j$ the function $g_{k_{j+1}}-g_{k_j}$ is bounded between $0$ and $1$, and supported in the set $\big\{ 2^{-k_{j+1}-1}\le \text {dist}(x,\partial\Omega)\le 2^{-k_j}\big\}$.
But then $g:=\sum _{j\ge1} \frac{g_{k_{j+1}}-g_{k_j} }j$ is a locally finite sum of smooth functions, hence smooth in $\Omega$; clearly $g(x)\to0$ for $x\to\partial\Omega$, and, again by Beppo Levi's theorem, $\int_\Omega fg\ge \sum_{j\ge1}\frac1j=+\infty$.
Note that no assumption on $\partial\Omega$ are needed
Very nice Pietro, thank you! May I ask a followup (which essentially comes down to the distinction you raised in your first comment)? Suppose I remove the assumption that $fg\in L^1$ for all $g\in C_0^1(\Omega)$, and replace it by the assumption that $fg\in L^1$ for all $g\in C_c^1(\Omega)$. Would the result still hold?
Here $C^1_c(\Omega)$ is the set of continuously differentiable functions with compact support in $\Omega$.
Isn't this the same as saying $f$ is $ L^1_{loc}(\Omega)$?
@BenCiotti, as a matter of fact, substituting $C^1_0(\Omega)$ with $C^1_c(\Omega)$ is equivalent to say that $f$ is a locally integrable function
Ok, so in that case, could the result be shown? Perhaps using the additional assumptions?
Excuse me, what result do you mean?
|
2025-03-21T14:48:32.018889
| 2020-09-12T19:22:36 |
371551
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Asvin",
"Marek Kryspin",
"Random",
"https://mathoverflow.net/users/140015",
"https://mathoverflow.net/users/58001",
"https://mathoverflow.net/users/88679"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632956",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371551"
}
|
Stack Exchange
|
Probability of satisfying the congruent mod equation
I'm wondering about the probability of picking three different numbers $x,y,z$ out of the set $[50]=\left\{ 1,2,3,...,50\right\}$ satisfying the equation: $$xyz\equiv \gcd(x,y,z)\mod 7$$ I started out by identifying a set of possible events $\Omega$ that is, the set of all three-element subsets. Cardinality $|\Omega|={50 \choose 3}$. But I have trouble determining the cardinality of the set of favorable events: $$\mathcal{M}=\left\{ \left\{ x,y,z\right\}\in\Omega: xyz\equiv \gcd(x,y,z)\mod 7 \right\} $$
My efforts and observations:
Some simplification of congruent mod equation.
Because $\gcd(x,y,z)|x \ \& \gcd(x,y,z)|y \ \& \gcd(x,y,z)|z$, it will take place: $$\text{gcd}^3(x,y,z)|xyz$$
So there is $x',y',z'$ such that $xyz=\text{gcd}^3(x,y,z)x'y'z'$ so the equation can be written: $$xyz\equiv \gcd(x,y,z)\mod 7$$ $$\text{gcd}^3(x,y,z)x'y'z'\equiv \gcd(x,y,z)\mod 7$$ $$\gcd(x,y,z)\left( \text{gcd}^2(x,y,z)x'y'z'-1\right)\equiv 0\mod 7$$ $7$ as it is prime, this should be equivalent:
$$\gcd(x,y,z)\equiv 0\mod 7 \ \ \text{OR} \ \ \text{gcd}^2(x,y,z)x'y'z'\equiv 1\mod 7$$ The first congruence is satisfied when $x,y,z$ are multiples of $7$. In the set $[50]$ is $\left\lfloor \frac{50}{7} \right\rfloor=7$ multiples of $7$. So the first congruence is satisfied for ${7 \choose 3}=35$ such threes $x,y,z$. How to calculate how many numbers satisfy the second congruence is more problematic for me.
Numerical calculations
I wrote a program that counts these triples that satisfy the congruence. The program counted: $$|\mathcal{M}|=2079$$ I have tried to notice a pattern in the solutions. For example, I researched what $\gcd$ are possible for threes that belong to $\mathcal{M}$:
3. Generalization of the problem
I was thinking about making a generalization to any set $[n]$. Then let $|\Omega_n|={n \choose 3}$ and: $$\mathcal{M}_n=\left\{ \left\{ x,y,z\right\}\in\Omega_n: xyz\equiv \gcd(x,y,z)\mod 7 \right\} $$ And we are looking for: $$\mathscr{P}(\mathcal{M}_n)= \frac{\left| \mathcal{M}_n\right| }{\left| \Omega\right| } = \frac{\left| \mathcal{M}_n\right| }{ {n \choose 3} } $$ Using the earlier program, I drew the distribution $\mathscr{P}(\mathcal{M}_n)$ vs $n$:
It turns out that local minima occur in arithmetic intervals and occur for $n=5,12,19,26,33,40,47,...$ rather, it is not a coincidence that this sequence differs $7$. In contrast, local maxima occur for $n=4,11,18,25,32,39,46,...$ the question of the limit of the sequence $\mathscr{P}(\mathcal{M}_n)$ also seems interesting: $$ \lim_{n \to \infty } \mathscr{P}(\mathcal{M}_n)\approx\frac{539}{5000}$$ although I'm more interested in how to solve this problem analytically.
Best regards MK
Clearly if one of $x,y,z$ is divisible by 7 then they all must be, so we can reduce to the case where all of them are coprime to 7. Denoting $k = gcd(x,y,z)$ we can write $x = a \cdot k, y = b \cdot k, z = c \cdot k$, and it is enough to solve the equation $abc = \frac{1}{k^2}$ where $gcd(a,b,c)=1$.
Yes, but that's an equation: $$abc = \frac{1}{k^2}$$ is $\mod 7$? What should I substitute for $a,b,c$?
Can anyone help please? I have no other ideas for counting the solutions.
PS This is not my homework, I ask out of curiosity :)
The way I would think about it is to condition on gcd(x, y, z) = const and then count how many solutions are possible. Under this condition, a could be in the range 50/k, b in the same range and then to count c you might need to condition on gcd(a,b) being something again. (x =ak, y = bk, z = ck).
|
2025-03-21T14:48:32.019115
| 2020-09-12T20:08:23 |
371552
|
{
"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:632957",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371552"
}
|
Stack Exchange
|
Calculus variation question
Assume that we have to minimize the integral $I[y]=\int_0^1 L(x,y,y'(x))dx$ for smooth diffeomorphic mappings $y:[0,1]\to [0,1]$ with $y(0)=0$ and $y(1)=1$, where $L\in C^\infty(R\times R\times R)$, and $L$ satisfies convex condition wrt to $y'$ and coercivity condition. Is the minimiser of $I$ the only diffeomorphic solution $y:[0,1]\to[0,1]$ of Euler-Lagrange equation $\partial_t (\partial_{y'} L)=\partial_y L$? Some reference I need.
|
2025-03-21T14:48:32.019173
| 2020-09-12T21:40:57 |
371557
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Emil Jeřábek",
"Hanul Jeon",
"https://mathoverflow.net/users/12705",
"https://mathoverflow.net/users/48041"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632958",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371557"
}
|
Stack Exchange
|
Is the collection scheme provable in ZF-Regularity?
By the axiom schema of collection in ZF, I mean: $$\forall A \exists B \forall x \in A (\exists y \phi(x,y) \to \exists y \in B \phi(x,y))$$, for every formula $\phi$ that doesn't use the symbol $B$.
On the other hand the axiom schema of replacement states that if $\phi$ is a first-order formula, and $A$ is a set such that for any $x\in A$ there exists a unique $y$ such that $\phi(x,y)$, then there exists a set $B$ such that $y\in B$ if and only if $\phi(x,y)$ for some $x\in A$. That is, $B$ is the "image" of $A$ under the "definable class function" $\phi$. [Quoted from here]
The proof that I know of uses Regularity. We replace each element $x$ of $A$ by the minimal stage of the cumulative hierarchy that has an element $y$ that satisfy $\phi(x,y)$, then we take the union of the resulting set.
I'm personally not aware of another proof of collection in ZF. Since the above proof clearly depends on Regularity, hence my question:
Is collection scheme provable in ZF-Regularity?
In other words, is there a model of ZF-Regularity in which collection fails?
I don’t know who invented the following model: consider a model of $\mathrm{ZFC}^-$ of the form $\bigcup_{\alpha\in\mathrm{Ord}}V_\alpha(A)$ where $A$ is a countable infinite set of Quine atoms, and let $M$ be its submodel consisting of sets whose transitive closure includes only finitely many atoms from $A$. Then $M$ is a model of $\mathrm{ZFC}^-$ where collection fails (specifically, the model satisfies $\forall n\in\omega,\exists x,|\mathrm{tc}(x)\cap A|\ge n$, but not $\exists X,\forall n\in\omega,\exists x\in X\dots$).
Your question is already answered in Mathematics Stackexchange. See the following question and answer.
|
2025-03-21T14:48:32.019322
| 2020-09-12T22:40:23 |
371560
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Andy Putman",
"Joseph O'Rourke",
"https://mathoverflow.net/users/317",
"https://mathoverflow.net/users/6094"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632959",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371560"
}
|
Stack Exchange
|
Untangling two simple closed curves on a surface
Let $S$ be a smooth surface and $\gamma_1, \gamma_2$ be two transversal simple closed curves on it. Suppose moreover that there exists a simple closed curve $\gamma_1'$ on $S$ isotopic to $\gamma_1$ and such that $\#(\gamma_1\cap \gamma_2)>\#(\gamma_1'\cap \gamma_2)$.
Question. Is it true that there is a disk on $S\setminus (\gamma_1\cup\gamma_2)$ whose boundary is composed of one arc of $\gamma_1$ and one arc of $\gamma_2$?
Note that in case such a disk exists, one can construct an isotopy of $\gamma_1$ that would decrease the number of intersections of $\gamma_1$ with $\gamma_2$ by two.
Yes. This is discussed in Farb-Margalit’s “Primer on mapping class groups”, where it is called the bigon criterion.
This MSE post may help: Proof of the bigon criterion, especially Lee Moser's answer.
This is also proved as Lemma 3.1 in
Joel Hass and Peter Scott,
Intersections of curves on surfaces, Israel Journal of Mathematics 51 (1985), 90–120. https://doi.org/10.1007/BF02772960
|
2025-03-21T14:48:32.019418
| 2020-09-13T00:57:04 |
371564
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Andrew",
"Giorgio Metafune",
"Ma Joad",
"Quarto Bendir",
"https://mathoverflow.net/users/121404",
"https://mathoverflow.net/users/14551",
"https://mathoverflow.net/users/150653",
"https://mathoverflow.net/users/156492"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632960",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371564"
}
|
Stack Exchange
|
Perturbation in the equation $u_t=\epsilon Pu$, where $P$ is an elliptic partial differential operator
Let $Pu=\sum_{ij} \partial_j(a_{ij}(x) \partial_{i} u)$ be an elliptic operator. Consider the equation
$$
(u=u_\epsilon)\\
\partial_t u=\epsilon Pu \text{ in } \mathbb R^+ \times \Omega,\\
u(0,x)=u_0(x), u(t,x)=0\text{ for }x \in \partial\Omega
$$
where $\Omega$ is an open set in $\mathbb R^n$ (not necessarily bounded).
A weak solution to this equation exists and is unique in $L^2(0,T,H_0^1(\Omega))$.
Let $u_\epsilon$ be the solution to the equation corresponding to the given value of $\epsilon$. I wish to show that when $\epsilon \to 0$, $u_\epsilon$ converges in $L^2$ to a limit $u$.
By checking the formulation of a weak solution ($\chi$ is a test function, for example, in $C^1$):
$$
\int u_\epsilon \partial_t \chi=-\epsilon\int (\nabla u) A (\nabla\chi),
$$
we could show that $u_\epsilon$ is bounded in the $H^1$ norm for $\epsilon<1/2$, and hence that
$$
\int (\partial_tu_{\epsilon_1}-\partial_tu_{\epsilon_2}) \chi \to 0
$$
as $\epsilon_1,\epsilon_2 \to 0$, for every test function $\chi$. Hence $u_\epsilon$ converges to a distributional limit.
Question: is it ture that $u_\epsilon$ converges almost everywhere to a limit $u$? This is true in the case of $Pu=\Delta u$ by direct computation, after writing out the solution explicitly as an integral (heat equation). But in the general case, how could I prove the same result?
Wouldn't a limit $u=u_0$ and in the case $u_0$ has no derivatives there would be no convergence of $\partial_j u_\epsilon$?
@Andrew I have now fixed the problem. In fact, I am just expecting convergence in some sense.
As said by @Andrew, if $u$ is the solution for $\epsilon=1$, then $u_\epsilon (t)=u(\epsilon t)$.
Depending on the particular applications you're interested in you might want to look at Eberhard Hopf "The partial differential equation $u_t+uu_x=\mu u_{xx}$" Comm. Pure Appl. Math. 3 (1950), no. 3, 201-230, or S.N. Kruzkov "First order quasilinear equations in several independent variables," which are concerned with these kinds of questions
|
2025-03-21T14:48:32.019579
| 2020-09-13T05:12:26 |
371567
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Federico Poloni",
"https://mathoverflow.net/users/1898"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632961",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371567"
}
|
Stack Exchange
|
Determinant of a certain Toeplitz matrix
Compute the following determinant
\begin{vmatrix} x & 1 & 2 & 3 & \cdots & n-1 & n\\ 1 & x & 1 & 2 & \cdots & n-2 & n-1\\ 2 & 1 & x & 1 & \cdots & n-3 & n-2\\ 3 & 2 & 1 & x & \cdots & n-4 & n-3\\ \vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\ n-1 & n-2 & n-3 & n-4 & \cdots & x & 1\\ n & n-1 & n-2 & n-3 & \cdots & 1 &x \end{vmatrix}
I tried the following. I subtracted the second row from the first, the third from the second, the fourth from the third, and so on. I got:
\begin{vmatrix}
x-1 & 1-x & 1 & 1 & \cdots & 1 & 1\\
-1 & x-1 & 1-x & 1 & \cdots & 1 & 1\\
-1 & -1 & x-1 & 1-x & \cdots & 1 & 1\\
-1 & -1 & -1 & x-1 & \cdots & 1 & 1\\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
-1 & -1 & -1 & -1 & \cdots & x-1 & 1-x\\
n & n-1 & n-2 & n-3 & \cdots & 1 &x \end{vmatrix}
I did the same thing with the columns. I subtracted the second row from the first, the third from the second, the fourth from the third, and so on. And I got:
\begin{vmatrix}
2x-2 & -x & 0 & 0 & \cdots & 0 & 1\\
-x & 2x-2 & -x & 0 & \cdots & 0 & 1\\
0 & -x & 2x-2 & -x & \cdots & 0 & 1\\
0 & 0 & -x & 2x-2 & \cdots & 0 & 1\\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
0 & 0 & 0 & 0 & \cdots & 2x-2 & 1-x\\
1 & 1 & 1 & 1 & \cdots & 1-x &x \end{vmatrix}
I hope I didn’t make a mistake somewhere. With this part I don't know what to do next. I don't know if I'm doing it right. Thank you in advance !
One could use the matrix determinant lemma to reduce it to $xL-2I$, where $L$ is the finite difference matrix, possibly with other boundary conditions. Since the eigenvalues of $L$ are known in closed form, it is possible to piece together a solution, but this does not look like an easy computation.
Let $A_{n+1}$ denote the determinant of the $(n+1)\times (n+1)$ matrix in the question. By Laplace expansion over the first row, we get:
$$A_{n+1} = (2x-2)A_n + xB_n + (-1)^nC_n,$$
where
$$B_n := \det \begin{bmatrix}
-x & -x & 0 & \cdots & 0 & 1\\
0 & 2x-2 & -x & \cdots & 0 & 1\\
0 & -x & 2x-2 & \cdots & 0 & 1\\
\vdots & \vdots & \vdots & \ddots & \vdots & \vdots\\
0 & 0 & 0 & \cdots & 2x-2 & 1-x\\
1 & 1 & 1 & \cdots & 1-x &x \end{bmatrix}
$$
and
$$C_n := \det\begin{bmatrix}
-x & 2x-2 & -x & 0 & \cdots & 0 \\
0 & -x & 2x-2 & -x & \cdots & 0 \\
0 & 0 & -x & 2x-2 & \cdots & 0 \\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & 0 & \cdots & 2x-2 \\
1 & 1 & 1 & 1 & \cdots & 1-x \end{bmatrix}
$$
Expanding both $B_n$ and $C_n$ over the first column, we get
$$B_n = (-x) A_{n-1} + (-1)^{n+1}C_{n-1}$$
and
$$C_n = (-x)C_{n-1} + (-1)^{n+1} D_{n-1},$$
where
$D_n$ is the determinant of the following $n\times n$ Toeplitz tridiagonal matrix:
\begin{bmatrix}
2x-2 & -x & 0 & 0 & \cdots & 0 \\
-x & 2x-2 & -x & 0 & \cdots & 0\\
0 & -x & 2x-2 & -x & \cdots & 0\\
0 & 0 & -x & 2x-2 & \cdots & 0 \\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots \\
0 & 0 & 0 & 0 & \cdots & 2x-2
\end{bmatrix}
It is expressed in terms of Lucas sequence and its generating function as follows:
\begin{split}
D_n &= U_{n+1}(2x-2,x^2) \\
&= \frac{(x-1+\sqrt{1-2x})^{n+1}-(x-1-\sqrt{1-2x})^{n+1}}{2\sqrt{1-2x}}\\
&=[z^n]\ \frac{1}{1-(2x-2)z+x^2z^2}.
\end{split}
Plugging the recurrence formula for $B_n$ into that for $A_{n+1}$, we get
$$(\star)\qquad A_{n+1} = (2x-2)A_n - x^2 A_{n-1} + (-1)^{n+1}xC_{n-1} + (-1)^n C_n.$$
We now have all ingredients to derive a closed-form expression for $A_n$.
Taking into account $C_1=1-x$, we unroll the recurrence for $C_n$ to get:
\begin{split}
C_n &= (-x)^{n-1}(1-x)+(-1)^{n+1} \sum_{i=1}^{n-1} x^{n-1-i} D_i \\
&= (-x)^n+(-1)^{n+1} [z^{n-1}]\ \frac{1}{(1-xz)(1-(2x-2)z+x^2z^2)}.
\end{split}
We will need the generating function:
\begin{split}
\mathcal{C}(z) &:= \sum_{n=1}^{\infty} (-1)^{n-1} C_n z^{n-1} \\
&= \frac{1}{(1-xz)(1-(2x-2)z+x^2z^2)} - \frac{x}{1-xz} \\
&= \frac{1-x+(2x-2)xz-x^3z^2}{(1-xz)(1-(2x-2)z+x^2z^2)}.
\end{split}
Define the generating function for $A_n$:
$$\mathcal{A}(z) := \sum_{n=1}^{\infty} A_n z^n.$$
Now, let's multiply the recurrence $(\star)$ by $z^{n-1}$ and sum over $n\geq 2$ to obtain:
$$\frac{\mathcal{A}(z) - A_1z - A_2z^2}{z^2} = (2x-2)\frac{\mathcal{A}(z) - A_1z}z - x^2\mathcal{A}(z) - xz\mathcal{C}(z) - (\mathcal{C}(z)-C_1),$$
implying that
$$\mathcal{A}(z) = \frac{z(x^5z^4-4x^4z^3+4x^3z^3+6x^3z^2-8x^2z^2-4x^2z+3xz^2+4xz+x-z)}{(1-xz)(1-(2x-2)z+x^2z^2)^2}$$
and thus
\begin{split}
A_n &= \frac12 D_n + \frac{n+1-x}2 D_{n-1} + \frac{x^n}2 \\
&=
\frac{(n+\sqrt{1-2x})(x-1+\sqrt{1-2x})^n-(n-\sqrt{1-2x})(x-1-\sqrt{1-2x})^n}{4\sqrt{1-2x}} + \frac{x^n}2.
\end{split}
UPDATE. Another way to proceed from the recurrences for $A_n$, $B_n$, $C_n$, and $D_n$ is to combine them into a single matrix recurrence:
$$
\begin{bmatrix} A_{n+1}\\ B_{n+1}\\ (-1)^{n+1}C_{n+1}\\ D_{n+1} \\ D_n\end{bmatrix}
= \begin{bmatrix}
2x-2 & x & 1 & 0 & 0\\
-x & 0 & 1 & 0 & 0\\
0 & 0 & x & -1 & 0\\
0 & 0 & 0 & 2x-2 & -x^2\\
0 & 0 & 0 & 1 & 0
\end{bmatrix}
\cdot
\begin{bmatrix} A_{n}\\ B_{n}\\ (-1)^{n}C_{n}\\ D_{n} \\ D_{n-1}\end{bmatrix}.
$$
From the characteristic polynomial of the matrix in r.h.s., we get a recurrence for $A_n$:
$$A_n = (5x - 4)A_{n-1} + (-10x^2 + 12x - 4)A_{n-2} + (10x^3 - 12x^2 + 4x)A_{n-3} + (-5x^4 + 4x^3)A_{n-4} + x^5A_{n-5},$$
from which it is easy to obtain an explicit formula.
PS. Up to the change of $x$ to $-x$, $A_n$ represents the characteristic polynomial of the distance matrix, also known as the distance polynomial, of the path graph $P_n$. Another formula (in terms of Chebyshev polynomials) for it is given in OEIS A203993.
|
2025-03-21T14:48:32.019869
| 2020-09-13T05:35:24 |
371569
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Alex Mine",
"Chris Gerig",
"Filippo Alberto Edoardo",
"Julian Rosen",
"Maxime Ramzi",
"Qiaochu Yuan",
"Steven Landsburg",
"Timothy Chow",
"Todd Trimble",
"bof",
"https://mathoverflow.net/users/102343",
"https://mathoverflow.net/users/10503",
"https://mathoverflow.net/users/12310",
"https://mathoverflow.net/users/164546",
"https://mathoverflow.net/users/18238",
"https://mathoverflow.net/users/290",
"https://mathoverflow.net/users/2926",
"https://mathoverflow.net/users/3106",
"https://mathoverflow.net/users/43266",
"https://mathoverflow.net/users/44191",
"https://mathoverflow.net/users/5263",
"user44191"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632962",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371569"
}
|
Stack Exchange
|
What, precisely, do we mean when we say that a f.d. vector space is canonically isomorphic to its double dual?
I've been reading the Xena Project blog, which has been loads of fun. In the linked post Kevin gives the natural isomorphism $V \to V^{\ast \ast}$ from a f.d. vector space to its dual as an example of a "canonical isomorphism":
People say that the obvious map from a vector space to its double dual is “canonical” but they could instead say that it is a natural transformation.
I think this is not a complete description. When we say that $V$ is canonically isomorphic to its double dual we don't just mean that there exists some natural isomorphism from the identity functor to the double dual functor - what we mean, and what we use in practice, is that a particularly canonical natural transformation $V \to V^{\ast \ast}$ is an isomorphism. We can multiply any such natural isomorphism by a scalar $c \neq 0, 1$ in the ground field $k$ and obtain a different natural isomorphism between these two functors, and these are not canonical!
But reading the linked post showed me I don't know precisely what I mean by "canonical" in the paragraph above. So:
Question: What, precisely, do we mean by "canonical" when we say that the usual natural transformation $V \to V^{\ast \ast}$ is the "canonical" one we want? What other "canonical" maps are canonical in the same way?
What follows are some off-the-cuff thoughts about this.
One thought is that when we write this map down we're "using as little as possible"; we're not even really using that we're working in vector spaces. If we define the dual to be the internal hom $[V, k]$ then ultimately all we're using is the evaluation map $V \otimes [V, k] \to k$ together with currying. In other words, one way to make precise what we're using is the closed monoidal structure on $\text{Vect}$. (We don't even need a closed structure if we define the dual to be the monoidal dual but I expect the internal hom to be more familiar to more mathematicians.)
So, here's what I think ought to be true: in the free closed monoidal category on an object $V$ (blithely assuming that such a thing exists - maybe I should have used monoidal duals after all, because I'm much more confident that the free monoidal category on a dualizable object exists), there ought to be a literally unique morphism $V \to [[V, 1], 1]$ where $1$ denotes the unit object. This is the "walking double dual map" and, by the universal property, reproduces all the other ones, and I think this is a candidate for what we might mean by "canonical map."
This trick of taking free categories is a really good trick actually, since it also cleanly describes the nonexistence of canonical maps. For example there is no canonical map $V \to V \otimes V$, and you could use representation theory to show that the only $GL(V)$-equivariant map is zero, but actually it's just true that in the free monoidal category on $V$ there are no maps $V \to V \otimes V$ whatsoever. Similarly there is no canonical map $V \to V^{\ast}$, and similarly you could use representation theory to show that the only $GL(V)$-equivariant map is zero, but actually it's just true that in the free closed monoidal category on $V$ (if it exists; take the free monoidal category with duals if not) there are no such maps whatsoever.
But I don't think this sort of reasoning is enough to capture other examples where two functors are naturally isomorphic and there's a particularly canonical natural isomorphism that we want in practice. For example, it's also true that de Rham cohomology $H^{\bullet}_{dR}(X, \mathbb{R})$ on smooth manifolds is canonically isomorphic to singular cohomology $H^{\bullet}(X, \mathbb{R})$ with real coefficients, and by this we don't mean just that there exists some natural isomorphism but that the particularly canonical natural transformation given by integrating differential forms over simplices is an isomorphism. We can multiply any such natural isomorphism by $c^n$ in degree $n$ where $c \in \mathbb{R} \setminus \{ 0, 1 \}$ and we'll get a different one, even one that respects cup products, and again these are not canonical (although I wonder if we could easily distinguish the usual one from the one obtained by setting $c = -1$ - do we have to choose how simplices are oriented or something like that somewhere?) But now I really don't know what I mean by that! It doesn't seem like I can pull the same trick of zooming out to a more general categorical picture. de Rham cohomology is a pretty specific functor defined in a pretty specific way. Maybe this one is a genuinely different sense of "canonical," closer to "preferred," I don't know.
Some previous discussion of "canonical" stuff on MO:
What is the definition of canonical? (2013)
Are there examples of families of objects which are canonically isomorphic, but where diagrams of canonical isomorphisms don't commute? (2014)
The meaning and purpose of "canonical" (2019)
Here's a fun sort-of-related question I was thinking about: does the point, as a $0$-dimensional manifold, have a canonical orientation?
I think this question, though closed, is probably also relevant; https://mathoverflow.net/questions/345136/is-a-vector-space-naturally-isomorphic-to-its-dual#comment863420_345136
-1 f.t.a.'f.d.'
A point does not have a canonical orientation, nothing stops the global choice of $+1$ or $-1$, and I think this hits to the heart of your question. There are "preferred choices" of what is to be the "canonical" object and it depends on context (I see this is also stated in one of your MO links). If I want to know the total number of points in my hand, the canonical sign is $+1$ based on the context "minus means 'taking away'".
Perhaps folks can clarify the definition of orientation they have in mind. The tangent space to a point is 0-dimensional, and a 0-dimensional vector space has a unique ordered basis, so in this sense there is a canonical orientation.
@Julian: that definition would give a unique orientation, which is not right. You can define an orientation to be a choice of generator for $H^n(-, \mathbb{Z})$ and then something funny happens when $n=0$ - this group is arguably canonically isomorphic to $\mathbb{Z}$ so has a distinguished generator, namely $1$ (distinguished abstractly by being the multiplicative unit).
Possibly also related: https://mathoverflow.net/questions/244131/nuances-regarding-naturality
To me, the essence of the "canonicity" that we care about is the fact that the construction is "defined the same way for every object", and this is something that the concept of a natural transformation nicely packages if you look at it the right way.
@Alex: I think the standard analysis of the double dual isomorphism $V \to V^{\ast \ast}$ does not capture the sense in which it is "defined the same way for every object." When we talk about it as a natural transformation we don't capture e.g. that we are using the same map over every field, in a way which is compatible with extension of scalars. On the other hand the free category argument I used implies this. Note that this is a strictly stronger statement with concrete mathematical consequences: e.g. it implies that the double dual map is compatible with twisting by automorphisms of $k$.
A card-carrying structuralist confessed that she secretly believed in sets. The priest replied, "Your sins are forgiven. To be honest, I secretly believe in urelements." Okay, dumb joke, but I'm having trouble seeing why this isn't a purely philosophical question. Does any substantive mathematical question hinge on it?
@Timothy: most concretely I am hoping that there is some nice category-theoretic thing to say here that is strictly stronger than just that there is some natural isomorphism of functors. For example, as I mentioned in the previous comment, the double dual map $V \to V^{\ast \ast}$ is compatible with extension of scalars, and this has concrete mathematical substance (e.g. it would be relevant to understanding the interaction of taking the double dual with Galois descent). There is at least one more kind of functoriality to incorporate, in the underlying field $k$ itself.
@QiaochuYuan : I guess what I mean is, if when trying to solve a problem, you have a good mathematical reason to distinguish between different "natural isomorphisms," then the context should suggest what categories to consider. But it sounds like you're trying to pre-emptively pin down the word "canonical" before a practical need arises. That sounds to me like premature theorizing. Or maybe I'm still not understanding the motivation behind your question?
@Timothy: the linked Xena project blog post gives some context. Kevin gives the example that Grothendieck freely calls two different localizations $R[1/f], R[1/g]$ "equal" if they are canonically isomorphic, e.g. if $g^2 = f^3$, and describes some trouble he ran into trying to formalize arguments about localizations in Lean involving showing that a bunch of diagrams commuted. That example has me curious about the whole "canonical isomorphism" thing as a whole, although I agree that this is perhaps zooming out too much.
There is this beautiful text https://perso.univ-rennes1.fr/matthieu.romagny/notes/canonicite.pdf by Mathieu Romagny which shows that there is no functorial isomorphism between a space and its double dual, which might be relevant.
Regarding the second paragraph after the fold, it's definitely true that the free closed monoidal category on a category exists.
@FilippoAlbertoEdoardo It looks to me that Romagny is talking about something else, about single duals.
A silly remark : there are canonical maps that aren't canonical isomorphisms; and whether or not they are isomorphisms depends on the base (for instance $2: M\to M$ is certainly defined canonically). Now if you're working over a field of char. $\neq 2$, would you say "the canonical isomorphism $2: M\to M$", or "the canonical map $2: M\to M$, which is an isomorphism" ?
|
2025-03-21T14:48:32.020593
| 2020-09-13T06:22:47 |
371570
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"GGT",
"Vivek Shende",
"https://mathoverflow.net/users/45170",
"https://mathoverflow.net/users/4707"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632963",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371570"
}
|
Stack Exchange
|
Connected relative Gromov Witten invariants
I am currently interested to compute relative Gromov Witten invariants(GW) over $\mathbb{P}^1$.
In the paper
https://arxiv.org/pdf/math/0204305.pdf
eq 3.1 gives the count of relative disconnected GW invariants let $\mu,\nu$ are partition of $d$ then gives the GW invariants $\mu$ is the ramification profile over say $0$ and $\nu$ over $\infty$.
$$\langle \mu,\prod_{i}^{n}\tau_{k_i}(w),\nu\rangle^{\bullet} $$
I could use the formula to count the disconnected GW over $\mathbb{P}^1$. Is it true that if partition $\mu$ if it's one part that is $\mu=[d]$ then the connected GW invariants and disconnected are the same?
We can group these invariants as genus $g$ where $g$ is a positive integer. We say a GW invariant is of genus $g$ if
$$\sum_i k_i = 2g-2+\ell{\mu}+\ell{\nu}$$
where $\ell{}$ denotes the length of the partition.
My calculation doesn't agree for the case calculating genus 1 GW invariants for the case say
Also I got different answer for
$$<1|\tau_{2}|1>^{\bullet} = (61009/66355200)$$ and
$$<1|\tau_{2}|1>^{\circ} = 1/24$$ and
My main aim, for now, is to compute connected relative GW invariants for small genus and few partitions. Is there is any reference where it's calculated or table?
Also, Theorem 3 in the paper gives a possible calculation but I am not getting what exactly the formula means some step of the calculation would help.
https://en.m.wikipedia.org/wiki/Exponential_formula
Thanks, I have the disconnected numbers so taking log might help let me count.
I guess I need some multivariable version. I guess Theorem 3 in the paper gives the answer but I don't understand it to compute.
|
2025-03-21T14:48:32.020731
| 2020-09-13T06:24:11 |
371571
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Andreas Blass",
"Asaf Karagila",
"D.S. Lipham",
"Gabe Goldberg",
"Martin Väth",
"bof",
"https://mathoverflow.net/users/102684",
"https://mathoverflow.net/users/165275",
"https://mathoverflow.net/users/43266",
"https://mathoverflow.net/users/6794",
"https://mathoverflow.net/users/7206",
"https://mathoverflow.net/users/95718"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632964",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371571"
}
|
Stack Exchange
|
Axiom of Countable Choice and meager sets
Let us recall that the Axiom of Countable Choice (denoted by ACC) says that the countable product $\prod_{n\in\omega}X_n$ of nonempty sets $X_n$ is nonempty.
It is easy to see that ACC implies that for any sequence of meager sets $(X_n)_{n\in\omega}$ in a Polish space $X$ the union $\bigcup_{n\in\omega}X_n$ is meager in $X$. Let us denote the latter statement by (UMM), abbreviated from "union of meager is meager".
So, (ACC)$\Rightarrow$(UMM).
On the other hand, it is consistent with (ZF) that the real line can be equal to the union of a countable family of countable sets, in which case (UMM) does not hold. This means that (UMM) cannot be proved in (ZF) alone.
Problem 1. What is the place of (UMM) among other weaker versions of AC?
Denote by (UCC) the statement: the union of a countable family of countable sets is countable.
Problem 2. Does (UMM) imply (UCC)?
Good questions.
And what about (UCM), the union of a countable family of countable sets is meager? Does (UCM) imply (UCC)?
@bof: I can't say if the answer is positive or not, but at least in Truss' model, the countable union of countable sets of reals is countable, but $\omega_1$ is singular.
Since (UMM) is only about Polish spaces, it should be preserved from the ground model to symmetric extensions that introduce no new sets of low rank. So I'd be inclined to try violating (UCC) among sets of very high rank and thus answer Problem 2 negatively.
@Andreas: That's a good point. If $\Bbb R$ can be well-ordered, but some strange sets exist above it, it would be a counterexample. I suppose that the right question, then, is restricting these choice principles to sets of reals.
What is your definition of meager in (ZF)? Being a countable union of nowhere dense sets? In that case, the given argument about the failure of (UMM) in (ZF) is not clear: It is not obvious (to me) that the real line cannot be be meager in (ZF). Recall that the general category theorem of Baire is in (ZF) equivalent to countable dependent choices (DC); perhaps a proof that the real line is not meager already requires more than ZF?
The Baire category theorem for separable spaces is provable in ZF since the dependent choices one makes can be restricted to a countable dense subset, which is wellordered at the outset so that one can choose canonically.
@GabeGoldberg But wouldn't that prove UMM in ZF?
@D.S. Lipham. Quite the opposite. From GabeGoldberg's (correct) observation, it follows even in ZF that the real line is not meager. Hence the argument in the original question shows indeed that UMM is unprovable in ZF.
Martin, ZF proves that $\Bbb R$ is not meager, it does not prove that it is a countable union of countable sets. Therefore it is consistent that the countable union of meager sets is not meager.
Asaf, yes that's what I meant exactly: It is consistent (with ZF) that UMM is unprovable. That's why I wrote "quite the opposite" (to the assertion that UMM is provable in ZF).
|
2025-03-21T14:48:32.020963
| 2020-09-13T08:08:45 |
371576
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Alec Rhea",
"Peter LeFanu Lumsdaine",
"daniel gratzer",
"https://mathoverflow.net/users/2273",
"https://mathoverflow.net/users/76636",
"https://mathoverflow.net/users/92164"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632965",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371576"
}
|
Stack Exchange
|
Filled natural transformations
According to the nlab, a commutative square
can be viewed as a 'lifting problem' between $f$ and $g$, and a solution to this lifting problem is a morphism $\ell:B\to C$ filling the above square, so
commutes. Taking this view, a natural transformation $\alpha:F\Rightarrow G$ is a collection of lifting problems between the images of $F$ and $G$ on Hom-sets.
Is the notion of a 'solved' or 'filled' natural transformation, such that for each arrow $f:X\to Y$ in the domain category we have a filled square, useful or studied anywhere?
By a filled natural transformation square for $f$ I mean a commutative square as below
This seems superficially related to the notion of a pseudonatural transformation, where $\ell_f$ would be replaced by an isomorphic $2$-cell satisfying coherence diagrams, but I'm not sure there's any actual connection. I would also be interested in the notion of an 'orthogonal transformation', where each $\ell_f$ is unique so $F(f)$ and $G(f)$ are orthogonal for all $f$.
Just a note: Lifting problems are almost always drawn with the diagonal filler going upwards-rightwards, not downwards-leftwards as you have here. Of course it’s formally the same, but the convention is very helpful for readability and distinguishing the roles of the morphisms involved, so I’d recommend following it except when there’s a clear reason not to.
@PeterLeFanuLumsdaine Will do, I’m walking my dog at the moment but I’ll edit when I get home.
This seems highly related to the notion of "coherent" or "natural" lifts that we use in the study of categorical models of type theory. I'm on mobile, but perhaps see "Natural models of homotopy type theory" by Steve Awodey and look at the discussion around the elimination rule for the intensional identity type.
@DanielGratzer I’ll check it out, thanks for the reference.
|
2025-03-21T14:48:32.021120
| 2020-09-13T09:38:31 |
371578
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"1ENİGMA1",
"YCor",
"https://mathoverflow.net/users/14094",
"https://mathoverflow.net/users/82036"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632966",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371578"
}
|
Stack Exchange
|
Characterizing centralizer of nilpotent self-maps
Let $\mathcal{C}_n$ be the monoid of self-maps $\alpha$ of $\{1\dots,n\}$ that are order-preserving ($\forall x,y$, $x\le y$ $\Rightarrow$ $\alpha(x)\le\alpha(y)$ and decreasing ($\forall x$, $\alpha(x)\le x$).
Let N($\mathcal{C_n})$ be the set of nilpotent elements of $\mathcal{C}_n$, i.e., $\alpha\in$ N($\mathcal{C_n})$ if and only if there exists positive integer $m$ such that $\alpha^m=0$, where $0$ is the constant map to $1$.
I wonder whether we can characterize the centralizer of any element (different zero) in N($\mathcal{C_n})$. Let $\alpha\in$ N($\mathcal{C_n})$ with $\operatorname{image}(\alpha)=\{1=a_0,a_1,\dots,a_r\}$. Then
$C(\alpha)=\{\beta:\alpha\beta=\beta\alpha\}$.
Decreasing usually means something else. You might call this "regressive" (suggested here at MathSE)
Centralizer in $\mathcal{C}_n$ or in $\mathsf{N}(\mathcal{C}_n)$?
@YCor I try to find in $\mathsf{N}(\mathcal{C}_n)$. Actually, to find the centralizer in $\mathcal{C}_n$ is so interesting.
|
2025-03-21T14:48:32.021216
| 2020-09-13T10:14:51 |
371580
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Lagrida Yassine",
"Sylvain JULIEN",
"https://mathoverflow.net/users/13625",
"https://mathoverflow.net/users/164630"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632967",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371580"
}
|
Stack Exchange
|
Symmetry in Hardy-Littlewood k-tuple conjecture
Assuming Hardy-Littlewood $k$-tuple conjecture, do the "dual" prime constellations $(0,h_1, h_2,\cdots, h_i,\cdots, h_{k-1}=d)$ and $(0, h_{k-1}-h_{k-2}, h_{k-1}-h_{k-3},\cdots,h'_i=h_{k-1}-h_{k-i},\cdots,h_{k-1})$ corresponding to reversed sequences of prime gaps have the same distribution?
If yes, does it imply that the function $f(n):=\dfrac{\log g_n}{\log\log p_n}$ and the function $f'(n)$ obtained through the substitution $g_n\mapsto g'_n:=\dfrac{\log^{2} p_n}{g_n}$ reach the same values an asymptotically equal number of times? Is it related to the functional equation of zeta with which it would then share the same type of symmetry?
(Edited after Lagrida's answer and accordingly)
Yes there is a symmetry, i will add an heuristic argument for that !
Let $k \in \mathbb{N}, k \geqslant 2$.
Let $q \in \mathbb{P}, \ q \geqslant 5 $ and :
$$N_q := \displaystyle{\small \prod_{\substack{p \leqslant q \\ \text{p prime}}} {\normalsize p}}$$
Let : $1 \leqslant b \leqslant N_q$.
We have :
$$\gcd(b, N_q) = 1 \iff \gcd(N_q-b, N_q)=1 \tag{1}$$
Then the numbers coprime to $N_q$ and less than $N_q$ are symetric to $\dfrac{1}{2}N_q$.
Consider the k-tuple : $\mathcal{H}_k := (0,h_1,h_2,\cdots,h_{k-1})$, with $0 < h_1 < \cdots < h_{k-1}$.
Using $(1)$, if $(b,b+h_1,b+h_2,\cdots,b+h_{k-1})$ is coprime to $N_q$ then we have also $(N_q-b-h_{k-1}, N_q-b-h_{k-2}, \cdots,N_q-b-h_2, N_q-b-h_1, N_q-b)$ is coprime to $N_q$, (name that property 1).
Consider the k-tuple : $\mathcal{H}^{'}_k := (0,(h_{k-1}-h_{k-2}),(h_{k-1}-h_{k-3}),\cdots,(h_{k-1}-0))$
Using property1, you can see that :
$$b+\mathcal{H}_k \text{ is coprime to } N_q \iff N_q-b-h_{k-1}+\mathcal{H}^{'}_k \text{ is coprime to } N_q $$
Example : Let $\mathcal{H}_3=(0,2,6)$ and $q=7$, for $b=11$ we have $11+(0,2,6)=(11, 13, 17)$ is coprime to $N_7=210$.
We have $N_7-b-h_{k-1}=210-11-6=193$ and $\mathcal{H}^{'}_3 = (0, 4, 6)$.
Then we have $193+(0, 4, 6) = (193, 197, 199)$ it coprime too to $N_7$.
Using Chineese Romander theorem we can prove that :
$$\#\{(b,b+h_1,b+h_2,\cdots,b+h_{k-1})\in\mathbb{N}^{k} \, | \, 1 \leqslant b \leqslant N_q \ , \gcd(b, N_q)=\gcd(b+h_i, N_q)=1\} = \displaystyle{\small \prod_{\substack{p \leqslant q \\ \text{p prime}}} {\normalsize (p-w(\mathcal{H}_k, p))}}$$
Where $w(\mathcal{H}_k, p)$ is the number of distinct residues $\pmod p$ in $\mathcal{H}_k$.
Let $x \in \mathbb{R}$.
Let $q(x)$ be the largest prime number verifiying $x \geqslant \displaystyle \Big({\small \prod_{\substack{p \leqslant q(x) \\ \text{p prime}}} {\normalsize p}}\Big)$.
Using prime number theorem we have $q(x) \sim \log(x)$.
Consider :
$$I_{\mathcal{H}_k}(x) := \#\{(b,b+h_1,b+h_2,\cdots,b+h_{k-1})\in\mathbb{N}^k \, | \, b \leq x, \ \gcd(b, N_{q(x)}) = \gcd(b+h_i, N_{q(x)})=1 \}$$
And :
$$\pi_{\mathcal{H}_k}(x) := \#\{(p,p+h_1,p+h_2,\cdots,p+h_{k-1})\in\mathbb{P}^k \, | \, p \leq x\}$$
We can prove as $x \to +\infty$ that:
$$I_{\mathcal{H}_k}(x) \sim \mathfrak{S}(\mathcal{H}_k) \, e^{-\gamma k} \, \dfrac{x}{\log(\log(x))^k}$$
With $\mathfrak{S}(\mathcal{H}_k) := \displaystyle\prod_{\text{p prime}}\frac{1-\frac{w(\mathcal{H}_k, p)}{p}}{(1-\frac1p)^{k}}$.
If $p \in \mathbb{P}, \ p > q(x)=(1+o(1)) \log(x)$ then $p$ is coprime to $N_{q(x)}$, this is the relation trivial between prime numbers less than $x$ and numbers coprime to $2,3,\cdots,q(x)$ and less than $x$. I give a non-trivial relation as :
$$I_{\mathcal{H}_k}(x) \sim \pi_{\mathcal{H}_k}(x) \big( \pi(q(x)) e^{-\gamma} \big)^k$$
If we prove this conjecture then we have :
$$\pi_{\mathcal{H}_k}(x) \sim \mathfrak{S}(\mathcal{H}_k) \dfrac{x}{\log(x)^k}.$$
We can find the same results with Goldbach's conjecture or primes of the form $n^2+1$, you can see my article : here
I accepted your answer to my first question but I'm of course still interested in answers to the remaining ones.
|
2025-03-21T14:48:32.021444
| 2020-09-13T12:11:20 |
371582
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Wilberd van der Kallen",
"https://mathoverflow.net/users/165273",
"https://mathoverflow.net/users/4794",
"nikhilesh dasgupta"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632968",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371582"
}
|
Stack Exchange
|
A question on $SK_1$ of rings
Let $B$ be a commutative ring with unity and $B/nil(B):=B_{red}$, where $nil(B)$ is the nilradical of $B$. Is $SK_1(B)=SK_1(B_{red}) ?$ In particular, is it true when $B$ is an affine algebra over an algebraically closed field ?
Yes. An element in the kernel of $SK_1(B)\rightarrow SK_1(B_{red})$ is represented by a matrix $M\in GL_n(B)$ for some $n$. Write $\overline{M}$ for the reduction of $M$ mod $nil(B)$. Then $\overline{M}$ is a product of elementary matrices, all of which lift to elementary matrices over $B$. Adjusting $M$ accordingly, we can assume that $\overline{M}$ is the identity.
It follows that the elements on the diagonal of $M$ are all $1$ mod $nil(B)$, hence all units in $B$. This allows us to use elementary operations to convert $M$ to a diagonal matrix, which therefore (by Whitehead's lemma) represents the zero element of $SK_1(B)$.
(The same argument works if $B_{red}$ is replaced by $B/I$, where $I$ is any ideal contained in the Jacobson radical.)
Thank you sir for the response. It is clear to me that the map $SK_1(B) \rightarrow SK_1(B_{red})$ is injective. But why that map is {\bf surjective } ? I understand the preimage of a determinant 1 matrix is invertible. But why the preimage has determinant 1 is not what is clear to me.
You can multiply with a diagonal matrix that is congruent to the identity and get the determinant equal to one.
Thank you sir very much for the clarification.
|
2025-03-21T14:48:32.021578
| 2020-09-13T12:36:32 |
371583
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Mathmo",
"Robert Israel",
"https://mathoverflow.net/users/108697",
"https://mathoverflow.net/users/13650"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632969",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371583"
}
|
Stack Exchange
|
Does a spectral theorem exist for linear operator pencils?
I was wondering if a version of the spectral theorem (the projection valued measure case) holds for linear pencils of the form
$$
A-\lambda B
$$
where $A,B$ are self-adjoint on some Hilbert space $\mathcal{H}$ (and possibly unbounded) but $B$ is strictly positive? If so, is there a good reference on this? Note that under "nice" conditions, $B^{-1}A$ is self-adjoint on a weighted inner product (put factors $B^{1/2}$ in the inner product of $\mathcal{H}$ - this is straightforward to prove rigorously), so it may be the case that $B^{-1}A$ is usually studied instead. I was also wondering if there are natural conditions for $B^{-1}A$ to be essentially self-adjoint on $\mathcal{D}(B^{1/2})$ (with the weighted inner product)? Certainly $B$ or $A$ bounded suffices but what's the most general condition?
Perhaps you mean $B^{-1/2} A B^{-1/2}$, which is not the same as $B^{-1} A$ when they don't commute, but has better properties.
Ah, no I did mean $B^{-1}A$. However, I have since convinced myself that I missed a subtle point - it looks like this is symmetric w.r.t. the new Hilbert space, the domain of $B^{-1/2}$ but not necessarily self-adjoint. I've amended the question accordingly.
|
2025-03-21T14:48:32.021699
| 2020-09-13T13:51:53 |
371588
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Jeremy Rickard",
"Pedro",
"Yemon Choi",
"https://mathoverflow.net/users/21326",
"https://mathoverflow.net/users/22989",
"https://mathoverflow.net/users/763"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632970",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371588"
}
|
Stack Exchange
|
Reference for an analogue of Goursat's lemma for noncommutative rings?
The following question might be a duplicate but I couldn't find anything when I searched just now. (I also would not object hugely if the question gets moved to MSE except that I do not have an account there and so would not be able to respond to comments or accept answers.)
It seems to be folklore that Goursat's lemma from group theory works for other kinds of algebraic structure. An answer to this older MO question gives precise conditions on a category that are sufficient and close-to-necessary for the natural "abstract Goursat lemma" to be valid.
In particular, it seems that this should work for rings. That is, suppose $T$ is a subdirect product of $R\times S$ where $R$ and $S$ are not-necessarily-commutative rings, and $\pi_R:R\times S \to R$, $\pi_S:R\times S \to S$ are the coordinate projections. Then we have ideals $R_0 = \pi_R( T\cap \ker\pi_S) \unlhd R$ and $S_0=\pi_S(T\cap \ker\pi_R)\unlhd S$ and if I am not mistaken the calculation shown in the MO question I linked to shows that $T$ "descends" to the graph of a ring isomorphism $R/R_0 \to S/S_0$.
In some work with my PhD student, we found ourselves needing a special case of this result which we proved by hand. Only later did it occur to me that what we had done should be a special case of a more general known result, and then this led me to look up Goursat's lemma.
My question is this: can someone supply a reference to the literature, such as one of the standard "books people in North American systems read to prefer for qualifier/comprehensive exams", which states Goursat's lemma for rings, and does not deduce it from some general result about e.g. exact Mal'cev categories?
I would prefer that my PhD student does not needlessly reinvent the wheel if there is a clear statement that can be quoted, but I also don't want to force them to suddenly start learning general category-theoretic language when the application we have in mind is much more concrete and specialized.
I don't have time right now to compare their theorems with your post, but does this paper of Anderson and Camillo help?
What kind of "more concrete and specialized" mathematics is your student doing? I am just surprised they would be so averse to category theoretic language.
@PedroTamaroff I don't really want to get into a discussion about my student's PhD, since it's not really relevant, but (speaking as someone whose own PhD happily employs category-theoretic language and POV, but who works in functional analysis), I am surprised at your surprise. E.g. it really isn't necessary to be thinking of comma categories when doing analysis on RKHS (this isn't the thesis topic, but just for sake of argument)
@JeremyRickard Thanks! based on a quick look, it does seem that Theorem 11 part 1(b) would do the job, so if all else fails I might pass that on to my student as a reference that could be used, after encouraging them to work through the proof themselves.
@YemonChoi I see. I won't be surprised at your surprise of my surprise, since that may lead us into an odd path. My point is that the requisite language orbits around kernels, subobjects and similar things, which your student may not be terribly unfamiliar with. Best of luck!
|
2025-03-21T14:48:32.022065
| 2020-09-13T14:00:56 |
371590
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Asaf Shachar",
"Iosif Pinelis",
"https://mathoverflow.net/users/36721",
"https://mathoverflow.net/users/46290"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632971",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371590"
}
|
Stack Exchange
|
How to show continuity and monotonicity of solutions to this parametrized equation?
Let $1 \le p <2$ be a parameter. Consider the equation
$$
\frac{2^{p/2} (1-\sqrt{s})^p-1}{\sqrt{s}}=-2^{p/2-1}p(1-\sqrt{s})^{p-1}. \tag{1}
$$
I am rather certain that for each $1 \le p <2$, there is unique solution $s=s(p)$ in $(\frac{1}{4},1]$.
Question: Is $ p\to s(p)$ monotonically decreasing in $p$? Is it continuous in $p$?
How can I prove this rigorously?
Mathematica doesn't give a closed-form formula for $s(p)$.
Motivation:
This question comes from trying to find a "point of contact" when a certain chord between $(0,H(0)), (s,H(s))$
coincides with the tangent to $H$ at $s$, where $H:=F^q$ and $$
F(s) :=
\begin{cases}
2(\sqrt{s}-1)^2, & \text{ if }\, s \ge \frac{1}{4} \\
1-2s, & \text{ if }\, s \le \frac{1}{4}
\end{cases}
$$
The equation $\frac{H(s)-H(0)}{s-0}=H'(s)$ is nothing but equation $(1)$ above.
One idea is to assume that $s(p)$ is differentiable, and differentiate equation $(1)$ w.r.t $p$. Doing that, one gets the following (details here)
This is a stream line plot of $s(p)$: the function must follow one of these lines, depending on its initial condition. The horizontal-axis is the $p$-variable and the vertical-axis is $s$-variable. $s(p)$ seems monotonically decreasing on the interval, as required. This seems to suggest that there is a unique solution for every initial condition.
Analysis of $p=1,2$:
Let's prove that $s(2)=\frac{1}{4},s(1)=(2-\sqrt 2)^2 \simeq 0.343$.
For $p=1$ the equation reduces to $ \sqrt 2(1-\sqrt s)-1=-\frac{\sqrt s}{\sqrt 2}$. Setting $x=\sqrt s$, we obtain $ 1-\sqrt 2=x(1/\sqrt 2-\sqrt 2) \Rightarrow x=2-\sqrt 2.$
For $p=2$ the equation reduces to
$2(1-\sqrt s)^2-1=-2(1-\sqrt s)\sqrt s \Rightarrow -1=-2(1-\sqrt s)\big((1-\sqrt s)+\sqrt s \big) \Rightarrow 2(1-\sqrt s)=1 \Rightarrow s=\frac{1}{4}.$
With $r:=p/2\in[1/2,1)$ and $y:=1-\sqrt s\in[0,1/2)$, rewrite your equation (1) as
$$G(r,y):=2^r y^{2 r-1} (r+y-r y)-1=0. \tag{2}$$
For any $r\in(1/2,1)$, $G(r,0)=-1\ne0$, so that $y=0$ is not a solution of equation (2). Also, $y^{2 r-1}$ is undefined for $r=1/2$ and $y=0$. So, in what follows let us assume $r\in(1/2,1)$ and $y\in(0,1/2)$ by default.
Clearly, $G(r,y)$ is strictly and continuously increasing in $y$ from $G(r,0+)=-1<0$ to $G(r,\frac12-)=2^{-r} (1+r - 2^r)>0$, for each $r$; here one may use the concavity of $1+r - 2^r$ in $r$. So, for each $r$, equation (2) has a unique root
$$Y:=Y(r)\in(0,1/2). \tag{3}$$
Moreover,
$$G'_y(r,y)=2^{r+1} r y^{2 r-2}(r-1/2 + (1-r)y)>0.$$
So, by the implicit function theorem, the function $Y$ is differentiable (and hence continuous). Moreover,
$$Y'(r)=-\frac{G'_r(r,y)}{G'_y(r,y)}\Big|_{y=Y}
\overset{\text{sign}}=H(r,Y)>H(1/2,Y)\overset{\text{sign}}=h(Y), $$
where $a\overset{\text{sign}}=b$ means $\text{sign}\, a=\text{sign}\,b$, $H(r,y):=-1 + y - (y + r (1 - y)) \ln(2 y^2)$, and
$$h(y):= -\frac{1 - y}{1 + y} - \frac12\,\ln(2 y^2).$$
Note that $h(1/2)>0$ and $h'(y)=-\frac{1+y^2}{y (1+y)^2}<0$, whence $h>0$ and hence $Y'>0$.
Thus, $Y(r)$ is continuously increasing in $r$, which means that the root $s$ of your equation (1) is continuously decreasing in $p$, as you conjectured.
This seems like a nice solution, thanks. I am sorry to bother you, but if I am not mistaken, the rewriting of the equation should be $G(r,y):=2^r y^{2 r-1} (r+y-ry)-1=0, $ instead of $G(r,y):=2^r y^{2 r-1} (r+y)-1=0. $ (you have a missing term of $ry$ coming from the multiplication of the original equation by $\sqrt s=1-y$). I was able to verify all the steps after this modification, except for the last one which concerns the sign of $G'_r$.
@AsafShachar : I did miss that term. However, now the solution is only a bit simpler. I have also added details.
Put $t=1-\sqrt{s}\in[0,1/2)$ so the equation writes
$$ \Big(1-\frac p2\Big)\, t^p+ \frac p2\, t^{p-1}=2^{-\frac p2}$$
Now if we put $u:=t^{p-1}$ the equation takes the form
$$u+\Big( \frac2p -1\Big)\,u^q =\frac {2^{1-\frac p 2 }} p$$
with $q=\frac p{p-1} >1$, that can be solved by series (see e.g. here) (this way one covers an interval $1.57<p\le2$ if I'm not wrong. To cover the other values of $p$, close to $1$, one needs to put the equation in other forms).
This can be done with Maple and Mathematica as follows. First, let us look at the plot done with Maple
plots:-implicitplot((2^(p/2)*(1 - sqrt(s))^p - 1)/sqrt(s) =
-2^(p/2 - 1)*p*(1 - sqrt(s))^(p - 1), p = 1 .. 2, s = 1/4 .. 1);
The result suggests that $s(p)$ changes from approximately $0.34$ to approximately $0.25$ as $p$ runs from $1$ to $2$. More exactly, making use of Mathematica, we have
NMaximize[{s, (2^(p/2)*(1 - Sqrt[s])^p - 1)/Sqrt[s] == -2^(p/2 - 1)*
p*(1 - Sqrt[s])^(p - 1) && p >= 1 && p <= 2}, {p, s}]
$\{0.343146,\{p\to 1.,s\to 0.343146\}\}$
and
NMinimize[{s, (2^(p/2)*(1 - Sqrt[s])^p - 1)/Sqrt[s] == -2^(p/2 - 1)*
p*(1 - Sqrt[s])^(p - 1) && p >= 1 && p <= 2}, {p, s}]
$\{0.25, \{p -> 2., s -> 0.25\}\}$
We can find the exact values by
solve(eval((2^(p/2)*(1 - sqrt(s))^p - 1)/sqrt(s) = -2^(p/2 - 1)*p*(1 - sqrt(s))^(p - 1), p = 1), s);
$-4\,\sqrt {2}+6$
and
solve(eval((2^(p/2)*(1 - sqrt(s))^p - 1)/sqrt(s) = -2^(p/2 - 1)*p*(1 - sqrt(s))^(p - 1), p = 2), s);
$\frac 1 4$
Now we find the implicit derivative of $s$ with respect to $s$ by
a := implicitdiff((2^(p/2)*(1 - sqrt(s))^p - 1)/sqrt(s) = -2^(p/2 - 1)*p*(1 - sqrt(s))^(p - 1), s, p):
and its maximum value when $p$ runs from $1$ to $2$ by
DirectSearch:-GlobalOptima(a, {(2^(p/2)*(1 - sqrt(s))^p - 1)/sqrt(s) =
-2^(p/2 - 1)*p*(1-sqrt(s))^(p - 1), 1<= p, p<=2, s<=-4*sqrt(2) + 6, 1/4 <= s}, maximize);
$[-0.0482867952575873, [p = 1.99999990682054, s = 0.250000000105689], 358]$
Because the default absolute error of the GlobalOptima command equals $10^{-6}$, this does the job.
|
2025-03-21T14:48:32.022651
| 2020-09-13T14:23:52 |
371591
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Iosif Pinelis",
"Yemon Choi",
"https://mathoverflow.net/users/36721",
"https://mathoverflow.net/users/763"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632972",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371591"
}
|
Stack Exchange
|
How to derive the asymptotic distribution under the alternative hypothesis?
In the class, my professor introduced the ADF test, and I suddenly realized that it seems that all tests are under the null hypothesis. I'm curious that if it is possible to make the judgement under the alternative hypothesis or find out the distribution under the alternative hypothesis.
I have found a question in Stackexchange which suggests that for the situation of $\alpha = 1$ and $\alpha \neq 0$, we can look the problem as $\alpha = \hat{\alpha} + \hat{\gamma}$ so that we can derive the distribution under the alternative hypothesis, but I don't know how to deal with the problem such as $H_0$: $\alpha = 1$ and $H_{A}$:$\alpha < 1$ like what ADF test states.
Is there any textbook or other materials which discuss this kind of problem?
This is a question for your professor or a classmate, not for this forum.
I’m voting to close this question for the reason given in a previous comment
|
2025-03-21T14:48:32.022752
| 2020-09-13T15:13:33 |
371596
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Sam Gunningham",
"https://mathoverflow.net/users/149206",
"https://mathoverflow.net/users/7762",
"mi.f.zh"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632973",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371596"
}
|
Stack Exchange
|
Reference: Irreducible components of the Steinberg variety are conormal bundles
The (partial) Springer resolution is defined as a map $\mu: T^*\mathcal{F} \to \mathcal{N}$, where $\mathcal{F}$ is the partial flag variety consisting of $n$-step partial flags of $\mathbb{C}^d$, and $\mathcal{N}$ a set of nilpotent operators $\mathcal{N} = \{X \in \text{End}(\mathbb{C}^d) : X^n = 0\}$. There's a characterisation of the cotangent bundle of the flag variety as the set of pairs $T^*\mathcal{F} = \{(X, F_\bullet) \in \mathcal{N} \times \mathcal{F} : XF_i \subset F_{i-1} \text{ for all } i\}$.
Using this characterisation, the Steinberg variety is the fibred product $Z = T^*\mathcal{F} \times_\mathcal{N} T^*\mathcal{F}$, alternatively written as the set of triples $Z = \{(X, F_\bullet, F_\bullet'):\dots\}$. There is a standard result that this variety is the union of conormal bundles to the $GL_d(\mathbb{C})$-diagonal orbits in $\mathcal{F} \times \mathcal{F}$, see e.g. Chriss-Ginzburg Prop 4.1.6 or 3.3.4.
On the other hand, the Bruhat decomposition tells us that these diagonal orbits are parametrised by elements of the corresponding Weyl group (in this case, $S_n$ the symmetric group). It is another standard fact that the irreducible components of the Steinberg variety are the (closures of the) conormal bundles to these orbits, see e.g. Chriss-Ginzburg Corollary 3.3.5.
I'm looking for either a proof of this result, or at the very least a good moral reason for why this should be true.
This is a repost of the question I asked on MSE here.
Does this not follow from the fact that these conormal bundles are connected (so their closures are irreducible) and all of the same dimension (so must be components)?
After thinking about it, why do we need the equi-dimension property? I don't think the conormal bundles aren't all of the same dimension in general.
The conormal bundle to a submanifold $Z \subseteq X$ is a lagrangian submanifold of $T^\ast X$. In particular it has dimension $=dim(X)$.
Ah, I missed the fact that you were looking at the space of all partial flags. This is a disjoint union of manifolds of different dimension. Of course, what I wrote above applies in the case of $X$ connected (e.g. $Fl_d$ for a fixed dimension vector $d$).
If you have a variety which is a finite union of locally closed subvarieties all of the same dimension then their closures will all be top dimensional irreducible closed subsets, and thus irreducible components. The equidimensionality is crucial here (think of the affine line as a union of the origin and the punctured affine line - certainly the origin is not an irreducible component of the affine line). I had in mind applying this for each connected component of $Fl \times Fl$ at a time. Perhaps I have totally misunderstood your question though - apologies if so!
|
2025-03-21T14:48:32.022955
| 2020-09-13T16:31:18 |
371602
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Asaf Karagila",
"Gabe Goldberg",
"Mirco A. Mannucci",
"https://mathoverflow.net/users/102684",
"https://mathoverflow.net/users/10671",
"https://mathoverflow.net/users/15293",
"https://mathoverflow.net/users/7206",
"jonasreitz"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632974",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371602"
}
|
Stack Exchange
|
Set Theoretic Geology II: The structure of the directed partial order of grounds
In my previous question Set-theoretic geology: controlled erosion?
and the great answer by Jonas Reitz, I have learned a few things, starting from the awareness that I understand the fine-grain structure of Set Theoretic Geology even less than I had assumed.
That is, of course, good news: more to learn!
The second thing I have learned is:
if I want to understand more, I have to start from the STRUCTURAL STANDPOINT, ie I have to grasp, given a transitive model M (I could do away with that, by starting from V, but I prefer concrete set models), the structure of the partial order of grounds of $M$.
To be more specific, Let us begin with $GROUNDS(M)$, and take a look at its structure: it is a partial order, and looks like that it is directed.
So, given two grounds, say $G_1$ and $G_2$, there is a third G which refines both.
Joel's Modal Logic of Forcing is $S4.2$ (please correct me if I am wrong!), which makes sense to me: this logic corresponds exactly to directed partial pre-orders.
But here is where things become quite hazy to me: what about actual meets?
QUESTIONS
When $GROUNDS(M)$ has the structure of a meet-semilattice?
When is $GROUNDS(M)$ equipped with a full lattice structure?
When $GROUNDS(M)$, assuming 1 and 2, is a complete (sups, infs) lattice?
More related questions:
$GROUNDS(M)$ is a subclass of $TM(M)$, ie the class (set) of transitive sub-models of $M$, so it makes sense to loosen the questions above by asking when the infs and sups asked for are not part of the directed order, but still exist in $TM(M)$.
Any answer to any or some of the questions is welcome.
I mean, the intersection of any set-many grounds is a ground, is there anything left to say?
If this is true, you have answered my first question: the partial order of grounds is a infinite meet semilattice. But, is it true? I was under the impression that it was only directed, in general, not a semilattice. If this is true, then I doubt that the logic of forcing is simply S4.2, but rather a somewhat stronger modal logic. Secondly, I do not think you read through my question carefully enough, that is ONE question
what about sups, or even iarbitrary sups?
Hmm. I could be misremembering, then. But if $W$ and $W'$ are both grounds, and $W\cap W'$ is a model of ZFC which contains a ground $U$, then it is a model of ZFC between $U$ and $V$, where $V=U[G]$ for some set-generic filter $G$. So by Vopenka's theorem $W\cap W'$ is a ground.
I suspect so. You are a (very) smart guy Asaf, and I do greatly esteem your set theory first-class competence, I would appreciate if you take your time and come up with an answer. I can guarantee you that you will get my vote, and if you come up to an answer to all of the above, even a negative one, you have my GREEN. L'hitraot, haveri
apologies, I had not read your full comment. THAT is interesting! Why don't you put it into a formal answer? It does not address all points, but it is very good start
No good deed goes unpunished. I'm sure someone will come soon that can write a much better answer to all of your questions. I, on the other hand, need to clean the kitchen and cook the hell out of some chicken.
LOL! Menial works has always priority. I will be waiting then, no worries.
@Asaf It's not true that the intersection of two grounds is a ground since it might not satisfy ZFC. You can get an example using the technique from my answer here by changing the word "generic" in the second sentence to "Cohen generic" and working in a big collapse extension: https://mathoverflow.net/questions/297756/intersection-of-two-generic-extensions/297769
See, now that's why I didn't post an answer. Thanks, @Gabe, I felt like this might "too easy".
@MircoA.Mannucci Another thing you can do is get a situation where you have two grounds whose intersection is $L(\mathbb R)$, and in this case assuming there is no $\omega_1$-sequence of reals in $L(\mathbb R)$, there cannot be a largest common ground, since any inner model of ZFC contained in $L(\mathbb R)$ has countably many reals, and hence has a forcing extension contained in $L(\mathbb R)$.
@Gabe: Although in that case $L(\Bbb R)$ is in fact a ground... Just not a ZFC one.
@AsafKaragila Mathematics is not ready for such problems
@GabeGoldberg thanks for chiming in! So, looks like geology is a complicated science... I would still love if you and/or Asaf posted a partial answer: listing what is known and what is not known (example: your counterexample cited above, but hopefully examples where the meet exists in GROUNDS(M), etc. We need to build a list of what is known...
Mirco, this is also a fantastic question - the structure of the grounds as a partial order seems to be a very basic aspect of forcing that is not entirely understood. Once again I don’t have a complete answer, but I can provide some background & a few observations.
Intersection of grounds. As pointed out in the comments, it is not the case that the intersection of grounds is (necessarily) a ground - the intersection may fail to satisfy ZFC. However, the intersection does contain a ground (see Directedness)
Directedness. Theorem (Usuba): The grounds are downward-set-directed (that is, the intersection of any collection of set-many grounds contains a further ground).
This fundamental result resolved a number of open questions in Set Theoretic Geology - not least of which is the lovely fact that the Mantle (the intersection of all grounds) is always a model of ZFC.
Meets. Since we can get below any set-indexed collection of grounds, it’s natural to ask whether there is a unique, largest such ground below such a collection. I believe the answer is no (I think the example linked in comments by @gabe-goldberg Intersection of two generic extensions may be a good candidate for a counterexample, but I haven’t thought it through).
The Ground Axiom. The ground axiom states “there are no grounds except V itself” -- if this is the case, then GROUND(V) is trivial.
Least element. The Mantle (the intersection of all grounds) is, by Usuba’s result, a model of ZFC. If GROUND(V) has a least element, it is equal to the Mantle - this will happen exactly when V is a set forcing extension of a model of the Ground Axiom. If GROUND(V) does not have a least element, it may be the case that moving from the Mantle to V can be accomplished by class forcing. Finally, it may be the case that V is not even a class forcing extension of the Mantle (by any class forcing definable in the Mantle) -- see http://jdh.hamkins.org/the-universe-need-not-be-a-class-forcing-extension-of-hod/ .
With regards to your final “more related questions”, I believe that if infs or sups exist then they must be ground models, by an argument similar to @asaf-karagila ‘s in the comments - we can use directedness to get below a collection of grounds, so if an inf exists it is an intermediate model of ZFC between a ground and an extension, hence it is also a ground.
None of the above resolves your three Questions - under what circumstances do we get a nice structure on GROUND(V)? For example, if we start in a model of the Ground Axiom and carry out some forcing, what is the relationship between properties of the forcing we choose and the structure of grounds in the resulting extension? I really like this line of thinking as an avenue for understanding structural properties of forcing.
I think that it's safe to assume that the user known as Asaf Karagila is in fact Asaf Karagila. No need to refer to them as "@asaf-karagila" (which will not notify them, and even if it would, there shouldn't be a hyphen there anyway)...
Noted! Thanks Asaf
|
2025-03-21T14:48:32.023483
| 2020-09-13T17:33:36 |
371606
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Claus",
"D.S. Lipham",
"HenryM",
"Mary Sp.",
"Mirco A. Mannucci",
"Ryan Budney",
"https://mathoverflow.net/users/1465",
"https://mathoverflow.net/users/15293",
"https://mathoverflow.net/users/156936",
"https://mathoverflow.net/users/161374",
"https://mathoverflow.net/users/165325",
"https://mathoverflow.net/users/40804",
"https://mathoverflow.net/users/5428",
"https://mathoverflow.net/users/95718",
"leonbloy",
"mme"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632975",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371606"
}
|
Stack Exchange
|
What are good mathematical models for spider webs?
Sometimes I see spider webs in very complex surroundings, like in the middle of twigs in a tree or in a bush. I keep thinking “if you understand the spider web, you understand the space around it”. What fascinates me, in some sense it gives a discrete view on the continuous space surrounding it.
I started to wonder what are good mathematical models for spider webs. Obvious candidates are geometric graphs embedded in surfaces, or rather in space. One could argue that Tutte’s Spring Theorem from 1963 is the base model: a planar geometric graph, given as the equilibrium position for a system of springs representing the edges of the graph. It is the minimum-energy configuration of the system of springs (see the picture for illustration). There are generalizations of such minimum-energy configurations for convex graph embeddings into space (Linial, Lovász, Wigderson 1988), where you place, for example, four vertices of the graph at the vertices of a simplex in $\mathbb R^3$.
I think such systems of springs are good models, because the threads of the spider web are elastic. However, when viewed as models for spider webs, I wonder whether these minimum-energy spring models are missing two aspects:
The purpose of spider webs is to catch prey, so I feel the ideal model should also consider
(A) maximizing the area covered (or the volume of the convex hull) and
(B) minimizing the distances between the edges.
To me, formalizing (A) and (B) and combining it with the minimum-energy principle for a system of springs would be the ideal mathematical model for spider webs.
Now, it is not obvious to me whether the minimum-energy principle alone determines a geometric graph satisfying (A) and/or (B)? Asking differently, if you add conditions like (A) or (B) to the minimum-energy principle, will this lead to different geometric graphs?
My second, broader question: Are you aware of any mathematical models developed explicitely to model spider webs? I checked MO and MSE and searched on the internet, but could not find anything. Maybe I am looking in the wrong fields, I wonder. Any help would be greatly appreciated!
References:
Tutte, W. T. (1963), "How to draw a graph", Proceedings of the London Mathematical Society, 13: 743–767, doi:10.1112/plms/s3-13.1.743
Linial, N.; Lovász, L.; Wigderson, A. (1988), "Rubber bands, convex embeddings and graph connectivity", Combinatorica, 8(1): 91–102, doi:10.1007/BF02122557
The picture is from Daniel Spielman’s lecture notes pdf on the web
OK I read it now. Guaranteed that I will come for an answer to your spiderquestion. Meanwhile, as an appetizer, are you familiar with the SPRING LAYOUT for graphs? That goes a long way...
Regarding your last points (structure and capturing), real webs often have different types of silk to attain those different purposes. In particular, in orbital webs the radial threads are structural (stronger) while spirals are designed for capture (only they are sticky, as you can test by touch). Webs are truly impressive.
In response to the second question (which I interpret as asking for math models of spider webs as they appear in Nature): There exist several distinct types of spider webs. The most common type, the orb web of araneids, has been modeled in Simple Model for the Mechanics of Spider Webs (2010).
A key property of the orb web model is that the web is free of stress concentrations even when a few spiral threads are broken. This is distinctly different from usual elastic materials in which a crack causes stress concentrations and weakens the material.
The model highlights the mechanical adaptability of the web: spiders can increase the number of spiral threads to make a dense web (to catch small insects) or they can adjust the number of radial threads (to adapt to environmental conditions or reduce the cost of making the web) – in both cases without reducing the damage tolerance of the web.
Left panel: Construction of the orb web described in the cited paper.
Right panel: Naturally occurring orb web (Wikipedia).
Great reference, thanks a lot! I have not thought about stability conditions yet, that’s a great add!
I have seen a spider web in Florida that looks like a vertical funnel, suspended in mid air. very large, maybe 4 feet tall. Unlike the funnel example in the link, the one I saw has walls you can easily see through.
@HenryM: there are a wide variety of web types. Most people think of the orb web, which this answer focuses on. There are funnel webs, cob webs (eg: black widows), sac webs, etc. Some spiders only use webbing for their egg sacs, or as a climbing apparatus (jumping spiders). There are an enormous number of web types. Florida has some spiders that make some quite novel web types, like the bolas spider or the ogre spider.
In topology there is the notion of an infinite spider's web in the (complex) plane $\mathbb C$ that was introduced in 2010 https://arxiv.org/pdf/1009.5081.pdf.
A set $E\subseteq \mathbb C$ is an infinite spider’s web if $E$ is connected and there
exists a sequence of bounded simply connected domains $(G_n)$ with
$G_n \subset G_{n+1},$
$\partial G_n\subset E,$ and
$\bigcup _{n\in \mathbb N}G_n = \mathbb C.$
In certain cases we also have that $E$ is closed and nowhere dense, and each $\partial G_n$ is a simple closed curve (Jordan curve), so that $E$ more closely resembles a traditional spider's web. These sets can be generated by iteration of entire functions such as $f(z)=\frac{1}{2}(\cos z^{1/4}+\cosh z^{1/4})$. The image below shows a spider's web consisting of the points $z\in \mathbb C$ such that $f^n(z)\to\infty$ at a certain rate (see https://arxiv.org/pdf/1009.5081.pdf for details).
Every locally connected Julia set of a transcendental entire function also has this form (shown in https://arxiv.org/pdf/1110.3256.pdf).
This seems to go the other direction from what the OP is asking. This is not a mathematical model for real-world spiderwebs; rather it's a mathematical object which we mentally model by thinking about spiderwebs. I do not think Julia sets of transcendental entire functions shed much light about the spiderwebs I see in the garden. Nice picture though.
@MikeMiller Yes the definition I gave is a pretty vague mathematical abstraction. However I think there are certain geometric regularities that we observe in actual spider's webs that may be be captured by simple iterative processes (iteration of entire functions is just an example). This could be considered when designing mathematical models.
Fair point! ${}$
So, I promised I would come up with some answer, but looks like there is already a great deal in the great answers above.
Anyway, I find it impossible to resist the temptation, especially because I think there is so much math that we can learn from Nature, particularly from our little friends, the spiders.
SPIDERWEBS AS SMART SENSORS (MORPHOLOGICAL COMPUTING)
Spider webs are not only for catching preys. They are, in a sense, an extension of their sensory apparatus, in that they help a spider detect at least three types of objects: a prey, a predator, and a potential mate.
How? Well, the web is kind of elastic, and it acts as a strange non linear filter: by "measuring" the perturbations on the web, our friends can isolate some frequencies which give them the clues.
See here and here and also here for details.
As far as I know, the Theory of Morphological Computation is still undeveloped, particularly from the mathematical standpoint. Perhaps some smart fellow here on MO can enlighten us. Meanwhile, just wish to point out that the changes of configurations mentioned by Carlo above are also done as "tuning" the morphological computing capabilities of the web
SPIDERWEB AS QUANTUM GRAVITY MODELS
As we all know, quantum gravity is the holy grail of modern physics. Among the most intriguing attempts so far, there is Fotini Markopoulou Kalamara's Quantum Graphity.
see here.
To summarise Fotini's brilliant idea is not easy, but here is the gist: start from a universe in which there is no space-time, and try to build it as a graph. Create a quantum system, that basically is a quantum superposition of many graphs, and associate to this beast an hamiltonian.
Set it to some default eigenvalue of energy, say HOT. That corresponds to a fully connected graph, where every point is one step from any other. Too many connections to make up our space time! But now suppose it "cools off": the edges get disactivated, till it settles into something like our ordinary space time (the full theory is, as far as I know, still undeveloped, needs some really good mathematician to work it out). Now, I would suggest you toy with that theory, precisely because , as you suggested, spiderwebs give you an insight into the nature of space (and time too). For quantum gravity, you do not simply need to replicate the topological properties of space-time, but also its metrics (say volumes, areas, etc. Essentially approximate general relativity).
So, perhaps the DEMIURGOS is a giant super inteligent spider!
SPIDER WEBS AS WEIGHTED SIMPLICIAL COMPLEXES
Now, after the double detour, back to the question and the answer: I suspect, in fact I am pretty sure, that the best way to model spider webs are Weighted Simplicial Complexes, ie simplicial complexes in which all simplices have a weight (either a real number, or even a complex one, in case we wanna formalize quantum spiders): see here.
Why weighted simplicial complexes? Because they generalize weighted graphs, and they have an entire artillery (laplacian, persistent homology, etc) which can be put to use here. Example: you ask for minimal area, that to me means that when you write your energy, you must add a term which tries to minimize the total area covered by the web. I also think that these complexes could add some light into spider webs as morphological computers (I have a small paper with some folks on diffusion of information on weighted simplicial complexes. The core idea is that higher simplices are hubs of information broadcasting, which in case of graphs are missing)
Mirco thanks for a fantastic answer with so much extra perspective. Are you able to give the link to your paper about weighted simplicial complexes as information hubs?Very interested in that. And will also follow on on Quantum Graphity. Plus, never heard of morphological computing, want to learn about it
+1 for "perhaps the DEMIURGOS is a giant super inteligent spider" and giving a reason for that! Your answer is a great read.
@Claus the article was meant for communication researchers, not mathematicians (though I have plans to morph it into a real math article) Here it is: https://digitalcommons.chapman.edu/cgi/viewcontent.cgi?article=1024&context=comm_articles
@MaryS. thanks! spiders are very smart beings...
@MircoA.Mannucci Just saw your article, really interesting!! I am a big fan of such applications of math. Please let us know when your math background paper Is ready. This whole area is fascinating
A biologist friend told me about this question on MathOverflow, so I wanted to contribute a useful link to a related article that appeared in NATURE.
Published: 01 February 2012
Nonlinear material behaviour of spider silk yields robust webs
Steven W. Cranford, Anna Tarakanova, Nicola M. Pugno & Markus J. Buehler
Nature volume 482, pages72–76(2012)
This is the link https://www.nature.com/articles/nature10739
The mathematically interesting feature investigated here is the nonlinear response of silk threads to stress:
From the abstract of this article: Here we report web deformation experiments and simulations that identify the nonlinear response of silk threads to stress — involving softening at a yield point and substantial stiffening at large strain until failure — as being crucial to localize load-induced deformation and resulting in mechanically robust spider webs. Control simulations confirmed that a nonlinear stress response results in superior resistance to structural defects in the web compared to linear elastic or elastic–plastic (softening) material behaviour. (...) The superior performance of silk in webs is therefore not due merely to its exceptional ultimate strength and strain, but arises from the nonlinear response of silk threads to strain and their geometrical arrangement in a web.
|
2025-03-21T14:48:32.024478
| 2020-09-13T17:43:02 |
371607
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Agustí Roig",
"anon",
"https://mathoverflow.net/users/1246",
"https://mathoverflow.net/users/161984"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632976",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371607"
}
|
Stack Exchange
|
Amitsur cohomology and limits
Let $R \longrightarrow S$ be a faithfully flat ring extension and $\{ G_\alpha\}$ an inverse system of affine group schemes. I would like to know if there is some hope for a bijection like
$$
H^1(S/R, \underleftarrow{\lim} G_\alpha ) = \underleftarrow{\lim} H^1(S/R, G_\alpha) \ .
$$
Here $H^1$ denotes the Amitsur cohomology.
For instance, this is obviously true when you take a product of affine group schemes:
$$
H^1(S/R, F \times G ) = H^1(S/R, F) \times H^1(S/R, G) \ .
$$
I think the obvious map $H^1(S/R, \underleftarrow{\lim} G_\alpha ) \longrightarrow \underleftarrow{\lim} H^1(S/R, G_\alpha)$ is always injective but doesn't look surjective to me, in general.
Is that true? Are there sufficient conditions to make it surjective? For instance, I would have no problem if the $G_\alpha$'s need to be algebraic.
Actually, the case $\mathbf{k} \longrightarrow \overline{\mathbf{k}}$, where $\mathbf{k}$ is a field (of zero characteristic, if this helps) and $\overline{\mathbf{k}}$ its algebraic closure, so that $\overline{\mathbf{k}} / \mathbf{k}$ is a finite Galois extension would be enough for me.
Hence, in this case, I'm asking if there is a bijection between Galois cohomologies:
$$
H^1(\Gamma, \underleftarrow{\lim} G_\alpha(\overline{\mathbf{k}}) ) = \underleftarrow{\lim} H^1(\Gamma, G_\alpha (\overline{\mathbf{k}}) ) \ .
$$
Here, $\Gamma $ is the Galois group of the extension $\overline{\mathbf{k}} / \mathbf{k}$.
Any hints or references will be much appreciated.
There's usually a $lim^1$ (or worse) to worry about. See the first section of Milne Compositio Math. 140 (2004), 1149--1175; arXiv:math/0209076
@anon Thank you! I'll watch your reference: indeed Milne sounds like the right person to talk about this.
|
2025-03-21T14:48:32.024610
| 2020-09-13T18:21:16 |
371612
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Carlo Beenakker",
"Daniel Li",
"https://mathoverflow.net/users/11260",
"https://mathoverflow.net/users/116621"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632977",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371612"
}
|
Stack Exchange
|
Minimal perturbation of a Wigner matrix needed to produce an orthogonal top eigenvector
The instructor proposed a the following statement in the passing and suggested that we think about it (although it is not required):
For any $N \times N$ Wigner matrix, we replace $k$ entries with resampled copy (i.e. the new entries have exact same distribution as the original but are resampled independently). Suppose that $k/N^{5/3}\to\infty$. Denote top eigenvector of original and resampled matrix as $v,v'$. Show that
$\mathbb{E}|\langle v,v' \rangle| \to_N0$.
Attempt: the intuition is of course that because of resampling too many entries, the two eigenvectors become less and less correlated. And in high dimensions, two independent vectors distributed uniformly on a ball tend to be orthogonal. But I struggle to explain the $N^{5/3}$ threshold or to formulate a rigorous proof.
Any directional or strategical suggestion is welcome.
this seems quite nontrivial to me; the top eigenvector refers to the largest eigenvalue, which has a Tracy-Widom distribution with a level spacing $\delta_N\propto N^{-2/3}$ -- larger than in the bulk of the spectrum, where the spacing scales $\propto N^{-1}$. The question would seem to ask for the minimal rank $k$ of a perturbation that shifts the largest eigenvalue by $\delta_N$, to obtain an orthogonal eigenvector.
Thank you but how shifting largest eigenvalue by $\delta_N$ gives orthogonality? Ps. this is not supposed to be trivial because we are asked to genuinely think about it.
the shift of an eigenvalue by the eigenvalue spacing creates independent eigenvectors, which would then become orthogonal for the reason mentioned in the question.
Oh oh ok. But spacing between first and second eigenvalue is $N^{-1/6}$, right? This means before reaching that level, changing k entries contributes roughly $\sqrt{k}/N$. This is where we get $N^{5/3}$?
|
2025-03-21T14:48:32.024800
| 2020-09-13T19:02:13 |
371614
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Alex M.",
"Gerald Edgar",
"Johannes Hahn",
"Nik Weaver",
"Pietro Majer",
"Robert Bryant",
"https://mathoverflow.net/users/112113",
"https://mathoverflow.net/users/13972",
"https://mathoverflow.net/users/161104",
"https://mathoverflow.net/users/23141",
"https://mathoverflow.net/users/3041",
"https://mathoverflow.net/users/454",
"https://mathoverflow.net/users/54780",
"https://mathoverflow.net/users/6101",
"lambda",
"mw19930312"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:632978",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371614"
}
|
Stack Exchange
|
Vector-Valued Stone-Weierstrass Theorem?
The standard statement of the Stone-Weierstrass theorem is:
Let $X$ be compact Hausdorff topological space, and $\mathcal{A}$ a subalgebra of the continuous functions from $X$ to $\mathbb{R}$ which separates points. Then $\mathcal{A}$ is dense in $C(X, \mathbb{R})$ in sup-norm.
Most materials that I can find on the extension of Stone-Weierstrass theorem discuss only the multivariate case, i.e., $X\in \mathbb{R}^d$. I wonder whether this theorem can be extended to vector-valued continuous functions. Specifically, let $\mathcal{A}$ be a subalgebra of continuous functions $X\to \mathbb{R}^n$, with the multiplication defined componentwisely, i.e., $\forall f, g\in \mathcal{A}$, $fg = (f_1g_1, \ldots, f_ng_n)$. Then shall we claim $\mathcal{A}$ is dense in $C(X, \mathbb{R}^n)$ in sup-norm if $\mathcal{A}$ separates points?
Any direct answer or reference would greatly help me!
Edit: As Nik Weaver points out, the original conjecture is false since the functions of the form $x\mapsto (f(x), 0, \ldots, 0)$ create a counter-example. I wonder whether there are non-trivial Weierstrass-type theorems on vector-valued functions. For instance, what if we further assume $\mathcal{A}$ is dense on each `axis'?
What is the multiplication in an "algebra of vector valued functions"? Or maybe you mean a structure of $C(X,\mathbb{R})$-module?
The statement is trivially false: the functions of the form $x \mapsto (f(x),0,\ldots, 0)$ with $f \in C(X)$ are sup-norm closed and separate points.
@NikWeaver Thanks for the counterexample here! I've added a short comment in the post.
@PietroMajer The componentwise multiplication induces an "algebra of vector valued functions", doesn't it?
I see, thanks! Then it is like the algebra $C(X\times{1,\dots,n},\mathbb{R})$?
Maybe it would be instructive to do the special case where $X$ is a single point.
@GeraldEdgar Thanks for the suggestion! I'm more interested in the case where $X$ is a compact interval like $[0, 1]$, which might be fundamentally different from the case where $X$ is a single point.
@PietroMajer I'm not sure whether I'm correct. I guess it is like the algebra $C(X, \mathbb{R}\times {1, \ldots, n})$?
I believe it is isomorphic to $C(X \times {1, \dots, n}, \mathbb R)$ at least as a topological algebra. Applying Stone--Weierstrass to that algebra it seems the correct condition is that for $x, y \in X$ and any pair of indices $i, j$ there exists $f \in \mathcal A$ with $f_i(x) \ne f_j(y)$.
no, just write $f(x,j)$ instead of $f_j(x)$. It's an algebra isomorphism wrto the one you said. So in fact it is a way to include your setting in the usual one.
@lambda yes, that what i was saying
@lambda This is inspiring! Thanks for the help!
@PietroMajer I really appreciate the detailed discussions here! Thanks again!
I think that you want something like this:
Let $E\to X$ be a (finite rank) vector bundle over a compact, Hausdorff topological space $X$, let $\mathcal{A}\subset C(X,\mathbb{R})$ be a subalgebra that separates points, and let $\mathcal{E}\subset C(X,E)$ be an $\mathcal{A}$-submodule of the $C(X,\mathbb{R})$-module of continuous section of $E\to X$. Suppose that, at every point $x\in X$, the set $\{\,e(x)\ |\ e\in\mathcal{E}\ \}$ spans $E_x$. Then $\mathcal{E}$ is dense in $C(X,E)$ with respect to the sup-norm defined by any norm on $E$.
Addendum: Here is a sketch of the argument: First, by an easy compactness argument, one can show that $\mathcal{E}$ contains a finite set $e_1,\ldots e_m$ such that $e_1(x),e_2(x),\ldots,e_m(x)$ spans $E_x$ for all $x\in X$. Then $\mathcal{E}$ contains all the sections of the form $$a_1\, e_1 + \cdots + a_m\,e_m$$ where $a_i\in\mathcal{A}$, and every section $e\in C(X,E)$ can be written in the form $$e = f_1\, e_1 + \cdots + f_m\,e_m$$ for some functions $f_i\in C(X,\mathbb{R})$. By the Stone-Weierstrass Theorem, for any given $\delta>0$, we can choose $a_i\in \mathcal{A}$ so that $\|f_i-a_i\|<\delta$ for all $1\le i\le m$. Now the equivalence of all norms in finite dimensional vector spaces can be applied (together with the compactness of $X$) to conclude that $\mathcal{E}$ is dense in $C(X,E)$ in any sup-norm derived from a norm on the (finite rank) vector bundle $E$.
Thanks for the addendum! While I'm digesting this, I have a quick question here. Does this result hold for matrix-valued subalgebras, i.e., $\mathcal{A} \subset C(X, \mathbb{R}^{n\times n})$ equipped with matrix multiplication?
@mw19930312: I'm confused by your question. Could you state what you mean by 'this result' more precisely?
@mw19930312: I'm sorry, but I don't know what you mean by assuming that $E\subset \mathbb{R}^n$. In my answer, $E\to X$ is a vector bundle over $X$ and $\mathcal{A}$ is a subset of $C(X,\mathbb{R})$. Maybe you mean something like "If $\mathcal{A}\subset C(X,\mathrm{End}(E))$ is an algebra of bundle maps from $E$ to $E$ and $\mathcal{E}\subset C(X,E)$ is a space of sections that is an $\mathcal{A}$-module, then under what conditions can we conclude that $\mathcal{E}$ is dense in $C(X,E)$ in the sup-norm induced by any norm on $E$"?
Hi Robert, sorry for coming back to this question after it is posted for a month. I believe you are interpreting this correctly, I'm looking for the case of $\mathcal{A}\subset C(X, \mathrm{End}(E))$. If the same result holds, would you mind referring me a detailed proof of this?
@mw19930312: That's OK. What I don't know is what you want to assume about $\mathcal{A}\subset C(X,\mathrm{End}(E))$. Is it a subalgebra? Are you asking for checkable conditions that would guarantee that $\mathcal{A}$ is $C^0$-dense in $C(X,\mathrm{End}(E))$?
I assume $\mathcal{A}$ is a subalgebra. We can assume $\mathcal{A}$ separate points first. Then I wonder whether your answer still holds, i.e., $\mathcal{E}$ is dense in $C(X, E)$, where $\mathcal{E}$ is an $\mathcal{A}$-module, if span of ${e(x) |e\in \mathcal{E}}$ equals to $E_x$. If it holds, would you mind referring me a proof?
Also, if $\mathcal{A}$ does not separate points, then can we identify the denseness of $\mathcal{E}$ through checking whether $\mathcal{E}\big\vert_{[x]}$ spans $E\big\vert_{[x]}$, where $[x]$ is the equivalent class of $x$ under $\mathcal{A}$?
@mw19930312: I'm confused by your bringing in $\mathcal{E}$ in a question about $\mathcal{A}$. The question about $\mathcal{A}$ being dense in $C(X,\mathrm{End}(E))$ makes sense without any reference to an $\mathcal{A}$-module $\mathcal{E}\subset \Gamma(X,E)$. Let's not lose sight of your objective: Is it $\mathcal{A}$ or $\mathcal{E}$. Also, I'm not sure what 'separates points' means in reference to $\mathcal{A}$.
@RobertBryant: I would like to use this in a work of mine, but in order to save space, it would be preferable to give a reference instead of a proof. Do you know of any, please?
@AlexM.: I'm sorry, but I don't know a reference where the above statement is proved explicitly. It's certainly a well-known result, though. For example, the statement that every continuous section of a smooth vector bundle over a compact manifold can be uniformly approximated by a smooth section, which is used all over the place in differential topology, is a special case, but I'm not sure that I've ever seen anyone give a reference. It's kind of a folk theorem. Maybe it's an exercise in a introductory topology course. Maybe, Spanier?
This is a comment, not an answer but I am, alas, not entitled. Vector valued Stone-Weierstraß theorems were studied in great detail in the second half of the last century and there is a comprehensive monograph on the subject by João Prolla ("Weierstraß-Stone, the theorem", 1993). Not on topic, but he also considered the case of bounded, continuous vector-valued functions on non-compact spaces, using the strict topology of R.C. Buck.
Great thanks for the references here! By any chance, could you refer me to some other materials on this topic? The book by João Prollaa appears very difficult for me to find. I've got a note by him named `Approximation of Vector-Valued Functions' in 1977. Is there any recent update on this notes?
The book title written by Yoda has been.
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.