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
|
|---|---|---|---|---|---|
2025-03-21T14:48:31.689266
| 2020-08-06T21:20:01 |
368509
|
{
"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:631778",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368509"
}
|
Stack Exchange
|
Does Barvinok's algorithm apply to convex integer program?
Barvinok provided a counting algorithm to count number of integer solutions to integer linear program that runs in polynomial time if the number of integer variables is fixed.
If we have convex constraints instead of linear is there any such fixed dimension polynomial time counting algorithm?
|
2025-03-21T14:48:31.689323
| 2020-08-06T23:12:06 |
368513
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Andy Putman",
"Ben W",
"Greg Martin",
"Jochen Glueck",
"Kapil",
"YCor",
"https://mathoverflow.net/users/102946",
"https://mathoverflow.net/users/124862",
"https://mathoverflow.net/users/14094",
"https://mathoverflow.net/users/317",
"https://mathoverflow.net/users/5091",
"https://mathoverflow.net/users/73784"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631779",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368513"
}
|
Stack Exchange
|
Is it the referee's responsibility to verify results from arXiv preprints used in the refereed paper?
I'm refereeing a Banach spaces paper and it looks pretty good. I'm about ready to recommend it for publication.
However, its main result depends crucially on some other results that are in preprints on the arxiv. Is it the referee's responsibility to verify those results too? Or should I just alert the editor that we should wait for publication of those preprints? Or something else?
It's not that I'm trying to be lazy exactly. I'm just super busy with other things, and so want to be efficient with my time.
Related. Also related.
@JochenGlueck Interesting link. Yeah, I'm definitely of the opinion that everything should be rigorously checked, and that is what I always do. But this is a little different. Presumably, there is a different referee who is also checking the arxiv preprints. The question is, is that good enough? And if it's good enough, should we wait for them, or is it okay to recommend and just alert the editor so he can withdraw it if the preprint ends up having a flaw? Etc. Either way the proofs will all be double-checked by a careful referee. I'm just wondering what the proper procedure is here.
Yes, I understand the question, of course - I merely added the links for the sake of completeness.
You can explain the situation to the editorial board. I think it's up to the editorial board to deal with the issue, using your feedback, and possibly doing further investigations.
@YCor Okay cool, that's what I'll do then. Thanks!
Does the editorial board accept papers referencing other papers where the latter papers have not yet been refereed and published? Some would not.
@Kapil: I do not know of any journal in pure math that does not allow references to preprints in their papers (and indeed, given that a number of important papers are never published, this would prevent them from publishing papers is certain areas).
I'm going to use the word "I" in this answer since there is no universally agreed-upon standard for what a referee should do.
I feel that the referee's only job is to make an informed recommendation to an editor as to whether or not a paper should be accepted. The extent to which that includes verifying that a paper is correct is a bit subtle -- you want to be able to vouch for the paper's correctness, but ultimately the correctness of the paper is the author's responsibility. My personal interpretation of that mandate is that I should understand and believe all the arguments of the paper, though there have been times when I have not verified things like enormous calculations, in which case I make that caveat in my report. I should also have no good reason to doubt the results of anything that is cited. If I have such a doubt (and I frequently have doubts about the correctness of both the published and the unpublished literature!), then my job is not to go and referee those papers as well. Instead, in my report I give an honest account to the editor of my concerns, and let them decide how they want to handle them.
+1. Ultimately, the author is trying to make a persuasive argument; if a good-faith referee is not persuaded, the author needs to do more work, not the referee.
|
2025-03-21T14:48:31.689723
| 2020-08-06T23:22:17 |
368515
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Connor Malin",
"Jens Reinhold",
"Michael Albanese",
"https://mathoverflow.net/users/134512",
"https://mathoverflow.net/users/14233",
"https://mathoverflow.net/users/21564"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631780",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368515"
}
|
Stack Exchange
|
If two smooth manifolds are homeomorphic, then their stable tangent bundles are vector bundle isomorphic
I am currently reading Kervaire-Milnor's paper "Groups of Homotopy Spheres I", Annals of Mathematics, and I am trying to prove (or disprove) the following result. The more elementary the proof, the better.
If two smooth manifolds are homeomorphic, then their stable tangent
bundles (i.e. the Whitney sum of the tangent bundle with the trivial
line bundle) are vector bundle isomorphic.
I am trying to prove this as an intermediate step to give an alternative proof for KM's Theorem 3.1: Every homotopy sphere is $s$-parallelizable.
The result you are hoping for is in fact false.
In section 9 of Microbundles: Part I, Milnor constructs an open set $U \subset \mathbb{R}^m$. With its standard smooth structure, the (stable) tangent bundle of $U\times\mathbb{R}^k \subset \mathbb{R}^{m+k}$ is trivial, while in Corollary 9.3, Milnor shows that it admits a smooth structure for which the tangent bundle has a non-zero Pontryagin class. As Pontryagin classes are stable, the stable tangent bundle of the latter manifold is not trivial, and hence not isomorphic to the stable tangent bundle of $U\times\mathbb{R}^k$ with its standard smooth structure.
Milnor, John W., Microbundles, Topology 3, Suppl. 1, 53-80 (1964). ZBL0124.38404.
Is this true for compact manifolds though? I think it should follow from the fact the Spivak normal bundle is unique. (I also see the claim that any homotopy equivalence of manifolds can be covered by fiberwise isomorphisms of normal bundles, which I believe implies it).
@ConnorMalin: I am not sure if it holds for compact manifolds, but I would be interested to know one way or the other.
If one can embed the mapping cylinder of a (smooth map) between compact manifolds, then I think it should be true. This will need some type of relative version of the Whitney embedding theorem, which I don't know to be true.
@ConnorMalin: It seems like this answer to your question provides an example of closed smooth manifolds which are homeomorphic but do not have stably equivalent tangent bundles.
Yes its nice to have a concrete example. It definitely felt like it should be true from fundamental results of smoothing theory (apparently not when I made those comments though), but for the easiest examples, spheres and tori, it fails.
Let me add something to Michael Albanese's great answer to see this question in a broader context.
Novikov proved that the rational Pontryagin classes are homeomorphism invariants (in fact, this was one of the achievements for which he received the Fields medal in 1970). The integral Pontryagin classes, however, are not invariant under homeomorphism, see Chapter 4.4 of "The Novikov Conjecture" by Kreck and Lück.
Some polynomials in the Pontryagin classes are even homotopy invariant: for instance, $p_1$ of a closed oriented $4$-manifold $M$ agrees (by Hirzebruch's signature theorem) with $3\sigma(M)$ times the fundamental class in cohomology (where $\sigma(M)$ denotes the signature of $M$), which is invariant under homotopy equivalence.
The famous Novikov conjecture asks whether certain so-called "higher signatures" are also invariant under homotopy equivalence. It is one of the most important open questions in topology.
I was under the impression that Novikov proved that the rational Pontryagin classes are homeomorphism invariants of a closed, orientable, smooth manifold. Can the closedness and orientability hypotheses be removed? It should be noted that the non-zero Pontryagin class I refer to in my answer is torsion.
I am not sure what exactly Novikov proved originally, but I think these extra conditions are not necessary for the statement to hold. I suggest having a look at the "epilogue" of Milnor-Stasheff, which explicity mentions Novikov's result. There are also other proofs using different methods: arxiv.org/pdf/0901.0819.pdf
|
2025-03-21T14:48:31.690007
| 2020-08-06T23:41:12 |
368518
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"D.-C. Cisinski",
"https://mathoverflow.net/users/1017"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631781",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368518"
}
|
Stack Exchange
|
Suspension Theorem in $\mathbb{A}^1$-homotopy
In algebraic topology, the suspension theorem tells us that for a topological space $X$, we have
$$\tilde{H}^n(X,F)\cong \tilde{H}^{n+k}(S^k\wedge X,F).$$
So I'm wondering if this has an analogue in the category of $\mathbb{A}^1$-homotopy, i.e. do we have
$$H^n_{ét}(X,F)\cong H^{n+a+b}_{ét}\left( (\mathbb{A}^1/\{0,1\})^{\wedge a}\wedge (\mathbb{G}_m)^{\wedge b}\wedge X,F\right)?$$
One thing that would make me think that this is true is the comparison theorem for étale cohomology and singular cohomology. Indeed, we have
$$\left(\mathbb{A}^1/\{0,1\}\right)(\mathbb{C})^{an}\cong S^1\cong \mathbb{G}_m(\mathbb{C})^{an}$$
and thus we have
$$H^{n+a+b}_{ét}\left( (\mathbb{A}^1/\{0,1\})^{\wedge a}\wedge (\mathbb{G}_m)^{\wedge b}\wedge X,F\right)\cong H^{n+a+b}((S^1)^{\wedge(a+b)}\wedge X(\mathbb{C}),F)\cong H^{n}(X(\mathbb{C})^{an},F)\cong H^n_{ét}(X,F)$$
if we assume that $F$ is finite so that we can apply the comparision theorem. Is this true? If this argument does not work, is there another proof?
In motivic homotopy theory, we have two spheres: the classical one (as in any homotopy theory, whatever that means), and the geometric one, given by the projective line pointed at infinity, say. This makes motivic cohomology bigraded: there is the classical degree and the weight.
In étale cohomology, if you restrict to a sheaf of coefficients $F$ which is $\mathbb{Z}/n\mathbb{Z}$-linear, with $n$ invertible in $\mathcal{O}_X$, the formula you seek takes the form:
$H^i_{\acute{e}t}(X,F)\cong H^{i+j}_{\acute{e}t}(X\wedge(\mathbb{G}_m)^{\wedge j},F(j))$
with $F(j)$ the $\mathbb{Z}/n\mathbb{Z}$-linear tensor product of $F$ with $\mu_n^{\otimes j}$, where $\mu_n$ is the sheaf of $n$-th roots of unity and $\mu_n^{\otimes j}$ its $j$-th tensor power.
If $X$ is defined over an algebraically closed field, we may choose a primitive $n$-th root of unity $\mu_n\cong\mathbb{Z}/n\mathbb{Z}$ and thus get a formula which looks like the one you have in classical topology.
This question has already been satisfactorily answered in the comments. To avoid it lingering around as "unanswered," I am providing a CW answer with a reference for the observations made by Denis-Charles Cisinski.
A complete treatment of etale cohomology operations is provided in Operations in étale and motivic cohomology, where, as the authors write in the abstract "We classify all étale cohomology operations." This includes lots of pointers to the early literature, by Epstein, Morel, Voevodsky, and others. For results of the form stated by Cisinski, see, e.g., section 7. For a result along the lines of the "suspension theorem" referenced in the OP, see page 5.
|
2025-03-21T14:48:31.690475
| 2020-08-06T23:54:20 |
368520
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Kakalot",
"https://mathoverflow.net/users/120300"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631782",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368520"
}
|
Stack Exchange
|
Construct a function with certain growth property
I have the following question:
Does there exist a non-negative function $g$ on $(0,1)$ such that
$$1\leq F(x):=\dfrac{\displaystyle\sum_{k=0}^{\infty}a_{k}\,(k+1)^{2}\,x^{k}}{\displaystyle\sum_{k=0}^{\infty}(k+1)\,x^{k}}\leq 2,\;\forall\;x\in (0,1),$$ and $$\displaystyle\lim_{x\rightarrow 1^{-}}F(x)\;\text{ does not exist?}$$
Here $$a_{k}:=\int_{0}^{1}g(x)\,x^{2k+1}dx.$$
Note that if $g\equiv2$ then the first condition satisfies, but the second condition doesn't.
Such a function $g$ exists.
Indeed, we have
\begin{equation*}
\sum_{k=0}^\infty(k+1)\,x^k=\frac1{(1-x)^2}
\end{equation*}
and
\begin{align*}
\sum_{k=0}^\infty a_k(k+1)^2\,x^k&=
\int_0^1 du\,g(u)\,\sum_{k=0}^\infty x^ku^{2k+1} \\
&= \int_0^1 du\,g(u)u\,\frac{1+xu^2}{(1-xu^2)^3} \\
&=\frac12\int_0^1 dv\,h(v)\,\frac{1+xv}{(1-xv)^3},
\end{align*}
where $h(v):=g(\sqrt v)\ge0$ for $v\in(0,1)$. So,
\begin{equation*}
F_h(x):=F(x)=(1-x)^2S(x)/2, \tag{-1}
\end{equation*}
where
\begin{equation*}
S(x):=S_h(x):=\int_0^1 dv\,h(v)\,\frac{1+xv}{(1-xv)^3}. \tag{0}
\end{equation*}
So, for $h(x)=h_2(x):=2$, we have $F_h=1$ and hence
the condition
\begin{equation*}
1\le F(x)\le2 \tag{1}
\end{equation*}
for $x\in(0,1)$ holds.
Consider now $\tilde h:=h_2+ph=2+ph$, where $p\in(0,\infty)$ and $h$ is a measurable nonnegative function such that $F_h$ is bounded (on $(0,1)$) and the limit $F_h(1-)$ does not exist. If the $p$ is small enough, then condition (1) will hold for $F_{\tilde h}=1+pF_h$ in place of $F$ and the limit $F_{\tilde h}(1-)$ will not exist.
Thus, it suffices to construct a measurable nonnegative function $h$ on $(0,1)$ such that $F_h$ is bounded and the limit $F_h(1-)$ does not exist. Let us show that the formula
\begin{equation*}
h(v):=\sum_{n\ge1}h_n1(a_n<v<b_n)
\end{equation*}
defines such a function, where
\begin{equation*}
a_n:=1-q_n,\quad 0<b_n-a_n=o(q_n),\quad h_n:=\frac{q_n}{b_n-a_n},\quad q_n:=\frac1{2^{2^n}}, \tag{2}
\end{equation*}
so that $q_n^2=q_{n+1}$.
Take now any $v_0\in(0,1)$. Since $h$ is bounded on $(0,v_0)$, we have $(1-x)^2\int_0^{v_0} dv\,h(v)\,\frac{1+xv}{(1-xv)^3}\to0$ as $x\uparrow1$. So, by (-1) and (0),
\begin{equation*}
F_h(x)\sim G(x):=G_h(x):=(1-x)^2\int_0^1 dv\,h(v)\,\frac1{(1-xv)^3}
\end{equation*}
as $x\uparrow1$.
Thus, it suffices to show that the function $G$ is bounded (on $(0,1)$) and the limit $G(1-)$ does not exist.
Note that
\begin{equation*}
G(x)=(1-x)^2\sum_{n\ge1} h_n \int_{a_n}^{b_n} \frac{dv}{(1-xv)^3}.
\end{equation*}
For any natural $n_0$, the sum $\sum_{n_0\ge n\ge1} h_n \int_{a_n}^{b_n} \frac{dv}{(1-xv)^3}\le\sum_{n_0\ge n\ge1} h_n \int_{a_n}^{b_n} \frac{dv}{(1-v)^3}$ is bounded and hence $(1-x)^2\sum_{n_0\ge n\ge1} h_n \int_{a_n}^{b_n} \frac{dv}{(1-xv)^3}\to0$ as $x\uparrow1$.
Also, in view of the first two conditions in (2), it is easy to see that $1-xv\sim1-xa_n$ as $n\to\infty$ uniformly in $x\in(0,1)$ and $v\in(a_n,b_n)$.
So, letting
\begin{equation*}
q:=1-x,
\end{equation*}
we see that
\begin{equation*}
G(x)\sim H(q):=\sum_{n\ge1}\frac{q^2 q_n}{(q+q_n)^3}
\end{equation*}
as $x\uparrow1$ or, equivalently, $q\downarrow0$.
Thus, it suffices to show that the function $H$ is bounded (on $(0,1)$) and the limit $H(0+)$ does not exist.
For natural $m\to\infty$ and $q\in[q_{m+1},q_m]$, write
\begin{equation*}
H(q)=H_{<m}+H_m+H_{m+1}+H_{>m+1},
\end{equation*}
where
\begin{equation*}
H_{<m}:=\sum_{n<m}\frac{q^2 q_n}{(q+q_n)^3}\sim\sum_{n<m}\frac{q^2}{q_n^2}
\sim\frac{q^2}{q_{m-1}^2}=\frac{q^2}{q_m}\le q_m\to0,
\end{equation*}
\begin{equation*}
H_{>m+1}:=\sum_{n>m+1}\frac{q^2 q_n}{(q+q_n)^3}\sim\sum_{n>m+1}\frac{q_n}q
\sim\frac{q_{m+2}}q\le\frac{q_{m+2}}{q_{m+1}}\to0,
\end{equation*}
\begin{equation*}
H_m:=\frac{q^2 q_m}{(q+q_m)^3}\le\frac{q^2}{q_m^2}\le1,
\end{equation*}
\begin{equation*}
H_{m+1}:=\frac{q^2 q_{m+1}}{(q+q_{m+1})^3}\le\frac{q_{m+1}}q\le1.
\end{equation*}
So, $H$ is indeed bounded.
Finally, for $q=q_{m+1}$
\begin{equation*}
H_m+H_{m+1}=\frac{q_{m+1}^2q_m}{(q_{m+1}+q_m)^3}+\frac18\to\frac18,
\end{equation*}
whereas
for $q=2q_{m+1}\big[\in[q_{m+1},q_m]\big]$
\begin{equation*}
H_m+H_{m+1}=\frac{4q_{m+1}^2q_m}{(2q_{m+1}+q_m)^3}+\frac4{27}\to\frac4{27},
\end{equation*}
which yields
$H(q_m)\to\frac18$ and $H(2q_m)\to\frac4{27}$. Thus, the limit $H(0+)$ does not exist.
@Pinelis Thks, I need time to digest your proof.
|
2025-03-21T14:48:31.690696
| 2020-08-01T20:29:28 |
368097
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Joseph O'Rourke",
"Penelope Benenati",
"https://mathoverflow.net/users/115803",
"https://mathoverflow.net/users/143907",
"https://mathoverflow.net/users/6094",
"mike"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631783",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368097"
}
|
Stack Exchange
|
Probability of intersecting a rectangle with random straight lines
We are given a rectangle $R$ with sides lengths $r_1$ and $r_2$, contained in a square $S$, with sides lengths $s_1=s_2\ge r_1$ and $s_2=s_1\ge r_2$. $R$ and $S$ are axis-aligned in a cartesian plane $P$. With the following recursive random process, we select straight lines orthogonal to the sides of $R$ (and $S$), until $R$ is cut.
At each time step, we select one of the two axes of $P$ with probability $\tfrac12$. Let $a$ the axis selected. Thereafter a straight line $L$ is selected uniformly at random from the ones cutting $S$ and orthogonal to $a$. Let $S'$ and $S''$ be the two parts of $S$ generated by the cut of $L$. These two random steps are repeated until $R$ is cut by $L$, and each time $R$ is not cut, $S$ is transformed by removing its part (either $S'$ or $S''$) that does not contain $R$.
Question: Given the coordinates of the vertices of $R$ providing its position within $S$, what is the probability $p_i$ that it is eventually cut (at the end of the random process) by a line orthogonal to its sides with length $r_i$ for $i\in\{1,2\}$?
(For the sake of clarity, we obviously have $p_1=1-p_2$.).
If you put the rectangle in the lower left hand corner, as in J O'Rourke's first picture, the process of cutting either side by itself is the well-known stick breaking process. The number of tries until you cut the side on the y axis is about -log(r_1/s_1), and similarly for the side on the x axis. It seems to me that the probability of hitting one side first is likely to depend on those quantities. I haven't tried to make a proof out of this.
Thank you for your comment @mike. I see your point. Would you have any suggestion about what happens then when the rectangle is not positioned in a corner of the square?
I wonder if the probability is dependent only on $r_i$, or also dependent on
the placement of $R$ within $S$?
In these two examples,
it takes an average of $2.70$ steps to reach slicing $R$ on the left, but
$3.16$ steps on the right.
I realize I'm ignoring your condition that $s_i \gg r_i$.
Added 4Aug2020.
I include below some simulation data that might help a theoretical investigation.
Here are two examples where $R = 0.2 \times 0.1$ in a unit square $S$.
On the left, after one million trials, the probability that the long
side of $R$ is sliced was $0.591$.
On the right, the probability was $0.622$.
Thank you for your comment Joseph. In my opinion the required probability is independent of the placement of $R$ within $S$, and is independent of the number of time steps to reach slicing R, but I might be wrong.
Furthermore, if the conjecture is true, an interesting generalization would be to extend it when $S$ is not a square, and it is a rectangle. In this case I think that we can obtain the same result if the probabilities to choose an axis at each time step are $\frac{s_i}{s_1+s_2}$ for $i\in{1,2}$, instead of $\tfrac12$.
@PenelopeBenenati: One aspect I don't understand. The probability of slicing $R$ is $1$: Eventually it will occur. So do you mean by "the probability," the probability that $R$ is cut on any given step? If so, then this depends on the number of steps.
@PenelopeBenenati: Re $S$ a rectangle: $S$ is a rectangle after the first step.
By writing "probability" in my comment, I mean the probability required in the problem question, namely "the probability that $R$ is sliced by a line orthogonal to its edges with length $r_i$", which I conjecture to be equal to $\frac{r_i}{\sum_i r_i}$.
Yes, $S$ is a.s. a rectangle after the first step, but I conjecture that only the shape of $S$ at time step $t=0$ matters, because all the probabilities for the events occurring a time $t\ge1$ are conditioned on the first cut(s) selected.
Starting from a very simple example, consider the special case where $r_1=s_1$ (forgetting for a moment the condition $s_1\gg r_1$ in the problem text, which I just changed because I am interesting in the more general setting where $r_1\le s_1$ and $r_2\le s_2$), and $r_2 < s_2$. Is it true in this specific case that the probability that $R$ is sliced by a line orthogonal to the edges with length $r_1$ (respectively $r_2$) is $\frac{r_1}{r_1+r_2}$ (respectively $\frac{r_2}{r_1+r_2}$)?
@PenelopeBenenati: I think I finally understand your question. $R$ will eventually be cut. You are seeking the probability that it is cut on the long side vs the short side, the $r_1$ side vs the $r_2$ side.
Thank you for your comment Joseph, yes, this is exactly what I meant. Do you think I should rephrase the question in the problem text, because it is not sufficiently clear? I accept any suggestion. Thank you.
@PenelopeBenenati: Ignoring the $s_i \gg r_i$ condition, simulations suggest: (1) the conjecture is not precisely true; (2) the probability depends on the position of $R$ within $S$. But (a) this is from simulations, and (b) still your conjecture may hold for $s_i \gg r_i$.
[The software is saying the comments should cease.]
Thank you again Joseph for your comments. Based on them I have now rephrased the question, which is finally much more significant.
|
2025-03-21T14:48:31.691069
| 2020-08-01T21:11:32 |
368101
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Francesco Polizzi",
"Kamel",
"https://mathoverflow.net/users/163078",
"https://mathoverflow.net/users/7460"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631784",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368101"
}
|
Stack Exchange
|
Mori's cone theorem
I need the proof (reference) of Mori’s theorem about this implication :
Let $X$ be a projective complex manifold. If $X$ contains no rational curves, then $K_K$ is nef.
A more precise form of this result is originally due to Miyaoka and Mori, see Theorem 3.6 page 67 in
Olivier Debarre: Higher-dimensional algebraic geometry, Universitext. New York, NY: Springer. xiii, 233 p. (2001). ZBL0978.14001.
Thank you, but the problem that
It is not available on the internet
Do you have a pdf version
I have a paper copy. At any rate, you can find the same result in these GAEL notes by Debarre: https://www.math.ens.fr/~debarre/NotesGAEL.pdf
Okay, thank you so much
|
2025-03-21T14:48:31.691158
| 2020-08-01T22:17:39 |
368104
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"5W1H",
"Asvin",
"Nulhomologous",
"https://mathoverflow.net/users/158462",
"https://mathoverflow.net/users/163079",
"https://mathoverflow.net/users/58001"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631785",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368104"
}
|
Stack Exchange
|
Finding $K$-rational points on $X_0(35)$
Let $K=\mathbb{Q}(\sqrt{-2})$. How can I compute the $K$-rational points on the modular curve $X_0(35)$?
Recall that $X_0(35)$ is a hyperelliptic curve of genus 3 and has the simplified affine model:
\begin{align*}
y^2&=x^8-4x^7-6x^6-4x^5-9x^4+4x^3-6x^2+4x+1\\
&=(x^2+x-1)(x^6-5x^5-9x^3-5x-1)
\end{align*}
My attempt at finding $K$-rational points on $X_0(35)$ is as follows: First I find a rational map $f$ from $X_0(35)$ to a quotient curve $E$ of $X_0(35)$ with $E$ an elliptic curve (which is induced by the involution $w_5$ by Kubert). Second, I determine the preimages of $E(K)$ under $f$. If $E$ is of rank 0, $E(K)$ is finite. However, in my case $E(K)$ is of rank 1. As a result it is computationally infeasible to determine the preimages of the infinitely many points of $E(K)$. Is there a way to work-around this issue?
Any help in finding $K$-rational points on $X_0(35)$ would be appreciated.
Can you try and brute force enough points and then hope that the Chabauty-Coleman bound is sharp?
I'm not familiar with the Chabauty-Coleman technique but I will try to make it work out. Thank you for your input.
The group $J_0(35)(\mathbb Q)$ (where $J_0(35)$ is the Jacobian
of $X_0(35)$) has rank 0 (as shown for example by a 2-descent
computation in Magma); it is isomorphic to
${\mathbb Z}/24{\mathbb Z} \times {\mathbb Z}/2{\mathbb Z}$,
with generators the difference of the two points at infinity
on $X_0(35)$ and the 2-torsion point corresponding to the
factorization of the polynomial on the right hand side of
the hyperelliptic equation. (One can check that the two points
generate a group of the given isomorphism type, and the
reduction of $J_0(35)$ mod 3 has a group of this type as its
group of ${\mathbb F}_3$-points.)
If $P$ is a point in $X_0(35)(K)$ with $x$-coordinate not
in $\mathbb Q$, then the sum of $P$ with
its Galois conjugate, minus the sum of the two points at
infinity, gives rise to a nonzero $\mathbb Q$-rational point on
$J_0(35)$. One can check that none of the points is of this
form.
This leaves the case when $x(P) \in \mathbb Q$. There are the
$\mathbb Q$-rational points (the two points at infinity
and the two points with $x(P) = 0$); for all other such points,
the Galois conjugate of $P$ must be the image of $P$ under
the hyperelliptic involution, so $y(P)$ must be $\sqrt{-2}$
times a rational number. Put differently, $P$ gives rise
to a $\mathbb Q$-rational point on the quadratic twist of
$X_0(35)$ by $-2$. But this twist does not even have
${\mathbb Q}_2$-rational points, let alone $\mathbb Q$-rational
ones.
The conclusion is that
$$ X_0(35)(K) = X_0(35)({\mathbb Q}) = \{\infty_+, \infty_-, (0,1), (0,-1)\} . $$
I wasn't sure how to deal with the case where x-coordinate is not in $\mathbb{Q}$. Thank you so much for clarifying my confusion.
So, the argument shows that for any quadratic field $K=\mathbb{Q}(\sqrt{d})$, where $d\in \mathbb{Z}$ is square free, $X_0(35)(K)=X_0(35)(\mathbb Q)$ if and only if the quadratic twist of $X_0(35)$ by $d$ does not have any rational point. Even more,$$X_0(35)(K)=X_0(35)(\mathbb Q)\cup X_0^{\langle d\rangle }(35)(\mathbb Q)$$ where the union is disjoint and $X_0^{\langle d\rangle }(35)(\mathbb Q)$ denotes the quadratic twist by $d$ of $X_0(35)(\mathbb Q)$
|
2025-03-21T14:48:31.691396
| 2020-08-01T22:55:17 |
368106
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Ali Taghavi",
"LSpice",
"https://mathoverflow.net/users/2383",
"https://mathoverflow.net/users/33741",
"https://mathoverflow.net/users/36688",
"leo monsaingeon"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631786",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368106"
}
|
Stack Exchange
|
Is it a sufficient condition for linearity?
Edit: According to the comment by LSpice we come back to the initial version of this question
Let $f:\mathbb{R}^n\to \mathbb{R}^n$ be a smooth function such that for every $x\in \mathbb{R}^n$ the derivative $A=Df(x)$ satisfies the condition $a_{ii}=1$ and $a_{ij}=k(x),\; \forall i\neq j$ where $k(x)$ depends only on $x$. Does it imply that $f $ is a linear map?
A some what similar form of this linear part is considered here
https://qr.ae/pNsOY1
If $f$ were linear its differential would be constant, so your $k(x)$ actually cannot depend on $x$ if you believe your claim to be true. Or am I missing smomething?
@leomonsaingeon you are right but in the question we do not assume $k(x)$ is constant. BTW please read the revised version since I insist that the linear part would be ientified to the original quora question.
@leomonsaingeon In the other word is there a non linear map whose pointwise linear part is in the form of linked quora question?
Questions should be self contained. I think it would be particularly desireable in this case to include whatever is relevant from the Quora question, because it's not clear, at least to me: are you imposing an extra hypothesis, or referring us (for some reason) to a computation of a determinant, or something else?
@LSpice I think the question is clear. The quora link is added just to keep a kind of copy rigth because that quora question leads me to arrive at this question. I do not understand why the question is not self contained?
It's not clear what the significance of the off-diagonal entries is; if you consider instead $g$, where the $i$th component function of $g$ is $x \mapsto f_i(x) - \sum_{j \ne i} x_j$, then $Dg(x)$ is everywhere scalar, so you might as well ask whether a function with everywhere-scalar derivative is linear.
I don't know if the question is self contained, because you say "I insist that the linear part would be i[d]entified to the original quora question". Since you asked for a re-reading on that basis, I assumed you were meaning to add additional information.
Also, notice that your hypothesis on the derivative is always satisfied when $n = 1$.
@LSpice yes you are right. May be this is the reason that in the first version of the question I considered the case $a_{ii}=1$ and $a_{ij}=k(x)$.
@LSpice The first version was the question originaly aroused me a few days ago
@LSpice Thank you for your attention.I revise the question
@LSpice If I would not refere to the quora question, I am almost sure that a participant would comment "from where you arrive at this form of linear part?"
I answer the version where you assume that the diagonal entries of $Df$ are constant (all $1$, in your problem), and the off-diagonal entries of each $Df(x)$ are all the same value $k(x)$.
I assume that, by linear, you mean affine linear. Then this is clear if $n = 1$. Otherwise, fix $i$, let $f_i$ be the appropriate component function of $f$, and choose any $j \ne i$. Since $k$ equals $\frac\partial{\partial x_j}f_i$ and $\frac\partial{\partial x_i}f_i$ equals $1$, we have that $\frac\partial{\partial x_i}k = \frac\partial{\partial x_j}1 = 0$. That is, $k$ is constant, so $Df$ is constant, so $f$ is affine linear.
thank you very much for your answer. Actually you answered both version as you indicated in your comment for n=1.
Is it easy to classify all smooth maps whose derivative A satisfies $a_{ii}=L(x)$ and $a_{ij}=K(x)$?
I don't know, but, at least for $n \ge 3$, we do have by the same trick that all partial derivatives of $L$ are equal, all partial derivatives of $K$ are equal, and they are all equal to each other.
(Note again that your new condition is no restriction when $n = 1$.)
@AliTaghavi, given a smooth map $f$ like that, replace $f$ by $f - f(0)$ so $f(0) = 0$, then consider $F$ with component functions $F_i \colon x \mapsto \frac1{n!}\sum_{\sigma \in \operatorname S_n} f_{\sigma i}(\sigma x)$. Then $DF$ equals $Df$ everywhere and $F(0) = 0 = f(0)$, so $F = f$. So I think: those functions that, after normalised by $f(0) = 0$, satisfy $f_i(x) = f_{\sigma i}(\sigma x)$, with appropriate quantification. (No condition for $n = 1$, as it should(n't) be.)
|
2025-03-21T14:48:31.691696
| 2020-08-01T23:50:07 |
368110
|
{
"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:631787",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368110"
}
|
Stack Exchange
|
if such counter example exists for Lehmer's totient problem could we prove that there are infinity of them or just finitely?
I asked this question one month Ago in MSE but no answer for existence of argument which show if such counter example exists we would have infinity of them or just finitely many examples
Lehmer's totient problem asks whether there is any composite number $n$ such that Euler's totient function $φ(n)$ divides $n − 1$. which it is unsolved problem or we may reformulate that question as : if $φ(n)$ divides $n − 1$ then $n$ must be a prime , Now my question here is :if a such counter example exists for Lehmer's totient problem could we prove that there are infinity of them or just finitely of them ?
|
2025-03-21T14:48:31.691770
| 2020-08-02T01:11:04 |
368111
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Will Sawin",
"curious math guy",
"https://mathoverflow.net/users/152554",
"https://mathoverflow.net/users/18060"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631788",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368111"
}
|
Stack Exchange
|
$p$-adic Galois representation and Étale homology
Let $X$ be a smooth proper scheme over some $p$-adic field $K$. The "usual" way to get a Galois representation out of this is to consider the étale cohomology (either $p$ or $\ell$-adic). Then $p$-adic Hodge theory tells me that these are "nice" by which I mean Hodge-Tate.
Another way that I'd think of would be to consider the étale homology $H_{n}(X_{\overline{K}},\mathbb{Q}_p)$. Do we know anything about these representations, for instance, are they Hodge-Tate? Are they "interesting" (i.e. non-trivial?).
Aren't these just the duals to étale cohomology?
As I understand it, yes. But are "niceness" properties preserved under taking duals?
The only one I can think of that isn't is that Frobenius eigenvalues are algebraic integers (On the duals, they're inverses of algebraic integers).
|
2025-03-21T14:48:31.691975
| 2020-08-02T02:15:02 |
368113
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Asvin",
"Harry Gindi",
"R. van Dobben de Bruyn",
"Yuhang Chen",
"https://mathoverflow.net/users/1353",
"https://mathoverflow.net/users/146366",
"https://mathoverflow.net/users/58001",
"https://mathoverflow.net/users/82179"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631789",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368113"
}
|
Stack Exchange
|
The coarse moduli space of a weighted projective line
Fix two positive integers $a$ and $b$. Consider a weighted projective line $\mathbb{P}(a,b)$ as a quotient stack
$$[(\mathbb{C}^2-\{0\})/\mathbb{C}^*]$$
where $\mathbb{C}^*$ acts on $\mathbb{C}^2-\{0\}$ by $$\lambda \cdot (x,y) = (\lambda^a x,\lambda^b y).$$
Then $\mathbb{P}(a,b)$ has its underlying quotient space $|\mathbb{P}(a,b)| = (\mathbb{C}^2-\{0\})/\mathbb{C}^*$ as a coarse moduli space. It's known that $\mathbb{P}(a,b)$ also has the usual projective line $\mathbb{P}^1$ as its coarse moduli space (which I believe is not true in higher dimensions).
What's the morphism $\mathbb{P}(a,b) \to \mathbb{P}^1$ corresponding to the coarse moduli space $\mathbb{P}^1$? In other words, how to show that $|\mathbb{P}(a,b)|$ is homeomorphic to $\mathbb{P}^1$?
Isn't the coarse moduli space of any weighted projective space just projective space?
I think you send $(x,y) \to (x^b,y^a)$? I believe the obvious generalization to higher dimensions works too.
@HarryGindi Do you have a reference for this result?
@Asvin This is a well-defined map. But I think it's not bijective in the level of quotient spaces.
@HarryGindi: this is not true in general (but ok in dimension $1$). They are all toric varieties, and so can be described by their fans. I believe that $\mathbf P(d_0,\ldots,d_n)$ is isomorphic to $\mathbf P^n$ if and only if it is smooth; but already things like $\mathbf P(1,1,2)$ are singular! See for example Dolgachev's Weighted projective varieties (available on his website), 1.2.3 and Prop. 1.3.3(iii), complemented with Fulton's Toric varieties, §2.2.
This is also the whole point of using stacky blowups in the new approach to resolution of singularities in characteristic $0$ by Abramovich–Temkin–Włodarczyk and McQuillan: unlike the course spaces, the stacky weighted projective spaces are smooth.
I think @Asvin's suggestion is right, except you need to divide by their $\gcd$ (which is often assumed $1$). The proof is by computing the (GIT) invariants, or computing the $\operatorname{Proj}$ of the obvious graded ring. There is no "obvious generalisation to higher dimension".
@R.vanDobbendeBruyn Do you have a reference for the proof?
It seems to be a special case of Prop. 1.3.1 in Dolgachev's paper (linked above). (Typo: I think it should be $q_0/a_0$, not $q_0/a$. The latter is not an integer.) Specifically, see the second corollary of 1.3.1.
@R.vanDobbendeBruyn Thanks!
@R.vanDobbendeBruyn Ah yeah, I think the thing I was thinking of was that the coarse moduli space is always Proj of the weighted polynomial ring.
|
2025-03-21T14:48:31.692165
| 2020-08-02T04:23:12 |
368116
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Emil Jeřábek",
"Mohsen Shahriari",
"https://mathoverflow.net/users/12705",
"https://mathoverflow.net/users/76416"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631790",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368116"
}
|
Stack Exchange
|
Disjunction in weakened Robinson arithmetic
Let $ T $ denote the theory obtained by removing the axiom $ \forall x ( x = 0 \lor \exists y \, S y = x ) $ and restricting double negation elimination to disjunction-free formulas of Robinson arithmetic. In other words, $ T $ is axiomatized over intuitionistic logic by arithmetical axioms of Robinson arithmetic other than $ \forall x ( x = 0 \lor \exists y \, S y = x ) $, and double negation elimination for disjunction-free formulas.
I've been playing around with $ T $ for quite a while (actually months, on and off), and it seems that no genuine theorems containing a positive disjunction can be proven in $ T $. By genuine I mean those theorems that are not proven by mere logic, like those of the form $ A \rightarrow A \lor B $, or $ A \lor B \rightarrow C \lor D $ where $ A $ implies $ C $ and $ B $ implies $ D $. For example, I couldn't find any way of proving $ \forall x ( x < 2 \rightarrow x = 0 \lor x = 1 ) $. As neither induction nor classical logic is at hand, non of the usual ways of thinking seems to work, at least as far as I could check. I'm really not sure about unprovability of this sentence, as I couldn't find any useful method to show that it's not provable.
So to avoid any possible complexity that could come from considering more general theorems containing positive disjunctions, I ask my question like this:
Is there a method for showing that $ \forall x ( \exists y \, x + S y = 2 \rightarrow x = 0 \lor x = 1 ) $ is provable/unprovable in $ T $?
EDIT:
After @MattF.'s comment, I'm inclined to add that I've been investigating theories other that $ T $, too. For example by adding to $ T $ the axiom $ \forall x ( x \ne 0 \rightarrow \exists y \, S y = x ) $, or other disjunction-free axioms. It's obvious that in the presence of classical logic, these extra axioms trivially imply the above sentence. I first thought that it would be better not to mention these other theories, to keep the question simpler. But if answering the question is easier for these theories than for $ T $, it is desired as well.
EDIT:
To make my thoughts even clearer, I've become suspicious that in the absence of classical logic and extra axioms containing positive disjunctions, the source of all genuine arithmetical theorems containing positive disjunctions is the axiom $ \forall x ( x = 0 \lor \exists y \, S y = x ) $. So I could ask my question like this:
Is there a set $ \Gamma $ of disjunction-free axioms such that $ T + \Gamma \vdash \forall x ( x < 2 \rightarrow x = 0 \lor x = 1 ) $?
@MattF. Yes, it includes that implication.
$\quad$
I don't know if I can prove it from these axioms using classical logic. The mention of classical logic in the "at hand" part of the post was due to the fact that I've been investigating theories other than $T$, too. For example, by adding to $T$ the axiom $x\ne0\to\exists y,Sy=x$ and other disjunction-free axioms. In the presence of classical logic, those trivially lead to proving the statement. I thought it would be better to keep things simpler by not mentioning those theories. But an answer addressing those theories is desired as well.
It is a general fact about intuitionistic logic that disjunction-free theories have the disjunction property. Thus, assuming $T+\Gamma\vdash\forall x,(x<2\to x=0\lor x=1)$, let $c$ be a fresh constant; the disjunction-free theory $T+\Gamma+c<2$ proves $c=0\lor c=1$, hence by the DP, it proves $c=0$ or $c=1$. Thus, $T+\Gamma$ proves $\forall x,(x<2\to x=0)$ or $\forall x,(x<2\to x=1)$. Hopefully, $T$ still proves $0<2$, $1<2$, and $0\ne1$, which means $T+\Gamma$ is inconsistent.
(More generally, the disjunction property holds for “$\lor$-Harrop” theories: axiomatized by formulas in which disjunction can only occur inside antecedents of implications.)
@EmilJeřábek Well, the logic of $T+\Gamma$ is not exactly intuitionistic logic, as it has axioms of the form $\neg\neg A\to A$ for disjunction-free $A$. These axioms are not readily realizable, and similar considerations seem to suggest that other usual techniques of proving disjunction property (Kleene-Aczels'slash, adding nodes as roots of Kripke models, an such) won't work either. Will you please illustrate how to prove disjunction property in this case?
The axioms $\neg\neg A\to A$ are $\lor$-free, hence the result applies to them. And the general statement can be proved in the same way as the more commonly stated fact that theories axiomatized by Harrop formulas have the DP: by induction on the size of a cut-free proof in the intuitionistic sequent calculs allowing at most one formula in the succedent. (Some variant of Kleene’s slash likely works, too, if you make $\exists$ “non-constructive”. That is, redefine $|\exists x,\phi$ as $\exists x,|\phi$.)
@EmilJeřábek I got it. Thanks.
On second thought, I’m not sure the Kleene slash argument will work. But in any case, the proof using cut-free sequent calculus is quite simple.
|
2025-03-21T14:48:31.692496
| 2020-08-02T05:31:13 |
368118
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"DSM",
"https://mathoverflow.net/users/155380"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631791",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368118"
}
|
Stack Exchange
|
Is the Loewner maximum uniquely defined?
Given 2 (symmetric) PSD matrices $A,B$, is the following set $S_{A,B}$ non-empty?
$$ S_{A,B} = \{ C: C\succeq A, C\succeq B, \text{ and }\forall D, D\succeq A, D\succeq B \implies D\succeq C \} $$
If $S_{A,B}$ is non-empty, it has to be unique due the Loewner order being antisymmetric. Such an element would, in some sense, would constitute a PSD maximum of $A,B$.
If $S_{A,B}$ is non-empty, is there a poly-time algorithm to compute it?
Wondering how would this work for matrices $A$ and $B$ which commute. In that case, one can do a simultaneous diagonalization. Say $A=UD_AU^H$ and $B=UD_BU^H$. Then would $C = U \max{D_A, D_B} U^H$ work?
Kadison in this paper proves that self-adjoint $A$ and $B$ will have a greatest lower bound if and only if they are comparable (meaning $A\leq B$ or $B\leq A$ or $A=B$). Another way to put this is that the Loewner order is an anti-lattice.
Since this is for all self-adjoint then one can flip the order by multiplying by $-1$ to get that $A$ and $B$ will have a least upper bound if and only if they are comparable.
|
2025-03-21T14:48:31.692608
| 2020-08-02T06:30:00 |
368121
|
{
"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:631792",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368121"
}
|
Stack Exchange
|
Comparing generic versions of $\mathbb{R}$
This question was previously asked and bountied at MSE, unsuccessfully.
I'm currently interested in the behavior of cardinalities in generic extensions of models of $\mathsf{ZF+\neg AC}$, and especially of $\mathsf{ZF+AD}$. The following seems both natural and much simpler than various other questions I've asked on the topic:
Suppose $V$ is a transitive model of $\mathsf{ZF+AD}$. Let $c,d$ be mutually Cohen generic over $V$; is there in $V[c,d]$ a bijection between $\mathbb{R}^{V[c]}$ and $\mathbb{R}^{V[d]}$?
"Obviously" the answer should be yes, but I don't see how to prove that. We have a set $\mathcal{X}\in V$ of names for reals such that every real in a Cohen extension is named by some element of $\mathcal{X}$, and each Cohen real $a$ induces an equivalence relation $\sim_a$ on $\mathcal{X}$ as $\nu\sim_a\mu\leftrightarrow\nu[a]=\mu[a]$. So really this question is asking for a bijection in $V[c,d]$ between $\mathcal{X}/\sim_c$ and $\mathcal{X}/\sim_d$. Intuitively this should exist since Cohen forcing is as homogeneous as one could hope; however, in the absence of choice in $V[c,d]$ I don't actually see how to build one.
I would also be very interested in a partial negative answer for $\mathsf{ZF}$-models, but the determinacy case is really my main point of focus.
To preempt one natural attempt, note that Cohen forcing kills determinacy so we can't use determinacy in $V[c,d]$ even though we have it in $V$. While at first glance this might appear to contradict (say) the generic absoluteness of the theory of $L(\mathbb{R})$ given large cardinals, there is no discrepancy since $(L(\mathbb{R}))^V[G]\not=(L(\mathbb{R}))^{V[G]}$ in general.
The answer seems to be no. Moreover: Suppose that every set of reals has the property of Baire. Let $\mathbb{C}$ be Cohen forcing and let $P$ be any wellorderable partial order. If $(c,d)$ is generic for $\mathbb{C} \times P$, then there is no injection in $V[c,d]$ from the Cantor space of $V[c]$ to any set in $V[d]$.
Here's a proof, which is a variation on the standard argument showing that if all sets of reals have the property of Baire, then there is no function choosing between complementary $\mathbb{E}_{0}$ degrees. We will call members of the Cantor space reals. Suppose that $\tau$ is a $\mathbb{C} \times P$ name for a function mapping the reals of $V[g_{left}]$ into $V[g_{right}]$, where $g_{left}$ and $g_{right}$ are the $\mathbb{C} \times P$-names for the left and right coordinates of the generic. We will find a condition forcing the realization of $\tau$ not to be injective.
Let $N$ be the set of nice Cohen names $\sigma$ for elements of the Cantor space which are also functions with domain $\mathbb{C}$, where each value $\sigma(p)$ has the form $\check{t}$, for $t$ an element of $\mathbb{C}$ of the same length as $p$. That is, $N$ is the set of nice $\mathbb{C}$-names for reals such that each condition of $\mathbb{C}$ decides the initial segment of the realization of the name up to the length of the condition (and maybe more).
Let $\langle q_n : n \in \omega \rangle$ be the natural enumeration of $\mathbb{C}$, where shorter sequences are listed before longer ones, and sequences of the same length are listed in lexicographic order (any enumeration in ordertype $\omega$ would work). Given $p \in \mathbb{C}$, let $n_p$ be such that $p = q_{n_{p}}$. There is a natural bijection $b$ between the Cantor space and $N$, where we let $x$ in the Cantor space correspond to the set of pairs $(p, \check{t})$, where $t = \langle x(n_{p \upharpoonright i}) : i \leq |p| \rangle$.
Applying the assumption that every subset of the Cantor space has the property of Baire, and the Kuratowski-Ulam theorem to deal with the wellorderable poset $P$, we get $s$ in $\mathbb{C}$ and $(p_0, p_1) \in \mathbb{C} \times P$ such that, for comeagerly many $x$ extending $s$, there is some $\mathbb{C}$-name $\rho$ with $(p_0,p_1)$ forcing $(b(x)_{g_{left}}, \rho_{g_{right}}) \in \tau$. That is, $(p_0,p_1)$ forces some $P$-name $\rho$ to represent via $g_{right}$ the $\tau$-value for the $g_{left}$-realization of $b(x)$. However, we can find $x$ and $x'$ extending s and in this comeager set such that $x$ and $x'$ disagree at exactly one value $n^*$ with $q_{n^*}$ compatible with $p_0$, and with this $q_{n^*}$ a proper extension of $p_0$. So $p_0$ doesn't decide whether or not the realizations of $b(x)$ and $b(x')$ will be the same.
Moreover, there exist $P$-names $\rho$ and $\rho'$ such that $(p_0,p_1)$ forces both
$(b(x)_{g_{left}}, \rho_{g_{right}}) \in \tau$
and
$(b(x')_{g_{left}}, \rho'_{g_{right}}) \in \tau$.
Let $p'_1$ be a strengthening of $p_1$ deciding whether or not $\rho$ and $\rho'$ have the same realization.
Since there are extensions of $p_0$ forcing that the realizations of $b(x)$ and $b(x')$ will be the same, and $\tau$ is a name for a function, it must be that $p'_1$ forces the realizations of $\rho$ and $\rho'$ to be the same. The condition $q_{n^*}$ however extends $p_0$ and forces that the realizations of $b(x)$ and $b(x')$ will be different, and therefore $(q_{n^{*}}, p'_1)$ forces that the realization of $\tau$ will not be an injection.
|
2025-03-21T14:48:31.692955
| 2020-08-02T06:49:12 |
368123
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Dry Bones",
"Noah Schweber",
"https://mathoverflow.net/users/8133",
"https://mathoverflow.net/users/84756"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631793",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368123"
}
|
Stack Exchange
|
Can a generic $\mathbb{R}$ have a new cardinality?
This question was asked and bountied at MSE, without success.
My main question is whether, starting with a model of determinacy, a "generic $\mathbb{R}$" could be different in cardinality from any ground model set:
Is there a well-founded $V\models\mathsf{ZF+AD}$, a forcing $\mathbb{P}\in V$, and a $G$ which is $\mathbb{P}$-generic over $V$ such that $V[G]\models$ "There is no bijection between $\mathbb{R}^{V[G]}$ and any set in $V$"?
In case the answer is yes, there is a natural follow-up question - whether the above can happen "canonically:"
Are there $V\models\mathsf{ZF+AD}$, $\mathbb{P}\in V$, and $G,H$ mutually $\mathbb{P}$-generic over $V$ such that $V[G\times H]$ satisfies "$\mathbb{R}^{V[G]}$ and $\mathbb{R}^{V[H]}$ are in bijection with each other but are not in bijection with any set in $V$?"
If the answer to this question is yes, that would give a very surprising answer to this old question of mine. I suspect that the first question has an affirmative answer and strongly suspect that the second question has a negative answer, but I don't see how to prove either point.
Here are a couple quick comments:
Since this is only interesting if we add reals, $\mathsf{AD}$ will not be preserved. So determinacy doesn't give us a lot of "leverage" in the forcing extension that I can see.
To preempt worries about triviality, we can have $V[G]\models$ "There is some $a$ which is not in bijection with any $b\in V$." For example, Monro showed that we can have an amorphous set in $V[G]$ even if there are no amorphous sets in $V$; since amorphousness is downwards-absolute, any amorphous set in such a $V[G]$ is not (in $V[G]$, anyways) in bijection with any set in $V$. (This reference was pointed out to me by Asaf Karagila.)
That said, questions of this type are always trivial over $\mathsf{ZFC}$ since forcing preserves choice and adds no new ordinals.
But that said, the previous bulletpoint is quite fragile and I don't see that it gives any insight into my question - there's no obvious replacement for $Ord$ that I see here to serve the same role as something simultaneously "invariant" and "universal," especially since it would have to be "generically universal."
What reference would you indicate for a beginner in Logic willing to understand what is being asked? =]
@DryBones Unfortunately this really isn't accessible without a significant amount of experience - specifically, comfort with forcing is the key ingredient (although you also need to be comfortable with the axiom of determinacy, that's somewhat a secondary concern - the question is still interesting if we replace $\mathsf{ZF+AD}$ with just $\mathsf{ZF}$). Forcing is treated for example in Kunen's old book "Set theory and independence proofs;" this and this are also good.
Forcing, however, relies on a mastery of the basics of set theory and model theory. So it's quite a long road.
Ok, I can imagine. Thanks, anyway!
@DryBones To be fair I might be overemphasizing the role of model theory. It's not really directly needed if you're willing to take a couple things on faith. However, the basic idea behind forcing is that we're building models of a certain very complicated theory (namely $\mathsf{ZF}$ or similar), and understanding basic properties of models will help dispel many (very reasonable) initial worries. So personally I wouldn't approach forcing without being comfortable with the downward Lowenheim-Skolem, compactness, completeness, and incompleteness theorems - the experience that confers will help.
I'm taking a look at Kunen's book and I like it! Though I'm still far from properly 'understanding' your question in detail, I'm compelled to ask if you know the answer for, say, ${\sf ZF}+\lnot{\sf AD}$. I mean, should the construction of $V[G]$ rely on some property of $V$ that depends on $V\models{\sf AD}$ ? Maybe this question doesn't make sense, but showing me why will probably help a bit on finding my way through this labyrinth! =]
@DryBones The construction of $V[G]$ makes sense as long as $V\models\mathsf{ZF}$ - or indeed vastly less ($\mathsf{KP}$ is more than enough). Per the end of my question, I know the answer - for silly reasons - if we consider $V\models\mathsf{ZFC}$. Meanwhile, I know basically nothing about what happens if we consider $V\models \mathsf{ZF+\neg AC}$. So why $\mathsf{AD}$? Well, many choiceless models are truly terrible (e.g. the reals can be a countable union of countable sets); amongst the $\mathsf{ZF+\neg AC}$-models, the $\mathsf{AD}$-models are in many ways very nicely behaved. (cont'd)
Moreover, determinacy models show up as natural submodels ("inner models") of choice models - e.g. assuming large cardinals, $L(\mathbb{R})$ satisfies $\mathsf{ZF+DC+AD}$. In a sense, determinacy is how choice fails "canonically." So looking at models of $\mathsf{ZF+AD}$ (or its ilk) is a natural thing to do here. Unfortunately, the question of how much we can conclude about $V[G]$ from knowing that $V\models\mathsf{AD}$ is a complicated one. In particular, unlike choice determinacy is not preserved by forcing. Forcing over $\mathsf{AD}$-models is as far as I'm aware still not well-understood.
Thanks, as expected your answer gives me a lot more to learn about! But if I got it partially, then it's not clear how $V\models{\sf AD}$ might help to choose $\Bbb P$ and $G$ with the desired properties, is that right?
@DryBones No it is not, at least to me. There are things one might try on the general grounds that they're usually relatively easy and instructive - e.g. Cohen forcing - but nothing stands out.
|
2025-03-21T14:48:31.693341
| 2020-08-02T07:17:19 |
368125
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Hannes",
"Jochen Wengenroth",
"LSpice",
"Pietro Majer",
"Zac",
"https://mathoverflow.net/users/157076",
"https://mathoverflow.net/users/21051",
"https://mathoverflow.net/users/2383",
"https://mathoverflow.net/users/6101",
"https://mathoverflow.net/users/85906"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631794",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368125"
}
|
Stack Exchange
|
$L^p$ compactness for a sequence of functions from compactness of cut-off
Fix $p \in [1,\infty)$. Let $f_n:[a,b] \to \mathbb R$, $n \in \mathbb N$, be a sequence of $C^1$ functions. For every fixed $m\in \mathbb N^*$, suppose that the sequence of functions $$\{f_{n}\psi_m(f_n)\}_{n \in \mathbb N}$$ has a strongly convergent subsequence in $L^p([a,b])$. Here $\psi_m$ is a smooth cut-off function such that
$$\psi_m(f) =
\begin{cases}
1 \qquad \text{ if } |f|\ge 1/m \\
0 \qquad \text{ if } |f|\le 1/(2m)
\end{cases}
$$
and $0 \le \psi_m \le 1$.
Suppose in addition that $\Vert f_n \Vert_{L^p} \le C$ (for a constant $C>0$ that does not depend on $n,m$ and for all $p \in [1,\infty]$) and suppose that every subsequence $\{f_{n_k}\psi_m(f_{n_k})\}_{n_k}$ is also compact in $L^p$ for any fixed $m$.
How can we prove that $\{f_n\}_{n\in \mathbb N}$ also has a strongly convergent subsequence in $L^p([a,b])$?
Under these assumptions, can we prove the result following this argument, which is rewritten below?
For any $f$,
$$ f - f \psi_m (f) = \begin{cases} f & \text{if } |f| \le 1/2m,\\ 0 &
\text{ if } |f| \ge 1/m.\end{cases}$$
In particular,
\begin{align*} \int |f - f \psi_m (f)|^p &= \int_{|f| <1/m} |f - f
\psi_m (f)|^p \\ &\le \int_{|f|\le 1/2m} |f|^p + \int_{1/2m \le
|f|<1/m} |f - f \psi_m (f)|^p \\ &\le \frac{b-a}{(2m)^p} +
\frac{(b-a)}{m^p}\\ \Rightarrow \|f - f \psi_m (f)\|_{L^p} &< C/m
\end{align*}
where $C$ depends on $b-a, p$ only. Note we used $|1-\psi_m|\le 1$.
Then using a diagonal argument, there is a subsequence $\{f_{n_k}\}$
of $\{f_n\}$ and $f\in L^p [a, b]$ so that for each $m$, the sequence
$\{ f_{n_k} \psi_m (f_{n_k})\}$ converges to $f$ in $L^p$. Now we show
that $\{f_{n_k}\}$ also converges to $f$ in $L^p$: for any
$\epsilon>0$, fix one $m\in \mathbb N$ with $C/m < \epsilon/2$. Since
$\{ f_{n_k} \psi_m (f_{n_k})\}$ converges to $f$ in $L^p$, there is
$K$ so that $\| f_{n_k} \psi_m (f_{n_k}) - f\| _{L^p} < \epsilon/2$
for all $k\ge K$. Then
\begin{align*} \|f_{n_k} -f\|_{L^p} \le \|f_{n_k} - f_{n_k} \psi_m
(f_{n_k}) \|_{L^p} + \| f_{n_k} \psi_m (f_{n_k}) -f\|_{L^p} <
\epsilon/2 + \epsilon/2 \end{align*}
for all $k\ge K$.
If the above fails, you can also add the assumption $\Vert D_x(f_n\psi_m(f_n))\Vert_{L^1} \le C_m$, where $C_m$ is a constant that depends only on $m$.
This question is motivated by two previous questions on Math Stack Exchange [A]
"how can we prove": we can't because it is not true. In your hypotheses , (f_n) can be unbounded in any L^p for p>1.
@PietroMajer Thanks, I've edited the question.
I think one shouldn't rely on content remaining indefinitely in the formatting sandbox. (Remember MO and other SE sites are meant for long-term archiving.) Probably better to move it here.
@LSpice I put the link just to give credit to the original writer of this proof, but I rewrote it in the body of the question to preserve it in case of changes to the Sandbox link
Sounds good to me. (I didn't realise you weren't the author.)
Let $$g^m_n := f_n \psi_m(f_n).$$ The assumptions mean that $(f_n)_n$ is a bounded sequence in $L^p(a,b)$ and that $(g_n^m)_n$ is relatively compact in $L^p(a,b)$ for each $m$. We use the Frechet-Kolmogorov theorem characterizing compactness in $L^p$ spaces to show that this transfers to $(f_n)_n$. (Then not only $(f_n)_n$ has a convergent subsequence, but also every subsequence $(f_{n_k})_k$.)
Split and estimate
\begin{align*}\|\tau_h f_n - f_n\|_{L^p(a,b-h)} &\leq \|\tau_h(f_n-g_n^m)\|_{L^p(a,b-h)} + \|\tau_h g_n^m - g_n^m\|_{L^p(a,b-h)} \\ & \qquad + \|g_n^m - f_n\|_{L^p(a,b-h)} \\ & \leq \|\tau_h g_n^m - g_n^m\|_{L^p(a,b-h)} + 2\|g_n^m - f_n\|_{L^p(a,b)}.\end{align*}
(I use $(\tau_h g)(x) := g(x+h)$ because it is more familar to me.)
Fix $\varepsilon > 0$. Choose an $m$ large enough such that $\|g_n^m - f_n\|_{L^p(a,b)} < \varepsilon/3$ for all $n$ (calculation in OP). Now, for the chosen $m$, choose $h_0$ small enough such that $\|\tau_h g_n^m - g_n^m\|_{L^p(a,b-h)} < \varepsilon/3$ for all $h \leq h_0$, uniformly for all $n$; this is possible by the Frechet-Kolmogorov theorem. ($(g_n^m)_n$ is clearly bounded in $L^p(a,b)$ if $(f_n)_n$ is.)
Then $\|\tau_h f_n - f_n\|_{L^p(a,b-h)} < \varepsilon$ for all $h \leq h_0$ uniformly in $n$ and again the Frechet-Kolmogorov theorem says that $(f_n)_n$ is relatively compact in $L^p(a,b)$.
Due to interest by the OP, here also a less abstract diagonal argument. Iteratively for $m=1,2, \dots$, choose nested subsequences $(f_{n_m(i)})_i$ such that $(f_{n_m(i)}\psi_m(f_{n_m(i)}))_i$ converges (to some $f^m$). Set $h_j := f_{n_j(j)}$. Then split $$h_k - h_\ell = (h_k - h_k\psi_m(h_k)) + (h_k\psi_m(h_k) - h_\ell\psi_m(h_\ell)) + (h_\ell\psi_m(h_\ell) - h_\ell).$$
For $\varepsilon > 0$ given, the norms of the first and last summands can be made smaller than $\varepsilon/3$ uniformly in $k,\ell$ by choosing $m$ large enough (calculation in OP). For $k,\ell \geq m$, $(h_k)$ and $(h_\ell)$ are subsequences of $(f_{n_m(i)})_i$, so $(h_j \psi_m(h_j))_j$ converges and is a Cauchy sequence. Thus, choosing $k,\ell$ large enough makes the middle summand smaller than $\varepsilon/3$. Hence $(h_j)_j$ is also a Cauchy sequence and, by completeness of $L^p(a,b)$, convergent.
This should not be a contradiction to the counterexample posted on MSE in the link in OP, since there the assumption was only that $(g_n^m)_n$ admits a convergent subsequence for each $m$, and it was shown that then $(f_n)_n$ need not admit a convergent subsequence. Here OP asked for $(g_n^m)_n$ relatively compact for each $m$ which the counterexample sequence is not, if I see it correctly.
@Zac I added a diagonal argument.
Thank you very much for your addition!
I have just one more curiosity: would all of this still work if we defined $$\psi_m(f) =
\begin{cases}
1 \qquad \text{ if } |f - 1|\ge 1/m \
0 \qquad \text{ if } |f - 1|\le 1/(2m)
\end{cases}
$$ instead of as in the original question?
Uh, you roughly would need to make sure that the $(f_n)$ stay away from $1$ in a uniform manner, I guess. (In the sense that the measure of the set where $|f_n - 1| \leq 1/m$ goes to zero uniformly in $n$ as $m \to \infty$.)
It seems fair. Why wasn't this needed in the previous case?
With the modified $\psi_m$, the constant one function is a nonzero fixed point of $f \mapsto f-\psi_m(f)$ for every $m$. For the original $\psi_m$, there is no nonzero fixed point for each $m$. Essentially, this makes the calculation in the OP possible.
I see. Thank you very much. What kind of assumption of $f_n$ would imply this condition on the measure of the set ${|f_n -1| \le 1/m}$? For example, a bound on the derivative? higher integrability?
I've put this in a separate post: https://mathoverflow.net/questions/368583/lp-compactness-for-a-sequence-of-functions-from-compactness-of-product-with-c
You should consider upvoting and accepting Hannes 'answer, Zac.
|
2025-03-21T14:48:31.693763
| 2020-08-02T07:57:52 |
368127
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Lord Shadow",
"https://mathoverflow.net/users/163091"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631795",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368127"
}
|
Stack Exchange
|
When annihilator of ideal and ideal is co maximal
Let $R$ be a commutative ring with identity. It is not always true that ideal $J$ and annihilator of ideal $ann(J)$ are co maximal (ex: integral domain)
Is there a sufficient (necessary) condition( or ring) under which this happen $J$ and $ann(J)$ are comaximal?
The necessary and sufficient condition is that $J$ be generated by an idempotent.
A. Assume $J=(e)$ is generated by an idempotent $e$ ($e^2=e$). Then the annihilator of $J$ contains $1-e$, so the sum of $J$ and its annihilator contains $e + (1-e)=1$. This shows that $J$ and its annihilator are comaximal.
B. Now suppose that the annihilator $K$ of $J$ is comaximal with $J$. That is, $J+K=R$. Then there exist $j\in J$ and $k\in K$ such that $j+k=1$. Multiplying through by $j$ one learns $j^2+jk = j^2 = j$, so $j\in J$ is an idempotent. Similarly, $k\in K$ is an idempotent. Since $J$ annihilates $K$, we have $kJ=0$, hence for $x\in J$ we have $x = x-0=x-kx = (1-k)x = jx$. We now have that $j\in J$ (i.e. $(j)\subseteq J$) and, for all $x\in J$ we have $x = jx$ (i.e. $J\subseteq (j)$), so we get that $J = (j)$. This shows that $J$ is generated by an idempotent.
August 3 Edit:
In response to the question below,
I wonder is this is a standard result (it seems to be a standard one)?
the answer is Yes, essentially. It is standard that the following are equivalent for unital rings:
(1) $J$ and $K$ are complementary ideals of $R$. $(J+K=R$, $J\cap K=(0))$
(2) $J$ is generated by a central idempotent $(J=(e))$ and $K$
is generated by the complementary idempotent $(K=(1-e))$.
(3) $R$ factors as $R\cong J\times K\cong R/ J\times R/K$.
Where does your condition fit into this picture? (Condition = $J$ is comaximal with its annihilator.)
It is a well-known fact that if $J$ and $K$ are comaximal, then their intersection equals their product. Therefore they are disjoint iff they annihilate each other. In particular, if $J$ annihilates $K$ and is comaximal with $K$, then $J$ is a complement to $K$.
Here is how you prove the well-known fact that if $J+K=R$, then
$J\cap K=JK$. It uses the fact that $JK\subseteq J\cap K$ for any two ideals.
$$
\underline{J\cap K}=R(J\cap K)=(J+K)(J\cap K)=J(J\cap K)+K(J\cap K)\subseteq \underline{JK}\subseteq \underline{J\cap K}.
$$
I remember idempotent are related to decomposition. Although I did got sufficient conditions, it's interesting as you noted that it is also necessary condition
I wonder is this is a standard result (it seems to be a standard one)? If yes, is there some place where I can learn more about comaximality and annhilators are related.
|
2025-03-21T14:48:31.694072
| 2020-08-02T08:47:30 |
368129
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Claus",
"Tony Huynh",
"https://mathoverflow.net/users/156936",
"https://mathoverflow.net/users/2233"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631796",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368129"
}
|
Stack Exchange
|
Algebraic graph invariant $\mu(G)$ which links Four-Color-Theorem with Schrödinger operators: further topological characterizations of graphs?
30 years ago, Yves Colin de Verdière introduced the algebraic graph invariant $\mu(G)$ for any undirected graph $G$, see [1]. It was motivated by the study of the second eigenvalue of certain Schrödinger operators [2,3]. It is defined in purely algebraic terms as the maximum corank in a set of generalized Laplacian matrices of $G$.
It turned out to be very powerful concept, linking algebraic with topological graph theory (and, by conjecture, with graph coloring). For example,
Four Color Theorem: Colin de Verdière conjectures $\chi(G)\leq\mu(G)+1$ where $\chi(G)$ is the chromatic number of $G$, see [4]. If true, this would prove the Four Color Theorem.
Graph minor monotone: The property $\mu(G)\leq k$ is closed under taking graph minors of $G$, meaning $\mu(g)\leq \mu(G)$ if $g$ is a minor of $G$, see [1]. So, by the Robertson–Seymour Graph Minor Theorem, the property $\mu(G)\leq k$ can be characterized by a finite number of excluded graph minors.
Embeddability: $\mu(G)$ characterizes this topological property for several families of graphs: embeddable in a line $(\mu\leq1)$, outerplanar $(\mu\leq2)$, planar $(\mu\leq3)$, or linklessly i.e. flat embeddable in ${\mathbb R}^3$ $(\mu\leq4)$, see [1,2].
Embeddings in more general surfaces: If $G$ embeds in the real projective plane or in the Klein bottle, then $\mu\leq5$. If it embeds in the torus, $\mu\leq6$. If it embeds in a surface $S$ with negative Euler characteristic $\psi$, then $\mu\leq 4−2\psi$, see [4]
Now, I have two questions, the first one being the main one:
MAIN QUESTION: Is someone aware of further embeddability characterizations based on $\mu(G)$ beyond the results in bullet point No. 3? In No. 3, we have full characterizations, while the results in No. 4 are just implications for $\mu(G)$ in case $G$ can be embedded, i.e. just in one direction.
Quoting [3]: "The parameter was motivated by the study of the maximum
multiplicity of the second eigenvalue of certain Schrödinger operators. These operators
are defined on Riemann surfaces. It turned out that in this study one can approximate
the surface by a sufficiently densely embedded graph $G$, in such a way that $\mu(G)$ is the
maximum multiplicity of the second eigenvalue of the operator, or a lower bound to it."
SECOND QUESTION: So it appears $\mu(G)$ was developed to resolve a problem in Schrödinger operator theory. I wondered when/how the idea emerged to study $\mu(G)$ as a graph invariant in its own right? I looked at [1] and [CV 1] but could not find an answer.
References
[1] Yves Colin de Verdiere (1990): Sur un nouvel invariant des graphes et un critère de planarité, J. Combin. Th. (B) 50, 11–21.
[2] L. Lovasz & A. Schrijver (1998): A Borsuk theorem for antipodal links and a spectral characterization of linklessly embeddable graphs, Proc. Amer. Math. Soc. 126, 1275–1285.
[3] H. van der Holst, L. Lovasz & A. Schrijver (1999): The Colin de Verdière graph parameter, pp. 29–
85 in: Graph Theory and Combinatorial Biology (L. Lovasz et al., eds.), János Bolyai Math.
Soc., Budapest.
[4] Andries E. Brouwer, Willem H. Haemers (2011): Spectra of graphs, Springer Monograph.
Earlier work that Colin de Verdière cites in his article [1]:
[CV 1] Y. COLIN DE VERDIÈRE, Spectres de variétés riemanniennes et spectres de graphes, Proc. Intern. Congress of Math., Berkeley 1986, 522-530.
[CV 2] Y. COLIN DE VERDIÈRE, Sur la multiplicité de la premiere valeur propre non nulle du laplacien, Comment. Math. Helv. 61 (1986), 254-270.
[CV 3] Y. COLIN DE VERDIÈRE, Sur une hypothèse de transversalité d’Arnold, Comment. Math. Helv. 63 (1988). 184-193.
[CV 4] Y. COLIN DE VERDIÈRE, Constructions de laplaciens dont une partie finie du spectre est donné, Ann. Sci. École Norm. Sup. 20 (1987), 599-615.
https://en.wikipedia.org/wiki/Colin_de_Verdi%C3%A8re_graph_invariant
Embeddability in any surface but the sphere (or plane) can probably not be characterized via the Colin de Verdière number.
Suppose that $K_n$ is the largest complete graph that embedds into a surface $S$.
This shows that the best we can hope for is "$G$ embedds in $S$ $\Leftrightarrow$ $\mu(G)\le\mu(K_n)= n-1$".
The following is still a bit hand-wavy (maybe someone can help):
I can imagine, that a disjoint union of sufficiently many $K_n$ can no longer be embedded into $S$ (except if $S$ is a sphere/plane). My intuition is that any additional $K_n$ must embedd in one of the regions given by the embedding of the previous $K_n$, and this region is probably "of a lesser genus" (if the genus is not already 0).
For example, this is true for $S$ being the projective plane: $K_5$ embedds in $\Bbb R P^2$, but $K_5+K_5$ does not (see here).
Also, a claim in this question seems to support this in the orientable case.
But we also have $\mu(K_n+\cdots +K_n)=\mu(K_n)=n-1$ (see [1]), contradicting the desired characterization.
[1] van der Holst, Lovász, Schrijver:
"The Colin de Verdière graph parameter", Theorem 2.5
Thanks for a great answer! It clarifies why we don't have full characterizations here except for the sphere (and linkless embeddability). The reason you are giving is clear, and the example with the projective plane is crucial. Very grateful for your answer.
The handwavy part follows from the additivity of Euler genus for graphs. In fact, the following more general result is true. If $G$ and $H$ are graphs such that $|V(G) \cap V(G)| \leq 2$, then the Euler genus of $G \cup H$ is at least the Euler genus of $G$ plus the Euler genus of $H$. The reference is An Additivity Theorem for the Genus of a Graph by Gary L. Miller.
Perhaps I can contribute to the history part the question, since I was quite close to the Institut Fourier at that time and was very interested in their work (I am a physicist). Grenoble now has several different research groups doing graph theory (like G-SCOP, Institut Fourier, GIPSA-lab, LIG) but I think L'Institut Fourier was the early one for graph theory.
Here are two original quotes from Yves Colin de Verdière about the time when $\mu(G)$ evolved; my translation. The quotes give a view about his collaboration with the graph theory team; and his view of graphs as singular Riemannian manifolds, in the context of his differential geometry work.
First quote from Yves Colin de Verdière 2004 [Theorem 5 is about graph minor monotonicity of $\mu$ and Theorem 6 is the characterization of planar graphs. Theorem 17 is from S. Cheng: Eigenfunctions and nodal sets. Comment. Math. Helv., 51:43-55,
1976]:
I discovered theorems 5 and 6, trying to understand Cheng's theorem (Theorem 17) and its possible extension to dimension 3. This theorem was stated in the context of partial differential equations and differential geometry. It took me many years and timely encounters to discover that graph theory was the natural framework for the study of these problems. I was fortunate to benefit in Grenoble from the availability of colleagues in graph theory, in particular François Jaeger (1947-1997), who helped me to discover this subject far away from my original background. It is one of the things I find fascinating in mathematics, these unexpected links between fields that are a priori very far away!
Second quote from Yves Colin de Verdière 1986:
Let $\Gamma_N$ be the complete graph with $N$ vertices ($N\geq4$): each pair of distinct vertices is joined by a single edge. $\Gamma_N$ is considered as a singular Riemannian manifold of dimension 1; if $\cal A$ is the set of $N(N-1)/2$ edges, a Riemannian metric on $\Gamma_N$ is entirely determined (up to isometry) by the length $l(a)$ of any edge $a$ of $\cal A$.
The original quotes are in French
First quote: in SUR LE SPECTRE DES OPÉRATEURS
DE TYPE SCHRÖDINGER SUR LES GRAPHES, Exposés à l’Ecole Polytechnique pour les
professeurs de Mathématiques Spéciales, Yves Colin de Verdière, 17 mai 2004:
J’ai découvert les théorèmes 5 et 6, en essayant de comprendre le théorème de Cheng (Théorème 17) et son éventuelle extension à la dimension 3. Ce théorème était énoncé dans le contexte des équations aux dérivées partielles et de la géométrie différentielle. Il m’a fallu de nombreuses années et des rencontres opportunes pour découvrir que la théorie des graphes était le cadre naturel pour l’étude de ces problèmes. J’ai eu la chance de bénéficier à Grenoble de la disponibilité des collègues de théorie des graphes, en particulier de François Jaeger (1947-1997), qui m’ont aidé à découvrir ce sujet loin de ma culture de base... C’est une des choses que je trouve fascinantes en mathématiques que ces liens imprévus entre des domaines a priori très lointains!
Second quote: in Sur la multiplicité de la première valeur propre non nulle du Laplacien, Yves Colin de Verdière, Comment. Math. Helv. 61, 254-270, 1986:
Soit $\Gamma_N$ le graphe complet à $N$ sommets ($N\geq4$): chaque couple de sommets distincts est joint par une arête unique. On considère $\Gamma_N$ comme une variété riemannienne singulière de dimension 1; si $\cal A$ est l'énsemble des $N(N-1)/2$ arêtes, une métrique riemannienne sur $\Gamma_N$ est entièrement déterminée (à isométrie près) par la longueur l(a) de toute arête a de si.
@gwynnethm.sc. This is a super useful quote! Really very helpful. Thank you very much!
The following embeddability characterization for $\mu(G)\le 5$ is merely a conjecture, but I think it fits the question:
Conjecture (van der Holst, [1]): $\mu(G)\le 5$ if and only if $G$ is 4-flat.
A graph is called 4-flat if the following 2-dimensional CW-complex $C(G)$ (PL-) embeds in $\Bbb R^4$: $C(G)$ is obtained from $G$ by attaching a 2-cell along each cycle of $G$.
The 4-flat graphs include all planar and linkless graphs, as well as all cones over such. In fact, these graphs have $\mu(G)\le 5$.
[1] van der Holst (2006). "Graphs and obstructions in four dimensions".
van der Holst also conjectures the full list of forbidden minors (for either 4-flat or $\mu(G)\le 5$) to consists of the 78 graphs of the Heawood family (graphs obtained from $K_7$ and $K_{3,3,1,1}$ by $\Delta Y$- and $Y\Delta$-tranforms). In fact, all Heawood graphs have $\mu(G)=6$.
Personally, I am somewhat sceptical of the conjectured characterization of $\mu(G)\le 5$, in the sense that if it turns out true, it is most likely a coincidence, in the same way as "linkless $\Leftrightarrow$ $\mu(G)\le 4$" feels like a coincidence that is "only" verified by comparing the list of forbidden minors for linkless and $\mu(G)\le 4$ (slight exaggeration). van der Holst and Pendavingh in [2] even introduced a somewhat more natural graph invariant $\sigma(G)$ that generalizes planarity, linklessness and 4-flatness, but that also diverges from $\mu(G)$ for sufficiently complex graphs. The question is just: are 4-flat graphs already "sufficiently complex"?
[2] van der Holst, Pendavingh, "On a graph property generalizing planarity and flatness".
|
2025-03-21T14:48:31.694820
| 2020-08-02T08:52:24 |
368130
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Geoff Robinson",
"Gerhard Paseman",
"https://mathoverflow.net/users/14450",
"https://mathoverflow.net/users/3402"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631797",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368130"
}
|
Stack Exchange
|
Primality test for $N=2^mp^n +1$
This question is related to my previous question.
Can you prove or disprove the following claim:
Let $N=2^mp^n+1$ , $m>0 , n>0$ and $p$ is an odd prime . If there exists an integer $a$ such that $$\displaystyle\sum_{i=0}^{p-1} (-1)^i \cdot a^{i \cdot(N-1)/2p} \equiv 0 \pmod{N}$$ then $N$ is a prime.
You can run this test here. I tried to mimic the proof given in this answer , but I didn't manage to adapt it for this generalization.
EDIT
Test implementation in PARI/GP without directly computing the sum.
I think this is true. Let $b = a^{\frac{N-1}{2p}} = a^{2^{m-1}p^{n-1}}$, and note that we
have $\frac{b^{p}+1}{b+1} \equiv 0$ (mod $N$).
Now $a$ and $N$ must be coprime, so that $b$ and $N$ are coprime. We have $b^{2p} \equiv 1$ (mod $N$).
Now $b^{p}-1$ and $b^{p} +1$ have gcd dividing $2$. However $\frac{b^{p}+1}{b+1}$ is always odd, so that $\frac{b^{p}+1}{b+1}$ and $b^{p}-1$ are relatively prime.
If $q$ is a prime divisor of $N$, then $b+1$ is not divisible by $q$, for otherwise $\frac{b^{p}+1}{b+1} \equiv p$ (mod $q$), whereas we should have $\frac{b^{p}+1}{b+1} \equiv 0$ (mod $q$). Also, $b \not \equiv 1$ (mod $q$), since $b^{p}-1$ is coprime to $N.$
Hence $b^{2}-1$ is coprime to $N.$
The multiplicative order of $b$ (mod $N$) is a divisor of $2p$, but is not equal to $1$,$2$ or $p$, since $b^{2}-1$ and $b^{p}-1$ are both coprime to $N$.
Hence $b$ has multiplicative order $2p$ (mod $N$).
Now the multiplicative order of $a$ (mod $N$) is a divisor of $2^{m}p^{n}$, but none of
$b = a^{2^{m-1}p^{n-1}}$, $b^{2} = a^{2^{m}p^{n-1}}$ or $b^{p} = a^{2^{m-1}p^{n}}$, are congruent to $1$ (mod $N$).
Thus $a$ has multiplicative order $2^{m}p^{n} = N-1$ in $\left(\mathbb{Z}/N\mathbb{Z}\right)^{\times}.$
But the multiplicative order of $a$ (mod $N$) is a divisor of $\phi(N)$, so we must have $\phi(N) = N-1$, and $N$ is prime.
Later edit: I point out that both this and the linked problem are implied by the following general theorem: If $m >1$ is an odd integer, then $m$ is prime if and only if there is an integer h with $\Phi_{m-1}(h) \equiv 0$ (mod $m$).
When $m$ is prime, there is such an integer $h$ since the multiplicative group of $\mathbb{Z}/m\mathbb{Z}$ is cyclic, and we may take $h$ such that $h +m\mathbb{Z}$ is a generator of that group.
For the other direction, assume that such an integer exists. Then $h$ is coprime to $m$ since $\Phi_{m-1}(x)$ has constant term $\pm 1.$
Also, the multiplicative order of $h +m\mathbb{Z}$ in $\mathbb{Z}/m\mathbb{Z}$ is a divisor of $m-1$ as $h^{m-1} \equiv 1$ (mod $m$)- because $h^{m-1}-1$ is divisible by $\Phi_{m-1}(h)$.
I claim that $h^{d}-1$ is coprime to $m$ whenever $d$ is a proper divisor of $m-1$. Let $q$ be a prime divisor of $m$. If $q$ divides $h^{d}-1$, then in $\mathbb{Z}[\omega]$, where $\omega$ is a primitive $m-1$-th root of unity, there is a primitive $m-1$-th root of unity $\alpha$, a $d$-th root of unity $\beta$, and a prime ideal $\pi$ (containing $q$) of $\mathbb{Z}[\omega]$ such that
$h- \alpha \in \pi$ and $h- \beta \in \pi$. Then $\alpha - \beta \in \pi$, so that $0 \neq 1 - \overline{\alpha}\beta \in \pi$.
But $1 - \overline{\alpha}\beta$ = $1- \omega^{k}$ for some $k$ with $0 < k < m-1$, and this is a factor of $m-1 = \prod_{j=1}^{m-2}(1-\omega^{j}) = m-1$ in $\mathbb{Z}[\omega].$ Hence $m-1 \in \pi$, so that $q|m-1$, a contradiction as $q$ is prime and we assumed $q|m$.
Thus $h^{d}-1$ is coprime to $m$ whenever $d$ is a divisor of $m-1$ with $d \neq m-1$.
Hence $h+ m\mathbb{Z}$ has multiplicative order $m-1$ in the group of units of $\mathbb{Z}/m\mathbb{Z}$, a group of order $\phi(m) \leq m-1$. Hence $\phi(m) = m-1$ and $m$ is prime.
Although I am not challenging the mathematics, I wonder about the utility. If m is not prime (but we don't know that yet), we also want a guarantee that there is no h with P(h) a multiple of m, where P() is the polynomial we would use to test (and would equal Phi_m() if m were prime, but doesn't because m is not prime). Can you give us a theorem using P()? Gerhard "Wants To Cover His Bases" Paseman, 2020.08.05.
If the question is addressed to me, I agree that this criterion seems likely to have no "practical" use. I partly wanted to point out that there is a whole family of examples of problems of this kind that could be posed of the same nature. These problems would not strictly be "duplicates" of each other, but are close to being duplicates.
Nevertheless, in theory the method does give a test for primality of $m$, knowing only the prime factorization of $m-1$. While calculating the polynomial $\Phi_{m-1}(x)$ may be unwieldy, it does not know or care whether or not $m$ is prime. Also, we may note that $(h+1)^{\phi(m-1) } \geq \Phi_{m-1}(h) \geq (h-1)^{\phi(m-1)}.$
It is addressed to you Geoff. I think my point is now that the problems have to do with a given P(), while your theorem does not talk about P(), but about something that is computationally harder to obtain (what is Phi_m when you know that m is not prime but also do not know the factors of m) (sorry, I need to use m-1 in some places)? I like the later edit, but in my view it has relevance more to general theory than to the proposed series of problems. Gerhard "Wants Theorems Talking About P()" Paseman, 2020.08.05.
Note that the polynomial I use is $\Phi_{m-1}(x)$, not $\Phi_{m}(x)$. Note also that both these questions test with integer of the form $b = a^{\frac{m-1}{f}}$, where $f$ is basically the radical of $(m-1),$ and we basically need to check whether $b^{\frac{f}{p} } \equiv 1$ (mod $m$) for any prime divisor $p$ of $f$. I think this is close to what happens in the new edit.
Perhaps. For even better bounds, check out https://mathoverflow.net/q/221357/ .Thanks for reminding me about Phi_{m-1}. Gerhard "Maybe Phi Is Good Enough" Paseman, 2020.08.05.
|
2025-03-21T14:48:31.695200
| 2020-08-02T09:52:24 |
368132
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Carlo Beenakker",
"Curt von Keyserlingk",
"Joako",
"Marcus M",
"Martin Hairer",
"Will Sawin",
"https://mathoverflow.net/users/11260",
"https://mathoverflow.net/users/163095",
"https://mathoverflow.net/users/18060",
"https://mathoverflow.net/users/303257",
"https://mathoverflow.net/users/36212",
"https://mathoverflow.net/users/38566",
"https://mathoverflow.net/users/69870",
"user36212"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631798",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368132"
}
|
Stack Exchange
|
Smooth functions that resemble random walks
If the Riemann hypothesis holds, then the Mertens function $M(n)\equiv\sum_{x\leq n} \mu(n)$ behaves much like a 1D random walk. This includes the statements that
$M(n)$ changes sign infinitely often
$M(n)=O(\sqrt{n})$ (ignoring subleading logarithmic corrections).
It is also believed that 3) $\mu(n)=M(n)-M(n-1)$ "looks random". This seems to be a topic of current research, but is sometimes phrased as the "Mobius randomness law" (Eq 5 here), which says that for functions of low complexity $\xi (n)$
$\sum_{n\leq N} \xi(n) \mu(n) = o(\sum_{n\leq N} |\xi(n)|)$
Some weaker analogue of this conjecture is proved in the linked note.
Now, the Mertens function can be extended to the reals through an integral expression
$$M(x) = \int^{c+\mathrm{i}\infty}_{c-\mathrm{i}\infty } \frac{ds}{2\pi \mathrm{i}} \, \frac{x^s}{s \zeta(s)}\,\,\,\,\, (*)$$
My question is: Does anyone know of a function $f:\mathbb{R}\rightarrow \mathbb{R}$, which is a deterministic$\dagger$ combination of known analytic functions (e.g., Eq. $(*)$) and which can also be proved to obey conditions 1., 2., and 3. above? Here 3. would mean that $f(n)-f(n-1)$ "looks random" in some sense, e.g., the sense described above. Perhaps there are many examples; if so, what's the simplest? I don't particularly care whether $f(n)$ takes integer values; I just want it to look like a random walk.
**EDITS/ I've updated the wording of this question. There is an obvious set of examples if I merely insist on 1. and 2. I should have emphasized the need for $f(n)-f(n-1)$ to "look random".
$\dagger=$ I want $f$ to be expressible as a deterministic combination of known functions; so I won't accept e.g., a fourier series with randomly chosen coefficients (see Carlo's answer below). Morally, I'm interested in the appearance of randomness from seemingly deterministic expressions (e.g., $(*)$).
Well, for example $\sqrt{n}\sin n$. In general it’s not useful to ask for soft-analytic properties like smoothness, since for any function from the integers to the reals, there is a smooth function that extends the domain to the reals.
@user36212 Does $\sqrt{n} \sin n$ obey (1)? Even if you add floor functions I'd be surprised if that's known (or even true).
Thank you both for your comments. What I really want is for $f(x)$ to "look like a random walk" at integer points. This will certainly require that 1 and 2 are satisfied. However, I obviously need to add an extra condition; something along the lines that $f(n+1)-f(n)$ and $f(m+2)-f(m+1)$ are uncorrelated i.e., steps at different times are uncorrelated.
does the function $M(x)$ need to return integers for integer $x$?
Hi Carlo. No, that is not important to me. Sorry for not making that clear. I've updated the question to clarify that.
Marcus M - no, I presume it doesn’t go through integers at integer values; but that you can fix by smoothly transforming $n$. Point is it looks nothing like a random walk.
If your interest is mainly in understanding the appearance of randomness (or something that 'looks like it') from compact deterministic expressions, then wouldn't you have more luck looking at pseudo-random number generators than anything to do with analytic functions?
Martin: A fair point, similar to the point Carlo makes below. I was hoping for something compact like $(*)$, but maybe that's too much to ask. Also "compact" is a vague term.
PS. Are there any random number generators that one can rigorously prove pass tests like that in (3)?
Here is a plot idea: as the lobes of the Fabius function follows a path that could be considered as one realization of a Random Walk, maybe integrating the function would look like random when zooming out enough. (...)
(...) Since, the path is deterministic due the Thue-Morse pattern, but the Fabius function is nowhere analytic, maybe changing the lobes for $\sin(x)^4$ function could lead to something that behaves as you want: I tried it with some approximations in Desmos (it got stuck calculating the integral, but errasing ot will show what I mean with the function that change signs pseudo-randomly).
Smooth random functions, random ODEs, and Gaussian processes (2018) describes an approach that takes a finite Fourier series on the interval $(0,1)$ with randomly chosen coefficients. The integral of this function approaches Brownian motion in the limit that the number $M$ of Fourier coefficients tends to infinity.
The plot shows three such functions, for $M=1/\lambda=5,25,$ and $125$.
For $M=1000$ the curve is a Brownian path within plotting accuracy, the plot below shows 10 realizations.
Carlo: Thank you for the example. It's not quite what I was looking for, although interesting. Your answer has led me to refine my question. In particular, I'd like $f(x)$ to have an expression as a deterministic combination of known functions. In your example, it seems to be important that the fourier coefficients are chosen randomly.
I'm not sure that I understand your point: this is a fully deterministic construction, the Fourier coefficients are just a fixed string of numbers, how I arrived at them should not matter. If you wish, you can construct the coefficients from the digits of $\pi$ --- wouldn't that satisfy your requirement of "appearance of randomness from seemingly deterministic expressions"?
Hi Carlo, the construction using the digits of pi almost certainly works. It's not quite as compact as $(*)$ but I'll grant that it satisfies my requirements. However, I think you'd struggle to prove this, as $\pi$ is only conjectured to be a normal number (https://en.wikipedia.org/wiki/Normal_number). I'll look into known examples of normal numbers and see whether they could be consistent with 3).
@curt Because Chowla's conjecture implies Sarnak's conjecture, every normal number (such as the Champernowne constant $.12345679101112\dots$) satisfies a form of 3.
@WillSawin : Thanks, but could you explain or provide a reference? My understanding is that both Chowla and Sarnak's conjecture are respectively statements about averaged moments of the mobius function and approximate orthogonality of the mobius function with "simple" functions. How does one connect those statements to one concerning general normal numbers?
@curt See here, https://terrytao.wordpress.com/2012/10/14/the-chowla-conjecture-and-the-sarnak-conjecture/ in particular "The argument does not use any number-theoretic properties of the Möbius function; one could replace $\mu$ in both conjectures by any other function from the natural numbers to ${-1,0,+1}$ and obtain the same implication." i.e. just apply the argument stated there to the digits of a normal number.
@WillSawin Thanks! That makes a lot of sense -- the statement of Chowla's conjecture is obviously true for normal numbers (with some massaging to map the digits to the set 1,0,-1 ). So, Carlo's example should satisfy 3) by e.g., taking the digits in Champernowne's constant mod 3. Shifting the results by -1, so that the result lives in {-1,0,1}. Great!
|
2025-03-21T14:48:31.695940
| 2020-08-02T11:02:10 |
368133
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Gilles Mordant",
"https://mathoverflow.net/users/125260",
"https://mathoverflow.net/users/153595",
"香结丁"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631799",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368133"
}
|
Stack Exchange
|
Asymptotic properties of ANOVA when the number of groups goes to infinity
Suppose
$$X_{ij} = \mu_j + \varepsilon_{ij}, \quad j = 1, \cdots, J, \quad i = 1, \cdots, N_j$$
ANOVA can allow us to test whether $\mu_1 = \cdots = \mu_J$.
In traditional ANOVA, however, the number of groups $J$ is fixed. But I want to know the proof technique in the case that $J$ goes to infinity.
Does anyone know some papers containing this question with detailed proof? Thanks so much!
The solution will depend on your assumptions on how the triangular array of $N_j$ behaves. If you keep the classical ANOVA assumptions (that is homoscedasticity and normality of errors), the calculations remain the same as in the classical case excepted that the first sum is indexed by J and J goes to infinity.
@GillesMordant Thank you so much! But I am still a little confused. Since $J \rightarrow \infty$, the number of parameters that need to be examined also goes to infinity, some extra assumptions may need to clarify. I will extremely appreciate you if you can provide me some articles or notes that involve this question!
The following article is what you are looking for:
Akritas, M., Arnold, S. (2000). Asymptotics for Analysis of Variance When the Number of Levels is Large, Journal of the American Statistical Association, 95:449, 212-226.
Thanks so much, it is exactly what I am finding.
Set $J$ the number of groups, and $N_j$ the number of observations for the group $j$, $1\le j\le N$. The test statistic is the ratio
$$
T_{n,J}:=\frac{A}{B},
$$
with
$$
A:=\frac{\sum_{j=1}^J N_j (\bar X- \bar{X}_{\cdot j})^2}{J-1}
$$
and
$$
B:=\frac{\sum_{j=1}^J \sum_{i=1}^{N_j}(X_{ij}- \bar{X}_{\cdot j})}{N-J},
$$
where $\bar{X}_{\cdot j}$ is the mean of the $j$th group and $N:= \sum_{j=1}^JN_j$.
Under the classical assumptions of normality of the residuals and homoscedasticity, $T_{n,J}$ has a Fisher distribution with d.f. $J-1,N-J$ under the null hypothesis that all group means are equal.
In your case, if you consider a sequence of $(J_k)_{k\in \mathbb{N}}$ with $J_k \to \infty$ as $k\to \infty$, the classical result still holds true for each $J_k$.
Now, if you want to go further and study the properties of the test, you should first make hypotheses about the sample sizes per group. If you want to study effect sizes, for instance, you will again require hypotheses on the differences of the means and so on.
Thanks so much~
|
2025-03-21T14:48:31.696153
| 2020-08-02T11:41:10 |
368135
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Alex Ravsky",
"https://mathoverflow.net/users/122662",
"https://mathoverflow.net/users/43954",
"Đào Thanh Oai"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631800",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368135"
}
|
Stack Exchange
|
Construct closed chain of $k$-gon around $n$ points-$n, k$ are odd primes number
Question 1: I am looking for a proof of the conjectures 1, 2, 3 as follows?
Question 2: In conjecture 3, in general case, I can not give a formula of $X$. But I think, If $n, k$ are odd primes number then $X=\frac{2nk}{gcd(k-2,2k)gcd(n,k)}$, I checked with some small case. Can You give a general formula of $X$?
Consider $n, k \ge 3$ be two integer numbers, given $n$ general points $P_1$, $P_2$,....,$P_n$ and $O$ is arbitrary point in the plane, let $P_{n+i}=P_i$ for $i=1,\ldots,....$. Construct a chain of $m$ regular $k$-gon:
Construct first $k$-gon: $A_{1\;1}A_{1\;2}....A_{1\;k}$ such that $A_{1\;1}=O$; $A_{1\;2}=P_1$, the centroid of the first $k$-gon is $A_1$
Second $k$-gon: $A_{2\;1}A_{2\;2}....A_{2\;k}$ such that $A_{2\;1}=A_{1\;3}$; $A_{2\;2}=P_2$ the centroid of the second $k$-gon is $A_2$
$.................................$
$i$ th $k$-gon: $A_{i\;1}A_{i\;2}....A_{i\;k}$ such that $A_{i+1\;1}=A_{i\;3}$; $A_{i+1\;2}=P_{i+1}$ the centroid of the $i$ th $k$-gon is $A_i$
$.................................$
$m$ th $k$-gon: $A_{m\;1}A_{m\;2}....A_{m\;k}$ such that $A_{m\;1}=A_{m-1\;3}$; $A_{m\;2}=P_{m}$ the centroid of the $m$ th $k$-gon is $A_m$
and all regular polygon is same direction.
Definition: The chain is closed if exist $m$ such that $A_{m\;3}=A_{1\;1}=O$. The chain is open if no exist $m$ such that $A_{m\;3}=A_{1\;1}=O$
Conjecture 1: If $n=\frac{2k}{gcd(k-2,2k)}$ then the chain is opened.
Conjecture 2: If $n\ne\frac{2k}{gcd(k-2,2k)}$ then the chain is closed.
Conjecture 3: If the chain is closed then $m=n.X$ and $X$ points $A_i, A_{n+i}, A_{2n+i},...,A_{nx+i}$ be form $X$-gon for $i=1, 2,...,n$ which the centroid of the $X$-gon is fixed when we moved $O$, these regular polygon equal.
UPDATE GEOGEBRA SOFTWARE APPLET
3-gon and 3 points
4-gon and 3 points
5-gon and 3 points
7-gon and 3 points
8 gon and 3 points
9-gon and 3 points
11-gon and 3 points
3-gon and 5 points
4-gon and 5 points
4-gon and 6 points
3-gon and 7 points
See also:
Napoleon theorem
Van Aubel theorem
Petr–Douglas–Neumann theorem
@AlexRavsky I corrected
By geogebra software applet, I see that the chain will closed or open that is not depent on $P_j$ and $O$, this result depends on $n, k$ @AlexRavsky
We can consider all given points as complex numbers, points of the complex plane. Then, as far as I understood, for all integers $m\ge 1$, $1\le j\le k$ we we have $$A_{m,j}=A_m+(P_m-A_m)\xi^{j-2},$$
where $\xi=\exp\frac{2\pi i}{k}$.
Since $A_{m+1,1}= A_{m,3}$, we have
$$A_{m+1}+(P_{m+1}-A_{m+1})\xi^{-1}=A_m+(P_m-A_m)\xi.$$
It follows
$$A_{m+1}=-A_m\xi+\frac{P_m\xi^2-P_{m+1}}{\xi-1}.$$
This formula suggests that the chain can be closed for a specific choice of $O$ and $P_m$’s, but I guess that you are looking for a stable general pattern. So let’s look when the sequence $\{A_m\}$ is periodic.
Putting $B_m=A_m(-\xi)^{-m}$, we obtain
$$B_{m+1}=B_m+\frac{P_m\xi^2-P_{m+1}}{\xi-1}(-\xi)^{-m-1}.$$
Since the sequence $\{P_n\}$ has a period $n$, we have
$$B_{m+2n}-B_{m+n}=(B_{m+n}-B_{m}) (-\xi)^{-n},$$
that is
$$A_{m+2n}-A_{m+n}(1+(-\xi)^{n})+A_m(-\xi)^{n}=0.$$
An equation $\lambda^2-(1+(-\xi)^{n})\lambda +(-\xi)^{n}$ has roots $1$ and $(-\xi)^{n}$. The following cases are possible.
1)) $(-\xi)^{-n}=1$. This holds iff ($n$ is even and $k|n$) or ($n$ is odd, $k$ is even and $k|2n$). The theory of recurrence relations implies that $A_{r n+m}=c_1(m) + c_2(m)r$ for each $r$ and some constants $c_1(m)$ and $c_2(m)$ depending on $m$. If all $c_2(m)$ are zeroes then the sequence $\{A_m\}$ has a period $n$ (or its divisor). Otherwise the sequence $\{A_m\}$ is not periodic. Thus the sequence $\{A_m\}$ is periodic iff for each $m$ we have $A_m=A_{m+n}$. This can happen iff the choice of $P_m$’s is specific. Namely,
1.1)) If $n$ is even and $k|n$ then
$$0=A_{m+n}(-\xi)^{-m-n}- A_{m}(-\xi)^{-m}=B_{m+n}-B_{m}=\frac{1+\xi }{1-\xi}\sum_{j=1}^n P_{m+j} (-\xi)^{-(m+j)},$$
that is, $(-\xi)^{-1}$ is a root of a polynomial $P(x)=\sum_{j=1}^n P_{j} x^j$.
1.2)) If $n$ is odd, $k$ is even, and $k|2n$ then
$$0=A_{m+2n}(-\xi)^{-m-2n}- A_{m}(-\xi)^{-m}=B_{m+2n}-B_{m}=\frac{1+\xi }{1-\xi}\sum_{j=1}^{2n} P_{m+j} (-\xi)^{-(m+j)},$$
that is, $(-\xi)^{-1}$ is a root of a polynomial $(1+x^n)P(x)=\sum_{j=1}^{2n} P_{j} x^j$
2)) $(-\xi)^{-n}\ne 1$. (This case holds, in particular, when both $n$ and $k$ are odd). The theory of recurrence relations implies that $A_{r n+m}=c_1(m) + c_2(m)(-\xi)^{nr}$ for each $r$ and some constants $c_1(m)$ and $c_2(m)$ depending on $m$. If all $c_2(m)$ are zeroes then the sequence $\{A_m\}$ has a period $n$ (or its divisor). Otherwise $-\xi$ is a primitive $q$-th root of unity, where
$$q=\cases{k, \mbox{ if }k\equiv 0\pmod 4\\
k/2, \mbox{ if }k\equiv 2\pmod 4\\
2k, \mbox{ if }k\equiv 1,3\pmod 4}.$$
Remark that $q=\frac{2k}{\gcd(k-2,2k)}=\frac{2k}{\gcd(k-2,4)}$. Thus $(-\xi)^n$ is a primitive $\tfrac{q}{\gcd(q,n)}$-th root of unity, and so the sequence $\{A_m\}$ has a period $\tfrac{qn}{\gcd(q,n)}=\operatorname{lcm}(q,n)$ (or its divisor). Moreover, for each $m$, points $\{A_{r n+m}: 0\le r\le q-1\}$ are vertices of a $q$-qon.
Finally, recall that for each $m\ge 1$, $1\le j\le k$ we have $A_{m,j}=A_m+(P_m-A_m)\xi^{j-2}$. It follows that if the sequence $\{A_m\}$ has a period $p$ then for each fixed $j$ a sequence $\{A_{m,j}\}$ has a period $\operatorname{lcm}(p,n)$ (or its divisor).
Can You help me give your conclude? @AlexRavsky
@ĐàoThanhOai The aim of my answer was to provide an analytic description of your construction and its properties. In particular, to describe when the sequence ${A_n}$ is periodic, and my calculations show that it depends on points $P_j$.
Can you see my new update and click there? note that $O$ is blue color, $P_j$ are red color. If the chain is closed, then it is not depend on $P_i$ or $O$ it depends on $n, k$
I update the construct Figure https://i.sstatic.net/JKk9r.png
@ĐàoThanhOai I updated my answer. Sorry for the delay. A generic situation is similar to your conjectures, see 2)).
Thank You very much for your answer, are You think this result nice?
@ĐàoThanhOai Thanks. Yes, I think the result is nice. If you wish, we can try to write a small joint paper with your discovery and pictures and my proofs. We can try to publish it, for instance, in journals which are local for me, "Carpathian Mathematical Publications" or "Matematychni Studii". I even know a potential referee (but I cannot guarantee a positive review).
I am very happy if You write the paper. You decide Co-author or only your auhthor. There are many result from this configuration. My English is not good. You can contact with me via<EMAIL_ADDRESS>
You can select some journal https://ijgeometry.com/ or http://geometry-math-journal.ro
|
2025-03-21T14:48:31.696676
| 2020-08-02T12:43:25 |
368139
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Alkan",
"Gerry Myerson",
"https://mathoverflow.net/users/158000",
"https://mathoverflow.net/users/161528"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631801",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368139"
}
|
Stack Exchange
|
On nontotient Fibonacci numbers
This question is related to sequence of numbers $t$ such that $F_{6t}$ is a nontotient where $F_n$ represents the sequence of Fibonacci numbers for $n\geq 0$.
The online encyclopedia Wikipedia has the articles Fibonacci number and Nontotient, respectively.
There are very few terms in https://oeis.org/A335976 and some prime numbers appear as initial terms as expected.
Conjecture. There are infinitely many numbers $t$ such that $F_{6t}$ is a nontotient.
Question. Can someone prove or disprove above conjecture?
Initial terms of sequence of composite numbers $t_{c}$ such that $F_{6t_{c}}$ is a nontotient are also very welcome as helpful comment. Additionally, I couldn't find that question in literature yet, but if one can find references that have these or strongly related results, I will be very grateful for this response.
Thanks.
Why in particular $F_{6t}$? Why not all $F_n$?
Because only $6t$ indices can be totient except beginning of sequence.
I've extended OEIS A335976 with many terms.
The numerical data so far is in favor of the conjecture, although I think it may be hard to prove it rigorously.
Still, we can note a few major factors that limit the possibility for $F_{6t}$ be totient:
the small value of $\nu_2(F_{6t})$;
the small value of $\Omega(F_{6t})$;
abundance of prime factors $\equiv 1\pmod3$.
Essentially $F_{6t}$ is totient iff we can partition the prime factors of $F_{6t}$ into $\nu_2(F_{6t})$ or smaller number of subsets (each including at least one prime $2$) such that the product in each subset is $\varphi(q)$ for some prime power $q$. Furthermore, if a subset has just one prime $\equiv 2\pmod3$ the corresponding $q$ (if exists) cannot be prime, which makes its existence less likely.
We have $\nu_2(F_{6t})=3$ for odd $t$ and $\nu_2(F_{6t})\geq 4$ for even $t$, and that explains why we mostly see odd terms in A335976. As for $\Omega(F_{6t})$ values, those from A335976 have these values in the interval $[7,24]$ with more than half concentrated in the "middle" subinterval [14,17].
|
2025-03-21T14:48:31.696834
| 2020-08-02T14:37:17 |
368144
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Carlo Beenakker",
"Harry Wilson",
"LSpice",
"https://mathoverflow.net/users/11260",
"https://mathoverflow.net/users/2383",
"https://mathoverflow.net/users/43395"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631802",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368144"
}
|
Stack Exchange
|
Is the DispersiveWiki website down?
Somewhat recently I learned about a website called the DispersiveWiki which is/was maintained by James Colliander. I have found a few different links to the website from around the web (e.g. http://math.mit.edu/~gigliola/ and https://terrytao.wordpress.com/ ) such as:
http://tosio.math.toronto.edu/wiki/index.php/Main_Page
https://kvm16.pims.math.ca/DispersiveWiki/index.php?title=Main_Page
https://dispersivewiki.org/DispersiveWiki/index.php?title=Main_Page
For a few weeks now I have been trying to access the DispersiveWiki without success – for all the links above, I have been getting an error message of ‘The website took too long to respond.’ I have tried this using various browsers and methods of connecting to the internet, and so I have begun to suspect that this may not be a problem on my end.
In summary, my questions are:
Do other people have difficulty accessing the website?
Is the DispersiveWiki still being maintained? If so, what is its current web address?
This does not seem like a question in research-level mathematics.
It is also down on my end, but I really don't think this question is appropriate.
it seems that the WayBackMachine has archived only a single page from that site, regrettably; since the DispersiveWiki was maintained (at least originally) by @TerryTao , perhaps that MO user knows more...
|
2025-03-21T14:48:31.696954
| 2020-08-02T14:53:13 |
368145
|
{
"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:631803",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368145"
}
|
Stack Exchange
|
Are a map with constant singular values and its inverse always conjugate through isometries?
Let $U \subseteq \mathbb R^2$ be open, connected and bounded, and let $0<\sigma_1<\sigma_2$ satisfy $\sigma_1 \sigma_2=1$.
Suppose that $f:U \to U$ is a diffeomorphism whose singular values (of $df$) are constant $\sigma_1,\sigma_2$.
Question: Do there exist smooth isometries $
\phi_1, \phi_2:U \to U$ such that $\phi_1 \circ f \circ \phi_2=f^{-1}$? ($\phi_i $ must be affine; I want them to map $U$ into $U$.)
Note that $df^{-1}=(df)^{-1}$ has singular values $\sigma_1,\sigma_2$ the same as $df$.
Here are two examples for this phenomena:
1. Affine maps on ellipses:
Let $0<a<b$, $ab=1$, and let
$$
U=U_{a,b}=\biggl\{(x,y) \,\biggm | \, \frac{x^2}{a^2} + \frac{y^2}{b^2} < 1 \biggr\}.
$$
Take $f(x,y)=A\pmatrix{x\\y}$, where $$\begin{align*} & A=A(\theta)= \begin{pmatrix} a& 0 \\ 0 & b \end{pmatrix} \begin{pmatrix} \cos\theta & -\sin\theta \\ \sin\theta & \cos \theta \end{pmatrix}\begin{pmatrix} 1/a& 0 \\ 0 & 1/b \end{pmatrix}=
\begin{pmatrix} \cos\theta & -\frac ab \sin\theta \\ \frac ba \sin\theta & \cos \theta \end{pmatrix}
\end{align*}.$$
Then $A(\theta)^{-1}=A(-\theta)=JA(\theta) J$, where $J=\begin{pmatrix} 1& 0 \\ 0 & -1 \end{pmatrix}$ is the reflection around the $y$ axis.
2. Non-affine maps on the disk:
Let $U=D\setminus\{0\}$ where $D \subseteq \mathbb R^2$ is the unit disk.
$f_c: (r,\theta)\to (r,\theta+c\log r )$. Then we have $f_{c}^{-1}=f_{-c}=Jf_{c}J$.
Note that
$
[df_c]_{\{ \frac{\partial}{\partial r},\frac{1}{r}\frac{\partial}{\partial \theta}\}}=\begin{pmatrix} 1 & 0 \\\ c & 1\end{pmatrix},
$
so the singular values of $f_c$ are constants which depend on $c$.
There are many local solutions to the PDE $\sigma_i(df)=\sigma_i$, so I don't expect $f$ and $f^{-1}$ to be the same up to isometries in general.
|
2025-03-21T14:48:31.697091
| 2020-08-02T15:29:27 |
368146
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Victor Ostrik",
"https://mathoverflow.net/users/4158"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631804",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368146"
}
|
Stack Exchange
|
additivity of trace with respect to short exact sequences
Let $\mathcal{C}$ be an abelian rigid symmetric monoidal category over a field $K$. Assume that the endomorphism ring of the tensor unit in $\mathcal{C}$ is $K$. If $X$ is an object in $\mathcal{C}$ and $f:X\to X$ is an endomorphism one can define the trace of $f$ as the composite
$$\textbf{1}\stackrel{coev_X}{\to} X\otimes X^*\stackrel{f\otimes 1}{\to} X\otimes X^*\stackrel{c_{X,X^*}}{\to} X^*\otimes X\stackrel{ev_X}{\to} \textbf{1}$$
I am looking for a reference to a proof of the following claim: Assume that we have a short exact sequence $$0\to A\to B\to C\to 0$$ in $\mathcal{C}$ and that $f:B\to B$ is an endomorphism. Assume moreover that $f(A)\subseteq A$, so that $f$ induces endomorphisms $g:A\to A$ and $h:C\to C$. Then
$$\text{Tr}(f) = \text{Tr}(g) + \text{Tr}(h).$$
This is a consequence of Proposition 4.7.5 in https://pages.uoregon.edu/vostrik/book/surv-205.pdf Unfortunately there is no proof..
|
2025-03-21T14:48:31.697194
| 2020-08-02T16:40:22 |
368150
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Andrej Bauer",
"Carl-Fredrik Nyberg Brodda",
"Hanul Jeon",
"James",
"Jim Kingdon",
"LSpice",
"aws",
"https://mathoverflow.net/users/1176",
"https://mathoverflow.net/users/120914",
"https://mathoverflow.net/users/163109",
"https://mathoverflow.net/users/2383",
"https://mathoverflow.net/users/30790",
"https://mathoverflow.net/users/48041",
"https://mathoverflow.net/users/489586"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631805",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368150"
}
|
Stack Exchange
|
Does $\forall x \forall y\ (x \in y) \lor \lnot (x \in y)$ imply excluded middle?
Suppose that we take constructive set theory and add the axiom $\forall x \forall y\ (x \in y) \lor \lnot (x \in y)$. Does this imply excluded middle, or are there still some formulas $\varphi$ for which $\varphi \lor \lnot \varphi$ isn't provable using this new axiom?
You are confounding excluded middle with "there is a formula $\phi$ such that neither $\phi$ nor $\lnot \phi$ is provable". Even if we remove the faulty use of "provable" and consider "there is $\phi$ such that neither $\phi$ nor $\lnot \phi$", which is expressed as $\exists \phi . \lnot \phi \land \lnot\lnot \phi$, that is not excluded middle! It's an outright false statement (constructively and classically).
Sorry, I wrote that without thinking.
No problem, I am just pointing it out so that you can edit the question, lest this sort of error spreads. I have seen experienced mathematicians make it (and conclude that constructive mathematics is nonsense).
It depends how much separation is available. If you can construct the set $\{ z \in \{ \emptyset \} \;|\; \varphi \}$ then you can show $\varphi \vee \neg \varphi$. So for theories with full separation, like IZF, you can derive excluded middle, whereas for CZF where you only have separation for bounded formulas, you can only get excluded middle for bounded formulas.
Edit: See Rathjen, Indefiniteness in semi-intuitionistic set theories: On a conjecture of Feferman for a set theory with bounded excluded middle, but in which $\mathbf{CH} \vee \neg \mathbf{CH}$ is unprovable.
Does the end of your sentence ("you can only get excluded middle for bounded formulas") mean that, say, CZF + @James's axiom + $\lnot\text{EM}$ is consistent, or does it only mean that your trick for proving EM doesn't work?
As far as I can remember CZF with bounded excluded middle is strictly weaker than ZF, but I'll have to think a bit to find a proof or reference.
I added a reference in an edit. @MattF. I agree, but I don't know a proof offhand.
@aws We may refer Constructive Zermelo-Fraenkel Set Theory, Power Set, and the Calculus of Constructions by Rathjen. According to this article, $\mathsf{CZF}(\mathcal{P})$ with the bounded LEM (which is a subtheory of Tharp's quasi-intuitionistic set theory) has the equal proof-theoretic strength of $\mathsf{KP}(\mathcal{P})$. Hence CZF with bounded LEM is weaker than ZF.
@HanulJeon Thanks! That works too.
Thank you very much!
This statement does imply excluded middle in a theory like IZF which has full separation. See https://us.metamath.org/ileuni/exmidel.html for a formalization of this in IZF. The notation there is that DECID $\phi$ is defined to be $\phi \vee ¬ \phi$ and EXMID is defined using a few technicalities but it implies $\psi \vee ¬ \psi$ for any proposition $\psi$.
Isn’t your answer a proper subset of the other answer?
I do also agree with that answer - my apologies if overlapping answers are frowned upon.
|
2025-03-21T14:48:31.697433
| 2020-08-02T17:16:10 |
368155
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Mare",
"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:631806",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368155"
}
|
Stack Exchange
|
Quiver and relations of Schur algebras
Assume that the Schur algebra $S(n,r)$ with $n \geq r$ is not representation-finite.
Question: For which $n$, $r$ is the quiver and relations of the blocks of $S(n, r)$ explicitly known?
I just found the cases $n=r=4$ and $n=r=5$ in Xi - On representation types of $q$-Schur algebras.
Erdmann's article Schur algebras of finite type shows that $S(n,r)$ has finite representation type in prime characteristic $p$ if and only if $n=2$ and $r < p^2$ or $n \ge 3$ and $r \le 2p$ or $p=2$, $n=2$ and $r=5$ or $7$. In these cases the quiver and relations for (the basic algebra Morita equivalent to) each block are found explicitly. Quivers and relations are also found for some blocks of Schur algebras of infinite representation type. For example, Proposition 5.2 gives a basic algebra Morita equivalent to the principal block of $S(2,p^2)$ or $S(2,p^2+1)$ when $p > 2$. There are further results in Section 5 on blocks of $S(3,r)$ of infinite type.
More recently Doty, Erdmann, Martin and Nakano have classified all the tame Schur algebras. As one would expect, their paper gives some information about the Ext quivers. For example, see page 153 for the quiver for the basic algebra of $S(2,6)$ in characteristic $2$: it has wild representation type.
I can't find any explicit results in these papers for $n \ge r$. Since the authors' main interest is in the finite/tame/wild distinction, they concentrate on the blocks at the `threshold', where typically $n < r$. The corollary on page 143 of the second paper outlines a method for embedding the module category of $S(n,d)$ in the module category of $S(n',d)$, whenever $n' \ge d$.
Thank you, that is an interesting survey. But note that my assumption were that the algebras are not representation-finite and $n \geq r$.
|
2025-03-21T14:48:31.697577
| 2020-08-02T19:41:15 |
368160
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"HIMANSHU",
"YCor",
"https://mathoverflow.net/users/101534",
"https://mathoverflow.net/users/14094"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631807",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368160"
}
|
Stack Exchange
|
Units in group rings
Let $F$ be any field with $p$ elements and $G$ be any finite $p$-group, combining together they form a group ring $FG$. And $V(FG)$ denotes group of units of coefficient-sum equal to 1 in $FG$. We have $V(FG)=N\rtimes G$ for some subgroup $N$, iff we can have a group homomorphism $$f:V(FG)\rightarrow G,$$ which fixes elements of $G$ and sends the complement part to $1_G$.
My question here is assuming that one of two above conditions hold for some finite $p$-group $G$, what can we say about any subgroup $H$ of $G$ ?
I mean, will it satisfy any of the above two equivalent conditions ? I am not able to show it . According to me, this condition will not get inherited for any general subgroup $H$. There must be some extra condition on the subgroup, which I am trying to get. Please help me out this situation.
What's a "$p$-element"?
@YCor I am sorry if it is bit confusing but it simply means field consists of p number of elements.
|
2025-03-21T14:48:31.697667
| 2020-08-02T20:10:26 |
368162
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Gerhard Paseman",
"Gerry Myerson",
"https://mathoverflow.net/users/158000",
"https://mathoverflow.net/users/3402"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631808",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368162"
}
|
Stack Exchange
|
A sequence reminiscent of Fibonacci's recursion
The sequence in question is A296768 in the Online Encyclopedia. It starts with
1, 3, 5, 9, 11, 17, 24, 32, 36, 46, ...
It is obtained by starting with the positive integers in order, (b(i)= i for all positive integers i),and permuting them again and again. On the k-th pass, we exchange b(k+1) with b(b(k+1) + b(k)). The sequence is the limit of these sequences as k goes to infinity.
Is it true that the number 2 gets carried off to infinity, and does not appear in the final sequence? This seems to be the case, based on the first 67 values. More generally, is it true that this sequence is increasing? If it is increasing, what is its growth rate?
Link: http://oeis.org/A296768 Author of the sequence was ... David S. Newman.
I believe a proof by induction exists that the sequence is increasing and that its growth is roughly quadratic. Once you can prove it is increasing the analysis should be straight forward. I think the key is to prove that when looking at moving b(k+1) that b(k) holds "the largest number moved so far", and that b(k+1) will thus be at most b(k)+k+1, and sometimes less. Gerhard "One Swap At A Time" Paseman, 2020.08.03.
Further, subtracting your sequence from the triangular sequence gives 0,0,1,1,1+3,1+3,1+3,1+3,1+3+5,..., which if you can prove this continues and how, may lead to a quickly estimated asymptotic. Gerhard "Good Luck With The Pattern" Paseman, 2020.08.03.
And if I am right about that, one gets something like b(n) stabilizes to n(n+1)/2 - O(n^{3/2}). Gerhard "Check My Lack Of Work" Paseman, 2020.08.03.
Consider the following set of conditions. Before one swaps $b(k+1)$,
$b(i)$ is an increasing with $i$ sequence for $i$ less than $k+1$. (And for large enough $k$, $b(k)$ is greater than $k$.)
For $i$ larger than $k$, $b(i)$ is at most $i$.
$b(i)$ is less than $i$ exactly when $i$ is greater than $k$ and $i$ is part of the increasing sequence ($i$ is one of $b(1)$ or $b(2)$ or ... or $b(k)$).
Thus $b(k+1)$ is a value at most $k+1$ which gets swapped with the value $b(k)+b(k+1)=b(b(k)+b(k+1))$.
We see the conditions 1,2, and 3 hold for $k+1$ in place of $k$ after the swap.
As a result, $b(k)$ stabilizes into an increasing sequence. It seems to be the case that the $k$th triangular number $T(k)=b(k)+b(1)+b(2)+\cdots+b(j)$ with $b(j)$ largest value less than or equal to $k$, or something close to that. If this is true, then $b(k)$ grows like $k(k+1)/2$ minus a term like $O(k^{3/2})$.
Edit 2020.08.05
Here is an alternate definition of the sequence. Define $pr(k)$ to be $b(i)$ when $k$ equals $b(i+1)$, otherwise $pr(k)$ is 0. Then
$$b(k+1)=b(k)+k+1-pr(k+1).$$
Of course $i$ is positive and $b(1)=1$, or define $b(0)$ to be 0 to take one step back.
End Edit 2020.08.05.
Gerhard "Maybe Like K Power Phi?" Paseman, 2020.08.03.
As we swap b(k+1), the stable version is b(k+1)=b(k)+old version of b(k+1), where the old version is either k+1=b(k+1), or it is some smaller value j=b(k+1) where k+1-j = b(i) for some relatively small i. The difference T(k+1)-stable b(k+1)=b(1)+...+b(i) with all these terms b(j) strictly smaller than k+1 is k to some fractional power, but I don't know whether the power is like 3/2. Gerhard "It Could Be Like Phi" Paseman, 2020.08.04.
I just noticed this. Let b(0) be 0, and take the definition above. Then pr(1) is used to define b(1) but depends on b(1). So better not to take a step back. To be safe, define b(2)=3 as well, then start using the definition using pr. Gerhard "Should Comprehend What I Write" Paseman, 2020.08.07.
|
2025-03-21T14:48:31.697927
| 2020-08-02T20:23:01 |
368163
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Carl-Fredrik Nyberg Brodda",
"Derek Holt",
"Dominic Petti",
"LSpice",
"https://mathoverflow.net/users/120914",
"https://mathoverflow.net/users/149915",
"https://mathoverflow.net/users/2383",
"https://mathoverflow.net/users/35840"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631809",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368163"
}
|
Stack Exchange
|
Torsion & Artin groups
I have seen it conjectured several times that Artin groups are torsion-free. This is a very basic question one could ask about these groups. Intuitively, to me it seems like it must be true however it seems impossible to prove. I am curious if anyone is actually working on this / what kind of methods people may have tried to prove such a thing.
I know S. Rees and D. Holt have worked on this and related areas not too long ago. I think these slides could be a good place to start. D. Holt is very active on here, so he can probably give a good answer to this, if anyone can! It seems to me that a large chunk of the research is focussed on solving the word problem for Artin groups (which remains open afaik), but I have not looked at this very extensively myself.
Thanks for the reply! I'll take a look
All of the results about Artin groups that I know of concern specific types of Artin groups (spherical type, large type, right angled, and variations of these). Very little is known about Artin groups in general - for example, it is not known whether the word problem is solvable. One example of a general result (due to Crisp and Paris) is that the subgroup generated by the squares of the generators is a right angled Artin group with $[x^2,y^2] = 1 \Leftrightarrow [x,y]=1$.
@DerekHolt's reference: Crisp and Paris - The solution to a conjecture of Tits on the subgroup generated by the squares of the generators of an Artin group (MSN).
@DerekHolt oh cool! I’ll take a look
This is a consequence of the $K(\pi,1)$ conjecture, stating that there is an explicit $K(\pi,1)$ for Artin groups which is the complement of a complex hyperplane arrangement whose real locus are the hyperplanes of the reflections in the associated Coxeter group quotient by the action of the Coxeter group. See Conjecture 2.2 in Paris' survey Lectures on Artin groups and the $K(\pi, 1)$ conjecture for the statement and the discussion after for known cases. Since this is a finite-dimensional space (whose complex dimension is bounded by that of the number of generators), this implies that there cannot be any torsion. Recently Paolini and Salvetti have announced a Proof of the $K(\pi, 1)$ conjecture for affine Artin groups. See also the introduction of their paper for a summary of previous results and McCammond's survey The mysterious geometry of Artin groups: the affine case of the torsion-free problem was solved by McCammond and Sulway in Artin groups of Euclidean type.
|
2025-03-21T14:48:31.698126
| 2020-08-02T21:30:19 |
368168
|
{
"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:631810",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368168"
}
|
Stack Exchange
|
What proportion of $n m \times n m$ positive-definite fixed-trace symmetric (Hermitian) matrices remain positive-definite under a certain operation?
Given the class of $n m \times n m$ positive-definite (symmetric or Hermitian) fixed-trace (say, 1), $n,m\geq 2$, what "proportion" of the class remains positive-definite if either the $n^2$ blocks of size $m \times m$ or the $m^2$ blocks of size $n \times n$ are transposed in place?
Further, how might the answer to the question depend upon $k$, where
the rank of the matrices under consideration is fixed at $k \leq n m$?
Let us denote the matrices in question by $\rho_{nm}$ and, first, consider the use of the measure generated by the Hilbert-Schmidt metric
\begin{equation}
\mbox{d} s^2_{HS}=\frac{1}{2} \mbox{Tr}[(\mbox{d} \rho_{nm})^2].
\end{equation}
Then, A. Lovas and A. Andai
LovasAndaiPaper MR3673324
have formally answered the question for the case of symmetric $\rho_{22}$, obtaining the value $\frac{29}{64}$ for the proportion ("separability probability") of this nine-dimensional set of "two-rebit density matrices" that remain positive-definite under the indicated operation of "partial transposition".
In their Conclusions, Lovas and Andai write: ``The structure of the unit ball in operator norm of $2\times 2$ matrices plays a
critical role in separability probability of qubit-qubit and rebit-rebit
quantum systems.
It is quite surprising that the space of $2\times 2$ real or complex matrices
seems simple, but to compute the volume of the set
\begin{equation*}
\Big\{\begin{pmatrix}a & b\\ c& e\end{pmatrix} \Big\vert\ a, b, c, e \in \mathbb{K},
\Big| \Big|{\begin{pmatrix} a & b\\ c& e\end{pmatrix}} \Big| \Big| <1,\ \
\Big| \Big|{\begin{pmatrix} a & \varepsilon b\\ \frac{c}{\varepsilon}& e
\end{pmatrix}} <1 \Big| \Big|\Big\}
\end{equation*}
for a given parameter $\varepsilon\in [0,1]$, which is the value of
the function $\chi_{d}(\varepsilon)$, is a very challenging problem.
The gist of our considerations is that the behavior of the function
$\chi_{d}(\varepsilon)$ determines the separability probabilities with respect
to the Hilbert-Schmidt measure.'' (The operator norm $ \Big| \Big| \hspace{.15in} \Big| \Big|$ is the largest singular value or Schatten-$\infty$ norm.)
The function $\chi_{1}(\varepsilon)$--found employing an auxiliary "defect function"--which is used for the determination of the $\frac{29}{64}$ is given by
\begin{equation} \label{BasicFormula}
\tilde{\chi}_1 (\varepsilon ) = 1-\frac{4}{\pi^2}\int\limits_\varepsilon^1
\left(
s+\frac{1}{s}-
\frac{1}{2}\left(s-\frac{1}{s}\right)^2\log \left(\frac{1+s}{1-s}\right)
\right)\frac{1}{s}
\mbox{d} s
\end{equation}
\begin{equation}
= \frac{4}{\pi^2}\int\limits_0^\varepsilon
\left(
s+\frac{1}{s}-
\frac{1}{2}\left(s-\frac{1}{s}\right)^2\log \left(\frac{1+s}{1-s}\right)
\right)\frac{1}{s}
\mbox{d} s .
\end{equation}
Let us note that
$\tilde{\chi}_1 (\varepsilon )$ has a closed form,
\begin{equation} \label{poly}
\frac{2 \left(\varepsilon ^2 \left(4 \text{Li}_2(\varepsilon )-\text{Li}_2\left(\varepsilon
^2\right)\right)+\varepsilon ^4 \left(-\tanh ^{-1}(\varepsilon )\right)+\varepsilon ^3-\varepsilon
+\tanh ^{-1}(\varepsilon )\right)}{\pi ^2 \varepsilon ^2},
\end{equation}
where the polylogarithmic function is defined by the infinite sum
\begin{equation*}
\text{Li}_s (z) =
\sum\limits_{k=1}^\infty
\frac{z^k}{k^s},
\end{equation*}
for arbitrary complex $s$ and for all complex arguments $z$ with $|z|<1$.
Lovas and Andai left unanswered the (two-qubit) matter of (15-dimensional) Hermitian $\rho_{22}$.
In MasterLovasAndai MR3767844
Slater was able to construct--though yet without formalized proof--the much simpler
\begin{equation} \label{BasicFormula2}
\tilde{\chi}_2 (\varepsilon ) = \frac{1}{3} \varepsilon^2 (4-\varepsilon^2)
\end{equation}
leading to the two-qubit separability probability of $\frac{8}{33}$.
Also, in this paper, counterparts were given for quaternionic [$\tilde{\chi}_4 (\varepsilon ) = \frac{1}{35} \varepsilon^4 (84-64\varepsilon^2+15 \varepsilon^4)$ yielding $\frac{26}{323}$],...density matrices.
Then, these three formulas were incorporated into a "Master Lovas-Andai" formula--the index $d$ being a form of "Dyson-index" of random matrix theory-
\begin{equation}
\tilde{\chi}_d (\varepsilon ) = \frac{\varepsilon^d \Gamma(d+1)^3 \, _3\tilde{F}_2\left(-\frac{d}{2},\frac{d}{2},d;\frac{d}{2}+1,\frac{3 d}{2}+1;\varepsilon^2\right)}{\Gamma(\frac{d}{2}+1)^2},
\end{equation}
where the regularized hypergeometric function is indicated.
For $\alpha=\frac{d}{2}$, the desired probabilities ($\frac{29}{64}, \frac{8}{33},\ldots$) are yielded by
\begin{equation} \label{Hou1}
P(\alpha) =\Sigma_{i=0}^\infty f(\alpha+i),
\end{equation}
where
\begin{equation} \label{Hou2}
f(\alpha) = P(\alpha)-P(\alpha +1) = \frac{ q(\alpha) 2^{-4 \alpha -6} \Gamma{(3 \alpha +\frac{5}{2})} \Gamma{(5 \alpha +2})}{6 \Gamma{(\alpha +1)} \Gamma{(2 \alpha +3)}
\Gamma{(5 \alpha +\frac{13}{2})}},
\end{equation}
and
\begin{equation} \label{Hou3}
q(\alpha) = 185000 \alpha ^5+779750 \alpha ^4+1289125 \alpha ^3+1042015 \alpha ^2+410694 \alpha +63000 =
\end{equation}
\begin{equation}
\alpha \bigg(5 \alpha \Big(25 \alpha \big(2 \alpha (740 \alpha
+3119)+10313\big)+208403\Big)+410694\bigg)+63000.
\end{equation}
(Qing-Hu Hou helped in the derivation of this formula, using "Zeilberger's algorithm" ["creative telescoping"] ConciseFormula.)
An equivalent formula--now employing $d$ rather than $\alpha=\frac{d}{2}$--was given by C. F. Dunkl in App. D of MasterLovasAndai
\begin{equation}
\mathcal{P}(d) =3456^{d}\frac{\left( \frac{1}{2}\right) _{d/2}%
^{3}\left( \frac{7}{6}\right) _{d/2}^{2}\left( \frac{5}{6}\right)
_{d/2}^{2}\left( 2d\right) !}{\left( \frac{d}{2}\right) !\left( 3\right)
_{5d}}\sum_{i\geq0,j\geq0}^{i+j\leq d/2}\frac{\left( -\frac{d}{2}\right)
_{i+j}\left( \frac{d}{2}\right) _{j}\left( d\right) _{j}\left(
2+3d\right) _{i}\left( 1+d\right) _{i}}{\left( 2+\frac{5d}{2}\right)
_{i+j}\left( 1+\frac{d}{2}\right) _{j}i!j!\left( -2d\right) _{i}}.
\end{equation}
However, for dimensions $n$ or $m$ greater than 2, no analogous formulas are yet available.
Extensive numerical (quasirandom estimation) investigations
NumericalExact
have led to conjectures that for $n=3,m=2$ (or $n=2,m=3$) for symmetric ("rebit-retrit") density matrices the Hilbert-Schmidt probability in question is $\frac{860}{6561} =\frac{2 \cdot 5 \cdot 43}{3^8}$, and for Hermitian ("qubit-qutrit" density matrices, the corresponding probability is $\frac{27}{1000}=\frac{3^3}{2^3 \cdot 5^3}$.
However, despite these limited results pertaining to small $n,m$, Szarek, Bengtsson and Zyczkowski
StructureBody MR2200422 (2006i:81029)
were able to formally establish--specifically in the case of the Hilbert-Schmidt measure--that for all dimensions, both in the symmetric and Hermitian scenarios, the probability for the class of rank-$nm-1$ matrices is one-half that for the class of full rank ($nm$) matrices. The proof was accomplished by showing that the set of full rank ($nm$) matrices is "pyramid-decomposable", and hence is a body of constant height.
Ruskai and Werner
RuskaiWerner MR2525543 (2010h:81031)
have established that the probability in question is zero if the rank of the $n m \times n m$ density matrix is less than or equal to $\mbox{max}(n,m)$.
For rank-4 ($6 \times 6$) qubit-qutrit density matrices, certain numerical evidence suggests that the associated probability might be $\frac{1}{34}$ that of the rank-6 probability (conjectured, as indicated above, to be $\frac{27}{1000}$).
Additionally, other choices of measures on the density matrices have been considered (in particular, the "Bures", an example of an operator monotone measure)
[GeometryOfQuantumStates][4] MR3752196 (extensive review of first edition MR2230995 (2007k:81001))
for which a two-rebit estimate of 0.15709623 has been obtained, and a two-qubit conjecture of $\frac{25}{341} =\frac{5^2}{11 \cdot 31}$ advanced.
For asymptotic aspects of this question, see. Chap. 9 of
[AliceBobBanach][4] MR3699754
Let us consider approaching the problem from a numerical (random-matrix) viewpoint,
denoting the matrices in question by $\rho_{nm}$ and, first, considering the use of the measure generated by the Hilbert-Schmidt metric
\begin{equation}
\mbox{d} s^2_{HS}=\frac{1}{2} \mbox{Tr}[(\mbox{d} \rho_{nm})^2].
\end{equation}
Say we are interested in the proportion of $n m \times n m$ matrices of rank $k$ that remain positive-definite under the indicated operation of partial transposition, and desire to generate a matrix in that class random with respect to the Hilbert-Schmidt measure.
A procedure to do so—based on the Wishart-Laguerre distribution—has been outlined (not yet published) by K. Zyczkowski and the group of A. Khevelidze, I. Rogojin and V. Abgaryan.
In the case of Hermitian matrices, one generates a $k \times k+ 2 (nm -k)$ matrix $G$, the entries of which are complex numbers, the real and imaginary parts being normal random variates. Then, $G G^\dagger$ is normalized to have trace 1. (If $k=n m$, this is the desired density matrix.
RandomMatrix) If $k<nm$, the normalized matrix is diagonalized fo obtain an $k \times k$ matrix which is extended to an $nm \times nm$ matrix, the additional entries being zeros. Rotating this expanded matrix by a random unitary, the desired $nm \times nm$ matrix $\rho_{nm}$, random with respect to the Hilbert-Schmidt measure on the $nm \times nm$ Hermitian, positive-definite, trace 1 matrices of rank $k$ is obtained.
In the case of symmetric matrices, one generates a $k \times k+ 2 (nm -k)+1$ matrix $G$, the entries of which are normal random variates. Then, $G G^T$ is normalized to have trace 1. (If $k=n m$, this is the desired density matrix.) If $k<nm$, it is diagonalized fo obtain an $k \times k$ matrix which is extended to an $nm \times nm$ matrix, the additional entries being zeros. Rotating this expanded matrix by a random orthogonal one, the desired $nm \times nm$ matrix $\rho_{nm}$, random with respect to the Hilbert-Schmidt measure on the $nm \times nm$ symmetric, positive-definite, trace 1 matrices of rank $k$ is obtained.
We currently are implementing this procedure in two cases.
In one, we are analyzing $6 \times 6$ (rebit-retrit) density matrices of rank 4. Based on 675 million randomly generated (in the indicated manner) such matrices, we obtain an estimate of a probability of 0.00774073 of having a positive partial-transpose (equivalent to separability in this case).
Relatedly, in a previous study,
NumericalExact
we had advanced a conjecture of $\frac{860}{6561} =\frac{2 \cdot 5 \cdot 43}{3^8} \approx 0.131078$ for the $6 \times 6$ symmetric density matrices of full rank (6). (It has been formally proven that in the $4 \times 4$ counterpart, the Hilbert-Schmidt probability is $\frac{29}{64}=\frac{29}{2^6}$ LovasAndai MasterLovasAndai
.)
Now, we find it intriguing to propose a rank-4 rebit-retrit separability probability conjecture of $\frac{387}{5000} =\frac{3^2 \cdot 43}{2^3 \cdot 5^{4}} = 0.00774$. Subject to these two conjectures, the ratio of the rank-4 to rank-6 probabilities would be
$\frac{59049}{1000000}=\frac{3^{10}}{2^6 \cdot 5^6} \approx 0.059049$, with the common factor 43 cancelling.
Our second ongoing study is concerned with the rank-6 $(4 \times 2) \times (4 \times 2) $ Hermitian density matrices, so far--based on 149 million random matrices--obtaining estimates of 0.00129085 and
0.000054624 for the full-rank and rank-6 Hilbert-Schmidt probabilities, respectively.
|
2025-03-21T14:48:31.698838
| 2020-08-02T21:54:35 |
368169
|
{
"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:631811",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368169"
}
|
Stack Exchange
|
Does the Plücker embedding topology agree with the quotient topology?
Let $\mathbb{F}$ be an algebraically closed field of characteristic zero, let $k\leq n$ be positive integers, and let $V=\mathbb{F}^n$. One can view the Grassmannian $G(k,V)$ of $k$-planes in $V$ as an algebraic variety via the Plücker embedding, which sends a $k$-dimensional subspace $W \in G(k,V)$ to $[w_1 \wedge w_2 \wedge \dots \wedge w_k] \subseteq \mathbb{P}(\bigwedge ^k (V))$, where $w_1,\dots, w_k \in V$ are any vectors such that $W=\text{span}\{w_1,\dots, w_k\}$. Indeed, it is easily seen that this map is injective, and its image is the set of decomposable elements of $\mathbb{P}(\bigwedge ^k (V))$, which forms a projective variety. In particular, this gives a Zariski topology on $G(k,V)$.
Another natural way put a topology on $G(k,V)$ is via a quotient topology. Note that $G(k,V)$ is naturally bijective as a set with $S/\equiv$, where $S \subseteq \mathbb{F}^{n k}$ is the set of full-rank $n \times k$ matrices, and $\equiv$ is the equivalence relation on $S$ given by: $A \equiv B$ if and only if there exists $X \in GL_k(\mathbb{F})$ such that $A=BX$. The bijection is given by $W \mapsto (w_1,\dots, w_k)$. The set $S$ forms an algebraic variety, and hence carries a (Zariski) topology.
Does the topology on $G(k,V)$ induced by the quotient topology on $S/\equiv$ agree with the topology on $G(k,V)$ induced by the Plücker embedding?
Cross-posted from Math.SE
|
2025-03-21T14:48:31.698975
| 2020-08-02T23:20:54 |
368172
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"D.-C. Cisinski",
"Denis Nardin",
"Peter Scholze",
"Z. M",
"curious math guy",
"https://mathoverflow.net/users/1017",
"https://mathoverflow.net/users/152554",
"https://mathoverflow.net/users/176381",
"https://mathoverflow.net/users/43054",
"https://mathoverflow.net/users/6074"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631812",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368172"
}
|
Stack Exchange
|
Homology of the étale homotopy type
$\DeclareMathOperator\Et{Et}$Let $X$ be a scheme and denote by $\Et(X)$ the associated étale homotopy type. Then by the work of Artin–Mazur, we know that for an abelian group $A$, we have
$$H^n(\Et(X),A)=H^n_{\text{ét}}(X,A)$$
and
$$\pi^1(\Et(X))\cong \pi^1_{\text{alg}}(X).$$
Therefore I wonder what $H_n(\Et(X),A)$ is? Naïvely I would hope that it is $H_{n,\text{ét}}(X,A)$, however, it seems that no survey or reference mentions this.
What's your definition of étale homology?
In the case that $A=\mathbb{Q}_p$ (which is the case I'm most interested in), I know it as the dual of the étale cohomology.
In that case, doesn't the theorem basically follow from the theorem for cohomology by taking duals?
It does absolutely, but I'm wondering if this agrees with more general coefficients. The way I've heard about constucting the étale homology is to consider étale cosheaves and their derived functor.
I'm sure there are easier and better ways to think about this, but here's how I like to think about it.
Work on the big pro-etale site on all schemes, which maps to the pro-etale site of a point, $\pi: \mathrm{Sch}_{\mathrm{proet}}\to \ast_{\mathrm{proet}}$. Also, let's consider (hypercomplete) sheaves of anima. Then $\pi^\ast$ has a left adjoint $\pi_\natural$, and (for our given scheme $X$) $\pi_\natural(X)$ is a condensed anima that "is" the etale homotopy type of $X$. Concretely, if $X_\bullet\to X$ is a proetale hypercover by w-contractible $X_\bullet$, then $\pi_\natural(X)$ is represented by the simplicial extremally disconnected profinite set $\pi_0(X_\bullet)$. By adjunction, there is a natural map $X\to \pi^\ast \pi_\natural (X)$.
Now giving a $\mathbb Q_\ell$-local system $\mathbb L$ (same for other coefficient rings) is the same as giving a map $X\to \pi^\ast B\mathrm{GL}_n(\mathbb Q_\ell)$, i.e. equivalently a map $\pi_\natural(X)\to B\mathrm{GL}_n(\mathbb Q_\ell)$, i.e. a $\mathbb Q_\ell$-local system on the condensed anima $\pi_\natural(X)$.
Now what is homology of $X$ with coefficients in $\mathbb L$? One definition uses the formalism of solid $\mathbb Q_\ell$-sheaves $D_\blacksquare(X,\mathbb Q_\ell)$ (no reference for them in the case of schemes yet, sorry!). In that setting, pullback $D_\blacksquare(\ast,\mathbb Q_\ell)\to D_\blacksquare(X,\mathbb Q_\ell)$ has a left adjoint, which takes $\mathbb L$ to the homology of $X$ with coefficients in $\mathbb L$. Here $D_\blacksquare(\ast,\mathbb Q_\ell)$ is the "usual" derived category of solid $\mathbb Q_\ell$-modules.
Working with the condensed anima $\pi_\natural(X)$ instead, one can apply a similar procedure, and the two answers will agree, so indeed the homology of $X$ with coefficients in $\mathbb L$ agrees with the homology of $\pi_\natural(X)$ with coefficients in $\mathbb L$.
In practice, if $X$ is sufficiently nice, then these homology groups will be finite-dimensional over $\mathbb Q_\ell$ and dual to cohomology, so this would indeed follow from the usual statements about cohomology. For general $X$, this statement about homology is however slightly finer (as cohomology is always the dual of homology, but not conversely).
Edit: Here is a way to phrase the answer so that it does not involve a reference to $D_\blacksquare(X,\mathbb Q_\ell)$. One can instead use the full derived category of pro-etale $\mathbb Q_\ell$-sheaves $D(X_{\mathrm{proet}},\mathbb Q_\ell)$; then, just like for sheaves of anima, the pullback along $f: X_{\mathrm{proet}}\to \ast_{\mathrm{proet}}$ has a left adjoint $f_\natural$, and $f_\natural \mathbb L\in D(\ast_{\mathrm{proet}},\mathbb Q_\ell)$ is a complex of condensed $\mathbb Q_\ell$-vector spaces that can be considered as the homology of $\mathbb L$. Again, a similar construction can be done for $\pi_\natural(X)$, and these two notions of homology agree.
On the other hand, I expect that it is extremely difficult to compute this notion of homology even for $X=\mathbb P^1_k$ for $k$ an algebraically closed field. However, passing to the solidification, one can compute it in practice, and I guess it usually agrees with the homology of the Artin--Mazur pro-(homotopy type).
I do like very much everything you do on analytic rings and condensed mathematics, but I have trouble to see why solid modules are necessary here. I mean that the derived category of pro-étale l-adic sheaves, as documented in your joint paper with Bhatt, is sufficient to make sense of the left adjoint of the pullback functor: we can simply apply the $l$-adic version of $\pi_\sharp$ to $\mathbb{L}$ to get homology with coefficients. Could you explain what to expect from the consideration of solid modules everywhere?
As long as $X$ is nice, you are right, the left adjoint exists on the usual level, but not for general $X$. If $X=\mathrm{Spec} \mathbb Q$ for example, $H^1(X,\mathbb F_2)$ is infinite, and dually $H_1(X,\mathbb F_2)$ ought to be an infinite product of $\mathbb F_2$'s. So you need some solid formalism, I think.
On the other hand, I wouldn't be completely sure how to define homology of the pro-etale homotopy type with coefficients. I guess usually one would use coefficients that come from stage in the pro-limit, and then take the projective limit of the homologies? I think this implicit passage to the projective limit is what's mirrored here by the use of solid modules, specifically $\mathbb Z[S]^\blacksquare = \varprojlim_i \mathbb Z[S_i]$ for profinite $S=\varprojlim_i S_i$.
I should maybe mention that one advantage of using the condensed anima $\pi_\natural(X)$ in place of the Artin--Mazur pro-homotopy type is that $\pi_\natural(X)$ can see general $\mathbb Q_\ell$-local systems (as in the argument above), which I think is not the case for the latter (whose $\pi_1$ is given by the SGA3 $\pi_1$ which is too small for say nodal curves of positive genus). But then to define the homology of a condensed anima, I think I really need the solid formalism.
The pullback functor $\pi^*$ is well defined for $l$-adic sheaves and it has a left adjoint exactly as (in fact because it exsists in the case of) sheaves of anima. I would define homology by applying this left adjoint $\pi_\sharp$. I do not think and finitess assumtion is needed.
This fits with homology à la Artin-Mazur: they define homology as the pro-group obtained by applying the (left derived) pro-adjoint of the constant sheaf functor.
Ah, I see what you say. The problem is that if you apply the left adjoint on the full derived category of pro-etale $\mathbb Q_\ell$-sheaves, I don't think you can compute the answer: You need to solidify to get a sensible answer. On the other hand, you are right that it's not necessary for the comparison! But if you want this notion of homology to agree with the Artin--Mazur one, you need to solidify.
The solidification is needed only in the absence of finiteness assumptions then, is that what you claim? I would expect that the "classical" left adjoint of $f^$ is well behaved whenever the formation of $f_$ is compatible with arbitrary base change: for instance, if $f$ is proper or if the target of $f$ is $0$-dimensional (and $f$ of finite presentation say).
No, I think it's impossible to control the left adjoint to $f^\ast$ even in the best possible cases, see my edited answer.
Sorry to insist, but the left adjoint of $f^*$ exists whever $f$ is a morphism of finite presentation with $0$-dimensional codomain, at least if we restrict to constructible $l$-adic sheaves (with $l$ prime to the residue characteristics). This left adjoint sends the constant sheaf to $f_! f^!$ of the constant sheaf.
Right -- this is the left adjoint I was referring to in the beginning of my comments. But if you take the one on general pro-etale $\mathbb Q_\ell$ (or even $\mathbb F_\ell$)-sheaves, it gets more complicated. So you have three options: Restrict to nice $X$ and nice coefficients, and get a good left adjoint; take general $X$ and solid sheaves, and get a good left adjoint; take general $X$ and all pro-etale sheaves, and get a weird left adjoint.
Ah, I think I follow you now: if we take $\mathbb{F}_l$ coefficients say, what I describe above would fit with the Artin-Mazur kind of construction (literally) whereas the left adjoint in the pro-étale setting would only fit with the actual Artin-Mazur construction, only after solidification (if your guess is correct!).
Sorry, is there a reason (or better, a reference) for the left adjoint $f_\natural$ being called "relative homology", i.e. how does that coincide with the classical notion of homology?
OK, this is answered here.
|
2025-03-21T14:48:31.699512
| 2020-08-02T23:40:34 |
368174
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Neal",
"Will Jagy",
"https://mathoverflow.net/users/20796",
"https://mathoverflow.net/users/3324"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631813",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368174"
}
|
Stack Exchange
|
On isospectral planar domains (and a paper by Buser, Conway, Doyle and Semmler)
I have never seen a short, elegant way (from the viewpoint of a non-topologist) which constructs isospectral planar domains from Sunada group triples, although essentially those triples live at the very origins of the first constructions of such pairs. In a paper by Buser, Conway, Doyle and Semmler, the authors try to easily describe such examples in the orbifold language, and here is an example.
I find the second paragraph pretty unclear, and especially the fact that those quotients are planar (the most important feature), is too vague for me, and just too short.
My question(s):
(1) can anyone explain this second paragraph in clear terms to me, knowing that I am not a topologist ?;
(2) is there a more elegant ("direct") way to pass from the isospectral hyperbolic examples, to the plane ?
For reference, here is a link to the paper hosted by Peter Doyle.
recommend Conway's little book, The Sensual Quadratic Form from about the same time, and usually downloadable after search. The big book, SPLAG, will have more
|
2025-03-21T14:48:31.699617
| 2020-08-02T23:59:01 |
368175
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Igor Belegradek",
"Mark Grant",
"https://mathoverflow.net/users/1573",
"https://mathoverflow.net/users/2082",
"https://mathoverflow.net/users/8103",
"shuhalo"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631814",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368175"
}
|
Stack Exchange
|
Extending smooth triangulation
Can one always extend a smooth triangulation from a smooth submanifold $S$ to the ambient manifold $M$? (For simplicity both $S$ and $M$ are compact without boundary). Is the extension possible when $S$ is a sphere?
What I know is summarized in this answer. Namely, in the PL category the answer is yes, by a result of Armstrong, i.e., one can always extend a triangulation from a locally flat PL submanifold. Also one can always extend a smooth triangulation from a codimension one separating hypersurface (this is due to Munkres). I think, the same is true for non-separating hypersurfaces. Thus it is enough to extend the smooth triangulation of $S$ to a tubular neighborhood of $S$.
Naively, I would expect that the extension is always possible. Otherwise, one can probably use it to define a potentially interesting invariant of smooth manifolds, and I have never heard of such an invariant.
Is this the same as this question https://mathoverflow.net/q/206212/8103?
@MarkGrant: it is not the same. An exercise in Munkres' "Elementary Differential topology", and probably, a theorem in Verona's book in your linked answer says that there is a smooth triangulation of $M$ in which $S$ is a subcomplex. What I ask for is to extend a smooth triangulation of $S$ to a smooth triangulation of $M$. I don't want to change the given triangulation of $S$. I do not want to subdivide it.
A good example is PL knotted ball pair $(B^4, B^2)$ that is the cone on a nontrivial knot in $S^3$. There is of course a triangulation of $B^4$ that restricts to a triangulation of $B^2$ but if we triangulate $B^2$ as a simplex, this triangulation does not extend to a triangulation of $B^4$ because it does not capture the local knottedness at the cone point.
You mention a result by Munkres on smooth separating hypersurfaces, which I presume is in his textbook. Do you know whether it extends to topological separating hypersurfaces?
|
2025-03-21T14:48:31.699784
| 2020-08-03T00:38:19 |
368178
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Gordon Royle",
"Jimmy",
"Matthieu Latapy",
"https://mathoverflow.net/users/1492",
"https://mathoverflow.net/users/158328",
"https://mathoverflow.net/users/8012"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631815",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368178"
}
|
Stack Exchange
|
Degree sequences after vertex removals
Consider a graph $G=(V, E)$ with $|V|=n$ vertices. Let $(v_1, \dots, v_n)$ be an ordered list of its vertices. Let $G_i=G[\{v_{i+1}, \dots, v_n\}]$ be the induced subgraph on the last $n-i$ vertices. Let $(d_1, \dots, d_n)$ be the degree sequence of $G_i$: $d_i$ is the degree of $v_i$ in $G_i$.
I wonder whether such sequences have been studied, as they raise natural and interesting questions.
For instance, my original motivation is to obtain, for a given graph, such a sequence that is maximum with respect to the majorization order. A natural strategy would be to choose $v_1$ with maximum degree in $G_0$, $v_2$ with maximum degree in $G_1$, and so on. But there is an ambiguity when several vertices have the maximum degree, and other sequences may make sense.
Thank you for any information and suggestions.
This is not the same, but has a similar feel - if you repeatedly remove a vertex of minimum degree, and take the maximum value you encounter during this process, then you get a well-defined graph parameter called the degeneracy of the graph. Maybe this gives an example you might try to generalise.
Thanks @GordonRoyle. This is indeed of similar feel and could give a good start. I will look into this notion and think further on it.
Two somewhat related questions were posted here: https://mathoverflow.net/questions/378933/random-subgraph-properties and here: https://mathoverflow.net/questions/381631/are-two-degree-sequences-compatible-for-random-simple-graph-generation
Several works study the degree sequence obtained when vertices are removed from random graphs with given degree sequence. Vertices are generally removed by decreasing order of degrees, or uniformly at random, but other variants are considered. Some empirical works also consider various kinds of real-world complex networks (like, e.g., the internet, web graphs, or biological networks).
The general goal of these works is to estimate the robustness of graphs to failures (random removals) and attacks (targeted removals), through the size of the largest connected component of the remaining graph. In the random setting, this is strongly related to the remaining degree sequence. And in all cases, it is shown that targeted attacks and failures have similar effects on purely random graph (ER model), whereas degree sequences met in practice make graphs much more sensitive to attack than to failures, and than purely random graphs.
Notice that this is strongly related epidemiology: vertex removals may be seen as vaccination campaigns, and the question of which individuals to vaccinate first certainly is the hottest topic these days.
I co-authored in 2009 a quite comprehensive survey of these works, but surely much progress has been made since then. This may be a good entry point, though, as we put some efforts in trying to shed light on the underlying assumptions and formal approaches, when available.
|
2025-03-21T14:48:31.700008
| 2020-08-03T01:01:45 |
368179
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Denis Nardin",
"curious math guy",
"https://mathoverflow.net/users/152554",
"https://mathoverflow.net/users/43054"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631816",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368179"
}
|
Stack Exchange
|
Essential Image of the Étale Homotopy type
For any scheme $X$ we can associate the étale homotopy type $Et(X)$, which is a pro-object in the homotopy category of CW-complexes. My question is, do we have a good understanding of the essential image, i.e., do we have a set of necessary and sufficient conditions on a pro-CW-complex for it to be the étale homotopy type of some scheme. Even more "ambitiously", do we have such a set of conditions such that a pro-CW-complex is the étale homotopy type of a smooth projective scheme $K$-scheme?
@vrz The étale homotopy type is perfectly well defined for any scheme, by taking the shape of the étale topos, although probably considering only the profinite shape would make the question more approachable.
@vrz I'm happy to assume that $K$ is spearably closed and characteristic zero, if that helps in any way.
|
2025-03-21T14:48:31.700096
| 2020-08-03T02:10:20 |
368182
|
{
"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:631817",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368182"
}
|
Stack Exchange
|
Question about generalizing Cauchy identity
One of the Cauchy identities says that
$$\prod_{i,j}(1+x_iy_j) = \sum_\lambda s_\lambda (x_1, \cdots,x_m) s_{\lambda'}
(y_1, \cdots,y_n) $$
Where $\lambda$ is a Young diagram, $\lambda'$ is the transpose diagram, and $s$ is the Schur polynomial.
I would like to elevate $x_i$ and $y_j$ to $d$ dimensional vectors $\vec x_i$ and $\vec y_j$. Is there any generalized Cauchy identity, or any other nice closed form, for $$\prod_{i,j}(1+\vec x_i\cdot \vec y_j),$$ with $d>1$?
|
2025-03-21T14:48:31.700277
| 2020-08-03T03:24:17 |
368183
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Descartes Before the Horse",
"GH from MO",
"Gerhard Paseman",
"https://mathoverflow.net/users/11919",
"https://mathoverflow.net/users/128941",
"https://mathoverflow.net/users/3402"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631818",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368183"
}
|
Stack Exchange
|
Updates on a least prime factor conjecture by Erdos
In the 1993 article "Estimates of the Least Prime Factor of a Binomial Coefficient," Erdos et al. conjectured that
$$\operatorname{lpf} {N \choose k} \leq \max(N/k,13)$$
With finitely many exceptional $(N,k)$. Here, $\operatorname{lpf}(x)$ denotes the smallest prime factor of $x$.
I am posting here to ask whether any progress has been made toward this conjecture.
As written, this does not hold.(Take 194 choose 192.) Perhaps you could state the conjecture precisely? Gerhard "Such Factors Are Often Small" Paseman, 2020.08.02.
@GerhardPaseman I have modified the post accordingly; what I missed in the original paper is that it was conjectures there were finitely many exceptional $(N,k)$.
There are at least twelve exceptions to the inequality stated above. These are listed in the paper by Erdős at al., under Definition 3.
The authors of the paper you mention, Erdos, Lacampagne, and Selfridge, define $p(m)$ to be the least prime divisor of $m$ and concern themselves what can be said about $p(\binom{n}{k}).$ I suspect that Selfridge wrote the article. It has his style of saying a lot in a succinct way which is puzzling but solvable with some thought on the part of the reader. The conjecture stated in the abstract is $$p(\binom{n}{k}) \leq \max(\frac{n}{k},29).$$ That shouldn't be thought of as their big conjecture but rather a terse and amusing way of capturing some of the the main points.
The short answer to your question is that they did a lot of computation, made some observations that had theoretical backing and strong computational support. No-one, as far as I know, has challenged or refuted them and perhaps it isn't especially attractive to try further computation. Or perhaps it is, but not to report "I didn't find anything else either."
Aside: They are perhaps more interested in the growth rate of $g(k),$ the minimal $n>k$ with $p(\binom{n}{k})>k.$ I feel compelled to quote a small stretch of the article:
That is a whole mess of conjectures, but not snappy enough for an abstract. That is the subject of section 1 of the paper. They and others explored $g(k)$ up to about $k=140$ and with more powerful computers the results were later extended to about $k=200.$ The current record lower bound is $$g(k) \geq exp(c(\log^3k/\log \log k)^{1/2}).$$
Getting back to the conjecture you ask about, the first puzzle is
The stated conjecture is clearly false. $\binom{n}{n-1}=n$ is prime when $n$ is. I can imagine Selfridge saying "Well of course we don't mean that." And if you read in further, the investigation is only for $k<n/2.$ The case $n=2k$ is a very small puzzle left to you.
Another puzzle is
Why $\frac{n}{k}?$ Is that best possible? Here is my take on that, read the article for a more elegant and general treatment: Suppose $p=q(k-1)!+1$ is prime. Then, for $n=pk,$ $$\binom{n}{k}=\prod_{i=0}^{k-1}\frac{n-i}{k-i}$$ where the $k$ factors are integers. If they all happen to be prime then $p(\binom{n}{k})=p=\frac{n}{k}.$ I can convince myself that for every $k$ we would expect that to happen infinitely often. Perhaps that is alluded to in the article or perhaps it is too obviously right (or wrong) to mention.
Another (small) puzzle is
How you could ever have $p(\binom{n}{k})>\frac{n}{k}?$ After all, there is a $0 \leq j <k$ with $\frac{n-j}{k}$ an integer, so $\frac{n}{k}$ seems a clear upper bound. And it is, for large enough $n.$
However, while $\binom{62}{6}$ is divisible by some divisor of
$\frac{60}{6},$ that divisor is $1$ !
$$\binom{62}{6}=\frac{62}{2}\cdot 61 \cdot \frac{60}{60}\cdot 59
\cdot \frac{58}{2} \cdot \frac{57}{3}$$
They conjecture that, for $n \gt k^2$, $p(\binom{n}{k}) \geq \frac{n}{k}$ with that one exception of $p(\binom{62}{6})=\frac{n-5}{3}=19$
They also conjecture that this one, along with $p(\binom{959}{6})=19,p(\binom{474}{66})=23$ and $p(\binom{284}{28})=29$ are the only cases with $p(\binom{n}{k}) \gt \min(\frac{n}{k},19).$
They find eight cases with $p(\binom{n}{k})=17>\frac{n}{k}.$
They say that they wrote a program to find all cases of $p(\binom{n}{k})=p>\frac{N}{k}$ with $p>5$ and $k \leq 12000.$ It must not have been entirely exhaustive because they say that there was only one output other than the twelve mentioned for $331 <k <625$ and that was $p(\binom{3574}{406})=13$. They continue "Thus, at this point in time, it is possible that $p(\binom{n}{k})\leq\max(\frac{n}{k},13).$" So that is short of making a conjecture, but I don't know that there are any exceptions known other than the thirten they mention.
One might wonder why they said $p>5.$ Anyone familiar with Pascal's Triangle $\mod 2$ will realize that for every $k>2$ there are lots of cases of $p(\binom{n}{k})=3$ with $2k<n<3k.$ The article gives a nice proof that there is always at least one case of $p(\binom{n}{k})=5$ with $3k<n<4k.$
There is much more in that article, but I will stop there.
The conjecture as written is false:
Let $N=194+(2*3*5*7*11*13)*2n$, $k=N-2$, where $n$ is a natural number.
Then $C(N,k)=C(N,2)=(97+2*3*5*7*11*13*n)(193+2*3*5*7*11*13*2n)$, having no prime factors $\leq 13$.
|
2025-03-21T14:48:31.700660
| 2020-08-03T05:08:05 |
368185
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Nick Addington",
"Sasha",
"https://mathoverflow.net/users/16914",
"https://mathoverflow.net/users/4428"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631819",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368185"
}
|
Stack Exchange
|
Reference request: excess normal bundle and derived pullback
Consider a fiber square
$\require{AMScd}$
\begin{CD}
X' @>i'>> Y'\\
@V g V V @VV f V\\
X @>>i> Y,
\end{CD}
where $i$ and $i'$ are regular immersions, and consider the excess normal bundle defined by the exact sequence
$$ 0 \to N_{X'/Y'} \to N_{X/Y} \to E \to 0, $$
which measures the failure of $f$ to be transverse to $i(X)$ in the sense of differential topology.
Does anyone know a reference for the fact that $L_j f^*(i_* \mathcal O_X) = \Lambda^j E^*$?
If $f$ is also a regular immersion then this is SGA 6, VII, Proposition 2.5, although that's not the friendliest reference. If need be I can derive the fact I want from that special case, but I'd rather just have it off the shelf.
Perhaps, Proposition 1.28 in https://arxiv.org/abs/1411.7994 may help.
Thanks Sasha, this is very nice too!
I have an idle dream of writing a sequel to Huybrechts' Fourier-Mukai book that includes all these little things that I need all the time. Or maybe you can do it, you're much energetic than I am.
I have no experience in writing books, so I would be afraid to start.
|
2025-03-21T14:48:31.700776
| 2020-08-03T05:21:49 |
368187
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"erz",
"https://mathoverflow.net/users/53155"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631820",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368187"
}
|
Stack Exchange
|
Is there a complex Katetov-Tong theorem?
For any bounded function $f: X \to \mathbb R$, not-necessarilly continuous, one can define for any $x$, the real functions
$$
\limsup f(x) = \inf_{U\in\mathcal V_x} \sup_{u\in U} f(u)
$$
and
$$
\liminf f(x) = \sup_{U\in\mathcal V_x} \inf_{u\in U} f(u),
$$
where $\mathcal V_x$ denotes the set of open neighborhoods of $x$. Then, we can measure the oscilation of $f$ in $x$ as $$o_f(x) = \limsup f(x)-\liminf f(x).$$ The greater the value of $o_f(x)$ the farther $f$ is from any function $g$ continuous at $x$.
The Katetov-Tong theorem guarantees that, provided $X$ is a normal space, there exists a continuous function $g: X \to \mathbb R$ such that $$\|f-g\|=\frac{1}{2}\sup_{x\in X} o_f(x),$$
where the norm considered is the usual supremum norm.
I would like to know whether it is possible to obtain an analogous result for functions $f: X \to \mathbb C$. The closest result I got from this relies on Michael's Selection Theorem. I was able to prove that, provided $X$ is a paracompact space, for any $f: X \to \mathbb C$, there exists a continuous $g: X\to \mathbb C$ such that
$$
\|f-g\|=\frac{1}{2}\sup_{x\in X} o_f(x),
$$
where $o_f(x)$ is defined as the diameter of the smallest ball containing all limit points of nets of the form $(f(x_i))_{i\in I}$ with $x_i\to x$. This definition coincides with the definition of $o_f$ for real functions.
I wonder if this result is valid not only for paracompact spaces but also for normal spaces, giving us a complex version of Katetov-Tong's theorem.
This gives a worse estimate, but you can just say that $f=(f_1,f_2)$, where $f_1$ and $f_2$ are real-valued, and then produce continuous $g_1$ and $g_2$ based on them.
|
2025-03-21T14:48:31.700921
| 2020-08-03T05:59:37 |
368188
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Alessandro Codenotti",
"aduh",
"https://mathoverflow.net/users/49381",
"https://mathoverflow.net/users/96899"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631821",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368188"
}
|
Stack Exchange
|
Does a generalization of Tietze's extension theorem hold for set-valued functions?
Let $X$ be a normal topological space. Tietze's extension theorem says that if $A \subset X$ is closed, then a continuous function $f: A \to \mathbb R^n$ can be extended to a continuous function whose domain is all of $X$. If $X$ is assumed to be a metric space, then the theorem holds for functions taking values in any locally convex linear space (see this).
I am wondering if the theorem holds for certain set-valued functions.
In particular, let $\phi$ be a function from closed $A \subset X$ into compact subsets of $\mathbb R^n$. To say that $\phi$ is continuous on $A$ means that the following conditions are met for every $x \in A$:
(1) For every neighborhood $U$ of $\phi(x)$, there is a neighborhood $V$ of $x$ such that $\phi(y) \subset U$ for all $y \in V$;
(2) For every open subset $U$ of $\mathbb R^n$ for which $\phi(x) \cap U \neq \emptyset$, there is a neighborhood of $V$ of $x$ such that $\phi(y) \cap U \neq \emptyset$ for all $y \in V$.
Can $\phi$ be extended to a continuous set-valued function whose domain is all of $X$?
In principle, I don't mind assuming that $X$ is actually a subset of $\mathbb R^m$.
Is this equivalent to asking whether $K(\Bbb R^n)$ with the Vietoris topology is an absolute extensor?
@AlessandroCodenotti After looking up "absolute extensor", I believe the answer is yes by Martin's comment to this answer. Does an affirmative answer to my question then follow from the result mentioned in the first paragraph?
|
2025-03-21T14:48:31.701062
| 2020-08-03T08:28:47 |
368189
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Mathsfreak",
"Piotr Hajlasz",
"Raziel",
"https://mathoverflow.net/users/121665",
"https://mathoverflow.net/users/13915",
"https://mathoverflow.net/users/163130"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631822",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368189"
}
|
Stack Exchange
|
Proof of Rashevskii-Chow theorem
I'm looking for a good quotation and comprehensive explaination of the theorem of Chow-Rashewski.
I'm writing my thesis on sub-Riemannian Geometry and a special control problem. Therefore I want to state the theorem of Chow–Rashewski in its sub-Riemannian version and prove it:
Let $M$ be a connected manifold and $\Delta$ a distribution on $M$ that is bracket generation then there is for every $p,q\in M$ a curve that is almost everywhere horizontal that connects $p$ with $q$.
And I also have an additional question sometimes it says that this curve has to be horizontal almost everywhere and sometimes it says everywhere, why?
Thank you all for your explainations so far. I was working now on the book of Agrachev and I understand everthing apart of the connection between the bracket generating condition and the differentials of the functions $\phi_{i}$ could once again please someone help me to understand this last but central step?
As a reference, in addition to the classical ones cited above, I can recommend the following:
Agrachev, Andrei; Barilari, Davide; Boscain, Ugo, A comprehensive introduction to sub-Riemannian geometry., ZBL07073879.
The proof of the Chow-Rashewski theorem is in Section 3.2. An electronic version of the book is also freely available online (https://www.imj-prg.fr/~davide.barilari/ABB-v2.pdf)
The idea is of course the same as the one in the proof given above by Piotr Hajlasz, but I think that the presentation in the book is more geometric and concise.
Concerning your last question (everywhere vs almost everywhere). Horizontal curves might not be differentiable at certain points (e.g. think at a curve with a corner). In order to define a length, the tangent vector of an horizontal curve $\gamma:[0,1]\to M$ should be defined almost everywhere on $[0,1$]. There are then several regularity classes of curves which one might use (all used in the literature):
$\gamma \in W^{1,1}$ that is absolutely continuous curves (the largest class one can think of)
$\gamma \in W^{1,2}$ that is absolutely continuous curves whose tangent vector is $L^2$ (slightly smaller, but natural in view of minimization of the energy functional, and furthermore the space of "admissible velocities" is Hilbert)
$\gamma \in W^{1,\infty}$ that is curves that are locally Lipschitz in charts (as I comment blow, also this class is natural as one can always reduce to this case when dealing with the length-minimization problem)
in any case, of course, the tangent vector, which is defined almost everywhere, is required to belong to the sub-Riemannian distribution. The proof of the Chow-Rashevskii theorem shows that connectivity is achieved by horizontal curves that are concatenation of a finite number of smooth curves, which belongs to all the classes above (so the choice of regularity class above is irrelevant).
It turns out that also the sub-Riemannian distance (defined as the infimum of the length of horizontal curves between two points) does not depend on the choice of the regularity class. This is due to the fact that, within a given regularity class ($W^{1,1}$, $W^{1,2}$ or $W^{1,\infty}$) one can always reparametrize the curve, without changing its length, in such a way that the reparametrized curve has constant speed. This is proved in Section 3.6 of the book by Agrachev, Barilari and Boscain.
I was working with the proof of Agrachev and I did not get yet the connection between the generated Lie Algebra and the differentials of $\phi_{i}$ could someone help me explain line 3 on page 80 why this is a contradiction? Apart of that I got the proof. Thank you all for your help
Assume that, on the contrary, for any $f \in \mathcal{F}$ we have that $f|_{\Sigma_2}$ is tangent to $\Sigma_2$. It follows that any iterated bracket, restricted to $\Sigma_2$, would still be tangent to $\Sigma_2$. This contradicts the bracket generating conditions.
does your argument mean since $f\vert_ {\Sigma_{2}}$ can not be tangent to $\Sigma_{2}$ everywhere on $\Sigma_{2}$ because then the bracket generation condition in $q_{1}$ would not hold we have to find $q_{2}$ where $f(q_{2})$ is not tangent? I'm just wondering why we can use the generation condition in a different point
Probably my question is why we have to take always different points $q_i$ in each step and can not stay in $q_0$?
I think the answer to my question is that is $|\mathcal{F}|<dim(M)$ we can not find enough linear independent vectorfields in $q_0$
The original references are:
W.-L. Chow, Über Systeme von linearen partiellen Differentialgleichungen erster Ordnung. Math. Ann. 117 (1939), 98–105
P. K. Rashevsky, Any two points of a totally nonholonomic space may be connected by an admissible line.
Uch. Zap. Ped. Inst. im. Liebknechta, Ser. Phys. Math. 2 (1938), 83–94 (in Russian).
There are several different proofs of this result. I learned it from Proposition III.4.1 in:
N. Th Varopoulos, L. Saloff-Coste, T. Coulhon, Analysis and Geometry on Groups. Cambridge University Press.
The proof is quite concise, but not too difficult. Here is my own version of the proof from that book.
Chow–Rashevsky theorem
Let $Z$ be a smooth vector field and $Z_{t}$ the local $1$-parameter
family of diffeomorphisms associated with $Z$. Fix $f\in C^{\infty}$
and a point $m$. Then the function $h(t) = f(X_{t}(m))$ is smooth
and $h^{(k)}(0) = (X^{k}f)(m)$. Hence the Taylor series for
$h$ at $t=0$ is given by
\begin{equation}
(1)\qquad \sum_{k=0}^{\infty} X^{k}f(m) \frac{t^{k}}{k!},
\end{equation}
which means
$$
h(t) = \sum_{k=0}^{i} X^{k}f(m) \frac{t^{k}}{k!} + O(t^{i+1})
\qquad {\rm as} \ t\to\infty.
$$
We will use the formal expresion $(e^{tX}f)(m)$ to denote (1).
Let $Z_{1},\ldots,Z_{k}$ be smooth vector fields. Let $f\in C^{\infty}$.
Fix a point $m$ and define
$$
H(t_{1},\ldots,t_{k}) =
f(Z_{1,t_{1}}\circ Z_{2,t_{2}} \circ\cdots \circ Z_{k,t_{k}}(m)).
$$
Note that
$$
\frac{\partial^{m_{1}}}{\partial t_{1}^{m_{1}}}
H(0,t_{2},\ldots,t_{k}) =
(Z_{1}^{m_{1}} f) (Z_{2,t_{2}} \circ\cdots\circ Z_{k,t_{k}}(m)).
$$
Taking then the derivatives with respect to $t_{2},\ldots,t_{k}$
yields
$$
\frac{\partial^{m_{1}+\ldots+m_{k}}}{\partial
t_{1}^{m_{1}}\ldots \partial t_{k}^{m_{k}}}
H(0,\ldots,0) =
(Z_{k}^{m_{k}}\ldots Z_{1}^{m_{1}}f)(m).
$$
Hence the Taylor series for $H$ is given by
$$
\sum_{m_{1}=0}^{\infty} \ldots \sum_{m_{k}=0}^{\infty}
\frac{t_{1}^{m_{1}}\ldots t_{k}^{m_{k}}}{m_{1}!\ldots m_{k}!}
(Z_{k}^{m_{k}}\ldots Z_{1}^{m_{1}}f)(m),
$$
which will be formally denoted by
$$
(e^{t_{k}Z_{k}}\ldots e^{t_{1}Z_{1}} f)(m).
$$
Before we prove the Chow--Rashevsky's theorem we show how to use
the above Taylor's formula to prove the following theorem.
Theorem. Let $G$ be a Lie group. Then $$ \exp(tX)\exp(tY) = \exp\Big( t(X+Y) + \frac{t^{2}}{2}[X,Y] +
O(t^{3})\Big). $$
Proof. Note that
$\exp(tX)\exp(sY)$ is the same as $Y_{s}\circ X_{t}(e)$
($e$ denotes the neutral element of $G$), because
$s\mapsto \exp(tX)\exp(sY)$ is the integral curve of $Y$
passing through $\exp(tX)$ at $s=0$. Thus the Taylor series for
$f(\exp(tX)\exp(sY))$ is
$e^{tX}e^{sY} f(e)$ and hence the Taylor series for
$h(t) = f(\exp(tX)\exp(tY))$ at $t=0$ is
\begin{eqnarray*}
e^{tX}e^{tY}f(e)
& = &
\Big(1+tX + \frac{t^{2}}{2}X^{2} + O(t^{3}) \Big)
\Big(1+tY + \frac{t^{2}}{2}Y^{2} + O(t^{3}) \Big)f(e) \\
& = &
f(e) + t(X+Y)f(e) +
t^{2}\Big(\frac{X^{2}}{2} + XY + \frac{Y^{2}}{2}\Big)f(e)
+ O(t^{3})
\end{eqnarray*}
Now there is a smooth function $t\mapsto Z(t)$, $Z(0) = 0$ such that
$$
\exp(tX)\exp(tY) = \exp(Z(t))
$$
for small $t$. We can write $Z(t) = tZ_{1}+t^{2}Z_{2} + O(t^{3})$.
Since $f(\exp(tW)) = f(e) + tWf(e) + \frac{t^{2}}{2}W^{2}f(e) + O(t^{3})$
and since obviously $f(A(t) + O(t^{3})) = f(A(t)) + O(t^{3})$,
we have
$$
f(\exp(Z(t)) =
f(\exp(t(Z_{1}+tZ_{2}))) + O(t^{3}).
$$
Fix $s$ and then
$$
f(\exp(t(Z_{1}+sZ_{2}))) =
f(e) + t(Z_{1}+sZ_{2})f(e) +
\frac{t^{2}}{2}(Z_{1}+sZ_{2})^{2}f(e) + O(t^{3}) = A
$$
Now substituting $s=t$ yields
$$
A = f(e) + tZ_{1}f(e) + t^{2}Z_{2}f(e)
+ \frac{t^{2}}{2}Z_{1}^{2}f(e) + O(t^{3})).
$$
Taking coordinate functions as $f$ and comparing the Taylor series yields
$$
Z_{1} = X+Y,\qquad Z_{2} + \frac{Z_{1}^{2}}{2} =
\frac{X^{2}}{2} + XY + \frac{Y^{2}}{2}.
$$
Hence $Z_{2} = \frac{1}{2}[X,Y]$, which implies
$$
Z(t) = t(X+Y) + \frac{t^{2}}{2}[X,Y] + O(t^{3}),
$$
and hence the theorem follows. $\Box$
As an immediate consequence we obtain
Corollary $\exp(-tX)\exp(-tY)\exp(tX)\exp(tY) = \exp(t^{2}[X,Y] + O(t^{3})).$
We will see now that the corollary holds for arbitrary smooth
vector fields, not necessarily on the Lie group.
Corollary $Y_{t}\circ X_{t}\circ Y_{-t} \circ X_{-t}(m)
= m+ t^{2}[X,Y]_{m} + O(t^{3})$.
Proof. The Taylor series for
$h(t) = f(Y_{t}(X_{t}(Y_{-t}(X_{-t}(m)))))$ is
\begin{eqnarray*}
e^{-tX}e^{-tY}e^{tX}e^{tY} f(m)
& = &
(1 - tX + \frac{t^{2}}{2}X^{2} + O(t^{3}))
(1 - tY + \frac{t^{2}}{2}Y^{2} + O(t^{3})) \times \\
& \times &
(1 + tX + \frac{t^{2}}{2}X^{2} + O(t^{3}))
(1 + tY + \frac{t^{2}}{2}Y^{2} + O(t^{3})) f(m) \\
& = &
(1 + t^{2}[X,Y] + O(t^{3})) f(m).
\end{eqnarray*}
Now we can turn to the main subject of the section, namely the connectivity
theorem of Chow and Rashevsky.
Theorem (Chow-Raschevsky) Let $\Omega\subset\mathbb{R}^{n}$ be an open domain and let
$X_{1},\ldots,X_{k}$ be smooth vector fields satisfying
H"ormander's condition i.e. for some positive integer $d$ comutators
of length less than or equal to $d$ span the tangent space
$\mathbb{R}^{n}$ at every point of $\Omega$. Then every two points in
$\Omega$ can be connected by an admissible curve. Moreover for any
compact set $K\subset\Omega$ there is a constant $C>0$ such that
\begin{equation}
(2)\qquad \rho(x,y) \leq C|x-y|^{1/d} \qquad
\mbox{for all $x,y\in K$}. \end{equation}
Remark. The estimate (2) is due to Nagel, Stein and Waigner.
Proof. Let $Y_{1},\ldots,Y_{p}$ be smooth vector fields.
Fix $m\in\Omega$. Define by induction
\begin{eqnarray*}
C_{1}(t) & = & Y_{1,t}(m) \\
C_{p}(t) & = &
C_{p-1}(t)^{-1}\circ Y_{p,-t}\circ C_{p-1}(t) \circ Y_{p,t}(m).
\end{eqnarray*}
Recall that $Y_{j,t}$ denotes the local family of diffeomorpisms associated to
$Y_j$.
Since both $C_{p}(t)$ and $C_{p}(t)^{-1}$ are compositions of diffeomorphisms
$Y_{j,\pm t}$ one easily obtaines that the Taylor series for
$f(C_{p}(t))$ and $f(C_{p}(t)^{-1})$ are given by
$\widetilde{c}_{p}(t)f(m)$ and $\widetilde{c}_{p}(t)^{-1}f(m)$ where
$\widetilde{c}_{p}(t)$ is a formal series defined by induction as follows
\begin{eqnarray*}
\widetilde{c}_{1}(t) & = & e^{tY_{1}} \\
\widetilde{c}_{p}(t) & = & e^{tY_{p}} \widetilde{c}_{p-1}(t)
e^{-tY_{p}} \widetilde{c}_{p-1}(t)^{-1}.
\end{eqnarray*}
It is easy to prove by induction that
\begin{equation}
(3)\qquad
\widetilde{c}_{p}(t) = 1 + t^{p} [Y_{p},[Y_{p-1},[\ldots,Y_{1}]\ldots]
+ O(t^{p+1}),
\end{equation}
and hence
$$
\widetilde{c}_{p}(t)^{-1} = 1 - t^{p} [Y_{p},[Y_{p-1},[\ldots,Y_{1}]\ldots]
+ O(t^{p+1}).
$$
Indeed, for $p=1$, (3) is obvious.
Assume it is true for $p$ and we prove it for $p+1$.
We have
\begin{eqnarray*}
\widetilde{c}_{p+1}(t)
& = &
e^{tY_{p+1}}\widetilde{c}_{p}(t) e^{-tY_{p+1}}\widetilde{c}_{p}(t)^{-1} \\
& = &
e^{tY_{p+1}} (\widetilde{c}_{p}(t) - 1)e^{-tY_{p+1}}
\widetilde{c}(t)^{-1} + \widetilde{c}_{p}(t)^{-1} \\
& = &
(1 + tY_{p+1})(\widetilde{c}_{p}(t)-1)(1 - tY_{p+1})
\widetilde{c}_{p}(t)^{-1} + \widetilde{c}_{p}(t)^{-1} + O(t^{p+2}) \\
& = &
(\widetilde{c}(t)-1)\widetilde{c}_{p}(t)^{-1} +
t^{p+1}[Y_{p+1},[Y_{p},[\ldots,Y_{1}]\ldots] +
\widetilde{c}_{p}(t)^{-1} + O(t^{p+2}) \\
& = &
1 + t^{p+1}[Y_{p+1},[Y_{p},[\ldots,Y_{1}]\ldots] + O(t^{p+2}).
\end{eqnarray*}
The claim is proved.
Hence the Taylor series of $f(C_{p}(t))$ at $t=0$ begins with
$$
f(m) + t^{p}[Y_{p},[Y_{p-1},[\ldots,Y_{1}]\ldots]f(m) + O(t^{p+1})
$$
and the Taylor series of $f(C_{p}(t)^{-1})$ at $t=0$ begins with
$$
f(m) - t^{p}[Y_{p},[Y_{p-1},[\ldots,Y_{1}]\ldots]f(m) + O(t^{p+1}).
$$
Now if $F_{1}$ and $F_{2}$ are two $C^{\infty}$ functions with Taylor series
$F_{1}(t) = a + bt^{p} +\ldots$ and
$F_{1}(t) = a - bt^{p} +\ldots$ then it is easy to see that the function
$$
G(t) = \left\{
\begin{array}{cc}
F_{1}(t^{1/p}) & \mbox{if $t\geq 0$} \\
F_{2}((-t)^{1/p}) & \mbox{if $t<0$}
\end{array}
\right.
$$
is $C^{1}$ in the neighborhood of $0$ and $G'(0)=b$.
Taking $F_{1}(t)=f(C_{p}(t))$ and $F_{2}(t) = f(C_{p}(t)^{-1})$,
where $f$ are all coordinate functions we conclude that the function
$$
\phi(t) = \left\{
\begin{array}{cc}
C_{p}(t^{1/p}) & \mbox{if $t\geq 0$} \\
C_{p}((-t)^{1/p})^{-1} & \mbox{if $t<0$}
\end{array}
\right.
$$
is a $C^1$ path passig through $m$ at $t=0$ with
$\phi'(0) = [Y_{p},[Y_{p-1},[\ldots,Y_{1}]\ldots]$.
Let $V_{1},\ldots,V_{n}$ be a basis of $\mathbb{R}^{n}=T_{m}\Omega$ arising from
H"ormander's condition i.e.,
$$
V_{i} = [X_{i,p_{i}},[X_{i,p_{i}-1},[\ldots,X_{i,1}]\ldots],
$$
where $i=1,2,\ldots,n$, $p_{i}\leq d$ and $X_{i,l}\in\{ X_{1},\ldots,X_{k}\}$.
Let $\phi_{i}(t)$ be a $C^1$ path defined as above for
$Y_{1}=X_{i,1},\ldots,Y_{p_{i}} = X_{i,p_{i}}$. Then $\phi_{i}'(0)=V_{i}$.
Finally define $\Phi$ by
$$
\Phi(\theta) = \phi_{1}(\theta_{1})\circ \cdots \phi_{n}(\theta_{n}),
\qquad \theta = (\theta_{1},\ldots,\theta_{n}).
$$
Then $\Phi$ is a $C^1$ mapping from a neighborhood of $0$ in $\mathbb{R}^{n}$
to $\Omega$. Since $\partial\Phi/\partial\theta_{i}(0)=\phi_{i}'(0)=V_{i}$
we conclude that $\Phi$ is a diffeomorphism in a neighborhood of $0$.
This implies that any point in the neighborhood of $m=\Phi(0)$ can be
connected to $m$ by an admissible curve.
More procisely $\phi_{i}(\theta_{i})$ is a composition of diffeomorphisms
of the form $X_{j,\pm|\theta_{i}|^{1/p_{i}}}$. Hence denoting the composition
by $\prod$ we may write
\begin{equation}
(4)\qquad
\Phi(\theta) = \left( \prod_{i=1}^{n} \prod_{\alpha=1}^{M_{i}}
X_{i,j_{\alpha},\pm|\theta_{i}|^{1/p_{i}}} \right)(m).
\end{equation}
Assume that $|\theta|\leq 1$. For any $x$,
the two points $x$ and $X_{i,j_{\alpha},\pm|\theta_{i}|^{1/p_{i}}(x)}$
can be connected by an admissible curve --- an integral curve of
$X_{i,j_{\alpha}}$ and hence the Carnot--Carath'eodory distance between
these two pints is no more than $|\theta_{i}|^{1/p_{i}}\leq |\theta|^{1/d}$.
Now we can move from $m$ to $\Phi(\theta)$ on such admissible curves and hence
\begin{equation}
(5)\qquad
\rho(\Phi(\theta),m) \leq C_{1}|\theta|^{1/d}
\approx C_{2}|\Phi(\theta)-m|^{1/d},
\end{equation}
where $C_{1}=\sum_{i=1}^{n}M_{i}$ equals the number of integral curver we use to
join $m$ with $\Phi(\theta)$ (see (4)). We employed also the fact
that $|\theta|\approx |\Phi(\theta) - m|$ which is a consequence of the fact
that $\Phi$ is a diffeomorphism.
Since we can connect all the points in a neighborchood of any point it easily
follows that we can connect any two points in $\Omega$.
The estimate (2) follows from (5).
$\Box$
thanks, could you provide a pdf of the proof, since it is not available form my institution?
@Mathsfreak I added details of the proof. It is an expanded version of the argument from the book that I quoted. In fact you can find the book online if you know how to search for books.
Sussmann, Hector J., Orbits of families of vector fields and integrability of
distributions, Trans. Amer. Math. Soc., 180, 1973, 171--188, gives a very easy explanation, using flows of vector fields.
The article is available free of charge.
It suffices to have the curve horizontal almost everywhere, because then it will stay tangent to any immersed submanifold whose tangent spaces contain the distribution; just write out local coordinates in which the submanifold is locally given by setting various coordinate functions to constants.
|
2025-03-21T14:48:31.702192
| 2020-08-03T10:05:14 |
368192
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"LSpice",
"Pol van Hoften",
"https://mathoverflow.net/users/2383",
"https://mathoverflow.net/users/56856"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631823",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368192"
}
|
Stack Exchange
|
When is $G(\mathbb{Z}_p)$ topologically generated by maximal tori?
Let $G/\mathbb{Z}_p$ be a connected reductive group, and let $T_1, \cdots, T_n \subset G_{\mathbb{Q}_p}$ be maximal tori. Suppose that we have precisely one maximal torus in each $G(\mathbb{Q}_p)$-conjugacy class of maximal tori; is it then true that $T_1(\mathbb{Z}_p), \cdots, T_n(\mathbb{Z}_p)$ (topologically) generate $G(\mathbb{Z}_p)$? Here by $T(\mathbb{Z}_p)$ I mean the $\mathbb{Z}_p$ points of the connected component of the identity of the Neron model $\mathcal{T}$ of $T$. As LSpice pointed out in the comments, we need the inclusions $\mathcal{T}_i \subset G$ to be defined over $\mathbb{Z}_p$.
If $G=\operatorname{GL_n}$, then this is true. Our conjugacy classes of maximal tori correspond to commutative semi-simple algebras $C/\mathbb{Q}_p$ of degree $n$ and $M_n(\mathbb{Z}_p)$ is (topologically) generated by $\{\mathcal{O}_C\}_{C \in \mathcal{C}}$, where $\mathcal{C}$ is the set of isomorphism classes of such $C$ (and we make some choice of $\mathcal{O}_C \to M_n(\mathbb{Z}_p)$ for each class). An argument using the $p$-adic exponential map now shows that the (topological) closure $H$ of the group generated by our tori contains $1+p \cdot M_n(\mathbb{Z}_p)$, and then all we have to do is show that $H$ surjects onto $\operatorname{GL}_n(\mathbb{F}_p)$.
More generally, I wonder if there is an analogous statement with $\mathcal{G}/\mathbb{Z}_p$ a parahoric group scheme.
I guess you want the $T_i$'s in $G$, not just in $G_{\mathbb Q_p}$, to make sense of $T_i(\mathbb Z_p)$. Is it always possible to choose a cross-section of the set of $G(\mathbb Q_p)$-conjugacy classes of tori in $G_{\mathbb Q_p}$, each of which is defined over $\mathbb Z_p$? I'd think that would involve some kind of strong splitness/tameness (on all tori) condition—maybe something like (as investigated by Fintzen) $p \nmid \lvert W\rvert$, or something?
Also, when speaking of $G = \operatorname{GL}n$, you seem to switch to speaking of $\operatorname{Lie}(G)$ instead, since you say "$M_n(\mathbb Z_p)$ is (topologically) generated by ${\mathcal O_C}{C \in \mathcal C}$". Is that intentional, or did I misunderstand?
That is not intentional, but the statement on Lie algebras can be used to prove the group theoretic statement using the exponential map (and some further arguments).
I guess I need my tori to be defined over $\mathbb{Z}_p$, I will edit the question. Also, could you explain why we would need all tori to split over tamely ramified extensions? Does it have something to do with properties of Neron models?
I don't know that we would need all tori to split over tamely ramified extensions; I just guessed, and may be wildly off base. My point was that I didn't think a random $\mathbb Q_p$-torus could be given a $\mathbb Z_p$-model, but you're probably right—the connected Néron model should do it, I guess. (I'm not really comfortable, despite @BCnrd's heroic efforts ("Reductive group schemes"), with what 'connected reductive group', or even 'torus', means over $\mathbb Z_p$.)
|
2025-03-21T14:48:31.702518
| 2020-08-03T10:06:36 |
368193
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"David Roberts",
"Dmitri Pavlov",
"Praphulla Koushik",
"https://mathoverflow.net/users/118688",
"https://mathoverflow.net/users/402",
"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:631824",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368193"
}
|
Stack Exchange
|
stacks that are not necessarily fibered in groupoids appearing in algebraic geometry and differential geometry
Question:
What are (some of) the stacks (occurring in algebraic/differential geometry) that are fibered in arbitrary categories and not necessarily in groupoids?
In the notes Notes on Grothendieck topologies,fibered categories and descent theory Angelo Vistoli introduce the notion of a stack over a site $(\mathcal{C},\mathcal{J})$ to be a fibered category (not necessarily fibered in groupoids) over $\mathcal{C}$ satisfying some "locally determined" condition.
But, examples of stacks of interest in algebraic geometry and differential geometry (a small set of examples I have seen) are always fibered in groupoids. So, what could be a justification or necessity for introducing the notion of stacks fibered over arbitrary categories, if "almost all" stacks that occur in Algebraic geometry (that I know) are fibered in groupoids.
There might be interesting examples of stacks outside algebraic geometry of differential geometry that are not necessarily fibered in groupoids. I would be happy to see such examples (please add them as answers if you wish) but as for this question, I would like to learn about situations in algebraic geometry or differential geometry.
The categories of continuous fields of Hilbert spaces, Banach spaces and $C^*$-algebras should all be stacks over the appropriate base category of spaces (for instance, locally compact Hausdorff). I've not seen this mentioned, since the people who deal with these don't usually think about stacks.
@DavidRoberts are you counting those as algebraic or differential geometry?
I would say differential topology, maybe.
@DavidRoberts fair enough..
What you are referring to are sometimes called stacks or sheaves of categories. A famously important example is the stack $\mathrm{QCoh}$ sending a scheme $U$ to the category of quasi-coherent sheaves over it (if you want to work with fibered categories, an object of this fibered category is a pair $(U,F)$ where $U$ is a scheme and $F$ a quasi-coherent sheaf on $U$).
Yes, I have seen this example in Angelo Vistoli's notes (Example 3.2.1 page 53). Do you have any personal choice of appearance of this stack in any construction or some concept..
@PraphullaKoushik: One obvious occurrence is the definition of quasi-coherent sheaves on an arbitrary stack S: these are defined as (derived) maps S→QCoh.
Virtually any kind of algebraic structure (e.g., group, ring, module, vector space, affine space, etc.) leads to a stack in categories
whose objects are bundles of such structures and morphisms are fiberwise
homomorphisms of such structures.
For example, the stack Vect of (finite-dimensional, say) vector bundles
is a stack in categories over the site of cartesian smooth manifolds.
Likewise, the stack BGrb^n_A of bundle n-gerbes with structure group A
is a stack in (n+1)-categories.
As a practical application, one can immediately define the category
of vector bundles or bundle n-gerbes on a given stack or ∞-stack S as the category
of (derived) sections S→Vect or S→BGrb^n_A.
This also captures the symmetric monoidal structure, and in the case of
vector bundles, the bimonoidal structure.
I understood most of first three paragraphs. I have difficulty in understanding last 2 lines.. I will read it again and ask if I have any specific question. Thanks for the answer..
|
2025-03-21T14:48:31.702753
| 2020-08-03T10:15:01 |
368195
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Aurel",
"LSpice",
"Will Sawin",
"https://mathoverflow.net/users/18060",
"https://mathoverflow.net/users/2383",
"https://mathoverflow.net/users/40821",
"https://mathoverflow.net/users/84165",
"lisyarus"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631825",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368195"
}
|
Stack Exchange
|
Explicit computation of spinor norm
I've asked this on math.stackexchange, unsuccessfully. I hope this question is appropriate for mathoverflow.
Let $V$ be a finite-dimensional vector space over a field $K$ with $\operatorname{char}K\neq 2$, and $Q$ a non-degenerate quadratic form on $V$. The spinor norm is a homomophism
$$sn: O(V,Q) \rightarrow K^*/(K^*)^2$$
defined as $Q(v)$ for reflections by a non-isotropic vector $v$.
Alternatively, for $g \in O(V,Q)$ let $a \in \Gamma(V,Q)$ be the element of the Clifford group that realizes $g$ via an inner graded automorphism. Then, $sn(g)$ is defined as $N(a)=a^t a$, which is a scalar if $a$ comes from the Clifford group.
I am interested in explicitly computing $sn(g)$ for a given $g\in O(V,Q)$. I know a bit in some special cases:
For the Euclidean space and the corresponding $O(n,\mathbb R)$ group the spinor norm is trivial $sn(g)=1$, since the group is generated by reflections by vectors of unit norm
For an algebraically-closed field $K$, the spinor norm is always trivial since $K^*/(K^*)^2$ is trivial
For a metabolic space $V = W \oplus W^*$ with the form $Q(w,f) = f(w)$, any $g \in \operatorname{GL}(W)$ gives rise to an orthogonal transformation on $V$ by the formula $g \cdot (w,f) = \left(gw, \left(g^{-1}\right)^*f\right)$. The spinor norm of this transformation is equal to $\det g$ (this is half-anecdotal: I've heard it in a Russian video lecture on Clifford algebras, presented without complete proof).
In particular, for any quadratic space that has a metabolic subspace as a direct (orthogonal) summand, the spinor norm is surjective.
Clearly, the spinor norm of $\Omega(V,Q)$ (the commutator subgroup of $O(V,Q)$) is trivial, since $K^*/(K^*)^2$ is abelian. This article states that $\Omega$ is precisely the kernel of the spinor norm, providing an injective morphism $O/\Omega \rightarrow K^*/(K^*)^2$, though I don't see how it helps in actually computing the spinor norm of a given orthogonal transformation.
I've done some calculations with the real hyperbolic plane with orthogonal basis $\{e_1, e_2\}$ such that $Q(e_1)=1$ and $Q(e_2)=-1$ by explicitly computing the elements of the Clifford group that represent certain orthogonal transformations. It seems that the spinor norm of a matrix $A$ (which is $\pm 1$ in the real case) in this basis coincides with the sign of $A_{2,2}$.
Having in mind the connected components of an indefinite real orthogonal group $O(p,q)$ and using that the spinor norm is a continuous map to a discrete space $\{\pm 1\}$, it has to be constant on connected components, thus it is enough to compute it for a single representative from each component. This gives a generalization of the previous result, namely the spinor norm is $+1$ iff the transformation preserves orientation of the negative-definite subspace, and the spinor norm equals the determinant of the lower-right $q\times q$ submatrix (in a basis where positive-definite vectors come before negative-definite ones). This is basically my own findings, and I would appreciate a reference that supports/disproves this claim.
In general, it feels that there should be some explicit (maybe polynomial?) formula $O(V,Q) \rightarrow K^*$ implementing the spinor norm, but I failed to find any references on this. In any way, I am happy with any explicit way of computing the spinor norm of an orthogonal matrix for a general quadratic form, or otherwise an explanation of why this isn't that straightforward or even possible.
Of course, the spinor formula itself arguably is an explicit formula, for some values of explicit, so I'll take explicit to mean polynomial in the entries, as you suggest—in which case I imagine one can prove rigorously that the answer is 'no'. If some special formulæ are instead of interest, Jessica Fintzen, Tasho Kaletha, and I recently found ourselves having to do some such computations, and found that, at least for semisimple elements, there's a reasonably easy, mostly explicit (in terms of eigenvalues) answer. (I hope self reference is OK if not, don't read next comment.) They're …
… described in §5.1 of Fintzen, Kaletha, and Spice - On certain sign characters … in the form that's of interest to us, but most of them come from Scharlau's book Quadratic and Hermitian forms.
Although it turned out not to be most useful for us, §2 of Zassenhaus - On the spinor norm, and specifically (2.1), might be closer to what you want. It's a nice paper, but not the last word on the subject; you may like to look at who cites it.
@LSpice Thank you a lot for this decent set of references. I've checked Scharlau's book and it already contains quite a lot, even if not exactly what I've hoped for. I'll accept if you post it as an answer.
I have done that. Unless you're sure this completely answers your question, you might want to give it a day or two before accepting to see if anyone gives a better-fitting answer. (I think questions with an accepted answer are less likely to attract a new answer.)
@LSpice Yes, was thinking of exactly the same. Thank you!
In terms of an algorithm (not a polynomial formula) to compute the spinor norm, you can decompose your orthogonal transformation as a product of reflections. Then each reflections has a simple lift to the Clifford group and you can take the corresponding product.
@Aurel In this case one doesn't even need the lift to Clifford group, since the door norm can be directly computed from such a decomposition. However, I failed to find an algorithmic way to do that.
I'm now thinking if it would be possible to adapt the QR decomposition to do this.
Posted from the comments (1 2 3), by request.
Of course, the spinor formula itself arguably is an explicit formula, for some values of explicit, so I'll take explicit to mean polynomial in the entries, as you suggest—in which case I imagine one can prove rigorously that the answer is ‘no’. If some special formulæ are instead of interest, Jessica Fintzen, Tasho Kaletha, and I recently found ourselves having to do some such computations, and found that, at least for semisimple elements, there's a reasonably easy, mostly explicit (in terms of eigenvalues) answer. It's assembled from some facts described in §5.1 of Fintzen, Kaletha, and Spice - On certain sign characters … in the form that's of interest to us, but most of them come from Scharlau's book Quadratic and Hermitian forms.
Although it turned out not to be most useful for us, §2 of Zassenhaus - On the spinor norm, and specifically (2.1), might be closer to what you want. It's a nice paper, but not the last word on the subject; you may like to look at who cites it.
If one lets $K$ be the field of rational functions on the entries of a matrix in $SO(n)$, then the spinor norm of the universal element of $SO(n)$ over $K$ is an element of $K^\times / (K^\times)^2$, and if we lift it do $K^\times$, doesn't that give a (probably very complicated) formula for the spinor norm as a rational function, which we can represent as a polynomial by multiplying the numerator by the denominator? (Of course, this polynomial formula would not be so useful at the points where it vanishes).
@WillSawin, the spinor norm is sensitive to field extensions (for example, passing from $\mathbb R$ to $\mathbb C$ loses all information), so we'd have to argue at least why square classes in the ground field don't collapse in $K$.
Do you know if the universal element of $\operatorname{SO}(n)(K)$ is semisimple?
I believe I can answer both your questions. The universal element is semisimple since it has distinct eigenvalues, because distinct eigenvalues is an open condition and we can write down at least one matrix with distinct eigenvalues. Classes in the ground field don't collapse in $K$ because $K$ is equal to the field of rational functions in $n(n-1)/2$ variables over the ground field, since $SO(n)$ is birational to affine space.
|
2025-03-21T14:48:31.703218
| 2020-08-03T10:36:09 |
368201
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"erz",
"https://mathoverflow.net/users/53155"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631826",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368201"
}
|
Stack Exchange
|
Do products of distance functions separate points?
Let $(X,d)$ be a metric space without isolated points and of diameter $1$. Let $Y=\{y_m\}_{m=1}^{\infty}$ be a dense subset of $X$.
Define $g_0\equiv 1$, and for $m>0$ let $g_m=d(\cdot,y_1)\dotsm d(\cdot,y_m)$.
I wonder if $\{g_0,g_1,\dotsc\}$ separates points of $X$ in the following strong sense:
If $x_1,\dotsc,x_n\in X$, $a_1,\dotsc,a_n\in\mathbb{R}$ are such that $a_1g_m(x_1)+\dotsb+a_ng_m(x_n)=0$, for every $m=0,1,\dotsc$, then $a_1=\dotsb=a_n=0$.
Some partial progress:
Let's show that WLOG $x_1,\dotsc,x_n\not\in Y$.
Assume that $x_1,\dotsc,x_k\in \{y_1,\dotsc,y_l\}$ and $x_{k+1},\dotsc,x_n\not\in Y$. Then $g_l(x_1)=\dotsb=g_l(x_k)=0$, and $g_l(x_{k+1})=\dotsb=g_l(x_n)\ne 0$.
Take $Z=\{z_m=y_{m+l}\}_{m=1}^{\infty}$, which is still dense in $X$, and let $h_m=d(\cdot,z_1)\dotsm d(\cdot,z_m)$, $h_0\equiv 1$. Then $g_{m+l}=g_l h_m$, for every $m$. For every $m$ we have
\begin{gather*}
0=a_1g_{m+l}(x_1)+\dotsb+a_ng_{m+l}(x_n)=a_1g_l(x_1)h_m(x_1)+\dotsb+a_ng_l(x_n)h_m(x_n) \\
=a_{k+1}g_l(x_{k+1})h_m(x_{k+1})+\dotsb+a_ng_l(x_n)h_m(x_n),
\end{gather*}
and since we "can" prove the statement with $x_{k+1},\dotsc,x_n\not\in Z$, it follows that $a_{k+1}g_l(x_{k+1})=\dotsb=a_ng_l(x_n)=0$, from where $a_{k+1}=\dotsb=a_n=0$. Now assume that $x_k$ is $y_m$ with the biggest $m$. Then $0=a_1g_{m-1}(x_1)+\dotsb+a_ng_{m-1}(x_n)=a_kg_{m-1}(x_k)$, from where $a_k=0$, and so on.
Let us prove the statement for $n=2,3$.
Let $x$, $y$, $z$ be distinct. First, assume that $ag_m(x)=cg_m(z)$, for every $m=0,1,\dotsc$. Since $Y$ is dense, there is $m$ such that $d(x,y_{m+1})<d(z,y_{m+1})$. But then either $a=0=c$, or $ag_m(x)=cg_m(z)$ and $ag_{m+1}(x)=cg_{m+1}(z)$ are incompatible. Now assume that $ag_m(x)+cg_m(z)=bg_m(y)$, where $a,b,c>0$. Again, using density, there is $m$ such that $d(y,y_{m+1})<d(x,y_{m+1})$ and $d(y,y_{m+1})<d(z,y_{m+1})$, which leads to $ag_{m+1}(x)+cg_{m+1}(z)>bg_{m+1}(y)$.
Similar arguments work if we assume that all of $a_k$ but one are of the same sign. But even if two are positive and two are negative I am at a loss.
@MattF. you are right, the problem is rather strange...
@MattF. you probaby mean a rectangle, right?
@MattF. I agree that this seems relevant, but I tried to build a counterexample out of it and failed
|
2025-03-21T14:48:31.703367
| 2020-08-03T10:50:52 |
368202
|
{
"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:631827",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368202"
}
|
Stack Exchange
|
RSA as a hidden subgroup problem
The Hidden Subgroup Problem (HSP) covers several known problems (e.g. Integer Factorization Problem, Discrete Logarithm Problem) as a special case:
Definition [Hidden Subgroup Problem (HSP)] Let $\mathbb{G}$ be a group and $\mathbb{H}$ an unknown subgroup of $\mathbb{G}$, i.e., $\mathbb{H} \leq \mathbb{G}$. Let $S$ be any set and $f$ be a function that maps the group elements of $\mathbb{G}$ to $S$, i.e., $f: \mathbb{G} \rightarrow S$. The function $S$ has the special property that it can distinguish cosets of $\mathbb{H}$: $$ f(e_1) = f(e_2) \Leftrightarrow e_1\mathbb{H} = e_2\mathbb{H} $$ The Hidden Subgroup Problem is, given $\mathbb{G}$ and (oracle-)access to the function $f$, to determine a generating set for the subgroup $\mathbb{H}$.
I found several papers which show how to instantiate the involved groups and functions, for example in case of the Integer Factorization Problem:
$$\mathbb{G} = \mathbb{Z}^+_{\varphi(N)}; \mathbb{H} = \langle r\rangle^+; S = \mathbb{Z}^*_N, f(x)=g^x\pmod{N}$$
whereof the integer $r$ is the order of the multiplicative group generated by $g$.
I am trying to find a direct HSP approach to the RSA problem, without to find the order of a
subgroup or to factorize the modulus by any other methods.
RSA Problem: Given an RSA modulus $N$ and a public exponent $e$ and an integer $C$, find $m$ such that $m^e \equiv C \pmod{N}$
I came up with the following solution, but i am not sure if it is valid according to the definition of the HSP, since it contains $m$, the target integer, in the description of the group operation.
Let $\mathbb{G} = (\mathbb{Z},\circ_m)$ and $\mathbb{H} = \{0, m, m+N, m+2N, m+3N,\ldots\}$, $S = \mathbb{Z}^*_N$ as well as $f(x) = x^e\pmod{N}$. The group operation that is defined on $\mathbb{G}$ is $\circ_m(a,b) := a+b-m$. So $\circ$ makes $\mathbb{H}$ a subgroup of $\mathbb{G}$ (neutral element is $m$ and inverse is $a^{-1} = 2m-a$). And $f$ can distinguish $N$ cosets of $\mathbb{H}$: $$c\mathbb{H} = \{m+c, m+c+N, m+c+2N,\ldots\}$$. Does the term "given a group $\mathbb{G}$" forbid the definition of a valid but non accessible group operation? Is it enough that someone could, even not knowing $m$, compute $\circ_m(a,b)-\circ_m(u,v) = a+b-u-v$?
--
(i asked this question also on cstheory.stackexchange (here) but did not get an answer)
Link to the question on cstheory?
|
2025-03-21T14:48:31.703540
| 2020-08-03T10:56:05 |
368203
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"H A Helfgott",
"M. Winter",
"https://mathoverflow.net/users/108884",
"https://mathoverflow.net/users/398"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631828",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368203"
}
|
Stack Exchange
|
From one eigenvector to many, in a very local graph?
Let $\Gamma$ be an undirected graph of bounded degree $d$ with $V = \{1,2,\dotsc,N\}$ as its set of vertices, and edges only between vertices that are at a distance $\leq M$ apart (where $M$ is much smaller than $N$). Let $\Delta$ be the graph Laplacian of $\Gamma$.
Define an inner product on functions $V\to \mathbb{C}$ by giving to $V$ the uniform probability measure, i.e., each vertex has measure $1/N$.
Let $f:V\to \mathbb{C}$ be such that $|f|_2^2=1$ and $\eta = |\langle f,\Delta f\rangle|$ is large. (For instance, $f$ could be an eigenvector with large eigenvalue.)
(a) If $|f(n)|=1$ for all $n$, then it follows easily that there is a large number of orthonormal vectors $w:V\to \mathbb{C}$ with $|\langle w, \Delta w\rangle|$ large: just take all $w$ of the form $w(n) = w_a(n) = f(n) e(a n/N)$ for $|a| \leq \eta N/10 M$, say.
(b) Under the weaker assumption that $|f(n)|\geq \epsilon$ for at least $\epsilon N$ elements $n\in V$, one can very likely obtain a similar conclusion. Here's a clumsy way: chop up $V$ into disjoint neighborhoods of size $100 M/\epsilon$ or so; then, for any neighborhood $U$ that is not too "poor" and does not have neighbors that are too "rich", it should be the case that $\langle w, \Delta w\rangle$ is large for the restriction $w=f|_U$. Is there a more elegant argument (perhaps along the lines of (a))?
(c) Somewhat orthogonal question: if no functions $f:V\to \mathbb{C}$ with $|f|_2^2=1$ and $\eta=\left|\langle f,\Delta f\rangle\right|$ large satisfy $|f(n)|\geq \epsilon$ for at least $\epsilon N$ elements $n\in V$, does it follow that there is a small subset $Y\subset V$ (with $|Y| = O(\epsilon N)$ elements, say) such that $\Delta|_{V\setminus Y}$ (defined as the operator $f\mapsto (\Delta f_{V\setminus Y})_{V\setminus Y}$) has no large eigenvalues? (Alternatively: if there are several functions $f_i:V\to \mathbb{C}$ with $|f_i|_2^2=1$ and $|\langle f,\Delta f\rangle|$ large such that, for many $n$, there is some $f_i$ such that $|f_i(n)|\geq \epsilon$, can we proceed as in (b) and obtain a large number of orthonormal $w$ such that $|\langle w, \Delta w\rangle|$ is large?
In your first paragraph, you define the edges of the graph by referring to a notion of distance between the vertices (which are natural numbers). What is this distance? The absolute value of the difference of the numbers?
Yes, the absolute value of the difference of the numbers.
Let me show how to do (b), in a more general context than I set out in (b).
Let $f:V\to \mathbb{C}$ with $|f|_2=1$ and $|\langle f, \Delta f\rangle|\geq \alpha>0$.
Consider a partition of $V$ inducing an equivalence relation $\sim$. Define the linear operator $\Delta_\sim$ on functions $v:V\to \mathbb{C}$ by $$(\Delta_\sim v)(n) = v(n) - \frac{1}{d} \mathop{\sum_{n': \{n,n'\}\in E}}_{n'\sim n} v(n').$$
The graph Laplacian $\Delta$ thus equals $\Delta_\sim$ for $\sim$ corresponding to the trivial partition.
Clearly
$$\begin{aligned}\left|\langle f, (\Delta_\sim-\Delta) f\rangle\right| &=
\left|\frac{1}{d} \sum_n v(n) \mathop{\sum_{n': \{n,n'\}\in E}}_{n'\not\sim n} v(n')\right|\leq
\frac{1}{d} \sum_n \mathop{\sum_{n': \{n,n'\}\in E}}_{n'\not\sim n}
\frac{|v(n)|^2 + |v(n')|^2}{2} \\ &=
\frac{1}{d} \sum_n \mathop{\sum_{n': \{n,n'\}\in E}}_{n'\not\sim n}
|v(n)|^2 \leq \sum_{n\in \partial_\sim} |v(n)|^2,\end{aligned}$$
where $\partial_\sim\subset V$ is the set of all $n$ such that $n'\sim n$ for
some edge $\{n,n'\}\in E$.
Consider partitions of $V$ into segments of the form
$$\{1,2,\dotsc,a M\}, \{a M + 1,\dotsc, (a + 2 C M)\},\dotsc, \{a + 2 C k M+1,\dotsc, N\},\;\;\;\;\;\;\;\;\;\;(*)$$
with all segments being of length $2 C M$ (except possibly for the first and the last one, which may be shorter). In fact, consider such partitions only for $a=1,3,\dotsc, 2 C - 1$. For the corresponding equivalence relation $\sim_a$,
$$\delta_{\sim_a} \subset \delta_a := \bigcup_{j=0}^{k}\; \{(2 C j + a-1) M + 1, (2 C j + a-1) M +2,\dotsc, (2 C (j+1) + a) M\},
$$
since all of our edges are of length $\leq M$.
It is clear that $\delta_1,\delta_3,\dotsc,\delta_{2 C - 1}$ are disjoint. Hence, by pigeonhole, there is an $a\in \{1,3,\dotsc,2 C - 1\}$ such that $\sum_{n\in \delta_a} |v(n)|^2 \leq 1/C$. We choose that $a$ and work with the corresponding partition (*). Then $|\langle f,\Delta_\sim f\rangle|\geq (\alpha - 1/C)$. Let $C = \lceil 3/\alpha\rceil$, so that $|\langle f,\Delta_\sim f\rangle|\geq 2 \alpha/3$. Write $P=\{I\}$ for our partition into intervals $I$.
Let $S$ be the set of intervals $I\in P$ such that $|\langle f|_I,\Delta f|_I\rangle| \geq \frac{\alpha}{3} \left| f|_I\right|_2^2$. It is clear that
$$\sum_{I\in P\setminus S} |\langle f|_I,\Delta f|_I\rangle| < \frac{\alpha}{3}
\sum_{I\in P\setminus S} \left| f|_I \right|_2 \leq \frac{\alpha}{3}
\sum_n |f(n)|^2 = \frac{\alpha}{3}$$ and so $\sum_{I\in S} |\langle f|_I,\Delta f|_I\rangle| > \alpha/3$. It remains to bound the size of $S$ from below.
Assume $|f|_\infty\leq K$.
Since the $L^2\to L^2$ operator norm of $\Delta$ is $\leq 2$, we see that, for every $I\in S$,
$$|\langle f|_I,\Delta f|_I\rangle| \leq 2 \left|f|_I\right|_2^2 \leq
2 K^2 \cdot \frac{2 C M}{N} \ll \frac{K^2}{\alpha} \frac{M}{N}.$$
Hence $$|S| \gg \frac{\alpha^2}{K^2} \frac{N}{M}.$$
Here is my self-answer to (c), based on my self-answer to (b).
For any $f:V\to \mathbb{C}$ with $|f|_2^2=1$ and $|\langle f,\Delta f\rangle|\geq \alpha>0$, we obtain, proceeding as in my self-answer to (b), that there is an interval of the form $$I = \{(2m-1)M + 1, (2m-1) M +2, \dotsc, (2m-1) M + 2 C M\}\cap [1,N]$$
such that $\left|\langle f|_I,\Delta f|_I\rangle\right|\geq \frac{\alpha}{2} \left| f|_I\right|_2^2$, where $C = \lceil 2/\alpha\rceil$.
Let $Y$ be the union of the set $\mathbf{I}$ of all such intervals for all such functions $f$. Then, for any function $f$ with $|f|_2^2=1$ and support on $V\setminus Y$, we know that $|\langle f,\Delta f\rangle|<\alpha$ (or else we would get a contradiction).
It is easy to see that we can choose a subset $\mathbf{I}'\subset \mathbf{I}$ consisting of $|\mathbf{I}'|\geq |Y|/4 C M$ disjoint intervals. For each $I\in \mathbf{I}'$, there exists, by construction, a function $g$ supported on $I$ with $|g|_2^2=1$ and $|\langle g,\Delta g\rangle|\geq \alpha/2$. Functions $g$ corresponding to different $I\in \mathbf{I}'$ are obviously orthogonal to each other.
Thus, for any $\epsilon>0$, we know that either $|Y|\leq \epsilon N$, or there are $\gg \epsilon \alpha N/M$ orthogonal functions $g$ with $|g|_2^2=1$ and $|\langle g,\Delta g\rangle| \geq \alpha/2$.
|
2025-03-21T14:48:31.703912
| 2020-08-03T11:19:44 |
368206
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Bipolar Minds",
"David White",
"Denis Nardin",
"Jake Wetlock",
"Robin Houston",
"https://mathoverflow.net/users/11540",
"https://mathoverflow.net/users/153228",
"https://mathoverflow.net/users/43054",
"https://mathoverflow.net/users/58211",
"https://mathoverflow.net/users/8217"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631829",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368206"
}
|
Stack Exchange
|
Rigid monoidal and closed monoidal categories
I am trying to understand the relationship between rigid monoidal categories and closed monoidal
categories. First every rigid monoidal category is closed, with an adjoint to the functor $X \otimes -$ given by $X^* \otimes -$.
Let $\mathcal{C}$ be a closed monoidal category (i.e., with internal homs), such that for all $X \in \mathcal{C}$, the functor $X \otimes -$ and its adjoint forms an equivalence of the category $\mathcal{C}$ with itself. Does it follow that $\mathcal{C}$ is rigid?
I don't know the answer, but does Warning 1.2 at this link help? https://ncatlab.org/nlab/show/dual+object+in+a+closed+category
I don't understand what your question is. What does "adjoints of all $X\in\mathcal{C}$ form an equivalence" mean? The right adjoint to $X\otimes -$ is a functor $\mathcal{C}\to \mathcal{C}$, are you asking whether these are equivalences? In that case the answer is no...
I think he means: "suppose $C$ is closed monoidal and for every $X$, the adjoint of $X\otimes -$ is an equivalence. Does it follow that $C$ is rigid?"
Yes this is what I mean. I have reworded to make things clearer.
Obviously, given such a closed monoidal category $C$, and given $X$, a candidate for $X^$ is $Hom(X,1)$. But as I commented before, it's not clear why there should be a coevaluation morphism $1 \to X\otimes X^$ in general. How does $Hom(X,-)$ being an equivalence help?
@David: The (naive) guess was that it might be constructable from the unit of the equivalence.
@JakeWetlock I think if you set $X^\vee := [X, \mathbb{I}]$, then $\eta_\mathbb{I}$ and $\epsilon_\mathbb{I}$ serve as coevaluation and evaluation of the duality, where $\eta$ and $\epsilon$ are unit and counit of the adjunction. The zig-zag equations are automatically fulfilled. You don't need an adjoint equivalence.
Are you assuming a symmetric monoidal category? It sounds as though you probably are, since you only mention the functor $X\otimes-$ and not $-\otimes X$, but you haven’t mentioned it so I’m not completely sure.
@BipolarMinds Isn't is the case that $\eta_1: 1\to Hom(X,X\otimes 1)$? (That is, first do $X\otimes -$ then do $Hom(X,-)$). Why should $Hom(X,X)$ be the same as $X\otimes Hom(X,1)$?
@DavidWhite Oh sorry, seems like I was a bit too fast! You're right, its completely unclear
How strong is the condition the OP is assuming? In homological algebra, this would be like asking every module to be faithful, right? If $X\otimes -$ and $Hom(X,-)$ are equivalences for all $X$, does the whole category collapse?
Let $1$ be the unit of $C$. For every $X$, we define $X^* = Hom(X,1)$. I will assume $C$ is strict closed symmetric monoidal. Further assuming the condition the OP specified, we can show $C$ is rigid.
Let's unpack the additional condition the OP wants to assume. For every $X$, the functors $F(-) = X\otimes -$ and $G(-) = Hom(X,-)$ form an adjoint equivalence. In particular, the counit of the adjunction, $\epsilon: FG \to 1_C$ is a natural isomorphism. So, $\epsilon_1: X\otimes X^* = X\otimes Hom(X,1) \to 1$ is an isomorphism. Call this morphism $ev_X$. Define the coevaluation as its inverse.
Following Section 2.10 of Tensor Categories, we must show that the compositions:
$X \stackrel{coev_X\otimes id}{\longrightarrow} (X\otimes X^*) \otimes X \stackrel{\alpha}{\longrightarrow} X\otimes (X^*\otimes X) \stackrel{id\otimes ev_X}{\longrightarrow} X$, and
$X^*\stackrel{id \otimes coev_X}{\longrightarrow} X^*\otimes (X \otimes X^*) \stackrel{\alpha^{-1}}{\longrightarrow} (X^*\otimes X) \otimes X^* \stackrel{ev_X\otimes id}{\longrightarrow} X^*$
are the identity morphisms. But the first is just $X\cong 1\otimes X$, followed by the associator, followed by $X\otimes 1 \cong X$, which is certainly the identity on $X$ (by strictness), and the second works the same way. So, indeed, $X^*$ is a left dual to $X$.
By symmetry, $X^*$ is also the right dual to $X$, and $X^*\otimes X \cong 1$, so the two morphisms in 2.10.2 of Tensor Categories are also identities. Hence, every $X$ has both a left and right dual, so $C$ is rigid.
Stupid question: Is an adjoint equivalence the same as an adjunction between equivalences?
I just meant that in the usual adjunction you get from being closed monoidal, the unit and counit are natural isomorphisms (like the OP wanted), so that the adjunction is in fact an equivalence. Hopefully the term "adjoint equivalence" doesn't have alternative meanings in the tensor category world. https://ncatlab.org/nlab/show/adjoint+equivalence
As it says here, every equivalence can be "improved" to an adjoint equivalence, and it just comes down to which morphisms you pick to be the unit and counit. So it is no loss of generality. https://math.stackexchange.com/questions/595482/is-an-equivalence-an-adjunction
|
2025-03-21T14:48:31.704347
| 2020-08-03T12:03:36 |
368210
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"David E Speyer",
"Gerry Myerson",
"abx",
"https://mathoverflow.net/users/158000",
"https://mathoverflow.net/users/163136",
"https://mathoverflow.net/users/297",
"https://mathoverflow.net/users/40297",
"worm"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631830",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368210"
}
|
Stack Exchange
|
For two polynomials, what is the relationship between the least linear combination and the resultants?
Let $k$ be a field and $k[x]$ be the ring of polynomials over $k$. Given two polynomis $m_1(x), m_2(x) \in k[x]$, I want to know the relationship between the resultants and the least linear combination.
(In Euclidean rings, the least linear combination is the greatest common divisor (GCD).)
What kind of relationship could there possibly be? The resultant is a scalar, not a polynomial.
If we simplify the sylvester matrix of m1(x) and m2(x) to a row echelon form, we can transform the non-zero row at the bottom of the matrix into polynomial form which is exactly the least linear combination.
If the resultant is $0$, then the GCD has positive degree; if the resultant is nonzero, then the GCD is $1$. I recall some more interesting questions about this where $k$ is a ring, but there isn't anything more to say about a field.
Let $n\in\mathbb{N}^*$ be a composite number. So $\mathbb{Z}_n[x]$ is not an integral domain. Through the experimental results, the above reply is still valid in $\mathbb{Z}_n[x]$ .
Could you please tell me some more interesting questions about this where $k$ is a ring?
Related questions, over rings: https://mathoverflow.net/questions/227227/reduced-resultant-of-monic-polynomials and https://mathoverflow.net/questions/17501/the-resultant-and-the-ideal-generated-by-two-polynomials-in-mathbbzx and https://mathoverflow.net/questions/248488/reduced-resultants-and-bezouts-identity and https://mathoverflow.net/questions/248574/ideal-generated-by-two-univariate-coprime-integer-polynomials and probably others. Search reduced resultant site:mathoverflow.net and related questions in the results you get.
I was remembering the question https://mathoverflow.net/questions/17501 and its many interesting answers.
Have you had a look, worm, at those links I gave?
Thank you very much. I have solved my problem well according to your suggestion.
The important notion here is that of subresultant. Suppose $P_1, P_2$ are two polynomials of degrees $d_1,d_2$ and suppose $d_1\ge d_2$. To compute the GCD you would typically use Euclid's division algorithm: you divide $P_1$ by $P_2$ and get a remainder $P_3$ then you divide $P_2$ by $P_3$ and get the remainder $P_4$ etc. The last nonzero remainder is the GCD. Now imagine doing that for generic polynomials
$$
P_1(x)=a_{1,d_1}x^{d_1}+a_{1,d_1-1}x^{d_1-1}+\cdots+a_{1,1}x+a_{1,0}
$$
and
$$
P_2(x)=a_{2,d_2}x^{d_2}+a_{2,d_2-1}x^{d_2-1}+\cdots+a_{2,1}x+a_{2,0}\ .
$$
The very first step would be to subtract $\frac{a_{1,d_1}}{a_{2,d_2}}x^{d_1-d_2}P_2(x)$ from $P_1(x)$. Don't do that. Instead multiply $P_1$ by $a_{2,d_2}$ and then subtract $a_{1,d_1}x^{d_1-d_2}P_2(x)$ so as not to produce fractions. Rince and repeat.
Generically the degree of the remainder only drops by one. The resultant essentially is the degree zero remainder, i.e., $P_{d_2+2}$. The previous $P$'s are the subresultants (up to one's choice of normalization convention, there may also be extraneous factors to peel off).
A good reference on the subject is the book "Algorithms in Real Algebraic Geometry" by Basu, Pollack and Roy.
|
2025-03-21T14:48:31.704574
| 2020-08-03T12:14:36 |
368211
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Geoff Robinson",
"Mare",
"Richard Stanley",
"https://mathoverflow.net/users/14450",
"https://mathoverflow.net/users/2807",
"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:631831",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368211"
}
|
Stack Exchange
|
Characteristic polynomials of Cartan matrices of Lie algebras
Let $L$ be a simple Lie algebra over $\mathbb{C}$ with Cartan matrix $C_L$ (as in https://de.wikipedia.org/wiki/Cartan-Matrix )
Question 1: Is is true that the characteristic polynomial $f$ of $C_L$ has the property that all irreducible factors of $f$ (over $\mathbb{Z}$) have abelian Galois group?
This is true for all Lie algebras of rank at most 15 and especially for all exceptional Lie algebras.
In case question 1 is true, is there an easy reason why this is true?
The number of irreducible factors in type A seems to be https://oeis.org/A069930 .
Question 2: In case question 1 has a positive answer, can the roots of the characteristic polynomial be explicitly written down in a nice way?
For example for type $E_7$ we have that the characteristic polynomial has irreducible factors
$x_1-2$ and $x_1^6-12x_1^5+54x_1^4-112x_1^3+105x_1^2-36x_1+1 $
and one can see the exact roots of the second polynomial here: https://www.wolframalpha.com/input/?i=x_1%5E6-12*x_1%5E5%2B54*x_1%5E4-112*x_1%5E3%2B105*x_1%5E2-36*x_1%2B1 .
In case question 1 has a positive answer we can also associate to every lie algebra $L$ with Cartan matrix $C_L$ a finite abelian group, namely the direct product of the Galois groups of the irreducible factors.
Can this group associated to $L$ be described in a nice way?
Here some examples:
Type $F$ has Klein four group.
Type $G$ has cyclic group of order 2.
Type $E_6$ has Klein four group.
Type $E_7$ has cyclic group of order 7.
Type $E_8$ has the direct product of cyclic groups of order 4 and 2.
Type $A_30$ has the direct product of two cyclic groups of order 15.
Question 3: Do these matrices belong to a certain class of special symmetric integer matrices whose characteristic polynomials all have irreducible factors with abelian Galois group?
In fact in all examples that I looked at for symmetric Cartan matrices of quiver algebras, the irreducible factors of the characteristic polynomials always had abelian Galois group.
A number of Japanese authors, such as T. Wada, have considered the eigenvalues of the Cartan matrices of blocks of finite groups.
@GeoffRobinson I just looked at Cartan matrices of quiver algebras with finite global dimension so far.
For type $A_n$ the eigenvalues are $2+2\cos(2\pi j/(n+1))$, $1\leq j\leq n$, so the Galois group of every irreducible factor is abelian. Perhaps there are similar formulas for other types. If true, this should be known.
As a follow-up on my previous comment, see https://arxiv.org/pdf/1110.6620.pdf.
|
2025-03-21T14:48:31.704771
| 2020-08-03T12:38:35 |
368213
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Carlo Beenakker",
"DKNguyen",
"Gabe K",
"LSpice",
"Nik Weaver",
"Peter - Reinstate Monica",
"Richard Montgomery",
"Robin Saunders",
"Sam Hopkins",
"https://mathoverflow.net/users/11260",
"https://mathoverflow.net/users/125275",
"https://mathoverflow.net/users/137024",
"https://mathoverflow.net/users/142929",
"https://mathoverflow.net/users/161947",
"https://mathoverflow.net/users/163341",
"https://mathoverflow.net/users/23141",
"https://mathoverflow.net/users/2383",
"https://mathoverflow.net/users/25028",
"https://mathoverflow.net/users/2906",
"https://mathoverflow.net/users/41862",
"https://mathoverflow.net/users/4336",
"user142929",
"vmist",
"zibadawa timmy"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631832",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368213"
}
|
Stack Exchange
|
Should water at the scale of a cell feel more like tar?
The Navier-Stokes equations are as follows,
$$\dot{u}+(u\cdot \nabla ) u +\nu \nabla^2 u =\nabla p$$
where $u$ is the velocity field, $\nu$ is the viscosity, and $p$ is the pressure.
Some elementary manipulations show that if you zoom in by a factor of $\lambda$, then you expect viscosity to scale as $\lambda^{\frac{3}{2}}$. So, for example, if you zoom in to the length scale of a cell, you expect viscosity to be around a million times larger than humans experience it.
This is not observed, however, which makes sense since we expect the components of a cell to move around extremely quickly. (EDIT: this is observed - see answer - my initial google searches were untrustworthy, damn google). Nonetheless, the calculation above suggests that they feel like they are moving through one of the most viscous fluids imaginable.
What then is the mechanism that prevents this? I have seen some explanations through the ideas of 'microviscosity' and 'macroviscosity' in the physics community, but I couldn't find much of a theoretic backing for them.
I'm wondering if there is a more mathematical explanation, perhaps directly from the Navier-Stokes equation itself (seems unlikely), or something from a kinetic theory point of view? For example some kind of statistical model of water molecules that reproduces the result?
My comment is unrelated to your question, but maybe this can be interesting for you. I've asked few months ago the post with title Fluid dynamics for the sepiolite on Physics Stack Exchange, post with identitficator 552396. On the other hand Wikipedia has an article dedicated to Sepiolite (if you're interested in this material). Isn't required a response/reply for this comment and I hope that it doesn't disturb to you.
@user142929's PSE question: https://physics.stackexchange.com/questions/552396/fluid-dynamics-for-the-sepiolite .
Have you ever heard of the Reynolds number? It's the reason insects fly so differently from birds and aircraft and can manuever the way they do whereas birds and aircraft cannot. They are swimming through the air more than flying through it. But we don't know the physics down at that level very well. If we did, we wouldn't have so much trouble simulating and building micro-UAVs (which are still enormous compared to insects).
"This is not observed, however": No?
There is a beautiful article (a write-up of a talk, actually), by E.M. Purcell, Life at low Reynolds number, that explains how bacteria swim.
Low Reynolds number is the technical way to phrase the statement in the OP that motion at that scale feels like moving in a tar pit. The governing equation is the linearized Navier-Stokes equation, a.k.a. the Stokes equation, which lacks the inertial $v\nabla v$ term. The linearity of the Stokes equation means that the swimming technique which we would use, moving arms or legs back and forth, will not work. Purcell calls this the "scallop theorem": opening and closing the shells of a scallop will just move the object back and forth, without net forward motion.
Inertia can still play a role on short time scales, as explained in Emergency cell swimming.
The way bacteria move in the absence of inertia is the way a corkscrew enters a material upon turning, the cork screw being the flagellum. In fact, any nonsymmetrical object, when turned will propagate in a tar pit. Typical velocities are $1$ mm/min, as Purcell says: "Motion at low Reynolds number is very majestic, slow, and regular."
Here is a visualization of a sperm cell moving by rotating its flagellum (published just this week in Science Advances).
Note that the rotation is only clearly visible in three dimensions. Two-dimensional projections suggest a beating motion (first reported by Van Leeuwenhoek in the 17th century), which is not an effective means of propagation at low Reynolds number.
So to be clear you’re saying the NS equations at that scale are still approximately valid?
yes, and in fact they are valid in the fully linearized form (without the $v\nabla v$ term)
What about paramecia?
paramecia work around the scallop theorem by beating their cilia in an asymmetric stroke, see https://www.pnas.org/content/108/18/7290
It's worth noting that the way humans swim is already very inefficient. Nearly all of the energy expended when swimming is wasted, and world class pace in the water is essentially a brisk walk for most people.
I think I understand the projections on the $x z$- and $x y$-planes, but what is the extra orange dot on the $y z$-plane tracing?
@LSpice If you look on the 3d-animated portion you will also see an orange dot, which is at the midpoint of the string/flagellum from what I can see.
I'm now wondering if it's actually faster/more efficient to move in a viscous fluid than a non viscous fluid. For example could you think of this as saying the cork screw motion is not efficient for humans, or boats (even though propellers still use it), and actually they might benefit from water being more viscous than it is?
@vmist For what it's worth, the paper this post starts out with mentions that the flagellum-propulsion is about 1% efficient. Which sounds awful, but the total amount of power they need to run the thing is still just a small fraction of their total energy intake, so the efficiency is pretty irrelevant. Good human swimmers are something like 3% efficient. Dolphins are more like 80%, I think.
You may be interested in Shapere, A., and F. Wilczek. 1987. Self-propulsion at low Reynolds number. Phys. Rev. Lett. 58: 2051–2054 where they use gauge theory to describe micro-swimming. Because the Stokes equation - the infinite viscosity limit of Navier-Stokes - is linear, it allows us to define a connection for the principal G bundle:
(located shapes) --> (unlocated shapes). Here G is the group of rigid motions of space, a located shape is (say) a volume-preserving embedding of the ball into usual 3-space, and the space of unlocated shapes is the quotient space of the space of located shapes by the action of G. Think of the ball as the cell (parmecium, E Coli, cyanobacterium, ..) which wants to move. A swimming stroke is then a loop in the space of unlocated shapes. The resulting holonomy for the Stokes connection is computed by solving the Stokes equation with zero boundary data at infinity. Shapere in his thesis estimates the curvature at the embedding which is a round ball, and thereby investigates ``infinitesimal swimming motions''. Some of this story can also be found in my book, A Tour of SubRiemannian Geometry.
The 1987 Phys.Rev.Lett. seems not to be freely downloadable, but this longer write-up from 1989 is.
Thank you Carlo.
If I am not mistaken, the Navier Stokes equations do not include random motion due to thermal fluctuations. Because of typical physiological temperatures, molecules bounce around vividly through random kicks in an overdamped viscous fluid, giving rise to Brownian dynamics.
This seems to be a possible explanation of random motion, but not volitional motion of the sort that @vmist seems to consider and that @CarloBeenakker's answer addresses.
@LSpice Perhaps it was meant as a response to the original question's comment that "we expect the components of a cell to move around extremely quickly".
|
2025-03-21T14:48:31.705305
| 2020-08-03T12:53:20 |
368214
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Christian Remling",
"Francois Ziegler",
"MaoWao",
"https://mathoverflow.net/users/19276",
"https://mathoverflow.net/users/48839",
"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:631833",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368214"
}
|
Stack Exchange
|
Canonical commutation relations-bounded vs. unbounded picture
Suppose that $Q,P$ are self-adjoint operators which satisfy the relation $$(1) \ \ \ \ \ [Q,P]=iI$$ One can easily show that in this case $P,Q$ cannot be bounded. However one can find unbounded operators (multiplication by $x$ and $\frac1i \frac{d}{dx}$) satisfying this relation. When dealing with unbounded operators one encounters problems with the domains so in order to avoid them one proceed as follows: as any self-adjoint operator gives rises to the one parameter group of unitary operators via $P \mapsto (e^{itP})_{t \in \mathbb{R}}$ one can formulate this problem in terms of this one-parameter groups. The canonical commutation relation takes the form $$(2) \ \ \ V(s)U(t)=e^{its}U(t)V(s)$$ where $U$ corresponds to $Q$ while $V $ corresponds to $P$. One form of the Stone-von Neumann theorem states that any irreducible representation of $(2)$ (say on the space $\mathcal{H}$) is unitary equivalent to the operators of multiplication by $x$ and $\frac1i \frac{d}{dx}$ (in more details: there exists a unitary $W:L^2(\mathbb{R}) \to \mathcal{H}$ such that $W^{-1}U(t)W=e^{itQ}$ and $W^{-1}V(s)W=e^{isP}$ where $P=\frac1i \frac{d}{dx}$ and $Q=M_x$. As far as I know for the relation $(1)$ this is no longer true (probably due to issues with the domains). So I would like to clarify what is the exact relation between $(1)$ and $(2)$.
Suppose that if $U(t)$ and $V(s)$ are one parameter groups of unitaries with generators $Q$ and $P$ resp. Is it true that $P,Q$ satisfy $(1)$ if and only if $U(t)$ and $V(s)$ satisfy $(2)$?
EDIT: Let me try to give some more details how I understand my question: suppose that $P,Q$ are densely defined self-adjoint operators with the property that there is a dense linear subspace $D$ contained in the intersection $\bigcap_{n,m}dom(Q^m P^n)$. Then the implication $(1) \Rightarrow (2)$ I understand as follows: we take $P,Q$ as above satisfying $(1)$ and form $U(t)=e^{itQ}$ and $V(s)=e^{isP}$ Is it true that $U(t)$ and $V(s)$ satisfy $(2)$? Regarding the opposite implication: let $U(t),V(s)$ be a one parameter groups of unitaries (strongly continuous) and take $Q,P$ to be their infinitesimal generators. Is it true that if $U(t),V(s)$ satisfy $(2)$ then $P,Q$ satisfy $(1)$? I don't see why this question is not well defined: in particular I'm not using any notion such that ,,irreducible representation of something (possibly unbounded)''. I just wanted to clarify how one can deduce $(1)$ from $(2)$ and vice versa
Does this answer your question? The Stone-von Neumann theorem without exponentials?
(1) has no immediate meaning for unbounded operators, so one has to give a precise interpretation, and one way to proceed would be (2) (until the last few lines of your post, I thought this is what you were doing). Of course, if this path is taken, then your question disappears. In any event, your question isn't precise until you clarify what (1) means for unbounded operators.
To reiterate what Christian Remling said: You have to define the operator $[Q,P]$. If $P$ and $Q$ are not everywhere defined, it is not obvious what their commutator is supposed to be. What is its domain? Do you want it to be closable and densely defined?
|
2025-03-21T14:48:31.705535
| 2020-08-03T13:14:56 |
368217
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Donu Arapura",
"Henri",
"Kamel",
"abx",
"https://mathoverflow.net/users/163078",
"https://mathoverflow.net/users/40297",
"https://mathoverflow.net/users/4144",
"https://mathoverflow.net/users/5659"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631834",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368217"
}
|
Stack Exchange
|
Corollary of Mori’s theorem
Is there a direct proof of the corollary of Mori's theorem which says: If a projective complex manifold does not contain a rational curve then $ K_X $ is nef
What do you mean by a direct proof? The reference given to your previous question gives a direct proof.
I just want to use the absence of a rational curve to show that $ K_X $ is nef
Do you mean that you are wondering whether one can prove such a statement without appealing to Mori theorem?
The result of Mori that you quote is deep. I don't know if, more than 30 years later, there is a proof that avoids some form of his original "bend and break" technique. I assume that's what you're asking for.
Yes that's it @Donu Arapura
|
2025-03-21T14:48:31.705633
| 2020-08-03T13:49:16 |
368219
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Lucia",
"Phil Tosteson",
"Terry Tao",
"https://mathoverflow.net/users/38624",
"https://mathoverflow.net/users/52918",
"https://mathoverflow.net/users/766"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631835",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368219"
}
|
Stack Exchange
|
Sequences for which $\prod (1-z^n)^{a(n)}$ is a polynomial
This is mostly a reference request.
I'm working with complex coefficients, although all I have in mind have integer coefficients.
Let $a=(a(n))_{n\ge 1}$ be a sequence, say of integers (I have non-negative integers in mind), although things would make sense with complex numbers. The infinite product $\Phi_a(z)=\prod_n(1-z^n)^{a(n)}$ naturally lies in the ring of power series $\mathbf{C}[\![z]\!]$, and $a$ is determined by $\Phi_a$. Say that $a$ is special if $\Phi_a$ is a polynomial.
Is there a standard terminology for the assignment $a\mapsto \Phi_a$ or the inverse map $\Phi_a\mapsto a$? for what I call "special" sequences? Is there a study of special sequences (they seem far from arbitrary), notably involving the Möbius function $\mu$, or general facts about their asymptotics? Relevant keywords/references would be welcome.
I've seen the assignment $\Phi$ at several places, e.g. related to partitions or modular forms, but the emphasis was not on special sequences, but rather simple-minded $a$ giving rise to interesting $\Phi_a$, while I'm rather interested in the opposite: interesting $a$ giving rise to simple-minded $\Phi_a$.
Examples of naturally occurring "special" sequences $a$ are notably the function $n\mapsto a_d(n)$, the dimension of the degree $n$ graded component of the free Lie algebra on $d$ generators, for which $\Phi_a(z)=1-dz$ (OEIS: for $d=2$ A001037: 2, 1, 2, 3, 6, 9, 18, 30, 56, 99, 186..., for $d=3$ A027376: 3, 3, 8, 18, 48, 116, 312, 810, 2184...), see also Wikipedia: Cyclotomic identity. Other examples (among others): A125951: 1, 1, 2, 2, 4, 5, 10, 15, 26, 42, 74, 121, 212, 357... and A020999 for which such a formula $\prod (1-z^n)^{a(n)}=$(polynomial) is explicitly mentioned in both cases.
When $a_n$ is negative I have seen this called the Euler transform before. See e.g. here: https://mathworld.wolfram.com/EulerTransform.html
If $R$ is a unique factorization domain of polynomials over some finite field, then the zeta function $\zeta_R(z) = \sum_{f \in R, \hbox{ monic}} z^{\mathrm{deg}(f)}$ is equal to the product $\prod_{n=1}^\infty ( 1-z^n)^{a_n}$ where $a_n$ is the number of irreducible polynomials in $R$ of degree $n$ (this is the generating function form of the fundamental theorem of arithmetic). In many cases the Weil conjectures apply and make $\zeta_R$ a rational function. With a bit of algebraic number theory one can generalise beyond the UFD case to Dedekind domains.
If one just takes logs and compares coefficients, then the condition for the sequence $a$ to be special is that $na(n)$ must equal $\sum_{j=1}^k \epsilon_j \sum_{d|n} \mu(d) \alpha_j^{n/d}$, where $\epsilon_j =\pm 1$ and $\alpha_j$ are complex numbers. (This is in the case that the product is a rational function. For a polynomial, the $\epsilon_j$ are all $1$.)
|
2025-03-21T14:48:31.705861
| 2020-08-03T14:14:01 |
368223
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Asaf",
"Benjamin Steinberg",
"GA316",
"Terry Tao",
"https://mathoverflow.net/users/15934",
"https://mathoverflow.net/users/33047",
"https://mathoverflow.net/users/766",
"https://mathoverflow.net/users/8857"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631836",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368223"
}
|
Stack Exchange
|
Suggestion for framing a course in Representation theory + Spectral graph theory
I am going to give a course in spectral graph theory to graduate students. I want to learn and teach the connection between the spectral graph theory and the representation theory of finite groups. I am good at both the areas but I am not sure where to start and what to include. It would be a great help to me if you can suggest what to add beyond the basics in both the areas. I haven't seen any book on the connection between these areas.
Some things in my mind are:
Representation theory of $Aut(G)$ where $G$ is a finite simple graph.
Reducible/Irreducible eigenspaces.
I am not aware of the literature where representation theory is used in spectral graph theory or vice versa. Kindly give your suggestions.
Thank you.
The book Elementary Number Theory, Group Theory and Ramanujan Graphs by Davidoff, Sarnak and Valette might be a good start. It does spectral graph theory and group representations, mostly applied to Cayley graphs.
The Gowers theory of quasirandom groups seems appropriate to include in this course. https://terrytao.wordpress.com/2011/12/16/254b-notes-3-quasirandom-groups-expansion-and-selbergs-316-theorem/
@BenjaminSteinberg Thanks for the reference. Also, it suggests that expanders are definitely had to be included.
@TerryTao Thank you. I am not much familiar with quasi randomness but Proposition 4 is very interesting. I will check the literature.
The book by Lubotzky about expanders is considered the standard reference for expanders, together with some notes by Hoory-Linial-Wigderson.
Update
I have since uploaded a preprint discussing this connection. This is probably not it‘s final form, but since I claimed writing on this some years ago, it is more than time to finally mention it here. Allow me to also point to my PhD thesis (follow the links in my profile) where I explore some of this in more detail.
Otherwise, the answer is unchanged and contains below some of the sources and main ideas I have used in the past.
I alway put my focus on the idea of the graph realization, because it gives the subject a geometric touch. A graph realization is simply a map assigning to each vertex $i\in V$ a point $v_i$ in Euclidean space.
And such a realization can be highly symmetric (related to representation theory) or it can be some sort of balanced configuation (related to spectral graph theory). These ideas are not independent.
For example, suppose you have a realization that satisfies some kind of self-stress condition:
$$(*)\qquad \sum_{j\in N(i)} v_j = \theta v_i\quad\text{for all $i\in V$}.$$
Let $M$ be the matrix in which the $v_i$ are the rows, then you can write $(*)$ as $AM=\theta M$ (where $A$ is the adjacency matrix of the graph). Immediately you see that $\theta$ must be an eigenvalue of $A$, and the columns of $M$ must be eigenvectors.
The columns need not span the whole eigenspace.
But if they do, then we call it a spectral realization (see also the link [1] below).
If you define the arrangement space $U:=\mathrm{span}(M)$ as the column span of $M$ (see also the link [3] below), then you have a handy way to define symmetric and spectral realizations:
a realization is symmetric if its arrangement space is $\mathrm{Aut}(G)$-invariant.
a realization is spectral if its arrangement space is an eigenspace of $A$.
And since eigenspaces are always invariant, we immediately find that spectral realizations are always as symmetric as the underlying graph.
In my opinion, it is this property of spectral realizations that tells us a lot about the structure of the graph (at least for highly symmetric graphs).
Others might use them on less symmetric graphs in graph drawing algorithms or optimization (but I feel this is less related to representation theory).
If you take the convex hull of the vertices in a spectral graph realization, you obtain the eigenpolytope of a graph.
The literature on these is quite scattered, but the initial source is probably "Graphs, groups and polytopes" by Godsil (I have since tried to organize the literature in this other (work in progress) preprint).
Godsil proved that the eigenpolytope is as symmetric as the initial graph. He also proves group theoretic properties of $\mathrm{Aut}(G)$ from these polytopes (which are just graph realizations in disguise).
You asked specifically about reducible/irreducible eigenspaces. In general, it is quite tricky to determine whether the eigenspaces of a graph are irreducible (without computing all irreducible subspaces). But there is one case for which it is easy: distance-transitive graphs. For these, the eigenspaces are exactly the irreducible subspaces of $\mathrm{Aut}(G)$. This basically follows from Proposition 4.1.11 (p. 137) in "Distance Regular Graphs" by Brouwer, Cohen and Neumaier.
Their proof is in a purely represenation theoretic language, but in the preprints I also discuss some more elementary approaches.
Finally, I can think about some connections to rigidity theory.
One might consider only the deformations of a graph realization that preserves the symmetry of the structure.
Whether such deformations exist depends on the decomposition of the permutation-representation of $\mathrm{Aut}(G)$ into irreducible representations (in particular, their multiplicities).
To connect this to spectral graph theory, one can observe that if a realization is rigid (i.e. it cannot be deformed without loosing symmetry), and irreducible, then one can show that it satisfies $(*)$ (it is not necessarily spectral, but almost).
Of course, for distance-transitive graphs, this implies that the realization is spectral.
Here are some older posts of mine that might be related:
[1] directly related: Representations of the automorphism group of graphs via spectral graphs theory
[2] how to get the irreducible subspaces when the eigenspaces are not irreducible: Determining the irreducible invariant subspaces of a permutation action by computing eigenspaces of a matrix
[3] a simple construction (the arrangement space) that I always found helpful for organizing my thoughts when working in spectral graph theory, representation theory and geometry at the same time (check in particular the two last bullet points): Where have you encountered "arrangement spaces"?
Very nice answer. Informative. Please update us once the paper is complete and thanks for the wonderful references.
A paper that might be suitable for your course, that touches on both the points you listed, is Graph Automorphisms from the Geometric Viewpoint. As the abstract says, it's concerned with the representation theory of $Aut(G)$ where $G$ is a finite graph.
Another collection of literature that might go along with the course you sketch is the theory of Quiver representations.
Lastly, Daniel Spielman wrote a book on Spectral and Algebraic Graph Theory. While not tied directly to representations of $Aut(G)$, this might have some ideas that could help as you plan your course. It sounds like a great course. Good luck!
|
2025-03-21T14:48:31.706523
| 2020-08-03T14:25:58 |
368224
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Ben Wieland",
"Jon Pridham",
"Phil Tosteson",
"curious math guy",
"https://mathoverflow.net/users/103678",
"https://mathoverflow.net/users/152554",
"https://mathoverflow.net/users/4639",
"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:631837",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368224"
}
|
Stack Exchange
|
Étale homotopy type of (derived) loop space
A feature of derived algebraic geometry is that we have internal homs. Furthermore, we can think of $B\mathbb{Z}$ as the derived algebraic geometric analogue of $S^1$. Thus we have an analogue of the loop space, the derived loop space
$$LX:=\mathbb{R}\operatorname{Map}(B\mathbb{Z},X)$$
using the notation of "Chern character, loop spaces and derived algebraic geometry" (https://hal.archives-ouvertes.fr/hal-00772859/document). My question is how does this interact with the étale homotopy type? Specifically I would like to know if the following is true
$$\operatorname{Et}(LX)=\Omega \operatorname{Et}(X)$$
where the right hand side denotes the loop space of the étale type of $X$. A naive hope for a proof would be
$$\operatorname{Et}(\mathbb{R}\operatorname{Map}(B\mathbb{Z},X))\cong \operatorname{Map}(\operatorname{Et}(B\mathbb{Z}),\operatorname{Et}(X)) \cong \operatorname{Map}(S^1,Et(X))$$
but I don't know if this makes sense.
The proof should be a bit simpler than that, as you can write $LX = X \times^h_{X \times X} X$. (You can always express maps from a simplicial set as a homotopy limit.) The etale homotopy type won't see any of the derived structure, though.
@JonPridham So you think this is true? If you add it in an answer, I'm happy to accept it.
By @Jon Pridham's comment you can see that the etale homotopy type of $LX$ is $Et(X)$, when $X$ is a scheme. This disagrees with the free loop space of $Et(X)$.
Could we at least hope that there exists some (derived) scheme $Y$ such that $\Omega Et(X)\cong Et(Y)$?
It seems unlikely you'd have a natural way of constructing $Y$. Reverting to the original question, Friedlander found a class of smooth morphisms which become fibrations in etale homotopy (after fairly mild completion), so it would be enough to take $X$ to be a higher stack whose diagonal takes lies in that class. In practice, that would mean $BG$ and its higher analogues, for $G$ smooth (no derived structure).
You could try to cheat and construct a scheme $Y(Z)$ with the homotopy type of an arbitrary simplicial set $Z$, by gluing algebraic simplices together. I'm not sure if this works /what finiteness conditions you need to impose on $Z$ to get a scheme versus an Ind scheme.
@PhilTosteson I don't know the "cheat" you are refering to, even though it sounds interesting. Given an arbitrary simplicial set, how do you construct an Ind scheme with that homotopy type?
Geometry and homotopy theory are different. Etale homotopy type is a functor from geometry to homotopy theory. Stacks and derived schemes are a vertical dimension of homotopy theory along a horizontal geometric space. The derived loop space is a purely homotopical operation acting on the vertical dimension. The etale homotopy type combines the horizontal and vertical data, but that is destructive. In other words, the derived loop space is about small loops in the vertical direction, not large horizontal/geometric loops. Whereas, the etale homotopy type knows about geometric loops.
|
2025-03-21T14:48:31.706747
| 2020-08-03T14:26:03 |
368225
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Frode Alfson Bjørdal",
"Hanul Jeon",
"https://mathoverflow.net/users/37385",
"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:631838",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368225"
}
|
Stack Exchange
|
Adjunction, infinity and hereditarily finite sets
Is
$$\mathrm{U}_{\omega}=\Big\{x\mid\forall z\Big(\big(\emptyset\in z\wedge \forall u, v\;(u,v\in z\rightarrow\{w\mid w\in u\vee w=v\}\in z)\big)\rightarrow x\in z\Big)\Big\}$$
identical with the set $\mathrm{V}_{\omega}$ of hereditarily finite sets, i.e. the level $\omega$ of the cumulative hierarchy?
Adjunction of two sets $u$ and $v$ are $u\cup{v}$, so I think the condition $w\in u\lor w=u$ is a typo (one of two $u$s should be $v$.) Am I correct?
@HanulJeon Indeed! Thanks, I correct.
I assume that the definition of $U_\omega$ has a typo: adjunction operation usually means the following binary operation:
$$u;v:=u\cup\{v\}$$
If you just assume $z$ in the definition of $U_\omega$ is closed under successor operator $u\mapsto u\cup\{u\}$, then $U_\omega$ would be $\omega$.
It suffices to show that every set which contains $\varnothing$ and is closed under adjunction contains $V_\omega$.
Let $z$ be a set such that $\varnothing\in z$ and is closed under adjunction.
Assume inductively that $V_n\subseteq z$. For $x\in V_{n+1}=\mathcal{P}(V_n)$, we have $x=\{a_0,\cdots, a_k\}$ for some $a_0,\cdots,a_k\in V_n$. By induction on $i\le k$, you can see that $\{a_0,\cdots, a_i\}\in z$ (In the case of $i=0$, $\{a_0\}$ is the adjunction of $\varnothing$ and $a_0\in z$.) Hence $x\in z$. This shows $V_{n+1}\subseteq z$.
|
2025-03-21T14:48:31.706874
| 2020-08-03T14:33:23 |
368226
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Gianfranco",
"Iosif Pinelis",
"https://mathoverflow.net/users/150973",
"https://mathoverflow.net/users/36721"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631839",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368226"
}
|
Stack Exchange
|
Uniform distribution on a simplex
In a context where I try to estimate some combinatorial sums, I'm faced with vector random variables $(x_1,...,x_n)$ uniformly distributed with $n \rightarrow \infty$. I want to know if the components have to behave in a wellknown fashion. I recently read the following. (Sourav Chatterjee math summer school 2012)
"
Classical example: Uniform distribution on the simplex $\{ (x_1,...,x_n) \ | \ \sum\limits_{k=1}^{n} x_k =1 \}$
In this example, it is known that for n large, the coordinates behave like i.i.d. $Exp(1)$ random variables
"
Where can I find a proof of that result or related?
Thanks for any answer/pointer
Do you have a response to the answers below?
Let $(X_1,\dots,X_n)$ be a random point uniformly distributed on your simplex. Then it is well known (cf. e.g. Remark 1.3 and formula (2.4)) that $(X_1,\dots,X_n)$ equals
$$\frac{(Z_1,\dots,Z_n)}{Z_1+\dots+Z_n}$$
in distribution,
where $Z_1,\dots,Z_n$ are iid $Exp(1)$ random variables. So, each $X_i$ equals
$$\frac{Z_1}{Z_1+\dots+Z_n}=\frac{Z_1}n\Big/\frac{Z_1+\dots+Z_n}n$$
in distribution. Also, $\frac{Z_1+\dots+Z_n}n\to1$ almost surely and hence in distribution (as $n\to\infty$), by the strong law of large numbers. Thus, for each $i$, the distribution of $nX_i$ (not of $X_i$) goes to $Exp(1)$.
Great, thanks a lot for the comment and the reference!
The simplex is, of course, the first quadrant of the $\ell_1$ sphere. This delightful article gives a simple formula for uniform sampling on the $\ell_p$ sphere for any $1 \leq p \leq \infty$.
Thanks you very much for your interesting answer
|
2025-03-21T14:48:31.707013
| 2020-08-03T14:42:15 |
368228
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Alexandre Eremenko",
"Paul",
"Willie Wong",
"https://mathoverflow.net/users/126827",
"https://mathoverflow.net/users/25510",
"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:631840",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368228"
}
|
Stack Exchange
|
A question of uniqueness
Let $u$ an harmonique function on $\Omega=(a,b)\times (0,+\infty)$ and boundary conditions :
$\displaystyle u(a,y)=u(b,y)=0,\quad\forall y\geq 0$
$\displaystyle u(x,0)=0,\,\lim_{y\to +\infty} u(x,y)=0 \quad \forall x\in (a,b)$
Can we conclude that $\quad u=0$ on $\Omega$ ?
My adempt
Let $$\Omega_{R}=(a,b)\times (0,R),\forall R>0$$By IBP, i show that
$$\int_{\Omega_R}u\Delta u=\int_a^{b}u(x,R)\frac{du}{dy}(x,R)dx-\int_{\Omega_R}|\nabla u|^2 $$ Thus
$$\forall R>0,\quad \int_{\Omega_R}|\nabla u|^2=\int_a^bu(x,R)\frac{du}{dy}(x,R)dx$$
I need help to cointinuous ( For example to show $\int_{\Omega}|\nabla u|^2=0$)
edit Continuing the initial reasoning, with $a=0$ and $b=\pi$ as suggested by A Ermenko
$\Big(\int_{\Omega_R}|\nabla u|^2\Big)^2=\Big(\int_0^{\pi}u(x,R)\frac{du}{dy}(x,R)dx\Big)^2\\
\leq \int_0^{\pi}u(x,R)^2dx\int_0^{\pi}\Big(\frac{du}{dy}(x,R)\Big)^2dx ,\mbox{by Cauchy–Schwarz inequality
}\\\\$
$\leq \int_0^{\pi}u(x,R)^2dx \int_0^{\pi}|\nabla u|^2(x,R)dx \\\\$
$= \int_0^{\pi}u(x,R)^2dx\int_{]0,\pi[\times\{R\}}|\nabla u|^2 \\\\$
$\leq \int_0^{\pi}u(x,R)^2dx \int_{\Omega_R}|\nabla u|^2,\mbox{because} ]0,\pi[\times\{R\}\subset\Omega_R$
Then $$\int_{\Omega_R}|\nabla u|^2\leq \int_0^{\pi}u(x,R)^2.$$ I can only conclude if $\displaystyle ||u(.,R)||_{L^2]0,\pi[}\to^{R\to\infty} 0$
Isn't it easier to just use maximum principle?
Anyway, for the method you choose: $u(x,R) \frac{du}{dy}(x,R) = \frac12 \frac{d}{dy} |u(x,y)|^2 \Big]_{y = R}$. If the lim of $u(x,y)$ tends to zero, there exists some $R$ for which the integral of this derivative is negative.
@Willie Wong the maximum principale is available if $\Omega$ a bounded domain theoreme 2.17 https://www.math.ucdavis.edu/~hunter/pdes/ch2.pdf
Let $\Omega_t = \Omega \cap { y < t}$. Apply the maximum principle for $u$ on $\Omega_t$. Take $t\to \infty$.
@Willie Wong I tried to apply the maximum principle with your indication with $ \Omega_t $ but without success. Can you give your answer because I do not see why it is easy and especially that A.Ermenko says that there is a counter example.
@cerise: I had taken your convergence $\lim_{y\to\infty} u(x,y) = 0$ to mean convergence with respect to some reasonable function norm (say $L^\infty$). If you only wished to have pointwise convergence and no more, then as Eremenko said there are counterexamples.
If you want the positive answer you should state your last condition more carefully.
For example, add that $u$ is bounded, or that $u(x+iy)$ tends to $0$ as $y\to\infty$
UNIFORMLY with respect to $x$.
As you presently stated, the answer is negative. I sketch the construction of a counterexample.
There exists a non-zero entire function, real on the real line and such that $f(re^{i\theta})\to 0$ as $r\to+\infty$ for every $\theta$. (See, for example, my answer to this question, which explains how to construct $f$.)
$v(z)=\Im f(z)$ is a non-zero harmonic function in the upper half-plane, equal to $0$ on
the real line and $u(re^{i\theta})\to 0$ as $r\to+\infty$ for every $\theta\in(0,\pi)$.
Take without loss of generality $a=0,\; b=\pi$, then $e^{-i(z-\pi)}$
maps your strip $0<z<\pi$ into the upper half-plane, with a removed half-disk.
So the function $w(z)=v(e^{-i(z-\pi)})$ is harmonic, zero on infinite sides
of your half-strip and satisfies the property at $\infty$:
$w(x+iy)\to 0$ as $y\to+\infty$ for every $x\in(0,\pi)$. Notice that this function is unbounded. To satisfy the last requirement, that $u(x)=0$ for $0<x<\pi$,
set $u(z)=w(z)-w_1(z)$, where $w_1(z)$ is the solution of Dirichlet problem matching
the boundary values of $w$ on the finite part of the boundary and bounded in your strip.
Remark. Since every simply connected domain other than the plane is conformally equivalent to the unit disk, your question is equivalent to the following. Suppose that $u$ is harmonic in the unit disk and $\lim_{z\to\zeta} u(z)=0$ for all $\zeta\in\{ \zeta:|\zeta|=1\}\backslash\{1\}$, and moreover $u(z)\to 0$ along any non-tangential segment ending at $1$, that is $u(1-re^{i\theta})\to 0,\; r\to 0$ for every $\theta\in(-\pi/2,\pi/2)$. Does it follow that $u=0$? The answer is NO.
@ A.Eremenko I think you are right, assumptions are missing, I edited my post. I have not yet understood the construction of a counter example
@cerise: I do not see how you edited your post, but if you add one of the conditions that I suggested, the counterexample will loose its meaning.
@ A.Eremenko I edited to continue my reasoning and didn't add anything as additional assumptions. I noticed that the answer is positive to the question if $\displaystyle ||u(.,R)||_{L^2]0,\pi[}\to^{R\to\infty} 0$
conflicting opinions of experts, does not help me.
@cerise: your conclusion that $| u(,.R)|_2\to 0,; R\to\infty$ is incorrect: it does not hold in my example. From your assumptions AS STATED it does NOT follow that $u=0$. Look in the literature about "passing to the limit under the integral sign".
@A.Eremenko I said I can only conclude if $\displaystyle ||u(.,R)||_{L^2]0,\pi[}\to^{R\to\infty} 0$. in the sense: if I assume this hypothesis of more
|
2025-03-21T14:48:31.707707
| 2020-08-03T14:50:55 |
368229
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Tim Campion",
"YCor",
"https://mathoverflow.net/users/14094",
"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:631841",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368229"
}
|
Stack Exchange
|
Examples of symplectic manifolds whose Betti numbers are not non-decreasing
I am looking for examples of closed symplectic manifolds $(M,\omega)$ whose Betti numbers do not satisfy a non-decreasing property. Meaning, it fails to satisfy $b_k(M) \leq b_{k+2}(M)$ for some $k < n=\frac{1}{2}\dim M$. (Edit: I've been told in the comments that this property for the Betti numbers is also called unimodal). It is possible there are not any known examples but I have not perused the literature enough to be sure.
If $M$ is a Kahler manifold, then a consequence of the Hard Lefschetz theorem shows that $M$ does satisfy this non-decreasing property. Outside of Kahler examples, there are symplectic manifolds which satisfy a Hard Lefschetz property. Lastly, there are also examples of symplectic manifolds which do not satisfy the Hard Lefschetz property but as far as I know, the known examples still satisfy the non-decreasing property.
This question was asked many years ago but the link in the comment leads to an unavailable page:
Examples of non-Kahler symplectic manifolds.
As the keyword might appear in this context, it's the same as the sequence of Betti numbers not being unimodal.
The broken link at the other question leads, according to the wayback machine, to Symplectic manifolds with no Kahler structure by Aleksy Tralle and John Oprea, Springer, Lecture Notes in Mathematics vol 1661. A concerning observation is that although the archived page includes the doi information for the book, the new webpage doesn't seem to prominently display this information. Is this standard for Springer now?
Note to self: the doi can be extracted from the modern Springer URL as the number, 10.1007 in this case, which appears before the % sign.
|
2025-03-21T14:48:31.707871
| 2020-08-03T15:02:48 |
368230
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Andreas Blass",
"Gerhard Paseman",
"https://mathoverflow.net/users/3402",
"https://mathoverflow.net/users/6794"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631842",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368230"
}
|
Stack Exchange
|
Map $f:\mathbb{N}\to\mathbb{N}$ such that every 2-set is a neighbor exactly once
Is there a map $f:\mathbb{N}\to\mathbb{N}$ such that for all $a,b\in\mathbb{N}$ with $a\neq b$ there is exactly one $n\in\mathbb{N}$ such that $\{a,b\} = \{f(n),f(n+1)\}$?
You are asking for an Euler tour of the countable complete graph. This should be doable with a random walk, as every node has infinite degree. Just remember to go back to every place you ever visited again and again. If you want a guarantee, consider completing a tour of K2n+1, and extend that to a tour of K2n+3. Gerhard "Take Lots Of Pocket Money" Paseman, 2020.08.03.
How about defining $f$ by induction? $f(0)=0$ and $f(n+1)=$ the smallest $k$ such that ${f(n),k}\neq{f(m),f(m+1)}$ for all $m<n$.
I think that is what I just said. ;) (Near equivalence left to the reader to demonstrate.) Upon reflection, that gives a different tour. Perhaps our walkers will meet? Gerhard "Using Different Words, Of Course" Paseman, 2020.08.03.
Aaron Meyerowitz's answer shows that my previous comment doesn't work. I'd delete it, but then what other people have written, referring to it, would no longer make sense. So I'll leave it up, but I hereby warn people not to rely on it.
I'm not sure that the descriptions so far work. They seem to result in things like
$ 0, 1, 2, 0, 3, 1, 4, 0, 5, 1, 6, 0, 7, 1, 8, 0, 9, 1, 10,\cdots$
Here is something that works.
The sequence starts $0, 1, 2, 0, 3, 2, 4, 3, 1, 4, 0, 5, 4, 6, 5, 3, 6, 2, 5, 1, 6$ so the pairs visited are
$ (0, 1), (1, 2), (0, 2), (0, 3), (2, 3), (2, 4), (3, 4), (1, 3), (1, 4), (0, 4), (0, 5), $
$(4, 5), (4, 6), (5, 6), (3, 5), (3, 6), (2, 6), (2, 5), (1, 5), (1, 6), (0, 6), (0, 7), (6, 7)$
The rule is that, after determining $f(1),\ldots, f(n)$, set $f(n+1)$ so that $\lbrace f(n),f(n+1)\rbrace $ is the least possible pair not yet used where the order is $\lbrace a,b \rbrace < \lbrace c,d \rbrace$ if $\max(a,b) <\max(c,d)$ and, for $a=\max(a,b,c),$ we have $\lbrace a,b \rbrace < \lbrace a,c \rbrace$ if $|a-b|<|a-c|$ or $|a-b|=|a-c$ and $b<c.$
I found a graphic representation helped. Just subtracting edges, one finds K_2n+1 - K_2n-1 is an edge shared by 2n-1 triangles, and has an Euler circuit. So one can progress along these differences to provide an Euler circuit for an arbitrarily large finite complete graph with an odd number of vertices. Gerhard "The Differences Are All Planar" Paseman, 2020.08.03.
|
2025-03-21T14:48:31.708060
| 2020-08-07T00:18:43 |
368523
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Aaron Bergman",
"David Ben-Zvi",
"Dustin Clausen",
"MathCrawler",
"Z. M",
"https://mathoverflow.net/users/161310",
"https://mathoverflow.net/users/176381",
"https://mathoverflow.net/users/3931",
"https://mathoverflow.net/users/582",
"https://mathoverflow.net/users/947"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631843",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368523"
}
|
Stack Exchange
|
Does there exist a GRR-like generalization of the AS Index Theorem?
The Hirzebruch Riemann-Roch Theorem (HRR) expresses an analytic/algebraic invariant, namely the Euler-Poincaré characteristic of a vector bundle $V$ over a compact complex/algebraic manifold $X$, as the evaluation of a cohomological expression. It has different manifestations in the analytic and algebraic categories; the general form in the analytic category is something like
$$
\chi(X,V) = T(X,V),
$$
where $X$ is a compact complex manifold, $V$ a holomorphic vector bundle, $\chi(M,V)$ $=$ $\sum_{i \ge 0} (-1)^i\dim_{\mathbb{C}}H^i(X,\mathcal{O}(V))$
the holomorphic Euler-Poincaré characteristic, and $T(X,V)$ a particular cohomological expression (see [1], § 21; [2]).
Subsequently, there were two important generalizations of this result:
The Grothendieck Rieman-Roch Theorem (GRR). This shifts the focus from objects $(X,V)$ to morphisms $f:X \rightarrow Y$, and to coherent sheaves, which are amenable to being pushed forward, and so its algebraic manifestion appears as a commutative diagram
$\require{AMScd}$
\begin{CD}\tag{1}
K_{\omega}(X) @>f_!>> K_{\omega}(Y)\\
@V\tau_XVV @VV\tau_YV\\
K_{\text{coh}}(X) @>>f_*> K_{\text{coh}}(Y)
\end{CD}
Here, $f:Y \rightarrow Y$ is a morphism of algebraic varieties, $K_{\omega} :=(K_{\omega}(-),(-)_*)$ an algebraic theory built from coherent sheaves, $K_{\text{coh}} := (K_{\text{coh}}(-), (-)_!)$ a cohomological theory, and $\tau:K_{\omega}\rightarrow K_{\text{coh}}$ a natural transformation of theories (see [3]; [4] Theorem 18.3; [5]; [6]). This generalizes (HRR) insofar as an appropriate instance of (HRR) is obtained from an appropriate instance of (GRR) by applying it to the morphism $X \rightarrow \text{pt}$.
The Atiyah-Singer Index Theorem (ASI). This shifts the focus from objects $(X,V)$ to (pseudo)elliptic operators $D:\mathcal{C}^{\infty}(X,E) \rightarrow \mathcal{C}^{\infty}(X;F)$ on real $\mathcal{C}^{\infty}$-vector bundles over a compact real $\mathcal{C}^{\infty}$-manifold $X$. It takes the form
$$
\text{index}(D) = T(X;E,F)
$$
where
$$
\text{index}(D) := \dim \ker(D) - \dim \text{coker}(D)
$$
is an analytical invariant and $T(X;E,F)$ a cohomological (topological) invariant (see [7], Theorem (6.8)). Then (HRR) arises by specializing $D$ to the Dolbeault operator $\overline{\partial}$ (see [8], Theorem (4.3).
Question:
Does there exist a generalization of (ASI) to a relative form, (GAS), say, in a similar fashion as (HRR) generalizes to (GRR), so that we have a square of generalizations
$\require{AMScd}$
\begin{CD}\tag{2}
(HRR) @>>> (ASI)\\
@VVV @VVV\\
(GRR) @>>> (GAS)
\end{CD}
so that (GAS) specializes, on one hand, to (ASI), and, on the other hand, to (GRR) (and, as the cherry on the cake, do everything equivariantly).
Addendum. Meanwhile, [9] has come to my attention, showing that I am with my question in very good company. Specifically, 6. Further remarks. (3) there refers to loc.cit. Theorem 1 (ASI) and loc.cit. Theorem 3 (HRR) as follows:
In view of Theorem 3 it is natural to hope that the Grothendieck-Riemann-Roch theorem for proper maps of complex manifolds will come out of a suitable generalization of Theorem 1.
Using the notation of loc.cit. one central issue might be, for the special case where the target theory is singular (co)homology, to give, for a proper map $f:X \rightarrow Y$ of real manifolds and an operator $D$ on $X$, an appropriate definition of $\text{ch}(f!D)$.
_
[1] Hirzebruch, F. --
Topological Methods in Algebraic Geometry
(Reprint of the 1978 Edition). Springer 1978.
[2] O'Brian, N.R. et al. --
Hirzebruch-Riemann-Roch for coherent sheaves, Amer. J. Math. 103, (1981), 253-271.
[3] Borel, A, & Serre, J.-P. --
Le théorème de Riemann-Roch (d'après Grothendieck), Bull. Soc. Math. France
86 (1958), 97 - 136.
[4] Fulton, W. --
Intersection Theory (2nd Ed.), Springer 1998.
(Reprint of the 1978 Edition). Springer 1978.
[5] Baum, P. et al. --
Riemann-Roch and topological K-theory for singular varieties, Acta Math. 43 (1979),155-192.
[6] O'Brian, N.R. et al. --
A Grothendieck-Riemann-Roch formula for maps of complex manifolds, Math. Ann. 271, (1985), 493-526.
[7] Atiyah, M.F & Singer, I.M. --
The index of elliptic operators I. Ann. Math. 87 (1968),
484-530.
[8] Atiyah, M.F & Singer, I.M. --
The index of elliptic operators III. Ann. Math. 87 (1968),
546-604.
[9] Atiyah, M.F & Singer, I.M. --
The index of elliptic operators on compact manifolds. Bull. Amer. Math. Soc. 69 (1963), no. 3, 422--433.
Isn’t this just the families index theorem, proved in Atiyah-Singer IV?
I guess the families index theorem corresponds to the special case of GRR where the map is smooth.
@AaronBergmann Rather not, I think, for else, (GRR) would arise just by specializing some parameters in its formulation, just as (HRR) arises from (GRR) by specializing the morphism and from (ASI) by specializing the operator, and I know of no way to formulate the Index Theorem for Families in such a way that specializing some parameters ieads to (GRR). By the way, this would make the appearance of elaborate papers like [6] 14 years later look somewhat strange.
@MathCrawler It seems to me that multiple generalizations are going on here. GRR both generalizes HRR to the relative case and to the case of coherent sheaves. The families index theorem, I believe, gives you the relative generalization you are looking for. However, for coherent sheaves you have to be a bit more careful because they can be trickier on analytic manifolds than on algebraic varieties, even in the smooth case. Cont’d.
If I remember correctly, the Toledo-Tong stuff deals with the fact that you may not have a resolution by locally frees on a complex analytic manifold. I suspect that if you grant such a resolution (taking the protective case, say), things would go through in a straightforward manner. I’m not an expert, however, so I’m hoping someone will chime in, and I don’t need to try to work through the details :)
Protective $\longrightarrow$ projective of course.
@AaronBergman Even restricting to the projective case over the complex numbers, I do not see how the Families' Index Theorem yields (GRR). Could you be more explicit (provide some details, cite sources, etc.)?
Are you ok with the case of a product?
@MathCrawler Nonetheless, from a little googling, this thesis seems to have a generalization: https://repository.upenn.edu/cgi/viewcontent.cgi?article=4325&context=edissertations
To amplify Dustin's comment, I believe the Families Index Theorem applies to maps that are "smooth" in the sense of algebraic geometry, i.e., submersions, while GRR applies to general proper maps, such as resolutions of singularities -- i.e. the families allowed in algebraic geometry are typically much less constrained than in differential topology. Then again maybe all that's missing (given general functoriality) is to know a version of Atiyah-Singer for closed embeddings, and maybe that's well known?
I'm sorry for the self-citation. But your question is largely answered in the monograph Coherent Sheaves, Superconnections, and Riemann-Roch-Grothendieck, or the arxiv version, joint work of Jean-Michel Bismut, Shu Shen, and me.
In the above work we proved the Grothendieck-Riemann-Roch theorem for coherent sheaves on compact complex manifolds. The target theory is the Bott-Chern cohomology.
We use antiholomorphic flat superconnections as a tool to study coherent sheaves on complex manifolds, on which we can build Chern-Weil theory. We then proved the Grothendieck-Riemann-Roch theorem as a version of the family index theorem.
You opt for real Bott–Chern cohomology. Does your argument also work for rational Bott–Chern cohomology? On the other hand, is there any prior proof (via a family version of index theorem) for Deligne cohomology?
|
2025-03-21T14:48:31.708575
| 2020-08-07T01:44:23 |
368530
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Benjamin Steinberg",
"Ehud Meir",
"Sam Hopkins",
"darij grinberg",
"https://mathoverflow.net/users/15934",
"https://mathoverflow.net/users/17773",
"https://mathoverflow.net/users/25028",
"https://mathoverflow.net/users/2530",
"https://mathoverflow.net/users/41644",
"kodlu"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631844",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368530"
}
|
Stack Exchange
|
Bijections between binary sequences and primitive elements in a finite field
Let $n>1$ be a natural number. We call a binary sequence $(b_1,\ldots, b_n)\in \{0,1\}^n$ $rigid$ if it is not a proper power of a sequence of shorter length. So for example $(0,1,0,1) = (0,1)^2$ is not rigid while $(0,0,1,1)$ is rigid. Differently phrased, we have an action of the cyclic group of order $n$, $C_n$, on the set of all binary sequences of length $n$, and a sequence is rigid if and only if $C_n$ acts freely on it.
If $n=p_1^{a_1}p_2^{a_2}\cdots p_k^{a_k}$ is the prime factorisation of $n$ then using the inclusion-exclusion principle the number of rigid sequences can be shown to be
$$\sum_{d| p_1p_2\ldots p_k} (-1)^{r(d)}2^{\frac{n}{d}}$$ where $r(d)$ is the number of prime divisors of $d$. The number of $C_n$-orbits is then the above expression dividied by $n$.
On the other hand, consider the number of primitive elements in $\mathbb{F}_{2^n}$. These are elements $x\in \mathbb{F}_{2^n}$ which generate the field $\mathbb{F}_{2^n}$ over $\mathbb{F}_2$.
A similar calculation using the inclusion-exclusion principle then shows that the number of primitive elements is equal to the number of rigid sequences in $\{0,1\}^n$. The Galois-group of $\mathbb{F}_{2^n}/\mathbb{F}_2$, which is also $C_n$, acts on the primitive elements. The orbits are in one to one correspondence with irreducible polynomials of degree $n$ over $\mathbb{F}_2$.
Question: Is there a known bijection between the set of rigid sequences of length $n$ and the set of primitive elements in $\mathbb{F}_{2^n}$? Is there a bijection between the corresponding $C_n$-orbit sets?
You probably should specify how you’re realizing the elements of the finite field.
Good point. We can think of this as $\mathbb{F}_2[x]/(f)$ for some irreducible $f$, but I am mostly interested in bijection between the orbit set and the set of irreducible polynomials of degree $n$ over $\mathbb{F}_2$.
What you call rigid words are normally called primitive words
Isn't the number of primitive elements $\varphi(2^n-1)$?
@kodlu: Wrong kind of primitive. You're looking at generators of the multiplicative group; Ehud is looking at generators of the field extension. Each of the former is one of the latter, but not vice versa.
@darjigrinberg, thank you
|
2025-03-21T14:48:31.708740
| 2020-08-07T03:27:02 |
368531
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"JamalS",
"Narasimham",
"Piyush Grover",
"Yemon Choi",
"https://mathoverflow.net/users/30684",
"https://mathoverflow.net/users/47973",
"https://mathoverflow.net/users/50447",
"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:631845",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368531"
}
|
Stack Exchange
|
Tips to organize a successful math workshop
I am a PhD student from India working on representations of quantum groups. I want to organize a workshop on Hopf Algebra and Quantum groups but there are only 2 or 3 specialists in India currently working on it. So I googled some professors outside India who are currently working on this field. Now the problem is how to approach them to participate as a speaker. Since I have no experience organizing such a workshop I want some help. I donot know what offers I have to make to them. I mean giving air fares both ways and lodging and broading will be enough or do I need to arrange some honorarium. Also there is another problem. To propose a workshop, I have to give details of the speakers, the topics they are covering and outcome of the proposed workshop. After recieving all the data the working committee is either going to accept or reject the proposal. If they accept it then they are going to give the funding. Now it may happen that the funds given may not be enough or in worse case they can reject the proposal. Now if some professor gets interested at the prospect and later find that the proposal is rejected, what they will think about me.
So I will be very grateful if someone comes up some piece of advise how to proceed.
Is there anyone at your institute who has had some experience organizing similar workshops in different areas of pure maths?
I highly doubt anyone will accept unless the workshop is 2+ years away in future (due to COVID). You might have better luck organizing something online.
@PiyushGrover 2 years? Lots of restrictions are being dropped. If the whole second "wave" thing happens maybe they'll come back but it's a safe bet in a year it'll be more or less resolved in terms of restrictions interfering with those not at risk. I think unless a professor is part of the vulnerable group they have no reason not to go.
@ Snehashish mukhejiee: A Zoom type conference with big screen display is perhaps a quicker option during covid9 time. I suppose after the conference you are planning to bring out notes and proceedings in a book form.
The Lorentz Center has some advice that you might find useful, I have organized several workshops there and followed a route similar to the one you describe.
Tentative answers to your specific questions:
Since you can only invite speakers conditional on the acceptance of the workshop proposal, it is crucial that you invite them well in advance, so that they can block their calendar without interfering with other commitments. I would invite at least a year in advance --- of course now with COVID a two-year advance seems more reliable. Don't worry about "what they will think of me" if the proposal gets rejected: if you explain in advance that the invitation depends on the acceptance by some agency they will understand.
I would initially offer to reimburse "local expenses" (= hotel and meals) and explain that you have some funds for travel expenses which you can distribute according to need. Some speakers may be able and willing to use their own funds for travel. You could further motivate this by explaining that if speakers can help in this way with the expenses you will be able to support junior participants.
Honorarium? No, for academic lectures that is unusual and not needed --- at least not for lectures delivered "in real life". The idea being that the hospitality which the host will offer (e.g. taking you out to a nice restaurant) is sufficient compensation. Since COVID I have seen a modest honorarium (a few hundred $) being offered for online lectures, mainly to secure commitment from the speaker.
|
2025-03-21T14:48:31.709156
| 2020-08-07T04:35:01 |
368533
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Mats Granvik",
"Tom Copeland",
"https://mathoverflow.net/users/12178",
"https://mathoverflow.net/users/25104"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631846",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368533"
}
|
Stack Exchange
|
Show that the ratio of limits converges to the nearest Riemann zeta zero except when the ratio is a singularity
Let $h(s,n)$ be:
$$h(s,n)=\lim_{c\to 1} \, \frac{(-1)^{n-2}}{(n-2)!}\zeta (c)^{n-2} \sum _{k=1}^{n-1} \frac{(-1)^{k-1} \binom{n-2}{k-1}}{\zeta ((c-1) (k-1)+s)}$$
and let $g(s,n)$ be:
$$g(s,n)=\lim_{c\to 1} \, \frac{(-1)^{n-1}}{(n-1)!} \zeta (c)^{n-1} \sum _{k=1}^n \frac{(-1)^{k-1} \binom{n-1}{k-1}}{\zeta ((c-1) (k-1)+s)}$$
Question:
Does the ratio $$\rho(s) = i s+\lim\limits_{n \rightarrow \infty}\frac{h(i s,n)}{g(i s,n)}$$ converge to the nearest Riemann zeta zero?
For $s=15$ and $n=12$, we get: $\rho(15) = 0.5 +14.1347 i$
The first plot is the real part of $\rho(s)$, which begins at the trivial zero $-2$ and then tends to be close to $1/2$ except at singularities. The Gram points appear to be a subset of the singularities.
The second plot is the imaginary part of $\rho(s)$, which has heights close to imaginary parts of Riemann zeta zeros.
(*start*)
(*Mathematica program for the plots*)
Clear[n, k, s, c, z, f, g];
n = 11;
ss = 40;
h[s_] = Limit[((-1)^(n - 2) Zeta[
c]^(n - 2) Sum[(-1)^(k - 1)*
Binomial[n - 2, k - 1]/Zeta[s + (k - 1)*(c - 1)], {k, 1,
n - 1}]/(n - 2)!), c -> 1];
g[s_] = Limit[((-1)^(n - 1) Zeta[
c]^(n - 1) Sum[(-1)^(k - 1)*
Binomial[n - 1, k - 1]/Zeta[s + (k - 1)*(c - 1)], {k, 1,
n}]/(n - 1)!), c -> 1];
Monitor[b = Table[s*I + h[s*N[I]]/g[s*N[I]], {s, 0, ss, 1/10}];, s*10]
ListLinePlot[Re[b], DataRange -> {0, ss}]
ListLinePlot[Im[b], DataRange -> {0, ss}]
(*end*)
(*start*)
(*Mathematica program for the first non trivial zeta zero*)
Clear[n, k, s, c, z, f, g];
n = 12;
h[s_] = Limit[((-1)^(n - 2) Zeta[
c]^(n - 2) Sum[(-1)^(k - 1)*
Binomial[n - 2, k - 1]/Zeta[s + (k - 1)*(c - 1)], {k, 1,
n - 1}]/(n - 2)!), c -> 1];
g[s_] = Limit[((-1)^(n - 1) Zeta[
c]^(n - 1) Sum[(-1)^(k - 1)*
Binomial[n - 1, k - 1]/Zeta[s + (k - 1)*(c - 1)], {k, 1,
n}]/(n - 1)!), c -> 1];
s = 15;
s*I + h[s*N[I]]/g[s*N[I]]
(*end*)
Clear[n, k, s, c];
n = 7;
s = N[14*I];
s - n*Limit[
1/Zeta[c]*
Sum[(-1)^(k - 1)*
Binomial[n - 1, k - 1]/Zeta[s + (k - 1)*(c - 1)], {k, 1,
n}]/
Sum[(-1)^(k - 1)*
Binomial[n, k - 1]/Zeta[s + (k - 1)*(c - 1)], {k, 1, n + 1}],
c -> 1]
For $n=7$ and $s=14i$:
$$0.5 + 14.1347i = s-n \left(\lim_{c\to 1} \, \frac{\sum _{k=1}^{n} \frac{(-1)^{k-1} \binom{n-1}{k-1}}{\zeta ((c-1) (k-1)+s)}}{\zeta (c) \sum _{k=1}^{n+1} \frac{(-1)^{k-1} \binom{n}{k-1}}{\zeta ((c-1) (k-1)+s)}}\right)$$
The conjecture is that as $n \rightarrow \infty$ the limit above converges to the Riemann zeta zero nearest to $s$.
Related:
https://mathoverflow.net/a/368105/25104
https://math.stackexchange.com/a/3735702/8530
Set $s=14.000000000000000000000000000000...i$ with 1000 zeros after the decimal point.
Set $n=21$ and set $c = 1 + 1/10^{40}$;
With those parameters compute this formula:
$$s-\frac{n \sum _{k=1}^n \frac{(-1)^{k-1} \binom{n-1}{k-1}}{\zeta ((c-1) (k-1)+s)}}{\zeta (c) \sum _{k=1}^{n+1} \frac{(-1)^{k-1} \binom{n}{k-1}}{\zeta ((c-1) (k-1)+s)}}$$
What you will get is the 25 first decimal digits of the first Riemann zeta zero:
0.50000000000000000000000055508907479219367612957050478295942858083862
3727033228398609021142110650620136997773667771872221905026127340639625
41218507480832131294005829437
+
14.134725141734693790457251915896759601972505820234600660252328557362
5629956990194271674005286735176937891872097245657731536209606798029380
8035224527780328742481096881866 I
Of course ideally:
$n \rightarrow \infty$ and $c \rightarrow 1$
(*Mathematica*)
(*start*)
Clear[n, k, s, c];
n = 21;
s = N[14*I, 1000];
c = 1 + 1/10^40;
s - n*(1/Zeta[c]*
Sum[(-1)^(k - 1)*
Binomial[n - 1, k - 1]/Zeta[s + (k - 1)*(c - 1)], {k, 1, n}]/
Sum[(-1)^(k - 1)*
Binomial[n, k - 1]/Zeta[s + (k - 1)*(c - 1)], {k, 1, n + 1}])
(*end*)
https://math.stackexchange.com/q/3789849/8530
Here is my derivation.
A very short Mathematica program for computing the zeta zeros is:
Clear[x, t, nn];
nn = 12;
t = 15;
a = Series[1/Zeta[x + t*I], {x, 0, nn}];
t*I + N[Coefficient[a, x^(nn - 1)]/Coefficient[a, x^nn]]
which for $t=15$ gives $0.5 + 14.1347i$
Tom Copeland has recorded what he calls "Coefficients of list partition transform: reciprocal of an exponential generating function (e.g.f.)." in the OEIS here: https://oeis.org/A133314 together with several links to papers.
This is the table starting:
1
[-1]
[-1, 2]
[-1, 6, -6]
[-1, 8, 6, -36, 24]
[-1, 10, 20, -60, -90, 240, -120]
[-1, 12, 30, -90, 20, -360, 480, -90, 1080, -1800, 720]
These numbers above appear to be the same as the coefficients in the power series expansion of $$\frac{1}{f(x)} \tag{1}$$:
Which is given by the Mathematica command:
Series[1/f[x], {x, 0, 6}]
or as a table:
TableForm[CoefficientList[Series[1/f[x], {x, 0, 4}], x]]
$$\begin{array}{l}
\frac{1}{f[0]} \\
-\frac{f'[0]}{f[0]^2} \\
\frac{f'[0]^2}{f[0]^3}-\frac{f''[0]}{2 f[0]^2} \\
-\frac{f'[0]^3}{f[0]^4}+\frac{f'[0] f''[0]}{f[0]^3}-\frac{f^{(3)}[0]}{6 f[0]^2} \\
\frac{24 f'[0]^4-36 f[0] f'[0]^2 f''[0]+6 f[0]^2 f''[0]^2+8 f[0]^2 f'[0] f^{(3)}[0]-f[0]^3 f^{(4)}[0]}{24 f[0]^5}
\end{array}$$
This is of course is essentially the same as repeated derivatives of $(1)$ if one discards signs and multiply with factorials.
In Mathematica for the Riemann zeta function this would be:
Clear[s];
D[1/Zeta[s], {s, 0}]
D[1/Zeta[s], {s, 1}]
D[1/Zeta[s], {s, 2}]
D[1/Zeta[s], {s, 3}]
D[1/Zeta[s], {s, 4}]
D[1/Zeta[s], {s, 5}]
Mathematica knows that the first derivative of $(1)$ is:
$$\frac{\partial \frac{1}{\zeta (s)}}{\partial s^1}=\frac{\zeta '(s)}{\zeta (s)^2}=\lim_{c\to 1} \, \left(\frac{\zeta (c)}{\zeta (s)}-\frac{\zeta (c)}{\zeta (c+s-1)}\right) \tag{2}$$
To get the second derivative we then recursively (as in repeated derivatives) nest the right hand side of $(2)$ into right hand side of $(2)$ to get:
$$\frac{\partial ^2\frac{1}{\zeta (s)}}{\partial s^2} = \frac{2 \zeta '(s)^2}{\zeta (s)^3}-\frac{\zeta ''(s)}{\zeta (s)^2}= \lim_{c\to 1} \, \left(\frac{\zeta (c)}{\frac{1}{\frac{\zeta (c)}{\zeta (s)}-\frac{\zeta (c)}{\zeta (c+s-1)}}}-\frac{\zeta (c)}{\frac{1}{\frac{\zeta (c)}{\zeta (c+s-1)}-\frac{\zeta (c)}{\zeta (c+c+s-1-1)}}}\right) \tag{3}$$
To get the third derivative we insert the right hand side of $(3)$ into the right hand side of $(2)$ to get:
$$\frac{\partial ^3\frac{1}{\zeta (s)}}{\partial s^3} = \frac{6 \zeta '(s)^3+\zeta ^{(3)}(s) \zeta (s)^2-6 \zeta (s) \zeta '(s) \zeta ''(s)}{\zeta (s)^4} = \lim_{c\to 1} \, \left(\frac{\zeta (c)}{\frac{1}{\frac{\zeta (c)}{\frac{1}{\frac{\zeta (c)}{\zeta (s)}-\frac{\zeta (c)}{\zeta (c+s-1)}}}-\frac{\zeta (c)}{\frac{1}{\frac{\zeta (c)}{\zeta (c+s-1)}-\frac{\zeta (c)}{\zeta (c+c+s-1-1)}}}}}-\frac{\zeta (c)}{\frac{1}{\frac{\zeta (c)}{\frac{1}{\frac{\zeta (c)}{\zeta (c+s-1)}-\frac{\zeta (c)}{\zeta (c+c+s-1-1)}}}-\frac{\zeta (c)}{\frac{1}{\frac{\zeta (c)}{\zeta (c+c+s-1-1)}-\frac{\zeta (c)}{\zeta (c+c+c+s-1-1-1)}}}}}\right) \tag{4}$$
and so on...
This should be possible to show with some insertion of variables into the nested derivative limits. But I don't know how to do induction to prove it. And I have not yet inserted the variables, which probably should be inserted where there are free standing integers (in this case maybe the ones in the numerators).
In Mathematica this would be:
Expand[Limit[(Zeta[c]/Zeta[s] - Zeta[c]/Zeta[s + c - 1]), c -> 1]]
Expand[Limit[(Zeta[
c]/((Zeta[c]/Zeta[s] - Zeta[c]/Zeta[s + c - 1]))^-1 -
Zeta[c]/((Zeta[c]/Zeta[s + c - 1] -
Zeta[c]/Zeta[s + c - 1 + c - 1]))^-1), c -> 1]]
Expand[Limit[(Zeta[
c]/((Zeta[c]/((Zeta[c]/Zeta[s] - Zeta[c]/Zeta[s + c - 1]))^-1 -
Zeta[c]/((Zeta[c]/Zeta[s + c - 1] -
Zeta[c]/Zeta[s + c - 1 + c - 1]))^-1))^-1 -
Zeta[c]/((Zeta[
c]/((Zeta[c]/Zeta[s + c - 1] -
Zeta[c]/Zeta[s + c - 1 + c - 1]))^-1 -
Zeta[c]/((Zeta[c]/Zeta[s + c - 1 + c - 1] -
Zeta[c]/Zeta[s + c - 1 + c - 1 + c - 1]))^-1))^-1),
c -> 1]]
Expand[Limit[(Zeta[
c]/((Zeta[
c]/((Zeta[
c]/((Zeta[c]/Zeta[s] - Zeta[c]/Zeta[s + c - 1]))^-1 -
Zeta[c]/((Zeta[c]/Zeta[s + c - 1] -
Zeta[c]/Zeta[s + c - 1 + c - 1]))^-1))^-1 -
Zeta[c]/((Zeta[
c]/((Zeta[c]/Zeta[s + c - 1] -
Zeta[c]/Zeta[s + c - 1 + c - 1]))^-1 -
Zeta[c]/((Zeta[c]/Zeta[s + c - 1 + c - 1] -
Zeta[c]/
Zeta[s + c - 1 + c - 1 + c - 1]))^-1))^-1))^-1 -
Zeta[c]/((Zeta[
c]/((Zeta[
c]/((Zeta[c]/Zeta[s + c - 1] -
Zeta[c]/Zeta[s + c - 1 + c - 1]))^-1 -
Zeta[c]/((Zeta[c]/Zeta[s + c - 1 + c - 1] -
Zeta[c]/Zeta[s + c - 1 + c - 1 + c - 1]))^-1))^-1 -
Zeta[c]/((Zeta[
c]/((Zeta[c]/Zeta[s + c - 1 + c - 1] -
Zeta[c]/Zeta[s + c - 1 + c - 1 + c - 1]))^-1 -
Zeta[c]/((Zeta[c]/Zeta[s + c - 1 + c - 1 + c - 1] -
Zeta[c]/
Zeta[s + c - 1 + c - 1 + c - 1 + c -
1]))^-1))^-1))^-1), c -> 1]]
Now we apply the Mathematica FullSimplify command to the expressions inside the limits:
FullSimplify[(Zeta[c]/Zeta[s] - Zeta[c]/Zeta[s + c - 1])]
FullSimplify[(Zeta[
c]/((Zeta[c]/Zeta[s] - Zeta[c]/Zeta[s + c - 1]))^-1 -
Zeta[c]/((Zeta[c]/Zeta[s + c - 1] -
Zeta[c]/Zeta[s + c - 1 + c - 1]))^-1)]
FullSimplify[(Zeta[
c]/((Zeta[c]/((Zeta[c]/Zeta[s] - Zeta[c]/Zeta[s + c - 1]))^-1 -
Zeta[c]/((Zeta[c]/Zeta[s + c - 1] -
Zeta[c]/Zeta[s + c - 1 + c - 1]))^-1))^-1 -
Zeta[c]/((Zeta[
c]/((Zeta[c]/Zeta[s + c - 1] -
Zeta[c]/Zeta[s + c - 1 + c - 1]))^-1 -
Zeta[c]/((Zeta[c]/Zeta[s + c - 1 + c - 1] -
Zeta[c]/Zeta[s + c - 1 + c - 1 + c - 1]))^-1))^-1)]
FullSimplify[(Zeta[
c]/((Zeta[
c]/((Zeta[
c]/((Zeta[c]/Zeta[s] - Zeta[c]/Zeta[s + c - 1]))^-1 -
Zeta[c]/((Zeta[c]/Zeta[s + c - 1] -
Zeta[c]/Zeta[s + c - 1 + c - 1]))^-1))^-1 -
Zeta[c]/((Zeta[
c]/((Zeta[c]/Zeta[s + c - 1] -
Zeta[c]/Zeta[s + c - 1 + c - 1]))^-1 -
Zeta[c]/((Zeta[c]/Zeta[s + c - 1 + c - 1] -
Zeta[c]/
Zeta[s + c - 1 + c - 1 + c - 1]))^-1))^-1))^-1 -
Zeta[c]/((Zeta[
c]/((Zeta[
c]/((Zeta[c]/Zeta[s + c - 1] -
Zeta[c]/Zeta[s + c - 1 + c - 1]))^-1 -
Zeta[c]/((Zeta[c]/Zeta[s + c - 1 + c - 1] -
Zeta[c]/Zeta[s + c - 1 + c - 1 + c - 1]))^-1))^-1 -
Zeta[c]/((Zeta[
c]/((Zeta[c]/Zeta[s + c - 1 + c - 1] -
Zeta[c]/Zeta[s + c - 1 + c - 1 + c - 1]))^-1 -
Zeta[c]/((Zeta[c]/Zeta[s + c - 1 + c - 1 + c - 1] -
Zeta[c]/
Zeta[s + c - 1 + c - 1 + c - 1 + c -
1]))^-1))^-1))^-1)]
This FullSimplify then gives us (to my surprise) for the right hand side of $(2),(3)$ and $(4)$:
$$\zeta (c) \left(\frac{1}{\zeta (s)}-\frac{1}{\zeta (c+s-1)}\right) \tag{from RHS of 2}$$
$$\zeta (c)^2 \left(\frac{1}{\zeta (s)}-\frac{2}{\zeta (c+s-1)}+\frac{1}{\zeta (2 c+s-2)}\right) \tag{from RHS of 3}$$
$$\zeta (c)^3 \left(\frac{1}{\zeta (s)}-\frac{3}{\zeta (c+s-1)}+\frac{3}{\zeta (2 c+s-2)}-\frac{1}{\zeta (3 c+s-3)}\right) \tag{from RHS of 4}$$
$$\zeta (c)^4 \left(\frac{1}{\zeta (s)}-\frac{4}{\zeta (c+s-1)}+\frac{6}{\zeta (2 c+s-2)}-\frac{4}{\zeta (3 c+s-3)}+\frac{1}{\zeta (4 c+s-4)}\right)$$
Apparently the numerators inside the parentheses are binomial coefficients with alternating signs and the denominators with the Riemann zeta function look like multiples of natural numbers. This leads us to the conjectured form:
$$g(s,n)=\lim_{c\to 1} \, \frac{(-1)^{n-1}}{(n-1)!} \zeta (c)^{n-1} \sum _{k=1}^n \frac{(-1)^{k-1} \binom{n-1}{k-1}}{\zeta ((c-1) (k-1)+s)}$$
when including signs and factorials. Because of the special limit for derivatives this formula works only for the Riemann zeta function. The Gamma function should give something similar.
n = 1;
Limit[((-1)^(n - 1) Zeta[
c]^(n - 1) Sum[(-1)^(k - 1)*
Binomial[n - 1, k - 1]/Zeta[s + (k - 1)*(c - 1)], {k, 1,
n}]/(n - 1)!), c -> 1]
1/Zeta[s]
n = 2;
Limit[((-1)^(n - 1) Zeta[
c]^(n - 1) Sum[(-1)^(k - 1)*
Binomial[n - 1, k - 1]/Zeta[s + (k - 1)*(c - 1)], {k, 1,
n}]/(n - 1)!), c -> 1]
-(Derivative[1][Zeta][s]/Zeta[s]^2)
n = 3;
Limit[((-1)^(n - 1) Zeta[
c]^(n - 1) Sum[(-1)^(k - 1)*
Binomial[n - 1, k - 1]/Zeta[s + (k - 1)*(c - 1)], {k, 1,
n}]/(n - 1)!), c -> 1]
(2 Derivative[1][Zeta][s]^2 -
Zeta[s] (Zeta^[Prime][Prime])[s])/(2 Zeta[s]^3)
n = 4;
Limit[((-1)^(n - 1) Zeta[
c]^(n - 1) Sum[(-1)^(k - 1)*
Binomial[n - 1, k - 1]/Zeta[s + (k - 1)*(c - 1)], {k, 1,
n}]/(n - 1)!), c -> 1]
OEIS A133314 is precisely the entry for the reciprocal for an e.g.f. or Taylor series for an function f(x) analytic at the origin for which f(0)=1, as the title says. I retained the phrase "list partition transform" only for historical reasons. It's how I discovered it.
Mathematica better give the same answer.
I added a picture and some Mathematica code for the limits, for comparison. Seems to agree with the series expansion for 1/f(x), sofar. I plan to add this as either a comment or formula to A133314.
May I suggest that you replace the codes with the corresponding mathematical statements in standard form and move the codes to github or other site with a link. (I used MathCad efficiently and effectively for years and don't relish learning yet another programming syntax/language/script).
Here is a programmatically exact explanation of the derivation:
First observe that the first derivative of: $$\frac{1}{\zeta(s)} \tag{1}$$
is:
$$\frac{\partial \frac{1}{\zeta (s)}}{\partial s^1}=-\frac{\zeta '(s)}{\zeta (s)^2}$$
Mathematica knows that the first derivative can be computed through the formula:
$$-\frac{\zeta '(s)}{\zeta (s)^2}=\lim_{c\to 1} \, \left(\frac{\zeta (c)}{\zeta (-(n-1)+(n-1)c+s)}-\frac{\zeta (c)}{\zeta (-n+nc+s)}\right)$$
for $n=1,2,3,4,5,...$
For $n=1$ the expression inside the limit is:
$$A1=\left(\frac{\zeta (c)}{\zeta (-0+0c+s)}-\frac{\zeta (c)}{\zeta (-1+1c+s)}\right)$$
For $n=2$:
$$A2=\left(\frac{\zeta (c)}{\zeta (-1+1c+s)}-\frac{\zeta (c)}{\zeta (-2+2c+s)}\right)$$
For $n=3$:
$$A3=\left(\frac{\zeta (c)}{\zeta (-2+2c+s)}-\frac{\zeta (c)}{\zeta (-3+3c+s)}\right)$$
For $n=4$:
$$A4=\left(\frac{\zeta (c)}{\zeta (-3+3c+s)}-\frac{\zeta (c)}{\zeta (-4+4c+s)}\right)$$
Then substitute to form the second derivative of $(1)$:
In $A1$ replace all $\zeta(-1+c+s)$ with $\frac{1}{A2}$ which results in:
$B1=\frac{\zeta (c)}{\zeta (s)}-\zeta (c) \left(\frac{\zeta (c)}{\zeta (c+s-1)}-\frac{\zeta (c)}{\zeta (2 c+s-2)}\right)$
In $B1$ replace all $\zeta(s)$ with $\frac{1}{A1}$:
$$B2=\zeta (c) \left(\frac{\zeta (c)}{\zeta (s)}-\frac{\zeta (c)}{\zeta (c+s-1)}\right)-\zeta (c) \left(\frac{\zeta (c)}{\zeta (c+s-1)}-\frac{\zeta (c)}{\zeta (2 c+s-2)}\right)$$
Here we FullSimplify $B2$ and get:
$$B3=\zeta (c)^2 \left(\frac{1}{\zeta (s)}-\frac{2}{\zeta (c+s-1)}+\frac{1}{\zeta (2 c+s-2)}\right)$$
which has binomial coefficients in the numerator. The limit:
$$\lim\limits_{c \rightarrow 1} B3 = \lim\limits_{c \rightarrow 1} \zeta (c)^2 \left(\frac{1}{\zeta (s)}-\frac{2}{\zeta (c+s-1)}+\frac{1}{\zeta (2 c+s-2)}\right)=\frac{2 \zeta '(s)^2-\zeta (s) \zeta ''(s)}{\zeta (s)^3} = \frac{\partial ^2\frac{1}{\zeta (s)}}{\partial s^2}$$
Mathematica puts it in a more readable form:
Clear[s, c];
A0 = 1/Zeta[s];
Limit[Zeta[c] A0 - Zeta[c]/Zeta[-1 + c + s], c -> 1];
A1 = Zeta[c]/Zeta[-0 + 0 c + s] - Zeta[c]/Zeta[-1 + 1 c + s];
A2 = Zeta[c]/Zeta[-1 + 1 c + s] - Zeta[c]/Zeta[-2 + 2 c + s];
A3 = Zeta[c]/Zeta[-2 + 2 c + s] - Zeta[c]/Zeta[-3 + 3 c + s];
A4 = Zeta[c]/Zeta[-3 + 3 c + s] - Zeta[c]/Zeta[-4 + 4 c + s];
A5 = Zeta[c]/Zeta[-4 + 4 c + s] - Zeta[c]/Zeta[-5 + 5 c + s];
B1 = ReplaceAll[A1, Zeta[-1 + 1 c + s] -> 1/A2];
B2 = ReplaceAll[B1, Zeta[-0 + 0 c + s] -> 1/A1];
C1 = ReplaceAll[B2, Zeta[-2 + 2 c + s] -> 1/A3];
C2 = ReplaceAll[C1, Zeta[-1 + 1 c + s] -> 1/A2];
C3 = ReplaceAll[C2, Zeta[-0 + 0 c + s] -> 1/A1];
D1 = ReplaceAll[C3, Zeta[-3 + 3 c + s] -> 1/A4];
D2 = ReplaceAll[D1, Zeta[-2 + 2 c + s] -> 1/A3];
D3 = ReplaceAll[D2, Zeta[-1 + 1 c + s] -> 1/A2];
D4 = ReplaceAll[D3, Zeta[-0 + 0 c + s] -> 1/A1];
E1 = ReplaceAll[D4, Zeta[-4 + 4 c + s] -> 1/A5];
E2 = ReplaceAll[E1, Zeta[-3 + 3 c + s] -> 1/A4];
E3 = ReplaceAll[E2, Zeta[-2 + 2 c + s] -> 1/A3];
E4 = ReplaceAll[E3, Zeta[-1 + 1 c + s] -> 1/A2];
E5 = ReplaceAll[E4, Zeta[-0 + 0 c + s] -> 1/A1];
FullSimplify[A0]
FullSimplify[A1]
FullSimplify[B2]
FullSimplify[C3]
FullSimplify[D4]
FullSimplify[E5]
B1 = ReplaceAll[A1, Zeta[-1 + 1 c + s] -> 1/A2] means:
B1 equals the result of: "In A1 replace all Zeta[-1 + 1 c + s] with 1/A2"
FullSimplify[A0]
$$\frac{1}{\zeta (s)}$$
FullSimplify[A1]
$$\zeta (c) \left(\frac{1}{\zeta (s)}-\frac{1}{\zeta (c+s-1)}\right)$$
FullSimplify[A2]
$$\zeta (c)^2 \left(\frac{1}{\zeta (s)}-\frac{2}{\zeta (c+s-1)}+\frac{1}{\zeta (2 c+s-2)}\right)$$
FullSimplify[A3]
$$\zeta (c)^3 \left(\frac{1}{\zeta (s)}-\frac{3}{\zeta (c+s-1)}+\frac{3}{\zeta (2 c+s-2)}-\frac{1}{\zeta (3 c+s-3)}\right)$$
FullSimplify[A4]
$$\zeta (c)^4 \left(\frac{1}{\zeta (s)}-\frac{4}{\zeta (c+s-1)}+\frac{6}{\zeta (2 c+s-2)}-\frac{4}{\zeta (3 c+s-3)}+\frac{1}{\zeta (4 c+s-4)}\right)$$
FullSimplify[A5]
$$\zeta (c)^5 \left(\frac{1}{\zeta (s)}-\frac{5}{\zeta (c+s-1)}+\frac{10}{\zeta (2 c+s-2)}-\frac{10}{\zeta (3 c+s-3)}+\frac{5}{\zeta (4 c+s-4)}-\frac{1}{\zeta (5 c+s-5)}\right)$$
I will continue editing later.
https://math.stackexchange.com/q/3789849/8530
This may be related
Hypergeometric-like Representation of the Zeta-Function of Riemann
where the Binomial expansion is used
|
2025-03-21T14:48:31.709790
| 2020-08-07T04:40:27 |
368536
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Carlo Beenakker",
"hasan ",
"https://mathoverflow.net/users/11260",
"https://mathoverflow.net/users/148500"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631847",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368536"
}
|
Stack Exchange
|
Verification of an Cauchy's contour Integral of Complementary Error function?
I tried to find an integral of the following,$\DeclareMathOperator{\erfc}{erfc}$
$\int\limits_0^{2\pi} \erfc(a + b\cos(\theta))\erfc(c + d\sin(\theta))\,d\theta $
Where, $a,b,c,d \in \Bbb R$
Now, $\cos(\theta) = \frac{e^{j\theta} + e^{-j\theta}}{2};\quad\sin(\theta) = \frac{e^{j\theta } - e^{-j\theta }}{2j}$
Let $z = {e^{j\theta }}$
Therefore, I can rewrite the integral as,
$\oint\limits_{|z| = 1} {\erfc\left( {a + b\left( {\frac{{z + {z^*}}}{2}} \right)} \right)erfc\left( {c + d\left( {\frac{{z - {z^*}}}{{2j}}} \right)} \right)\frac{{dz}}{{iz}}} $
Again, let $f(z) = \erfc\left(a + b\left(\frac{z + z^*}{2}\right)\right)\erfc\left(c + d\left(\frac{z - z^*}{2j}\right) \right)$
Hence , the final integral can be written as,
$\oint\limits_{|z| = 1}f(z)\frac{dz}{iz} = 2\pi f(0) = 2\pi\erfc(a)\erfc(c)$
As $0$ falls inside $|z| = 1$.
Can you tell me if I am correct or wrong or it needs more reasoning? Thank you.
the $z^\ast$ in your integrand is $1/z$ on the unit circle, so you have poles at the origin, the integrand is not analytic inside the unit circle; and indeed, the answer is incorrect, for example, for $a=b=c=d=1$ the integral equals $0.6268$ while $2\pi ,\text{erfc}(1)\text{erfc}(1)=0.1555.$ --- also crossposted at https://math.stackexchange.com/q/3781857/87355
@CarloBeenakker does that mean the integration is not possible in close from? Can you give me a hint to do that. Yes I posted this in overflow as no one answered in MSE Thank you.
indeed, no closed form expression as far as I can see.
@CarloBeenakker I think I am wrong. It depends on $a,b,c,d$
|
2025-03-21T14:48:31.709937
| 2020-08-07T06:01:01 |
368537
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Alapan Das",
"Gerry Myerson",
"Giorgio Metafune",
"Gjergji Zaimi",
"LSpice",
"YanChen",
"https://mathoverflow.net/users/150653",
"https://mathoverflow.net/users/156029",
"https://mathoverflow.net/users/158000",
"https://mathoverflow.net/users/163164",
"https://mathoverflow.net/users/2383",
"https://mathoverflow.net/users/2384",
"https://mathoverflow.net/users/64302",
"user2520938"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631848",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368537"
}
|
Stack Exchange
|
The complex trigonometric function degenerates to the positive integer
For any integer $N \geq 2$, we have the identity:
$$\frac{\ \prod _{n=1}^{N-1}\ \left(2+2\sum _{m=1}^{n\ }\cos \frac{\ m\pi \ }{N}\ \right)\ }{\prod _{n=1}^{N-1}\ \left(1+2\sum _{m=1}^{n\ }\cos \frac{\ m\pi \ }{N}\ \right)}=N$$
So how to prove it? Any help and suggestion will be appreciated, thank you!
If you don't know how to prove it, how do you know it's true?
I believe the identity holds because it comes from a real physical problem about fracture mechanics. And I also verified the identity for N<20 via Mathematica. The left side expression of the equation is neat and orderly, so I wonder if there's some kind of rigorous proof in mathematics.
For large, $n$, we can approximate the expression as $$A=\prod_{k=1}^{N-1} \left(1+\frac{1}{(1+\frac{2}{\pi}N\sin(\frac{\pi k}{N}))}\right)$$. And this $A$ can be further simplified as $$\ln(A)=\frac{\pi}{2N}\sum_{k=1}^{N-1} \frac{1}{\sin(\frac{\pi k}{N})}$$. Which is very very close to $\ln(N)$. Hence, $A≈N$, at least for $N$ sufficiently larger than zero.
I have verified this identity up to $N=10000$.
The sum $$\frac{\pi}{N}\sum_{k=1}^{N-1} \frac{1}{\sin(\frac{\pi k}{N})} ≈\ln(N) +\gamma +\ln\left(\frac{\alpha}{\sin(\alpha)}\right)+\ln(1+\cos(\alpha)) =\ln(N)+B , \gamma \leq B \leq \gamma+ \ln(\frac{\pi}{2})+\ln(2) $$, as $0<\alpha<\frac{\pi}{2}$ and $\gamma$ is the Euler-Mescheroni constant.
Following Johann Cigler's suggestion, set $q=e^{\frac{i\pi}{N}}$. We will need the two evaluations
$$\prod_{n=1}^{2N-1}(1-q^n)=\left.\frac{x^{2N}-1}{x-1}\right|_{x=1}=2N \tag{1}$$
$$\prod_{n=1}^{N-1}(1-q^{2n})=\left.\frac{x^{N}-1}{x-1}\right|_{x=1}=N \tag{2}$$
In your expression we can write things in the form
$$2+2\sum _{m=1}^{n\ }\cos \frac{\ m\pi \ }{N}=2+\sum_{m=1}^n(q^m+q^{-m})=\frac{(1-q^{n+1})(1+q^n)}{q^n(1-q)}$$
and
$$1+2\sum _{m=1}^{n\ }\cos \frac{\ m\pi \ }{N}=1+\sum_{m=1}^n(q^m+q^{-m})=\frac{(1-q^{2n+1})}{q^n(1-q)}.$$
Putting everything together, we see that your expression evaluates as
\begin{gather*}
\prod_{n=1}^{N-1}\frac{(1-q^{n+1})(1+q^n)}{q^n(1-q)}\cdot \frac{q^n(1-q)}{(1-q^{2n+1})}=\prod_{n=1}^{N-1}\frac{(1-q^{n+1})(1+q^n)}{1-q^{2n+1}} \\
=\frac{(1-q^N)}{1-q}\prod_{n=1}^{N-1}\frac{(1-q^n)(1+q^n)}{1-q^{2n+1}}=(1-q^N)\frac{\prod_{n=1}^{N-1}(1-q^{2n})^2}{\prod_{n=1}^{2N-1}(1-q^n)}=2\cdot\frac{N^2}{2N}=N
\end{gather*}
as desired.
@T.Amdeberhan I rolled back your edit, because the variable in equations (1) and (2) plays a different role from $q$ which is a fixed root of unity. For example, you would say $1,q,q^2$ are the roots of $x^3-1$ rather than those of of $q^3-1$.
What is $x$ in the first two identities, and what does the evaluation mean? It seems undefined.
@user2520938: $x$ is a variable, and the notation $P(x)|_{x=a}$ is the evaluation of the polynomial $P$ at $a$.
The LHS of your identities don't involve an $x$ though, and the RHS is a rational function, not a polynomial, and in fact is a rational function the denominator of which is zero in the point your are evaluating in.
@user2520938: I should have said that more generally it denotes the limit of the expression, rather than evaluation. In this case, if you cancel the $(x-1)$ from numerator and denominator it coincides with the evaluation of the polynomial that's left at $x=1$.
Thanks for your answer. It's better to present more details about identities 1 and 2 although the conclusions hold.
@YanChen I suggest working them out as simple exercises, but if anything is unclear, feel free to ask.
If $t=q^2$ in equation (2), the monic polynomial $(x^N-1)/(x-1)$ has roots $t, t^2, \dots, t^{N-1}$ hence factors as $(x-t)(x-t^2) \dots...(x-t^{N-1})$. Setting $x=1$, (2) follows.
I have no answer, but it seems that the problem is equivalent with
$\frac{(-1;q)_{N-1}(-q;q)_{N-1}}{\binom{2N-2}{N-1}_q}=N$ for $q=e^{i \pi/N}.$
Shouldn't this be a comment, since it's not an answer?
|
2025-03-21T14:48:31.710181
| 2020-08-07T06:44:32 |
368540
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Carlo Beenakker",
"Rajesh D",
"https://mathoverflow.net/users/11260",
"https://mathoverflow.net/users/14414"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631849",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368540"
}
|
Stack Exchange
|
Asymptotic expansion of the inverse of a matrix valued function
A matrix valued function is of the form $\psi:\mathbb{R}_+\to\mathbb{R}^{n\times n}$ and it is known that $\psi(\lambda)$ is always a positive definite matrix. The asymptotic exapnsion of $\psi(\lambda)$ is given as $$\psi(\lambda) = A + \frac{B_1}{\lambda}+ \frac{B_2}{\lambda^2}+ \frac{B_3}{\lambda^3}+...$$ where $A$ is an $n\times n$ all ones matrix and $B_1,B_2,B_3...$ are nonsingular $n\times n$ matrices.
I want to get an asymptotic expansion for $\psi(\lambda)^{-1}$. I don't want to compute the exact coefficient matrices but want to show that it is of the form $$\psi(\lambda)^{-1} = \lambda D_{-1} + D_0 + \frac{D_1}{\lambda} + \frac{D_2}{\lambda^2} + \frac{D_3}{\lambda^3} +...$$ where $D_{-1},D_0,D_1,D_2,..$ are $n\times n$ matrices and $D_{-1}$ is a non zero matrix.
What I know
Neumann's series and this post.
PS : if this isn't a research level problem, please let me know.
If we disregard the positivity constraint, this is not true in general, the leading order term can be of order $n-1$ rather than of order 1.
The problem is treated in Laurent expansion of the inverse of perturbed, singular matrices. The leading order term is of the order of the singularity in $A$, which for $A$ an $n\times n$ all-1 matrix is of order $n-1$.
Here is an example for $n=3$:
$$\psi=\left(
\begin{array}{ccc}
1 & 1+\frac{1}{\lambda} & 1+\frac{4}{\lambda} \\
1+\frac{1}{\lambda} & 1 & 1+\frac{1}{\lambda} \\
1+\frac{4}{\lambda} & 1+\frac{1}{\lambda}& 1 \\
\end{array}
\right)$$
has inverse of order $\lambda^2$:
$$\psi^{-1}=\left(
\begin{array}{ccc}
-\frac{1}{8} {\lambda} (2 {\lambda}+1) & \frac{1}{2} {\lambda} ({\lambda}+1) & -\frac{1}{8} {\lambda} (2 {\lambda}-1) \\
\frac{1}{2} {\lambda} ({\lambda}+1) & -{\lambda} ({\lambda}+2) & \frac{1}{2} {\lambda} ({\lambda}+1) \\
-\frac{1}{8} {\lambda} (2 {\lambda}-1) & \frac{1}{2} {\lambda} ({\lambda}+1) & -\frac{1}{8} {\lambda} (2 {\lambda}+1) \\
\end{array}
\right).$$
I do not know how/if the positivity constraint modifies the order.
I notice I did not use your additional condition that $\psi$ is positive definite for all $\lambda$.
Yes, that's something need to be resolved. To see if it can make the leading $n-2$ matrices zero.
Oh the paper clearly says $H_{-s}$ is not zero. So thats about it?
I am confused by that statement; it does not hold, for example, for the matrix $A+I/\lambda$ for $n=3$, then the leading order term is of order $\lambda$, not $\lambda^2$.
In some of my computations too I observed the same. I get $\lambda$ increase and not any higher powers.
Also clearly the statement in that paper that $H_{-s} \ne 0$ is wrong as per your example $A+I/{\lambda}$ when $n = 3$.
The crucial for the answer is the value $s$ and it is not the order of singularity of $A$. It depends on first $p$ matrices in the series and there is a procedure to determine $s$. https://pdfs.semanticscholar.org/106b/31e8a3020c9c9820c3fb8da0e3269b88fc70.pdf?_ga=2.206622291.1745264539.1596903256-1361304415.1586702362
|
2025-03-21T14:48:31.710475
| 2020-08-07T09:27:08 |
368552
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Mikhail Borovoi",
"https://mathoverflow.net/users/4149"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631850",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368552"
}
|
Stack Exchange
|
Fixed space of maximal torus and Weyl group
Let $G$ be a compact connected Lie group and $T\subset G$ a maximal torus. Let $V$ be a representation of $G$ and $U=\{v\in V: tv=v\textrm{ for all }t\in T\}$. For any $g\in N(T)$ we have for all $t\in T$ and $v\in U$ that $g^{-1}tgv=v \Rightarrow t(gv)=gv$. This shows that for all $v\in U$ we have $gv\in U$ as well. From this we can define a representation of the Weyl group $W$ on $U$. I have the following two questions:
Does this $W$-module structure on $U$ depend on the choice of the maximal torus $T$?
Assume that $V$ is irreducible and $U$ nontrivial. Is there a way to understand when $U$ is irreducible as $W$-module? Is it always the case?
For (1). We consider $U=U(G,T)$ and $W=W(G,T)$. If $T'\subset G$ is another maximal torus, then there exists $g\in G$ such that $T'=gTg^{-1}$. The isomorphism $g\colon (G,T)\to (G,T')$ induces compatible isomorphisms $U(G,T)\overset{\sim}{\to} U(G,T')$ and $W(G,T)\overset{\sim}{\to} W(G,T')$. Therefore, the answer to Question (1) is No.
This paper of Humphreys addresses your second question (the first is answered in the comments - the $W$-module structure is independent of the choice of torus): https://people.math.umass.edu/~jeh/pub/zero.pdf
Here is a quote from the paper (section 1.4):
Indeed, it is usually unclear how to determine directly whether or not
the W-module $L_\lambda(0)$ [i.e. $U$] is simple, even if its dimension is compatible with
simplicity.
The $W$-module $U$ is not always irreducible. Indeed in type $A_2$ (i.e. $G=SU(3)$), there is a formula for the dimension of $U$ in Section 2.2, which shows that the dimension may be unbounded.
|
2025-03-21T14:48:31.710594
| 2020-08-07T10:22:21 |
368555
|
{
"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:631851",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368555"
}
|
Stack Exchange
|
Primitive collections as lattice generators for the Mori cone
I am looking at the following theorem from "Toric Varieties" by Cox, Little and Schenk:
Theorem 6.4.11: If $X_{\Sigma}$ is a simplicial toric variety, then
$\overline{NE}(X_{\Sigma}) = \large{\Sigma}_P \mathbf{R}_{\geq
0}r(P)$, where the sum is over all primitive collections $P$.
Such primitive collections $P$ gives rise to a 1-cycle in $N_1(X_{\Sigma})$ represented by an element $r(P) = (b_{\rho})_{\rho \in \Sigma(1)} \in \mathbf{R}^{\Sigma(1)}$, which can be shown to represent an effective class. Assuming that $X_{\Sigma}$ is a smooth projective variety, it is easy to show that the numbers $b_{\rho}$ are integral, in which case $r(P)$ sits inside $\overline{NE}(X_{\Sigma}) \cap \mathbf{Z}^n$.
Apparently, it is not known whether the $r(P)$'s always generate $\overline{NE}(X_{\Sigma}) \cap \mathbf{Z}^n$ as a semigroup.
However, I'm interested in the low-dimensional cases, specifically dimension 3.
My question is: Is it known, or can it be shown that the $r(P)$'s generate $\overline{NE}(X_{\Sigma}) \cap \mathbf{Z}^n$ for $X_{\Sigma}$ smooth, projective and of dimension $3$?
|
2025-03-21T14:48:31.710681
| 2020-08-07T10:26:15 |
368556
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Chris Wuthrich",
"Gerry Myerson",
"https://mathoverflow.net/users/158000",
"https://mathoverflow.net/users/41291",
"https://mathoverflow.net/users/5015",
"მამუკა ჯიბლაძე"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631852",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368556"
}
|
Stack Exchange
|
The smallest primitive root modulo powers of prime
I wrote a program to calculate the minimal primitive root modulo $p^a$ where $p > 2$ is a prime, by enumerating $g$ from $2$ and checking whether it's a primitive root, but I forgot to check $\gcd(g, p) = 1$. However, it still worked in all the test cases.
So is it true that the smallest primitive root modulo $p^a$ is smaller than $p$?
P.S. I think this should be right because the smallest primitive root modulo $p$ is $O(\log^6 p)$ (assuming the generalized Riemann hypothesis), which is much smaller than $p$ when $p$ is large enough. But I have no idea how to prove this.
Note that the smallest primitive root modulo $p=40487$ is $5$, but $5$ is not a primitive root modulo $p^2$. See also https://primes.utm.edu/curios/page.php/40487.html
There are $(p-1)\varphi(p-1)$ primitive roots mod $p^2$ and $(p-1)\varphi(p-1)$ elements mod $p^2$ are primitive elements mod $p$. Assume your claim is wrong: all primitive roots mod $p$ between $1$ and $p$ are not primitive roots mod $p^2$, but all between $p$ and $p^2$ are. Now take any, say $g$ below $p$, then its inverse $h$ mod $p^2$ has the same property, so $1<h<p$. But then $gh\equiv 1$ mod $p^2$ is not possible.
oops, the second $(p-1)\varphi(p-1)$ should be a $p,\varphi(p-1)$, sorry.
Adding to the information by @GerryMyerson - the smallest (?) example of some primitive root modulo $p$ which is not a primitive root modulo $p^2$ seems to be: $14$ is a primitive root modulo $29$ but not modulo $29^2$.
|
2025-03-21T14:48:31.710796
| 2020-08-07T10:50:44 |
368559
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"D.S. Lipham",
"David Roberts",
"Rivers McForge",
"Rodrigo Freire",
"Taras Banakh",
"bof",
"https://mathoverflow.net/users/155425",
"https://mathoverflow.net/users/4177",
"https://mathoverflow.net/users/43266",
"https://mathoverflow.net/users/61536",
"https://mathoverflow.net/users/95718",
"https://mathoverflow.net/users/9825"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631853",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368559"
}
|
Stack Exchange
|
Foundational results dependent on/equivalent to the continuum hypothesis or its negation?
I remember at a certain point early in my mathematical studies learning that the Axiom of Choice is equivalent to the following statement on Cartesian products:
If $\{ X_i \}_{i \in I}$ is any collection of nonempty sets indexed by an index set $I$, then $\prod_{i \in I} X_i$ is nonempty.
To me, this settled the question of whether to use Axiom of Choice in practical contexts (although it's still interesting to consider systems of math where it doesn't hold, and the interdependence of various other theorems/results/lemmas/axioms on $AC$).
My first question is:
Question 1--Is there any similarly fundamental lemma or theorem which depends on the continuum hypothesis or its negation? That is, are there any basic facts in set theory, topology, measure theory, etc. which are (a) "self-evident" and (b) equivalent to $CH$ or $\lnot CH$?
I would also be interested in hearing if such a statement existed for $GCH$ or its negation $\lnot GCH$, although to me $GCH$ seems "less likely" to be true than $CH$ just because it makes a much broader statement over the class of all cardinals, whereas $CH$ is a relatively narrow statement about the relationship of two cardinals $2^{\aleph_0}$ and $\aleph_1$.
Currently, the two "simplest" results (that I know of) in this vein that would directly depend on $CH$ or $\lnot CH$ are:
Wetzel's problem
Whether or not $\Bbb{R}^\omega$ is normal in the box topology
But neither of these seem intuitively true or false, much less so essential that we had better accept them one way or another if we want to get any serious math done in the related field.
I'm aware that attempts have been made to resolve $CH$ one way or another (e.g. Freiling's axiom of symmetry) that are basically trying to reduce $CH$ to such an obviously true/false statement of general set theory/topology/measure theory. So I have a follow-up:
Question 2--What seem to be the obstacles to finding such a resolution of $CH$ or $\lnot CH$? That is, why is it so difficult to make concrete and testable statements (i.e. not trivial things like "There exists an element of $2^{2^{\aleph_0}}$ which is neither countable nor of size $\mathfrak{c}$") dependent on $CH$'s truth or falsity? And, should this difficulty be taken as evidence one way or the other for $CH$? Slash, is it actually considered evidence one way or the other for $CH$?
For instance: every Borel set is either of size $\aleph_0$ (if countable) or of size $2^{\aleph_0}$ (if uncountable). Is our difficulty in constructing a set of intermediate cardinality (as opposed to the ease with which we can construct a non-measurable set) evidence that no such intermediate-cardinality set exists?
I'll also mention that I take the "Platonic view" of $CH$. That is, I believe that despite the existence of models of set theory where either $CH$ or $\lnot CH$ holds, the statement
"If $S = 2^\Bbb{N}$ is the set of all subsets of $\Bbb{N}$, then for $A \subset S$ any subset of $S$, either $A$ is countable, or there exists a 1-1 correspondence between $A$ and $S$"
has a canonical and demonstrable true/false answer.
Matt, I actually pored over a number of related questions before asking this to make sure my questions weren’t answered there, including the one you suggested. If you’re not sure how my questions are distinct, I’m happy to edit for clarification.
https://www.amazon.com/Hypothese-Du-Continu-W-Sierpinski/dp/B000JUOD6S)
It's not at all evident to me that $\prod_{i\in I} X_i$ is non-empty because there is no basic process by which to define an element when $I$ is uncountable (unless you assume that $I$ is well-ordered which assumes AC in the first place). Countable Choice and even Dependent Choice, however, are totally obvious. Some evidence for the truth of CH may be that it is implied by AD+DC. CH is also equivalent to the statement that every free ultrafilter on $\omega$ induces the same ordering on $\omega^\omega$. This seems unlikely, but is not a problem under AD because free ultrafilters do not exist.
@D.S.Lipham If all the $X_i$ are nonempty then there must exist a point $x_i \in X_i$ for all $i$, right? Then $(x_i){i \in I} \in \prod{i \in I} X_i$ and the product is nonempty. I don't even need to write down a choice function to see that there should be an element of the product.
@RiversMcForge That element of the product is a choice function.
@D.S.Lipham Yes, I'm aware that it's a choice function, lol. What I meant was, I don't need to "define an element [of the product] when $I$ is uncountable." All I need is that there exists a point $x_i \in X_i$ for all $i$ because the $X_i$ are nonempty, so some element of the product exists. Do you accept that if ${ X_i }_{i \in I}$ is any collection of nonempty sets, then there exists a point $x_i \in X_i$ for all $i$? If not, why not?
I like the result of Seierpinski saying that GCH implies AC. So, if you believe in AC, then this gives reasons to believe in GCH.
Concerning your first question, there is a simple, if not "self-evident", order-theoretic statement equivalent to $CH$ admitting a generalization equivalent to $GCH$:
If $L$ is a linear ordering of size $2^{\omega}$, then $L$ embeds every cardinal less than $2^{\omega}$ or $L^*$ ($L$ reversed) embeds every cardinal less than $2^{\omega}$.
This statement can be read in general terms as follows: In order to arrange $2^{\omega}$ points in a line one cannot bypass a smaller cardinal (in the sense that it will appear directly or reversed).
The following generalization is equivalent to $GCH$:
For every cardinal $\lambda$, if $L$ is a linear ordering of size $\lambda$, then $L$ embeds every cardinal less than $\lambda$ or $L^*$ embeds every cardinal less than $\lambda$.
It says, in general terms, that for every cardinal $\lambda$, in order to arrange $\lambda$ points in a line one cannot bypass a smaller cardinal.
It is important to remark that the regularity thus stated is trivially true in the finite realm (for finite $\lambda$), contrary to $GCH$ (there are, in general, many numbers between $n$ and $2^n$). Therefore, it is, at least, a more uniform statement generalizing a fact of the finite realm for which, presumably, we have a more reliable "intuition". This feature is present in the usual axioms of set theory.
Addendum:
I have just recalled the paper below which seems to be relevant to your question. It connects $CH$ with a more or less concrete problem of machine learning:
Shai Ben-David, Pavel Hrubeš, Shay Moran, Amir Shpilka and Amir Yehudayoff, Learnability can be undecidable, Nat. Mach. Intell. 1 (2019) 44–48, doi:10.1038/s42256-018-0002-3.
($CH$ is equivalent to a version of learnability)
See also the freely-accessible paper: William Taylor, Learnability Can Be Independent of ZFC Axioms: Explanations and Implications https://arxiv.org/abs/1909.08410
Thanks for your help.
@rodrigo In your answer, when you say $2^{\omega}$, you meant $2^{\aleph_0}$, right? Under ordinal exponentiation, $2^{\omega} = \sup_n 2^n = \omega$.
Yes, I mean cardinal exponentiation, that is, $2^{\aleph_0}$.
As it is stated in the comments, one reference is Sierpinski's book, Hypothese Du Continu, though it is not in English.
Another reference is Propositions Equivalent to the Continuum Hypothesis.
See also Some propositions equivalent to the continuum hypothesis and The continuum hypothesis (CH) and its equivalent.
You may be also interested in Eliminating the Continuum Hypothesis.
Let me also state one equivalent of CH. I have taken it from Interactions between (set theory, model theory) and (algebraic geometry, algebraic number theory ,...):
Let $R$ be a ring and $D(R)$ its unbounded derived category. Let $D^c(R)$ be the full subcategory of compact objects (in the explicit example below it is spanned by bounded complexes of f.g. projective modules). We say that $D(R)$ satisfies Adams representability if any cohomological functor $D^c(R)^{op}\rightarrow Ab$, i.e. additive and taking exact triangles to exact sequences, is isomorphic to the restriction of a representable functor in $D(R)$ (in particular it extends to the whole $D(R)$), and any natural transformation between restrictions of representable functors $D^c(R)^{op}\rightarrow Ab$ is induced by a morphism in $D(R)$ between the representatives.
Let $\mathbb C\langle x,y\rangle$ be the ring of noncommutative polinomials on two variables. The statement '$D(\mathbb C\langle x,y\rangle)$ satisfies Adams representability' is equivalent to the continuum hypothesis.
For another interesting equivalent of CH see: Reductions between certain incidence problems and the continuum hypothesis.
|
2025-03-21T14:48:31.711275
| 2020-08-07T11:52:14 |
368563
|
{
"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:631854",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368563"
}
|
Stack Exchange
|
Functional derivative of differential entropy
I have trouble finding the derivative of the differential entropy w.r.t the probability density function, i.e. what is $\frac{\delta F[p]}{\delta p(x)}$, where $F[p] = \int_X p(x)\ln(p(x))dx$, and $p(x)$ is a probability density function.
I tried finding the derivative with the definition of the Frechet derivative, but was not able to solve it. Also, there already exist many posts on the derivative of the related Shannon Entropy, but I was not able to find anything on the differential entropy.
Any ideas?
Thanks,
Jan
Looking at your "variational" $\delta$-notation, it appears that it will be enough for you to consider the Gateaux derivative, which is a weaker version of the Fréchet derivative. For $F(p):=\int_X p\ln p\,dx$, the Gateaux derivative of $F$ at a point $p$ in the direction $h\in L^1(X)$ is $f'(0)$, where
$$f(t):=F(p+th).$$
For variational purposes, it would be even better to consider just the (right) directional derivative of $F$ at point $p$ in the direction $h\in L^1(X)$:
$$(D_+F)(p)(h):=f'_+(0):=\lim_{t\downarrow0}\frac{F(p+th)-F(p)}t$$
whenever this limit exists.
Even this right directional derivative will not in general exist for all $h\in L^1(X)$; e.g., it will not exist if the set on which $h<0$ and $p=0$ is of nonzero measure, because on that set $(p+th)\ln(p+th)$ will not be even be defined for any real $t>0$. However, noting that $(u\ln u)'=1+\ln u$ for real $u>0$, we see that, by the dominated convergence theorem, $(D_+F)(p)(h)$ will exist and be given by
the formula
$$(D_+F)(p)(h)=\int_X (1+\ln p)h\,dx$$
for any $h\in L^1(X)$ such that $|h\ln(p+th)|\le g$ for some $g\in L^1(X)$ and all small enough $t>0$.
|
2025-03-21T14:48:31.711407
| 2020-08-07T13:01:22 |
368566
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Will Sawin",
"https://mathoverflow.net/users/18060",
"https://mathoverflow.net/users/58056",
"random123"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631855",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368566"
}
|
Stack Exchange
|
Relation between characteristic cycle and singular support of constructible sheaf
Let $M$ be a real analytic manifold. Let $F$ be an object of the bounded derived category of sheaves on $M$ with real constructible cohomology sheaves. Let $CC(F)$ denote the characteristic cycle of $F$ and $SS(F)$ be its singular support (I follow the terminology and notation of the book “Sheaves on manifolds” by Kashiwara and Schapira). If I understand correctly one has the inclusion:
$$supp(CC(F))\subset SS(F)\,\,\,\,\, (1)$$
by formula (9.4.10) in the above book by Kashiwara-Schapira.
Does one have equality in the inclusion (1)?
No.
Consider $M = \mathbb R$, $F$ the direct sums of the constant sheaves on the positive real numbers, negative real numbers, and $0$, extended by zero to the whole space.
Then $F$ is the associated graded of a filtration on the constant sheaf, hence has the same characteristic cycle as the constant sheaf, which doesn't contain the cotangent space of $0$.
But the singular support of $F$ clearly contains the singular support of the skyscraper sheaf at $0$, which is the cotangent space of $0$.
I apologize for asking this question here. We know that the inclusion is an equality for l-adic perverse sheaves as has been proved by Saito. Is there a reference for a similar result for perverse sheaves over complex analytic manifolds?
@random123 I think this follows from basic results on the D-module side, i.e. that the support of the associated graded of a good filtration on a holonomic $D$-module is pure of dimension the dimension of the variety, that the characteristic cycle is the cycle class of this module, and that the singular support is the support of this module. I didn't find this statement in Kashiwara-Schapira, I'm not sure if it follows quickly from results there.
Thank you very much. I too had a similar thought in mind. I suppose one will need that Riemann-Hilbert correspondence commutes with characteristic cycle and singular support. That it does as described in an article of Schmid-Vilonen (Ann. Math. (2000)) on page 1115-1116, and [Theorem 11.3.3, Kashiwara-Schapira] respectively.
|
2025-03-21T14:48:31.711572
| 2020-08-07T13:08:10 |
368568
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Ben McKay",
"Deane Yang",
"RaphaelB4",
"Robert Bryant",
"https://mathoverflow.net/users/13268",
"https://mathoverflow.net/users/13972",
"https://mathoverflow.net/users/613",
"https://mathoverflow.net/users/99045"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631856",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368568"
}
|
Stack Exchange
|
Is it possible to calculate the parallel transport on a loop from the Riemann curvature?
I admit I am not a differential geometer (a probabilist actually). However recently I get interested and I would like to have more intuitions and insight of what is the Riemann curvature.
This is the way I see it so far (please correct me if I am wrong):
We start from a connection $\nabla$.
This defines a parallel
transport along paths $\gamma$ that is a linear application on the
vector bundle.
If $\gamma$ is a loop, this application may be different to the identity.
The Riemann curvature is different from the identity when $\gamma$ is a very small loop (at first order).
This kind of definition seems very similar to the one of the rotational (as presented in physics classes).
We start from a vector field (one form) $u$,
If $\gamma$ is loop, $I=\oint_{\gamma}u\cdot d\gamma$ can be different to $0$.
The rotation $\operatorname{rot}(u)$ is this value $I$ when $\gamma$ is a very small loop (at first order).
and we have the wonderful Stokes Theorem which for the $\operatorname{rot}$ follows very naturally from this definition (we glue small loops together to get a big loop) $$\oint_\gamma u \cdot d\gamma = \iint_\mathcal{S}\operatorname{rot}(u)\cdot d\sigma $$ with $\mathcal{S}$ a surface delimited by $\gamma$.
So here is my question: Does there exist an equivalent for the Riemann curvature? That is: can one calculate the parallel transport of $\nabla$ along a loop $\gamma$ from the Riemann curvature on $\mathcal{S}$?
Consider a connection on a vector bundle over a 1-dimensional manifold: the circle. You can easily find all such connections up to isomorphism, and see that the parallel transport does not have to be trivial, but the curvature must be zero.
Is this something like what you’re looking for? https://www.math.nyu.edu/~yangd/papers/holonomy.pdf
In the case that the Riemannian manifold $M$ in question has dimension $2$ and is oriented and $\gamma([0,1])\subset M$ is the piecewise-$C^1$ oriented boundary of a compact domain $S\subset M$, we have the famous Gauss-Bonnet Theorem, which asserts that the holonomy around $\gamma$ is equal to rotation by the angle
$$
\theta = \int_S K\,dA.
$$
Thus, yes, in this case, the holonomy around $\gamma$ can be 'computed' from the Riemann curvature tensor.
This may seem a little unsatisfying because the parallel transport around $\gamma$ is defined using only information in an open neighborhood of $\gamma$ (and, of course, one can get away with less than that), but the above formula uses information that could, a priori, come from very far away from the image $\gamma([0,1])$. However, simple examples show that, even for surfaces, one can have a closed curve with arbitrary holonomy for which the metric is flat on a neighborhood of the curve. Thus, the holonomy cannot be computed purely locally from the Riemann curvature tensor.
(By the way, Ben's cautionary remark is based on a $\gamma$ that is not the boundary of any surface, so there couldn't be a formula of the kind you are seeking that would address his 'counterexample'.)
Once one goes to higher dimensions, even for Riemannian $3$-manifolds$(M,g)$, there is no formula known that would start with the Riemann curvature tensor and construct 'something' that could be integrated over every oriented compact surface $S$ with connected boundary $\partial S$ so as to yield the element of $\mathrm{SO}(T_{p}M)$ that is the holonomy of the oriented curve $\partial S$ at the boundary point $p\in \partial S$.
Actual nonexistence would be hard to prove without making some assumptions about what form the 'something' constructed out of the Riemann curvature tensor might take. However, if you make reasonable assumptions, you can rule things out.
For example, it is not hard to show that there is no universal $2$-form on $(M^3,g)$ constructed polynomially out of the Riemann curvature tensor (even when one is allowed to use $g$ and $\nabla$ as well) that has the property that its integral over any oriented compact surface $S\subset M$ with circular boundary $\partial S$ gives even the angle of the rotation of the Riemannian holonomy around $\partial S$. (Note that this angle in $[0,\pi]$ does not depend on the point $p\in\partial S$ that one choses as the 'initial point'.)
Thank you a lot for your answer! I still have two simple question to make it clear :
1- if the Riemann curvature is 0 on $S$ then the parallele transport of $\nabla$ on the loop $\partial S$ is the identity.
2-If $\nabla_1$ and $\nabla_2$ are two connections with the same Riemann curvature tensor on $S$, does the two parallele transport of $\nabla_1$ and $\nabla_2$ on $\partial S$ are the same?
@RaphaelB4: Yes, if $S$ is a compact, oriented surface with $\partial S$ being a circle and with a smooth Riemannian metric $g$ whose Riemannian curvature is zero, then the parallel transport of $\nabla$ around the entire (oriented) loop $\partial S$ is the identity. Meanwhile, if $g_1$ and $g_2$ are two metrics on $S$ with vanishing Riemannian curvature, then the parallel transports of $\nabla_1$ and $\nabla_2$ along a proper subinterval $I\subset \partial S$ from one end of $I$ to the other need not be the same.
I think the interesting case is $g_1$ and $g_2$ with non vanishing Riemannian curvature $R_1 = R_2$ and $I=\partial S$ the entire loop. Can we still say something here?
@RaphaelB4: It wasn't clear what you meant. Remember that the $2$-form integrand $K,dA$ is constructed from the Riemann curvature tensor, the metric $g$ and the orientation of the surface by an algebraic formula. As long as $K_1,dA_1 = K_2,dA_2$, then the holonomies of the two metrics around the boundary circle $\partial S$ will be the same. Meanwhile, note that, if one is, as is standard, regarding the Riemann curvature tensor as a tensor of type $(1,3)$, then one can easily have $R_1 = R_2\not=0$ without having $g_1=g_2$.
|
2025-03-21T14:48:31.711960
| 2020-08-07T15:24:16 |
368578
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Vlad Matei",
"Zhou",
"https://mathoverflow.net/users/110368",
"https://mathoverflow.net/users/41010"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631857",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368578"
}
|
Stack Exchange
|
Restricted partition problem into parts with a given set of prime factors
I need a reference for the following question:
Let $\mathcal{P}$ be a finite set of $k$ primes and let $f(n)$ be the number of partitions of $n$ into parts whose prime factors are restricted to the set $\mathcal{P}$. Then $f(n)<Ce^{\ln(n)^{k+1}}$ for some constant $C$.
It feels very classical, but I have not managed to track it down.
Let $\mathcal{P}=\{p_1,p_2,\ldots,p_k\}$. Then the generating function of $f(n)$ is
\begin{align}
G_{f}(t):=\sum_{n\ge 0}f(n)e^{-nt}=\prod_{\ell_1,\ell_2,\ldots,\ell_k\ge 0}\frac{1}{1-\exp\left(-t\prod_{1\le j\le k}p_j^{\ell_j}\right)},
\end{align}
where $t>0$. This means
$$f(n)\le G_f(t)e^{nt}$$
for any integer $n>0$ and any real number $t>0$. On the other hand, using part integration,
\begin{align*}
\log G_{f}(t)&=-\sum_{\ell_1,\ell_2,\ldots,\ell_k\ge 0}\log\left(1-\exp\left(-t\prod_{1\le j\le k}p_j^{\ell_j}\right)\right)\\
&=-\int_{1-}^{\infty}\log\left(1-\exp\left(-t x\right)\right)\,dS(x)\\
&=\int_{1}^{\infty}\frac{tS(x)}{e^{xt}-1}\,dx,
\end{align*}
where
$$S(x)=\sum_{\substack{\prod_{1\le j\le k}p_j^{\ell_j}\le x,\;\ell_1,\ell_2,\ldots,\ell_k\ge 0}}1.$$
Clearly,
$$S(x)\le \prod_{1\le j\le k}\left(1+\left\lfloor\frac{\log x}{\log p_1}\right\rfloor\right)=\frac{(\log x)^k}{\prod_{1\le j\le k}\log p_j}+O\left((\log x)^{k-1}\right).$$
Thus one can from above obtain
\begin{align*}
\log G_{f}(t)
&\le \frac{\log^{k+1}(1/t)}{(k+1)\prod_{1\le j\le k}\log p_j}+O(\log^k(1/t)).
\end{align*}
Taking $t=1/n$ then there exist a constant $C_k>0$ such that
$$f(n)\le G_f(t)e^{nt}\le \exp\left(1+\frac{\log^{k+1}n}{(k+1)\prod_{1\le j\le k}\log p_j}+O(\log^kn)\right)\le C_k e^{\log^{k+1}n}$$
by note that $(k+1)\prod_{1\le j\le k}\log p_j>1$.
Thank you so much for your answer! The argument you made is along my lines of thought also, and definitely feels that it should be out there in the literature somewhere.
If there is only a prime number of $p $, there will be a lot of literature. For example, see [Mahler, K. (1940), On a Special Functional Equation. Journal of the London Mathematical Society, s1-15: 115-123. doi:10.1112/jlms/s1-15.2.115]. Otherwise, there seems to be no such literature. Of course, using [Richmond, L. B.
Asymptotic relations for partitions.
J. Number Theory 7 (1975), no. 4, 389–405.], we can also get a more accurate asymptotic formula than your result through more complicated calculation.
|
2025-03-21T14:48:31.712243
| 2020-08-07T15:34:40 |
368579
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Otto",
"https://mathoverflow.net/users/119731"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631858",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368579"
}
|
Stack Exchange
|
Consistency of an intersection property
If $\kappa$ is an infinite cardinal then we denote by $[\kappa]^\kappa$ the collection of subsets of $\kappa$ that have cardinality $\kappa$. We say that $\kappa$ is intersectionally strange if there is a cardinal $\lambda \in \kappa$ such that:
there is ${\cal A}\subseteq [\kappa]^\kappa$ with $|{\cal A}| >\kappa$, and $|X\cap Y|<\lambda$ for all $X\neq Y\in {\cal A}$.
Is it consistent that for every infinite cardinal $\kappa$ there is an intersectionally strange cardinal $\alpha$ with $\alpha > \kappa$?
Yes. This was studied extensively by Baumgartner (https://www.sciencedirect.com/science/article/pii/0003484376900188)
|
2025-03-21T14:48:31.712321
| 2020-08-07T15:42:35 |
368581
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Q_p",
"https://mathoverflow.net/users/480516"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631859",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368581"
}
|
Stack Exchange
|
Does this function have a holomorphic continuation in $\sigma > \frac{1}{2}$?
Define
\begin{equation}
F(\sigma) = \Re \sum_{h=1}^{\infty} \sum_{n=1}^{\infty} \frac{b_{n, h}}{n^{2\sigma}}(1+h/n)^{-\sigma}\Bigg( \frac{e^{i\log(1+h/n)}-1}{i\log(1+h/n)} \Bigg)
\end{equation} where $\sigma \in \mathbb{R}$ and $|b_{n, h}|\ll \log n$. What is the minimal real number $c$ such that $F(\sigma)$ has a holomorphic continuation in $\sigma>c$ ?
Heuristically, it seems to me that $c \leq 1/2$, considering the fact that the inner sum is dominated by terms with small $h/n$, so that convergence of the whole series is assured for $2\sigma>1$.
Observe first that
$$\lim_{x\to 0}\frac{e^{i\log(1+x)}-1}{i\log(1+x)}=1,$$
hence there exists an absolute constant $C>0$ such that
$$\Re\frac{e^{i\log(1+h/n)}-1}{i\log(1+h/n)}>\frac{1}{2}\qquad\text{for}\qquad n>Ch.$$
Now assume that $b_{n,h}$ is the indicator function of $n\in(Ch,2Ch)$. Then for $\sigma\in[1,2]$ we have
$$F(\sigma)>\frac{1}{2}\left(1+\frac{1}{C}\right)^{-2}\sum_{h=1}^\infty\,\sum_{n\in(Ch,2Ch)}\frac{1}{n^{2\sigma}}\gg\sum_{h=1}^\infty\frac{1}{h^{2\sigma-1}}.$$
It follows that $F(1)$ diverges, and $F(\sigma)\to\infty$ under $\sigma\to 1+$. The second statement shows that $F(\sigma)$ has no continuous extension to $\sigma\geq 1$.
Thanks ! I i guess will be interesting to see what happends modulo $\sigma=1$...That is, whether or not $\sigma=1$ is an isolated singulaity...
|
2025-03-21T14:48:31.712691
| 2020-08-07T15:53:49 |
368583
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Alex Ravsky",
"Giorgio Metafune",
"Zac",
"https://mathoverflow.net/users/150653",
"https://mathoverflow.net/users/157076",
"https://mathoverflow.net/users/43954"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631860",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368583"
}
|
Stack Exchange
|
$L^p$ compactness for a sequence of functions from compactness of product with cut-off
Fix $p \in [1,\infty)$. Let $f_n:[a,b] \to \mathbb R$, $n \in \mathbb N$, be a sequence of $C^1$ functions. For every fixed $m\in \mathbb N^*$, suppose that the sequence of functions $$\{f_{n}\psi_m(f_n)\}_{n \in \mathbb N}$$ has a strongly convergent subsequence in $L^p([a,b])$ and that every subsequence $\{f_{n_k}\psi_m(f_{n_k})\}_{n_k}$ is also compact in $L^p$ for any fixed $m$. Here $\psi_m$ is a smooth cut-off function such that $0 \le \psi_m \le 1$ and
$$\psi_m(f) =
\begin{cases}
1 \qquad \text{ if } |f - 1|\ge 1/m \\
0 \qquad \text{ if } |f -1 |\le 1/(2m)
\end{cases}
$$
Assume also that $\Vert f_n \Vert_{L^p} \le C$ (for a constant $C>0$ that does not depend on $n,m$ for all $p \in [1,\infty]$). . If necessary, also asssume that $\Vert D_x(f_n\psi_m(f_n))\Vert_{L^1} \le C_m$, where $C_m$ is a constant that depends only on $m$.
How can we prove (or disprove) that $\{f_n\}_{n\in \mathbb N}$ also has a strongly convergent subsequence in $L^p([a,b])$?
If the result is not true, what additional assumption would make it so?
This question is motivated by my previous question $L^p$ compactness for a sequence of functions from compactness of cut-off.
I guess the definition of the functions $\psi_m(f)$ should be something like
$$\psi_m(f)(x) =
\begin{cases}
1 \qquad \text{ if } |f(x) - 1|\ge 1/m \
0 \qquad \text{ if } |f(x) -1 |\le 1/(2m)
\end{cases}
$$
for each $x\in [a,b]$.
@AlexRavsky Yes, exactly. I was just trying to use a shorter notation.
The condition “every subsequence ${f_{n_k}\psi_m(f_{n_k})}{n_k}$ is also compact in $L^p$” is very strong.
It can be shown that it is equivalent to ${f{n}\psi_m(f_{n})}{n\in\Bbb N}$ has finitely many values. I guess you mean that
a set ${f{n}\psi_m(f_{n})}_{n\in\Bbb N}$ is totally bounded instead.
@AlexRavsky You're probably right
Then a usual condition providing the existence of a strongly convergent subsequence of ${f_n}$ is a total boundedness of ${f_n}$. It can be easily assured by the following natural assumption: there exists a subsequence ${c_k}$ of positive numbers converging to zero and an increasing subsequence ${m_k}$ of natural numbers such that $|f_n- f_{n}\psi_{m_k}(f_{n})|_{L^p}<c_k$ for each $n$ and $k$.
@AlexRavsky Thank you. What other (more explicit) assumption on $f_n$ would make this requirement hold?
For each $n,m$ put $X_{n,m}={x\in [a,b]: |f_n(x)-1|<1/m}$. Let $\mu$ be the standard Lebesgue measure on $[a,b]$
I guess
$$|f_n- f_{n}\psi_{m}(f_{n})|^p_{L^p}=\int_{X_n,m} |f_n(x)|^p|1-\psi_m(f_n)(x)|^p d\mu\le (1+1/m)^p\mu(X_{n,m}).$$
Thus the required condition should be $\lim_{m\to\infty}\sup_{n}\mu(X_{n,m})=0.$
@AlexRavsky Thanks! Is there anything that we can assume on $f_n$ (smoothness, integrability, a condition on the derivatives?) that would make this condition on $\mu(X_{n,m})$ automatically true?
@AlexRavsky This property on $\mu(X_{n,m})$ has essentially to do with level sets, so maybe continuity helps?
Let us continue this discussion in chat.
The question is not very clear, as stated, but the following can be proved. Let $B$ be a bounded set of $L^p(0,1)$ and assume that for every $\epsilon>0$ the set $$B_\epsilon=\{f \chi_{\{|f| \ge \epsilon\}}, \ f \in B\}$$ is relatively compact in $L^p$, then $B$ is relatively compact, too. In fact, given $\epsilon >0$, a finite $\epsilon/2$-net for $B_\epsilon$ is a $\epsilon$-net for $B$, since $(0,1)$ has finite measure. Changing characteristic functions with smooth cut-off (around 1, as in the original problem), should not change the conclusion.
Thank you very much for your answer. How can one prove the claim in your last sentence?
Also, I'm not sure about the argument with nets (because I'm not very familiar with them). How can one make it more explicit?
Sorry for the language which was a bit too cryptic. For $B$ I want to prove it is totally bounded, that is for every $\epsilon>0$ there exists a finite number of balls $B(f_i, \epsilon)$ covering it. Do it first for $B_\epsilon$ (with $\epsilon/2$ balls), by assumption, and then note that the double balls cover $B$.
Ok, thanks. How does compactness follow from the covering?
This is a standard and powerful equivalence which holds in metric spaces.
Ok, now I see. I'm still just a bit confused about the last claim: Consider $B_m = {f\psi_m(f): f \in B }$ as in the body of the question. If $B_m$ is totally bounded, how can I prove that $B$ is also totally bounded?
You should try to follow the argument I gave above.
It seems not to be working because, in the notation above, $f\chi_{{|f-1|\ge \epsilon}}$ is not arbitrarily small (it's actually arbitrarily close to $1$). But maybe I'm missing something.
Let us continue this discussion in chat.
|
2025-03-21T14:48:31.713004
| 2020-08-07T15:55:13 |
368584
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Gilles Mordant",
"Learning math",
"https://mathoverflow.net/users/125260",
"https://mathoverflow.net/users/35936"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631861",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368584"
}
|
Stack Exchange
|
First and last order statistics and their ratio for $\chi^2_{m}$ random samples
Let $X_1, \dots, X_n \sim_{iid} \chi^2_{m}$ be a random sample from a chi-squared distribution with $m$ degrees of freedom (d.f.). I was wondering if there's any known result for the order statistics
$$\max_{1 \le i \le n} X_i, \min_{1 \le i \le n} X_i $$
respectively as a function of $m$ and also $n$?
And finally, is there any known result for the ratio of this two order statistics:
$$ \frac{\max_{1 \le i \le n} X_i}{\min_{1 \le i \le n} X_i }?$$
P.S. I'm primarily interested how these three quantities above behave w.r.t. increasing d.f. $m$ and w.r.t. increasing sample size $n$.
In this regard, I've looked into this question on stats.SE, but couldn't make it helpful.
Any references would also be appreciated.
A book that may contain answers to all the questions you have is 'Order Statistics' by David and Nagaraja.
For you first question, the easiest way is to rely on the fact that order statistics can be written as the inverse cdf of the corresponding order statistic of a uniform sample, i.e.
$$X_{(1)}, \ldots, X_{(n)} = (F^{-1}(U_{(1)}),\ldots, F^{-1}(U_{(n)})),$$
where the equality sign is meaning equality in distribution.
Also, the joint distribution for two order statistics is well known and an interesting exercise.
Hi, thanks for your answer. I agree that joint distribution of the min and max can be calculated, and hence their ratio too. But the thing is, the variables being $\chi_m^2,$ the ratio distribution is mathematically very complicated. The real question is how they depend as a function of the d.f. $m?$ I'll check the book, but scouring the internet didn't produce a useful answer to this.
Ok, I see the edit to you question that stresses the points that you want. I do not really know what you want to do with it, but this reference (https://arxiv.org/pdf/1207.7209.pdf) could be helpful. Further, as your data is i.i.d. the book 'extremes and related properties of random sequences and processes' may also help you.
|
2025-03-21T14:48:31.713180
| 2020-08-07T16:15:25 |
368586
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Alex Kruckman",
"LSpice",
"bambamfox",
"https://mathoverflow.net/users/163441",
"https://mathoverflow.net/users/2126",
"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:631862",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368586"
}
|
Stack Exchange
|
Searching for an early, highly theoretical, even philosophical, math paper on models or small-world networks
All I can remember is that it was very high-level / abstact and kind of philosophical, explaining (the discovery or interdependence of) small world networks. I assume that it was +50 years old and 'might' be an iconic paper, but maybe not - surely it was by far not as popular as the, already mentioned, paper "small world networks and their dynamics" by Duncan J. Watts & Steven H. Strogatz, but also it was on a completely different level of math.
I think I found it once on Azimuth Blog from John Carlos Baez. But it is very large and I don't even find "small world network" in the search (it might have also been mentioned by someone in the comment section).
'Theoretical' pretty much describes all of pure mathematics, so I guess you mean it here more or less as a synonym for 'philosophical', or perhaps for 'speculative'?
That is true, 'theoretical' is not well suited. Let's say, 'very high-level, abstract' mathematics. Like proof theory or universal algorithms.
I replaced the model-theory tag with mathematical-modeling. Model theory is a branch of mathematical logic, which doesn't actually have anything to do with mathematical models in the applied math sense.
Surely this is an intersting point, but could you elaborate? The title of the paper is "On models and modelling" and I would say, it is not applied.. (well it was released in an applied mathematics journal xD)
Stanley Milgram, The Small World Problem, Psychology Today 2, 60 (1967)
seems to fit the bill: +50 years old, "kind of philosophical", and yes, iconic -- cited more than 9,000 times. There are a few related papers in that time frame, listed here, but the paper from Psychology Today had the largest impact.
Oh, a Milgram paper. Thank you very much for this! (Unfortunately it's not the one I am looking for)
I answered my question below, but I'll mark your answer as correct, for being a very valueable addition ;)
I am answering my own question, because I actually did find it again (after 5 years and many tries).
It was not even close to being iconic nor +50y (it just looked old) nor specific on small networks :)
Rosen, R. (1993). On models and modeling. Applied mathematics and computation, 56(2-3), 359-372. PDF
Thanks @Carlo Beenakker for bringing up the Milgram paper!
|
2025-03-21T14:48:31.713371
| 2020-08-07T17:43:47 |
368592
|
{
"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:631863",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368592"
}
|
Stack Exchange
|
(Pro-)representable functors and full subcategories in homotopy theory
$\DeclareMathOperator\Ab{Ab}\DeclareMathOperator\Ho{Ho}\DeclareMathOperator\Hom{Hom}\DeclareMathOperator\Hotc{Hotc}\DeclareMathOperator\Sm{Sm}$Let $\mathcal{C}\overset{\iota}{\longrightarrow} \mathcal{D}$ be the inclusion of a full subcategory. Consider a functor
$$F:\mathcal{C}^{op}\rightarrow \Ab.$$
I've often seen examples where this functor might not be representable, but represented by a functor in the larger category, i.e. that there exists a $D\in \mathcal{D}$ such that
$$F\cong \Hom_{\mathcal{D}}(-,D).$$
This is of course the idea behind say Schlessinger's criterion, which tells us when a deformation functor is pro-representable, or it is also the philosophy of stacks (extend the category of schemes with stacks to get representable functors).
My question is if there exists a nice set of criteria to study the representabilitly of $F$ in $\mathcal{D}$? A naïve approach I had was to Kan extend the functor to $\mathcal{D}$ and then to study the representability of the Kan extension. However this seems to be a bad idea in general.
A concrete situation I've encountered this is the following: consider the full subcategory $\Ho(\Sm)\subset \Hotc$ of smooth manifolds in the homotopy category of pointed CW-complexes. In the larger category, checking whether a functor is representable is "easy" by Brown representability. Consider for instance
$$F:\Ho(\Sm)^{\text{op}}\rightarrow \Ab,\quad X\mapsto H^n(X,F)$$ the singular cohomology of degree $n$. In $\Hotc$, it is represented by the $n$-th Eilenberg-Maclane space, which is a pointed CW-complex but not a smooth manifold. Without using the fact that the singular cohomology on $\Ho(\Sm)$ is the restriction functor of singular cohomology on topological spaces, can we show that it is representable?
This illustrates also why it is not a good idea to study this via Kan extension, as the Kan extension of singular cohomology does not agree with singular cohomology.
This is a partial answer. Broadly speaking, representability theorems break down into two types. In both cases, the functor $F$ has to satisfy some exactness condition. For Freyd type theorems, $F$ must satisfy some set-theoretic condition such as accessibility or a solution set condition. For Brown type theorems, the domain category must satisfy some set-theoretic conditions, such as local presentability. This is nicely explained in a recent paper by Blanc and Chorny. I think you'd be interested in this paper. Like much of Chorny's work, it makes use of small functors, meaning functors that are the left Kan extension of some functor whose domain is a small category. Equivalently, a small functor is a small colimit of representable functors.
Brown type theorems further break down into two types. Cohomological Brown representability says, essentially, "any contravariant cohomological functor $F:\mathcal{T} \to Ab$ that takes coproducts to products is representable as $Hom(-,c)$. Homological Brown representability is about covariant functors $F:\mathcal{T}\to Ab$ being representable as $Hom(c,-)$, but more conditions are required, very much related to your question. For example, in Brown's original theorem, homological functors from the category of finite spectra to $Ab$ are represented by objects $c$ that are (not necessarily finite) spectra. This result was generalized in 1992 by Neeman.
You can also study representability for cohomological functors $F: \mathcal{T}^{op}_0 \to Ab$ defined on the full subcategory of small objects in $\mathcal{T}$. An excellent reference is Neeman's book, or this 2005 paper by Rosicky, which carries out a nice generalization that gets away from the need for triangulated categories and into the land of combinatorial model categories. Again, small functors play a crucial role. So this essentially answers your questions for situations where $\mathcal{C}$ is the full subcategory of small objects in $\mathcal{D}$, or where $\mathcal{D}$ is generated from $\mathcal{C}$ under filtered colimits.
Lastly, there are plenty of counterexamples where Brown representability fails, meaning the hypotheses are truly needed. Three such examples are mentioned in the Blanc and Chorny paper above, others are mentioned in Neeman's book, and another is in the paper Brown representability does not come for free by Casacuberta and Neeman.
|
2025-03-21T14:48:31.713643
| 2020-08-07T17:52:03 |
368593
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Anthony Quas",
"LSpice",
"Manuel Saavedra",
"https://mathoverflow.net/users/11054",
"https://mathoverflow.net/users/163445",
"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:631864",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368593"
}
|
Stack Exchange
|
Balls in minimal systems
If $(X,T)$ is a minimal system uniquely ergodic with $\mu$, is there $p\in X$ such that $\mu(\partial B(p,t))=0$ for all $t>0$?
For $X={a,b}$, $\partial B(p,t)$ is empty
Let $T$ be an irrational rotation of the circle. We modify the metric on the circle as follows, letting $d(\cdot,\cdot)$ be the standard metric on the circle; and for $C$ a non-empty closed subset of the reals, let $D(x,C)$ denote the distance from $x$ to $C$. Let $C$ be a Cantor set of positive measure contained in $[\frac 14,\frac 12]$ and containing $\frac 14$ and
define a new metric by
$$
\rho(x,y)=
\begin{cases}
d(x,y)&\text{if $d(x,y)<\frac 14$}\\
\tfrac 14+D(d(x,y),C)&\text{if $d(x,y)\ge \frac 14$.}
\end{cases}
$$
This generates the same topology as the original metric.
But $\partial B(p,\frac 14)=p\pm C$.
I believe you have changed the question in response to my answer to your question. Please don't do this! Instead, leave the original question (you can correct any errors if there are any), but add a section labeled "ADDITIONAL QUESTION IN RESPONSE TO ANSWER"; or ask a completely new question referencing this question.
Thanks for the suggestions.
@ManuelSaavedra's modified follow-up question: https://mathoverflow.net/questions/368608/balls-in-minimal-systems-ii .
|
2025-03-21T14:48:31.713760
| 2020-08-07T19:15:15 |
368596
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Denis Nardin",
"JoeyBF",
"Tim Campion",
"Yang",
"https://mathoverflow.net/users/2362",
"https://mathoverflow.net/users/43054",
"https://mathoverflow.net/users/529138",
"https://mathoverflow.net/users/56392"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631865",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368596"
}
|
Stack Exchange
|
Intuition for categorical fibrations?
I think I have a pretty good intuitive understanding of most types of fibrations of quasicategories:
a (trivial) Kan fibration is a bundle of (contractible) spaces with equivalent fibers,
a left/right fibration is a bundle of spaces with covariant/contravariant functors between fibers,
a (co)Cartesian fibration is the same as left/right but now the fibers are $\infty$-categories,
an inner fibration is bundle of $\infty$-categories with correspondences between fibers.
One major exception is the class of categorical fibrations. I know they are the fibrations in the Joyal model structure on sSet but that description isn't very illuminating to me. I feel this is problematic since categorical fibrations are central to the theory of $\infty$-operads, which I am trying to learn at the moment.
What would be the best way to describe categorical fibrations in a similarly intuitive way?
I would disagree with your characterization of an inner fibration: bundles of ∞-categories with correspondences are better represented by flat fibrations (you can find them treated in appendix B in HA). Inner fibrations, like categorical fibrations, are not particularly meaningful on their own, they are just a convenient point-set condition which can always be enforced up to homotopy.
That description came from the paragraph following HTT.<IP_ADDRESS>. I don't know much about flat inner fibrations, but would it be fair to say the difference is that flatness allows you to compose correspondences, which you couldn't do otherwise?
According to Barwick and Shah, just as a flat categorical fibration $E \to B$ corresponds to a 2-functor $B \to Corr$, an isofibration $E \to B$ corresponds to some kind of lax 2-functor $B \to Corr$, where $Corr$ is the 2-category whose objects are categories and 1-morphisms are correspondences / profunctors / distributors / bimodules. A proof is expected to appear in work of Peter Haine.
Categorical fibrations are not particularly meaningful in their own right. Luckily, there is a characterization in the most interesting case, of categorical fibrations $p:Q\to R$ between quasicategories. Namely such a map $p$ is nothing more than an inner fibration and an isofibration, that is, it is weakly orthogonal to the inclusion of either endpoint into $E[1]$, the nerve of the isomorphism category. Alternative characterizations are that the restriction of $p$ to the cores is a Kan fibration, which might be the most intuitive description, or that $p$ induces an isofibration on homotopy categories. This approach to categorical fibrations is used throughout Riehl and Verity's work and can also be found in Rezk's notes "Stuff about Quasicategories."
It may be worth remarking here that every functor of $\infty$-categories is equivalent to an inner fibration with the same codomain, so that the most invariant and conceptual way of thinking about a categorical fibration may be simply as an isofibration, full stop. This is in close analogy to the canonical model structure on the category of small categories.
Would you please have a short remark on why every functor of $\infty$-categories is equivalent to an inner fibration with the same codimain? I thought this probably has something to do with the factorization axiom of the Joyal model structure on simplicial set, which replaces a functor between simplicial sets $F:A\to B$ by $F': A'\to B$ which is a fibration in $sSet_{Joy}$, but isn't this a categorical fibration (p 90 in HTT, where he defines the categorical fibration as the fibration in $sSet_{Joy}$)? Thanks so much!
|
2025-03-21T14:48:31.714008
| 2020-08-07T19:46:31 |
368600
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Deane Yang",
"Michael Albanese",
"https://mathoverflow.net/users/21564",
"https://mathoverflow.net/users/613"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631866",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368600"
}
|
Stack Exchange
|
Open neighbourhood of a point of space of Riemannian metrics
Let $M$ be a finite-dimensional compact smooth manifold and
$$\mathcal{M}et(M) = \{ g : g \text{ is a Riemannian metric on }M \}.$$
Q1-a: What metrics $g$ are very close to the given metric $g_0$? I.e. Is it possible $g\in B_\varepsilon(g_0,M)$ and $g$ has completely different curvature for sufficiently small $\varepsilon>0$?
Q1-b: For example, Is that true that there exists $\varepsilon>0$, such that all metrics in $B_\varepsilon(g_{can},\Bbb S^n)\subset \mathcal{M}et(\Bbb S^n)$ are of positive curvature?
Any references about understanding the structure of $\mathcal{M}et(M)$ would be helpful.
Q1-b is false because $g_0$ could be a metric which does not have positive curvature.
The answer is "it depends". You have to decide what topology you want to use on the space of metrics, It is completely analogous to the different possible topologies for the space of functions on $M$. Possible topologies include $C^k$, Sobolev, Holder, and $C^\infty$. You get to choose, depending on the specific requirements needed.
@MichaelAlbanese, if the metric $g$ is $C^2$-close to the standard metric on the sphere, then it has positive curvature.
@DeaneYang: As far as I can tell, $g_0$ is an arbitrary metric on $S^n$. Maybe C.F.G intended for $g_0$ to be the round metric. They should make this clear.
$\mathcal{M}et(M)$ carries many natural (= invariant under the action of the group of diffeomorphisms of $M$) Riemannian metrics. See the following papers (and references therein):
Martin Bauer, Philipp Harms, Peter W. Michor: Sobolev metrics on the manifold of all Riemannian metrics. Journal of Differential Geometry 94, 2 (2013), 187-208. (pdf)
In particular, for the Sobolev order $\ge 2+\frac{\dim(M)}2$ metric the curvature is continuous so Q2 has a positive answer.
A more quite recent paper concentrating on the well-posedness of the geodesic equations for Sobolev metrics is this one.
More details:
In the $C^\infty$-topology, where $\mathcal{M}et(M)$ is an open subset of a Frechet space, Q2 is has always a positive answer. See here for a detailed description of this topology. Namely, the mapping $g\mapsto R^g$ which maps a metric $g$ to its curvature, is smooth $C^\infty$, and thus continuous since all is Frechet. Then, choose a finite open atlas $U_\alpha$ for $M$ a compact $K_\alpha \subset U_\alpha$ such that $\bigcup_\alpha K_\alpha = M$, and a frame $(s^i_\alpha)$ on each chart. Then, if for all $i < j$ sectional curvature
$$
k(\text{span}(s^i_\alpha,s^j_\alpha)) = -\frac{g_0(R^{g_0}(s^i_\alpha,s^j_\alpha)s^i_\alpha,s^j_\alpha)}{g_0(s^i_\alpha,s^i_\alpha) g_0(s^j_\alpha,s^j_\alpha) - g_0(s^i_\alpha,s^j_\alpha)^2} \ge \epsilon_\alpha
$$
for each $\alpha$, then we have $>\epsilon_\alpha/2$ for each $g$ near $g_0$.
This also holds for the $C^2$-topology, or for the Sobolev metric of the order given above, involving the Sobolev lemma.
|
2025-03-21T14:48:31.714220
| 2020-08-07T21:37:45 |
368605
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Carlo Beenakker",
"David Handelman",
"Denis Serre",
"Mick",
"Victor Kleptsyn",
"https://mathoverflow.net/users/11260",
"https://mathoverflow.net/users/153125",
"https://mathoverflow.net/users/31371",
"https://mathoverflow.net/users/42278",
"https://mathoverflow.net/users/8799"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631867",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368605"
}
|
Stack Exchange
|
Largest eigenvalue of finite band random matrices
Let $\mathbf{M}_n$ be an $n \times n$ symmetric matrix
$$
\mathbf{M}_n = \begin{cases}
X_{j-i,i}\ &\text{if }i\leq j\leq r+i\\
0\ &\text{if }r+i< j\leq n\end{cases}
$$
for some fixed $r>0$, and the random variables $\{X_{i,j}\}$ are assumed real, positive, i.i.d., and have finite mean and variance.
As an example, for $r=1$ and $n=4$ we have,
$$\mathbf{M}_4 =
\begin{pmatrix}
X_{0,1} & X_{1,1} & 0 & 0\\
X_{1,1} & X_{0,2} & X_{1,2} & 0 \\
0 & X_{1,2} & X_{0,3} & X_{1,3} \\
0 & 0 & X_{1,3} & X_{0,4}
\end{pmatrix}$$
I was wondering if something is known about the asymptotic of $\lambda_1(\mathbf{M}_n)$, i.e., the largest eigenvalue of $\mathbf{M}_n$, in the limit $n \to \infty$. In particular, is something is known about the deviation of $\lambda_1(\mathbf{M}_n)$ from its mean, i.e.,
$$
\Pr\left[|\lambda_1(\mathbf{M}_n)-\mathbb{E}\lambda_1(\mathbf{M}_n)|\geq t\right]\leq ?
$$
I was wondering whether there is a general concentration bound, e.g., for non-identical matrices, which subsumes the above case.
https://www.google.com/url?sa=t&source=web&rct=j&url=http://math.mit.edu/~edelman/homepage/papers/sturm.pdf&ved=2ahUKEwiq9-3L84rrAhUeBhAIHZtyAo4QFjAFegQIAhAB&usg=AOvVaw182xiOD97OcJBU4FrOdsdA&cshid=1596865816503 maybe this helps
Obvious comment: $\max_i X_i + X_{i+1}$ is an upper bound for the spectral radius, as it is the maximum row sum.
Just to mention a related (though different) question : https://mathoverflow.net/q/51848 .
one answer at https://math.stackexchange.com/a/1603016/87355
Just as an easy remark: if $X_i$'s are i.i.d. with bounded support, the largest eigenvalue (= the norm) will converge to $2*\max supp X_i$. Upper bound is from the $\ell_1$ (or $\ell_{\infty}$)-norm (see David Handelman's remark), on the other hand, having a big submatrix with all $X_i$ close to the maximum provides a lower bound, and by independence such a submatrix sooner or later will occur.
Do you have any particular law of $X_i$ in mind?
@J.John : Well, what I wanted to use here is a simple remark that as the matrix is symmetric, its (spectral) norm exceeds the norm of any of its principal minors. So if the law is upper-bounded by $M$, for a large enough matrix there will be somewhere a large length $k$ subsequence $X_j\approx M, , j=m,...,m+k$, and the norm of such a submatrix is $\approx 2M$. But this is not what you want to use for unbounded distributions...
@J.John: yes, there is. As the matrix is a symmetric one with positive elements, its maximal eigenvalue = its spectral radius = its $\ell_2$-norm, and the corresponding eigenvector has nonnegative elements. So take such a unit-length eigenvector $v$ for $L_k$, and you get $$\lambda_{\max}(M_k)\ge (v,M_k v) \ge (v,L_k v) = \lambda_{max}(L_k)$$
I start with this simple remark: the tridiagonal matrix $$A_k=\begin{pmatrix}0 & 1 & & & \\ 1 & 0 & 1 & & \\ & 1 & 0 & \ddots & \\ & & \ddots & & 1 \\ & & & 1 & 0\end{pmatrix}$$, $A_k\in \mathbb{R}^{(k+1)\times (k+1)}$ have largest eigenvalue $\lambda_\max (A_k) =2\cos{\frac{\pi}{k+2}}$.
We will focus on the submatrices with large entries of $M_n$. When there are $k$ consecutive large entries :$ \forall i\leq k$ $X_{a+i}\geq C $ for some $a$, we will assume that $X_{a+i} = C$ for all $i$ and write $CA_k$. This is obiously not true but it is just to simplify the discussion.
We then have $$ M_n = \begin{pmatrix}\ddots & \\ & C_1A_{k_1} \\ & & \ddots \\ & & & C_2 A_{k_2} \\ & & & & \ddots \\ & & & & & . \end{pmatrix} $$
where $\ddots$ have small entries (let say $\mathcal{O}(1)$) and $C_i\gg 1$. The largest eigenvalue will come from these submatrices
$$\lambda_\max (M_n) \approx \max_j \lambda_{\max}(C_j A_{k_j})=\max_j 2 C_j\cos(\frac{\pi}{k_j+2})$$
For large $n$ the behaviour will depend on the tail of the random variable $X_1$.
We first consider the case of polynomial tail : $\mathbb{P}(X \geq K)\sim \frac{1}{K^\alpha}$.
For any $k$, $\lambda_{\max}(C A_{k})\geq K\Leftrightarrow C \geq \frac{K}{2\cos(\frac{\pi}{k+2})}$ and we estimate $$\mathbb{P}(\forall i\leq k, X_k \geq \frac{K}{2\cos(\frac{\pi}{k+2})}) = \Big(\frac{2\cos(\frac{\pi}{k+2})}{K} \Big)^k$$ For $K\rightarrow \infty$, one can see that the case $k=1$ have the much larger probability and we deduce that in this situation it is enougth to consider only $k=1$ submatrices.
Conclusion for polynomial tail we have
$$\lambda_\max (M_n) \approx \max_j X_j \sim n^{1/\alpha}$$ (Because there are $n$ iid $X_j$, we set $K=n^{1/\alpha}$ such that $\mathbb{P}(X_1 \geq K)=\frac{1}{n}$).
We now consider the case of exponential tail : $\mathbb{P}(X \geq K)\sim \exp(-\gamma K)$.
We estimate $$\mathbb{P}\Big(\forall i\leq k, X_k \geq \frac{K}{2\cos(\frac{\pi}{k+2})}\Big) = \exp\Big(-\frac{\gamma k K}{2 \cos(\frac{\pi}{k+2})} \Big)$$ Still here for $K\rightarrow \infty$, the case $k=1$ have the much larger probability.
Conclusion for exponential tail we have
$$\lambda_\max (M_n) \approx \max_j X_j \sim \frac{\log(n)}{\gamma}$$ (we set $K$ such that $\mathbb{P}(X_1 \geq K)=\frac{1}{n}$).
We continue with the case of sup-exponential tail : $\mathbb{P}(X \geq K)\sim \exp(-K^\gamma)$.
We have $$\mathbb{P}\Big(\forall i\leq k, X_k \geq \frac{K}{2\cos(\frac{\pi}{k+2})}\Big) = \exp\Big(-\frac{ k }{2^\gamma \cos(\frac{\pi}{k+2})^\gamma}K^\gamma \Big)$$ Here there is a $k^*$ that maximize $\frac{k}{\cos(\frac{\pi}{k+2})^\gamma}$ which have the much larger probability for $K\rightarrow \infty$.
We also set $K$ such that this event is of order $1/n$ and then for sup-exponential tail we have
$$\lambda_\max (M_n) \sim \frac{2\cos(\frac{\pi}{k^*+2})}{(k^*)^\frac{1}{\gamma}}\log(n)^{\frac{1}{\gamma}}$$
Finally in case of bounded $X$, for any $\epsilon>0$, and $k$, we can find $a$ such that $\forall i\leq k, X_{a+i}\geq \|X\|_\infty-\epsilon$ with probability that goes to $1$ as $n\rightarrow \infty$. Then $$2 \|X\|_\infty \geq \lambda_\max (M_n) \geq 2 (\|X\|_\infty-\epsilon) \cos(\frac{\pi}{k+2}) $$ and we get $\lambda_\max (M_n) \rightarrow 2 \|X\|_\infty$.
|
2025-03-21T14:48:31.714717
| 2020-08-07T22:13:15 |
368606
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Alexander Betts",
"Brian Hopkins",
"Fedor Petrov",
"Gerry Myerson",
"Jon Bannon",
"Kimball",
"Lars H",
"Wolfgang",
"aorq",
"https://mathoverflow.net/users/1079",
"https://mathoverflow.net/users/126183",
"https://mathoverflow.net/users/126481",
"https://mathoverflow.net/users/14807",
"https://mathoverflow.net/users/158000",
"https://mathoverflow.net/users/159461",
"https://mathoverflow.net/users/29783",
"https://mathoverflow.net/users/4312",
"https://mathoverflow.net/users/6269",
"https://mathoverflow.net/users/64302",
"https://mathoverflow.net/users/6518",
"pancini",
"user2520938"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631868",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368606"
}
|
Stack Exchange
|
Counterexamples against all odds
What are some examples of conjectures proved to be true generically (i.e. there is a dense $G_{\delta}$ of objects that affirm the conjecture) but are nevertheless false?
Also, it would be cool to see examples where the conjecture was proved true with probability 1, but was nevertheless false.
Of course one can manufacture statements from the spaces themselves, but I'm mostly interested in actual conjectures from contemporary mathematics whose resolution exhibited this pattern.
I am curious about this situation because sometimes, although we cannot find them easily, the "counterexample space" for a given conjecture may be quite large, yet inaccessible due to the limitations of existing techniques. For example, Tsirelson's conjecture and the Connes Embedding Conjecture were recently proven false, and although we cannot yet concretely construct a counterexample I see no reason to believe that counterexamples will necessarily be terribly rare objects...once the techniques are available to construct them. (These may be fighting words.)
The present question inquires about a distinct situation where it has been proved that a randomly selected object will not provide a counterexample. As dire as this sounds, the situation may be advantageous in that the construction of a counterexample may have to be much more surgical, and so one may see more clearly a way to build one. I'm wondering if my intuition about this is valid, based on recent history.
The question is just a passing curiosity, really, but I think someone may have a good story or two that will educate.
Many things are true generically... e.g. the “conjecture” ‘all algebraic varieties are smooth’ would satisfy your condition.
Every real number is irrational. :p
@ElliotG every real number is rational
Right! I'm interested in situations where there was a reticent conjecture, someone proved its holding generically, and then the conjecture was ultimately false, most likely by a targeted construction. For example, the "conjecture" you mention above could be blown up by coming up with a single example of a non-smooth variety. I am curious if history contains other kinds of plot twists...
I love these tongue in cheek ones...but am looking for actual serious conjectures in contemporary mathematics.
Alas, both examples with rational and irrational numbers fail: the first is not true generically, the second never was a conjecture.
Closely related: https://mathoverflow.net/q/15444/6518
It doesn't quite fit the pattern you are looking for, but the Grunwald--Wang Theorem is a good example. It is a local-to-global result in number theory, "known" to hold over all number fields by a 1933 proof of Grunwald. But in 1948 Wang published a paper titled "A counter-example to Grunwald's theorem", which showed that in fact Grunwald's result could be false for certain very specific number fields, and classified exactly for which fields the theorem failed.
So this is something which was "known" to be true in general, until a flaw was found and a handful of exceptional cases identified.
@FedorPetrov The rationals are dense in the reals, so according to the definition of "generic" in the question, it does satisfy the conditions.
@user2520938 dense but not $G_\delta$
@FedorPetrov Ah, I took that $G_\delta$ as just being a name for the set of objects satisfying the conjecture. I did not realise that this indicates a property of a subset. Now I know.
The most famous example is the so-called Riemann-Hilbert problem, which has a long and complicated history which I don't explain in detail. As it happens Hilbert's own formulation was not very exact, this was rather a program of research than an exact formulation with a yes/no answer. This was Problem 21 in his famous list. Hilbert believed that the question has a positive answer, and even that he solved it.
The most common version of the problem was whether there exists a Fuchsian system, that is a differential equation of the form
$$w'=A(z)w=\left(\sum_{j=1}^m\frac{A_j}{z-a_j}\right)w$$
on the Riemann sphere, with arbitrary prescribed singularities $a_j$ and prescribed monodromy representation. Here $A_j$ are constant $n\times n$ matrices, $w$ is a solution vector, and $w'=dw/dz$.
It was solved for $A$ in general position by Josip Plemelj in 1908, who obtained a positive answer, and for a long time it was assumed that the statement is true in general. It is true in dimension $2$, and it is true in higher dimensions under various very mild conditions which are violated on the set of large codimension. For example, the answer is positive if at least one $A_j$ is diagonalizable. However in 1989 Andrei Bolibrukh constructed a $3\times 3$ counterexample with $m=4$. Such counterexamples exist for every $n\geq 3$.
Ref. D. Anosov and A. Bolibruch, The Riemann-Hilbert problem
Bolibrukh, A. A.
The Riemann-Hilbert problem on the complex projective line. (Russian)
Mat. Zametki 46 (1989), no. 3, 118–120.
Let $S$ be a finite set of (reduced) points in the projective plane
and let $I$ be the (saturated) homogeneous ideal of $S$.
Recall that $I^{(m)}$ is the $m$th symbolic power of $I$,
consisting of polynomials that vanish to order at least $m$ at each point of $S$
(in characteristic $0$).
Evidently the ordinary power $I^m$ satisfies $I^m \subseteq I^{(m)}$.
This is just the statement that if each one of $F_1,\dotsc,F_m$ vanishes at a point $P$,
then every $(m-1)$th derivative of the product $F_1 \dotsm F_m$ vanishes there also.
In fact if $n \geq m$, then $I^n \subseteq I^{(m)}$.
Conversely, if $I^n \subseteq I^{(m)}$, then $n \geq m$.
So there's a pretty simple classification for when ordinary powers are contained in symbolic powers.
There's no obvious reason that $I^{(m)} \subset I^n$
should ever hold for any $m$ and $n$, beyond the trivial $n=1$, $m \geq 1$.
However, following work of Swanson,
the containment $I^{(2n)} \subseteq I^n$ was shown around 2000 or 2001
by Ein-Lazarsfeld-Smith, using asymptotic multiplier ideals,
and also by Hochster-Huneke, using tight closure methods.
More generally, for ideals of height $h$ (on smooth varieties),
$I^{(hn)} \subseteq I^n$ holds;
for points in the plane the height is $2$.
It's not the case, though, that if $I^{(m)} \subseteq I^n$,
it must be $m \geq 2n$ (or $hn$).
For example, complete intersections have $I^{(m)} \subseteq I^n$ as soon as $m \geq n$.
One can show that if $I^{(cn)} \subseteq I^n$ for all $I$ and all $n$, then it must be $c \geq h$.
But what about small values of $n$, or subleading terms, that is $m = hn + o(n)$?
In particular Huneke asked whether the containment
$I^{(4)} \subseteq I^2$ could be improved to $I^{(3)} \subseteq I^2$.
You can check computationally and it works for lots of examples,
so there's some plausibility.
This was very much a question,
but some people (not including Huneke) started calling it "Huneke's conjecture".
Around 2010,
Bocci-Harbourne showed that $I^{(3)} \subseteq I^2$ holds
for points in general position.
That's a dense $G_\delta$: "general position" means it holds on
a Zariski open, dense subset of the $(\mathbb{P}^2)^k$ that parametrizes
sets of $k$ points in the plane (ignoring order and collisions of points).
(On the other hand, this is very much like "all algebraic varieties are smooth".)
But, around 2013, a counterexample was found by
Dumnicki-Szemberg-Tutaj-Gasińka.
It's a collection of $12$ points and easy to verify,
once you know what to try.
It was even a previously known arrangement of points
(a dual Hesse arrangement).
Since then people have found families of counterexamples,
counterexamples in higher dimension,
higher-dimensional counterexamples consisting of positive-dimensional components (instead of points), and so on;
they look for families where $m=cn$ works with $1 \leq c < h$...
You can find literature on this
with keywords like "containment problem for symbolic powers",
resurgence, and Waldschmidt constant.
Why wasn't the counterexample found earlier?
For one thing, it's a bit of a niche subject.
The space of arrangements of $12$ points is $24$-dimensional
(and nobody knew whether $12$ was the right number of points).
And finally, the counterexample is not over the rationals
(it's over $\mathbb{Q}[\omega]$, $\omega$ a cube root of unity),
which means there's an extra step needed to enter it into Macaulay2.
In hindsight that might seem a bit trivial,
but this counterexample wasn't going to be found
by just guessing random field extensions and guessing some points.
Sorry for rambling on.
I think that algebraic geometry must have many versions of this story,
where something was known to hold generally (meaning, on a Zariski open, dense set),
conjectured to hold universally,
but found to have counterexamples.
This particular story is a favorite for me just because it relates to the motivation for my thesis problem.
(I studied the multiplier ideals that appeared in the Ein-Lazarsfeld-Smith proof.)
George Andrews and Cristina Ballantine's 2019 Almost partition identities builds on classical results to prove that various pairs of integer partition statistics are equal asymptotically 100% of the time yet are not equal infinitely often. One example:
The total number of parts in all self-conjugate partitions of $n$ =
the number of partitions of $n$ in which no odd part is repeated and there is exactly one even part (possibly repeated)
is true for almost all $n$. Yet equality fails infinitely often and the error grows without bound.
This may not satisfy your question, though, since no one ever claimed that the statistics were always equal---they skipped that step.
This list if examples may be started with "the number of partitions of $n$ onto an even number of distinct parts equals to the number of partitions of $n$ onto an odd number of distinct parts", with exceptions being pentagonal numbers $n=k(3k\pm 1)/2$, due to Euler himself. Well, this of course never was a conjecture.
@FedorPetrov Yes, and they start from that in the article. I was trying to satisfy the "contemporary" part of the question, at least.
(More of a comment than an answer, I suspect, but anyway…)
There are a bunch of results in graph theory that have to make exceptions for the Petersen graph, how do you rank that kind of counterexample? It's very far from unknown, you only have to remember to check it. And remarkably, it often stands as the only counterexample, so the theorem proper only has to include an exception.
Edit: An analogy, to help readers not familiar with graph theory get the gist of the situation: it's a bit like basic arithmetic, where you every now and then have to remember to exclude zero (e.g. "for all nonzero $a$ the equation $ax=b$ has a unique solution"), except that the thing you're making an exception for is more like $\dfrac{1+\sqrt{19}}{5}$ — complicated enough that most people having been shown the example and then asked to recreate it from memory will get it wrong. And it is the same exception in a bunch of unrelated theorems.
Edit 2: Examples of particular results provided in comment by @aorq. In each statement, the * means "except one particular graph".
Theorems:
Every* connected vertex-transitive graph of order 2p is Hamiltonian.
Every* bipartite Kneser graph is Hamiltonian.
Every* generalized Petersen graph has a 1-factorization and chromatic index 3.
Conjectures:
Every* bridgeless cubic graph admits a 2-bisection.
Every* connected Kneser graph is Hamiltonian.
Every* cyclically 4-edge connected cubic graphs has an even cycle double cover.
Every* connected metacirculant is Hamiltonian.
Every* connected strongly regular graph is Hamiltonian.
Every* bridgeless cubic graphs has circular chromatic index at most 7/2.
This might work in the context in the question if there is a natural sample space of graphs for which a particular statement is true with probability one but the Peterson graph belongs to a set of measure zero. However, you can see how such constructions might be reverse engineered so that they are in essence like the "every number is irrational" example...
@JonBannon, turning the set of graphs into a sample space is Standard Operating Procedure in probabilistic graph theory, but probably not in the way you seem to seek; maybe I'll turn that into another (not-quite-)answer. However, why require a set to have measure zero – doesn't that tend to exclude countable mathematics from consideration? In the results under consideration in my answer, the set of counterexamples has cardinality one, which should be as good as it gets.
Sure, Lars. Whatever is natural to an area is fine with me! Asymptotically probability zero may be nice in the discrete setting, maybe.
Reminds me of the famous saying: "if ever you have a conjecture in graph theory, check first that the Petersen graph is not a counterexample..."
I'd like this answer more if it gave even one example of a result for which the Petersen graph is the only counterexample ("$G$ is not isomorphic to the Petersen graph" doesn't count).
Known: Every* connected vertex-transitive graph of order 2p is Hamiltonian. Every* bipartite Kneser graph is Hamiltonian. Every* generalized Petersen graph has a 1-factorization and chromatic index 3. Open: Every* bridgeless cubic graph admits a 2-bisection. Every* connected Kneser graph is Hamiltonian. Every* cyclically 4-edge connected cubic graphs has an even cycle double cover. Every* connected metacirculant is Hamiltonian. Every* connected strongly regular graph is Hamiltonian. Every* bridgeless cubic graphs has circular chromatic index at most 7/2. [* = except a certain graph.]
The generic oracle hypothesis is false. In particular, $\mathsf{IP}^G \ne \mathsf{PSPACE}^G$ for a generic oracle $G$, but $\mathsf{IP} = \mathsf{PSPACE}$ in real life. Similarly, the random oracle hypothesis is false.
By the way, I gave this answer in response to a related MO question. There are a couple of other related MO questions, e.g., Heuristically false conjectures and Examples where physical heuristics led to incorrect answers.
It might be questionable if this example fits the bill, but in my opinion it does and it is a quite remarkable phenomenon.
The prime number theorem yields the asymptotic
\begin{equation}
\pi(x+\Phi(x))-\pi(x) \sim \frac{\Phi(x)}{\log(x)},
\end{equation}
as long as $\Phi(x) \gg x$. Now one might ask how small one is allowed to make $\Phi(x)$ so that this asymptotic remains true. For example Huxley showed that $\Phi(x) = x^{\frac{7}{12}+\epsilon}$ is admissible.
Assuming the Riemann hypothesis Selberg showed that the desired asymptotic holds for almost all $x$ as long as $\frac{\Phi(x)}{\log(x)^2} \to \infty$. It is natural to wonder if Selberg's result might be true without exception and it is here where the 'counterexamples against all odds' appear. Indeed Maier (in his paper 'primes in short intervals') showed the following:
\begin{equation}
\liminf_{x\to\infty} \frac{\pi(x+\Phi(x))-\pi(x)}{\Phi(x)/\log(x)} < 1 < \limsup_{x\to\infty} \frac{\pi(x+\Phi(x))-\pi(x)}{\Phi(x)/\log(x)}
\end{equation}
for $\Phi(x) = \log(x)^B$ with $B>1$. This shows the existence of exceptions in a quite spectacular manner.
Ursula Martin proved that when $p$ is prime, almost all finite $p$-groups have outer automorphism group a $p$-group, yet it follows from a Theorem of G. Glauberman that for any prime $p > 3$, and any (non-trivial) Sylow $p$-subgroup $P$ of a non-Abelian simple group $G$, it is never the case that ${\rm Out}(P)$ is a $p$-group. Note also that by Burnside's $p^{a}q^{b}$-theorem, every non-Abelian finite simple group $G$ has order divisible by some prime $p > 3$.
Let $\mathsf{A} = (A_1,\dots,A_m)$ be a tuple of $d \times d$ matrices. The joint spectral radius (JSR) of $\mathsf{A}$ is $\mathrm{JSR}(\mathsf{A}) := \lim_{n\to\infty} \sup_{i_1,\dots,i_n} \|A_{i_1} \dots A_{i_n}\|^{1/n}$, where $\|.\|$ is any norm on $\mathrm{Mat}(d\times d) = \mathbb{R}^{d^2}$.
The JSR was introduced by Rota and Strang in 1960. In the case of a single matrix ($m=1$), the JSR is equal to the spectral radius, that is, the biggest modulus of an eigenvalue of the matrix.
For equivalent definitions of the JSR, see e.g. Jungers' monograph.
The finiteness conjecture of Lagarias and Wang (1995) asserted that for any tuple $\mathsf{A} = (A_1,\dots,A_m)$, there is a product $A_{i_1} \dots A_{i_n}$ of some finite length $n$ whose spectral radius is exactly equal to $[\mathrm{JSR}(\mathsf{A})]^n$. This conjecture was disproved in 2001 by Bousch and Mairesse. More counterexamples were constructed later, e.g. here, here, and here.
However, it is conjectured (see Conjecture 8 by Maesumi) that if $m \ge 2$ and $d\ge 2$, then the counterexamples to the finiteness conjecture form a subset of $\mathbb{R}^{d^2m}$ of zero Lebesgue measure, so the finiteness conjecture is almost always true. This conjecture is supported by numerical evidence (see e.g. here), but so far remains entirely open.
|
2025-03-21T14:48:31.715788
| 2020-08-07T22:37:09 |
368608
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Anthony Quas",
"https://mathoverflow.net/users/11054"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631869",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368608"
}
|
Stack Exchange
|
Balls in minimal systems II
If $(X,T)$ is a minimal system uniquely ergodic with $\mu$. Is there $p\in X$ such that $\mu(\partial B(p,t))=0$ for all $t>0$ for some metric $d$ (with the same topology)?
This question is motivated by the answer of Anthony Quas in
Balls in minimal systems.
Is there a specific case you have in mind? I would not be surprised if the answer to your question is “yes” as there is a lot of freedom amongst metrics generating the same topology, but it doesn’t seem so easy to make a general argument.
This is not a complete answer, but too long for a comment.
If you remove unique ergodicity as an assumption, then I think there exist systems where it's not possible to find points where all boundaries of balls are universally null, i.e. have $0$ measure for all $T$-invariant measures. This is because in the paper "Can one always lower topological entropy?" by Shub and Weiss, they use the hypothesis "there exists aperiodic $p \in X$ so that for a dense set of $t$, $\{x : d(p, x) = t\}$ has measure $0$ for every $T$-invariant measure" to prove that $(X, T)$ has nontrivial factors of arbitrarily small entropy.
But, it's known that not every minimal system has factors of arbitrarily small entropy; for instance, there are minimal systems of infinite entropy (due to Lindenstrauss) for which every nontrivial factor has infinite entropy. So it seems that for his systems, you can't even get universally null boundaries for a dense set of radii, let alone all radii.
On the other hand, I don't think Lindenstrauss's systems are uniquely ergodic. So I guess in theory unique ergodicity could be enough. But I don't immediately see why. And I would be surprised, since if unique ergodicity did imply your property, it should imply that all uniquely ergodic systems have nontrivial factors of strictly smaller entropy, which I don't believe to be known.
EDIT: Of course I was being stupid; in fact Shub-Weiss explicitly note that their proof implies lowerability of entropy for uniquely ergodic systems (and even systems with countably many ergodic measures) for a clear reason: the boundaries of balls with fixed center are disjoint, so of course only countably many can have nonzero measure. So I guess these observations don't help much with your original question, sorry.
|
2025-03-21T14:48:31.715960
| 2020-08-07T23:28:04 |
368609
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"GabS",
"Math604",
"https://mathoverflow.net/users/127663",
"https://mathoverflow.net/users/66623"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631870",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368609"
}
|
Stack Exchange
|
Smooth dependence of solution to elliptic pde depending on parameter
I have a question in mind but let me generalize it slightly.
Suppose I am looking at some pde like
$$-\Delta u + t u = f(u)$$ in $B_1$ (here $u=u(x)$) with $u=0$ on $ \partial B_1$ where $B_1$ is the unit ball centred at the origin in $R^N$. For concreteness lets assume $f(u)=u^p$ where $p>1$. Lets assume I can show for all $t>0$ there is a bounded positive radial solution of the given pde.
My goal is to prove that for $t=1$ the solution is nondegenerate; meaning the kernel of the linearized operator is trivial. Now since this equation is not exactly the one I have in mind
I don't want to prove directly that the solution is nondegenerate since this might not extend to my case.
Using some other tricks I believe I can show that the solution for $t=1$ is nondegerate provided these solutions indexed by $t$ are sufficiently smooth in $t$. I believe the usual method to prove smoothness in $t$ is to use the implicit function theorem (or something close) but of course I can't do that here since I am really trying to prove one of the hypothesis. Any comments would be greatly appreciated.
Did you look at the paper of Zhang Liqun (Communications in PDE 1992) titled "Uniqueness Of Positive Solutions In A Ball". I believe there is a section on non-degereneracy.
i did not look at that exact paper. My question is not very well posed since I really have a different example in mind. The above example the radial solution is decreasing in $r$ and then I know how to prove non-degeneracy. So even though my end goal is to prove nondegeneracy I am really more asking about this dependence on $t$ and smoothness in $t$.
|
2025-03-21T14:48:31.716112
| 2020-08-08T01:34:47 |
368615
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Jori",
"https://mathoverflow.net/users/151697"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631871",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368615"
}
|
Stack Exchange
|
The connections between Kolmogorov complexity and mathematical logic
We know that Kolmogorov Cmplexity (KC) has connections to mathematical logic since it can be used to prove the Gödel incompleteness results (Chaitin's Theorem and Kritchman-Raz). Are there any other striking application of Kolmogorov complexity to mathematical logic (outside KC itself of course)?
Relatively simple examples like the ones I mentioned are preferred, but more complicated illustrations are also very welcome!
(This is cross posted from MSE, because I wasn't getting any useful replies there)
Four pointers to the literature from the last 25 years on applications of Kolmogorov complexity to mathematical logic:
Applications of Kolmogorov complexity to computable model theory (2007).
In this paper we answer the following well-known open question in
computable model theory. Does there exist a computable not
ℵ$_0$-categorical saturated structure with a unique computable
isomorphism type? Our answer is affirmative and uses a construction
based on Kolmogorov complexity.
Logical operations and Kolmogorov complexity (2002).
Conditional Kolmogorov complexity can be understood as the complexity
of the problem $Y\rightarrow X$, where $Y$ is the problem “construct
$y$” and $X$ is the problem “construct $x$”. Other logical operations
($\wedge,\lor,\leftrightarrow$) can be interpreted in a similar way,
extending Kolmogorov interpretation of intuitionistic logic and Kleene
realizability.
The Kolmogorov expression complexity of logics (1997).
We introduce the Kolmogorov variant of Vardi's expression complexity.
We define it by considering the value of the Kolmogorov complexity
$C(L[{\cal A}])$ of the infinite string $L[{\cal A}]$ of all truth
values of sentences of $L$ in ${\cal A}$. The higher is this value,
the more expressive is the logic $L$ in ${\cal A}$.
Kolmogorov complexity and the second incompleteness theorem (1995).
It is well known that Kolmogorov complexity has a close relation with
Gödel’s first incompleteness theorem. In this paper, we give a new
formulation of the first incompleteness theorem in terms of Kolmogorov
complexity, that is a generalization of Kolmogorov’s theorem, and
derive a semantic proof of the second incompleteness theorem from it.
Thank-you for your answer. So before Chaitin's work in '71, were there any significant applications of KC to mathematical logic that you are aware of?
|
2025-03-21T14:48:31.716333
| 2020-08-08T05:00:37 |
368619
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"A rural reader",
"Alex M.",
"Carlo Beenakker",
"Fei Cao",
"Willie Wong",
"https://mathoverflow.net/users/11260",
"https://mathoverflow.net/users/14069",
"https://mathoverflow.net/users/163454",
"https://mathoverflow.net/users/3948",
"https://mathoverflow.net/users/54780"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631872",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368619"
}
|
Stack Exchange
|
Modified energy method for transformed Fokker-Planck equation (tricky integration by parts…)
I came across Villani's paper titled "Hypocoercive diffusion operators" and couldn't figure out a computation that is skipped in that paper. Specifically, consider the following transformed Fokker-Planck equation, where $h(t,x,v)$ is the unknown, $(x,v) \in \mathbb{R}^n \times \mathbb{R}^n$, $V(x)$ is some potential force: $$\partial_t h + v\cdot \nabla_x h - \nabla V(x)\cdot \nabla_v h = \Delta_v h - v\cdot \nabla_v h.$$ Notice that the Laplacian $\Delta_v$ is only a partial Laplacian in the sense that it only acts on the velocity variables $v$, and for the usual $L^2$ energy $\int h^2 d\mu$, where $d\mu = f_\infty(x,v) dxdv$ and $f_\infty(x,v) = \frac{\mathrm{e}^{-\left(V(x)+\frac{|v|^2}{2}\right)}}{Z}$ with $Z$ a normalization constant making $f_\infty$ a probability density in $(x,v) \in \mathbb{R}^n \times \mathbb{R}^n$, and we easily have $\frac{1}{2} \frac{d}{dt} \int h^2 d\mu = -\int |\nabla_v h|^2 d\mu$. Then the author says under suitable assumptions on $V$, we can find suitable constants $a,c, K>0$ so that $$\frac{d}{dt}\left(\int h^2 d\mu + a\int |\nabla_x h|^2 d\mu + c\int |\nabla_v h|^2 d\mu \right) \leq -K\left(\int |\nabla_v h|^2 d\mu + \int |\nabla_v\nabla_x h|^2 d\mu + \int |\nabla_v\nabla_v h|^2 d\mu\right). $$ However, I have no clue why the above inequality holds (and justifying it in 1D should be enough for me, i.e., in the case $(x,v) \in \mathbb{R}\times\mathbb{R}$). What I did is the following (in 1D setting). Set
$$I(t):=\left(a\int |\nabla_x h|^2 d\mu + c\int |\nabla_v h|^2 d\mu \right).$$
Then
\begin{align*}
\frac 12\frac{dI}{dt} &= -a\int |\partial_v\partial_x h|^2 d\mu - c\int |\partial_v\partial_v h|^2 d\mu - c\int |\partial_v h|^2 d\mu\\
&\quad \color{red}{+ a\int \partial_x h \partial_x\left(V'(x)\partial_v h\right) - v\partial_xh\partial_{xx}h~d\mu} \\
&\quad \color{red}{+c\int V'(x)\partial_vh\partial_{vv}h - \partial_vh\left(\partial_x h+v\partial_v\partial_xh\right)~d\mu}
\end{align*}
But I have no clue as to the treatment of the terms in red. Any help would be greatly appreciated!
Edit: I have also asked this question on Math Stack Exchange (the link is https://math.stackexchange.com/questions/3782421/modified-energy-method-for-transformed-fokker-planck-equation-tricky-integratio) and no satisfying answer is given as well.
Multiply through by $h$, integrate, and use integration by parts.
@Aruralreader Can you elaborate more? For instance, multiply "who" through $h$?
There is a worked-out proof in page 10 and following of Hérau's lecture notes. The detailed steps are for $n=1$, $V=0$, but I assume once that is understood, the more general case would follow smoothly.
As a short-hand notation we write $ \|\partial_x h \|^2=\int (\partial h/\partial x)^2\,d\mu$ and $\langle\partial_x h,\partial_v h\rangle=\int (\partial h/\partial x)(\partial h/\partial v)\,d\mu$. We will make use of the identities
$$\langle f,\partial_v g\rangle=\langle(-\partial_v+v)f,g\rangle,$$
$$\langle v\partial_x f,f\rangle=0$$
$$[\partial_v ,v\partial_x]=\partial_x,$$
and the Cauchy-Schwartz + Young inequality
$$2|c\langle\partial_v f,\partial_x f\rangle|\leq 2c \|\partial_v f \|\, \|\partial_x f \|\leq c^2 \|\partial_v f \|^2+ \|\partial_x f \|^2.$$
The Fokker-Planck equation for $n=1$, $V=0$ reads
$$\partial_t h+Lh=0,\;\;L=v\partial_x + (-\partial_v+v)\partial_v .$$
The adjoint of $L$ is
$$L^\ast=-v\partial_x + (-\partial_v+v)\partial_v.$$
The resulting derivatives are
$$-\frac{1}{2}\frac{d}{dt} \|h \|^2= \|\partial_v h \|^2,$$
$$-\frac{1}{2}\frac{d}{dt} \|\partial_x h \|^2= \|\partial_v\partial_x h \|^2,$$
$$-\frac{1}{2}\frac{d}{dt} \|\partial_v h \|^2=\langle\partial_x h,\partial_v h\rangle+ \|(-\partial_v +v)\partial_v h \|^2=\langle\partial_x h,\partial_v h\rangle+ \|\partial_v^2 h \|^2+ \|\partial_v h \|^2.$$
So the derivative $dI/dt$ in the OP reduces to
$$\frac{dI}{dt}=-2a \|\partial_v\partial_x h \|^2-2c\biggl(\langle\partial_x h,\partial_v h\rangle+ \|\partial_v^2 h \|^2+ \|\partial_v h \|^2\biggr)$$
$$\qquad\leq -2a \|\partial_v\partial_x h \|^2-2c \|\partial_v^2 h \|^2-(2c-c^2) \|\partial_v h \|^2+ \|\partial_x h \|^2.$$
It remains to bound $ \|\partial_x h \|^2$.
In the lecture notes they do this by adding the mixed term $b\langle\partial_x h,\partial_v h\rangle$ to the left-hand-side of the inequality, which then gives a term $-b \|\partial_x h \|^2$ on the right-hand-side to dominate. Let me work that out, using the derivative$^{\ast}$
$$\frac{d}{dt}\langle\partial_x h,\partial_v h\rangle=- \|\partial_x h \|^2-2\langle\partial_v\partial_x h,\partial_v\partial_v h\rangle-\langle\partial_x h,\partial_v h\rangle,$$
hence
$$\frac{d}{dt}J\equiv\frac{d}{dt}\biggl( \|h \|^2+a \|\partial_x h \|^2+b\langle\partial_x h,\partial_v h\rangle+c \|\partial_v h \|^2\biggr)=$$
$$\qquad=-2(c+1) \|\partial_v h \|^2-2a \|\partial_v\partial_x h \|^2-b \|\partial_x h \|^2-2c\|\partial_v\partial_v h\|^2-2b\langle\partial_v\partial_x h,\partial_v\partial_v h\rangle-(b+2c)\langle\partial_x h,\partial_v h\rangle.$$
The first four terms on the right-hand-side have a fixed sign, the last two terms can be bounded by the Cauchy-Schwartz + Young inequality.
$^\ast$ The result for $(d/dt)\langle\partial_x h,\partial_v h\rangle$ given on page 10 of the cited lecture notes is mistaken. Here is a derivation:
\begin{align}
\frac{d}{dt}\langle\partial_x h,\partial_v h\rangle&=-\langle\partial_x Lh,\partial_v h\rangle-\langle\partial_x h,\partial_v Lh\rangle\\
&=-\langle\partial_x h,(L^\ast\partial_v +\partial_v L)h\rangle,\\
L^\ast\partial_v +\partial_v L&=\bigl(-v\partial_x+(-\partial_v+v)\partial_v\bigr)\partial_v+\partial_v\bigl(v\partial_x+(-\partial_v+v)\partial_v\bigr)\\
&=\partial_x+\partial_v+2(-\partial_v+v)\partial_v\partial_v,\\
\Rightarrow\frac{d}{dt}\langle\partial_x h,\partial_v h\rangle&=-\|\partial_x h\|^2-\langle\partial_x h,\partial_v h\rangle-2\langle\partial_x h,(-\partial_v+v)\partial_v\partial_v h\rangle\\
&=-\|\partial_x h\|^2-\langle\partial_x h,\partial_v h\rangle-2\langle\partial_v\partial_x h,\partial_v\partial_v h\rangle.
\end{align}
Thanks! But my problem is to derive inequality (11) in details. (If you read my problem statement carefully...)
Thanks! I will digest them and examine in details of your calculation, I will finish this within today
The only difference introducing the $V$ term is that $dI / dt$ includes additionally the term $+2a \int \partial_x h \partial_v h V'' ~d\mu$; if you have good control on the Hessian of $V$ then this term can be treated in the same way as the $-2c \langle \partial_x h, \partial_v h\rangle$ term.
@CarloBeenakker Actually I have several questions. (1) I do not see why $\langle v\partial_x f,f\rangle=0$ (2) the meaning of $[\partial_v v,\partial_x]=1$ is not clear (3) for $ n = 1$ and $V = 0$ the PDE reads as $\partial_t h+v\partial_x h + v\partial_v h - \Delta_v h=0$, not $\partial_t h+v\partial_x h+(-\partial_v+v)h=0.$
(1) $\langle v\partial_xf,f\rangle=-\langle f,v\partial_x f\rangle=\langle f,v\partial_x f\rangle\Rightarrow \langle v\partial_x f,f\rangle=0$; (2) $[\partial_v v,\partial_x]\equiv \partial_v v\partial_x-\partial_x\partial_v v =\partial_x$; (the "1" was a typo); (3) that was also a typo; there may be more typo's, it's quite a lengthy calculation.
@CarloBeenakker sorry I know think your answer to (1) is correct, it is not well-explained somehow
sorry, I don't understand the calculations $[\partial_v v,\partial_x]\equiv \partial_v v\partial_x-\partial_x\partial_v v =\partial_x$, $\partial_v v = 1$ right?
@CarloBeenakker can you add more details about the computation behind $\frac{d}{dt}\langle\partial_x h,\partial_v h\rangle=-||\partial_x h||^2+2\langle(-\partial_v+v)\partial_v h,\partial_v\partial_x h\rangle-\langle\partial_x h,\partial_v h\rangle$?
@FeiCao: I don't know why you have performed that edit, because there was no typo there: the two formulas (before and after the edit) are equal.
@AlexM. Originally CarloBeenakker wrote the wrong PDE
@FeiCao: Carlo Beenakker had written $(-\partial_v+v)\partial_v h$ and you have changed that into $v\partial_v h - \partial_v\partial_v h$, which is a bit useless given that the two are equal. This has nothing to with the PDE being right or wrong.
@AlexM. Ok now I see your points...
@CarloBeenakker If you don't mind, I really want to ask your the question on the computation of $\frac{d}{dt}\langle\partial_x h,\partial_v h\rangle$, in page 10 of the reference notes, after the sentence "The time derivative of the third term can be computed as follows", I cannot get the same expression as the author, my last term is $\langle -\partial_v\partial_v h,~\partial_x\partial_v h\rangle$ and not $\langle (-\partial_v + v)\partial_v h,~\partial_x\partial_v h\rangle$ and I cannot correct this part...
@CarloBeenakker Thanks! A small request as well: please be careful about the computations and type-ohs, your single type-oh may waste more than an hour of your reader, that's why I always triple triple triple check the stuff I wrote, I personally never want my reader to find a mistake or type-oh that will waste their time. Best.
you are right, that is an error in Hérau's lecture notes; I presume it does not break the inequality, but you'll want to check that.
@CarloBeenakker thanks! i checked the computations and now it is correct. A final question, the author of the notes mentioned a use of the Poincare inequality in space and velocity, how does that inequality work (it's my first time encountering Poincare inequality in multiple-variables...)?
Poincaré inequality: the analysis in the lecture notes applies to functions of zero mean in both $x$ and $v$, then the left-hand-side of the Poincaré inequality, the variance with respect to $x$ or $v$, is just the $L_2$ norm; the right-hand-side then bounds that by $|h|^2\leq c_x |\partial_x |^2$ and $| h|^2\leq c_v|\partial_v|^2$; Hérau sets $c_v=1$ (unit spectral gap) and denotes $c_x\equiv c_p$; on page 11 these two inequalities are applied, I would think $c_p$ should be $c_p+1$, but that only makes the inequality stronger.
|
2025-03-21T14:48:31.717013
| 2020-08-08T06:14:18 |
368622
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Antoine Labelle",
"GraphX",
"M. Winter",
"Thomas Sauvaget",
"https://mathoverflow.net/users/108884",
"https://mathoverflow.net/users/160416",
"https://mathoverflow.net/users/161328",
"https://mathoverflow.net/users/469"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631873",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368622"
}
|
Stack Exchange
|
Topological objects associated to Steinerberger's 4-regular graphs
Very recently, in arXiv:2008.01153, Steinerberger has associated to any sequence $(x_n)_{n\in\mathbb{N}}$ of distinct real numbers a 4-regular graph.
In the case irrational multiples, like $x_n=n\sqrt{2} \pmod{1}$, the plots in $\mathbb{R}^2$ seem to show the projection of a certain genus-g surface (see page 2 of the preprint). [edit:06-sept-2020: I had written that these were plots in $\mathbb{R}^3$, which is actually not the case, apologies.]
is that indeed the case, i.e. does a limit shape as $n$ goes to infinity exist ? What type of literature (e.g. keywords, theorems) one should be looking at to establish it ?
Isn't the paper saying that these surfaces are generated by the "van der Corpus sequence" rather than anything with $\sqrt 2$?
For the plots shown, yes, but it is written that it also happens for these other sequences. I am intertested by any sequence producing a limit space with that construction.
@ThomasSauvaget I think Béart answer points to the right field. More specifically, I think the area where graph theory intersects with differential topology is relevant for you. Not directly related, but to give you an idea of this area and the machinery/techniques, look at the references in this post https://mathoverflow.net/q/368129/161328
How do you embed the graph in $\mathbb{R}^3$?
@AntoineLabelle : thanks for the question, I now realise that these are actually plots of the graph in the plane (I'll edit the question). It still suggests that this is the projection of some higher dimensional manifold.
@ThomasSauvaget I still don't get how you canonically embed the graph in the plane.
The Steinerberger article seems to be aimed at tests about randomness. The question you are raising, which is given as an initial observation in this article, is a topic in topological graph theory.
The following link should give you helpful references for your question: Reference for topological graph theory (research / problem-oriented) .
Yes, I actually work in stats and was curious about the randomness test, but the geometric plot is hard to ignore. Thanks for the link.
Here is a short Mathematica script that computes the graph and plots it with some standard function
f[n_] := Mod[n * Sqrt[2]//N, 1];
n = 200;
seq = f /@ Range[1,n];
map = PositionIndex[seq];
sort = map[#][[1]] & /@ (Sort@seq);
edge1 = Partition[Range[1,n], 2, 1] ~ Join ~ {{n,1}};
edge2 = Partition[sort, 2, 1] ~ Join ~ {{sort[[-1]], sort[[1]]}};
G = Graph[Join[edge1, edge2]]
GraphPlot3D[G, GraphLayout->"SpectralEmbedding"]
GraphPlot3D[G, GraphLayout->"SpringElectricalEmbedding"]
It seems to resemble some kind of genus 1 surface.
But seems to have nothing to do with $\sqrt2$. If I replace $\sqrt 2$ with $\pi$, the result still looks like a torus:
Apparently, all we need is that the number is irrational.
Thanks for this code and plots, food for thought!
|
2025-03-21T14:48:31.717235
| 2020-08-08T06:16:04 |
368623
|
{
"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:631874",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368623"
}
|
Stack Exchange
|
Recover approximate monotonicity of induced norms
Let $A$ some square matrix with real entries.
Take any norm $\|\cdot\|$ consistent with a vector norm.
Gelfand's formula tells us that $\rho(A) = \lim_{n \rightarrow \infty} \|A^n\|^{1/n}$.
Moreover, from [1], for a sequence of $(n_i)_{i \in \mathbb{N}}$ such that $n_i$ is divisible by $n_{i-1}$, we also know that the sequence $\|A^{n_i}\|^{1/n_i}$ is monotone decreasing and converges towards $\rho(A)$. I am interested in what happens when this divisibility property is not verified.
If the matrix has non-negative entries, it seems the general property holds: For integers $n$ and $m$ such that $m > n$, it is the case that $\|A^m\|^{1/m} \leq \|A^n\|^{1/n}$.
If the matrix can have positive and negative entries, this more general observation does not seem to hold. I am trying to understand why it fails, how worse can the inequality become, and if it is possible to recover an inequality up to some function of $A$: $\|A^m\|^{1/m} \leq f(A)\cdot\|A^n\|^{1/n}$.
Any references to 1., or pointers for understanding 2. would be much appreciated.
[1] Yamamoto, Tetsuro. "On the extreme values of the roots of matrices." Journal of the Mathematical Society of Japan 19.2 (1967): 173-178.
Note that if (1) holds, then it holds also for all matrices that are similar to one with non-negative entries (since $v \mapsto |Xv|$ is a norm).
This is not a complete answer: If you allow positive and negative entries then this monotonicity will not hold in general. Consider
$$
A = \left[\begin{matrix} 0 & 1 & -1 & 0 & 0 \\ 0& 0 & 1&1 & 1 \\ 0 & 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 0 & -1 \\ 0& 0&0 &0&0 \end{matrix}\right]
$$
then
$$
A^2 = \left[\begin{matrix} 0 & 0 & 1 & 0 & 0 \\ 0& 0 & 0&1 & 0 \\ 0 & 0 & 0 & 0 & -1 \\ 0 & 0 & 0 & 0 & 0 \\ 0& 0&0 &0&0 \end{matrix}\right] \ \ \ \textrm{and} \ \ \
A^3 = \left[\begin{matrix} 0 & 0 & 0 & 1 & 1 \\ 0& 0 & 0&0 & -1 \\ 0 & 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 & 0 \\ 0& 0&0 &0&0 \end{matrix}\right].
$$
Thus, $\|A^3\|^{1/3} > 1= \|A^2\|^{1/2}$.
|
2025-03-21T14:48:31.717388
| 2020-08-08T07:00:03 |
368624
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Christian Clason",
"SS342",
"https://mathoverflow.net/users/147309",
"https://mathoverflow.net/users/30516"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631875",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368624"
}
|
Stack Exchange
|
the subdifferential at points of differentiability in infinite dimensional space
Let $ f: X\to (-\infty,+\infty]$ that $ X$ is an infinite dimensional space.
What are the conditions for $f$ and space $X$ to have the following equality correct?
$$\partial f(x)=\{\nabla f(x)\}$$ for all $x\in X$
I know when space X is finite dimensional and $f$ is proper convex function then
$$\partial f(x)=\{\nabla f(x)\}$$ for all $x\in X$
Thanks.
That depends on what you mean by $\nabla f$. If $X$ is a normed vector space and $f:X\to \mathbb{R}$ is Gâteaux differentiable, then $\partial f(x) = \{DF(x)\}$ for the Gâteaux derivative $DF(x)\in X^*$ of $F$ at $x$; the proof is virtually identical to the finite-dimensional one using the definition of the Gâteaux derivative and the characterization of the convex subdifferential using the directional derivative (shameless plug: Theorem 4.5 of https://arxiv.org/abs/1708.04180).
If $\nabla F(x)\in X$ is the gradient, then this of course requires $X$ to be a Hilbert space.
Christian Clason $\nabla f(x)$ is the gradient. Is the proof similar in this case?
The proof is the same, you just need to use the Riesz representation theorem to identify $DF(x)\in X^*$ with $\nabla F(x)\in X$; this is standard and not at all specific to convex analysis.
Thanks. I saw (shameless plug: Theorem 4.5 of https://arxiv.org/abs/1708.04180). it is very useful.
Can you introduce another useful resource in the field of optimization?
Well, that depends on what specifically you're interested in. If it's first-order algorithms, there's an expanded version https://arxiv.org/abs/2001.00216 (due for an update soon). The Preface there has a good number of references to textbooks and monographs in this area. I think the books by Schirotzek and Bauschke/Combettes in particular are very good.
It's great. Thanks
|
2025-03-21T14:48:31.717535
| 2020-08-08T07:34:31 |
368626
|
{
"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:631876",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368626"
}
|
Stack Exchange
|
Best bound of complex Hilbert transform
It is well-known (see Grafakos' Classical Fourier Analysis, Exercise 5.1.12) that if $f$ is a real valued $L^p(\mathbb R)$ function and $1<p<2$ , then we have the following inequality:
$$
\|Hf\|_{L^p(\mathbb R)}\leq A_p\|f\|_{L^p(\mathbb R)},
$$
where $A_p=\tan \frac {\pi}{2p}$. Moreover, this bound is sharp.
A nice and elegant proof and also be found in https://faculty.missouri.edu/~grafakosl/preprints/pichorides.pdf.
However, I was wondering if this bound is also true for complex valued functions $f$. The proof above used the fact that if $f$ is real valued, then the real part and the imaginary part of the integral
$$
\frac i \pi\int_{-\infty}^\infty \frac {f(t)}{z-t}dt
$$
converges to $f(x)$ and $Hf(x)$ respectively, as $y\to 0^+$, where $z=x+iy$. For complex valued $f$, this is no longer true.
I have found the answer myself; it is a result of Theorem 5.5.1 of Grafakos’ Classical Fourier Analysis. The result we need is that if a linear operator maps real valued functions to real valued functions and has real L^p-L^p operator norm bounded by A, then its complex operator norm is also bounded by A.
|
2025-03-21T14:48:31.717648
| 2020-08-08T08:34:43 |
368628
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Derek Holt",
"Francesco Polizzi",
"Ian Agol",
"Moishe Kohan",
"https://mathoverflow.net/users/1345",
"https://mathoverflow.net/users/35840",
"https://mathoverflow.net/users/39654",
"https://mathoverflow.net/users/7460"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:631877",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/368628"
}
|
Stack Exchange
|
Isomorphism of semidirect products of surface groups
Recall that the fundamental group of a closed Riemann surface of genus $h$ has the presentation $$\Pi_h= \langle a_1, \,b_1, \ldots, a_h,\, b_h \; | \; [a_1, \, b_1]\ldots [a_h, \, b_h]=1 \rangle.$$ Assume now that we have two split short exact sequences of groups
\begin{equation*}
\begin{split}
& 1 \longrightarrow \Pi_g \longrightarrow G_1 \longrightarrow \Pi_b \longrightarrow 1 \\
& 1 \longrightarrow \Pi_g \longrightarrow G_2 \longrightarrow \Pi_b \longrightarrow 1 \\
\end{split}
\end{equation*}
Question. Is there a theoretical or computational way to check whether $G_1$ and $G_2$ are isomorphic or not?
Note that I'm talking about abstract isomorphism of the middle groups, not of isomorphism of sequences. In my specific situation (where $g=41$ and $b=2$), I know the conjugacy action of $\Pi_b$ on $\Pi_g$ in both cases, so I can obtain explicit semidirect-type presentations for $G_1$ and $G_2$ and I can feed them to GAP4.
In this way, I can check that $G_1$ and $G_2$ have the same abelianization. But the problem about their isomorphism type eludes me for the moment.
Every answer or reference to the relevant literature will be greatly appreciated.
Computationally, since the groups have solvable word problem, it is possible to check whether a sequence of generator images induces an isomorphism, so one can search for an isomorphism. One can also compute varionus invariants (such as the abelian invariants) and try and prove that they are not isomorphic. Both of those processes are much more difficult when there are large numbers of generators, which appears to be the case in this example.
@DerekHolt: Thank you for the suggestions. Actually, as I said, in my example the abelianizations (and so the abelian invariants) are the same in both cases. As you remarked, there is a large number of generators, so trying to construct an isomorphism by hand seems to be difficult (at least, for me).
Nick Salter has found examples of surface-by-surface groups that fiber in different ways. However, I don't know if these are split.https://msp.org/gt/2015/19-5/p10.xhtml
It is likely to be undecidable for the same reason that the braid factorization is undecidable, check https://arxiv.org/pdf/math/0511153.pdf
If the action of $\Pi_b$ on $\Pi_g$ via conjugation in $G$ has non-trivial
kernel and infinite image in the outer automorphism group $Out(\Pi_g)$ then the extension structure for $G$ is unique.
If the image of $\Pi_b$ in $Out(\Pi_g)$ is finite then $G$ is virtually a product, and there are at most two such extension structures. (This is clear when $G\cong\Pi_g\times\Pi_b$ is a product!)
Otherwise, if the action is injective there are finitely many
such extension structures.
These results are in (or flow from) ``A group-theoretic analogue of the Parshin-Arakelov Theorem", by F. E. A. Johnson, Archiv Math. (Basel) 63 (1994), 354-361.
Let $d=(2-2g)(2-2b)$ be the Euler characteristic of $G$. Johnson's work implies an upper bound of the order of $d^d$ on the number of such extension structures
when the action is injective. On the other hand, Nick Salter showed that the number of such extension structures can be exponential in $d$.
This is very interesting, thank you.
|
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.