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2025-03-21T14:48:32.049106
| 2020-09-17T06:52:30 |
371877
|
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"Alec Rhea",
"Emil Jeřábek",
"bof",
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}
|
Stack Exchange
|
Consistency strength of iterated classes
Adding classes into a set theory like ${\bf ZFC}$ to get a theory like ${\bf MK}$ adds some consistency strength, but less than even a single inaccessible cardinal since $\kappa$ being inaccessible means $(V_\kappa,\in,V_{\kappa+1})$ is a model of ${\bf MK}$.
It's pointed out by Andreas Blass in the answer to this MO question that even iterating this process infinitely, with sets followed by classes followed by hyper classes followed by..., yields more consistency strength than ${\bf MK}$ but still less than a single inaccessible, which presumably means that the above construction can be tweaked to model this situation. He also mentions that the intuition for this situation is closer to an inaccessible than ${\bf MK}$, but still weaker.
I'm curious about the consistency strength of the following situation. Let us refer to sets as $0$-sets, classes as $1$-sets, hyperclasses as $2$-sets, and so on. For each pair of ordinals $\beta\leq\alpha$, let ${\bf ZFC}_{\alpha,\beta}$ be the theory obtained by adding the existence of $\gamma$-sets for all $\gamma<\alpha$ plus full comprehension on $\gamma$ sets for all $\gamma<\beta$ to ${\bf ZFC}$. The first index indicates how far up the collection hierarchy we are, and the second index indicates how far up comprehension applies; for example, $${\bf ZFC}_{2,2}={\bf MK},$$ $${\bf ZFC}_{2,1}={\bf GBC},$$ $${\bf ZFC}_{1,1}={\bf ZFC}.$$
It is unclear to me if, when Andreas says to 'continue this process indefinitely', he is talking about ${\bf ZFC}_{\omega,\omega}$ or ${\bf ZFC}_{O_n,O_n}$. Assuming the former, my question is:
Does there exist a large cardinal axiom $\phi$ and ordinals $\alpha,\beta$ such that the consistency strength of ${\bf ZFC}_{\alpha,\beta}$ is greater than or equal to the consistency strength of ${\bf ZFC}+\phi$?
If the latter is true, that ${\bf ZFC}_{\alpha,\beta}$ is weaker in consistency than a single inaccessible for all $\alpha,\beta$, my question is:
Are there other known 'consistency strength axioms' $\phi$ such that ${\bf ZFC}+\phi$ is equivalent in consistency strength to ${\bf ZFC}_{\alpha,\beta}$?
Joel Hamkins mentions in this MO answer that these iterated classes can be coded by well-founded class relations on the ordinals, and there is a PhD dissertation by Kameryn Williams exploring this encoding process and closely related topics including the consistency strength of ${\bf GBC}$ plus various transfinite recursion principles yielding theories with a consistency strength between ${\bf GBC}$ and ${\bf MK}$. The ${\bf ZFC}_{\alpha,\beta}$ hierarchy jumps from ${\bf GBC}$ to ${\bf MK}$ immediately, though, and is thusly adding more consistency strength than the transfinite recursion principles investigated in the Williams paper.
If you are treating classes as objects and forming classes of classes etc., doesn't that mean that your so-called "proper classes" are really just sets?
@bof There are no classes of classes, only hyperclasses of classes. In general there are no $\alpha$-sets containing other $\alpha$-sets, only $\alpha$-sets containing $\beta$-sets for $\beta<\alpha$.
@bof Actually we do have $0$-sets (regular sets) containing other sets, but anywhere farther up the hierarchy $x$ being a member of an $\alpha$-set implies that $x$ is a $\beta$-set for some $\beta<\alpha$.
It’s not exactly clear how you formulate $\mathrm{ZFC}{On,On}$, but a single inaccessible should be enough for that: let the collection of $\alpha$-sets be $V{\kappa+\alpha}$. Or am I missing something?
@EmilJeřábek That looks correct, thank you; indexing with $O_n$ would mean we use $V_{\kappa+\alpha}$ for all $\alpha\in O_n$ and allow comprehension over all of them, which would be the new full universe with an inaccessible correct?
|
2025-03-21T14:48:32.049352
| 2020-09-17T07:12:25 |
371880
|
{
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"Sasha",
"https://mathoverflow.net/users/41650",
"https://mathoverflow.net/users/4428",
"user41650"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371880"
}
|
Stack Exchange
|
Are two versions of Kuznetsov components equivalent?
Let $X$ be a degree $10$, genus $6$ index $1$ and Picard number $1$ smooth Fano threefold. There are two versions of Semi-orthogonal decompositions. The First version is $$D^b(X)=\langle\mathrm{Ku}(X),\mathcal{E}_X,\mathcal{O}_X\rangle$$ where $\mathcal{E}_X$ is the stable bundle of rank 2 with character $2-H+L+\frac{1}{3}P$, it is a pull back of tautological rank 2 sub-bundle of $\mathrm{Gr}(2,5)$. The second version of semi-orthogonal decomposition is $$D^b(X)=\langle\mathcal{A}_X,\mathcal{O}_X,\mathcal{E}_X^{\vee}\rangle$$ where $\mathcal{E}_X^{\vee}$ is the dual vector bundle of $\mathcal{E}_X$. I understand that both of them are coming from full exceptional collections of vector bundles on $Y\rightarrow\mathrm{Gr}(2,5)$ where $Y$ is a codimension 3 linear section. $$D^b(Y)=\langle\mathcal{U}_Y,\mathcal{O}_Y,\mathcal{U}_Y(1),\mathcal{O}_Y(1)\rangle.$$ This is a rectangular Lefschetz decomposition with $\mathcal{B}=\langle\mathcal{U}_Y,\mathcal{O}_Y\rangle$. Now since $X\rightarrow Y$ is either an embedding or a double cover, Then $D^b(X)$ TEMPhas the first type semi-orthogonal decomposition. If one consider the "rotate" of full exceptional (or by consecutively right mutations) $$D^b(Y)=\langle\mathcal{O}_Y,\mathcal{U}_Y(1),\mathcal{O}_Y(1),\mathcal{U}_Y(2)\rangle,$$ this is exactly $$\langle\mathcal{O}_Y,\mathcal{U}_Y^{\vee},\mathcal{O}_Y(1),\mathcal{U}_Y^{\vee}(1)\rangle$$ This is another rectangular Lefschetz decomposition with $\mathcal{B}'=\langle\mathcal{O}_Y,\mathcal{U}_Y^{\vee}\rangle$. Then from this, we TEMPhas $D^b(X)$ with the second type semi-orthogonal decomposition. By a result of A.Kuznetsov, either $\mathrm{Ku}(X)$ or $\mathcal{A}_X$ TEMPhas the same shape Serre functor, which is given by $\tau\circ[2]$, where $\tau$ is an involution functor, if $X$ is special, $\tau$ is induced by double cover.
I was wondering whether $\mathcal{A}_X$ is equivalent to $\mathrm{Ku}(X)$? I think they should be equivalent by several mutation functors (as in $Y_5$), but when I do right mutation $\mathrm{R}_{\mathcal{O}_X}\mathcal{E}_X$, I get a rank 3 vector bundle, I do not know how to proceed from here.
But it looks like the subcategory $\langle\mathcal{E}_X,\mathcal{O}_X\rangle$ is anti-equivalent to $\langle\mathcal{O}_X,\mathcal{E}^{\vee}\rangle$ by derived dual functor. So $\mathrm{Ku}(X)$ is anti-equivalent to $\mathcal{A}_X$?
The two categories are equivalent, see section 2.8 in A. Kuznetsov, A. Perry, "Derived categories of Gushel-Mukai varieties", Compositio Mathematica, V. 154 (2018), N. 7, pp. 1362--1406, available at http://www.mi-ras.ru/~akuznet/publications/2018-Derived%20categories%20of%20Gushel-Mukai%20varieties.pdf
@Sasha, thanks so much! That would be very nice! In fact, for the computation, I found that Ku(X) is better for special Gushel-Mukai while A_X is better for ordinary Gushel-Mukai in some sense. But they are equivalent, this is so nice.
|
2025-03-21T14:48:32.049639
| 2020-09-17T07:33:36 |
371883
|
{
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"url": "https://mathoverflow.net/questions/371883"
}
|
Stack Exchange
|
Partitioning an infinite cardinal $\kappa$ into pairwise neighboring subsets
We say that two disjoint, non-empty subsets $S, T$ of an infinite cardinal $\kappa$ are neighboring if there is $\alpha\in \kappa$ such that $$S\cap\{\alpha,\alpha+1\} \neq \varnothing \neq T\cap\{\alpha, \alpha+1\}.$$
Given an infinite cardinal $\kappa$, is there a partition ${\cal B}$ of $\kappa$ with $|{\cal B}|=\kappa$ and whenever $B_1\neq B_2 \in {\cal B}$ we have that $B_1, B_2$ are neighboring?
Yes. List the pairs $(\alpha,\beta)$ with $\alpha<\beta<\kappa$ as $(\alpha_\lambda,\beta_\lambda), \lambda<\kappa$.
Then construct the sets $B_\alpha\in\mathcal B, \alpha<\kappa$ as follows:
At stage 0, all $B_\alpha=\emptyset$.
At limit stages just take unions.
At successor stages $\lambda+2n, n\in\omega, n\ge 1$, choose the least $\lambda$ such that $(B_{\alpha_\lambda},B_{\beta_\lambda})$ are not yet neighboring. Put $\lambda$ into $B_{\alpha_\lambda}$ and $\lambda+1$ into $B_{\beta_\lambda}$.
|
2025-03-21T14:48:32.049734
| 2020-09-17T07:47:03 |
371886
|
{
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"authors": [
"Alexandre Eremenko",
"M.H.Hooshmand",
"https://mathoverflow.net/users/25510",
"https://mathoverflow.net/users/40520"
],
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371886"
}
|
Stack Exchange
|
General solution of a linear functional equation
As we know, general solution of the linear functional equation $f(x+1)-f(x)=g(x)$ ($g$ is a known function) is $f=f_0+\phi$, where $f_0$ is an its special solution and $\phi$ any 1-periodic function.
Now, does any one know general solution of the linear functional equation: $$rf(ax+b)+sf(cx+d)=g(x),$$ where $a,b,c,d, r$ and $s$ are real constants with $rasc\neq0$?
($g$ is a given real function, the
case $g=0$ and the equation $f(ax+b)=cf(x)$ is of special interest).
Are there any papers or books regarding the problem?
It is the same as you wrote in the first paragraph: the general solution is the sum of a single solution plus the general solution of homogeneous equation (with g=0).
As in the special case, the general solution is the sum of a particular solution and the general solution of the homogeneous equation.
To describe he general solution of the homogeneous equation
$$rf\phi_1+sf\circ\phi_2=0,$$
where $\phi_1(x)=ax+b$ and $\phi_2(x)=cx+d$ are affine functions, you use the fact that affine functions form a group. So there is an affine function $\phi_3$ such that
$\phi_2=\phi_3\circ\phi_1$, and making the change of the variable $t=\phi_1(x)$
we obtain the equation
$$rf(t)+sf\circ\phi_3(t)=0.$$
To further simplify this, consider two cases: a) $\phi_3(t)=at+b,\; a\neq 1$,
and b) $\phi_3(t)=t+b$. In the first case, by a conjugation in the affine group,
the equation is reduced to
$$rg(t)+sg(at)=0.$$
In the second case, it is reduced to
$$rg(t)+sg(t+1)=0.$$
The general solutions of these two standard equations must be clear.
Thanks. Can one obtain an explicit formula for the general solution from the above form?
@M.H.Hooshmand: yes, of course. This is a simple exercise.
|
2025-03-21T14:48:32.049865
| 2020-09-17T08:14:26 |
371892
|
{
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],
"authors": [
"Ilya Bogdanov",
"Mastrem",
"YCor",
"https://mathoverflow.net/users/14094",
"https://mathoverflow.net/users/165478",
"https://mathoverflow.net/users/17581"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/371892"
}
|
Stack Exchange
|
Fibonacci-like sequences in $\mathbb{F}_q$ where each element only depends on the previous one
Given a prime power $q$, consider all sequences $(a_n)_{n\in\mathbb{Z}}$ in $\mathbb{F}_q$ for which $a_{n+1}=a_n+a_{n-1}$ for all $n\in\mathbb{Z}$. Call such a sequence simple if there exists a function $f:\mathbb{F}_q\to\mathbb{F}_q$ such that $a_{n+1}=f(a_n)$ for all $n\in\mathbb{Z}$.
There are some trivial simple sequences. The null sequence is simple, as is $(cr^n)_{n\in\mathbb{Z}}$ for $c\in\mathbb{F}_q^*$ and $r$ a root of $X^2-X-1$. My questions are about nontrivial simple sequences.
I've asked a more specific version of this question on Math.Stackexchange. There, computations by the user @Servaes show that nontrivial simple sequences exist in $\mathbb{F}_p$ for $p\in\{199,211,233,281,421,461,521,557,859,911\}$
Questions:
Are there 'easy' conditions on primes $p$ such that no nontrivial simple sequences exist in $\mathbb{F}_p$ when $p$ satisfies these conditions? (and there are a large number of primes satisfying these conditions)
Are there infinitely many primes $p$ such that nontrivial simple sequences exist in $\mathbb{F}_p$?
Given a prime $p$, does there always exist a positive integer $n$ such that nontrivial simple sequences exist in $\mathbb{F}_{p^n}$?
In case the answer to the previous question is affirmative, let $n(p)$ be the smallest such positive integer. Is $n(p)$ bounded? If not, do there exist integers $m$ such that $n(p)=m$ for infinitely many primes?
I'm not sure there's a specific reason to stick to finite fields. If the characteristic of the field $F$ is $5$, the sequence $n\mapsto 2^n(a+bn)$ is non-trivial as soon as $a,b$ are nonzero with $b/a\notin\mathbf{F}_5$ (so exists as soon as $|F|> 5$).
(sorry, I meant $(-2)^n(a+bn)$; $-2$ being the double root of $x^2-x-1$ in char. 5)
$\def\ord{\mathop{\mathrm{ord}}}$Let $q=p^s$ for a prime $p$.
Let $\phi$ and $\psi$ be the roots of $X^2-X-1$; they may lie either in $\mathbb F_p$ (when $\left(\frac p5\right)=1$, call this case simple) or in $\mathbb F_{p^2}$. The case $\phi=\psi$, i.e. $p=5$, is covered by @YCor in the comments (1 2), so let us assume $\psi\neq \phi$. Notice that $\phi\psi=-1$.
The general form of a linear recurrence is then $a_n=a\phi^n+b\psi^n$; where $a,b\in\mathbb F_q$ if $\sqrt5\in\mathbb F_q$, and $a$ and $b$ are two conjugate elements in $K=\mathbb F_q[\sqrt5]$, otherwise (here, conjugate means that they are swapped by the nontrivial automorphism of $K$ over $\mathbb F_q$). Surely, this sequence is periodic with period $T=\ord \phi=\ord\psi$ (where $\ord$ means the multiplicative order in $\mathbb F_{p^2}$ which does not depend on $s$); so we need the terms $a_1,a_2,\dotsc,a_T$ to be distinct, while $a$ and $b$ are nonzero.
If two such terms are equal, we have
$$
a\phi^n+b\psi=a\phi^{n+k}+b\psi^{n+k}
\iff a\phi^n(\phi^k-1)=b\psi^n(\psi^k-1)
\iff \frac ba=\phi^{2n}(-1)^n\frac{\phi^k-1}{\psi^k-1}.
$$
For every prime $p$, the right-hand part attains finitely many values ($\leq T^2<p^4$), so, say, for $s=6$ there exist $a$ and $b$ which violate all equalities above and thus fit. This answers the third question.
Moreover, if the order $T$ of $\phi$ is relatively small comparative to $p$ (say, $T\leq \sqrt p$), then the required $a$ and $b$ will be found even in $\mathbb F_p$. But I am not sure whether this is a good condition to answer the second question.
A few more words on the fraction under consideration
$$
\phi^{2n}\frac{\phi^k-1}{\psi^k-1}.
$$
If, say, $\sqrt5\in\mathbb F_p$, and we want to have no desired sequence, we want this expression to take all values in $\mathbb F_p^*$. If $k$ is even, the expression is $-\phi^{k+2n}$, but for odd $k$ it is more complicated. If, say, $\phi$ is a generator of $\mathbb F_p^*$, then the whole $\mathbb F_p^*$ will be covered. Again, this is a condition for question 1, but it is too strong.
If $k$ is even, don't we have $\psi^k=(-\phi^{-1})^k = \phi^{-k}$ and $\frac{\phi^k-1}{\psi^k-1}=\frac{\phi^k-1}{\phi^{-k}-1}=\phi^k\cdot \frac{\phi^k-1}{1-\phi^k}=-\phi^k?$
@Mastrem: Oh, right! I was too fast... will try to correct now. Thanks!
|
2025-03-21T14:48:32.050142
| 2020-09-17T08:24:06 |
371893
|
{
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"authors": [
"George K",
"Igor Belegradek",
"Moishe Kohan",
"YCor",
"https://mathoverflow.net/users/14094",
"https://mathoverflow.net/users/1573",
"https://mathoverflow.net/users/159356",
"https://mathoverflow.net/users/39654",
"https://mathoverflow.net/users/40804",
"mme"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/371893"
}
|
Stack Exchange
|
Spherical space form conjecture
The spherical space form conjecture (now theorem) asserts that every finite group acting on the 3-sphere is conjugate to a subgroup of $\mathrm{SO}(4)$. Is the action assumed to be smooth? Can anything similar be said for topological actions? Also, I would appreciate a reference about the spherical space form conjecture.
I think it's been known for a while (long before Hamilton–Perelman and even Thurston) that in dimension 3 the smooth and topological Poincaré conjectures are equivalent. Unfortunately I don't even seen this mentioned in Wikipedia's Poincaré conjecture page. Hopefully somebody will provide a reference and more precise statement.
The SFS conjecture is about free finite group actions. If you drop the freeness assumption then in the smooth category every finite group action is still conjugate to an orthogonal action, but in the topological category this is no longer true (Bing's examples).It is a rather old theorem (Moise+Munkres) that in dimension 3 TOP=DIFF: every topological manifold admits a smooth structure and this structure is unique up to diffeomorphism. See the references I gave here.
@MoisheKohan Thank you. Does that mean that the spherical space form conjecture holds for topological, non-smooth actions over S^3 as well?
No. To rephrase Moishe: If your action is free, then it is topologically conjugate to a linear action (smoothly conjugate if the action was smooth to begin with). Non-free topological actions need not be smoothable. Eg, the double of the (non-manifold) Alexander horned ball is homeomorphic to $S^3$, and hence $S^3$ carries an involution with non-locally-flat fixed set, so this action is not even topologically conjugate to a smooth action.
I wonder if every locally linear (not necessarily free) action of a finite group on $S^3$ is conjugate to a linear action. The Ricci flow proof seems to fail in locally linear setting.
@IgorBelegradek Yes, one can prove that topological 3d orbifolds are smoothable. I am not sure if this is in the literature though. But it should be a good separate question.
@GeorgeK: Do you understand that your question is based on a false premise?
@Moishe Kohan I do. I hoped that the fact that the topological and the smooth category coincide in 3 dimensions might yield something good, so I thought to ask. Meanwhile, I found a few papers on the matter and I see that this is not the case. You only have that smooth actions are smoothly conjugate to subgroups of O(4) and that locally linear actions are topologically conjugate to subgroups of O(4) (and O(4) of course further reduces to SO(4) for actions which preserve orientation). At least these are the only results that I found. Thank you all again.
|
2025-03-21T14:48:32.050336
| 2020-09-17T09:22:41 |
371895
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Yangong Wu",
"https://mathoverflow.net/users/153484"
],
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"url": "https://mathoverflow.net/questions/371895"
}
|
Stack Exchange
|
How to prove this integral inequality in a 2-D region?
Let $\Omega$ be a 2D region. Now we have a partial differential equation system describing the characteristics of the region:
\begin{align*} \nabla \cdot (h_0^3 P_0 \nabla P_0) &= 0 \\ \nabla \cdot [h_0^3 \nabla (P_0 P_1)] &= 3 h_1 \nabla \cdot (h_0^2 P_0 \nabla P_0) \end{align*}
where $ P_0 $ is determined by a boundary condition, $ P_0>0 $, also, $P_1=0$ on the boundary $\partial \Omega$. $h_0$ is an arbitrary distribution function, $ h_0>0 $, while $h_1$ is a constant value and $h_1 \ll h_0$.
How to prove the following integral inequality:
$$ \frac{\iint P_1 \mathrm{d} \Omega }{h_1} < \iint \frac{ P_0 }{h_0} \mathrm{d} \Omega $$
$h_0$ is just an arbitrary function, nothing to do with Probability theory.
Can we prove this by constructing intermediate variables like this:
\begin{align}
\nabla \cdot (h_0^3 \nabla P_{0c}) &= 0 \
\nabla \cdot [h_0^3 \nabla ( P_{1c})] &= 3 h_1 \nabla \cdot (h_0^2 \nabla P_{0c} )
\end{align} and we try to prove the following integral inequality:
$$ \frac{\iint P_1 \mathrm{d} \Omega }{h_1}
< \frac{\iint P_{1c} \mathrm{d} \Omega }{h_1}
< \iint \frac{ P_{0c} }{h_0} \mathrm{d} \Omega
< \iint \frac{ P_0 }{h_0} \mathrm{d} \Omega $$
The middle P1c is less than P0c can be explained by physical meaning.
|
2025-03-21T14:48:32.050425
| 2020-09-17T10:25:07 |
371902
|
{
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"gmvh",
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"url": "https://mathoverflow.net/questions/371902"
}
|
Stack Exchange
|
block diagonal approximation of (SPD) matrix
I am interested in approximating a symmetric matrix in a block diagonal form, i.e. compute just some entries of the matrix located in blocks around the diagonal. Are there any theoretical guarantees about this approximation? I would be interested to approximate the difference in eigenvalues of the two matrices, i.e. the L2 norm of their difference.
My setting is in second order optimization, where we use the hessian of a function in order to minimize it. Since the problem can be very high-dimensional and the hessian very large, practitioners usually compute just a block-diagonal form of the hessian. However, I was not able to find any theory behind that. If it helps, we can also assume that our matrix is positive definite (happens when the function that we wish to minimize is strongly convex).
Any literature around this problem would be very helpful.
Is your matrix diagonally dominant? In that case, at least the positive-definiteness should not be spoiled by the approximation, because the approximating matrix just has fewer non-zero off-diagonal elements and hence is still diagonally dominant.
The Gershgorin Circle Lemma could be helpful in bounding how much the eigenvalues of the full and block-diagonal matrices can differ.
|
2025-03-21T14:48:32.050534
| 2020-09-17T12:01:35 |
371911
|
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"dohmatob",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/371911"
}
|
Stack Exchange
|
Lower-bound on smallest singular-value of rectangular random matrix
Let $X$ be a random $N \times n$ matrix with iid entries from $\mathcal N(0, 1)$ and with $n/N =: \lambda(N,n) \le \lambda_0$, for some $\lambda_0 \in (0, 1)$. That is, $X$ is genuinely rectangular (including the case where $X$ is an $N \times 1$ vector).
Question 1. Is it true that for every $C>0$, there exists $c,A>0$ (only depending on $C$ and $\lambda_0$) such that
$P(s_\min(X) \le c\sqrt{N}) \le Ae^{-CN}$ ? What about the particular case when $C=1$ ?
In case the answer to the above is negative,
Question 2. Find $A,c,C>0$ such that $P(s_\min(X) \le c\sqrt{N}) \le Ae^{-CN}$.
Note that the case of Rademacher entries is solved in Theorem 2.7.1 this document (by T. Tao).
Update
Question 1 (and therefore, Question 2) has an affirmative answer with $A=2$ and without any constraint on $\lambda$ under than $\lambda < 1$. See https://mathoverflow.net/a/372119/78539
The answer to your Question 2 is positive, in view of Theorem 3.1 and Fact 2.4 (for general subgaussian entries) in the paper by Litvak et al (referred to in the book by Tao that you linked).
Thanks. Concerning the Tao et al. paper reference, I was referring to Q2. I read the paragraph just before 3.4 of this paper https://www.math.uci.edu/~rvershyn/papers/rv-ICM2010.pdf (Rudelson and Vershynin), and the authors say one can have $C=1$ in the subGaussian case, more precisely they claim: "if $n/N<1$, then there exists $c>0$ such that $ P(s_\min(A) \ge c\sqrt{N}) \ge 1 - 2e^{-N}$. Is this a typo ?
OK, I see. I'm moving the discussion to a separate question https://mathoverflow.net/q/372015/78539. I really need the exponent to be $C$ (in order to absorb an entropy cost in the proof of something else I'm doing). Any input welcome!
Actually, Question 1 has an affirmative answer, $A=2$, for arbitrary $C>0$, and without any restriction on $\lambda$ other than $\lambda < 1$. See https://mathoverflow.net/a/372119/78539
|
2025-03-21T14:48:32.050695
| 2020-09-17T12:42:27 |
371913
|
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|
Stack Exchange
|
Discretizing a line segment with pixels which satisfies the Pythagorean theorem
There are plenty of line drawing algorithms to discretize line segments using pixels.
The Bresenham's algorithm gives a line where the number of pixels in the segment is the same as its width (in x-direction) or height (y-direction), whichever is largest.
One can also imagine an algorithm where one starts in one of the points, and
choose the lattice path between start and end point which minimizes total
distance squared of pixel centers to the true geometric line.
The number of pixels produced is the width+height, as we have a lattice path.
Note that the (geometric) length of the line segment is somewhere between the number of pixels produced by the two approaches above.
My question is, is there some (standard) algorithm where the number of pixels in the constructed line segment is equal to the the (rounded to nearest integer) length of the line segment? We want the line-segment to be connected, in the sense that every x-coordinate between the endpoints are covered by at least one pixel (and same for y-coordinates).
Of course, one can take the lattice path approach above, and iteratively remove pixels furthest from the true geometric line, but this seems inefficient, and might not guarantee connectednes.
I think this question makes sense if we extend the planar grid with a certain collection of diagonals, and require that the drawn segment between two points always use a shortest path in this graph. It was posed in this form by Pach, Pollack and Spencer 30 years ago, and it is still open.
I might not really understand the question, but one very simple-minded idea to plot a line segment using $n$ pixels is to find $n$ points evenly spaced along the line segment, and then replace each one with the nearest lattice point. As long as $n$ is greater than the horizontal distance plus one, successive points will be less than $1$ unit apart horizontally and therefore the result will be horizontally connected, and similarly for vertical connectedness.
Of course, if $n$ is too large then multiple points could be sent to the same pixel. But if $n$ is the length of the segment then non-adjacent points will be two units apart and cannot have the same closest lattice point. There could be adjacent pairs that go to the same pixel; in that case, if we really want exactly $n$ pixels we could adopt a rule for moving the second point. E.g., if the line goes from $(0,0)$ to $(a,b)$ with $0 \leq b \leq a$, then whenever the closest lattice point has already been taken, plot the pixel just above it, or whatever.
The case where $a = b$ might be good to think about. There are only $a + 1$ lattice points on the $x = y$ diagonal between $(0,0)$ and $(a,a)$, so if you insist on plotting $\sqrt{2}a$ points you'll have to do something like what I said above.
That sounds neat, but it could be that several points are sent to the same pixel, right?
Good point, I will address it in an edit.
Not certain this answers your question, but it might be a step toward your goal:
Tobias Christ, Dömötör Pálvölgyi, Miloš Stojaković.
"Digitalizing line segments."
Electronic Notes in Discrete Mathematics
Volume 38, 1 December 2011, Pages 273-278.
DOI link.
Preliminary arXiv abs.
Abstract. We introduce a novel and general approach for digitalization of line segments in the plane that satisfies a set of axioms naturally arising from Euclidean axioms. In particular, we show how to derive such a system of digital segments from any total order on the integers. As a consequence, using a well-chosen total order, we manage to define a system of digital segments such that all digital segments are, in Hausdorff metric, optimally close to their corresponding Euclidean segments, thus giving an explicit construction that resolves the main question of [J. Chun, M. Korman, M. Nöllenburg, and T. Tokuyama. Consistent digital rays. Discrete Comput. Geom., 42(3):359–378, 2009].
Here not the number of pixel but, as you write, the Hausdorff metric is studied. More precisely, the cited paper of J. Chun, M. Korman, M. Nöllenburg, and T. Tokuyama shows that the Hausdorff distance of some digital segment will be at least the log of the length of the original segment, which can be chosen to be arbitrarily long. So for this metric the answer to the original question is negative.
let the pixels be unit squares with integral corner coordinates, and the line segments be defined by two points $\lbrace p_0:=(x_0,\,y_0),\ p_1:=(x_1,\,y_1)\,|\,x_i,y_i\in\mathbb{R}\rbrace$.
The extremal cases are then $\lbrace p_0=(1.0-\varepsilon,\,0),\ p_1=(n+\varepsilon,\,0)\rbrace$ requiring $n+1$ pixel to cover a line-segment of length $n-1+2\varepsilon$; in that case the rounded nearest integer length would be $n-1$ for sufficiently small positive $\epsilon$ requiring $n+1$ pixels for covering, thus disproving the existence of a line-drawing algorithm with the sought properties.
The other extreme are line-segments with slope 1:
let $\lbrace p_0,\,p_1\rbrace = \lbrace (+\varepsilon,\,+\varepsilon),\ (n+1-\varepsilon,\,n+1-\varepsilon)\rbrace$ requiring $n+1$ pixel for covering a rounded length of $(n+1-2\varepsilon)\sqrt{2}$ which implies that the absolute error can grow linearly with the length of the line-segment.
Addendum:
one aspect of the question, that hasn't been made explicit, is the set of pixels generated by Bresenham's algorithm are 8-way connected, meaning that removing from the a discrete set of points may disconnect them and it also means the set of generated pixels need not cover the line entirely.
Now to the question for a line-rasterization that simultaneously approximates euclidean length via the number of generated pixels:
whenever the a pixel generated by the Bresenham algorithm only shares a corner with the previously generated one, the difference between partial segment-length and number of generated pixels is compared and, whenever that error exceeds $1$ we add of the two pixels that are adjacent to the current one and the previous one, the one whose center is closer to the line-segment.
Well, that's formally true, but I believe that the OP won't be much dissatisfied with an algorithm giving the error of $5$ instead of $1$ in the first case and definitely you are not prohibited to add some more pixels in the second case: the drawing is not assumed to be the most economical one :-)
|
2025-03-21T14:48:32.051118
| 2020-09-17T13:02:47 |
371914
|
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|
Stack Exchange
|
Weak topology on spaces of measures and Borel sets
Let $K$ be a compact Hausdorff space (not necessarily metric or even separable). Let $M(K)$ be the space of all Radon measures on $K$ (that is, finite signed regular Borel measures) endowed with the variation norm. By the Riesz representation theorem $M(K)$ is isometrically isomorphic to the dual space $C(K)^*$ of the space $C(K)$ of all continuous real-valued functions on $K$ endowed with the supremum norm. By the weak topology on $M(K)$ I mean the topology given by the subbase consisting of the sets $V(\mu;x^*;\varepsilon)=\{\nu\in M(K)\colon\ |x^*(\mu)-x^*(\nu)|<\varepsilon\}$, where $\mu\in M(K)$, $x^*\in M(K)^*$ ($\cong C(K)^{**}$), and $\varepsilon>0$.
Let $(\mu_n)$ be a sequence in $M(K)$. It is well-known that $(\mu_n)$ converges to some $\mu\in M(K)$ with respect to the weak topology if and only if $\lim_{n\to\infty}\mu_n(B)=\mu(B)$ for every Borel set $B\subseteq K$. My question is thus the following:
QUESTION: Can we define the weak topology on $M(K)$ by means of the subbase given by the sets of the form $W(\mu;B;\varepsilon)=\{\nu\in M(K)\colon\ |\mu(B)-\nu(B)|<\varepsilon\}$, where $\mu\in M(K)$, $B\subseteq K$ Borel, and $\varepsilon>0$?
(If it helps, one may assume that $K$ is totally-disconnected.)
Quick thoughts: the topology you describe is the weak one defined by the space of step functions with Borel subsets as sets of constancy. Hence its dual is precisely this linear space. But the dual of $M(K)$ with the weak topology is, as you say, $C(K)^{\ast \ast}$.
Oh, I see, so it actually means that the new topology (given by the sets $W(\mu;B;\varepsilon)$) is strictly weaker than the weak topology (given by the sets $V(\mu;x^*;\varepsilon)$...
|
2025-03-21T14:48:32.051257
| 2020-09-17T13:18:20 |
371917
|
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|
Stack Exchange
|
Continuous selection parameterizing discrete measures
Let $\mathcal{P}_n(\mathbb{R})$ denote the set of probability measures on $\mathbb{R}$ for the form $\sum_{i=1}^n k_i \delta_{x_i}$. Then any measure in $\mathcal{P}_n(\mathbb{R})$ is in the image of the map on $\Delta_n \times \mathbb{R}^n$, where $\Delta_n$ is the $n$-simplex, taking $(k_1,\dots,k_n)\times (x_1,\dots,x_n)$ to $\sum_{i=1}^n k_j \delta_{x_i}$. Clearly this map is continuous, when $\mathcal{P}_n(\mathbb{R})$ is equipped with the Prokhorov metric.
However, is it clear that it admits a continuous selection? Ie.: a continuous right inverse (definitely not unique of course)?
$\newcommand\de\delta\newcommand\De\Delta$
The answer is no. Suppose for simplicity that $n=2$ (the case $n>2$ is handled similarly). Suppose that $g$ is a right inverse in question. Then $g(\de_0)=((p,q),(0,0))$ for some $(p,q)\in\De_2$. Take now any $(s,t)\in\De_2\setminus\{(p,q),(q,p)\}$. Then for each natural $k$ we have $g(s\de_{1/k}+t\de_{2/k})=((s,t),(1/k,2/k))$ or $g(s\de_{1/k}+t\de_{2/k})=((t,s),(2/k,1/k))$, so that $g(s\de_{1/k}+t\de_{2/k})\not\to((p,q),(0,0))=g(\de_0)$ (as $k\to\infty$), whereas $s\de_{1/k}+t\de_{2/k}\to\de_0$. So, no such right inverse $g$ can be continuous.
|
2025-03-21T14:48:32.051360
| 2020-09-17T13:50:55 |
371920
|
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|
Stack Exchange
|
Mixed partial derivatives of planar functions converging to delta distribution
Given a sequence $(f_k)_{k\in\mathbb{N}}\subset C^2(\mathbb{R}^2)$ of strictly positive functions $f_k\equiv f_k(x,y)$ with $\|f_k(x,\cdot)\|_{L^1}=1$ for all $x\in\mathbb{R}$ and such that for each $x\in\mathbb{R}$ we have
$$\tag{1} \lim_{k\rightarrow\infty} f_k(x,\cdot) = \delta_x \qquad\text{ in the distributional sense},$$
with $\delta_x$ the delta distribution at $x$.
(The 'heat kernel' $f_k:=\frac{k}{4\pi}e^{-k(x-y)^2/4}$ is an example of such a sequence.)
Question: Can we infer that for infinitely many $k\in\mathbb{N}$, the mixed derivatives $\partial_y\partial_x\log(f_k)$ vanish (almost) nowhere on the diagonal $\Delta:=\{(x,x)\mid x\in\mathbb{R}\}$ (or any compact subset thereof)?
(Vague) Intuition: Let $k\in\mathbb{N}$ and $x_0\in\mathbb{R}$ be fixed, and $\delta>0$ be small. Considering the restriction of $f_k$ to the square $\mathcal{R}:= B_\delta(x_0)\times B_\delta(x_0)$ with $y$-sections $\mathcal{R}_y:= B_\delta(x_0)\times\{y\}$, we by $(1)$ find the functions $\phi_k^y := \left.f_k\right|_{\mathcal{R}_y}$ to 'bulk' increasingly at the point $(y,y)$ and 'flatten sharply' on $\mathcal{R}_y\setminus\{(y,y)\}$ as $k$ increases. Consequently, the (monotonic) transformations $\varphi_k^y:=\log(\phi^y_k)$ show a 'rapid decay below zero' on $\mathcal{R}_y\setminus\{(y,y)\}$ (as $k\rightarrow\infty$).
This suggests that (for $k$ large enough) we have $\psi_k^y:= \left.\partial_x(\varphi_k^y)\right|_{x=x_0}>0$ for $y>x_0$, and $\psi_k^y<0$ for $y<x_0$, so that $\left.\partial_y[\partial_x\log(f_k)]\right|_{(x,y)=(x_0, x_0)} = \partial_y(\left.\psi_k^y)\right|_{y=x_0} > 0$, provided that $(1)$ guarantees that
$$\tag{2}\lim_{h\rightarrow 0^+}\frac{\psi_k^{x_0+h} - \psi_k^{x_0-h}}{2h}>0.$$
Do you see a way to put this intuition into a rigorous proof? (Or is it wrong altogether and the claim doesn't hold?)
That would be way too much to ask for: take any $\psi$ with integral $1$ supported on $(-1,1)$ with integral $1$ and put $f_n(x,y)=n\psi(nx-ny+2)$, say (you do not need to be right in the middle to converge to the delta-measure; you can approach the diagonal "from the side" instead). Perhaps you will be happy with some weaker property?
(What I would really need is that: for every (non-empty) open (wrt. the subspace topology of $\mathbb{R}^2$) subset $U$ of $\Delta$ there exists a (non-empty) open subset $U'\subseteq U$ and an $n$ such that the restriction $\left.[\partial_y\partial_x\log(f_n)]\right|_{U'}$ is nowhere zero.
[In my case, the$f_n$ are fundamental solutions to the parabolic (backward/forward) Kolmogorov equations.])
(In the above, instead of one $n=n_{U'}$ I would in fact need to have infinitely many such $n$.)
Just make $\psi$ a positive constant on $(1,3)$ then and descend a bit later staying positive, if you wish..
@fedja If you don't mind me asking: Do you happen to have any recommendations on how to proceed to verify the weaker condition (given in the above comment) for sequences $(f_k)$ arising as fundamental solutions of a parabolic PDE? (Or would you "feel" this is also infeasible?)
Alas, no. You will, probably, need to use the equation somehow, but how exactly to do it in general (i.e., without playing with the particular (assumptions about) coefficients) is unclear to me.
|
2025-03-21T14:48:32.051709
| 2020-09-17T13:57:36 |
371921
|
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|
Stack Exchange
|
Covariant splittings of Hopf algebra projections
What is an example of a pair of Hopf algebras $(A,B)$ with a surjective Hopf algebra map $\phi:A \to B$ such that $\phi$ does not admit a $B$-bi-comodule splitting $s:B \to A$? To be clear, the right $B$-comodule structure on $A$ is given by
$$
(\textrm{id} \otimes \phi) \circ \Delta_A: A \to A \otimes B,
$$
where $\Delta_A$ is the coproduct of $A$, and the left coaction is defined similarily.
Did you tried with group algebras? (And a surjective, non split, group homomorphism )
At a glance, it looks like group algebras might actually not yield the example; if you identify the quotient group $Q = G/N$ with a subset $\tilde Q$ of $G$ via an arbitrary lifiting, the inherited coalgebra structure on the lift $k \cdot \tilde Q$ is extended linearly from $\Delta(q) = q \otimes q$, so there's a coalgebraic splitting, just not an algebraic one. Or am I missing something?
It's unlikely to be what you want, but in case it's not an accident that you haven't specified a fixed base ring, there are examples that fail to admit even group-theoretic splittings, e.g., $\mathbb Z[x] \to \mathbb F_2[x]$.
I'll give an example "occuring in nature." It's not the simplest possible, but you can get a simpler one by removing the generators of degrees 3, 5, and 7, which don't feature in the argument.
According to results of Borel from 1954, the mod-2 homology Hopf algebra $$H_9 = H_* (\mathrm{Spin}(9);\mathbb F_2)$$ is the exterior algebra on one generator each of degrees 3, 5, 6, 7, and 15. The standard inclusion $i$ of $\mathrm{Spin}(9)$ in $\mathrm{Spin}(10)$ preserves this exterior algebra structure, but $H_{10} = H_*(\mathrm{Spin}(10);\mathbb F_2)$ has a new generator $u_{9}$ of degree 9 such that $H_{10}$ is a free module of rank two over $i_* H_9$ with basis $\{1,u_9\}$,
following from the collapse of the Serre spectral sequence of the fiber bundle $\mathrm{Spin}(9) \to \mathrm{Spin}(10) \to S^9$. The new generator $u_9$ doesn't anticommute with the old ones as one might expect: giving the other generators the obvious names, one has
$$u_6 u_9 + u_9 u_6 = u_{15}$$
in $H_{10}$.
Particularly, the injection of left $H_9$-modules $H_9 \to H_{10}$ does not split. For degree reasons, $u_9$ would have to go to $u_3 u_6$ or zero under the splitting, but the product $H_9 \otimes H_{10} \to H_{10}$ sends $$u_6 \otimes u_9 \mapsto u_{15} + u_9 u_6,$$ while the product $H_9 \otimes H_9 \to H_9$ sends $$u_6 \otimes 0 \mapsto 0 \neq u_{15} + 0 u_6$$ and also $$u_6 \otimes u_3 u_6 \mapsto 0 \neq u_{15} + (u_3 u_6) u_6.$$
A $H^*(\mathrm{Spin}(9);\mathbb F_2)$-comodule splitting of the cohomological Hopf algebra map $$i^*\colon H^*(\mathrm{Spin}(10);\mathbb F_2) \to H^*(\mathrm{Spin}(9);\mathbb F_2)$$ would lead on dualization to a forbidden module splitting of the sort we ruled out in the previous paragraph, so $i^*$ is an example of the type you wanted.
|
2025-03-21T14:48:32.052162
| 2020-09-17T14:14:13 |
371925
|
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"Mateusz Kwaśnicki",
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|
Stack Exchange
|
A Riemann Hilbert problem on the unit square
Let $p_0=0$, $p_1=1$, $p_2=1+i$ and $p_3=i$ be the four vertices of a square $Q$ on the complex plane $\mathbb C$.
Let $f \in C^{\infty}_c((0,1))$ and consider the following Riemann-Hilbert problem on $Q$:
$$ \frac{\partial}{\partial \bar{z}}F=0,\quad \text{on $Q$},$$
subject to
$$ \textrm{Re}\{F\}(t,0)=f(t) \quad \text{for $t \in (0,1)$}$$
and $$\textrm{Re}\{F\}_{|_{\partial Q\setminus (0,1)}}=0.$$
Does there exist a solution to this problem?
This does not look like a true Riemann–Hilbert problem: simply extend the boundary values to a harmonic function in the unit square to get the real part of $F$, and set the imaginary part of $F$ to be equal to the harmonic conjugate of the real part. Am I missing something?
|
2025-03-21T14:48:32.052253
| 2020-09-17T15:26:28 |
371926
|
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|
Stack Exchange
|
Covering of discrete probability measures
Let $\mathcal{P}_{n:+}(\mathbb{R})$ denote the set of probability measures on $\mathbb{R}$ for the form $\sum_{i=1}^n k_i \delta_{x_i}$ where $k_i>0$. Then any measure in $\mathcal{P}_{n:+}(\mathbb{R})$ is in the image of the map on $\Delta_n \times \mathbb{R}^n$, where $\Delta_n$ is interior of the $n$-simplex (i.e.: $k_1,\dots,k_n \in (0,1)$ with $\sum_{i=1}^n k_i =1$, taking $(k_1,\dots,k_n)\times (x_1,\dots,x_n)$ to $\sum_{i=1}^n k_j \delta_{x_i}$. Clearly this map is continuous, when $\mathcal{P}_{n:+}(\mathbb{R})$ is equipped with the Prokhorov metric.
However, is it a covering map? I have not been able to disprove it so I'm thinking maybe it is...?
This map is not a covering one, because the preimages of singleton sets under this map are not of the same cardinality. E.g., the cardinality of the preimage of the singleton set $\{\frac1n\,\sum_{j=1}^n\delta_j\}$ is $n!$, whereas the preimage of the singleton set $\{\delta_0\}$ is of infinite cardinality.
From the site you linked to: "[for a covering map $f\colon X\to Y$,] the cardinal number of $f^{-1}(y)$ (which is possibly infinite) is independent of the choice of $y$ in $Y$."
So if we replace $\mathcal{P}{n:+}$ with the set $\left{ \sum{i=1}^n k_i\delta_{x_i}:k_i>0,, x_i\neq x_j \mbox{ if } i\neq j\right}$ then this should work?
@James_T : I think so.
|
2025-03-21T14:48:32.052360
| 2020-09-17T15:37:41 |
371927
|
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|
Stack Exchange
|
Is there a closed-form expression for these integrals?
I am computing the following integrals by numerical integration and this takes a lot of time, although I'm sure there is a general closed-form formula but I can't find it.
Let $t$ be a vector of $\mathbb R_{+}^{d}$. For any integer $d \ge 1$, define $K_d$ as a convex subset of $\mathbb R^{d}$ by :
$$x \in K_d \iff \forall\, j \in {1,..,d}, \;x_j \ge 0 \text{ and }\sum\limits_{j=1}^d t_j x_j \le 1$$
I think $K_d$ is usually called a simplex, but I am not sure.
Let now $i$ be a vector of integers in $\mathbb{N}^{d}$, and consider the integral :
$$I_{i}^{d} = \int\limits_{K_{d}} \;\;\prod_{j=1}^{d} x_j^{i_j} \;\;\partial x_1,...,\partial x_d$$
Can we compute an expression for $I_{i}^{d}$? Maybe some recursion on $d$ or on $i$ can be found?
Edit:
I founded a paper that solves the problem, Lasserre - Simple formula for integration of polynomials on a simplex. it gives a formula a little more general, that reduces to the following :
$$\text{If and only if t_j = 1 for all j, }I_{i}^{d} = \frac{\prod\limits_{j=1}^{d} i_j}{(d+\lvert i \lvert)!}$$
Then, as a comment showed, generalisation can be done via
$$I_i^d(t) = \bigl(\prod_{j = 1}^d t_j^{-(i_j + 1)}\bigr)I_i^d(1)$$
I think that we have $I_i^d = \int_0^{1/t_d} x_d^{i_d}(1 - t_d x_d)^{\sum_{j = 1}^{d - 1} i_j}I_i^{d - 1}\mathrm dx_d$, but, if not that, then something very like it.
Nevermind, there is a formula there for exactly that : https://arxiv.org/pdf/1908.06736.pdf
Thanks anyway !
Glad you found it. You might consider posting it as an answer, if it totally addresses your question. \ Your arXiv link, de-PDFd (and with the pun in its title unfurled for us to appreciate): Lasserre - Simple formula for integration of polynomials on a simplex.
Actualy i'm still stuck on the change of variables... I'll Edit, and maybe you could help ?
If my formula is correct (but it's just off the top of my head, so you should doubt it), then it gives the recurrence $I_i^d = c_d I_i^{d - 1}$ with coefficient $c_d = \int_0^{1/t_d} x_d^{i_d}(1 - t_d x_d)^{\sum_{j = 1}^{d - 1} i_j}\mathrm dx_d$, so you'd just have to compute that integral. If I'm right, I think you won't be able to avoid some such computation.
Note that, if $t_d$ is $1$, then this becomes $c_d = B(i_d + 1, \sum_{j = 1}^{d - 1} i_j + 1) = \dfrac{i_d!(\sum_{j = 1}^{d - 1} i_j)!}{(\sum_{j = 1}^d i_j)!}$, and that doesn't seem to simplify to your formula, so probably I made a mistake. Anyway it seems that in general $c_d(t_d) = t_d^{-(i_d + 1)}c_d(1)$.
Your last proposition, $c_d(t_d) = t_{d}^{-(i_d+1}} c_d(1)$ is indeed wht i tried already, but my computer says that it's not right (compared to the brutal integration)
Edit: You ere right, i checked with wolframalpha and it seems that the integrator i was using is not that good... a lot of time wasted beacuase of a stupid piece of code that does not do what it seems to do. Thanks anyway ! I'll add the info to the main post
I think your last line should be $I_i^d(t) = \bigl(\prod_{j = 1}^d t_j^{-(i_j + 1)}\bigr)I_i^d(1)$.
If $\Sigma_t=\{x_i \geq 0: x_1+\cdots +x_d=t\}$and $d\sigma_t$ is its surface measure, then $I_t=\int_{\Sigma_t}\prod_{i=1}^d x_i^{\alpha_i}\, d\sigma_t=t^{|\alpha|+d-1}I_1$ (change variable $x_i=ty_i$). Then for $\alpha_i>-1$
$$
\prod_{i=1}^d \Gamma (\alpha_i+1)=\int_{[0,\infty[^d} \prod_{i=1}^d x_i^{\alpha_i} e^{-(x_1+\cdots +x_d)}dx=\int_0^\infty \frac{e^{-t}}{\sqrt n} I_tdt=I_1\int_0^\infty t^{|\alpha|+d-1}\frac{e^{-t}}{\sqrt n} dt.
$$
This gives $I_1=\frac{\sqrt{n} \prod_{i=1}^d \Gamma (\alpha_i+1)}{\Gamma (|\alpha|+d)}$ and $$\int_{\{x_i \geq 0, x_1+\cdots +x_d \leq 1\}} \prod_{i=1}^d x_i^{\alpha_i}dx=\int_0^1 \frac{dt}{\sqrt n}I_t=\frac{\prod_{i=1}^d \Gamma (\alpha_i+1)}{\Gamma (|\alpha|+d+1)}.$$
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2025-03-21T14:48:32.052606
| 2020-09-17T15:47:24 |
371929
|
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|
Stack Exchange
|
Double-diagonalisation of nxn matrices?
I've come up with the following piece of Python code (using the library Sympy):
def double_diagonalize(m1, m2):
V, _ = (m1.T * m2).diagonalize()
U, _ = (m1 * m2.T).diagonalize()
return U, V
What I've found is that given many (but not all) random pairs of $
n \times n$ matrices $M_+$ and $M_-$, it produces $U$ and $V$ such that both $U^{-1} M_+ V^{-T}$ and $U^T M_- V$ evaluate to diagonal matrices.
My question is, is this result known? And by what name? It looks similar to Generalised Schur Decomposition except that the matrices $U$ and $V$ are not necessarily unitary, and the (pair of) "normal forms" $U^{-1} M_+ V^{-T}$ and $U^T M_- V$ are usually fully diagonal instead of merely upper triangular.
It might be related to this as well: https://en.wikipedia.org/wiki/Eigendecomposition_of_a_matrix#Generalized_eigenvalue_problem
It's the solution of the product eigenvalue problem $M_1^T M_2$, once you transpose the second relation.
I'm reading the paper you linked to
I'm not sure you will find that exact decomposition there (since it does not always exist, and orthogonal decompositions are preferred in computational practice), but that is the framework under which your problem falls in today's research terms in computational linear algebra.
|
2025-03-21T14:48:32.052718
| 2020-09-17T17:19:32 |
371937
|
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|
Stack Exchange
|
Lusztig correspondence and unipotent support
In "Unipotent support for irreducible representations", Lusztig proves the existence of a unique maximal unipotent class $\mathscr{O}_\pi$, called unipotent support, where the character of a representation $\pi$ is not trivial.
I would like first to ask if the relationship between the Lusztig correspondence $\mathfrak{L}$ and the unipotent support is known. More precisely, take two representations $\rho$ and $\sigma$ in the same Lusztig series such that $\mathscr{O}_\rho\subset\overline{\mathscr{O}_\sigma}$ (this defines the so-called "closure order"). Can something be said about the unipotent support of $\mathfrak{L}(\rho)$ and $\mathfrak{L}(\sigma)$? For instance, would one be contained in the closure of the other?
Also, it is known that unipotent representations of unitary groups are parametrized by partitions of the order of the group. Is there a combinatorial way to determine the unipotent support of such a representation in terms of the corresponding partition? I know that for symplectic groups there is indeed a combinatorial way.
Lastly, is there a package in Magma or GAP that calculates the unipotent support for (small) finite groups of Lie type?
I think you mean "rank of the group" where you say "order of the group."
@SamHopkins, I mean $n$ for $U_n(q)$.
That is indeed the (relative) rank. The order is a polynomial in $q$ of degree $n$.
|
2025-03-21T14:48:32.052852
| 2020-09-17T17:45:34 |
371940
|
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"David Roberts",
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|
Stack Exchange
|
Does every category with a subobject classifier embed into a topos?
I've never seen an example of a category with a subobject classifier which didn't embed nicely into a topos. Is there a good reason for this?
Question 1: Let $\mathcal C$ be a category with a subobject classifier $\Omega$ (and whatever finite limits this entails -- namely, a terminal object and pullbacks along monomorphisms). Does there exist a fully faithful functor $\mathcal C \to \mathcal E$, where $\mathcal E$ is an elementary topos, which preserves the subobject classifier and the aforementioned finite limits?
Question 2: Same as Question 1, but assuming that $\mathcal C$ has all finite limits, and requiring that $\mathcal C \to \mathcal E$ preserves them.
Question 3: Same as Question 2, but throwing in finite colimits as well.
Question 4: Now assume that $\mathcal C$ is locally presentable and has a subobject classifer $\Omega$. Does it follow that $\mathcal C$ is a (necessarily Grothendieck) topos?
Question 4 may be the most heavy-duty-sounding formulation, but it also gives me the most reason to think the answer might be "yes" -- after all, in order for a category $\mathcal C$ with finite limits and a subobject classifier to be a topos, it just needs to additionally be cartesian closed. And if $\mathcal C$ is locally presentable, then by the adjoint functor theorem, to verify this one just needs to check that the functors $X \times (-)$ preserve colimits. Plausibly, the subobject classifier might force this. As partial progress, I think I can show that in this case, coproducts are disjoint.
On Q4: The category of pointed sets is locally finitely presentable and has a subobject classifier. Yet it is not a topos.
@IvanDiLiberti Oh wow -- that's a great example! And the universal subobject is a map $1_+ \to 2_+$, where the domain $1_+$ is not the terminal object. I did not realize this was possible!
Hum, no the subobject classifier is $0_+ \to 1_+$ and $0_+$ is terminal.
Ah, right! In fact, this example seems to be general -- for any topos $\mathcal E$ with universal subobject $true: 1 \to \Omega$, the category $\mathcal E_\ast$ of pointed objects in $\mathcal E$ has a subobject classifier given by $(\Omega, true)$. But $\mathcal E_\ast$ is only a topos if it is the terminal category, which only happens if $\mathcal E$ is the trivial topos.
Not quite what you were asking, but there are pretoposes with subobject classifier (well-pointed, Boolean, and satisfying IAC, even) that are not cartesian closed, and very far from being (elementary) toposes.
Ivan's example in the comment actually proves that all the questions have negative answers.
As observed by Ivan, in the category of pointed set, there is a subobject classifier given by $\{*\} \to \{*,\bot \}$, where $*$ is the special point.
Indeed, a subobject of $X$, is just a subset of $X$ containing $*$ so it is classified by a unique map $X \to \{* = \top,\bot\}$ : the usual classifier of the map in Set.
Now, in a topos, you always have at least two maps from the terminal object to the sub-object classifier: the map $\top$ and the map $\bot$. If they are equal, the topos is degenerated. But in pointed set, there is only one map from $\{*\} \to \{*,\bot \}$, so there can't by a fully faithful functor to an elementary topos that preserve the subobject classifier and its universal subobject.
Maybe one comment. While the symmetric topos construction shows that finite limits can be formally added "lexely", providing a left adjoint to the forgetful functor Topoi $\to$ LP, this counterexample shows that cartesian closedness cannot be forced without destroying everything.
@Ivan "Lexely" or "laxly"?
@DavidRoberts I think Ivan did meant "lexely" as the lex colimits of https://arxiv.org/abs/1107.0778
That is precisely what I meant.
|
2025-03-21T14:48:32.053132
| 2020-09-17T18:12:35 |
371944
|
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|
Stack Exchange
|
Exhaustion of a noncompact manifold by domains with mean convex boundary
Let $(M,g)$ be a complete noncompact Riemannian manifold. Can we find an exhaustion $M=\bigcup_{i \ge 1} U_i$ such that each $U_i$ is a bounded domain with smooth boundary $\partial U_i$ which is mean convex? Here, mean convex means the mean curvature of $\partial U_i$ with respect to the outer normal vector is positive everywhere.
If not, what condition can we impose on $(M,g)$ to guarantee this? For instance, if $M$ has positive sectional curvature, then the conclusion is true.
Could you define "mean convex"?
In the context of the question, "mean convexity" means that the mean curvature vector $H_i$ of $\partial U_i$, dotted with the outward unit normal $\nu_i$ to $\partial U_i$ is positive.
Take a flat cylinder.
I think nonnegative Ricci curvature may imply what you want!
In another direction, a simply connected non positively curved manifold (aka Hadamard manifold) satisfies this, you can even replace "mean convex" by "convex" and take balls around any point.
I guess that a ball with an added cusp does not satisfy this, since in the postively curved part near the cusp you should need some concavity, but I do not have a proof of this.
To the best of my knowledge this is an open problem, formulated by Blaine Lawson. It is recorded as Problem 1.7 on page 459 in Proceedings of Symposia of Pure Mathematics Volume 44 (1986). Precisely he asks:
Characterize those complete Riemannian manifolds which admit an exhaustion by compact domains $\Omega_1 \subset \subset \Omega_2 \subset \subset \Omega_3 \subset \subset \cdots$ such that each boundary $\partial \Omega_k$ is smooth and of positive mean curvature with respect to the outer normal; e.g., concentric balls in $\mathbf{R}^n$. Some results in this direction are in [BR].
The paper that Lawson refers to here is Robert Brooks's 'The fundamental group and the spectrum of the Laplacian', which appeared in Comment. Math. Helvetici in 1981.
|
2025-03-21T14:48:32.053301
| 2020-09-17T19:24:30 |
371946
|
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|
Stack Exchange
|
Arithmetical function comparable to sine function
I was wondering if there exists or can we construct (using known arithmetic functions) an arithmetical function that has the same behaviour of the function sine or comparable to it (I mean that oscilates, changes regularly signs, periodic if it is necessary...).
Thanks in advance.
you may want to be explicit in what you call an arithmetic function; do you follow this definition: https://en.wikipedia.org/wiki/Arithmetic_function
Yes Sir, I mean that definition.
$f(n) = 1 - 2\cdot (n \text{ mod } 2) $
Thanks for your example.
The Liouville function ($-1$ to the power of the number of prime divisors of $n$) could be a candidate for a "sine-like" function. It's not periodic, but it does oscillate, changing sign infinitely often, and has the same range $[-1,1]$.
Thanks Sir for your answer.
|
2025-03-21T14:48:32.053402
| 2020-09-17T19:25:25 |
371947
|
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"François G. Dorais",
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|
Stack Exchange
|
Is equational logic in universal algebra a proof system not a logic system?
As far as I know a logic system defines its own semantics (e.g. $\models$), but not a proof calculus/system on its language. See p261 in Ebbinghaus et al's Mathematical Logic:
In universal algebra, it seems to me that "equational logic" is defined as a proof system, so is it not a logic system in the above sense, and is a counterexample of the use of "logic" in "logic system"?
See p94 in Burris et al's A Course in Universal Algebra:
and p42 of Baader et al's Term Writing and All That
Thanks.
Birkhoff's Completeness Theorem for equational logic proves the equivalence. However, I think the main issue is applying one book's definitions to another book by different authors, which is generally ill advised.
@RodrigoFreire Thanks. (1) From Burris' book and a new source, "equational logic" doen't seem be defined the way you suggest me to think of. (2) What other definitions of logic systems have you seen, and which ones are popular?
@FrançoisG.Dorais What other definitions of logic systems have you seen, and which ones are popular?
The definition of Tarskian operator (finitary closure operator) is closer to what you want. https://en.wikipedia.org/wiki/Closure_operator#Closure_operators_in_logic
@RodrigoFreire What is "what you want"? I think the link you gave is for introducing a proof system.
There is a general definition of logic according to which a logic is a finitary closure operator. The link I gave is for the corresponding wikipedia page. The proof system of equational logic defines an example of a finitary closure operator.
@Tim: I have deleted my first comment because it may be confusing. The answer to your question is no, equational logic is not a counterexample to the general definition of logic given in Ebbinghaus et al. The reason is that equational logic can also be defined model-theoretically, as it is complete. However, this equivalence is not trivial. On the other hand, equational logic as you have presented is trivially seen as a tarskian logic, that is, a finitary closure operator, which is another general definition of logic.
@RodrigoFreire Is "tarskian logic, that is, a finitary closure operator, which is another general definition of logic" the same thing as a proof system?
@Tim No, Tarskian logic is not the same thing as proof system.
@RodrigoFreire Is "tarskian logic, that is, a finitary closure operator, which is another general definition of logic" something whose definition defines $\vdash$ relation between sets of formulas and formulas? Is a proof system a rule based system whose definition defines a set of inference rules and axioms? Does a proof system induce a "tarskian logic, that is, a finitary closure operator, which is another general definition of logic"?
@Tim Yes, a proof system induces a finitary closure operator.
@RodrigoFreire (1) Is "tarskian logic, that is, a finitary closure operator, which is another general definition of logic" something whose definition defines ⊢ relation between sets of formulas and formulas? (2) On the semantics part, $\models$ between structures and formulas induces $\models$ between sets of formulas and formulas. Is the reverse true: does $\models$ between sets of formulas and formulas induce $\models$ between structures and formulas? Or is $\models$ between structures and formulas, which induces $\models$ between sets of formulas and formulas, unique?
@Tim (1) Yes. (2) No, but there is a conection between general valuations and tarskian logics. Maybe I should write an answer.
@RodrigoFreire Thanks. Another question, can we define the finitary closure operator in a tarskian logic system, without being induced from a proof system?
I will write a more complete answer.
The model-theoretic definition of logic quoted in the question is given in that book to be used in Lindström's theorem. Nevertheless, equational logic is a model-theoretic logic and the induced consequence relation is equivalent to the one defined via proof theory (for this logic is complete).
Now, other questions about other general definition of logic were raised in the comments. In order to address those questions we need some preliminary remarks and notation.
Let $X$ be a set (think of $X$ as a set of formulas). A logic on $X$ is a subset of $\wp(X)\times X$ (think of such a subset as a consequence relation). We use $\Gamma$ and $\Delta$ to denote general subsets of $X$ and $\phi$ and $\psi$ to denote general elements of $X$.
A valuation on $X$ is a subset of $X$ (think of such a subset as the formulas which are true according to the valuation).
We say that a valuation $w$ is compatible with a logic $l$ iff for every $(\Gamma,\phi)\in l$, if $\Gamma\subseteq w$, then $\phi\in w$.
If $l$ is a logic on $X$, then $l$ defines a set of valuations $G(l)$, those compatible with $l$.
Conversely, if $m$ is a set of valuations, then $m$ defines a logic $L(m)$: the set of the pairs $(\Gamma,\phi)$ such that for all $w\in m$, if $\Gamma\subseteq w$, then $\phi\in w$.
$G$ and $L$ constitute a Galois connection.
For a given logic $l$, the following are equivalent:
There is a set of valuations $m$ such that $l=L(m)$
$l=L(G(l))$
$l$ is reflexive, idempotent and monotonic.
A logic $l$ is reflexive if for every $\phi\in\Gamma$, $(\Gamma,\phi)\in l$.
A logic $l$ is idempotent if whenever $\Delta, \Gamma\subseteq X$, $\phi\in X$, $(\Gamma, \phi)\in l$ and for every $\psi\in\Gamma$ , $(\Delta,\psi)\in l$, we have $(\Delta,\phi)\in l$. A logic $l$ is monotonic if whenever $\Gamma\subseteq\Delta$, if $(\Gamma,\phi)\in l$, then $(\Delta,\phi)\in l$.
Now, we can consider the question:
Is the reverse true: does $\models$ between sets of formulas and formulas induce $\models$ between structures and formulas?
Yes, in some trivial sense: We can look at the structures inducing valuations compatible with $l$ (assuming that structures induce valuations in that case). But the difficulty is to recover your first consequence relation from the induced relation. This is not always possible:
We can look at the structures inducing valuations compatible with $l$, then we can look at all the consequence relations preserved by those structures. In general, we will not recover $l$ this way, even if $l$ is a finitary, reflexive, idempotent and monotonic relation. We can start with a set $m$ of valuations which are not induced by structures. We can take $l=L(m)$, and this will be finitary if $X$ is finite, for example.
But if we allow those general valuations, not only those induced by structures, then yes, from a tarskian logic $l$ we can define a canonical set of valuations $G(l)$ such that $l$ can be recovered from that set.
I have a little trouble with the quantifiers in "idempotent": "whenever $(\Gamma, \phi) \in l$ and for every $\psi \in \Gamma$, $\Delta, \psi \in l$, we have $(\Delta, \phi) \in l$". First, $\Delta, \psi$ should be $(\Delta, \psi)$, right? Second, should it be (with extra parentheses hopefully for clarity) "whenever (($(\Gamma, \phi) \in l$ and $\Delta \subseteq X$), and, (for every $\psi \in \Gamma$, we have $(\Delta, \psi) \in l$)), we have ($(\Delta, \phi) \in l$)"?
Right, there is a missing parentheses, thanks. Your reformulation is great.
@LSpice I have edited the answer. The idea is simple (the consequences of consequences of $\Delta$ are consequences of $\Delta$ directly), but it is a bit cumbersome to write it informally. Formally, $\forall\Delta\forall\Gamma\forall\phi(((\Gamma,\phi)\in l\wedge \forall\psi\in\Gamma(\Delta,\psi)\in l)\rightarrow (\Delta,\phi)\in l)$
Thanks, that's what I thought! I'm teaching an intro to proof course right now, and one of the things I struggle with is how to instill in them the proper balance between, on one hand, "(almost?) nobody learns best from stacked quantifiers, so try to say it in (mathematical) English"; and, on the other hand, "there are some things, noteably subtle grouping, that non-symbolic English just doesn't handle very well."
|
2025-03-21T14:48:32.053988
| 2020-09-17T19:26:19 |
371948
|
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|
Stack Exchange
|
Separability of $\mathbb{C}[x]$ over its $\mathbb{C}$-subalgebras
For commutative rings $R \subseteq S$,
recall that $S$ is separable over $R$, if $S$ is a projective $S \otimes_R S$-module, via $f: S \otimes_R S \to S$ given by: $f(s_1 \otimes_R s_2)=s_1s_2$.
Question 1: Is $\mathbb{C}[x]$ separable over $\mathbb{C}[x^2,x^3]$?
More generally,
Question 2: Is it possible to characterize all $\mathbb{C}$-subalgebras $\mathbb{C} \subset R \subset \mathbb{C}[x]$ such that $\mathbb{C}[x]$ is separable over $R$?
According to wikipedia: "Moreover, an algebra $S$ is separable if and only if it is flat when considered as a right module of $S \otimes_R S$ in the usual way".
Here $\mathbb{C}[x^2,x^3] \subset \mathbb{C}[x]$ is not flat; I am not sure if there is a connection between flatness or non-flatness of $\mathbb{C}[x^2,x^3] \subset \mathbb{C}[x]$ and $\mathbb{C}[x] \otimes_{\mathbb{C}[x^2,x^3]} \mathbb{C}[x] \subset \mathbb{C}[x]$.
If, for example, $\mathbb{C}[x^2,x^3] \subset \mathbb{C}[x] \otimes_{\mathbb{C}[x^2,x^3]} \mathbb{C}[x]$ is flat (I do not know if this is true or false), then
flatness of $\mathbb{C}[x] \otimes_{\mathbb{C}[x^2,x^3]} \mathbb{C}[x] \subset \mathbb{C}[x]$ would imply flatness of $\mathbb{C}[x^2,x^3] \subset \mathbb{C}[x]$, which is false.
(This question may be relevant. Also asked in MSE).
Edit: After receiving a comment that "it's unlikely that you can characterise all $R$ for which $R \subseteq \mathbb{C}[x]$ is separable, I would like to change Question 2 to the following question:
Question 3: Is it possible to characterize all $h \in \mathbb{C}[x]$, such that $\mathbb{C}[x]$ is separable over:
(i) $A=\mathbb{C}[h]$.
(ii) $B=\mathbb{C}+(h)$, where $(h)$ denotes the ideal of $\mathbb{C}[x]$ generated by $h$.
Example: If $h=x^2$, then $B=\mathbb{C}+(x^2)=\mathbb{C}[x^2,x^3] \subseteq \mathbb{C}[x]$ is not separable (first comment below).
Remark:
Denote $h(x)=c_nx^n+c_{n-1}x^{n-1}+\cdots+c_1x+c_0$.
According to Corollary 8 with $h(Z)=c_nZ^n+c_{n-1}Z^{n-1}+\cdots+c_1Z+c_0-h$, we obtain an answer to Question 3(i): $\mathbb{C}[x]$ is separable over $\mathbb{C}[h]$ iff $\deg(h)=1$ (namely, $B=\mathbb{C}[x]$).
Still, I am not sure what can be said about Question 3(ii); could it be that for any $h$ of degree $\geq 2$, $B= \mathbb{C}+(h) \subseteq \mathbb{C}[x]$ is inseparable? If not, could one present a counterexample of minimal $\deg(h)$?
Notice that by this question we have: $B=\mathbb{C}+(h)=\mathbb{C}[h,xh,\ldots,x^{n-1}h]$.
Lemma 4.1 seems relevant; however, here $B$ is integrally closed in its field of fractions $\mathbb{C}(x)$ iff $B=\mathbb{C}[x]$, so we are not able to apply Lemma 4.1.
Any help is welcome! Thank you very much!
Because $\mathbf C[x]$ is finitely generated over $\mathbf C$ (in particular over $R$), separable is just another word for unramified. This can be checked either using differentials, or fibrewise (it is equivalent to each fibre being a finite product of separable field extensions). So $\mathbf C[x^2,x^3] \subseteq \mathbf C[x]$ is inseparable since the fibre over $(x^2,x^3)$ is $\mathbf C[x]/x^2$, but it's unlikely that you can characterise all $R$ for which $R \subseteq \mathbf C[x]$ is separable.
@R.vanDobbendeBruyn, thank you very much for your helpful comment! Just to make sure, please what is the definition of being unramified that you use? Is it being separable and flat? (I apologize for my ignorance). Thank you.
No, unramified means finite type and $\Omega_{S/R} = 0$. If $R \to S$ is moreover flat, then it is called étale. See for example Tags 00UT and 08WD. The equivalence with separability is explained in Thm. 8.3.6 of Ford.
Thank you very much again! (ok, I recall now. I was confused and forgot about the kernel).
@R.vanDobbendeBruyn, please, what if we restrict Question 2 to $R=\mathbb{C}[h]$ or $R=\mathbb{C}+(h)$, where $h \in \mathbb{C}[x]$ with $\deg(h) \geq 2$? By $(h)$ I mean the ideal of $\mathbb{C}[x]$ generated by $h$. Thank you very much! (Should I ask this as a new question?).
In other words, my new question is: Let $h=h(x) \in \mathbb{C}[x]$ with $\deg(h) \geq 2$. Is it possible to characterize all $h$'s such that $\mathbb{C}[x]$ is separable over: (1) $\mathbb{C}[h]$ (2) $\mathbb{C}+(h)$ ?
As I wrote in my comment, these extensions are known in algebraic geometry as unramified, which can be tested computationally by the vanishing of $\Omega_{S/R}$. (In differential geometry, this means $Y \to X$ is an immersion, i.e. injective on tangent spaces.)
Lemma. Let $h \in \mathbf C[x]$ be a polynomial of degree $n \geq 2$. Then $\mathbf C+(h) \subseteq \mathbf C[x]$ is unramified is and only if $h$ is squarefree.
Proof. The subalgebra $\mathbf C + (h)$ is generated as $\mathbf C$-algebra by $h, xh, x^2h, \ldots, x^{n-1}h$, since $x^n h$ is a linear combination of $h^2$ and $x^ih$ for $i < n$. So the question is whether the map
\begin{align*}
\mathbf A^1 &\to \mathbf A^n\\
x &\mapsto \big(h(x),xh(x),\cdots,x^{n-1}h(x)\big)
\end{align*}
is unramified. The Jacobian is
$$\Big(\begin{matrix}h'(x) & xh'(x)+h(x) & x^2h'(x)+2xh(x) & \cdots & x^{n-1}h'(x) + (n-1)x^{n-2}h(x)\end{matrix}\Big),$$
and the ideal $I$ generated by the entries is just $(h,h')$. So $\Omega_{\mathbf A^1/\mathbf A^n} = k[x]/I$ is trivial if and only if $h$ and $h'$ have no factors in common, i.e. if and only if $h$ is squarefree. $\square$
A similar argument easily computes that the map
\begin{align*}
\mathbf A^1 &\to \mathbf A^1\\
x &\mapsto h(x)
\end{align*}
is never unramified if $n \geq 2$, since $h'(x)$ has at least one root. Thus, $\mathbf C[h] \subseteq \mathbf C[x]$ is never unramified if $\deg h \geq 2$.
Thank you very much!
I really apologize for interrupting you again, but what about separability of $\mathbb{C}[x,y_1,\ldots,y_r]$ over $\mathbb{C}+\langle h,y_1,\ldots,y_r \rangle$, $r \geq 1$.
If you have more questions, it's probably best to post it as a new question, so that everyone can see it.
Thank you for your response. I agree with you and will soon post it.
(I have posted it here: https://mathoverflow.net/questions/382422/separability-of-mathbbcx-y-1-ldots-y-r-over-mathbbc-h-y-1-ldots/382424#382424 Thank you!).
|
2025-03-21T14:48:32.054360
| 2020-09-17T19:39:06 |
371949
|
{
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"authors": [
"Gerry Myerson",
"https://mathoverflow.net/users/158000"
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"url": "https://mathoverflow.net/questions/371949"
}
|
Stack Exchange
|
division of polynomials
Consider the set of polynomials $k[x^1,\cdots,x^n]$ over any field $k$. Now, given $p,q\in k[x^1,\cdots,x^n]$, what are the necessary and suficient conditions in order to q divide p? That is: $\exists r \in k[x^1,\cdots,x^n]$ such that $p = rq$. I know that necessary condition is that every root of $q$ is also a root to $p$, but I don't know if this is a sufficient condition since I only can prove it when the field is algebraic close $k=\bar{k}$.
Is this even a good question to be done? If not please let me know why
The ring of polynomials over a field is a unique factorization domain, and so divisibility is just a matter of the exponents in the factorizations into irreducibles.
|
2025-03-21T14:48:32.054442
| 2020-09-17T20:15:48 |
371951
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"authors": [
"Bipolar Minds",
"Noah Snyder",
"https://mathoverflow.net/users/102390",
"https://mathoverflow.net/users/22",
"https://mathoverflow.net/users/58211",
"skd"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371951"
}
|
Stack Exchange
|
Does the functor $\mathcal{C} \to \mathcal{Z}(\mathcal{C})$ have adjoints?
Let $\mathcal{C}$ be a braided monoidal category. We have a canonical functor $\mathcal{C} \to \mathcal{Z}(\mathcal{C})$ from $\mathcal{C}$ to the Drinfeld center $\mathcal{Z}(\mathcal{C})$ sending an object $V$ in $\mathcal{C}$ to $(V,c_{V,\,\_})$. Here, $c$ is the braiding in $\mathcal{C}$.
When does this functor admit left/right adjoints, and how do they look like? You are free to assume as much as you want on the category $\mathcal{C}$ (abelian, finite, factorizable, etc).
Abstractly: C is an E_2-monoidal category, so by the universal property of the Drinfeld center (as the Hochschild cohomology of C, at least if one works in the derived setting), the identity C -> C factors through the canonical functor Z(C) -> C. (The functor Z(C) -> C sends a pair (x, phi) to x.)
@skd Sure, but does this mean that the forgetful functor Z(C) -> C is an adjoint of the above? Sorry, maybe this is trivial..
No, the forgetful functor Z(C) --> C is defined even when C is not braided, it has a right adjoint (often called induction) which again doesn't depend on the braiding. For example, it sends 1 to $\bigoplus_x x \otimes x*$ in the semisimple case (and a "canonical coend" in general). It's adjoint can't be the inclusion C --> Z(C) which depends on the braiding for its construction.
@Noah Snyder Thx, that's what I thought. I wasn't sure what skd was trying to tell me
Short answer: Yes, it can possibly have an adjoint.
Longer answer:
Assume that $\mathcal{C}$ is rigid, and that the coend $L = \int^{X \in \mathcal{C}} X^* \otimes X$ exists.
It is a coalgebra.
Your assumptions on $\mathcal{C}$ were that it is braided, and in that case, it is well-known that $L$ is even a bialgebra.
Moreover, we know that ${}_L\mathcal{C} = \mathcal{Z}(\mathcal{C})$, i.e. the center of $\mathcal{C}$ is isomorphic to the category of modules over $L$.
Under this isomorphism, your "free central object" $(V, c_{V, -})$ is sent to the trivial $L$-module on $V$, i.e. the action is $\varepsilon \otimes V \colon L \otimes V \to V$, where $\varepsilon \colon L \to 1$ is the counit of $L$.
It is an algebra morphism.
Thus, walking everything through the isomorphisms, the inclusion functor can actually be interpreted as the pullback functor
\begin{align}
\varepsilon^* \colon {}_1\mathcal{C} = \mathcal{C} \to {}_L\mathcal{C}
\ .
\end{align}
A sufficient condition for pullbacks along algebra morphisms to have adjoints was identified in my answer to my own question over on M.SE.
Translating to our situtation, $\varepsilon^*$ has a left adjoint if $\mathcal{C}$ has coequalizers and $L$ is coflat (i.e. $L \otimes - $ preserves coequalizers).
Then the left adjoint sends an $L$-module $(V, r)$ to the coequalizer of
$$
r,\ \varepsilon \otimes id_V \colon L \otimes V \to V
\ .
$$
So for a particular situation where it works:
take $\mathcal{C}$ to be a braided finite tensor category in the sense of EGNO.
Then in particular, $\mathcal{C}$ is abelian, so it has coequalizers, and the tensor product is exact, so every object is coflat.
Moreover, it's well-known that for these kinds of categories, the coend $L$ indeed does exist.
|
2025-03-21T14:48:32.054676
| 2020-09-17T20:54:38 |
371955
|
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371955"
}
|
Stack Exchange
|
Conditions on the inequality with a gauge norm
Let $\Phi(x)=\int_0^x \phi(y)\,dy$, $x \in \mathbb{R}_+$, be an N-function, and let $u$ be locally inferable on $\mathbb{R}_+$. Consider the gauge norm
$$
\rho_{\Phi,u}(f)=\inf\{\lambda>0: \int_{\mathbb{R}_+}\Phi\left(\frac{|f(x)|}{\lambda}\right)u(x)\,dx\leq 1\},
$$where $f \in M_+(R_+)$.
Suppose that $\Phi(2t)\approx \Phi(t)$ for $t\gg1$ and $u(\infty)=\infty$.
Question: Under which conditions would the following inequality hold:
$$
\rho_{\Phi, u}\left(t^{-1}\int_0^t f^*\right)\leq C \rho_{\Phi,u}(f^*),
$$
where $f^*$ is a nonincreasing rearrangement of $f$ on $\mathbb{R}_+$.
|
2025-03-21T14:48:32.054746
| 2020-09-17T21:02:35 |
371956
|
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/371956"
}
|
Stack Exchange
|
What is the importance of "small doubling" in the theory of approximate groups?
One question I have is "why are approximate groups important?". If the small doubling constant is $1$ then it's definitely a group. If I read Green's note correctly. (1, 2)
To be more specific let's look at Freiman's theorem.
Thm Let $G$ be a group and $A \subset G$ be a finite subset such that $|A^2| < \frac{3}{2}|A|$. There exists a subgroup $H$ with $|H| = |A^2|$ such that for every $a \in A$, we have $A \subset aH = Ha$.
One motivation I could see for approximate groups is that the objects we are dealing with are not quite perfectly symmetric. Perhaps the object is not quite a perfect circle, so that when we rotate it doesn't quite map to itself $A \cap R_\theta A \subset A$. This might have a name in the literature already. Such a shape might appear in the Number Theory or Fourier Series or something.
So why are theorems like this important? Or why can we already look at this as objects of pure study? Also what's so special about the fraction $\frac{3}{2}$ that is making the proof easier?
The lemma in the textbook doesn't look any better. (Book)
Lemma Let $G$ be a group and let $A \subset G$ be a finite subset such that $|A^2| < \frac{3}{2}|A|$ then $H = A^{-1}A$ is a subgroup of $G$. Moreover $H = AA^{-1}$ and $|H| < 2|A|$.
So how "close" are we to proving the theorem here?
I would say that it is not the approximate groups per se, but rather the idea of set addition which is important. This idea is not an invention of one single person; it goes back to Fermat (1640(!), the two-square theorem), Lagrange (1770, the four-square theorem), Goldbach (1742, Goldbach's conjecture), Cauchy and Davenport (1813 / 1935, the Cauchy-Davenport theorem), Minkowski (around year 1890, Minkowski addition), Schnirelmann (around year 1930, Schnirelmann's density theorem) and many, many, many others.
Concerning "what's so special about the fraction $\frac32$ that is making the proof easier?" It is not that the proof gets easier; the structure breaks at $\frac32$, and the assertion becomes wrong if you replace the coefficient $\frac32$ with, say, $1,50001$. Maybe, the right way to understand this is to concentrate first on the commutative case and study Kneser's theorem first.
|
2025-03-21T14:48:32.054930
| 2020-09-17T21:36:26 |
371959
|
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"authors": [
"Ben McKay",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/371959"
}
|
Stack Exchange
|
What is the Lipschitz constant of the differential of the matrix exponential $\mathfrak{so}(3)\to \mathrm{SO}(3)$
I'm interested in numerical methods on $\mathrm{SO}(3)$ manifold, and working on a particular problem using the exponential coordinates:
$$
R(u) := \exp(u_\times)
$$
with $u\in \mathbb{R}^3$ and where $u_\times \in \mathfrak{so}(3)$ is the cross-product matrix of vector $u$.
The directional derivative of $R(u)$ in the direction $Y$ is:
$$
[D_u R]Y = [T(u)Y]_\times R(u)
$$
for any vector $Y\in \mathbb{R}^3$, where
$$
T(u) := \int_0^1R(su)ds
$$
Both $R$ and $T$ are Lipschitz continuous with constants $1$ and $\tfrac{1}{2}$ respectively:
$$
\|R(u)-R(v)\| \le|u-v| \\
\|T(u)-T(v)\| \le \tfrac{1}{2}|u-v|
$$
for any $u$ and $v$, where I use the operator norm (subordinate norm) of the Euclidean norm.
To find the convergence bounds of Newton's iterations for the numerical method I'm using (conditions of Kantorovich) I need to estimate a bound on the second derivative (which is really hard to compute explicitly as far as I know), or the Lipschitz constant of the differential. Experimentally (using a program) I found that it is $1$. How to prove this?
It is probably easier to work with the double cover $SU(2)$, i.e. the unit quaternions, since the exponential is then the usual one applied to imaginary quaternions. Then use the fact that $SU(2)\to SO(3)$ is a local isometry.
Found the proof! It's done using the integral definition of $T$:
$$
T(v) = \int_0^1 R(su) ds = \lim_{n\rightarrow \infty} \frac{1}{n}\sum_{i=1}^n R\left(\tfrac{i}{n}v\right)
$$
So for any vectors $X$ and $Y$:
\begin{align*}
&\biggl|\left[\mathrm{D}_v \left(R(v)X\right)\right]Y - \left[\mathrm{D}_u \left(R(u)X\right)\right]Y\biggr| = \biggl|\left[R(u)X\right] \times \left[T(u)Y\right] - \left[R(v)X\right] \times \left[T(v)Y\right]\biggr| \\
&\le \lim_{n\rightarrow \infty} \frac{1}{n} \sum_{i=1}^n \biggl|\left[R(u)X\right] \times \left[R(\tfrac{iu}{n})Y\right] - \left[R(v)X\right] \times \left[R(\tfrac{iv}{n})Y\right]\biggr|
\end{align*}
and we only need to prove that each summand is less than $|u-v| |X| |Y|$:
\begin{align*}
& \biggl|\left[R(u)X\right] \times \left[R(\tfrac{iu}{n})Y\right] - \left[R(v)X\right] \times \left[R(\tfrac{iv}{n})Y\right]\biggr| \\
&= \biggl|R\left(\tfrac{i}{n}u\right)\left[\left(R\left(\tfrac{n-i}{n}u\right)X\right) \times Y\right] - R\left(\tfrac{i}{n}v\right) \left[\left(R\left(\tfrac{n-i}{n}v\right)X\right) \times Y\right]\biggr| \\
& \le \biggl|\left[ R\left(\tfrac{i}{n}u\right) - R\left(\tfrac{i}{n}v\right) \right]\left[\left(R\left(\tfrac{n-i}{n}u\right)X\right) \times Y\right]\biggr| \\
&+ \biggl| R\left(\tfrac{i}{n}v\right) \left[\left(\left(R\left(\tfrac{n-i}{n}v\right) - R\left(\tfrac{n-i}{n}u\right)\right)X\right) \times Y\right]\biggr| \\
&\le \bigl|\tfrac{i}{n}u - \tfrac{i}{n}v\bigr| \bigl|X\bigr| \bigl|Y\bigr| + \bigl|\tfrac{n-i}{n}u - \tfrac{n-i}{n}v\bigr| \bigl|X\bigr| \bigl|Y\bigr| \\
&= \bigl|u - v\bigr|\bigl|X\bigr| \bigl|Y\bigr|
\end{align*}
where we used the invariance of vector products under rotations, the triangle inequality and that $R$ is $1$-Lipschitz (see this question).
Since $X$ and $Y$ are arbitrary,
$$
\left\|\mathrm{D}_v R(v) - \mathrm{D}_u R(u)\right\| \le |u-v|
$$
in the subordinate norm.
|
2025-03-21T14:48:32.055116
| 2020-09-17T21:59:00 |
371960
|
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"Anthony Quas",
"David Roberts",
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|
Stack Exchange
|
Subset of $\mathbb R$ with equal Fourier, Hausdorff and Minkowski dimensions
It is a standard fact that for $0\leq s\le1$, there is a compact set $C\subseteq [0,1]$ with Hausdorff and Minkowski dimensions $s$ (by modifying the construction of a Cantor set).
It is also a standard fact that for $0\leq s\le1$, there is a compact set $S\subseteq [0,1]$ with Fourier and Hausdorff dimensions $s$.
My question is: for an arbitrary $0\leq s\le1$, can we find a subset of $\mathbb R$ so that all three dimensions are equal?
Can you remind us of the definition of Fourier dimension? (Or give a link)?
The Fourier dimension of a bored set in R^n is the supremum of real numbers $s\in [0,n]$ such that there is a Borel probability measure supported on A, with decay of the order |x|^{-s/2}.
Here's a reference: Thomas William Körner, Hausdorff and Fourier dimension
Studia Mathematica 206, Issue 1 (2011) pages 37-50, https://doi.org/10.4064/sm206-1-3
Yes. We just use a Baire Category argument (a similar technique also works in high dimensions). Consider the complete metric space $X$ of pairs $(E,\mu)$, where $\mu$ is a probability measure supported on $E$ such that
$$ \sup_{\xi \in \mathbf{Z}} |\widehat{\mu}(\xi)| |\xi|^{s/2} < \infty, $$
and $E$ is a compact subset of $[0,1]$. We define a distance function
$$ d((E_1,\mu_1),(E_2,\mu_2)) = \max \left( d_H(E_1,E_2), \sup_{\xi \in \mathbf{Z}} |\widehat{\mu_1}(\xi) - \widehat{\mu_2}(\xi)| |\xi|^{s/2} \right) $$
where $d_H$ is the Hausdorff metric between two sets. It is a useful heuristic that a generic set is as `thin as possible' with respect to the Hausdorff metric. It is simple to see that for any $(E,\mu)$ in $X$, the Fourier dimension of $E$ is at least equal to $s$, so we should expect quasi-all elements of $X$ have dimension $s$.
For each $t > s$, $\delta > 0$, and $\varepsilon > 0$, set
$$ A(t,\delta,s) = \{ (E,\mu) \in \mathcal{X} : |E_\delta| < \varepsilon \cdot \delta^s \} $$
where $E_\delta$ is the $\delta$ thickening of $E$. Then $A(t,\delta,s)$ is an open subset of $X$, and
$$ \bigcap_{n = 1}^\infty \bigcap_{m = 1}^\infty \bigcap_{k = 1}^\infty A(s+1/n,1/m,1/k) $$
is the set of all pairs $(E,\mu)$ in $X$ where $E$ has Minkowski dimension $s$. Thus it suffices to argue that $A(t,\delta,\varepsilon)$ is dense in $X$ for all required parameters. It is slightly technical to argue this, but the basic idea is to consider a random construction which, given a pair $(E_0,\mu_0)$, considers the random measure
$$ \mu = \mu_0 \cdot \sum_{k = 1}^K \phi_{\varepsilon_0}(x - x_k) $$
where $x_1,\dots, x_K$ are uniformly distributed on $[0,1]$, $\varepsilon_0 = K^{-1/s}$, and $\phi_{\varepsilon_0}$ is a smooth bump function supported on a ball radius $\varepsilon_0$. One then shows that with high probability that
$$ \sup_{\xi \in \mathbf{Z}} |\widehat{\mu}(\xi) - \widehat{\mu_0}(\xi)| = o(1) $$
as $K \to \infty$, and that $d_H(\text{supp}(\mu), \text{supp}(\mu_0)) \to 0$.
|
2025-03-21T14:48:32.055321
| 2020-09-17T23:26:14 |
371961
|
{
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"Deane Yang",
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|
Stack Exchange
|
Christoffel symbols as the expansion coefficients of covariant/contravariant derivatives
Page 155 of Vector and Tensor Analysis with Applications, by A.I Borishenko and I.E. Tarapov, the authors state that Christoffel symbols of the second kind are expansions of $\frac{\partial {\bf e}_j}{\partial x_k}$ with respect to the basis ${\bf e}_1 , {\bf e}_2 , {\bf e}_3$, whereas Christoffel symbols of the first kind are expansions of $\frac{\partial {\bf e}_j}{\partial x_k}$ with respect to the basis ${\bf e}^1 , {\bf e}^2 , {\bf e}^3$. Yet different derivatives subsequently appear in the arguments that follow, as shown below.
Would someone kindly shed a light on what's going on here?
This question was originally posted on math.stackexchange but I wasn't able to find any help. Also, I am self-taught and not a mathematician.
I'd say there's definitely a typo on the lhs of (5.10), and the usage of "expansion coefficients" is a bit surprising, but consistent if we fix the aforementioned typo.
Agreed, all cases of $\partial {\bf x}_k$ must be a typo, and should be $\partial {\bf x}^k$. That aside, have you got an inkling of how (5.9) and (5.10) imply (5.11)?
I don't see why one would need (5.9) - I would have just multiplied (the corrected) (5.10) by ${\bf e}{l} $ and used ${\bf e}^{i} \cdot {\bf e}{l} = \delta^{i}_{l} $, unless this book doesn't define the dual basis by the latter relation (which would also be very surprising).
@MichaelEngelhardt I thought so too but how does one get $g_{il}$ from $\delta_l^i$? -- see equation (5.12)
For the first equation, take the dot product of (5.9) with $e_i$. For the second, take the dot product of (5.10) with $e^i$.
@DeaneYang Many thanks! Can you confirm that (5.11) follows from (5.10) and the recovery theorem.
Not sure what the recovery theorem is. It follows from the fact that $e_i\cdot e^j = 1$ if $i=j$ and $0$ otherwise.
@DeaneYang That's what I meant. If you care to, please post an answer and I will mark it as such.
Thanks. I'm a bit reluctant to, since I'm just adding details @MichaelEngelhardt's correct answer. So I'll let him.
With a hat tip to Deane Yang, who supplied half of the discussion in the comments to the OP, the conclusion is:
The three instances of $\partial / \partial x_k $ in the displayed book excerpt are typos and should read $\partial / \partial x^k $.
(5.11) follows from (5.10) by taking the dot product with ${\bf e}_{l} $,
$$
{\bf e}_{l} \cdot \frac{\partial {\bf e}_{j} }{\partial x^k } = [ij,k] \ {\bf e}_{l} \cdot {\bf e}^{i} = [ij,k] \ \delta_{l}^{i} = [lj,k]
$$
(5.12) (left) follows from (5.9) by taking the dot product with ${\bf e}_{l} $, and invoking (5.11),
$$
[lj,k] = {\bf e}_{l} \cdot \frac{\partial {\bf e}_{j} }{\partial x^k } = \left\{ \begin{array}{c} i \\ j \ \ \ k\end{array} \right\} {\bf e}_{l} \cdot {\bf e}_{i} = \left\{ \begin{array}{c} i \\ j \ \ \ k\end{array} \right\} g_{li}
$$
(5.12) (right) follows from (5.10) by taking the dot product with ${\bf e}^{l} $, and invoking the definition (5.7),
$$
\left\{ \begin{array}{c} l \\ j \ \ \ k\end{array} \right\} =
{\bf e}^{l} \cdot \frac{\partial {\bf e}_{j} }{\partial x^k } = [ij,k] \ {\bf e}^{l} \cdot {\bf e}^{i} = [ij,k] g^{li}
$$
|
2025-03-21T14:48:32.055530
| 2020-09-17T23:33:44 |
371962
|
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"authors": [
"Drew Heard",
"Piotr Pstrągowski",
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|
Stack Exchange
|
$p$-completeness of the function spectrum $F(\Sigma^{\infty} BS, \Sigma^{\infty} BK)$
Let $S$ be a finite $p$-group and $K$ a compact Lie group, in the paper A Segal conjecture for $p$-completed classifying spaces, it is said that the function spectrum $F(\Sigma^{\infty} BS, \Sigma^{\infty} BK)$ is $p$-complete, but I have not succeeded in proving it. I hope this remains true when, more generally, $S$ is a $p$-toral group (replacing $\Sigma^{\infty} BS$ by $(\Sigma^{\infty} BS)^{\wedge}_p$, because $\Sigma^{\infty} BS$ is no longer $p$-complete). Any suggestion or idea?, please.
Ragnarsson's paper links to 'Classifying G-spaces and the Segal conjecture' - did you check there?
@Drew Yes, I did but found no mention of that result so far. I might have to take a closer look at that article.
Here is an argument (not as clean as Piotr's below). Use with caution; it's possible that I've made a mistake. We don't use any properties of $\Sigma^\infty BK$ -- this could be an arbitrary spectrum. But it's crucial that that we use $\Sigma^\infty BS$ with $S$ a finite $p$-group.
Proposition: If $S$ is a finite $p$-group and $U$ is any spectrum, then $F(\Sigma^\infty BS, U)$ is $p$-complete.
We will prove this using the following lemmas:
Lemma 1: If $S$ is a finite $p$-group, then $\tilde H_\ast(BS; A) = 0$ if $p: A \to A$ is an isomorphism.
As Piotr points out in his answer, this is a consequence of Maschke's theorem. We provide a proof using basic algebraic topology and finite group theory.
Proof: We reduce to the case where $S$ is abelian, where this is a standard calculation. For if $S$ is nonabelian, then there is always a nontrivial short exact sequence $Z(S) \to S \to S / Z(S)$ where $Z(S)$ is the center of $S$. Because $Z(S)$ is the center of $S$, the action of $S/Z(S)$ on $Z(S)$ is trivial. Thus in the Serre spectral sequence $H_\ast(B(S/Z(S)); \underline{H_\ast(B(Z(S))}) \Rightarrow H_\ast(BS)$ we have trivial coefficients. So we can induct on the order of $S$.
Corollary 2: Let $S$ be a finite $p$-group. Then $\Sigma^\infty BS$ is $p$-local.
Proof: Let $\ell \neq p$ be a different prime. The claim is that $\ell: \Sigma^\infty BS \to \Sigma^\infty BS$ is an equivalence of spectra. It suffices to show that $\ell: \Sigma BS \to \Sigma BS$ is an equivalence of spaces. By the homology Whitehead theorem for field coefficients, it suffices to show that $\ell: \tilde H_\ast(BS;k) \to \tilde H_\ast(BS;k)$ is an isomorphism for $k = \mathbb Q$ or $k = \mathbb F_q$ with $q$ a prime. But if $k = \mathbb Q$ or $k = \mathbb F_q$ with $q \neq p$, both sides are zero by Lemma 1, while if $k = \mathbb F_p$, then this follows from $\ell$ being prime to $p$.
Lemma 3: Let $T$ be a $p$-torsion spectrum -- i.e. $T$ is $p$-local and has trivial rationalization -- and let $U$ be an arbitrary spectrum. Then $F(T,U)$ is $p$-complete.
Proof: Let $X$ be such that $X \wedge M(p) = 0$; the claim is that $Map(X,F(T,U)) = 0$. Equivalently, the claim is that $Map(T, F(X,U)) = 0$. Because $T$ is $p$-local, it is equivalent to claim that $Map(T, F(X,U)^{(p)}) = 0$, where we have taken a $p$-colocalization (i.e. we have applied the right adjoint $(-)^{(p)}$ to the inclusion of the $p$-local spectra into all spectra). Then,
$$F(X,U)^{(p)} \wedge M(p) = (F(X,U)\wedge M(p))^{(p)} = F(X \wedge \Sigma^{-1} M(p), U)^{(p)} = 0$$
The first equivalence comes because $(-)^{(p)}$ commutes with finite colimits, the second by Spanier-Whitehead duality, and the third by the hypothesis that $X \wedge M(p) = 0$. Since $F(X,U)^{(p)}$ is by definition $p$-local, this means that it is rational. So
$$Map(T, F(X,U)^{(p)}) = Map_{H\mathbb Q}(H\mathbb Q \wedge T, F(X,U)^{(p)}) = 0$$
because by hypothesis $H\mathbb Q \wedge T = 0$.
Proof of Proposition: By Lemma 1, $\Sigma^\infty BS$ has trivial rationalization, and by Corollary 2, $\Sigma^\infty BS$ is $p$-local. So by Lemma 3, $F(\Sigma^\infty BS, U)$ is $p$-complete.
Tim's argument is correct, and here's a different way to see this.
To say that $\Sigma^{\infty} BS$ is a finite $p$-group has trivial rationalization and is $p$-local is the same as to observe that it is $p$-torsion, i.e. has $p$-torsion homotopy groups.
To see this without involving the Serre spectral sequence, observe that both the rational and $\mathbb{Z}/l$ cohomology (for $l \neq p$) of $\Sigma^{\infty} BS$ vanishes by Maschke's theorem. For a connective spectrum, this forces it to be $p$-torsion, by Hurewicz (applied to $\Sigma^{\infty} BS \otimes M(l)$).
We also have that $F(T, U)$ is p-complete for any $p$-torsion $T$ and arbitrary $U$.
To see the latter, observe that the subcategory of $p$-torsion spectra is generated under colimits and desuspensions by $M(p)$ (this is the same as saying that any such non-zero spectrum admits a non-zero map from $M(p)$, which is clear). Since $F(-, U)$ takes colimits to limits, and $p$-complete spectra are closed under limits, we then just need to know $F(S^{0}/p, U)$ is $p$-complete, but this is the same as $\Sigma^{-1} U \otimes S^{0}/p$, so we're done.
I added an argument using Maschke's theorem.
(Sorry I deleted my comment -- I was trying to edit it to say thanks for the addition, but went longer than 5 minutes and then ended up deleting it!) Interesting that both arguments for $T$ $p$-torsion $\Rightarrow F(T,U)$ $p$-complete seem to rely on the fact that $M(p)$ is a shift of its own Spanier-Whitehead dual -- especially in your formulation, this seems like the least "formal" fact used.
A very slightly different point of view: A spectrum $C$ is $p$-complete iff $F(A,C)=0$ for every $S/p$-acyclic spectrum $A$, i.e. every $A$ such that $A/p=0$. A spectrum $T$ is $p$-torsion if and only if $F(T,A)=0$ for every $S/p$-acyclic $A$. So if $T$ is $p$-torsion then for every $U$ the spectrum $F(T,U)$ is $p$-complete since, for every such $A$, $F(A,F(T,U))=F(T,F(A,U))=0$. The key point is that $F(A,U)$ is always $S/p$-acyclic if $A$ is $S/p$-acyclic. ($F(A,U)/p$ is a shift of $F(A/p,U)$.)
|
2025-03-21T14:48:32.056006
| 2020-09-17T23:49:13 |
371963
|
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|
Stack Exchange
|
Beilinson-Bernstein for nonintegral levels
If one wants to understand representations of $\mathfrak{g}$ (a finite dimensional semisimple Lie algebra) of weight $\lambda$, the happiest you could be is if $\lambda+\rho$ is (integral) regular dominant, i.e. it's an element of the weight lattice whose product $\langle \lambda+\rho,\check{\alpha}\rangle$ with every simple coroot $\check{\alpha}$ is a negative integer. In this case there's an equivalence between the category of these representations and twisted $\mathscr{D}$ modules on the flag variety $U(\mathfrak{g})_\lambda\text{-mod}\simeq \mathscr{D}_\lambda\text{-mod}$. See the book by Hotta, Toshiyuki, Tanisaki, section 11.2.
If $\lambda+\rho$ is just (integral) regular (nothing in $W$ stabilises it), there is an equivalence of derived categories. I've not worked with this, but presumably along with translation functors (tensoring with $\mathscr{O}(\mu)$ to make the twist of the $\mathscr{D}$ module dominant) this allows you to use standard Beilinson-Bernstein above to get a handle on such modules.
My question is: what is currently known when $\lambda+\rho$ is non-integral? If it splits up into rational and irrational case (where all $\langle \lambda+\rho,\check{\alpha}\rangle\in\mathbf{Q}$ or otherwise), I am more interested in the rational case.
IIRC the story is very similar to the integral case, except that instead of $W$ you need to consider the subgroup $W'$ generated by those simple reflections $s_{\alpha}$ that send $\lambda$ to $\lambda+n\alpha$ for some integer $n$. Then if $\lambda$ is $W'$-dominant then Beilinson-Bernstein holds and there is a theory of reflection functors that takes care of the other weights. But I might be misremembering/confusing with something else.
(I suspect that your rational/irrational bifurcation is motivated by the case of affine Kac-Moody algebras, where rationality of the level controls what category O looks like. There it appears because you have the entire affine Weyl group to play with, but here you only have the finite Weyl group and so the possibilities are more limited.)
Integrality doesn't really seem to play a role in the statement of Beilinson-Bernstein as far as I can tell. For a general weight $\lambda \in \mathfrak h^\ast$ it makes sense to talk about the notions of dominant and regular.
If $\lambda$ is regular, then there is a derived localization:
$D(D_{G/B}^\lambda-mod) \simeq D(U_\lambda -mod)$
If $\lambda$ is both regular and dominant, there is an abelian category localization:
$D_{G/B}^\lambda-mod \simeq U_\lambda-mod$
There are no integrality assumptions in the statement of Theorem 3.3.1 in the paper "A Proof of the Jantzen Conjecture", for example.
I suppose that defining the ring of twisted differential operators requires a bit more discussion in the non-integral case.
Thanks! I think my confusion came from the fact that in Hotta et al, integrality is assumed.
|
2025-03-21T14:48:32.056219
| 2020-09-18T00:19:32 |
371964
|
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"authors": [
"Bjørn Kjos-Hanssen",
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"url": "https://mathoverflow.net/questions/371964"
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|
Stack Exchange
|
When is a function on symmetric positive definite matrices an expectation of Gaussian?
Is there some characterization of real-valued functions of the form $\Phi(C)=\mathbb{E}F(X)$, where $X$ has the Gaussian $N(0,C)$ distribution, on the space of symmetric positive semidefinite $n\times n$ matrices $C$? In other words, given $\Phi\colon \mathrm{SPD}\to \mathbb{R}$, is there a way to tell whether such $F$ exists?
Edit: For example, is it clear that $\det(C)$ is not of this form?
Edit 2: Answer below shows that $\det(C)$ is not of this form even for $n=2$.
Here's an approach to determine if $\det(C)$ is of this form with $F$ analytic and $n=2$.
By Sylvester's criterion a matrix $\begin{pmatrix} a& b \\ b & c\end{pmatrix}$ is symmetric positive semidefinite iff $a\ge 0$ and $c\ge 0$, and $ac-b^2\ge 0$.
Let $X=(X_1,X_2)$ be normal with mean 0 and $E(X_1X_2)=b$, $E(X_1^2)=a$, $E(X_2^2)=c$.
The Pearson correlation coefficient is $\rho=b/\sqrt{ac}$.
Suppose $F(x,y)=\sum c_{mn}x^my^n$.
Is there an $F$ with $E(F(X))=ac-b^2$?
We have $X=\sqrt{a}Z_1$, $Y=\sqrt{c}(\rho Z_1+\sqrt{1-\rho^2}Z_2)$ where $Z_i$ are independent standard normal, so
using
$$E\left[(\rho Z_1+\sqrt{1-\rho^2}Z_2)^{2k}\right]=E\sum_{t=0}^{2k}\binom{2k}{t}\rho^t(1-\rho^2)^{(2k-t)/2}Z_1^t Z_2^{2k-t}$$
$$=\sum_{t=0}^{2k}\binom{2k}{t}\rho^t(1-\rho^2)^{(2k-t)/2}E[Z_1^t Z_2^{2k-t}]
=\sum_{u=0}^{k}\binom{2k}{2u}\rho^{2u}(1-\rho^2)^{(2k-2u)/2}E[Z_1^{2u} Z_2^{2k-2u}]$$
$$=\sum_{u=0}^{k}\binom{2k}{2u}\rho^{2u}(1-\rho^2)^{(2k-2u)/2}(2u-1)!!(2(k-u)-1)!!$$
(where $(-1)!!=1$)
we have
$$E(X^{2\ell}Y^{2k})=a^\ell c^k\sum_{u=0}^{k}\binom{2k}{2u}\rho^{2u}(1-\rho^2)^{k-u}(2(\ell+u)-1)!!(2(k-u)-1)!!$$
When $k=\ell=1$, it is $ac((1-\rho^2)+3\rho^2)=ac+2b^2$
since $\rho^2=b^2/ac$.
When $k=0$ and $\ell=2$, it is $3a^2$.
When $k=2$ and $\ell=0$, it is $3c^2-4b^4/a^2$.
I guess we should also do the case $E(X^3Y^1)$ and a couple of others.
A Taylor series in variables $a,b,c$ defines the zero function only if all coefficients are zero.
So now by calculating these expectations $E(X^pY^q)$ we can determine whether
$ac-b^2$ is obtainable.
Thank you, good idea. Only fourth order polynomials in $(X,Y)$ will give second order polynomials in $(a,b,c)$ so together with $\mathbb{E}X^3 Y=3ab$ your observations above means that such analytic $F$ does not exist. I think this also means that such non-analytic $F$ does not exist either, because otherwise $\det(C+I)=\mathbb{E} F(X+z_1,Y+z_2) = \mathbb{E}\tilde F(X,Y)$ where $\tilde F$ is now analytic but $\det(C+I)=ac-b^2+a+c$ has the same second order terms. So $\det$ is not of this form then.
@D_809 Can you explain how you get $\tilde F$?
$\tilde F(x,y) = \mathbb{E}F(x+z_1,y+z_2)$ where $(z_1,z_2)$ is standard Gaussian. In the above comment $(X,Y)$ was your $(X_1,X_2)$ and $(z_1,z_2)$ was independent of $(X,Y)$. By the way, this result gives the identities faster: https://en.wikipedia.org/wiki/Isserlis%27_theorem (I can not write @ your name for some reason).
Great. Well, you automatically notify the author when you comment on their answer.
|
2025-03-21T14:48:32.056427
| 2020-09-18T01:10:53 |
371966
|
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|
Stack Exchange
|
What is the name for Boolean algebra's version of $\models$ between sets of identities and identities?
On p62 in Schaum's Outline of Theory and Problems of Boolean Algebra and Switching Circuits by Elliott Mendelson (1970),
Part (b) of the corollary says that if an identity is satisfied by some algebra which satisfies the axioms of Boolean algebras, then it is satisfied by any algebra which satisfies the axioms of Boolean algebras.
In model theory, logic, and universal algebra, consider the following statement
Given a set of formulas or identities, if a formula or identity is satisfied by some (algebraic) structure which satisfies the set of formulas or identities, then it is also satisfied by any (algebraic)
structure which satisfies the set of formulas or identities.
Is it true that the statement isn't true in general?
Is the statement true in a context other than or more general than Boolean algebra?
Is there a name for the concept or property represented by the statement, either in Boolean algebra, or more generally in universal algebra, logic, or model theory?
Is the concept or property represented by the statement mentioned in some standard books in Boolean algebra, logic (e.g. Ebbinghaus'), universal algebra (e.g. Burris'), or model theory (e.g. Chang's)? (Burris' universal algebra book has Chapter IV on Boolean algebra, which I wonder if has relevant coverage, and the chapter is not as easy to access as Mendelson's book for me now.)
Thanks.
It seems the name for this idea is equationally complete theory, see page 30 of Walter Taylor's Equational Logic survey.
Not every theory is like that:
for example in the theory of lattices, which is defined by axioms such as
$$x\wedge (y\wedge z)=(x\wedge y)\wedge z$$
(the full list is at Wikipedia)
some are distributive and some are not.
So
$$x\wedge (y\vee z)=(x\wedge y)\vee x(\wedge z)$$
is an identity that holds in some but not all lattices.
Thanks. Is your example a counter-example to part (b) of the corollary?
@Tim yes but anyway (a) and (b) seem to be equivalent by the Completeness Theorem.
Your example is a counter-example to the Completeness Theorem?
No, because the distributive law is not provable from the axioms defining lattices
|
2025-03-21T14:48:32.056594
| 2020-09-18T02:13:55 |
371972
|
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"Acccumulation",
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|
Stack Exchange
|
Is it possible to create a polynomial $p(x)$ with this relation between $p(0)$ and $p(c)$?
Given $b$ and $c$ with $b,c>1$, is it possible to construct a polynomial $p(x)$, whose degree is $n$ for all $c$ and $b$, such that:
$|p|$ is strictly increasing on $[1,c]$
and $|b \cdot p(c)| < |p(0)|$?
This might be satisfied by an interpolating polynomial, but how to actually construct it is beyond me.
It doesn't make sense to say that $n$ depends on neither $c$ nor $b$. If you were to word it as "Does there exist an $n$ such that for all $b,c$, there exist a polynomial of degree $n$ such that ..." Also, your first condition can almost be simplified to "the absolute value is strictly increasing".
@Acccumulation Noted and changed accordingly.
No, it's not possible to construct such a $p$ whose degree $n$ is bounded independently of $b$ and $c$. In fact, it's not possible even if we fix the value of $c$. I'll prove this for $c=2$ below, but the same argument works in general.
Suppose to the contrary that it were possible. Then for every $b>1$ we could choose a polynomial $p_b$ such that
$p_b$ has degree $n$;
$p_b$ takes non-negative values and is strictly increasing on $[1,2]$;
$b\cdot|p_b(2)|<|p_b(0)|=1$.
(We take a polynomial satisfying the desired conditions and multiply by an appropriate scalar.)
Key claim: The coefficients of a polynomial $p_b$ satisfying these three conditions are bounded independently of $b$. That is, there is a constant $B$ such that the absolute value of every coefficient of every polynomial $p_b$ is at most $B$.
Proof of claim: Fix any $n+1$ distinct real numbers $x_0,x_1,\dots,x_n$ between $1$ and $2$ inclusive, and for $i=0,1,\dots,n$ define the polynomial $f_i(x)$ by$$f_i(x)=\prod_{j\neq i}\frac{x-x_j}{x_i-x_j} \,,$$where the product is taken over all indices $i=0,1,\dots,n$ except $i=j$. Thus, $f_i$ is the unique degree $n$ polynomial such that $f_i(x_i)=1$ and $f_i(x_j)=0$ for $j\neq i$.
Now the theory of Lagrange interpolation says that for any polynomial $p$ of degree at most $n$, we have
$$p(x) = \sum_{i=0}^np(x_i)f_i(x) \,.$$
But in our case, we know that we have
$$|p_b(x_i)|\leq|p_b(2)|\leq b^{-1}\cdot|p(0)|<1$$
for every $b$, since $p_b$ is strictly increasing on $[1,2]$. Thus, the absolute value of every coefficient of every polynomial $p_b=\sum_{i=0}^np(x_i)f_i$ is at most $(n+1)B'$, where $B'$ is the largest absolute value of any coefficient of any of the $f_i$. This gives our bound independently of $b$, and proves the key claim.
Now by a compactness argument (a.k.a. the Bolzano--Weierstraß Theorem), our key claim implies that we may choose an increasing sequence of integers $b_1<b_2<\dots$ such that the polynomials $p_{b_i}$ converge coefficientwise to a polynomial $p$. What can we say about this limiting polynomial $p$? Well, by taking an appropriate limit of the above properties of the $p_{b_i}$, we find:
$p$ has degree $n$;
$p$ takes non-negative values and is weakly increasing on $[1,2]$;
$|p(0)|=1$; and
$|p(2)|\leq b_i^{-1}$ for every $i$.
Since the integers $b_i$ increase without bound, this final condition implies that actually $|p(2)|=0$. Since $p$ is non-negatively valued and weakly increasing on $[1,2]$, we find that $p$ actually has to be equal to $0$ on all of $[1,2]$. This implies that $p$ must be the zero polynomial. But this contradicts the assumption that $|p(0)|=1$.
Your first condition yields $$|p(c)|=|p(1)|+\int_{1}^c |p'(x)|dx:=\|p(x)\|.$$
All linear functionals on a finite-dimensional space are bounded, so if $\deg p\leqslant n$, we get $|p(0)|\leqslant C_n \|p(x)\|$ for certain $C_n$. Thus, if $b>C_n$, the second condition is not achievable.
I think this answer really gets to the point of what's going on here. I think it might be helpful to remark that both Fedor's solution and the one I gave revolve around similar ideas: using the first condition to bound the polynomial $p$, and deriving a contradiction by playing off this bound against the value of $p(0)$ using the second condition. However, in the proof I gave, these two steps are bound up together in the choosing to normalise everything such that $|p(0)|=1$, so I think Fedor's proof makes this underlying structure clearer.
@AlexanderBetts Fedor is a genius. Your answer is good too.
Let me present a more explicit version of Fedor’s argument.
Choose distinct $x_0,\dots,x_n\in[1,c]$.
By Lagrange’s interpolation formula, there exist constants $a_0,\dots,a_n$ such that
$$
p(0)=\sum_{I=0}^n a_ip(x_i)
$$
for each polynomial $p$ of degree not exceeding $n$. Therefore,
$$
|p(0)|\leq \sum_{I=0}^n |a_i|\cdot |p(x_i)|\leq
|p(c)|\cdot \sum_{I=0}^n |a_i|.
$$
|
2025-03-21T14:48:32.056896
| 2020-09-18T04:43:48 |
371978
|
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"Anonymous",
"Simon Wadsley",
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|
Stack Exchange
|
$R/I\cong R/\text{Ann}_R(R/I)$ but $I\neq\text{Ann}_R(R/I)$
I originally asked this on Stack Exchange but with no luck. Upon doing research with some noncommutative rings, I thought of a curious question. Does there exist a noncommutative unital ring $R$ and left ideal $I$ such that $R/I\cong R/\text{Ann}_R(R/I)$ as left $R$-modules but $I\neq \text{Ann}_R(R/I)$? I suspect that such an $R$ and $I$ do exist but are awkward to construct. Note that
$$\text{Ann}_R(R/I)=\{r\in R:\forall x\in R,\ rx\in I\}\subseteq I$$
and that $\text{Ann}_R(R/I)$ is always a two-sided ideal. As such, if $I$ is not two-sided then we already get $I\neq \text{Ann}_R(R/I)$. Any ideas would be appreciated.
My question is equivalent to this question with the additional constraint that $J$ must be two-sided.
That map is not surjective is it? What would map to the coset $1+I$? I was thinking about using free rings as well, but couldn't think of a way to make it work. Maybe there's a way though.
You might as well assume that $\mathrm{Ann}_R(R/I)=0$ since if $S=R/\mathrm{Ann}_R(R/I)$ then $R/I\cong S$ as $R$-modules if and only if $S/(I/\mathrm{Ann}_R(R/I))\cong S$ as $S$-modules.
So now the question is whether you can have a ring $R$ and a non-zero left ideal $I$ such that $R/I\cong R$. Equivalently can there be $r\in R$ such that (i) $\mathrm{ann}_R(r)=I\neq 0$ and (ii) there is some $s\in R$ with $sr=1$; if $\varphi\colon R/I\to R$ is the isomorphism then take $r=\varphi(1+I)$ and $\varphi(s+I)=1$. To further rephrase we want an element $r$ of $R$ that is a right zero-divisor and a right unit.
This is indeed possible. Here is one example: let $R$ be the $\mathbb{C}$-linear endomorphisms of the polynomial ring $\mathbb{C}[X]$ and let $r$ be the element of $R$ given by multiplication by $X$. It is easy to construct a left inverse for $r$ (in fact one can easily find infinitely many). But if $x$ is the element of $R$ that sends a polynomial to its constant term viewed as the constant polynomial then $xr=0$.
The punchline If you just wanted to pull a rabbit out of a hat you could simply say that for $R=\mathrm{End}_{\mathbb{C}}(\mathbb{C}[X])$ and $r\in R$ given by $r(f)=Xf$, $I=\mathrm{ann}_R(r)\neq 0$ and there is an isomorphism of left $R$-modules $R/I\to R$ given by $x+I\mapsto xr$. It is surjective since there is $s\in R$ with $sr=1$ (Moreover $\mathrm{Ann}_R(R/I)=0$).
That is not an isomorphism. It is not injective.
Sorry. You are right. I think it is all correct except the punchline doesn't correctly reinterpret the previous discussion. I'll edit.
Thanks! This is really useful!
|
2025-03-21T14:48:32.057090
| 2020-09-18T07:35:20 |
371984
|
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"Jakob Möller",
"gerw",
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"url": "https://mathoverflow.net/questions/371984"
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|
Stack Exchange
|
Sobolev embedding if $p=3/2$, $n=3$ and $k=2$
Usually, Sobolev's embedding theorem goes as follows. Suppose $1\leq p<q < \infty$ and $k> \ell$. If
$$ \frac{1}{p} - \frac{k}{n} = \frac{1}{q} - \frac{\ell}{n} ,$$
then $W^{k,p}(\mathbb{R}^n) \subset W^{\ell,q}$. In the case $p=n$, $k=1$ we have $W^{1,n}(\mathbb{R}^n) \subset BMO$. I was wondering what happens in the case $p=\frac{3}{2}$, $n=3$ and $k=2$? This, applied to the formula above, yields
$$0 = \frac{1}{q} - \frac{\ell}{3}$$
So, intuitively, $W^{2,3/2}$ should be contained in something close to $L^{\infty}$ (being the case $\ell=0$, $q= \infty$). But what is it exactly?
I am confused. To get $0$ on the lhs, we need (for $n = 3$ and $k = 2$) $p = 3/2$. And then, $W^{2,3/2} \subset W^{1,3} \subset BMO$.
@gerw Sorry, I phrased the question in a way that made it incomprehensible. It should be clearer now.
My answer is still valid.
Yes and I'm very grateful for it ;)
|
2025-03-21T14:48:32.057193
| 2020-09-18T07:38:28 |
371985
|
{
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"authors": [
"Dirk",
"Mark L. Stone",
"Rodrigo de Azevedo",
"https://mathoverflow.net/users/165553",
"https://mathoverflow.net/users/75420",
"https://mathoverflow.net/users/91764",
"https://mathoverflow.net/users/9652",
"见得哆啦"
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"url": "https://mathoverflow.net/questions/371985"
}
|
Stack Exchange
|
Minimizing the Frobenius norm of a quadratic matrix expression
Given matrices $R \in \mathbb R^{m \times n}$ and $Y \in \mathbb R^{p \times n}$, where $R$ is full rank, how can I solve the following optimization problems?
$$\min_{X \in \mathbb R^{p \times m}} \| X R R^T X^T -Y Y^T \|_F \tag{1}$$
or
$$\min_{X \in \mathbb R^{p \times m}} \| X R R^T X^T -Y Y^T \|_2 \tag{2}$$
Also, $R$ is the matrix of the m rows in $Y$.
I am guessing the solution of $(1)$ could be $X=YR^+$, but I cannot prove it.
Are you assuming the solution is unique? If you square the norm and find where the gradient vanishes, you should obtain a cubic matrix equation.
Thank you! Details about dimensions added!
What does it mean to say that a non-square matrix is full rank? Fat? Full row rank? Tall? Full column rank?
I always understand full rank as "rank as large as possible".
@Dirk Doesn't everyone?
What is $R^+$ in $X=YR^+$?
@MarkL.Stone I assume it is the pseudoinverse of $R$, but I am not the OP.
|
2025-03-21T14:48:32.057292
| 2020-09-18T08:50:50 |
371988
|
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"site": "mathoverflow.net",
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"url": "https://mathoverflow.net/questions/371988"
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|
Stack Exchange
|
Hessian generating functions
I am looking for a characterization of functions $\Phi: \mathbb{R}^n \to \mathbb{R}^{n \times n}$ such that $\Phi(\mathbf{x}) = \nabla^2 f(\mathbf{x})$ for a function $f$ which is twice continuously differentiable. Is there any composition rules for $\Phi$?
For example, $\Phi(\mathbf{x}) = \mathbf{A}$, where $\mathbf{A}$ is any constant symmetric matrix, is a trivial example for $\Phi$. However, I look for non-trivial examples which depend on $\mathbf{x}$. In other words, I want to find rules for $\Phi$ which can generate valid Hessian matrices.
Is this even possible? Can someone provide my correct references and pointers?
|
2025-03-21T14:48:32.057362
| 2020-09-18T09:11:16 |
371992
|
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"ABIM",
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"url": "https://mathoverflow.net/questions/371992"
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|
Stack Exchange
|
Measurable selection for argmin to distance
Let $Y$ be a Banach space and equip $Y$ with the weak topology. Now, let $X$ be a closed, bounded, and convex subset of $Y$ and suppose that the relative (weak) topology on $X$ is metrizable with metric $d_X$. Let $x_1,\dots,x_n \in X$ for some natural $n>0$.
Let $\emptyset\neq Z\subseteq X$ be a compact subset of $X$. Then, does the map
$$
x\mapsto \min_{1\leq i\leq n}\, d_X(x_i,x)
$$
admit a measurable selection? I.e.: Does there exist a measurable function
$$
S\mapsto \{1,\dots,n\} \mbox{ s.t. }
d_{X}(x_{S(x)},x)=\min_{i=1,\dots,n} d_{X}(x_i,x)
$$
for all $x\in Z$ and $S$ is measurable as a function from $Z$ to $\{1,\dots,n\}$ (where the latter has the $\sigma$-algebra $2^{\{1,\dots,n\}}$?)
And would viewing the map $S$ as a function into $\mathbb{R}$ still be measurable? (yes unless im missing smething obvious)
https://math.stackexchange.com/a/491538/21674
Here is a simple direct argument. For $i=1,\ldots,n$, let $$C_i=\{z\in Z\mid d(z,x_i)\leq d(z,x_j), j=1,\ldots,n\}.$$
Clearly, each $C_i$ is closed and hence measurable. Let $M_i=C_i\setminus\bigcup_{l=1}^{i-1}C_l$. The nonempty sets of the form $M_i$ form a finite measurable partition of $Z$. Now let $S$ map each $M_i$ to $x_i$.
|
2025-03-21T14:48:32.057815
| 2020-09-18T09:36:53 |
371993
|
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"authors": [
"Mathmo",
"Michael Renardy",
"Nik Weaver",
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"url": "https://mathoverflow.net/questions/371993"
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|
Stack Exchange
|
For self-adjoint $A$ and $B$, when is $(A+iB)^*$ the closure of $A-iB$?
Suppose that I have two self-adjoint operators $A$ and $B$ such that $\mathcal{D}(A)\cap\mathcal{D}(B)$ is dense and $B$ positive. Then $A\pm iB$ (with domains $\mathcal{D}(A)\cap\mathcal{D}(B)$) are closable. What are generic conditions so that $(A+iB)^*$ is the closure of $(A-iB)|_{\mathcal{D}(A)\cap\mathcal{D}(B)}$? $A$ or $B$ bounded suffices, but does this hold in general?
NB: What I'm really interested in is whether $0$ is in the resolvent set of the closure of $A+iB$, which follows easily from the above condition.
In general this is not true. Let $\Omega$ be a smooth bounded domain, let $A$ be $-\Delta$ with Neumann conditions and let $B$ be $-\Delta$ with Dirichlet conditions. The $(A+iB)$ is $(-1-i)\Delta$ with domain $H^2_0(\Omega)$. Its adjoint is $(-1+i)\Delta$ with domain $H^2(\Omega)$, which is not the closure of $A-iB$.
Is this $B$ self-adjoint?
@NikWeaver: Oops, I did not read carefully enough. Edited accordingly.
Nice! Thanks for the example @MichaelRenardy
Yeah, neat example!
|
2025-03-21T14:48:32.057933
| 2020-09-18T10:39:09 |
371996
|
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"authors": [
"Anonymous",
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"https://mathoverflow.net/users/72288",
"user237522"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/371996"
}
|
Stack Exchange
|
$k[h(x),y] \subseteq k[h(x),y] + \langle h(x),y \rangle \subseteq k[x,y]$
Let $k$ be a field of characteristic zero.
Let $h=h(T) \in k[T]$ with $\deg(h)=d \geq 2$ and $h(0)=0$ (namely, $h$ has zero constant term).
Consider the following chain of $k$-algebras:
$$k \subseteq k[h(x),y] \subseteq k[h(x),y] + \langle h(x),y \rangle_{k[x,y]} \subseteq k[x,y]$$
where $\langle h(x),y \rangle_{k[x,y]}$ is the ideal in $k[x,y]$ generated by $h(x)$ and $y$.
Observe that
$k[h(x),y] \subsetneq k[h(x),y] + \langle h(x),y \rangle_{k[x,y]}$,
since, for example, $xy \in k[h(x),y] + \langle h(x),y \rangle_{k[x,y]} - k[h(x),y]$.
Recall the Artin–Tate Lemma: "Consider the inclusions of rings $R \subset B \subset A$. Suppose that $R$ is Noetherian, that $A$ is a finitely generated algebra over $R$ and that $A$ is a finitely generated module over $B$. Then $B$ is a finitely generated algebra over $R$".
Here $R=k[h(x),y]$, $B=k[h(x),y] + \langle h(x),y \rangle_{k[x,y]}$, $A=k[x,y]$.
Here $A$ is moreover a finitely generated module over $R$ (with $d$ generators:
$1,x,\dotsc,x^{d-1}$).
By the Artin–Tate Lemma, $k[h(x),y] + \langle h(x),y \rangle_{k[x,y]}$
is a finitely generated algebra over $k[h(x),y]$.
Could we find the algebra generators of $k[h(x),y] + \langle h(x),y \rangle_{k[x,y]}$ over
$k[h(x),y]$? Are they just $yx,yx^2,\dotsc,yx^{d-1},yx^d$?
Same question with $h(0) \neq 0$.
The Artin-Tate Lemma is somewhat overkill for this. Since $R$ is a Noetherian ring and $A$ is a finitely generated $R$-module, $A$ is a Noetherian module, so the $R$-submodule $B$ is automatically a finitely generated module.
@Anonymous, thank you very much for your comment.
You have the right idea. Every element of $B$ is of the form $r+ph+qy$ for some $r\in R$ and $p,q\in A$. Since $A$ is generated by $1,x,\ldots,x^{d-1}$ as an $R$-module, we see $$p=\sum_{i=0}^{d-1}p_ix^i$$ $$q=\sum_{i=0}^{d-1}q_ix^i$$ for some $p_i.q_i\in R$. But then $$r+ph+qy=r\cdot 1+\sum_{i=0}^{d-1}p_ihx^i+\sum_{i=0}^{d-1}q_iyx^i.$$ So, conclude that $B$ is generated by $1,h,hx,\ldots,hx^{d-1},y,yx,\ldots,yx^{d-1}$ as an $R$-module.
|
2025-03-21T14:48:32.058215
| 2020-09-18T10:39:48 |
371997
|
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|
Stack Exchange
|
What is the quotient (pseudo)metric $d_\sim$ and how do I identify the infimum of possible sequences in this instance?
Let $Z$ be the the set of dyadic and ternary rationals in the interval $\left[\frac12,1\right)$ whose 3-adic valuation is either $-1$ or $0$, with the standard absolute value topology inherited from the real line.
Let $X=\{z\in Z:\nu_3(x)=0\}$.
Let $Y=\{z\in Z:\nu_3(x)=-1\}$.
Now define an equivalence relation $\sim$ which partitions $Z$ into equivalence classes $\{x, y\}$ of cardinality $2$ with $x \in X$, $y \in Y$, and $f(x) = y$, where $f:X\to Y$ is given by
$$f(x)=\begin{cases}\frac{4x}3 &\text{if}& x<\frac34\\
\frac{2x}3& \text{if}& x>\frac34.\end{cases}$$
So for example $\frac5{8}\sim\frac56$ and $\frac{7}{8}\sim\frac{7}{12}$.
The quotient (pseudo)metric $d_\sim$ on $Z/{\sim}$ is the infimum distance by which one can traverse from any equivalence class to another, by stepping on up to infinitely many equivalence classes in-between and summing only the distance between classes and not the distance travelled within classes. More formally this is defined at @EricWofsey's answer to "Is there a conceptual reason why topological spaces have quotient structures while metric spaces don't?".
Question
I seek an explicit definition of $d_\sim$ in this instance. Of course, in a sense Eric's answer does give that, but it leaves me with a requirement to somehow iterate over all possible sequences of equivalence classes and determine the shortest path, something well beyond my capabilities.
Also Note
While this question stands alone without reference to the Collatz conjecture, I feel more comfortable declaring that identifying this metric is a component of my study of the conjecture, partly in the spirit of full-disclosure, but also becase it may be material to the answer, to be mindful of the following observations:
$g(x)=x+\frac132^{\nu_2(x)}$ is both a surjection $X\to Y$ and a surjection $Z/{\sim}\to Z/{\sim}$ and seen as a map $g:Z/{\sim}\to Z/{\sim}$ it is essentially the Collatz graph and its graph is connected if and only if the Collatz conjecture is true. One should not be surprised therefore, if some proof that $d_\sim$ is the trivial metric, $\forall [z_0],[z_1]:d_\sim([z_0],[z_1])=0$, were related to the claim that the graph of the orbit of $g$ through $Z/{\sim}$ is connected.
The $n$-indexed sequences of the form $s_n(x)=x+(1-2^{-6n})\cdot2^{\nu_2(x)}\cdot3^{\nu_3(x)-1}$ form an exact cover of $X$ (up to subsequences) and $g$ is their infinite limit. Moreover, for every $y\in Y$ there are precisely two $s_n$ (up to subsequences) whose union is the level set by $g$ of $y$.
If $d_\sim$ is not the trivial (pseudo)metric, it seems likely a proof that some sequence gives the infimum for the metric may use the sequences given in the two bullets above.
This question has a "dual", if you like, which is to enquire into the quotient pseudometric obtained when setting $x\sim' g(x)$ instead of $x\sim f(x)$ and moreover the conjecture is equivalent to the claim that $Z/({\sim}\cdot{\sim'})$ is a singleton having $Z$ as its only element.
One gets the trivial semi-distance. Let $d$ denote the semi-distance on $Z$ that defines $d_\sim$ on $Z/\sim$, that is $d_\sim([z],[z']):=d(z,z')$ for all $z$ and $z'$ in $Z$. Thus $d(z,z')\le |z-z'|$ for all $z$ and $z'$ in $Z$ and $d(x,f(x))=0$ for all $x\in X$.
It is convenient to extend the definition of $f$ to a self-map $[1/2,1)\to[1/2,1)$, still denoted $f$:
$$f(x)=\begin{cases}\frac{4x}3 &\text{if}& 0\le x<\frac34\\
\frac{2x}3& \text{if}& \frac34\le x <1.\end{cases}$$
The key point is that all orbits of $f$ are dense. So for every $z$ and $z'$ in $Z$, and for every $\epsilon>0$, there is $m\in\mathbb N$ such that $\lvert f^m(z)-z'\rvert< \epsilon$. Since $X$ is also dense, and since $f$ is everywhere (right) continuous, for all indices $ 0\le j\le m$ there are $x_j\in X$ such that
\begin{align*}
\bigl\lvert f^j(z)-x_j\bigr\rvert<\frac\epsilon m, & \qquad (0\le j\le m) \\
\bigl\lvert f(x_j)-f^{j+1}(z)\bigr\rvert<\frac\epsilon m, & \qquad (0\le j\le m-1).
\end{align*}
Therefore for $0\le j\le m-1$
$$\bigl\lvert f(x_j)-x_{j+1}\bigr\rvert \le \bigl\lvert f(x_j)-f^{j+1}(z)\bigr\rvert+\bigl\lvert f^{j+1}(z)-x_{j+1}\bigr\rvert\le \frac{2\epsilon}m$$
and
$$d(z,z')\le\bigl\lvert z-x_0\bigr\rvert+ \bigg(\sum_{j=0}^{m-1} \bigl\lvert f(x_j)-x_{j+1}\bigr\rvert \bigg) +\big|x_m-z'\big|\le 4\epsilon,$$
which proves that $d$ vanishes identically.
All orbits of $f$ are dense: Indeed, having all orbits dense is a property invariant by conjugation, and $f$ is conjugate to the irrational translation (modulo $1$) on $[0,1)\sim\mathbb{R/Z }$ given by $T_c:x\mapsto {x -c}$, where $c:=\frac {\log 3}{\log 2}-1$ (all orbits of a translation $T_c$ are dense iff $c$ is irrational).
Consider the homeomorphism $h:[0,1)\to[1/2,1)$ defined by $h(x)=2^{x-1}$. Then it is easy to check that $h\circ T_c=f \circ h$.
Thank-you most kindly. This will take some time for me to digest properly to my lack of skill but I think the gist is that the pseudometric is $d_\sim=0$. Does this necessarily push through to the quotient space not being Hausdorff? I assume not because the quotient space is richer than the quotient metric.
P.S. was this fairly instantly obvious to a better mathematician than I, or did it take some degree of thought? It seems fairly obvious the way you put it about every orbit being dense - must imply the distances between them are zero.
I think so (quotient being Hausdorff) but I do not see immediately.
On the PS: To me, it was not obvious, and I wasn't even sure that the distance was trivial. How I proceeded, it's nothing special, but here it is. 1) It seemed that $X$ and $Y$ do not play a special role in studying these objects, and that it is healthier to have $f$ as a self-map, to make iterations. 2) one works better on the interval, than on the quotient.
in this semi-distance, $1/2$ is close to the right-end, for $f(3/4-\epsilon)= 1-4\epsilon/3$ and $f(3/4+\epsilon)= 1/2-2\epsilon/3$. So nothing changes if one starts from a circle, identifying the endpoints of $[1/2,1]$, instead using $[1/2,1)$, which makes $f$ a homeomorphism. This allows to use the tools of homeomorphims of the circle. I wrote $f$ as a time-1 flow to compute its rotation number and find a conjugation with a rotation: only at the end I realized the conjugation was so easy.
PS: The quotient (semi)distance $\tilde d$ on $\tilde X$ for a quotient $\pi:X\to\tilde X$ is the maximum (semi)distance on $\tilde X$ that makes $\pi$ a $1$-Lipschitz map. This leads to the construction of $\tilde d$ via chains of jumps from a class to anothe (which I would call the construction rather than the definition of $\tilde d$ ).
Yes, it was meant to be a proof (sorry if it wasn't clear). Having every orbit dense is a conjugation invariant (recall that $f=hgh^{-1}\Rightarrow f^m=hg^mh^{-1}$ for all $m$), and the translation $x\mapsto x+c\mod 1$ on $\mathbb{ T:=R/Z}$ does have every orbit dense if (and only if) $c$ is irrational (for the orbit of $x$ is $x+c \mathbb Z\mod 1 $), so every orbit of $f$ is dense too.
Don't worry, it is perfectly normal that we sometimes miss something, and we have the right to ask any question --even because sometimes "trivialities" hide not-at-all-trivial facts.
It seems LSpice overlooked the condition in the first sentence: whose $3$-adic valuation is either $−1$ or $0$. This constrains the equivalence classes to have cardinality two. I'm just checking you didn't overlook the same, and whether your answer still stands notwithstanding this?
You don't really need to quotient, as your quotient set $Z/\sim$ can be identified with $X$. Alternatively, you may even consider $f$ on all rationals, and quotient on the $\mathbb Z$ action, so that equivelence classes are the whole orbits of $f$. Any class then contains exactly one element of $X$.
I'm sorry to push for more - please feel free to ignore. This is to do with the motivation behind this question: Consider the $n$-indexed sequence $\dfrac{2^{6n+2}-1}6=\frac12,\frac{85}{128},\frac{5461}{8192}\ldots$. This converges in $\Bbb R$ to $\frac23$ and in $\Bbb Q_2$ to $-\frac16$. In the trivial pseudometric here it trivially converges as it's the identity sequence. But it also converges to $[\frac12\sim\frac23]$ in the quotient topology (non-pseudometric). Any pointers how I might attempt to derive/construct this quotient topology and show it is Hausdorff or nontrivial?
ok but which quotient topology?
Simply $Z/{\sim}$ in the quotient topology as per https://math.stackexchange.com/a/3810857/334732 which may be stronger than the quotient pseudometric. I.e. the standard absolute value metric from $\Bbb R$ reduced by the equivalence relation $\sim$. Or maybe it's no stronger and it coincides with the pseudometric topology in this particular case?
So you mean the quotient topology on $Z/!\sim$ for the topological space $Z$ (as a subspace of $\mathbb R$) and the equivalence relation, as defined in your post here?
Yes, that's it. That is the ultimate motivation of this question... I hoped this question would either be it, or be a stepping stone towards it. The convergent sequences or open sets I'm interested have the recurrence relation $x_{n+1}=x_n+21\cdot2^{\nu_2(x_n)-6}\cdot3^{\nu_3(x_n)}$
Thank-you. This will provide a good month or two of food for thought for me. I am much obliged to you.
Topologic conjugacy, as it is written in the last line.
@samerivertwice, I agree: $(f \circ h)(x) = \frac{2^{x + 1}}3 = 2^{x - \log_2(3) + 1} = h(x - c + 1)$ for $x < c$, and $(f \circ h)(x) = \frac{2^x}3 = 2^{(x - \log_2(3) + 1) - 1} = h(x - c)$ for $x \ge c$.
You say I don't really need to quotient as the quotient can be identified with $X$. I want to show $g=x+\frac{21}{64}\cdot2^{\nu_2(x)}$ converges on transfinite iteration.
For some $x:\lvert x\rvert_3=1$, the infinite limit of $g^n(x)$ satisfies $\lvert x\rvert_3=3$. Fine, so apply $f^{-1}$ and I'm back in $X$. Therefore I can just use $j(x)=f^{-1}\circ \lim_{n\to\infty}g^n:X\to X$. Does your $h$ help me with the proof that $j$ converges on iteration? The claim that it does converge is equivalent to the Collatz conjecture.
|
2025-03-21T14:48:32.058857
| 2020-09-18T10:47:48 |
371998
|
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"Overflowian",
"Steven Sivek",
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|
Stack Exchange
|
May this slice disk for the unknot be pushed into the boundary?
Write the 4-ball as $\mathbb{D}^4=\mathbb{D}^2\times \mathbb{D}^2$.
Then its boundary $\mathbb{S}^3\simeq \mathbb{S}^1\times \mathbb{D}^2\cup \mathbb{D}^2\times \mathbb{S}^1$. We will use implicitely this homeomorphism.
Consider the unknot $K=\mathbb{S}^1\times\{0\}$ in the boundary $\partial \mathbb{D}^4\simeq\mathbb{S}^3$ of the four ball.
Notice that the disk $D = \mathbb{D}^2\times \{0\}$ is a smooth slice disk for $K$.
Is $D$ boundary parallel? I.e. is it obtained by pushing an unknotting disk $D'\subset \mathbb{S}^3,$ $\partial D' = K$ inside $int(\mathbb{D}^4)?$
A possible approach: take as $D'\subset \partial \mathbb{D}^4$ the PL disk obtained by gluing the annulus $\mathbb{S}^1\times ([-1,1]\times \{0\})\in \mathbb{S}^1\times \mathbb{D}^2\subset\partial \mathbb{D}^4$ to the disk $\mathbb{D}^2\times (\{-1\}\times \{0\})\in \mathbb{D}^2\times \mathbb{S}^1\subset\partial \mathbb{D}^4$.
Then gluing $D'$ to $D$ we get an embedded 2-sphere. If we manage to prove that this sphere is unknotted in $\mathbb{D}^4$, i.e. it bounds a 3-ball then we can use the latter to push $D'$ to $D$.
Relevance of this problem
In studying Kirby calculus, one finds often the claim that when you attach a $2$-handle, the cocore of a $2$-handle is an unknotted $2$-disk, i.e. boundary parallel.
The above problem is a possible way to prove this.
I don't understand the -1. I've posted this question on MSE before trying MO, in 2 days I've got +3 votes and no hints. It is surely something well known to people in the field but that's not a good reason for a down-vote.
There's a continuous family of disks $S^1\times[0,t] \cup (D^2 x {t})$ with boundary $K$. At $t=0$ you have $D^2 \times {0}$, and at $t=1$ the whole disk lies in $S^3$.
|
2025-03-21T14:48:32.059005
| 2020-09-18T10:52:02 |
371999
|
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"Michael Renardy",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/371999"
}
|
Stack Exchange
|
About solutions of Klein-Gordon equation
I wonder how to solve the Klein-Gordon equation
$$\left\{\begin{split}&\partial_t^2u-\Delta u+u=f\\&u(0,x)=u_1(x),\quad \partial_t u(0,x)=u_2(x)\end{split}\right.$$
where $u(t,x)$ defined on $\mathbb{R}^+\times\mathbb{R}^d$.
By applying the partial Fourier transform respect to the $x$ variable, I obtain the following representation of $u$:
$$\begin{split}u(t,x)=&\mathcal{F}^{-1}\left(\cos\sqrt{1+|\xi|^2}t\right)*u_1(x)+\mathcal{F}^{-1}\left(\frac{\sin\sqrt{1+|\xi|^2}t}{\sqrt{1+|\xi|^2}}\right)*u_2(x)\\&+\int_0^t\mathcal{F}^{-1}\left(\frac{\sin\sqrt{1+|\xi|^2}(t-s)}{\sqrt{1+|\xi|^2}}\right)*f(s,x)\,ds\end{split}$$
However, I don't know how to solve the inverse Fourier transforms in the formula: do there exist closed form expressions for such integrals?
A google search for "Klein Gordon Green's function" should point you in the right direction.
|
2025-03-21T14:48:32.059081
| 2020-09-18T11:42:17 |
372003
|
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/372003"
}
|
Stack Exchange
|
Akbulut's cork involution
Akbulut's cork is the Mazur manifold $W$ shown in the picture below,
This manifold carries an involution of it's boundary $f:\partial W\to \partial W$ that exchanges a meridian of the 0-framed curve with a meridian of the dotted curve.
In Akbulut - 4-manifolds, the author says that $f$ is related to performing surgeries to exchange the dot with the 0-framed knot.
I don't understand how this relates to $f$ and how $f$ is defined.
We can surely perform two surgeries (one along a circle and one along a 2-sphere) to get a new 4-manifold $\tilde{W}$ that has a Kirby diagram obtained by exchanging dot and zero in the above picture. However surgery does not affect the boundary so I do not see how this could induce a nontrivial map between the boundaries.
Precisely how is $f$ defined?
The boundary of W may be described as 0-framed surgery on both components of the link you drew. The link can be drawn in a more symmetric fashion, so that it is clear that there is an involution interchanging the two components. See for instance Figure 4 of Akbulut's paper, A solution to a conjecture of Zeeman, Topology 30 (1991, 513-515. This involution induces an involution on the surgered manifold that interchanges the two meridians.
If you don't care that the diffeomorphism on the boundary is an involution, you can just do an isotopy interchanging the components.
|
2025-03-21T14:48:32.059195
| 2020-09-18T12:39:01 |
372006
|
{
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"authors": [
"Rob Arthan",
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/372006"
}
|
Stack Exchange
|
Diagonal terms in the Kochen Stone inequality
In a paper in Lecture Notes in Mathematics vol. 1874, Yan states the Kochen-Stone theorem in the following form, where $A_n$ is a sequence of events such that $\sum_{n=1}^\infty P(A_n) = \infty$:
$$
P(A_n~\mbox{i.o.})
\geq \limsup_{n \to \infty} \frac{(\sum_{k=1}^n P(A_k))^2}{\sum_{i,k=1}^n P(A_i A_k)}
= \limsup_{n \to \infty} \frac{\sum_{1 \le i < k \le n} P(A_i)P(A_k)}{\sum_{1 \le i < k \le n}P(A_iA_k)} \tag*{$(2)$}
$$
((2) is his numbering).
I understand his proof of the inequality in (2), but I can't make sense of his proof of the equation (which states that the diagonal terms in the sums in the first fraction in (2) are negligible).
He says "Since $\sum_{k=1}^\infty P(A_k) = \infty$ and
$$
\left(\sum_{k=1}^n P(A_k)\right)^2 \le 2\sum_{1 \le i < k \le n}P(A_i)P(A_k) + \sum_{k=1}^n P(A_n) \tag*{$(3)$}
$$
we have
$$
\lim_{n \to \infty}\frac{\sum_{k=1}^n P(A_k)}{\sum_{1 \le i < k \le n}P(A_i)P(A_k)} = 0 \tag*{$(4)$}
$$
Thus equality (2) holds." ((3) and (4) are my numbering.)
I see how to get from (3) to (4) (by dividing by $\sum_{k=1}^n P(A_k)$ and noting that the left-hand side of the resulting inequality diverges), but I can't see how to get the equality in (2) from (4). Any suggestions on how to see this or any alternative proofs will be gratefully received as I am doing some work that relies heavily on the equation and am reluctant to use a result that I don't understand. Any relevant references would also be much appreciated.
Let
$$S_n=\sum_{k=1}^n P(A_k),\quad T_n:=\sum_{1\le i<k\le n}P(A_i)P(A_k),$$
$$R_n=\sum_{i,k=1}^n P(A_iA_k),\quad U_n:=\sum_{1\le i<k\le n}P(A_iA_k).$$
Then $2T_n\le S_n^2\le2T_n+S_n$, and $S_n<<S_n^2$ (because $S_n\to\infty$); we write $a<<b$ or, equivalently, $b>>a$ to mean $a=o(b)$; all the limits are taken as $n\to\infty$. Thus,
$$S_n^2\sim2T_n.\tag{*}$$
Also, $2U_n\le R_n\le2U_n+S_n$, $S_n<<S_n^2$, and $S_n^2\le (1+o(1))R_n$ (by the inequality in (2)). Thus,
$$R_n\sim2U_n.\tag{**}$$
From (*) and (**), we get
$$\frac{S_n^2}{R_n}\sim\frac{T_n}{U_n},$$
which immediately yields the equality in (2).
Thanks. So Yan was begin rather coy! What I have called (4) says $S_n/T_n \to 0$, which leads to your $(*)$. But Yan was leaving your $(**)$ entirely to the reader.
|
2025-03-21T14:48:32.059344
| 2020-09-18T13:23:05 |
372012
|
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/372012"
}
|
Stack Exchange
|
Fano surface of conics on Gushel-Mukai threefolds
Let $X$ be a smooth Gushel-Mukai threefold, there are following four cases:
$X_1$ is a special Gushel-Mukai with branch locus $\mathcal{B}$ on $Y_5$ general, i.e, it does contain any line or conic. $\pi:X\rightarrow Y_5$ is a covering map.
$X_2$ is a special Gushel-Mukai with branch locus $\mathcal{B}$ contain lines or conics.
$X_3$ is a general ordinary Gushel-Mukai.
$X_4$ is a non-general ordinary Gushel-Mukai.
Now, I consider their Fano surface of conics, denoted by $F_c(X_i)$.
$F_c(X_1)$ is a reducible surface with two irreducible components and their intersection is a curve parameterized by $\pi$-preimage of a smooth rational curves $\rho$ parametrizing the $(-1,1)$-lines on $Y_5$ and its singular locus is $\pi$-preimage of $\rho$.
$F_c(X_2)$ has extra singularities given by $\pi$-preimage of finitely many lines in $\mathcal{B}$ and finitley many conics in $\mathcal{B}$ .
$F_c(X_3)$ is a smooth irreducible surface.
My question
What about $F_c(X_4)$?, it is still irreducible, but can it be smooth? The normal bundle of a conic on $X_4$ can only be $(0,0), (-1,1)$ and $(-2,2)$-type. If it is singular, is it possible that the singular locus of $F_c(X_4)$ contains a curve?
|
2025-03-21T14:48:32.059441
| 2020-09-18T13:37:46 |
372014
|
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"authors": [
"MaoWao",
"https://mathoverflow.net/users/95776"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/372014"
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|
Stack Exchange
|
Lax CD(K, $\infty)$ space in the sense of Sturm
In K.T. Sturm's "On the geometry of metric measure spaces. I", Definition 4.5, he introduces a "lax" version of the usual CD(K,$\infty$) lower bound. Namely, one allows for $\epsilon$-approximate length-minimizing curves, so that a metric measure space satisfies "lax CD(K,$\infty$)" (in Sturm's notation, $\underline{Curv}_{lax} (M,d,m)\geq K$) provided that, for any $\epsilon > 0$, any two measures $\rho_0 $ and $\rho_1$ absolutely continuous with respect to $m$, there exists an absolutely continuous curve $\rho_t$ connecting these measures with length $\leq W_2 (\rho_0, \rho_1)+\varepsilon$, with $$ Ent(\rho_t \mid m) \leq (1-t) Ent( \rho_0 \mid m) + t Ent (\rho_1 \mid m) - 0.5 K t (1-t) W_2^2(\rho_0 , \rho_1 ) + \epsilon.$$
Sturm goes on to show this coincides with the usual CD(K,$\infty$) lower bound if the underlying space is compact.
My question is: has anyone else used this lax CD(K,$\infty$) notion, in the literature?
I cannot rule out that it was used by someone else, but at least it does not seem to be widely studied.
|
2025-03-21T14:48:32.059539
| 2020-09-18T13:46:20 |
372015
|
{
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"authors": [
"Terry Tao",
"dohmatob",
"https://mathoverflow.net/users/766",
"https://mathoverflow.net/users/78539"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/372015"
}
|
Stack Exchange
|
Prove / disprove: If $1 \le n < N$ and $A$ is an $N \times n$ matrix with iid from $\mathcal N(0,1)$, then $s_\min(A) \ge c\sqrt{N}$ w.p $1-2e^{-N}$
Let $1 \le n < N$ be integers and $A$ be a random $N\times n$ matrix with iid entries from $\mathcal N(0,1)$. This paper (Rudelson and Vershynin) claims in the paragraph just before formula (3.4) that, there exists a constant $c>0$ such that
$$
P(s_\min(A) \ge c\sqrt{N}) \ge 1 - 2e^{-CN},\text{ with }C=1! \tag{1}
$$
I doubt the validity of the above statement with $C=1$.
Note. I know that the statement can be made true with a smaller value of $C$ (and some mild conditions on the aspect ratio $n/N$).
Question. Can someone kindly rollout the sketch of a proof of (1), or else disprove it ?
Thanks in advance!
I believe Rudelson and Vershynin are only claiming this bound in the regime of tall matrices (in which the ratio $n/N$ is assumed to be sufficiently small).
Thanks for the input. Could you kindly provide a hight-level justification why this ought to be true for all $n/N < \lambda_0$, for some constant $\lambda_0 \in (0,1)$ ?
It's true (for any $C$, if $c$ is sufficiently small depending on $C$) for each individual column (i.e., for $n$=1) by direct calculation of the chi-squared distribution (or evaluating the measure that an $N$-dimensional gaussian gives to a small ball around the origin), and the epsilon net argument (see e.g., https://terrytao.wordpress.com/2010/03/05/254a-notes-7-the-least-singular-value/ ) then extends this bound to tall matrices (again for any $C$, if $c,\lambda_0$ are small enough depending on $C$.)
@TerryTao Thanks for the crucial hint (namely, "small-ball probabilities"). Below, I've posted an answer putting every thing together.
Below, I provide an answer inspired by the comments of user Terry Tao.
Let $n/N =: \lambda \in (0, 1)$ be the aspect ratio of $A$. We will prove the following.
Claim. For every $C>0$, there exists $c>0$ depending only on $\lambda$ and $C$ such that $s_\min(A) \ge c\sqrt{N}$ w.p $1-2e^{-CN}$.
Technical tools
For any positive integer $k$, let $\mathbb B_k := \{x \in \mathbb R^k \text{ s.t } \|x\| \le 1\}$ be the unit-ball in $k$-dimensional euclidean space $\mathbb R^k$ and let $\mathbb S_{k-1} := \{x \in \mathbb R^k \text{ s.t } \|x\| = 1\}$ be the corresponding unit-sphere.
Fact 1 (Gaussian small-ball probability). If $X = (X_1,\ldots,X_k) \sim \mathcal N(0,I_k)$, then there exists $C_1>0$ such that $P(\|X\| \le u \sqrt{k}) \le (C_1u)^k$ for all $u \ge 0$.
Proof. For every $u \ge 0$, one computes
$$
P(\|X\| \le u \sqrt{k}) = (2\pi)^{-k/2}\int_{u\sqrt{k} \mathbb B_k}e^{-\|x\|^2}dx \le (2\pi)^{-k/2}\mbox{vol}(u\sqrt{k}\mathbb B_k) \le (C_1u)^k,
$$
for some $C_1>0$ (which can be made explicit).
Fact 2 (Spectral-norm upper bound). For every $C>0$, there exists $C_0>0$ such that $s_{\max}(A) \le C_0\sqrt{N}$ w.p $1-e^{-CN}$.
Proof. See Fact 2.4 of this paper by Litvak, Pajor, and Rudelson.
Proof of the main claim
We are now ready to proof the main claim.
Proof of the main claim. Let $\epsilon \in (0, 1)$, to be prescribed later, and let $\mathcal N_\epsilon$ be a maximal $\epsilon$-net for $\mathbb S_{n-1}$. Note that $|\mathcal N_\epsilon| \le (3/\epsilon)^n = (3/\epsilon)^{\lambda N}$. Now, for each $x \in \mathbb S_{n-1}$, there exists $z \in \mathcal N_\epsilon$ such that $\|x-z\| \le \epsilon$. Writing $Ax = Az + A(x -z)$, the triangle inequality gives us $\|Ax\| \ge \|Az\| - \|A(x-z)\| \ge \|Az\|-\epsilon s_\max(A)$. Minimizing both sides, we obtain
$$
s_\min(A) = \inf_{x \in \mathbb S_{n-1}}\|Ax\| \ge \min_{z \in \mathcal N_\epsilon}\|Az\|-\epsilon s_\max(A).\tag{1}
$$
Let $C_1$ be as in Fact 1 with $k=N$. For arbitrary $C>0$, let $C_0$ be as in Fact 2 with $k=N$, and let $c > 0$, to be carefully chosen later. By (1), we know that if $s_\max(A) \le C_0\sqrt{N}$ and $\|Az\| \ge 2c \sqrt{N}$ for all $z \in \mathcal N_\epsilon$, then $s_\min(A) \ge 2c\sqrt{N}-\epsilon C_0\sqrt{N} \ge c\sqrt{N}$ when $\epsilon = c/C_0$ with $c < C_0$. Thus,
$$
\begin{split}
&P(s_\min(A) < c\sqrt{N}) \\
&\quad= P(s_\min(A) < c \sqrt{N},s_\max(A) > C_0\sqrt{N})\\
&\quad\quad\quad+ P(s_\min(A) > C_0\sqrt{N},s_\max(A) \le C_0\sqrt{N})\\
&\quad\le P(s_\max(A) > C_0\sqrt{N}) + P(s_\min(A) < c \sqrt{N},s_\max(A) \le C_0\sqrt{N})\\
&\quad \le e^{-CN} + P(\min_{z \in \mathcal N_\epsilon}\|Az\| < 2c\sqrt{N}) \le e^{-CN} + |\mathcal N_\epsilon|\cdot\max_{z \in \mathcal N_\epsilon}P(\|Az\| \le 2c\sqrt{N})\\
&\quad \le e^{-CN} + (3/\epsilon)^n(C_1 \cdot 2c)^N \le e^{-CN}+((3/\epsilon)^\lambda\cdot C_1\cdot 2c)^N\\
&\quad\le e^{-CN} + (2C_1(3C_0/c)^\lambda c)^N \le e^{-CN} + e^{-CN}=2e^{-C N},
\end{split}
$$
for sufficiently small $c \in (0,C_0)$ such that $2C_1(3C_0/c)^\lambda c < e^{-C}$.
Therefore, given arbitrary $C>0$, the bound $P(s_\min(A) \le c\sqrt{N}) \le 2e^{-CN}$ is guaranteed by taking $c \in (0,c_\lambda(C))$, where
$$
\begin{split}
c_\lambda(C) := \min(C_0,(2C_1e^C(3C_0)^\lambda)^{\frac{-1}{1-\lambda}})>0.
\end{split}
\tag{2}
$$
This completes the proof of the claim. $\quad\quad\Box$
Note. An important highlight the above proof is that it works for every aspect ratio $\lambda \in (0,1)$.
Going beyond Gaussian matrices
A careful inspection of the proof of main ingredients Fact 1 and Fact 2, reveals that we can replace the base distribution $N(0,1)$ of the coefficients of $A$ by any symmetric unit-variance $\sigma^2$-subGaussian such that $0 \le \sigma \le 1$.
|
2025-03-21T14:48:32.059861
| 2020-09-18T14:06:16 |
372017
|
{
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|
Stack Exchange
|
Polynomial time for a quadratic equation and linear inequalities?
Does anyone know how to find a feasible solution (or the infeasibility of any solution) in a polynomial time to the following problem:
\begin{align*}
xAx^t = 0, \\
Bx^t = c, \\
x_i \ge 0,
\end{align*}
where A is not definite positive nor definite negative, but it is symmetric.
Proving there is a polynomial-time solution to this problem will give a proof of P=NP
|
2025-03-21T14:48:32.060040
| 2020-09-18T14:18:06 |
372018
|
{
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"J.Mayol",
"Terry Tao",
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/372018"
}
|
Stack Exchange
|
Failure of Strichartz estimates for the wave equation: elaboration of a counter-example
One can read in Oh - Probabilistic perspectives in nonlinear dispersive PDEs (Proposition 64, p. 60) that there exists a function $F \in L^2_tL^{1}_x (\mathbb{R}_t\times \mathbb{R}^3_x)$ which is supported on frequencies $\lvert\xi\rvert \sim 1$ and such that $\int_{\mathbb{R}} e^{it\lvert\nabla\rvert}F(t')dt' \notin L^2_x$. Define $F(t,x)=\eta(t/T)\psi(x-te_3-\beta_te_1)$ where $\beta_t$ is a standard Brownian motion, $\psi$ is localized in frequency around $\lvert\xi\rvert \sim 2$ and $\eta$ is a smooth cutoff.
Then it is not difficult to see that $\|F\|_{L^2_xL^1_x} \lesssim \sqrt T$ and $\left\|\int_{\mathbb{R}} e^{it|\nabla|}F(t')dt'\right\|_{L^2_x}$ has infinite expectation (we take expectation with respect to the Brownian motion), hence the result. For the latter the proof uses Fourier–Plancherel and a lower bound on the frequency set $\lvert\xi\rvert \sim \xi_3+\xi_1^2$, which seems to work fine with the $\psi (\cdot -te_3-\beta_te_1)$.
However I was surprised by such a non-inuitive counterexample. If one takes $F(t,x)=\eta(t/T)\psi(x)$ then it is not a counter-example. Can someone explain to me the idea behind the clever substitution $\psi \to \psi (\cdot - te_3-\beta_te_1)$? I do not understand why this is a natural attempt. There might be some "singularity propagation on a cone" thing behind but it is not clear to me. For the use of the Brownian motion I do not get the point of why it is useful. Otherwise, is it possible to give a more intuitive counter-example?
@YCor is it better now? If there is need for more details about the proof in the document I can provide them.
The Brownian motion term is needed to destroy an "unwanted" cancellation in the Duhamel integral that would otherwise occur if this randomization was not present. I discuss this a little in my own unpublished notes on this example at http://www.math.ucla.edu/~tao/preprints/Expository/stein.dvi
@TerryTao thank you, your notes are nice, at least I understand a little more why the Brownian motion is needed. It seems like we can replace the Brownian motion with some $C^{1/2}$ function (with appropriate norm) and try to work with it, although this basically reduces to using the Brownian motion.
One last thing: that the $-te_3$ is needed is because we want to "kill" the oscillatory factor $e^{it|\xi|}$ in order to force the integral to diverge? Am I right?
Yes. In physical space this corresponds to trying to create a "sonic boom" effect by having the light cones from the sources at different times all line up.
|
2025-03-21T14:48:32.060242
| 2020-09-18T14:18:45 |
372019
|
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"url": "https://mathoverflow.net/questions/372019"
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|
Stack Exchange
|
An application of Leray-Schauder degree theory for Nirenberg problem on the 2-sphere
I'm studying the article "The scalar curvature equation on 2- and 3-spheres" by Chang, Gursky and Yang and I'm particulary interested in the 2-sphere case.
They prove that if $K:S^2\rightarrow \mathbb{R}$ is a positive smooth function satisfying some nondegeneracy condition, then $K$ can be obtained as Gaussian curvature from a conformal change of metric.
The idea of the proof is to consider the one parameter family of maps:
$$
K_s=sK+(1-s)
$$
and looking for zeroes of the maps
$$
\psi_s(w)=w-\Delta^{-1}\Psi_s(w)
$$
where $\Psi_s(w)=1-K_se^{2w}$.
They apply the Leray-Schauder degree theory and some apriori estimates on the $C^{2,\alpha}$ norm of zeroes of $\psi_s$ for $s\geq s_0$, for any given $s_0>0$
It is not clear to me, however, how to apply degree theory in this context. More precisely, from what I know, Leray-Schauder degree theory can by applied to maps $Id-K:\overline{U}\rightarrow X$ where $X$ is a Banach space, $U$ is a bounded open set in $X$, and $K$ is a compact map such that $0\notin(I-K)(\partial U)$
In the case at hand, it is not clear me how to choose $X$ and $U$.
The authors of the article consider the set $X=\{w\in C^{2,\alpha}(S^2)| \frac{1}{4\pi}\int Ke^{2w}=1\}$
and the bounded set $\Omega_C=\{w\in X| |w|_{C^{2,\alpha}}< C\}$ where $C$ is given by the apriori estimates.
I have a few doubts about this choice:
$\Delta^{-1}\Psi_s(w)$ is not well defined if $w\in X$, since for this to make sense I need that $\frac{1}{4\pi}\int K_se^{2w}=1$
X is not a Banach space, and therefore it is not clear to me how to apply degree theory.
Is there something I am missing?
|
2025-03-21T14:48:32.060374
| 2020-09-18T14:36:25 |
372020
|
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"Iosif Pinelis",
"OmarR",
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"url": "https://mathoverflow.net/questions/372020"
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|
Stack Exchange
|
Dual problem with integrals
I am reading a paper where the author derives the following Lagrangian dual problem :
$\min_v \int_R \frac{1}{4} \frac{\beta^2}{v-2\|x\|}dx+v\;\;\;\text{s.t.}\;\;\;v\geq 2\|x\|\;\;\;\forall x \in R$
from the primal problem :
$\max_{f(.)} \int_R (2\|x\| f(x) + \beta \sqrt{f(x)})dx\;\;\;
\text{s.t.}\;\;\;\int_R f(x) dx= 1 \;\;\;\text{and}\;\;\;f(x) \geq 0\;\;\;\forall x \in R$
where $f(.)$ belongs to the Banach space $L^2$ over a compact set $R$ (a distribution function).
Do you know how to construct a dual problem in case the objective and constraints of the primal include integrals. It was said that standard techniques of vector space optimization could be used to approach the function $f(.)$, but this is maybe not obvious. I could not pinpoint the starting point.
$\newcommand\R{\mathbb R}$
A convenient way to derive the dual problem from a primal one is by using the minimax duality for the Lagrangian, which is given here by the formula $$L(f,v):=\int_R\big[2|x|f(x)+b\sqrt{f(x)}\big]\,dx-v\Big(\int_Rf(x)\,dx-1\Big),$$
where $|x|:=\|x\|$ and $b:=\beta$. Clearly,
$$\sup_{f\ge0}\inf_{v\in\R} L(f,v)
=\sup\Big\{\int_R\big[2|x|f(x)+b\sqrt{f(x)}\big]\,dx\colon f\ge0,\int_R f(x)\,dx=1\Big\},$$
which is value of the primal problem.
The value of the dual problem is
$$\inf_{v\in\R}\sup_{f\ge0} L(f,v)
=\inf_{v\in\R}\Big(v+\sup\Big\{\int_R\big[2|x|f(x)+b\sqrt{f(x)}-vf(x)\big]\,dx\colon f\ge0\Big\}\Big)
=\inf_{v\in\R}\Big(v+\int_R s(|x|,v)\,dx\Big),
$$
where
$$s(a,v):=\sup\{2at+b\sqrt t-vt\colon t\ge0\}.$$
For any real $a>0$, it is easy to see that
$$s(a,v)=\frac{b^2}{4 (v-2 a)}$$
if $b>0$ and $v>2a$, $s(a,v)=0$ if $b\le0$ and $v\ge2a$, and $s(a,v)=\infty$ otherwise; in particular, $s(a,v)=\infty$ if $v<2a$. Thus, with $|R|:=\max\{|x|\colon x\in R\}$, the value of the dual problem is
$$\inf_{v>2|R|}\Big(v+\int_R \frac{\max(0,b)^2}{4 (v-2|x|)}\,dx\Big).
$$
The equality in the interchanging of $sup_{f}$ and $\int_R$ comes from the continuity and the measurability of the function $f(.)$, right.
@OmarR. : The inrterchange of $\sup_f$ and $\int_R$ is possible because, for $v>2a$, the maximizer of $2at+b\sqrt t-vt$ in $t\ge0$ is $\max(0,b)^2/(4(v-2a)^2)$, which is a measurable function of $a$ ($a$ here representing $|x|$).
Thank you very much.
|
2025-03-21T14:48:32.060542
| 2020-09-18T17:25:04 |
372026
|
{
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|
Stack Exchange
|
The space of skew-symmetric orthogonal matrices
Let $M_n \subseteq SO(2n)$ be the set of real $2n \times 2n$ matrices $J$ satisfying $J + J^{T} = 0$ and $J J^T = I$. Equivalently, these are the linear transformations such that, for all $x \in \mathbb{R}^{2n}$, we have $\langle Jx, Jx \rangle = \langle x, x \rangle$ and $\langle Jx, x \rangle = 0$. They can also be viewed as the linear complex structures on $\mathbb{R}^{2n}$ which preserve the inner product.
I'd like to understand $M_n$ better as a topological space, namely an $(n^2-n)$-manifold.
$M_1$ is just a discrete space consisting of two matrices: the anticlockwise and clockwise rotations by $\pi/2$.
For $n \geq 2$, we can see that $M_n$ is an $M_{n-1}$-bundle over $S^{2n-2}$. Specifically, given an arbitrary unit vector $x$, the image $y := Jx$ must lie in the intersection $S^{2n-2}$ of the orthogonal complement of $x$ with the unit sphere $S^{2n-1}$. Then the orthogonal complement of the space spanned by $x$ and $y$ is isomorphic to $\mathbb{R}^{2n-2}$, and the restriction of $J$ to this space can be any element of $M_{n-1}$.
Since the even-dimensional spheres are all simply-connected, it follows (by induction) that $M_n$ has two connected components for all $n \in \mathbb{N}$, each of which is simply-connected. For instance, $M_2$ is the union of two disjoint 2-spheres: the left- and right-isoclinic rotations by $\pi/2$. The two connected components of $M_n$ are two conjugacy classes in $SO(2n)$; they are interchanged by conjugating with an arbitrary reflection in $O(2n)$.
Is [each connected component of] $M_n$ homeomorphic to a known well-studied space? They're each an:
$S^2$-bundle over an $S^4$-bundle over $\dots$ an $S^{2n-4}$-bundle over $S^{2n-2}$
but that's not really very much information; can we say anything more specific about their topology?
Your $M_n$ is the Riemannian symmetric space $\mathrm{SO}(2n)/\mathrm{U}(n)$. I believe that its topology is quite well studied from that point of view.
Thanks! Yes, it looks like the space is called DIII in Cartan's classification of compact Riemannian symmetric spaces. (I'll accept this if you post it as an answer.)
yes, this is class DIII, the two connected components are distinguished by the Pfaffian, which equals $\pm 1$; in the physics context this matrix $M_n$ is the scattering matrix of a superconductor with preserved time-reversal symmetry but broken spin-rotation symmetry. The sign of the Pfaffian then distinguishes topologically trivial from topologically nontrivial superconductors.
In Morse Theory, Milnor has proved that $\mathrm{O}(2n)/\mathrm{U}(n)$ can be identified with orthogonal linear transformations $J$ with $J^2=-I$ i.e. complex structure.
Your $M_n$ is (two copies of) the Riemannian symmetric space $\mathrm{SO}(2n)/\mathrm{U}(n)$ (which is $DI\!I\!I$ in Cartan's nomenclature). Its topology is well-studied from that point of view.
|
2025-03-21T14:48:32.060751
| 2020-09-18T18:24:21 |
372029
|
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|
Stack Exchange
|
Definition of homogeneous Sobolev spaces
As we know the inhomogeneous Sobolev space (we only consider $s>0$)
$${H}^{s}\left(\mathbb{R}^{n}\right)=\left\{f \in L^2(\mathbb{R}^n):\int_{\mathbb{R}^{n}}|\xi|^{2 s}|\hat{f}(\xi)|^{2} \mathrm{d} \xi<\infty\right\}$$
I am quite confused by the homogeneous one $\dot H^s$ which consists of functions with the following quantity is bounded
$$
\|f\|_{\dot{H}^{s}}=\left(\int_{\mathbb{R}^{n}}|\xi|^{2 s}|\hat{f}(\xi)|^{2} \mathrm{d} \xi\right)^{1 / 2}\label{1}\tag{*}$$
There are several definitions of $\dot H^s$ (I use a subscript to distinguish them).
In L. Grafakos, Modern Fourier analysis, he defines
$$\dot{H}_G^{s}\left(\mathbb{R}^{n}\right)=\left\{f \in \mathscr{S}^{\prime}/\mathscr{P}: \int_{\mathbb{R}^{n}}|\xi|^{2 s}|\hat{f}(\xi)|^{2} \mathrm{d} \xi<\infty\right\}$$
Here $\mathscr{S}^{\prime}/\mathscr{P}$ is the equivalent class of distributions modulo polynomials (that is, we identify two distributions whose difference is a polynomial). Then \eqref{1} is a norm.
Maybe there is a more natural one by
$$\dot H^s_N(\mathbb{R}^n)=\text{completion of }\left\{f\in \mathscr{S}:\int_{\mathbb{R}^{n}}|\xi|^{2 s}|\hat{f}(\xi)|^{2} \mathrm{d} \xi<\infty\right\}\text{ under the norm \eqref{1}}$$
My questions are
Is $\dot H_G^s$ complete under \eqref{1}?
Is $\dot H^s_N=\dot H_G^s$?
Add after Michael's comment: There is another version in H. Bahouri, J.-Y. Chemin, R. Danchin, Fourier analysis and nonlinear partial differential equations.
They define
$$
\dot{H}_B^{s}\left(\mathbb{R}^{n}\right)=\left\{f \in \mathscr{S}^{\prime}: \hat{f} \in L_{\operatorname{loc}}^{1}\left(\mathbb{R}^{n}\right) \text { and } \int_{\mathbb{R}^{n}}|\xi|^{2 s}|\hat{f}(\xi)|^{2} \mathrm{d} \xi<\infty\right\}$$
The reason why they include $ \hat{f} \in L_{\operatorname{loc}}^{1}\left(\mathbb{R}^{n}\right)$ arise from problems with understanding the meaning of $|\xi|^s \hat f$ if one only knows that $\hat f\in \mathscr{S}'$
This definition makes more sense by restricting $\hat f\in L_{loc}$. However, it is known that for $s < n/2$, $\dot H^s_B$ is a Hilbert space with $(*)$, while, when $\dot H^{s}_B$ is not complete when $s\geq n/2$. I am trying to avoid this definition because it is not complete.
Your first definition does not seem to quite make sense, at least not without further explanation. If all you know is that f is a tempered distribution, how would you define the integral (even allowing for an infinite value)? You would need to know $\hat f$ is locally integrable or something like that.
@MichaelRenardy I add one paragraph. I hope it can address your concern. My question is still there.
I suggest to have a look at Chapter 2 of "An Introduction to the mathematical theory of the Navier-Stokes equation" by P. Galdi. He defines and study homogenuous Sobolev spaces for general $p$ but withouth using the Fourier transform.
@GiorgioMetafune I did not find the definition of fractional derivative. It seems he only deals with s being some integer.
Yes, that is true. But he explains well the density results and the need for the polynomial correction.
|
2025-03-21T14:48:32.060968
| 2020-09-18T18:52:00 |
372031
|
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|
Stack Exchange
|
Is there a CAS that can solve a given system of equations in a finite group algebra $kG$?
Let $k$ be a finite field with char$(k)=p>0$. Let $G$ be a finite group.
Consider the group algebra $kG$.
I would like to solve a given system of equations in $kG$.
Question:
Is there a computer algebra system that can solve a given system of equations in $kG$?
Any help is appreciated.
Thank you very much.
Is there a particular small example that you might like to see?
Hm...for example finding all idempotent elements of the group algebra
As you say you have a given system of equations, you can do this by brute-force in GAP. The following constructs a group algebra $kG$ in GAP -- this one constructs $kG$ when $k = GF(4)$ and $G = C_2$.
G := CyclicGroup(2);
k := FiniteField(4);
kG := GroupRing(k,G);
The output is then
gap> kG;
<algebra-with-one of dimension 2 over GF(2^2)>
and, as expected, one can run commands such as
gap> Size(kG);
16
gap> Elements(kG);
[ <zero> of ..., (Z(2)^0)*<identity> of ..., (Z(2)^0)*<identity> of ...+(Z(2)^0)*f1,
(Z(2)^0)*<identity> of ...+(Z(2^2))*f1, (Z(2)^0)*<identity> of ...+(Z(2^2)^2)*f1, (Z(2^2))*<identity> of ...,
(Z(2^2))*<identity> of ...+(Z(2)^0)*f1, (Z(2^2))*<identity> of ...+(Z(2^2))*f1, (Z(2^2))*<identity> of ...+(Z(2^2)^
2)*f1, (Z(2^2)^2)*<identity> of ..., (Z(2^2)^2)*<identity> of ...+(Z(2)^0)*f1,
(Z(2^2)^2)*<identity> of ...+(Z(2^2))*f1, (Z(2^2)^2)*<identity> of ...+(Z(2^2)^2)*f1, (Z(2)^0)*f1, (Z(2^2))*f1,
(Z(2^2)^2)*f1 ]
Now to solve your given system of equations is easy by brute force.
Thank you. Unfortunately, a problem is that this approach might take years, if the groups are slightly bigger. Therefore, I was wondering if there is some CAS or GAP - package which has already implemented some short cuts / fast algorithms.
@BernhardBoehmler Maybe you can work in a concise matrix representation by switching to FaithfulModule(kG) - for example in this particular case it provides a faithful representation with 2x2 matrices
A partial answer in a very special case, but still possibly useful to record.
The search for idempotents in a group algebra can be simplified by using GAP's MeatAxe: https://www.gap-system.org/Manuals/doc/ref/chap69.html
For example, if $G$ is the dihedral group with eight elements, and $F$ the field with eight elements, the following code verifies that $kG$ is indecomposable as a $G$-module, which means that the only idempotents are zero and one.
G := DihedralGroup(8)
k := FiniteField(8);
cayley := Action(G, AsList(G), OnRight);
reg := PermutationGModule(cayley, k);
MTX.IsIndecomposable(reg);
More generally, if $M$ is a MeatAxe module, you can compute a direct sum decomposition using
MTX.Indecomposition(M)
Of course, this doesn't solve for all the idempotents---it only returns a few. You can also use
MTX.HomogeneousComponents(M)
to break the search into smaller algebras before using a more brute-force method.
As a representation theorist, I'm embarrassed to have to ask, but I'm only used to thinking of representations over fields of characteristic $0$. Why does indecomposability imply that the only idempotents are trivial?
@LSpice Any idempotent in the group algebra induces an idempotent endomorphism of the regular representation (multiplication on the side opposite the action). The image and kernel of this endomorphism would then be a direct sum decomposition.
Got it, thanks. I was trying to think in terms of eigenspaces, which doesn't work in characteristic 2, and it didn't occur to me to substitute image and kernel.
|
2025-03-21T14:48:32.061187
| 2020-09-18T19:09:01 |
372033
|
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|
Stack Exchange
|
A homological algebra approach to the Union-closed sets conjecture
I noted a while ago that there is a nice homological formulation using incidence algebra of the Union-closed sets conjecture (https://en.wikipedia.org/wiki/Union-closed_sets_conjecture). It might just be a simple reformulation but it looks cool and might give a new perspective. Since I made no progress for some time and my experience with lattices is very little, I thought to share this here to see how far one can get and maybe prove the conjecture in some special cases.
Let $L$ be a finite lattice with minimum $m$ and maximum $M$ and $m \neq M$.
The incidence algebra $A=A_L$ of $L$ is by definition the $K$-algebra with basis $p_y^x$ for each $x \leq y$ (where we write $e_i :=p_i^i$) with multiplication $p_{y_1}^{x_1} p_{y_2}^{x_2}= \delta_{x_2,y_1} p_{y_2}^{x_1}$ where $\delta_{i,j}$ is the Kronecker delta. $A$ is isomorphic to the quiver algebra $KQ/I$ where $Q$ is the Hasse quiver of $L$ and the relations $I$ are such that any two paths with the same start and ending points get identified.
Then the Union-closed sets conjecture is equivalent to the statement that there exists a join-irreducible element $x \in L$ with $|[x,M]| \leq \frac{|L|}{2}$ (see conjecture 1 in https://www.uni-ulm.de/fileadmin/website_uni_ulm/mawi.inst.081/Henning/UCSurvey.pdf ).
Now let $A$ be the incidence algebra of $L$ and note that to every point $x \in L$ we have a unique indecomposable injective module (we use right modules) $e_x A$, injective indecomposable module $D(Ae_x)$ and simple module $S_x=top(e_x A)=socle(D(Ae_x))$. For example, as a vector space $e_x A$ is just the span of the elements $p_y^x$ of paths starting at $x$.
Then the conjecture is equivalent to the following statement (in case I made no mistake):
There exists an indecomposable projective module $e_x A$ with $injdim(S_x)=1$ (this is equivalent to $x$ being join-irreducible) that has a minimal injective coresolution $I_i$ such that $\dim I_0 \leq 2 (\dim I_1 - \dim I_2 + \dim I_3 - ....)$.
(To prove the equivalence, note that $|[x,M]| \leq \frac{|L|}{2}$ is equivalent to $dim(e_x A) \leq dim ( \Omega^{-1}(e_x A))$ and then replace $e_x A$ and $ \Omega^{-1}(e_x A)$ by the terms $dim(I_i)$ using the euler characteristic of the minimal injective coresolution, which is finite since $A$ has finite global dimension).
Now this seems to be a nice formulation, but calculating the minimal injective coresolution seems to be extremely complicated for lattices. This was done in https://arxiv.org/pdf/2009.07170.pdf in the distributive case and the same method just gives a non-minimal injective coresolution in general for a non-distributive lattice, which does not help much.
Main question: Can this homological approach work to give a proof of the Union-closed sets conjecture in some special cases?
It might be interesting to also study the terms $I_i$ for other properties. For example I wonder whether the following is true:
Question: Let $x$ be join-irreducible (equivalently $injdim(S_x)=1$) with a minimal injective coresolution $I_i$ of $e_x A$. Do we have that the sequence $dim(I_i)$ is weakly decreasing, that is $dim(I_i) \geq dim(I_{i+1})$ for all $i$?
Some random tests and small cases found no counterexample to this yet.
|
2025-03-21T14:48:32.061397
| 2020-09-19T00:01:18 |
372040
|
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|
Stack Exchange
|
Reference on internal categories and externalization
I'm looking for a reference on internal categories and externalization of internally defined notions.
The nlab has a stub on externalization (more details are available under small fibration) and the page on internal categories gives enough of an introduction that I can sketch most internal notions, but I could really use a concise introduction to internal categories and externalization, and if possible the relationship between internalization and externalization. Are they adjoint in some sense?
I'm fine assuming a background of $2$-category theory, so talking about the $2$-category of internal categories in a category with pullbacks etc. would make sense, but ideally the reference would assume no familiarity with internal category theory or externalization. Any assistance is appreciated.
I don't think such a reference exists. As observed in the abstract of these notes, much of the foundational work of Bénabou on fibred categories is unpublished. But perhaps you might start by reading those notes, and maybe chapter B2 of Sketches of an elephant.
@ZhenLin Thank you, I’ll take a look at those notes. My uni library is currently closed (and doesn’t have a copy to boot), and I don’t have half a g to drop on Sketches right now — are you aware of anywhere I can access that portion online?
I don't think Johnstone has made any part of the Elephant online. Personally I'm holding out for volume 3 (and a corrected edition of the first two volumes) before buying a personal copy...
@ZhenLin Cest la vie, that's a good idea anyway -- I'll see if any nearby universities or public libraries have copies available on request.
I would try Bart Jacobs' book Categorical logic and type theory. It has a very good intro to fibred categories and internal categories and the relation between them.
@DavidRoberts Will do, thank you; for any interested parties, the book David mentions can be found here: https://people.mpi-sws.org/~dreyer/courses/catlogic/jacobs.pdf.
@Alec good find. Also, PTJ once handed out paper versions of a preprint at a talk of his I attended. Putting his magnum opus online is a much more extreme option than just using the arXiv like a normal person.
@AlecRhea: Johnstone's Sketches of an Elephant is available here: http://libgen.rs/book/index.php?md5=22AECD1E74BE933CBA966B1396122B77
@DmitriPavlov Excellent, thank you! (I still intend to buy a copy eventually, for the record)
For what it's worth, I should add my comment as an answer. Chapter 1 of Bart Jacobs' book Categorical logic and type theory (Studies in Logic and the Foundations of Mathematics 141 (1999), (author's page, publisher page, pdf)) is a good intro to fibred category theory, and chapter 7 has an intro to internal category theory, and it links the two. Chapter 9 does some more advanced fibred category theory.
This ended up being exactly what I was looking for; I’ll have to learn about the internal logic of fibrations to fully understand what he’s saying, but that topic is also covered earlier on. Thank you!
|
2025-03-21T14:48:32.061633
| 2020-09-19T01:17:43 |
372041
|
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|
Stack Exchange
|
How do you make an accurate, integrable approximation of $a \operatorname{mod} \left(\frac xb,1 \right)$ with a scaling constant $N$?
I'm working on a project where I'm working with modulo functions. However, to continue, I need to integrate integral powers of a weighted sum of them (e.g of the form $\left(c+\operatorname{weighted sum} \right)^p$, with $c$ a real, positive constant, and $p \in \mathbb{Z}$). So, I first tried Fourier series. However, since the weights are pretty high, the errors blew up, and the amount of terms necessary to correct them are prohibitively high. So, I need another way to create the modulo functions.
This leads me to my question: just like the title says, is there some approximation of $a \operatorname{mod} \left(\frac xb,1 \right)$, $f(x)$ (with $a,b, \geq 1, \in \mathbb{R}$), that has an elementary, closed form antiderivative, has a scaling constant $N$ so that as $N \to \infty$, $|a \operatorname{mod} \left(\frac xb,1 \right)-f(x)| \to 0$ (hopefully $\sim \mathcal{O} \left(10^{-\operatorname{|poly(N)|}}\right)$, but not neccessary) at least on $\{ bk+0.1 \leq x \leq bk+0.9, k=\{0,1,2,3..,\lceil \frac nb \rceil\}\}$, with $n \in \mathbb{R}$ (but hopefully over all $x \in [0,n]$), and has a constant number of terms $k$ that independent of $N,n,a,b$.
Did you try to use the substitution $x=1/y$?
@მამუკაჯიბლაძე How would that help me?
There's something I don't understand here - $a\ \mbox{mod} (\frac{x}{b} )$ itself is elementary and has a closed form antiderivative, so why approximate it? Restricted to intervals between jumps, it's linear, and a square of a weighted sum thereof consists of pieces of parabolas.
@MichaelEngelhardt While that is true, the square (and other integral powers) of the function does not have such a closed form antiderivative (however, if they do, I would like to know).
Isn't the modulo function linear in each interval between jumps? Isn't the square of a linear function a parabola?
@MichaelEngelhardt Yes, but it isn't parabolic over $[0,b]$-- there are cuts, made more complicated by multiplying another mod function to it.
Sure, but it's a purely algebraic problem - you catalog where the jumps are, assemble the parabolas in between, and integrate them on each interval.
@MichaelEngelhardt Is there a way to determine where the jumps are, and the period for a product of $p$ mod functions? Or should that be another question?
I don't think it's a good strategy to multiply out the $p$ factors. Just construct the weighted sum on each interval between jumps, which gives a linear function, and then take the $p$-th power of that. Of course you know where the jumps are, they occur spaced by $b_i $ for the $i$-th term in your weighted sum, just put them all in a big list. Your function is not periodic if the $b_i $ are not commensurate.
@MichaelEngelhardt I wouldn't consider that efficient, as there are, at most $\prod_{i=1} b_i$ jumps, and cataloguing all of them would take a while, and probably require manual analysis. Hence, the original request for approximations.
I don't understand your reluctance. The jumps are at $b_1, 2b_1, 3b_1 ,\ldots,b_2, 2b_2, 3b_2,\ldots, b_3,\ldots $. This is trivial to program and to sort. With any approximation, you'll invest orders of magnitude more effort per jump to achieve convergence, and your integrations also won't be simpler.
Oh sorry, by some reason I imagined it was $a\operatorname{mod}(1,x/b)$ :)
Then, as @MichaelEngelhardt explains, it is just plain integrable, with the antiderivative$$\frac{ab}2\left(\left\lfloor\frac xb\right\rfloor+\operatorname{mod}\left(\frac xb,1\right)^2\right)$$
@მამუკაჯიბლაძე Ok, for example, what's the process to integrating this: $$\left(4\operatorname{mod}\left(\frac x5,1\right)+10\operatorname{mod}\left(\frac x{33},1\right)\right )^p$$ (say $p=2$ for now).
Sorry again, seems I suffer from attention deficit - did not notice you need powers of these things
Let's consider the concrete example given by the OP in comments,
$$
f(x) = \left( 4\ \mbox{mod} \left( \frac{x}{5} ,1\right) + 10\ \mbox{mod} \left( \frac{x}{33} ,1\right) \right)^{p} \ .
$$
$f(x)$ is periodic with period $5\cdot 33 = 165$. It is discontinuous at the points $\{ 5n : n\in \mathbb{Z} \} \cup \{ 33n : n\in \mathbb{Z} \} $; these can be easily listed and sorted in ascending order within any integration range one may be interested in (and for a periodic example such as this, it's of course sufficient to be able to treat one period). The entire integral of $f$ can be assembled by summing up the integrals over individual intervals between consecutive discontinuities, for all such intervals contained in the integration range one is interested in.
Consider an arbitrary such interval, $[d_i ,d_{i+1} ]$, where the $d_i $ denote the discontinuities. On this interval (caveat - for negative $x$, one might have to specify exactly how one interprets the mod function),
$$
f(x) = \left( 4\ \mbox{mod} \left( \frac{d_i }{5} ,1\right) + 10\ \mbox{mod} \left( \frac{d_i}{33} ,1\right) + \left( \frac{4}{5} + \frac{10}{33} \right) (x-d_i ) \right)^{p} \ .
$$
or, to streamline the notation,
$$
f(x) = (s+tx)^p
$$
and one has the integral
$$
\int_{d_i}^{d_{i+1} } dx\, f(x) = \frac{(s+td_{i+1} )^{p+1} }{t(p+1)} - \frac{(s+td_i )^{p+1} }{t(p+1)}
$$
It remains to sum up these contributions; note that the first and the last interval may be only partially integrated over and then the lower or upper integration limits, respectively, have to be appropriately adjusted.
While not exactly answering the question, it is the best (and only) answer given (at least so far), so as is my nature, I'll wait a day before accepting to allow for any new answers. But in the meantime, here's a +1 from me. Thanks!
Since no one else responded, accepted!
@DUO - I hope it helps in practice - this should be straightforward to implement computationally.
|
2025-03-21T14:48:32.062152
| 2020-09-19T01:49:25 |
372042
|
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|
Stack Exchange
|
Determining if a quadratic form is non-negative if variables are non-negative
Let $f(x_1,\dots,x_n) = \sum_{1 \le i \le j \le n} c_{i,j}x_ix_j$ be a homogeneous quadratic form. Is there a quick-ish way to determine whether $f(x_1,\dots,x_n) \ge 0$ for all $x_1,\dots,x_n \ge 0$?
I have a specific homogeneous quadratic form, where $n=44$. I am wondering whether I have to use a super computer to prove that it is non-negative if all of the variables are. I prefer not to disclose my quadratic form.
In general, I know how to figure out whether a given quadratic form is non-negative if the variables are, in $2^n$ time, since $f$ has at most $2^n$ (quickly computable) local minima on the set $\{(x_1,\dots,x_n) \in \mathbb{R}^n : x_1+\dots+x_n = 1, x_1,\dots,x_n \ge 0\}$ (choose a certain subset of the variables to be $0$, and then we get a bunch of linear equations, from looking at derivatives, that determine the rest). But I'm wondering if there's a quicker way, in general.
will not diagonaliztion of symmetric matrices algorithm work here?
@vidyarthi if you diagonalize, it is very hard to use the non-negativity condition, which might be (and is, in my particular case) crucial. I think you might be thinking about showing the quadratic form is positive definite, which is stronger.
Isn't this just a form of Quadratic Programming? Since you have no constraints on your $c_{ij}$ I believe it'll be NP-hard, so you shouldn't expect a much quicker way.
You want to test whether a matrix is "copositive".
You might find the criteria that you're looking for in this article Hannu Väliaho - Criteria for copositive matrices (1986)
It's mentioned in the introduction here that the problem of determining whether a matrix is copositive is NP-complete:
http://www-ljk.imag.fr/membres/Roland.Hildebrand/c6classification/cop_cert.pdf
Checking copositivity is a special case of (nonconvex) quadratic programming. There's a paper which reduces nonconvex quadratic programming to mixed-integer linear programming (with promising results) by using the KKT conditions:
https://arxiv.org/abs/1511.02423
I'd recommend doing the same for your particular quadratic program, and then feeding the resulting mixed-integer linear program into a state-of-the-art solver. These solvers use branch-and-bound techniques to prune the search tree, making it much more efficient in practice than trying to brute-force all $2^{44}$ subsets.
Thank you for your answer. On page 4 of the first link you gave, do I have to solve (6) or can I solve (5)? I.e., will the branch-and-bound techniques apply to and be quick enough to handle (5)?
Also, if the quadratic form had $46$ variables instead of $44$, how long do you think it would take for a state-of-the-art solver to solve the resulting mixed-integer linear program? Thanks!
@mathworker21 Equation (5) contains a nonlinear constraint $x^T \lambda = 0$ (because neither $x$ nor $\lambda$ is a constant), so it's not a MILP. Equation (6) has purely linear constraints and objective, so it's in the correct form for you to input it into a MILP solver.
Thank you! May you please address my second comment when you get a chance? It's my last question to you.
I'm afraid I can't answer the comment [about the 46-variable quadratic form]: the runtime will depend on both the quadratic form itself (not just the number of variables) and the solver you're using, so the best way to answer that question is to experiment yourself.
Maybe I should clarify that I just seek an upper bound. I can tell you that the coefficients of the quadratic form are integers between $0$ and $10$. Is the runtime at most one year on a standard computer? Is it at most $10$ hours? How about on a super computer? If you say it might be a couple of years, then of course I cannot just "experiment myself"...
You might consider the "sum-of-squares" approach. The idea is to find a set of polynomials so that your expression is the sum of squares of the elements in the region of interest. For your case, you could replace each $x_i$ with a new variable $z_i^2$; you are now asking if the corresponding 4-th degree unconstrained polynomial is non-negative.
This restatement may not sound like an improvement, but it turns out that SOS problems can be attacked using semidefinite programming techniques (e.g., see this page). You can use a freely available SDP solvers.
This is a sufficient approach, i.e., it may prove that your original quadratic form is positive, but it can't disprove it. Since you're trying to solve a specific problem, though, it may be worth the gamble.
I'm asking about copositivity. As I already said in the comments, my quadratic form is sometimes negative (if some of the variables are).
Oh, I see my answer was confusing; I've tried to edit it for clarity. Does it make sense now?
Oh, I'm just stupid. I understand now. I'll think more and get back to you. Thank you for your input.
|
2025-03-21T14:48:32.062487
| 2020-09-19T06:52:27 |
372051
|
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|
Stack Exchange
|
How large is the smallest ordinal larger than any “minimal ordinal parameter” for any pair of an Ordinal Turing Machine and a real?
In this question, the notation $P^x(\alpha)$ denotes a situation where a particular OTM-program $P$ performs a computation on input $x$ with an ordinal parameter $\alpha$, assuming that $x$ is written on the initial segment of length $\omega$ (the smallest limit ordinal) of the tape of $P$ at time $0$. That is, $x$ is the input for $P$ written in cells indexed by finite ordinals $(0, 1, 2, \ldots)$ before the start of computation, yet all cells indexed by all ordinals greater than or equal to $\omega$ are initially blank, except one cell indexed by $\alpha$ (this cell is marked by a non-zero symbol.)
Let $\beta$ denote the smallest ordinal such that for any pair of an OTM-program $P$ and a real $x$ (that is, $P$ quantifies over all programs and $x$ quantifies over all reals) exactly one of the following statements is true:
There does not exist an (uncountable or countable) ordinal $\alpha$ such that $P^x(\alpha)$ halts;
If there exists at least one (uncountable or countable) ordinal $\alpha$ such that $P^x(\alpha)$ halts, then, assuming that $\alpha_0$ is the smallest such ordinal, $\alpha_0 < \beta.$
How large is $\beta$?
So your question is what is the least non-OTM-computable with a real parameter, am I correct?
@HanulJeon I didn't understand the question. Regarding what you wrote (just to be sure) did you mean: "what is the least non-OTM-computable ordinal with any arbitrary real parameter allowed"? Admittedly, I don't understand the second part of your answer though (w.r.t. upper-bound). For V=L, my personal reasoning goes as follows: Given any arbitrary ordinal parameters less than a countable $\alpha$ the sup of values clocked (with parameters $< \alpha$) can be shown to be countable. And hence the upper-bound follows (because any real can be computed with some countable ordinal parameter).
@SSequence: "I didn't understand the question" — can you please specify which part of the question is unclear?
@SSequence: "You could just have $\omega$ as a parameter and a certain machine would halt regardless of what real input was placed on it" — yes, of course, a particular program $P_1$ will halt. But a particular program $P_2$ will not (with the same input.) I have emphasized that we take into account all programs, all inputs and all ordinal parameters (assuming that the parameter is minimal, as is written in the question).
@SSequence: [1/2]: yes, of course, but these facts do not affect the definition of $\beta$ at all. Consider the following game. I pick an arbitrary ordinal $\tau_0$. You pick an arbitrary OTM $P$ and an arbitrary real $x$, then write $x$ on the cells indexed by finite ordinals.
@SSequence: [2/2]: If there exists an ordinal $\tau_1 \geq \tau_0$ such that $P^x$ halts if and only if you write "1" in the $\tau_1$-th cell and does not halt if any ordinal parameter less than $\tau_0$ is used, you win. Does there exist an ordinal such that I always win, no matter which pair of $P$ and $x$ you choose? If yes, then $\beta$ is the smallest such always-winning ordinal. If no, I need an explanation.
@lyricallywicked I think I get what you are saying. For V=L, the answer is still $\omega_1$ (for reasons very similar to mentioned in second comment). Given the answer by by Hanul Jeon, it seems that the answer is $\omega_1$ in general too.
Here is your question as I understand it. You mean $P^x(\alpha)$ to denote a real number $x$ in first $\omega$ cells and a single $1$ at the position $\alpha \geq \omega$. And you are asking for the smallest ordinal $\beta$ (if it exists) such that for all possible choices for programs $P$ and real numbers $x$, one of the following is true: $(1)$ The smallest ordinal $\alpha_0$ such that $P^x(\alpha_0)$ halts is less than $\beta$. $(2)$ There exists no ordinal $\alpha \in Ord$ such that $P^x(\alpha)$ halts.
@lyricallywicked I am deleting some of my previous comments (before your comments about the game) as they make the discussion harder to follow. You might also delete some of your comments (before that point) if you want to.
@SSequence: yes, the interpretation above is correct (assuming that $\alpha$ is allowed to be uncountable.) I know that $0^{\sharp}$ is recognizable from $\omega_1$, if $0^{\sharp}$ exists, but it is not recognizable from any countable ordinal. Can the existence of such reals affect the answer to the question?
Yes, $\alpha$ is allowed to be any ordinal. I do not know the fully general answer. The interpretation written in the answer seems to be fairly close to what you have written. So unless that small difference in interpretation changes the answer, the general also seems to be $\omega_1$ (based on the given answer).
Since there is disputation on how to interpret the problem, I think it would be better to clarify my interpretation:
Let $P(x,\alpha)$ be a program, which takes a binary sequence $x\in 2^\mathbb{N}$ (also called a real, which is standard terminology in set theory) and an ordinal $\alpha$. Consider the set
$$H = \{\alpha\mid \text{$\alpha$ is the least ordinal such that $P(x,\alpha) $ halts for some $x$, $P$} \}.$$
Then $H$ is a set. What is the value of $\sup H$?
If I understand your problem correctly, then the answer is $\omega_1$. Please feel free to comment if there is an error in my proof.
For the lower bound, we will find an OTM-program with a parameter $x\in 2^\mathbb{N}$ that computes a countable ordinal. Assume that $x$ codes a well-order over $\omega$ whose order-type is $\alpha$. Consider the following procedure: decode $x$ and enumerate ordinals less than the order-type of $x$ by brute force. (This is possible since there are only countably many members in $x$ and we have infinite time.) In this way, we can compute $\alpha$ from $x$. Now take $P(\beta)$ as follows: if $\beta=\alpha$, it halts. If not, it does not halt.
For the upper bound, assume that we have a program $P$ of real parameter $x$. By Lemma 2.6 of Koepke's Turing Computations on Ordinals, the ordinal computation by $P$ is absolute between $V$ and $L[x]$. Assume that $P$ halts with an input $\alpha_0$, and $\alpha_0$ is the smallest such an ordinal. Moreover, assume that we take time $\theta$ to compute $P(\alpha_0)$.
Now consider the Skolem hull $M$ of sufficiently large $L_\gamma[x]$ generated by $\{\theta,\alpha_0,x\}$. By condensation, there is an isomorphism $\pi:M\to L_\beta[x]$ for some countable $\beta$. Then $L_\beta[x]$ thinks $P$ halts with an input $\pi(\alpha_0)$ and does not halt if we plug in ordinals smaller than $\pi(\alpha_0)$.
By $\pi(\alpha_0)\le \alpha_0$, Lemma 2.6 of Koepke and minimality of $\alpha_0$, we have $\pi(\alpha_0)=\alpha_0$. Hence $\alpha_0$ is countable.
[1/2]: I am not sure what you mean by "real parameter $x$" and why you wrote "$x$ codes a well-order" and "$P$ halts with an input $\alpha_0$", so in order to avoid ambiguity, I have to clarify that $x$ is not a parameter, $x$ is an input: it is an arbitrary real (the word "real" here implies an infinite binary sequence) written on all cells indexed by natural numbers. Note that since $x$ is arbitrary, it is not required to code a well-order.
[2/2]: Then $\alpha_0$ is not an input, it is always the smallest ordinal greater than or equal to $\omega$ such that a program $P$ halts given $x$ as the input and a single "1" written on the $\alpha_0$-th cell of the tape. Do these clarifications affect the answer?
@lyricallywicked The second half of your question is difficult to understand. You don't seem to specify any relation between $\alpha$ and $\beta$ in your possibilities-(1),(2) in the second half of your question. The answer given is for the following question: "what is the least ordinal not reachable by any OTM (with no ordinal parameters and any arbitrary real input allowed)".
@SSequence: [1/2] "You don't seem to specify any relation between $\alpha$ and $\beta$ in your possibilities-(1),(2)" — why? I have provided the full definition. If there exists at least one ordinal parameter such that an arbitrary OTM halts with $\alpha$ as the parameter and $x$ as the input, then there is the smallest such parameter, denoted by $\alpha_0$. Then $\beta$ is the smallest ordinal greater than any $\alpha_0$ under the assumption that $x$ is an arbitrary real.
@SSequence: [2/2] Regarding [The answer given is for the following question: "what is the least ordinal not reachable by any OTM (with no ordinal parameters and any arbitrary real input allowed)"] — if this is so, then no, this is not what this question is about. Definition of $\beta$ is not related to being (non-)reachable, (non-)computable or (non-)writable. It is related to ordinal parameters. Note that ordinal parameters for OTMs may be uncountable (for example, $0^{\sharp}$ is recognizable from $\omega_1$, if $0^{\sharp}$ exists, but it is not recognizable from any countable ordinal.)
@lyricallywicked I use the term 'a real parameter' for an input $x\subseteq \omega$, coded by a binary sequence of length $\omega$. I do not think the answer is changed even if $x$ is arbitrary (just take $x$ that codes a well-order of a given countable ordinal, we can make this choice as $x$ is arbitrary.)
I also know that $\alpha_0$ is not an input. My first part of the proof suggests there is a program $P$ such that if we put an appropriate binary sequence $x$, then $P(\alpha)$ halts but $P(\xi)$ does not halt for $\xi\neq\alpha$.
@lyricallywicked It would be better to give my interpretation of your problem in my answer. Could you check that I am correct?
@HanulJeon: Regarding "Let $P(x, \alpha)$ be a program, which takes a real $x$ and an ordinal $\alpha$" — I am still not sure what you mean by "a program takes an ordinal". The real $x$ is written on the cells indexed by natural numbers, then, additionally, a single "1" is written on the $\alpha$-th cell. All other cells are blank. If this is what you mean, this interpretation is correct. Regarding the definition of the set $H$ — if it takes all pairs of $P$ and $x$ into account, this interpretation is correct.
@lyricallywicked I meant a code of an ordinal a la Koepke: he coded ordinals to a binary sequence with a single 1. Since we are working with an ordinal greater than $\alpha$, we may replace $(x,\alpha)$ to a single binary sequence, which codes $x$ for the first $\omega$ cells and a single 1 at the $\alpha$th cell.
@lyricallywicked I mean the transfinite binary sequence (i.e., binary seqeuence of arbitrary ordinal length.) Sorry for possibly confusing terminology.
@HanulJeon: If "the transfinite binary sequence" means "the entire contents of the tape", the interpretation is correct. The reason of my confusion was that I thought that "a single binary sequence" implies a real whose bits are indexed by natural numbers.
|
2025-03-21T14:48:32.063132
| 2020-09-19T07:16:44 |
372054
|
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|
Stack Exchange
|
The union-closed sets conjecture for finite dimensional algebras
Say a finite dimensional algebra $A$ satisfies the right UC-condition if there exists an indecomposable projective module $P$ of $A$ such that $\operatorname{injdim}(\operatorname{top}(P))=1$ and $P$ has a minimal injective coresolution $(I_i)$ such that $$\dim I_0 \leq 2 (\dim I_1 - \dim I_2 + \dim I_3 - \dots).$$
Dually say that $A$ satisfies the left UC-condition if the dual statement is true (or equivalently, if the opposite algebra of $A$ satisfies the right UC-condition). And $A$ satisfies the UC-condition if it satisfies the left and right version.
For $A$ being the incidence algebra of a finite lattice $L$ this condition is equivalent to the Union-closed sets conjecture, see A homological algebra approach to the Union-closed sets conjecture .
But already for incidence algebras of general posets the UC-condition is not always true anymore.
Question: Can one expect that there is a more general class of finite dimensional algebras containing the incidence algebras of lattice such that the UC-condition is always true?
(one might ask: When does a finite dimensional algebra behave like "a lattice"?)
Of course my generalisation of the conjecture to a condition for general algebras might not be the best. For example for lattices being join-irreducible for a point $x$ is equivalent to $\operatorname{injdim}(\operatorname{top}(e_xA))=1$ and to $\dim(\operatorname{Ext}_A^1(A/J,\operatorname{top}(e_x A)))=1$, but for general algebras those two conditions are not equivalent in general and I choose the condition $\operatorname{injdim}(\operatorname{top}(e_xA))=1$ over $\dim(\operatorname{Ext}_A^1(A/J,\operatorname{top}(e_x A)))=1$, since it is more "homological". Maybe someone has a suggestion for other generalisations.
Finite dimensional algebras do in general not even have an indecomposable projective module $P$ with $\operatorname{injdim}(\operatorname{top}(P))=1$ but such modules exist when the quiver of $A$ is acyclic.
One can for example look at the UC-condition now for other combinatorial objects such as Dyck paths which are in natural bijection to linear Nakayama algebras (see for examaple https://arxiv.org/abs/1811.05846 ). This motives the following question:
Question: Which Dyck paths satsify the UC-condition?
(see http://www.findstat.org/StatisticsDatabase/St001594 for the statistic)
It seems very hard to check the condition in general but for some special subclasses there seem to be nice patterns.
Here one example for bouncing Dyck paths, that are in natural bijection to integer compositions (see for example What are the periodic Dyck paths? for the definition of bouncing Dyck paths):
For $n \geq 3$ the number of bouncing Dyck paths that satisfy the right UC-condition starts with 1,2,5,11,24,51,107,222,457 and might be given by https://oeis.org/A027934 and the sequence of bouncing Dyck paths that satisfy the two-sided UC-condition starts with 1,4,9,21,46,99,209 for $n=4,...,10$ and might be given by https://oeis.org/A027973 .
|
2025-03-21T14:48:32.063579
| 2020-09-19T07:33:11 |
372055
|
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"Riku",
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|
Stack Exchange
|
Prove convergence of whole sequence $f_n$ of solutions to a differential problem to a limit $f$ (without uniqueness assumptions)
Let $\{f_n\}_n \subset C^\infty \cap L^2(\mathbb R^N)$ be a sequence of functions that solves a linear differential equation $F_n(f_n, \nabla f_n) = 0$. Suppose that there exists a subsequence $n_k$ such that $f_{n_k} \to f$ weakly in $L^2$ and that is the weak solution (not necessarily unique) of the limit problem $F(f,\nabla f) =0$.
If we had a uniqueness result for the limit problem, it would be trivial that the whole sequence $f_n \to f$. What strategy can one use to show this kind of result without relying on the uniqueness for the limit problem?
To fix ideas, consider
$$u^n_t + f(u^n)_x = \frac{1}{n}\Delta u^n$$
and
$$u_t + f(u)_x = 0.$$
How do you show that the whole sequence $u^{n}$ converges if we only have convergence up to subsequences and we don't want to exploit the uniqueness of the limit problem?
Can you be more explicit on the regularity for $F$ and $F_n$? In particular, what type of converge do you assume? Are these operators local?...
@RomainGicquaud To fix ideas, choose $u_t +g(u)_x = 0$ and $u_t + g(u)_x - \frac{1}{n}\Delta u = 0$
I was tempted to post this as a comment since it is not more than a suggestion: however it is too long, thus here it is. The only paper I am aware of dealing with a problem of this kind is the following old paper [2] by Calogero Vinti. He deals with the following Cauchy problem for a single 1st order PDE in two variables:
$$
\begin{cases}
u_t (t,x)= f(u_x(t,x))\\
u(0,x)=u_0(x)
\end{cases}.
\label{1}\tag{1}
$$
The initial data $u_0$ belongs to $C^1_b(\Bbb R)$ (the spaces of bounded $C^1(\Bbb R)$ functions) and satisfies the condition
$$
|u_0^\prime(x+h)-u_0^\prime(x-h)|\le |h| M_{u_0}(x)\quad\forall x, h\in\Bbb R,
$$
where $M_{u_0}(x)$ is a non-negative $L^1_\text{loc}(\Bbb R)$ function: let's call $\mathscr{M}$ the class of these functions.
He uses an approximation method for constructing solutions to \eqref{1} proposed by Emilio Bajada in [1] which works also for $f\in C_b^{0,1}(\Bbb R)$, the class of bounded Lipschitz functions where uniqueness of solution lacks, as shown by an example of Mauro Pagni (given in the same paper in an enhanced form): called $\mathscr{K}$ the class of $C^{0,1}([0,a]\times[0,b])$ ($a,b>0$) solutions to problem \eqref{1}, with initial data belonging to $\mathscr{M}$, he proves the following result:
Theorem ([2], "teorema" in §3, pp. 254-262) If $u(t,x)\in\mathscr{K}$ for a given $u(0,x)=u_0(x)\in\mathscr{M}$, however you choose an approximating sequence $\{u_0^n\}_{n\in\Bbb N}\subset\mathscr{M}$ satisfying the following supplementary conditions
$|u_0^n-u_0|\underset{n\to\infty}{\longrightarrow} 0$ uniformly on the interval $\left[-\dfrac{a}{2}, b+ \dfrac{a}{2}\right]$ and
$\displaystyle\int\limits_{-\frac{a}{2}}^{b+\frac{a}{2}} |{u_0^n}^\prime(x)-u_0^\prime(x)|\mathrm{d}x \underset{n\to\infty}{\longrightarrow} 0$
it is possible to find at least a sequence $\{u^n(t,x)\}_{n\in \Bbb N}\subset\mathscr{K}$ of solutions to \eqref{1}, $u^n(0,x)=u^n_0(x)$ for all $n\in \Bbb N$ such that
$|u^n-u|\underset{n\to\infty}{\longrightarrow} 0$ uniformly on $[0,a]\times[0,b]$ and
$\displaystyle\int\limits_{0}^{b} |u^n_x(t,x)-u_x(t,x)|\mathrm{d}x \underset{n\to\infty}{\longrightarrow} 0$ uniformly on the interval $[0,a]$.
The proof goes on by checking that the sequence of solutions to \eqref{1} with the initial data $\{u_0^n\}_{n\in\Bbb N}$ constructed by using Bajada's method satisfies the statement of the theorem.
Appendix: the construction of solutions to the Cauchy problem \eqref{1}.
Here I recall the procedure for constructing the solutions introduced by Bajada in ([1], §2-4, pp. 5-10): obviously, for the proof, this last reference should be consulted.
Let's consider a strip $S= [0,a]\times\Bbb R$ and let $\{m_n\}_{n\in\Bbb N}$ be a monotone sequence of positive integers: define the positive real numbers
$$
\begin{split}
d_n &= \dfrac{a}{2^{m_n}}\\
a_{r_n}& = r_n d_n\quad\text{for all }r_n=0,1,2,3,\ldots, 2^{m_n}
\end{split}\quad n\in\Bbb N
$$
and use them in order to define the substrips
$$
S_{r_n}=[a_{r_n-1},a_{r_n}]\times \Bbb R\quad \text{for all }r_n=0,1,2,3,\ldots, 2^{m_n}
$$
Finally, define the sequence $\{v_{m_n}\}_{n\in\Bbb N}$ of functions $v_{m_n}:[0,a]\times \Bbb R\to\Bbb R$ as
$$
v_{m_n}(t,x)=\left\{
\begin{split}
\varphi_0(t,x)=&\dfrac{1}{2}u_0\left(x+\dfrac{t}{2}\right)\\
& +\dfrac{1}{2}u_0\left(x-\dfrac{t}{2}\right) \\
& \quad + \displaystyle\int\limits_{x-\frac{t}{2}}^{x+\frac{t}{2}}\!\!\!f\big(u_0^\prime(y)\big)\mathrm{d}y &\quad (t,x)\in S_0\\
& \vdots \\
\\
\varphi_{r_n}(t,x)=&\dfrac{1}{2}\varphi_{r_n-1}\left(a_{r_n-1},x+\dfrac{t-a_{r_n}}{2}\right)\\
& +\dfrac{1}{2}\varphi_{r_n-1}\left(a_{r_n-1},x-\dfrac{t-a_{r_n}}{2}\right) \\
& \quad + \displaystyle\int\limits_{x-\frac{t-a_{r_n}}{2}}^{x+\frac{t+a_{r_n}}{2}}\!\!\!f\big(\partial_x\varphi_{r_n-1}(a_{r_n-1},y)\big)\mathrm{d}y & \quad(t,x)\in S_{r_n}\\
&\vdots
\end{split}\right.
$$
Bajada proves that $\{v_{m_n}\}_{n\in\Bbb N}$ is a sequence of bounded continuous functions, and using a compactness argument he is able to apply the classical Ascoli-Arzelà theorem and conclude that there is a subsequence $\{v_{m_n^s}\}_{n\in\Bbb N}\subseteq\{v_{m_n}\}_{n\in\Bbb N}$ such that $\lim_{n\to\infty} v_{m_n^s}(t,x)=u(t,x)$ is a solution to \eqref{1}.
Notes on the appendix
I decided to explicitly add a description of Bajada's algoritm, since it seems to me that the brief sketch given by Vinti in ([2], §2, pp. 252-253) is flawed by many typos or at least written using an unfortunate notation: of course I hope that the "modernized" notation I used above does not hide the procedure in the same way.
Bajada says that its method is "an application of the method of successive approximations": as a matter of fact, this is more similar to a multidimensional variant of Euler's method or more generally of Runge-Kutta methods.
Notes
Applying to the question Vinti's result implies defining the approximating sequence as a sequence of solutions to particular Cauchy problems: while this is certainly possible in some (perhaps many) cases (including the example equation not involving the laplacian), I do not know if this holds true in general.
The topology used by Vinti et al in order to define their concept of continuity is the basically the topology of uniform convergence applied to subspaces of the space of Lipschitz functions: is it possible to extend the result to more general/weak topologies? I do not know.
Is it possible to extend the method of Bajada and Vinti in order to deal with higher dimension and/or higher order PDEs? My answer is unfortunately the same give to the preceding questions.
Finally note that the paper [2] is not an easy read: apart from the fact that it is written in Italian, its notation is also not modern. in the above presentation I attempted to update the notation, but in the paper we could say it follows Gaspard Monge's style: for example he uses $z$ as the independent variable and $p=z_x$, $q=z_y$.
References
[1] Emilio Baiada, "Considerazioni sull'esistenza della soluzione per un'equazione alle derivate parziali, con i dati iniziali nel campo reale (Considerations on the existence of a solution to a certain partial differential equation, with real valued initial data)" (in Italian), Annali di Matematica Pura e Applicata (IV), vol. XXXIV (1953), pp. 1-25, MR55541, Zbl 0051.07303.
[2] Calogero Vinti "Su una specie di dipendenza continua delle soluzioni dal dato iniziale, per l’equazione $p=f(q)$, in una classe ove manca l’unicità [On a kind of continuous dependence of solutions from the initial data, for the $p=f(q)$ equation, in a class where uniqueness lacks]" (in Italian), Annali della Scuola Normale Superiore di Pisa - Classe di Scienze, Serie 3, Volume 19 (1965) no. 2, p. 251-263, MR185249, Zbl 0133.04602.
|
2025-03-21T14:48:32.064182
| 2020-09-19T08:10:36 |
372057
|
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"Dirk",
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|
Stack Exchange
|
Prove integral inequality for divergence-free vector fields
Let $u$ be a divergence-free vector field $u:\mathbb R^n \to \mathbb R^ n$. Does the following inequality hold?
$$\Big( \int_{\mathbb R^n} |u|^2 dx\Big)^2 \le C\Big(\int_{\mathbb R^n} |u|^2|x|^2 dx \Big) \Big(\int_{\mathbb R^n} |\nabla u|^2 dx \Big). $$
How can it be proved? Any reference is appreciated.
The inequality does not seem to scale well for functions $A\cdot u$ and $A\to 0$.
@Dirk Sorry for the typo: I wrote two extra exponents
Looks pretty much like the Heisenberg uncertainty principle for the Fourier transform to me. Am I missing some subtlety here?
|
2025-03-21T14:48:32.064266
| 2020-09-19T09:56:13 |
372066
|
{
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"authors": [
"Mark Wildon",
"Max Lonysa Muller",
"Random",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/372066"
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|
Stack Exchange
|
Do identities exist for the binomial series $\sum_{k=m+1}^{n+1} \binom{k}{m} \binom{n+1}{k-1} $?
While examining the product of two upper triangular matrices, I've found that the $(m,n)$'th entry of the resulting matrix amounts to: $$\sum_{k=m+1}^{n+1} \binom{k}{m} \binom{n+1}{k-1} $$ when $n \geq m$ (all other entries are zero).
Although I have found some summations of products of binomial coefficients here, identities for the sum-product as described above have so far eluded me. Do you know whether identities for this series -- or perhaps even for generalizations of it -- exist?
Can't you kinda disprove that a closed form exists by considering $n,m$ close together. Like if $n-m = 1$, you get something. If $n-m=2$, you get something. If $n-m = 3$, you get something. I bet finding a closed form / identity that covers these three cases is already very nontrivial (or maybe if you also consider $n-m=4$).
$$\sum_{k}{{k\choose m} {n+1\choose k-1}}=\frac{1}{m}\sum_{k}k{k-1\choose m-1}{n+1\choose k-1} = \frac{n+1 \choose m-1}{m}\sum_{k}k{n-m+2 \choose k-m} = \frac{n+1 \choose m-1}{m}(\sum_{k}(k-m){n-m+2 \choose k-m}) + {n+1 \choose m-1}\sum_{k}{n-m+2 \choose k-m} = \frac{{n+1 \choose m-1}(n-m+2)}{m}(\sum_{k}{n-m+1 \choose k-m-1}) + {n+1 \choose m-1}2^{n-m+2} = \frac{{n+1 \choose m-1}2^{n-m+1}(n+m+2)}{m}$$
The identities I have used are ${n \choose k} = \frac{n}{k}{n-1 \choose k-1}, \sum_{k}{n \choose k} = 2^n, {a \choose b}{b \choose c} = {a \choose c}{a-c \choose b-c}$.
The summation in the question excludes $k=n+2,m$, so your answer is actually
$$\frac{{n+1 \choose m-1}2^{n-m+1}(n+m+2)}{m}-{n+2\choose m}-{n+1\choose m-1} $$
Nice. I had wrongly thought that nothing could be done with this type of sum.
@Random thank you! When you write $\sum_{k}$, could you please indicate what you mean by that? What exactly are the lower and upper bounds of such a sum?
@Max Muller Technically, I am summing over $k$ between $m$ and $n+2$. I prefer to sum over all integer $k$ and just say that the binomial coefficient ${a \choose b}$ vanishes if $b > a$ or $b < 0$. I find that summing over all integers gives more natural looking results.
Indeed as you can see from Mark's comment, the "unnatural" bounds obscured the fact that the answer has only small prime factors.
|
2025-03-21T14:48:32.064429
| 2020-09-19T10:20:36 |
372067
|
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|
Stack Exchange
|
Frobenius-Schur indicator and character table of finite groups
Let $G$ be a finite group and $\pi$ an irreducible complex representation. The Frobenius-Schur indicator of $\pi$ is defined as:
$$ \nu_2(\pi):=\frac{1}{|G|} \sum_{g \in G} \chi_{\pi}(g^2) $$
with $\chi_{\pi}$ the character of $\pi$.
Note that the map $s: g \mapsto g^2$ is well-defined on the conjugacy classes as $\tilde{s}: C(g) \mapsto C(g^2)$ because $(hgh^{-1})^2 = hg^2h^{-1}$. Let $\chi_1, \cdots, \chi_r$ be the irreducible characters of $G$ (with $\chi_i = \chi_{\pi_i}$), and $C_1, \cdots, C_r$ be the conjugacy classes, with $\chi_1$ the trivial and $C_1 = C(1)$. The character table of $G$ is given by the matrix $(\chi_{i,j})$ with $\{ \chi_{i,j} \} = \chi_i(C_j)$. The map $\tilde{s}$ induces a map $m$ on $\{1,2, \cdots, r \}$ such that $\tilde{s}(C_j) = C_{m(j)}$. It follows that the Frobenius-Schur indicator $\nu_2$ is completely determined by the character table $(\chi_{i,j})$ and the map $m$ as follows:
$$ \nu_2(\pi_i):=\frac{1}{|G|} \sum_{j} |C_{j}|\chi_{i,m(j)} = \sum_j \frac{\chi_{i,m(j)}}{\sum_i |\chi_{i,j}|^2}$$
because $|C_j| = |G|/\sum_i |\chi_{i,j}|^2$.
Note that the character table alone is not sufficient to determine $\nu_2$. For example, the quaternion group $Q_8$ and the dihedral group $D_4$ have same character table, but the first admits an irreducible complex representation with Frobenius-Schur indicator $-1$ (in fact it is the smallest such finite group) whereas the second not. But these do not have the same class type $(1,2,4A,4B,4)$ for the first and $(1,2A,2B,2C,4)$ for the second (a class is of type $nX$ if its elements has order $n$).
Question: Is the Frobenius-Schur indicator $\nu_2$ completely determined by the character table including the class types? If so, what is the formula?
It is "suggested" true by the section 71.12-5 in GAP manual, as GAP seems to need these data only to compute $\nu_2$.
I would guess the answer is no, I guess you need to know the power map $g \mapsto g^2$ to compute $\nu_2(\chi)$. About GAP, aren't the power maps usually included in the character table? With Display(CharacterTable("A5"));; you see the character table of $A_5$ along with the power maps. Perhaps with a computer search you could find two groups with same character tables and class types, but different Frobenius-Schur indicators.
@spin: two non-isomorphic groups with same character table and class types are called a Brauer pair. Among the $2$-groups, the smallest order of a group in a Brauer pair is $2^8$, and among the $56092$
groups of order $2^8$, there are exactly ten Brauer pairs (see MR2680716 Theorem 2.6.2 page 136). Perhaps the counter-example you expect is among these ten pairs.
Is that the usual definition of a Brauer pair? I thought Brauer pairs also have the same power map, hence they would have the same FS-indicators as well.
@spin: You are right, thanks!
I don't see a way at the moment to a full answer, but I mention the following in case someone can make use of it : the class function Sqr defined by Sqr(g) = the number of square roots of G in G is always a generalzied character: we have
${\rm Sqr}(g) = \sum_{ \chi \in {\rm Irr}(G)} \nu_{2}(\chi) \chi(g)$ for each $g \in G$, and clearly Sqr contains the trivial character with multiplicity one.
It follows that a necessary and sufficient condition for there to be no irreducible character $\chi$ of $G$ with $\nu_{2}(\chi) = -1$ is that the function Sqr is a genuine character of $G$, that is, a non-negative integer (not all zero) combination of irreducible characters of $G$.
(continued..) In general, I don't think it's clear how to calculate the function Sqr just from the class types and character table: in the examples given of the quaternion and dihedral groups of order 8 it is easy, because elements of order $4$ have no square root (in these groups), while in $Q8$ the identity has only two square roots and an element of order $2$ has six square roots , and in $D8$ the central involution has two square roots, the identity has $6$ square roots, and the non-central involutions have no square roots.
In general, when there are many classes of elements of the same order, I think it is less clear how to calculate just from class types and character table how many square roots elements have.
Conversely, in groups $G$ where Sqr is determinable from this information, we can tell whether Sqr is a character by calculating $\langle {\rm Sqr}, \chi \rangle $ for each irreducible character $\chi$.
From the source code of GAP:
#############################################################################
##
#M IndicatorOp( <ordtbl>, <characters>, <n> )
#M IndicatorOp( <modtbl>, <characters>, 2 )
##
InstallMethod( IndicatorOp,
"for an ord. character table, a hom. list, and a pos. integer",
[ IsOrdinaryTable, IsHomogeneousList, IsPosInt ],
function( tbl, characters, n )
local principal, map;
principal:= List( [ 1 .. NrConjugacyClasses( tbl ) ], x -> 1 );
map:= PowerMap( tbl, n );
return List( characters,
chi -> ScalarProduct( tbl, chi{ map }, principal ) );
end );
So in the computation performed on a character table, GAP computes the power map $g \mapsto g^n$ on the conjugacy classes of $G$, and with this for each character $\chi$ the scalar product $\nu_n(\chi) = \langle \chi^{(n)}, 1 \rangle$, where $\chi^{(n)}(g) = \chi(g^n)$ for all $g \in G$.
This shows that the suggested informal justification doesn't justify, rather than answering the main question ("Is $\nu_2$ determined by …") yes or no, right?
@LSpice: Yes, to be more clear my answer doesn't settle the main question posed by Sebastien. Point is that GAP does use more information than just the character table and class types to compute Frobenius-Schur indicators.
|
2025-03-21T14:48:32.064817
| 2020-09-19T10:22:06 |
372068
|
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|
Stack Exchange
|
Finite simple groups and negative Frobenius-Schur indicator
Let $G$ be a finite group and $\pi$ an irreducible complex representation. The Frobenius-Schur indicator of $\pi$ is defined as:
$$ \nu_2(\pi):=\frac{1}{|G|} \sum_{g \in G} \chi_{\pi}(g^2) $$
with $\chi_{\pi}$ the character of $\pi$. Recall that $\nu_2(\pi) \in \{-1, 0,1\}$. The only finite simple groups $G$ with $|G|<10^7$ and having an irreducible complex representation $\pi$ with $\nu_2(\pi) = -1$ are $\mathrm{PSU}(3,q)$ with $q=3,4,5,7,8$ and $\mathrm{PSU}(4,3)$.
gap> it:=SimpleGroupsIterator(10,10000000);; for g in it do if -1 in Indicator(CharacterTable(g),2) then Print([g]); fi; od;
[ PSU(3,3)][ PSU(3,4)][ PSU(3,5)][ PSU(4,3)][ PSU(3,8)][ PSU(3,7)]
gap>
Question: What are the finite simple groups known to have no irreducible complex representation $\pi$ with $\nu_2(\pi) = -1$?
What are those known to have an irreducible complex representation $\pi$ with $\nu_2(\pi) = -1$?
I am specifically interested in the groups $\mathrm{PSL}(2,q)$ for which I ckecked by GAP that for $q<500$, there is no irreducible complex representation $\pi$ with $\nu_2(\pi) = -1$:
gap> for q in [2..500] do if -1 in Indicator(CharacterTable( "PSL", 2, q),2) then Print([q]); fi; od;
gap>
So I expect that it is true for all $q$. Is there a proof using the character table (including class type) of $\mathrm{PSL}(2,q)$ (see this post)?
I think M. Geck has done quite a bit of work on this question for groups of Lie type
In the particular case of PSL(2,q), I think it is true for odd q because the Sylow 2-subgroups are dihedral (including possibly Klein 4) when q is odd. When q is a power of 2, there is one conjugacy class of involutions, and the number of solutions of x^2 = 1 is q^2, which is also the sum of the irreducible character degrees of PSL(2,q), and the FS-indicator formula for the number of solutions of x^2 = 1 forces all indicators to be 1.
For sporadic $G$, you can check the only case with indicator $-1$ is $G = McL$, which has two irreducible characters $\chi$ with $\nu_2(\chi) = -1$. When $G$ is an alternating group, you always have $\nu_2(\chi) \geq 0$ since all irreducible characters of $S_n$ have indicator $+1$.
@spin Yes: https://mathoverflow.net/q/54800/34538
Further to my earlier comment, it is definitely true that when $q$ is a power of $2$, all indicators are $+1$, so you only need to check $q$ odd. When I find time, I will try to make the rest of my comment into a formal answer.
@GeoffRobinson: when there is an irreducible character whose square is the sum of all the irreducible characters (with multiplicity one, or odd) then all the FS indicators are 1, by the result considered here. This applies to PSL(2,2^n) with the Steinberg character.
But I already explained in my September comment why it worked for $PSL(2,2^{n})$ (for a different reason than that given in the answer to yesterday's question).
@GeoffRobinson: Yes. This answer should help for $q$ odd also.
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2025-03-21T14:48:32.065003
| 2020-09-19T11:04:50 |
372069
|
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|
Stack Exchange
|
Is this definition of a Fuchsian operator correct?
In Bjork, Analytic D-modules and applications, the following definition of a Fuchsian operator is given:
Here, I believe, $D(0)=\mathcal{O}$, the zeroth filtered piece of the ring of germs of differential operators at $0$, i.e. the ring of regular functions at $0$ (the alternative interpretation of $D(0)$ of the stalk of $D$ at $0$ does not make much sense I think, since then the $\alpha_i$ would completely mess up the form of the differential operator written above).
Now according to this definition, the operator
$$P=(t-e)\partial_t-a = \frac{t-e}{t}\Big(t\partial_t-\frac{t}{t-e}a\Big),$$
where $e$ is some non-zero complex number, is not Fuchsian, since $(t-e)/t\not\in \mathcal{O}$. However, in "D-modules, perverse sheaves, and representation theory", the following fragment, I believe, is also supposed to serve as a definition of a Fuchsian operator:
The operator $P$ does satisfy the equivalent conditions of this theorem. In fact, point (iii) of this theorem makes it clear that if $Q$ is Fuchsian, then so is $g(t)Q$, for any rational function $g(t)$. This invariance under rescaling by a rational function is not satisfied by the definition in Bjork, so the two are clearly not the same.
So my question is, is this a misprint in Bjork? Or is Bjork using some alternative notion of Fuchsian operators, and if so, why?
|
2025-03-21T14:48:32.065123
| 2020-09-19T12:07:42 |
372071
|
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|
Stack Exchange
|
Why are financial markets modeled by càdlàg processes?
When opening a book or reading an article on mathematical finance, financial markets (e.g. stock prices) are always modeled by càdlàg semimartingales. I was wondering why it is that these processes are assumed to be càdlàg in the first place since there also exists a theory of stochastic integration for optional semimartingales that are not at all assumed to be càdlàg?
Some process that has a double jump. For example, $X_t := 1_{[0,t_0)} + 2 \cdot 1_{(t_0,T]}$ for $0 < t_0 < T < \infty$.
The initial wording of this question https://quant.stackexchange.com/questions/27763/why-does-jump-process-has-to-be-cadlag-and-not-the-other-way-around has a striking similarity
|
2025-03-21T14:48:32.065282
| 2020-09-19T13:45:25 |
372076
|
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"Bjørn Kjos-Hanssen",
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|
Stack Exchange
|
Generalising the union-closed sets conjecture from lattice to a larger class of posets
(edit: I decided to simplify the question and only pose it for bounded posets first)
The Union-closed sets conjecture is equivalent for lattices P to:
There exists a join-irreducible element $a$ with $|[a,M]| \leq |P|/2$, when $M$ is the maximum of $P$.
Recall that an element a of a poset is join-irreducible if there is no subset $X \subseteq P$ with $a\not\in X$ and $a=\bigvee X$.
Call a (finite) bounded poset $P$ lattice-like in case an element $x \in P$ is join-irreducible iff $x$ is covers a unique element.
Every lattice is lattice-like but not every bounded poset is lattice-like.
Question 1: Is the above conjecture also true for lattice-like posets?
This is true for all such posets with at most 8 points. I would think there is a counterexample but I have not found one yet.
Question 2: Are there attempts in the literature already to generalise the Union-closed sets conjecture from lattices to a larger class of posets?
Here is a counterexample of size 23.
Let $m=6$ and let $$P=\{0,a_1,\dots,a_m,1\}\cup\{b_{ij}: 1\le i<j\le m\}$$
where $0<a_i<b_{jk}<1$ whenever $i$ is distinct from $j$ and $k$.
The cardinality of $P$ is $|P|=m+2+\binom{m}{2}=6+2+15=23$.
The join-irreducible elements are only the $a_i$ and $0$, since
each $b_{ij}=\bigvee\{a_k:k\ne i, k\ne j\}$.
Each $|[a_i,1]|=2+\binom{m}{2}-(m-1)>|P|/2$ as long as
$$2+\binom{m}2 - (m-1) > \frac12\left(m+2+\binom{m}2\right)$$
$$4+\binom{m}2>3m$$
which is true for $m=6$ but not for $m=5$.
I wonder if 23 is actually the minimum cardinality of an example...
|
2025-03-21T14:48:32.065397
| 2020-09-19T14:28:57 |
372080
|
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|
Stack Exchange
|
Order of ultradistribution
I know that the order of any distribution of compact support is finite. Is this true in the case of ultra distribution of compact support ( dual of Denjoy-Carleman space)?
If an ultradistribution is of finite order, then it is automatically a distribution.
Thank you. I see it now.
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2025-03-21T14:48:32.065448
| 2020-09-19T14:46:47 |
372082
|
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"Nicolas Hemelsoet",
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|
Stack Exchange
|
A gentle introduction to harnack curves
I am starting to study Harnack curves and I would like to read a very introductory and intuitive text, if it's possible with a lot of examples and with a clear exposition of general ideas.
Did you look at articles by Mikhalkin ?
|
2025-03-21T14:48:32.065509
| 2020-09-19T15:09:02 |
372084
|
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|
Stack Exchange
|
Is strictly harder than NP-hard cryptography possible?
Looks like there is cryptography based on NP-hard problem, e.g. McEliece cryptosystem. The algorithm is an asymmetric encryption algorithm and is based on the hardness of decoding a general linear code (which is known to be NP-hard)
We are trying to strengthen this result.
Q1: Is strictly harder than NP-hard cryptography encryption or signature algorithm possible?
We don't allow One-Time Pads (OTP) and similar external secrets.
Conjecture J1: the answer is negative via generic attack of
symbolic execution and then solve SAT with NP-oracle.
J1 implies that if a C language program implements some cryptographic
algorithm and runs in time $X$ milliseconds, then the symbolic execution size of the CNF formula that breaks the algorithm is polynomial in $X$.
XXX make this more rigorous.
It may be a good idea to unroll the loops by hand and ask about loopless
programs.
Counterexample to J1 might be candidate for hard cryptography.
The main problem with J1 is that the resulting CNF might be of exponential
size. We did some experiments with CBMC: Bounded Model Checker with factorization and the hash function SHA256 and the
CNF were small enough.
Here is toy RSA example with zero knowledge of integer factorization:
void main() {
int nondetint();/* can be anything */
int p,q,n;
p=nondetint();
q=nondetint();
n=p*q;
__CPROVER_assert(!(n==13*17 && 1 <p && p <n && 1 < q && q <n),"factor");
}
$cbmc --trace factor1.c
This approach might be used to mine bitcoins SAT solving - An alternative to brute force bitcoin mining.
Also this appears consistent with the fact that if P=NP all crypto will break.
Potential candidates are $\Sigma_2^p$-hard problems.
Is there existing NP-hard cryptography?
@Cechco Yes, but the scheme was broken, see https://en.wikipedia.org/wiki/Merkle%E2%80%93Hellman_knapsack_cryptosystem
McEliece's cryptosystem is based on the hardness of decoding a general linear code, which is NP-hard. It might not be the most efficient cryptosystem today, but it isn't "broken" in any sense.
@BenSmith Thanks. Do you have opinion about the conjecture (in case it makes sense)?
Could you point to a definition of "NP-hard crypto" for the outsiders, please? Also, is there a way of making "we don't allow One Time Pads (OTP) and similar external secrets" less hand-wavy?
@AndrejBauer Thanks, I edited trying to clarify. Feel free to edit. OTP is unbreakable crypto by definition.
NP Hard is a lower bound on hardness, not an upper bound. "Harder" than NP Hard implies NP Hard
Also, a cipher by definition involves an external secret: Namely, the key. In a OTP, the key is necessarily the same size as the message
Perhaps you mean that the secret is "small" but the message can be reasonably "large"; in which case, a OTP can't work
@AndrejBauer A cipher indistinguishability problem for a given cipher is NP-Hard if an oracle that can win the cipher indistinguishability game in polynomial time can also be used to solve an NP-Hard problem in polynomial time
Take your favourite class of difficult decision problems (say, provably exponential time). Encode a message as a sequence of problem statements, one with FALSE answer for a 0 and one with TRUE answer for a 1.
I don't have an opinion on the conjecture, I'm not even sure it's well-posed (cf. @ogogmad 's comments). I'm a bit confused by the inclusion of OTP here, to be honest: with OTP you're looking at information-theoretic security, not algorithmic complexity.
@BenSmith The conjecture is true and confirmed by Daniel Kröning,
the CBMC author.
I think I may not understand your model of cryptography. My model would be that encryption is a polynomial time computable, injective, function from plaintexts of length $m$ to cipher texts of length $n$, and decryption is inverting this function. In that case, such a problem will always be in NP.
Indeed, we must have $m \leq n$, since we require that encryption be injective. Given a coded message of length $n$, the plaintext is then a witness of length $m \leq n$ that shows that the decoding can be found. So an NP-oracle can always just guess the message, and then check in polynomial time that its guess is correct.
Which aspect of your model am I missing?
Thanks. This makes sense, you are not missing anything. I though at very abstract level of breaking the underlying math: since there are harder than NP problems, can crypto benefit from this.
Since the discussion is concerned with asymmetric encryption, shouldn't "polynomial time" have a dependence on key length?
@S.Carnahan Yes, but the trouble is that now we get into modeling issues. Quite likely, the map (key, message) to cipher text is not invertible. Rather, as a code breaker, my goal is to find a preimage (key, message) where the message is plausible English text. I wanted to make the OP spell out their choices here instead of trying to guess.
For any model I can think of, decryption is in NP.
@David E Speyer what is your idea about keyless cryptography such as hash function? In this case the encrypted message is shorter than the message.
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2025-03-21T14:48:32.065857
| 2020-09-19T16:31:09 |
372089
|
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|
Stack Exchange
|
Writing Euler's equations in a different combination of variables? without explicit appearance of the variable $p$
The Euler equations are given as $$ \pmb{u}_t +\pmb{u}\cdot D\pmb{u} = Dp$$ $$div\mbox{ }\pmb{u} = 0$$
Where $$u = [u_1,u_2,\ldots u_n]^T$$
Now I want to rewrite these same equations but with a new combination of variables $$\pmb{v} = [u_1,u_2,\ldots u_n,p]^T $$
I want to do this just for the sake of getting some mathematical advantage and not for any physical inference. Is there any such formulation of equation already done? If not I'd like to know if it can be done? In this new combination, the equations will be only in terms of $\pmb{v}$ and some constants and no appearance of $p$ explicitly.
The construction of adding $p$ as an additional element to the vector $\mathbf u$ only hides it in the vector $\mathbf v$, without actually eliminating it from the Euler equation. This can be easily done with a vector $\mathbf e=(0,0,\ldots 0,0,1)$, and a diagonal matrix $\Delta={\rm diag}\,(1,1,,\ldots 1,1,0)$ so that $\Delta\cdot\mathbf v=(\mathbf u,0)$ and $\mathbf e\cdot\mathbf v=p$, but such a construction seems rather pointless:
$$\partial_t (\Delta\cdot\mathbf v)+(\Delta\cdot\mathbf v)D(\Delta\cdot\mathbf v)=D(\mathbf e\cdot\mathbf v).$$
To really eliminate the pressure from the Euler equation you can use the Leray projection, which maps a vector field to its zero-divergence part: If $\mathbf v=\mathbf w+\nabla\phi$ with $\text{div}\,\mathbf w=0$, then the Leray projection is $\mathbb P(\mathbf v)=\mathbf w$. Formally, this construction can be written as
$$\mathbb P(\mathbf u) = \mathbf u - \nabla \Delta^{-1} (\nabla \cdot \mathbf u).$$
The Euler equation then reads
$$\frac{\partial}{\partial t}\mathbf u+\mathbb P[\mathbf u\cdot\nabla\mathbf u]=0.$$
More explicitly, this can be written as (see for example page 6 of these notes):
$$\frac{\partial}{\partial t}\mathbf u(\mathbf x,t)+\mathbf u(\mathbf x,t)\cdot\nabla\mathbf u(\mathbf x,t)=\frac{1}{C_d}\int_{\mathbb R^d}\frac{\mathbf x- \mathbf y}{|\mathbf x-\mathbf y|^d}\sum_{i,j}\frac{\partial u_i(\mathbf y,t)}{\partial y_j}\frac{\partial u_j(\mathbf y,t)}{\partial y_i}\,d\mathbf y,$$
with $C_d$ the surface area of the $d$-dimensional unit sphere.
Request you to give me the equations in terms of $\pmb{e}$ and $\pmb{v}$ as you have mentioned that $p = \pmb{e}\cdot\pmb{v}$, eliminating both $\pmb{u}$ and $p$.
What is $\Delta^{-1}$. Should I simplify $ \mathbb{P}[ \pmb{u}\cdot\nabla\pmb{u}]$ by substituting $\pmb{u}\cdot\nabla \pmb{u}$ in $\mathbb{P}(\pmb{u}) = \pmb{u} - \nabla\Delta^{-1}(\nabla\cdot\pmb{u})$? I dont want $\mathbb{P}$ appearing in my final equation.
Or is there any direct formula or Fourier multiplier for $\mathbb{P}[\pmb{u}\cdot \nabla \pmb{u}]$?
I added the requested equation that does not contain $p$ explicitly, but really, that is just hiding the pressure in the notation; the Leray projection is the way to go if you want to truly eliminate the pressure.
Thank you very much for the help with that equation @CarloBeenakker. In your method, you have a term that is a Leray projection of a vector field, but how will you do furthur analysis without explicitly having a expression for the Leray projection in terms of $\pmb{u}$. Just utilize the fact that the divergence of the Leray projection is 0 and thats it? there has to be some way to express it in terms of $\pmb{u}$ ?
I have added an explicit formula for the Leray projection, as an integral over derivatives of the vector field $\mathbf u$.
|
2025-03-21T14:48:32.066234
| 2020-09-19T18:07:07 |
372094
|
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|
Stack Exchange
|
A possible characterization of the category of finite $p$-groups
Let $\mathcal{FG}$ be the category of finite groups. Let $S$ be a full subcategory of $\mathcal{FG}$.
Assume that $G\in \mathcal{FG}$ and $P\in S$ is a subgroup of $G$. We say that $P$ is $S$-maximal if there is no object $P'\in S$ with $P\subset P' \subset G$. Assume that $S$ satisfies the following conditions:
The subcategory $S$ is closed under taking subgroups, taking extensions" and isomorphy. That is: $P\in S, Q\subset P$ implies that $Q \in S$. Moreover for every short exact sequence $1\to P\to Q\to R\to 1$ we have $Q\in S$ if $P,R \in S$. Moreover every group isomorphic to an object of $S$ lies in $S$.
For every $G\in \mathcal{FG}$, every two $S$-maximal subgroups of $G$ are conjugate. Moreover for every maximal $S$-subgroup $P$ of $G\in \mathcal{FG}$ we have $N(N(P))=N(P)$ and $\lvert G\rvert/\lvert N(P)\rvert$ is coprime to $|P|$.
Does it imply that $S$ is the category of $p$-groups for some prime number $p$?
"$|G|/N(P)$ is coprime to $|G|$"?
@JeremyRickard I am really sorry I revise it.
@YCor Thank you for your revision
You probably want to rule out the trivial cases where $S$ consists of trivial groups or of all finite groups.
The three conditions in (1) can be collapsed to just: for every exact sequence $P \to Q \to R$, we have that $Q \in S$ if $P, R \in S$. (Oh, maybe you have to add the assumption that the trivial group belongs to $S$.)
Actually there's nothing categorical in the question, which is about a certain subclass of the class of finite groups.
@LSpice It seems to me you’ve just restated the closure under extensions condition, which does imply the closure under isomorphy condition, but not the closure under subgroups condition I think.
@YCor I agree that there is no formal reliance on category theory. But I think it’s fair to say that the closure conditions on $S$ are naturally motivated and thought about from a categorical perspective. (Perhaps you were replying to a since-deleted comment and I’m missing some context, in which case my apologies)
@TimCampion, re, given the containment $Q \subseteq R$ with $R \in S$, we have $R \to 1 \to R$, so use the condition once to conclude from $R \in S$ that $1 \in S$; and then $1 \to Q \to R$, and so use it again to conclude from $P, R \in S$ that $Q \in S$.
@LSpice Oh I see -- you don't require your exact sequences to be short exact! I'm a little hazy on what the definition of an "exact sequence" ought to be in a category like $Grp$ where not every monomorphism is regular, but I suppose that makes sense. On the other hand, your condition would also imply that $S$ is closed under quotients, which we didn't need to assume.
@TimCampion, re, I thought: $A \xrightarrow i B \xrightarrow j C$ is exact if $j \circ i$ factors through $1 \to C$ and if $\operatorname{im}(i) \to \ker(j)$ is an isomorphism. (I state it that way because I think it fits into a more general framework, but I'm no category theorist.) You're right about quotients; I first read the closure condition in (1) as: given $1 \to P \to Q \to R \to 1$, we have $Q \in S$ if and only if $P, R \in S$. I eventually realised that's not what it said, but forgot about that when re-stating.
We can get most of the way to an answer, without using all the conditions.
Definition: Say that $S \subseteq \mathcal{FG}$ is nice if it is nonempty and closed under subgroups, extensions, and isomorphy, and has the property that for any $G \in \mathcal{FG}$, the $S$-maximal subgroups of $G$ are all conjugate.
Proposition: Suppose that $S \subseteq \mathcal{FG}$ is nice. Then there is a set $\mathcal P$ of primes such that $P \in S$ if and only if the prime divisors of $|P|$ are all in $\mathcal P$.
This implies that any $\mathcal P$-Hall subgroup of $G$ is a maximal $S$-subgroup of $G$. Conversely, if $G$ has a maximal $S$-subgroup $H$, then assuming niceness, since $H$ contains the Sylow $p$-subgroups for each $p \in \mathcal P$, we have that $H$ is a $\mathcal P$-Hall subgroup. That is, if $S$ is nice, then a maximal $S$-subgroup is precisely a Hall $\mathcal P$-subgroup.
Proof: Let $\mathcal P$ be the set of primes $p$ such that there exists $P \in S$ with $p$ dividing $|P|$ (as it must be). By passing to subgroups, we have $C_p \in S$ for $p \in \mathcal P$, and by taking extensions we have that every finite $p$-group is in $S$ for every $p \in \mathcal P$. Let $G$ be a finite group such that the prime divisors of $|G|$ are in $\mathcal P$. Now, there are $S$-maximal subgroups of $G$ containing each Sylow subgroup of $G$, and by hypothesis these are all conjugate and in particular have the same order. So if $|G| = p^n m$ with $p$ not dividing $m$, then $p^n$ divides the order of any $S$-maximal subgroup $P \subseteq G$. Since this is true for all prime divisors $p$ of $|G|$, we have that $|G|$ divides $|P|$, so that $P = G$, and thus $G \in S$.
It's known that not all $\mathcal P$-Hall subgroups of a finite group $G$ are conjugate in general, but I don't happen to know whether the known counterexamples cover all possible sets $\mathcal P$ of primes other than the subsingletons and the set of all primes. If $G$ is solvable, then its $\mathcal P$-Hall subgroups are conjugate -- so if we replace $\mathcal{FG}$ with just the solvable finite groups we get an expanded set of possibilities for $S$.
(Many of the comments relate to an earlier version of this post which attempted to whittle down the set of $\mathcal P$ allowed. Thanks to Richard Lyons below for pointing out the mistake.)
EDIT: I just wanted to add another way to close the argument, alternate to Richard Lyons' above, which relies on the original literature on Hall subgroups.
Hall showed in Theorems Like Sylow's, discussion after Theorem A4, that if $\mathcal P$ contains two primes smaller than $n$, then any Hall $\mathcal P$-subgroup of $\Sigma_n$ cannot be solvable. Then Thompson showed in Hall subgroups of the symmetric groups that the only nonsolvable Hall subgroups of $\Sigma_n$ are the trivial group, $\Sigma_n$ itself, and (if $n$ is prime) $\Sigma_{n-1}$. So if $\mathcal P$ contains $p,q$ and omits $r$, then take $G = \Sigma_n$ for any composite $n > p,q,r$. The only candidate Hall subgroups of $G$ are the trivial group and $G$ itself; the former is not a $\mathcal P$-Hall subgroup because $p | n!$ and the latter is not because $r | n!$.
Why does your condition $\operatorname{trace}(x y) = 2\Re(\zeta_r)$ imply that $x y$ has order $r$?
@LSpice the determinant is 1 so the trace determines the characteristic polynomial, and hence the eigenvalues must be $\zeta_r,\bar \zeta_r$
Thanks! Why are distinct eigenvalues needed? If all you care about is the order, then it seems to me that $\begin{pmatrix} -1 & 0 \ 0 & -1 \end{pmatrix}$ is as good an order-$2$ element as any.
@LSpice I'm using distinct eigenvalues simply to ensure that the matrix is diagonalizable -- the matrix $\begin{pmatrix} -1 & 1 \ 0 & -1 \end{pmatrix}$ has the same characteristic polynomial as $\begin{pmatrix} -1 & 0 \ 0 & -1 \end{pmatrix}$ but doesn't even have finite order! We can avoid talking about eigenvalues by observing here that the characteristic polynomial $(z+1)^2$ doesn't divide the polynomial $z^2 - 1$, so doesn't ensure that that $z$ has order 2.
Let me try to say it once and for all, cleanly. If $M \in SL_2(\mathbb C)$, then its characteristic polynomial is $p(z) = z^2 - \operatorname{trace}(M)z + 1$. The fact is that for any $r \in \mathbb N_{\geq 3}$, if $\operatorname{trace}(M) = 2 \Re(\zeta_r)$, then $p(z)$ divides the $r$th cyclotomic polynomial, and so $M$ has order $r$; if $r = 2$, then $p(z)$ only divides the square of the $r$th cyclotomic polynomial, so we can't conclude this. In order to see this fact, I tend to think in terms of eigenvalues and Jordan normal forms, but the fact can be stated without referring to these.
Got it, thanks (and sorry for the several questions). (I wonder if the unipotent element you mention can be written as a product of two finite-order elements? It's probably straightforward to see either way, but it's not obvious to me.)
No, thanks for the questions! Especially because I was still confused myself.
@LSpice It occurs to me that since that matrix is in $SL(2,\mathbb Z)$, and since $PSL(2,\mathbb Z) = C_2 \ast C_3$ (I forget the exact generators), and since the kernel of $SL(2,\mathbb Z) \to PSL(2,\mathbb Z)$ is torsion, it follows that it can be written as a product of finite-order elements. But I don't know if two of them suffice!
@TimCampion Is the reduction correct to just finding $x$ and $y$? For example, let $P={3,5}$ and $G=SL_2(16)$, of order $2^<IP_ADDRESS>$. Then $G$ has a cyclic subgroup $H$ of order $15$ and any subgroup of $G$ of order dividing $15$ is conjugate to a subgroup of $H$. (Theorem. If a group has a nilpotent Hall $P$-subgroup $A$, then any $P$-subgroup of it is conjugate to a subgroup of $A$.) But $G$ contains $SL_2(4)\cong A_5$ and so it contains $x$ of order $3$ and $y$ of order $5$ such that $xy$ has order $2$.
@RichardLyons Thanks, I'm definitely out on a limb here, not really being a finite group theorist -- so I'm really glad to look at an actual example! And I think you're absolutely right -- Let $S$ be defined by $\mathcal P = {3,5}$ and let $G,H$ be as you defined them; if we assume that $S$ is nice, then we can conclude that every $x \in G$ of order 3 is contained in a subgroup isomorphic to $H$ and similarly for every $y \in G$ of order 5, but we can't conclude that for every such $x,y$ that they are both contained in the same copy of $H$, which is what my argument wrongly assumed.
If I understand correctly, the initial answer to the question is equivalent to the following assertion: for all distinct primes $p,q,r$, there exists a finite group $G$ such that every subgroup of $G$ of index coprime to $pq$ possesses an element of order $r$.
$\DeclareMathOperator\GL{GL}\newcommand\card[1]{\lvert#1\rvert}$Here is a proof of "yes," using Tim Campion's proposition below.
Let $p$ be the smallest prime in $P$. For any prime $q\ne p$, let $o_q(p)$ be the multiplicative order of $p$ modulo $q$, or equivalently the least $n$ such that $q$ divides $\card{\GL_n(p)}$. Assuming $P\ne\{p\}$, let $q\in P-\{p\}$ and $G=\GL(V)$ where $V$ is a vector space of order $p^n$ and $n=o_q(p)$. As $p$ is smallest, $n\ge2$. By assumption, $G$ contains a Hall $P$-subgroup $X$. Then $X$ contains a Sylow $p$-subgroup $U$ of $G$ as well as an element $x$ of order $q$. If $U$ is normal in $X$, then $\card X$ divides $\card U(p-1)^n$, the order of the full upper triangular group, which is not divisible by $q$ as $q>p$. So $U$ is not normal in $X$. The theory of $B$-$N$ pairs then implies that $X$ contains a copy of $\operatorname{SL}_2(p)$. Hence every prime divisor of $p-1$ lies in $P$, which forces $p=2$. Then $U$ is a Borel subgroup of $G$, so $X$ must be a parabolic subgroup of $G$. But because of $x$, $X$ stabilizes no proper subspace of $V$. The only such parabolic subgroup is $X=G$. Hence $P$ contains all prime divisors of $\card G$.
In particular, $3\in P$.
Now suppose that $P$ is not the set of all primes and choose a prime $r\not\in P$ to minimize $m=o_r(2)$. Since $3\in P$, $m\ge3$. Let $H=\GL(W)$, where $W$ is a vector space of order $2^m$. Then $H$ contains an element $y$ of order $r$. Let $U$ be a Sylow $2$-subgroup of $H$. Let $W_1$ and $W_{m-1}$ be $U$-invariant subspaces of $W$ of respective dimensions $1$ and $m-1$. Let $H_1$ and $H_{m-1}$ be their respective stabilizers in $H$. Then $H_1$ and $H_{m-1}$ are maximal parabolic subgroups of $H$ containing $U$, and each is an extension of an elementary abelian $2$-group by $\GL_{m-1}(2)$. Hence $H_1$ and $H_{m-1}$ are $P$-groups, by our choice of $r$. However, they are maximal subgroups of $H$ and they are not conjugate in $H$, being distinct parabolic subgroups containing $U$.
By assumption, $\langle H_1^g, H_{m-1}\rangle$ must be a $P$-group for some $g\in H$. Since $H_{m-1}$ is maximal and not equal to $H_1^g$, $H$ must be a $P$-group. But $y\in H$ has order $r\notin P$, contradiction.
|
2025-03-21T14:48:32.067004
| 2020-09-19T18:31:22 |
372096
|
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|
Stack Exchange
|
Analogue of Schur indicator for reps in p-adic fields
Suppose we have a finite group $G$, a division algebra $X$ over $\Bbb Q_p$ with (some) maximal subfield $L$ and an irrep $V$ of $G$ over $L$ (one may assume that everything is separable to make life easier). We can compute invariants $$\nu_s := \frac{1}{|G|} \sum_{g \in G} \chi_{V}(g^s).$$
Can we tell whether $V$ is induced from some $L' \subset L$, or admits an $X$-structure looking at those numbers?
P. S. I tried to prove easy part of Frobenius-Schur theorem (i. e. that those numbers can only be roots of unity or zero) but failed to do so; it's just false for $s \neq [L:\Bbb Q_p]$, if my computations are correct, but I may have failed at checking irreducibility. If it holds in case they are equal, or in general, I'd be happy to know why.
Just to check terminology: $V$ is induced from $L'$ means $V = V' \otimes_{L'} L$ for some $G$-rep'n $V'$ over $L'$, and $V$ has an $X$-structure means that the $L$-action on $V$ extends to an $X$-action by $G$-morphisms of $V$? \ Also, do you assume anything about $p$ and the order of $G$?
Yes for meaning of induced, by $X$-structure I just mean that $V$ is obtained from $X[G]$-module by restriction to $L[G] \subset X[G]$, probably it means same as you written. I do not want to assume that $G$ has no $p$-elements.
We mostly mean the same thing. I just wanted to check that, in $X[G]$, $X$ commutes with $G$.
For $p>2$, there exist finite groups of exponent $p$ with characters of arbitrary large degree, for example extraspecial $p$-groups of exponent $p$. Then $\nu_p$ is just the character degree for these groups, and can be arbitrarily large. I doubt that these numbers tell you much about irreps over $p$-adic fields.
@FriederLadisch, are the representations you mention (or enough of them to make sure that $\nu_p$ is unbounded) realised over $p$-adic fields?
@LSpice: well, they can be realized over any field containing the $p$-th roots of unity, so over $\mathbb{Q}_p(\varepsilon_p)$. If $V$ is one of my representations, I can also view it as representation over $\mathbb{Q}_p$ of dimension $(p-1) \operatorname{dim} V$, it is still irreducible then, but not absolutely irreducible.
|
2025-03-21T14:48:32.067180
| 2020-09-19T18:33:49 |
372097
|
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|
Stack Exchange
|
Resonance arising when harmonic oscillator is excited using sawtooth
Solutions to the differential equation $my'' + ky = F \sin \omega t$ show resonance when the driving frequency $\omega$ equals the natural frequency $\sqrt{k/m}$. That is, solutions are unbounded when $\omega = \sqrt{k/m}$ and periodic for all other frequencies. It seems that when the sine function is replaced by a sawtooth function, there are more resonant frequencies. Numerical experiments here with $m = k = F = 1$ seem to be resonant when $\omega = 1/n$ for any integer $n$. Are there any theoretical results that prove this conjecture?
Doesn't this follow from the Fourier decomposition of the sawtooth function?
It is easy (for a physicist/engineer!) to see why physically. The discontinuity in the sawtooth function gives the oscillator a "kick". Since there is no damping in the system, it resonates just as well if you "kick" it once every $n$ cycles of its natural frequency of vibration.as if you "kick" it at every cycle.
This question shows why mathematicians should take more engineering classes ;-)
The kick argument makes sense. However, it would imply that you always get resonance, regardless of the driving frequency, but you only get resonance for special frequencies.
The sawtooth function $f$ has Fourier decomposition
$$
f(t) = \frac{1}{2}-\frac{1}{\pi}\sum_{n=1}^\infty \frac{1}{n} \sin(n\omega t)
$$
Therefore, if $\omega=\frac{\omega_0}{n}$, the $n$-th harmonic of $f$ will have angular frequency $n\omega=\omega_0$, resulting in resonance.
|
2025-03-21T14:48:32.067312
| 2020-09-19T18:55:01 |
372099
|
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"Alapan Das",
"Max Lonysa Muller",
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|
Stack Exchange
|
Do generalizations of the identity $\sum_{n=k+2}^{\infty} \binom{n-1}{k} (\zeta(n) -1) = 1 $ exist?
On p. 263 of Borwein's paper entitled “Computational Strategies for the Riemann zeta function”, the following identity is stated: $$\sum_{n=k+2}^{\infty} \binom{n-1}{k} (\zeta(n) -1) =1 . \qquad \qquad (*)$$It is valid for all nonnegative integer $k$.
Can the identity be generalised and/or are there variations of it involving binomial sums which are also valid for infinitely many nonnegative lower bounds $k$?
I suppose one can rewrite the formulas of Landau: $$ \frac{1}{s-1} = \sum_{n=0}^{\infty} \binom{s+n-1}{n-1} \frac{\zeta(s+n)-1}{n} $$ and Ramaswami: $$ (1-2^{1-s})\zeta(s) = \sum_{n=1}^{\infty} \binom{s+n-1}{n} \zeta(s+n)$$ in a manner that is similar to $(*)$, but I'm looking for both more and more general examples of such sums. Preferably, all sums in the collection amount to the same constant and involve binomial coefficients.
These later identities are wrong. The first sum should be be $$\frac{\zeta(s)}{s}=\sum_{n=0}^{\infty} \binom{n+s-1}{n-1} \frac{(\zeta(s+n)-1)}{n}$$. And the second identity isn't converging. That may be $$\zeta(s)=\sum_{n=1}^{\infty} \binom{s+n-1}{n}(\zeta(s+n)-1)$$.
@AlapanDas Hmm, these identities were retrieved from p. 286 of Borwein's paper. So if they're wrong, you've found an error in a peer-reviewed article. Do you have sources for your expressions?
These identities can be proved by the identity $\zeta(n)=\frac{1}{\Gamma(n)}\int_{0}^{\infty} \frac{x^{n-1}}{e^x-1} dx$. Also, I have checked these identities in Wolfram alpha.
In my first comment, I have made a small error in the second identity. If we take $n$ from $1$, then the value is $1$, if the limit is from $n=0$, then the value will be $\zeta(s)$.
|
2025-03-21T14:48:32.067455
| 2020-09-19T18:58:04 |
372100
|
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"authors": [
"David Roberts",
"Jeff Strom",
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|
Stack Exchange
|
Products of cones and cones of joins
The join of $A$ and $B$ is the pushout of the diagram
$$
CA \times B \gets A\times B \to A\times CB,
$$
which can be formulated in either the pointed or unpointed topological
category. This pushout is naturally a subspace of $CA\times CB$.
In the unpointed context, there is a homeomorphism $C(A * B) \cong CA \times CB$
extending the inclusion of the join into the product.
I have also seen it asserted in the pointed
context, with less convincing arguments.
In the pointed context, the case in which one
of the spaces is the one-point space $P$ seems to raise problems:
since $CP = P$ and $A*P = CA$,
$$
C(A * P) = C(CA)
\qquad
\mbox{while}
\qquad
CA \times CP \cong CA.
$$
So I am reluctantly inclined to believe that the pointed version is false.
Question:
Is there a statement about pointed cones, products and joins that
does roughly the same job as the unpointed
homeomorphism $C(A* B) \cong CA\times CB$?
EDIT: Perhaps a solution is to restrict to CW complexes (or cell complexes), and prove that the cone and the reduced cone of such a space are homeomorphic, except for the one-point space (it's true for spheres, and probably induction will work).
@DavidRoberts The pointed cone is $CP = (\times I)/(\times I)$.
ah, of course! :-S
|
2025-03-21T14:48:32.067586
| 2020-09-19T19:01:46 |
372101
|
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"Joseph Gordon",
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"url": "https://mathoverflow.net/questions/372101"
}
|
Stack Exchange
|
Minimal data required to determine a convex polytope
Let $P\subset \Bbb R^d$ be a convex polytope.
Suppose that I know
its combinatorial type (aka. the face-lattice),
the length $\ell_i$ of each edge, and
the distance $r_i$ of each vertex from the origin.
Question: Does this already determine $P$ (up to orthogonal transformation)?
This is the case if all $\ell_i$ are the same, and all $r_i$ are the same (see this question). But what if they are not the same? What if I do not know the combinatorial type but only the edge-graph?
Update
I am not sure whether the formulation of my question was too vague, so below I added a second equivalent version of what I am asking:
Given two combinatorailly equivalent polytopes $P_1,P_2\subset\Bbb R^d$, and a corresponding face-lattice isomorphism $\phi:\mathcal F(P_1)\to\mathcal F(P_2)$.
Now suppose that each edge $e\in\mathcal F_1(P_2)$ has the same length as $\phi(e)\in\mathcal F_1(P_2)$, and that each vertex $v\in\mathcal F_0(P_1)$ has the same distance from the origin as $\phi(v)\in\mathcal F_0(P_2)$.
Is it then true that $P_1$ and $P_2$ are congruent (related by an orthogonal transformation)?
Do you assume the origin is inside of the polytope?
Why doesn't the same proof as in the question you mentioned apply? Base for $d\le 2$ is still true: in any dimension the $2$-face and the origin generate $3$-dimensional pyramid with fixed edge lengths. It should be rigid by Cauchy lemma about sign changes.
@SamHopkins Does this make a difference? But yes, we can assume that it is inside.
@JosephGordon I can imagine that the same technique applies if all the $r_i$ are the same, because then you know that the facets are inscribed and you know all edge lengths. But if the $r_i$ of $P$ are distinct, then "the $r_i$ of a facet" are not predetermined uniquely (up to some common factor or so), at least not obviously. So the shape of a facet might not be uniquely determined.
@JosephGordon What is Cauchy's lemma about sign changes? I know Cauchy's rigidity theorem which needs the shape of facets rather than 2-faces.
My mistake. I meant the lemma used in the proof of Cauchy's theorem, i.e. Lemma 26.4 here. But it can't be applied in this case.
A 2-dimensional counterexample
In the image below, the white dot represents the origin and is located outside the polygon. And it has to be: if the origin were inside, the shape would be unique, as shown here.
One can imagine to build higher dimensional counterexamples from this, e.g. prisms over these shapes.
Proof of a special case
Suppose that for each 2-face $\sigma\in\mathcal F(P)$ the (perpendicular) projection of the origin onto $\mathrm{aff}(\sigma)$ ends up in the relative interior of $\sigma$. Then the polytope is uniquely determined by its edge-lengths and vertex-origin-distances.
Proof.
Let $P$ be a $d$-polytope.
Each 2-face of $P$ and the origin form a (possibly degenerate) pyramid, in which all edge-length are known, and the apex projects to the interior of the base.
This case is discussed in this question where it is also proven that the base face of the pyramid is uniquely determined.
If $d=2$, we are done.
If $d\ge 3$, we can apply Cauchy's rigidity theorem in the 2-face version, that is, the last version mentioned over here, to obtain that $P$ is uniquely determined.
$\square$
In fact, using much stronger tool it is possible to show that it suffices that the origin lies strictly inside the polytope (or in a facet). Then the combinatorial type, edge lengths and vertex-origin distances determine the polytope up to isometry. A proof of this is given in
M. Winter, "Rigidity, Tensegrity, and Reconstruction of Polytopes Under Metric Constraints", International Mathematics Research Notices, https://doi.org/10.1093/imrn/rnad298.
Not an answer, just an additional illustration for the existing answer by @MattF. (which I find exhaustive).
|
2025-03-21T14:48:32.067908
| 2020-09-19T19:08:20 |
372103
|
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"Gerry Myerson",
"Random",
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"url": "https://mathoverflow.net/questions/372103"
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|
Stack Exchange
|
Recursive random number generator based on irrational numbers
Here $\{\cdot\}$ and $\lfloor \cdot\rfloor$ denote the fractional part and floor functions respectively. For a negative, non-integer number $x$, we use the following definition: $\{x\}=1-\{-x\}$. If $x$ is a negative integer, $\{x\} =0$. We are dealing with the following recurrence:
$$X_{k+2}=\{b_2 X_{k+1}+b_1 X_k\}$$
where $X_1$ is a uniform random variable on $[0,1]$ and $X_0\in [0,1]$ is a constant. Thus all the $X_k$'s are in $[0,1]$. Also, $b_1, b_2$ are integers, called bases; they represent bases in a numeration system.
The simple case: $b_1=0$
I extensively studied the case $b_1=0, b_2 > 1$ corresponding to a first-order recurrence, see here. The main results are:
The sequence $\lfloor b_2X_k \rfloor$ corresponds to the digits of $X_1$ in base $b_2$. These digits behave as independently and identically distributed discrete uniform variables on $\{0, 1,\cdots,b-1\}$.
The sequence $X_k$ behaves as identically distributed continuous uniform variables on $[0, 1]$. The correlation between $X_k$ and $X_{k+m}$ is equal to $b_2^{-m}$.
For a specific value of $X_1$, say $X_1=c$ with $c$ a normal number (say $c=\log 2$), the empirical process of observed $X_k$'s (corresponding to a specific realization of the theoretical stochastic process) satisfies the same properties for the empirical statistics: convergence of the empirical distribution to uniform on $[0, 1]$, convergence of the empirical auto-correlations to the theoretical values mentioned above, etc.
This happens because the sequence is ergodic. Note that almost all numbers are normal, though no one knows if any of $e,\pi,\sqrt{2},\log 2$ is normal. They are believed to be normal.
The general case, and my question
The general case is when both $b_1$ and $b_2$ are non zero. For simplicity, we can focus on the following specific case, which seems to behave very nicely: $X_0=\frac{1}{2}, b_1=-3, b_2 = -5$. More specifically, it now looks like the $X_k$'s are not only uniformly distributed on $[0, 1]$, but also asymptotically independently distributed. Thus we can use that sequence as a random number generator, with $X_1$ being the seed. This is a big contrast with the simple case discussed in the first section.
For instance (this is an illustration of what I mean by asymptotic independence), if $X_1=\frac{\sqrt{2}}{2}$, the empirical probabilities satisfy
$$\hat{P}\Big[\bigcap_{i=0}^m (X_{k+i}<\alpha_i)\Big]\rightarrow \prod_{i=0}^m \hat{P}\Big[X_{k+i}<\alpha_i\Big]\rightarrow\prod_{i=0}^m \alpha_i$$
regardless of $m$ and $0\leq \alpha_0,\cdots,\alpha_m\leq 1$, when more and more terms (that is more and more $k$'s) are used to estimate these probabilities. I thus assume (maybe erroneously) that it must also be true for the theoretical probabilities. This is illustrated further in the Appendix (last section).
My question is whether my conjecture (independence of the $X_k$'s) is true. It was verified empirically when $X_0=\frac{1}{2}, X_1=\frac{\sqrt{2}}{2}, b_1=-3, b_2=-5$, as well as for many other parameter sets. The generated deviates seem to approach randomness better than those generated using Excel, based on various statistical tests. Note that not any parameter set works; there are plenty of exceptions, and identifying these exceptions would be a bonus.
Computational considerations
No need to read this section, only if you are interested, but it is not directly related to my question.
When you compute the successive $X_k$'s, you lose one bit of precision at each iteration. This is not an issue thanks to ergodicity, it's like re-starting the sequence with new seeds every 45 or so iterations. It's only an issue if you look e.g. at long-range auto-correlations.
Also, it is possible to carry the computations very efficiently. You start with getting and storing several billions of binary digits of $X_1=\frac{\sqrt{2}}{2}$. See here how you can get these digits. Then you only need to perform simple additions and bit shifting with a big number library. For instance, $5x = 4x + x$, and computing $4x$ is just a bit shifting operation (no multiplication involved). Likewise with $3x=2x + x$. In my Perl code, if I use this little trick, it runs 10 times faster than doing an actual multiplication.
Appendix
I estimated the probability $P(X_k<\alpha_0, X_{k+1}<\alpha_1, X_{k+2}<\alpha_2)$ for a thousand randomly selected triplets $(\alpha_0,\alpha_1,\alpha_2)$ in $[0, 1]^3$ and 100,000 $(X_k,X_{k+1},X_{k+2})$'s. Assuming uniform distribution and independence between $X_k, X_{k+1}$ and $X_{k+2}$, the theoretical value is always $\alpha_0\cdot \alpha_1\cdot \alpha_2$. The data and source code is available in an Excel spreadsheet, here. It is very easy to replicate my results. The observed and theoretical values are extremely close, supporting the conjecture of stochastic independence and uniformity. Below is a scatter plot where each point corresponds to one of the $(\alpha_0, \alpha_1, \alpha_2)$'s, with the X-axis being the observed (estimated) probability, and the Y-axis being the theoretical probability (the product of $\alpha_0,\alpha_1,\alpha2$).
By contrast, here is an example where the independence assumption is violated, corresponding to $b_1=0, b_2=3$ with $\mbox{Correl}(X_k,X_{k+1}) = \frac{1}{3}$ and $\mbox{Correl}(X_k,X_{k+2}) = \frac{1}{9}$:
In short, in the example with independence (first chart) you seem to have, assuming $X_1$ is uniformly distributed on $[0,1]$:
$$P(X_{k+2}\in A_2 | X_{k+1}\in A_1, X_k\in A_0)=\mu(A_2)$$ where $A_0, A_1, A_2$ are Borel subsets of $[0, 1]$ and $\mu(\cdot)$ is the Lebesgue measure. Yet it is obviously true that $X_{k+2}-b_2X_{k+1}-b_1X_k$ can only take on $|b_1|+|b_2|$ distinct integer values. Note that if you consider the sequence $Y_k=X_{3k}$ instead, then the triplets $(Y_k,Y_{k+1},Y_{k+2})$ don't lie in such a small number of planes, unlike the $X_k$'s.
Rephrased differently, my question is whether or not the uniform distribution with independence is the main fixed point (also called equilibrium distribution or attractor) of the stochastic / chaotic system in question. Usually finding the attractor requires solving a stochastic integral equation, yet here if we suspect Uniform/Independent might be the solution, you just need to plug that presumed solution in the integral equation and see if it solves it.
Update on 9/25/2020
Based purely on pattern recognition techniques, I've found this:
$$X_k=\{A(k)X_1\}, \mbox{ with } A(k)=b_2A(k-1)+b_1A(k-2)$$
with $A(0)=\frac{X_0}{X_1}$ and $A(1)=1$. I don't have a proof, but this looks like something very easy to prove. In addition, it helps prove whether or not the stochastic uniform/independence solution is correct or not. More about this next week.
Version 14 of this question.
@ Gerry: I am only editing my answer, not sure if this counts in the 14 versions. Of course I want to minimize edits, but still have to make a lot of effort in that direction. For instance now I have found the exact number of planes for the sequence $Y_k$ in my answer, as a function of $b_1,b_2$..
No, it's 14 versions of the question. Each edit brings the question back to the front page, bumping off some other question. Each edit makes the question more of a moving target, more difficult to answer, more work for the reader to see what has really changed. Each edit may cause some readers to wonder why the author couldn't get it right the first time, or the second, or ... or the 13th.
Is there a way to make the question invisible / hidden for (say) 2 weeks so you can make many edits without bothering anyone until it's finalized? If I delete it, can I still edit it in private, and then restore it 2 weeks later? I suppose MO works differently, but on my blog you can save as draft (nobody but you see it) until you decide to publish. And you can un-publish / re-publish. Wondering if there is such a feature here.
I'm not sure whether you can delete a question such as this one that has an answer with many upvotes. There's probably something about that in an FAQ somewhere.
I will check the FAQ or ask my question about "private posting" in the appropriate forum.
Well, I found a couple of things at https://mathoverflow.net/help/deleted-questions It says, Users can delete their own questions if the question: has zero answers;
has only one answer, but that answer has no upvotes;
has no bounties that were awarded to any answer that isn't already deleted.
It also says, "Self-deleted questions can be viewed and undeleted by their original authors. If you delete your own question, you must undelete it before you are able to edit it." So, I think your suggestion won't work.
The workaround I have is to edit the question, save the content in Notepad on my desktop, and not publish the edit on MO. Do that 15 times, bring the notepad version back to MO for a "private edit" without saving it on MO, and instead save the edits in Notepad each time, until after a while it's good enough that I can actually publish the final version on MO. Not ideal, but that way nobody but me will see the intermediate edits.
Now version 16.
And the very last one, I promise.
I think there are still typos, but not in the formulas anymore. I won't fix them, they are of no concern. I now also have a closed form for $X_k$, but no more editing. Actually, so much more to add, but that will be in another question, or in an article I plan to write.
Of course the $X_k$ are not independent as random variables. So I assume you are referring to some notion of asymptotic independence, and it would help if you state your conjecture more precisely. One natural guess is equidistribution (see [1]) of r-tuples $(X_k,\ldots,X_{k+r-1})$. However, The triples $(X_k, X_{k+1},X_{k+2})$ will lie on a bounded number of planes in $[0,1]^3$; there will be at most 9 such planes in your example. This can be seen if one graphs these triples in 3D. A similar problem arose in the classical RANDU random number generator, see [3].
[1] Kuipers, L.; Niederreiter, H. (2006) [1974]. Uniform Distribution of Sequences. Dover Publications.
[2] https://en.wikipedia.org/wiki/Equidistributed_sequence#
[3] https://en.wikipedia.org/wiki/RANDU
Thank you, I clarified my conjecture according to your comment.
@ Yuval: I will look into that in more details. I noticed that for the simple case ($b_1=0$) you have the same problem in 2D, with couples $(X_k, X_{k+1})$ fitting on exactly $b_2$ parallel lines. If you consider the sequence $X_{2k+1}$ instead of $X_k$, the problem is gone. I would expect in the general case (3D issue) the number of planes should depend on $b_1, b_2$. The larger $b_1, b_2$, the more planes, and thus the better. But I haven't checked this yet, just a wild guess at this point.
@ Yuval: how did you come up with number of at must 8 planes? Is it by visual inspection or based on some logic? That number 8 may just be $|b_1|+|b_2|$. Maybe just a coincidence, maybe not. Maybe working with the sequence $X_{3k+1}$ would fix this at least in 3D.
Notice that $X_{k+1} - b_2 X_k - b_1 X_{k-1}$ is by definition an integer, and you can check that the expression can acheive at most $|b_1| + |b_2|$ different values.
Suppose $b_1$ and $b_2$ are positive. As Random wrote, by definition $X_{k+2}-b_2 X_{k+1}−b_1X_k$ is an integer between $0$ and $- b_1-b_2$. So that actually gives up to 9 planes in your example. No triple can lie outside these planes.
Here $b_1, b_2>0$ are integers. I investigated the sequence $Y_k=X_{3k}$, which has far more communal planes, and thus more useful to build a random generator. Of course, choosing large values for $b_1,b_2$ will further drastically improve the generator by adding a lot more planes. I suggest choosing values larger than (say) $2^{30}$ for $b_1,b_2$.
There are $M=b_2^3+3b_1b_2+b_1^3$ communal planes and they all have an equation of the form
$$b_1^3\cdot Y_k+b_2(b_2^2+3b_1) \cdot Y_{k+1}-Y_{k+2} =d.$$
The possible values for $d$ are $0,1,\cdots,M-1.$ Each plane (identified by $d$) contains a different proportion of triplets $(Y_k,Y_{k+1},Y_{k+2})$. The empirical distribution for these proportions is featured in the histogram below (corresponding to $b_1=5,b_2=3$), where the X-axis represents $d$, and the Y-axis the proportion of triplets lying in plan $d$.
Of course it is easy by looking at this chart to guess what the exact theoretical distribution is. To identify these planes, I used the program below and some experimental math.
# Compute equations of planes containing 3 random vectors
# P(k) = (x[k], x[k+1], x[k+2])
# P(l) = (x[l], x[l+1], x[l+2])
# P(m) = (x[m], x[m+1], x[m+2])
# (k, l, m) are randomly selected (M triplets)
#
# Equation of planes is x + s*y + t*z = intercept
# For each (k,l,m) output the coefficients s, t, intercept
#
# Goal: Find communal planes absorbing many (P(k), P(l), P(m))
# Once the planes are computed, sort them by s, t, intercept
$n=100000;
$b1=5;
$b2=3;
# xx[] is the original sequence
$xx[0]=0.5;
$xx[1]=sqrt(2)/2;
for ($k=2; $k<$n; $k++) {
$xx[$k]=$b2*$xx[$k-1]+$b1*$xx[$k-2]-int($b2*$xx[$k-1]+$b1*$xx[$k-2]);
if ($xx[$k]<0) { $xx[$k]=1+$xx[$k]; }
}
# we actually use 1 out of 3 consecutive terms from original sequence xx[]
# to see if it the new sequence x[] also has a small number of communal planes
for ($k=0; $k< $n/3; $k++) {
$x[$k]=$xx[3*$k];
}
$M=10000; # must be < n/3
open(OUT,">coplanes2.txt");
for ($iter=0; $iter<$M; $iter++) {
$k=int($M*rand());
$l=int($M*rand());
$m=int($M*rand());
# in case k=l or k=m or l=m, an ERROR message is reported
$a=$x[$k]; $b=$x[$k+1]; $c=$x[$k+2];
$d=$x[$l]; $e=$x[$l+1]; $f=$x[$l+2];
$p=$x[$m]; $q=$x[$m+1]; $r=$x[$m+2];
$u=($e-$b)*($r-$c)-($f-$c)*($q-$b);
$v=-($d-$a)*($r-$c)+($f-$c)*($p-$a);
$w=($d-$a)*($q-$b)-($e-$b)*($p-$a);
if ($u != 0) {
$s=$v/$u;
$t=$w/$u;
$intercept=($u*$a + $v*$b + $w*$c)/$u;
print OUT "$k\t$l\t$m\t";
print OUT "$s\t$t\t$intercept\n";
} else {
print OUT "$k\t$l\t$m\tERROR (u=0)\n";
}
}
close(OUT);
|
2025-03-21T14:48:32.069282
| 2020-09-19T19:53:07 |
372106
|
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"Tim Campion",
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"https://mathoverflow.net/users/2362",
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"https://mathoverflow.net/users/4639",
"https://mathoverflow.net/users/5792",
"jdc",
"overcaffeinated"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/372106"
}
|
Stack Exchange
|
Is Lie group cohomology determined by restriction to finite subgroups?
Consider the restriction of the group cohomology $H^*(BG,\mathbb{Z})$, where $G$ is a compact Lie group and $BG$ is its classifying space, to finite subgroups $F \le G$. If we consider the product of all such restrictions
$$H^*(BG,\mathbb{Z}) \to \prod_F H^*(BF,\mathbb{Z}),$$
is this map injective?
Note, according to McClure - Restriction maps in equivariant K-theory a similar result holds in equivariant K-theory. Perhaps there is a way to derive the above from McClure's theorem?
I asked this question on stackexchange and didn't manage to get a (complete) answer nor solve it myself. However, Qiaochu Yuan offered a proof for the non-torsion elements.
As Qiaochu suggests, $H^(BG)$ injects into $H^(NT)$ (normalizer of torus). A splitting is given by the Becker-Gottlieb transfer. The composite of a map and the transfer is the Euler characteristic of the fibers and $\chi(G/NT)=1$, so the transfer yields a splitting.
The Becker–Gottlieb observation can be found in the appendix to this paper—https://link.springer.com/chapter/10.1007/BFb0070640—which unfortunately points out on its second page, in Theorem 2, that the sequence $1 \to T \to N_G(T) \to W_G \to 1$ does not typically split. The last page suggests that it may still be an open question whether the Serre spectral sequence of $BT \to BN \to BW$ collapses.
There is something maddening about this, actually: there's a string of results from Borel (and Serre, in one paper) that for compact, connected Lie groups $G$, it is equivalent for $G$ to have $p$-torsion, for $BG$ to have $p$-torsion, for the order of $W$ to be divisible by $p$, and for $G$ to contain an elementary $p$-group (which can be taken of order $\leq p^3$ not contained in any maximal torus. But his proof, which he apologizes for, doesn't use the existence of some such subgroup $P$ to detect the $p$-torsion in $H^*BG \to H^*BP$. Rather, he reduces to the simply-connected case...
...through a sequence of lemmas and then verifies that the conditions simultaneously do hold or don't hold by proceeding $A$ to $G$ through the Killing–Cartan classification. His quote is wonderful:
"Les résultats de cet article mettent ainsi en évidence quelques liens entre la structure de groupe d'un groupe de Lie et sa topologie. Cependant, l'auteur regrette
vivement d'en avoir été réduit à établir plusieurs propositions d'énoncé général
à l'aide de vérifications utilisant la classification.
'You have a low shopkeeping mind. You think of things that would never come into a gentleman's head.'
'That's the Swiss national character, dear lady.'
(B. Shaw)."
I'm not sure if this is really helping, but there's another restriction. There's a paper of Feshbach showing that it is equivalent for $H^*BG$ to have a non-nilpotent $p$-torsion element $y$ and an elementary abelian $p$-group $P$ not contained in any maximal torus, and that in this case the image of $y$ in $H^*BP$ is nonzero. So it seems if the answer to your question were no, then there would be a nilpotent $p$-torsion element whose image vanishes under restrictions to all $H^*BP$.
@jdc's reference: Curtis, Wiederhold, and Williams - Normalizers of maximal tori.
After the heavy lifting done by people on MSE and in the comments, I think it's not too bad to finish off the proof that the answer is yes.
As argued by Ben Wieland in the comments, we reduce to showing that for any short exact sequence of topological groups
$$U(1)^n \to G \to W $$
where $W$ is finite, we have that $H^\ast(BG;\mathbb Z)$ injects into the product of $H^\ast(BF;\mathbb Z)$ over all finite subgroups $F \subseteq G$. The jist of the argument is going to be replacing $U(1)^n$ with $(Q/\mathbb Z)^n$, and then arguing that every finitely-generated subgroup of the resulting extension $G_\ast$ is finite.
The first thing to note is that for some $m \in \mathbb Z$, there exist maps of exact sequences
$$\require{AMScd} \begin{CD} (\mathbb Z/m)^n @>>> G_m @>>> W \\ @VVV @VVV @VV{=}V\\(\mathbb Q /\mathbb Z)^n @>>> G_\ast @>>> W \\ @VVV @VVV @VV{=}V\\ U(1)^n @>>> G @>>> W \end{CD}$$
The middle line exists and maps to the bottom line because (1) $\mathbb Q/\mathbb Z$ is the torsion subgroup of $U(1)$, and so must be preserved by the action of $W$, and (2) the discrete group quotient $U(1)^\delta / (\mathbb Q/\mathbb Z)$ is a rational vector space, so no matter the action of $W$, the cohomology of $W$ with values in this quotient vanishes by Maschke's theorem. Thus $H^\ast(BW; \underline{(\mathbb Q/\mathbb Z)^n}) \to H^\ast(BW; \underline{U(1)^{\delta,n}})$ is an isomorphism and in particular the class classifying the extension is hit.
The top line exists and maps to the middle line because when we choose a cocycle $W \times W \to (\mathbb Q/\mathbb Z)^n$, we see that because $W$ is finite, the cocyle has finite image, and every finitely-generated subgroup of $(\mathbb Q/\mathbb Z)^n$ is finite — thus the cocycle lives in some finite, $W$-invariant subgroup $(\mathbb Z/m)^n \subseteq (\mathbb Q / \mathbb Z)^n$ (using that $(\mathbb Z/m)^n \subseteq (\mathbb Q/\mathbb Z)^n$ is the $m$-torsion subgroup and so must be $W$-invariant).
By similar reasoning, we see that every finitely-generated subgroup of $G_\ast$ is finite. Therefore, because homology commutes with filtered colimits, we have $H_\ast(BG_\ast) = \varinjlim_{G' \subseteq G_\ast} H_\ast(BG')$, where the colimit is over finite subgroups (or even just those finite subgroups $G' = G_m$ of the form given above), and we use any constant coefficients.
Now, because $B(\mathbb Q/\mathbb Z)^n \to BU(1)^n$ is a homology isomorphism with any finite coefficients (this can be checked in several ways), we see that $BG_\ast \to BG$ is likewise a homology isomorphism with finite coefficients by the Serre spectral sequence. So the composite map $\varinjlim H_\ast(BG_m) \to H_\ast(BG_\ast) \to H_\ast(BG)$ is an isomorphism with finite coefficients. By the universal coefficient theorem, the map $H^\ast(BG) \to \prod H^\ast(BG_m)$ is injective with finite coefficients. Since Qiaochu has already shown that it is an injection on nontorsion elements, it follows that this map is an injection on integral cohomology (noting that these things are sufficiently finite to be safe).
Note that most of the above boiled down to facts about the extension $U(1)^n \to G \to W$, not really facts about (co)homology — the only thing we really needed was the isomorphism $H_\ast(B(\mathbb Q/\mathbb Z)) \cong H_\ast(BU(1))$ with finite coefficients.
In fact, Theorem 5.7 of Becker and Gottlieb's original paper The transfer map and fiber bundles (DOI) is actually stated for general cohomology theory, and implies by Ben Wieland's argument that $\Sigma^\infty BG$ splits off of $\Sigma^\infty BN(T)$, so the reduction to extensions $T \to N \to W$ with $T$ a torus and $W$ finite holds for an arbitrary homology or cohomology theory $E$.
It's not hard to show that if $E$ is a spectrum with trivial rationalization, then $B(\mathbb Q/\mathbb Z)^n \to BU(1)^n$ is and $E$-homology or cohomology equivalence. So the above argument shows that in this case, we have that $N_\ast \to N$ is an $E$-homology and $E$-cohomology equivalence, where $N_\ast$ fits in the extension $(\mathbb Q/\mathbb Z)^n \to N_\ast \to W$ as above. Moreover, since $BN_\ast = \varinjlim BN'$ where the colimit is over finite subgroups, and since this is a homotopy colimit, we have $E_\ast(BN) = \varinjlim E_\ast(BN')$, so that $\bigoplus E_\ast(BN') \to E_\ast(BN)$ is surjective. For cohomology, there are potential $\varprojlim^1$ issues.
Thus the statement we get is:
Theorem: Let $E$ be a spectrum with trivial rationalization, and let $G$ be a compact Lie group. Then $\bigoplus_{F \subseteq G} E_\ast(BF) \to E_\ast(BG)$ is surjective, where the sum is over finite subgroups $F \subseteq G$.
It would be nice if this could be upgraded to a statement about all spectra by considering also the rationalization, but that seems unpromising because of Maschke's theorem — Qiaochu's argument for nontorsion classes is more subtle, it seems.
Probably also some statement about cohomology is possible ….
In fact, it's not hard to extend the statement to arbitrary $G$-spaces. That is:
Theorem: Let $E$ be a spectrum with trivial rationalization, let $G$ be a compact Lie group, and let $X$ be a $G$-space. Then $\bigoplus_{F \subseteq G} E_\ast(X_{hF}) \to E_\ast(X_{hG})$ is surjective, where the sum is over finite subgroups $F \subseteq G$.
Thanks Tim! So I'm curious, can you derive McClure's theorem from this argument?
@overcaffeinated That's an interesting question. Certainly some modification would be needed -- I think that if you could show $(K/m)^\ast(BG) \to \prod_F (K/m)^\ast(BF)$ to be injective for each $0 \neq m \in \mathbb Z$, this would imply that $K^\ast(BG) \to \prod_F K^\ast(BF)$ is injective too. This reduces to the case where $E$ has trivial rationalization as above. But I'm not sure how to translate from the filtered-colimit statement to the cohomology-injectivity statement because I don't know how to control the $\varprojlim^1$ term.
I tried to add links to the comments, but couldn't find Qiaochu's comment (which @BenWieland also references).
Thanks! I added a link to Qiaochu's MSE answer.
Just a note that partly in response to overcaffeinated's comment here, I worked out a second answer which, as noted in a comment on that answer, suffices to recover McClure's result, as well as answering the question independently, without using Qiaochu's MSE answer.
$\DeclareMathOperator\Image{Image}$Here's the most general result I think I can muster. I've split it out into a second answer in order to keep the answer to the original question more self-contained.
Theorem 1: Let $G$ be a compact Lie group, let $X$ be a $G$-space, and let $E$ be a spectrum. Then the following hold, where $F$ ranges over finite subgroups of $G$:
The image of $\bigoplus_F E_\ast(X_{hF}) \to E_\ast(X_{hG})$ contains all of the torsion;
The kernel of $E^\ast(X_{hG}) \to \prod_F E^\ast(X_{hF})$ is contained in the subgroup of divisible elements.
This follows from the following two more precise theorems:
Theorem 2: Let $G$ be a compact Lie group, and let $X$ be a $G$-space. Let $N \subseteq G$ be the normalizer of a maximal torus $T \subseteq G$, and let $W = N / T$ be the Weyl group. Then $\Sigma^\infty_+ X_{hG}$ splits off of $\Sigma^\infty_+ X_{hN}$.
Proof: The splitting is given by the Becker-Gottlieb transfer: the fiber of $X_{hN} \to X_{hG}$ is $G/N$, the same as the fiber of $BN \to BG$, which has Euler characteristic 1.
Theorem 3:
Let $N$ be an extension of a finite group $W$ by a torus $T$, and let $E$ be a spectrum and $m \in \mathbb N_{\geq 2}$. Then the following hold, where $F$ ranges over finite subgroups of $N$:
$\varinjlim_F (E/m)_\ast(X_{hF}) \to (E/m)_\ast(X_{hN})$ is an isomorphism;
$(E/m)^\ast(X_{hN}) \to \varprojlim_F (E/m)^\ast(X_{hF})$ is an isomorphism.
Proof of Theorem 1 from Theorems 2 and 3: By Theorem 2, it will suffice to consider the case where $G = N$ is an extension of a finite group by a torus. Theorem 3 establishes Theorem 1 for $E/m$. Then (1) follows by considering the natural short exact sequence $0 \to E_{\ast}(-)/m \to (E/m)_\ast(-) \to E_{\ast-1}(-)^{\text{$m$-tor}} \to 0$, and the argument for (2) uses a similar exact sequence.
The proof of Theorem 3 will follow from a series of lemmas. For the remainder, we let $U(1)^n \to N \to W$ be an extension of a finite group by a torus, and we let $(\mathbb Q/\mathbb Z)^n \to N_\ast \to W$ and $(C_q)^n \to N_q \to W$ be the subextensions which exist by the analysis in my other answer. We fix a spectrum $E$, $m \in \mathbb N_{\geq 2}$, and an $N$-space $X$.
Lemma 4: The fiber of $X_{hN_\ast} \to X_{hN}$ is $B\mathbb Q^n$, and in particular this map is an $(E/m)_\ast$ and $(E/m)^\ast$ equivalence.
Proof: This comes via a diagram chase from the fiber sequence $B\mathbb Q^n \to B(\mathbb Q/\mathbb Z)^n \to BU(1)^n$.
Lemma 5: We have $(E/m)_\ast(X_{hN_\ast}) \cong \varinjlim_q (E/m)_\ast(X_{hN_q})$ canonically, and a canonical short exact sequence $0 \to \varprojlim^1 (E/m)^{\ast+1}(X_{hN_q}) \to (E/m)^\ast(X_{hN_\ast}) \to \varprojlim (E/m)^\ast(X_{hN_q}) \to 0$.
Proof: By the analysis in my other answer, we have $N_\ast = \varinjlim N_q$. Therefore $BN_\ast = \varinjlim BN_q$, and it follows that $X_{hN_\ast} = \varinjlim X_{hN_q}$. The lemma follows by the usual formulas for homology and cohomology of a filtered colimit.
Lemma 6: If $E$ is $m$-torsion, then $\varprojlim^1 E^\ast(X_{hN_q}) = 0$.
Proof of Theorem 3: This follows from Lemmas 4, 5, and 6, once we note that $E/m$ is $m^2$-torsion.
It remains to prove Lemma 6.
Lemma 7: Let $q,r \in \mathbb Z$, and consider the inclusion $C_{qr}^n \xrightarrow i (\mathbb Q/\mathbb Z)^n$. Consider also the inclusion $C_q^n \xrightarrow j C_{qr}^n$ with quotient $C_r^n$. Let $A$ be an $r$-torsion and $q$-torsion abelian group. Then $H^\ast(ij;A)$ is injective and $\Image(H^\ast(ij;A)) = \Image(H^\ast(j;A))$.
Proof: Direct calculation. More precisely, $H^\ast(BC_q;A)$ and $H^\ast(BC_{qr};A)$ both have $A$ in all degrees; the inclusion of $H^\ast(B(\mathbb Q/Z);A)$ is an isomorphism onto the even degrees, and $H^\ast(j;A)$ kills the odd classes while being an isomorphism on even classes. Then one extends this analysis to $n > 1$.
Lemma 8: Let $A \xrightarrow i B \xrightarrow j C$ be maps of chain complexes. Suppose that $ji$ is injective and $\Image(ji) = \Image(i)$. Then the sequence of homologies $H_\ast(A) \xrightarrow{i_\ast} H_\ast(B) \xrightarrow{j_\ast} H_\ast(C)$ has $i_\ast$ injective and $Image(j_\ast i_\ast) = Image(i_\ast)$.
Proof: Diagram chase.
Corollary 9: Fix $s \in \mathbb Z$, and consider the maps $H^s(BN_\ast;A) \xrightarrow i H^s(BN_{qr};A) \xrightarrow j H^s(BN_q;A)$. For $q$, $r$ sufficiently divisible by $m$ and $A$ $m$-torsion, we have that $ji$ is injective and $\Image(ji) = \Image(j)$.
Proof: Using Lemma 7 as a base case, use Lemma 8 to induct through the pages of the Serre spectral sequences for the fibrations over $BW$. This is a first-quadrant spectral sequence, so for fixed $s$ it stabilizes at a finite page. The statement can be tested on associated gradeds, so there are no extension problems.
Corollary 10: Assume that $E$ is bounded below and $m$-torsion, and fix $s \in \mathbb Z$. Consider the maps $E^s(X_{hN_\ast}) \xrightarrow i E^s(X_{hN_{qr}}) \xrightarrow j E^s(X_{hN_q})$. For $q,r$ sufficiently divisible by $m$, we have that $ji$ is injective and $\Image(ji) = \Image(j)$.
Proof: Using Corollary 9 as a base case, use Lemma 8 inductively to walk through the Atiyah–Hirzebruch spectral sequences for the fibrations over $BN_\ast$, $BN_{qr}$, and $BN_q$ respectively (which all have fiber $X$). Since we are assuming that $E$ is bounded below, this is essentially a first-quadrant spectral sequence so the argument goes in the same way as before.
Proof of Lemma 6: That this follows from Corollary 10 in the case where $E$ is bounded below can be seen in two ways — either from the eventual injectivity of $E^\ast(X_{hN_\ast}) \to E^\ast(X_{hN_q})$, or from the fact that sequence is Mittag–Leffler. When $E$ is not bounded below, we simply pass to a suitable connective cover of $E$, since we are always taking the cohomology of a suspension spectrum, which is bounded below.
$H^\ast(BG;\mathbb Z)$ is finitely-generated in each degree. In particular, it contains no nonzero divisible elements. So we recover the result of the my first answer, without relying on Qiaochu's answer. Similarly, $ku^\ast(BG)$ is finitely-generated in each degree by the Atiyah-Hirzebruch spectral sequence. For a space $X$, $ku^\ast(X) = KU^\ast(X)$ for $\ast = 0,1$, so $KU^\ast(BG)$ is finitely-generated in each degree. In particular, it contains no nonzero divisible elements, and we recover McClure's result.
Here's a criterion which covers most of the cases I'd care to think about: Suppose that $E$-cohomology refines to a graded $R$-module-valued functor for some PID $R$ such that $E_\ast$ is finitely-generated over $R$ in each degree. Then $E^\ast(X)$ is finitely-generated over $R$ in each degree if $X$ has finitely many cells in each dimension. If each residue field of $R$ has finite characteristic, then any finitely-generated $R$-module is free of nonzero divisible elements. So in this case, $E^\ast(X_{hG}) \to \prod_F E^\ast(X_{hF})$ is an injection.
I might be missing something obvious, but can you expand on the second part of lemma 4 ? The identification of the fiber is clear, and the fact that it has trivial $E/m$-co/homology is too; but the point about $E/m$-co/homology equivalence is slightly less so (to me). In case $E/m$ is bounded below I think I can make it work with some Serre-Atiyah-Hirzebruch spectral sequence game, but without that assumption I (think I) lose my nice convergence properties so I'm not sure how that works
I guess for cohomology you can actually get away with it by the same trick as in your last paragraph, but it seems trickier for homology
@MaximeRamzi I don't seem to recall thinking carefully about convergence of any spectral sequences here beyond "first-quadrant" arguments, so I probably overlooked this. As you say, for cohomology it's ultimately safe for us to assume that $E$ is bounded below, so it ends up not being an issue. It's nice to know somebody's reading at least parts of this argument carefully! I fear it gets more and more telegraphic. I should probably do a proper literature search to see if this is worth publishing / figure out some collaboration to do so...
I'd be curious how much you have already / will end up reading in detail. Let me know if there are other questions / difficulties. I didn't get a notification btw, just happened by bc of that other recent question, so ping me next time.
I'll have to think about the homology case... The weird thing is that here you don't have to worry about $\varprojlim^1$ at all so the argument is much shorter. This ends up being basically the last step. I wonder if some subtle weak convergence criterion from Boardman will suffice.
@MaximeRamzi Better argument: If $F \to X \to B$ is a fiber sequence with $B$ connected say, then it exhibits a colimit decomposition $X = \varinjlim_{b \in B} F$. This colimit is preserved by $E \wedge \Sigma^\infty_+(-)$, so $E \wedge \Sigma^\infty_+ X = \varinjlim_{b \in B} E \wedge \Sigma^\infty_+ F$. There is a comparison map to the fiber sequence $\ast \to B = B$ which gives a comparison map of colimit diagrams. If $E \wedge \Sigma^\infty F= 0$, the comparison is a levelwise equivalence, so descends to an equivalence on the colimit. The obvious variation also works for cohomology.
that's a very nice argument, thanks ! One thing further I'd like to know where there doesn't seem to be an immediate fix is how much of this works in a genuine setting. Indeed $E_(X_{hG})$ is begging to be $E^G_(X)$ for the Borel equivariant $E$, but what about more general $G$-spectra ? This theorem is very much a descent statement that looks slightly like some things I've seen elsewhere (e.g. in Mathew-Naumann-Noel "Derived restriction and induction theory")
(apparently MO doesn't let me tag you in answers to your own post...)
@MaximeRamzi I agree, it begs to be genuine-ified. The first question is what the statement should be in the genuine case. Of course, in equivariant homotopy theory, the "degree of genuineness" is parameterized by the choice of a suitable family of subgroups. The extreme generalization of this theorem might say something like "if $G$ is compact Lie, then all non-divisible cohomology of $G$-spaces in the category given by a family containing the finite subgroups injects into the corresponding cohomology in the category given by the family of finite subgroups." Maybe that sounds a bit strong?
OTOH, the most important genuine equivariant category over an infinite Lie group is cyclotomic spectra, which is defined basically with respect to the family of finite subgroups of $S^1$. I wonder if there might be something to say about "why the Nikolaus-Scholze approach to cyclotomic spectra suffices in the bounded-below case"... (I must have just missed the first ping)
|
2025-03-21T14:48:32.070788
| 2020-09-19T20:17:11 |
372108
|
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|
Stack Exchange
|
Truncation of infinity-categories
If we have a category $\mathcal{C}$, then we can see it as an $\infty$-category. Furthermore, we can truncate and $\infty$-category $\mathcal{X}$ to get a category $\mathcal{X}_{\leq 1}$. My question is if these functors are adjoint, i.e. if we have
$$\text{Hom}_{\mathfrak{Cat}}(\mathcal{X}_{\leq 1}, \mathcal{Y})\cong \text{Hom}_{\infty\text{-}\mathfrak{Cat}}(\mathcal{X},\mathcal{Y})$$
where $\mathcal{Y}$ is a category (which on the right is seen as an $\infty$-category).
I think you need to edit a bit your question. What is Y?
There is a bit of notation to be careful about here:
$\mathcal{X}_{\leqslant 1}$ is often used to denote the full subcategory of $\mathcal{X}$ of set-truncated object. For example if $\mathcal{X}$ is an $\infty$-topos, then $\mathcal{X}_{\leqslant 1}$ is its $1$-topos reflection.
with this definition, $\mathcal{X}_{\leqslant 1}$ is a $1$-category, but it is not the one that will have the property you want (it will be a right adjoint instead of a left adjoint, and only when restricted to finite limit preserving functor).
The $1$-category you want to consider is the homotopy category $h \mathcal{X}$ of $\mathcal{X}$, sometimes also denoted $\tau \mathcal{X}$, which is the category with the same objects as $\mathcal{X}$, and with the morphism sets
$$ h\mathcal{X}(a,b) \simeq \pi_0 ( \mathcal{X}(a,b) ) $$
Which does satisfies the property you ask.
A rigorous proof of this of course depends on what model of $\infty$-category you use, but if you use quasi-categories, this follows from points 1.2 and 1.8 in Joyal notes on quasi-categories.
Just to clarify: the functor $(-){\leq 1}$ which restricts the fundamental $\infty$-groupoid to the fundamental groupoid of a space has no right adjoint? I'm asking because I would like to use this to give a simple proof of Seifert-van Kampen: For an open cover $U,V$ of $X$, we have a pushout $U\cap V\rightarrow U\times V\rightarrow X$. If we use the homotopy hypothesis, we get a pushout $\Pi(U\cap V)\rightarrow \Pi(U)\times \Pi(V)\rightarrow \Pi(X)$. If the truncation functor $\tau{\leq 1}(-)$ were adjoint, then the pushout would be preserved, yielding a nice proof of S-vK.
The $\pi_1$ functor taking an $\infty$-groupoid to its fundamental groupoid is a left adjoint functor and this makes the groupoid version of S-vK. somehow trivial. (or at least it hides all the difficulty under the rug)
@curious: you do not get Seifert-van Kampen this way. What you get is that the fundamental groupoid functor preserves homotopy pushouts, but Seifert-van Kampen is a genuinely point-set result telling you when a point-set pushout is "close enough" to a homotopy pushout that the fundamental groupoid is what it should be.
@QiaochuYuan I'm sorry, what is wrong with my argument? Or rather why does it prove a property of homotopy pushouts instead of pushouts?
@curious: you're working with $\infty$-categories and adjunctions between them and in $\infty$-category land you only have $\infty$-limits and colimits and those are what right adjoints resp. left adjoints preserve.
@QiaochuYuan Ah yes, thank you!
|
2025-03-21T14:48:32.071034
| 2020-09-19T21:25:57 |
372112
|
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"Christian Remling",
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|
Stack Exchange
|
Determining what happens to the spectrum of Schrödinger operator as boundary condition changes
I recently came across a problem in research, and I'm asking about it here after trying in math stack exchange with no luck.
Suppose I have a metric graph $G$ (or even a closed interval, to make things simpler if needed) and a Schrödinger operator $L=-\frac{\partial ^2}{\partial x^2}+V$ (you can even ignore $V$ if that helps). Suppose also that I have a finite set of points $N\subset G$ (you can even think of $N$ as a single point if that helps). On each $x_0\in N$, I impose the following boundary condition -
$$f'(x_0^+)=f'(x_0^-)=:f'(x_0)$$
$$f(x_0^+)-f(x_0^-)=\alpha f'(x_0)$$
This means that $f'$ is 'continuous' at $x_0$ (not really continuous in the usual sense, just that the two sided limits exist and are equal, the function itself doesn't have to be differentiable), and the jump of $f$ at $x_0$ is proportional by a factor $\alpha$ to the value of the one sided limits of $f'(x_0)$ (although $f'$ need not exist).
The eigenvalue problem $Lf=\lambda f$ with the imposed boundary condition has a unique spectrum - $\{\lambda_n\}_{n=1}^{\infty}$.
Now, I choose to change the boundary conditions to something similar, but different:
$$f'(x_0^+)=f'(x_0^-)=:f'(x_0)$$
$$f(x_0^+)+f(x_0^-)=\alpha f'(x_0)$$
The change is in the minus sign in the second formula.
My question is – can I say something meaningful about how the spectrum of the problem changes due to the change in boundary conditions? The boundary conditions are (at least intuitively) very similar, so can I maybe somehow find a relation between the spectra of the two problems?
You can try to suggest ideas for the more general case of $G,L,N$, or even just think about some of the simplified cases I suggested. Ideally, I'd like an answer in the style of 'the change in boundary condition translates the spectrum by a fixed number' or something of this type - something which gives me a concrete relation between the two problems. But I'm aware that the answer might be much more complicated (if there even is an answer), so feel free to share any idea.
One thing I thought of is – given an eigenfunction $f$ of the first problem, maybe I can do some 'procedure' to it in order to turn it into an eigenfunction of the second problem (I thought about adding a constant to it, to match the new boundary condition, but then it won't necessarily be an eigenfunction of $L$, but maybe something else will work). But I don't know if this is a good way to go.
Thanks in advance!
If you change $-$ with $+$ the problem is no longer self-adjoint and the spectrum can be the whole complex plane. For example take $+$ and $\alpha=0$ in $I=[0,1]$ and $c=1/2$ and consider also Dirichlet bc at $0,1$. For a given $\lambda$, let $u$ solve $\lambda u-u''=0$ in $I$ and $u(0)=0$. Then $v(x)=-u(1-x)$ solves the same equation with $v(1)=0$. Now take $w=u$ in $[0,1/2]$ and $w=v$ in $[1/2,1]$. Left and right derivatives are equal at $1/2$ but $w$ has opposite values from both sides of $1/2$.
@GiorgioMetafune Why does $v$ solve the eigenvalue equation $v''=-\lambda u$?
$$v''=(-u+ux)''=(-u'+u'x+u)'=u''(-1+x)+2u'$$
But:
$$-\lambda v = \lambda u(1-x) = -u''(1-x)$$
And so if I'm not mistaken, the two expressions differ by $2u'$ which is generally non-zero.
I think that the $+$ problem is indeed self adjoint since it can be represented by the symmetric bilinear form:
$B(u,v)=\int (u'v'+Vuv)+\frac{1}{\alpha}\sum_{x\in N}(u(x_+) + u(x_-))(v(x_+)+v(x_-))$
$v$ Is $u$ evaluated at the point $1-x$ and I mean that $v"=\lambda v$.
@GiorgioMetafune Oh, that makes more sense, thanks. But why do you conclude that the problem is not self adjoint? I don't see why it says anything about the spectrum.
Doesn't the representation of the boundary condition via the symmetric bilinear form automatically gives that the problem is self adjoint? From what I know, this boundary condition is related to what's known in the literature as $\delta '$ boundary conditions, which are indeed self adjoint.
Actually I took only $\alpha=0$ which Is excluded in your form. I only used that the spectrum of a selfadjoint operatori Is real.
@GiorgioMetafune It's actually defined for $\alpha=0$ as well when you take the limit. Still, I'm not sure why your argument gives that the spectrum is not real, could you explain further?
Let us continue this discussion in chat.
If I may chime in on this, the problem is self-adjoint with any such interior condition. This follows, for example, by thinking of this condition as a point transfer matrix for the solution vector $(f,f')$, and any $SL(2,\mathbb R)$ matrix is fine here.
As for the question itself, you are dealing with a rank one perturbation (of the resolvent). There is some machinery available for this, but the statements are probably not nearly as specific as you were hoping for.
|
2025-03-21T14:48:32.071393
| 2020-09-19T22:27:27 |
372115
|
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|
Stack Exchange
|
Random matrices having all real eigenvalues: uniform vs gaussian distributions
Let $P_n$ be the probability that a $n \times n$ real random matrix with independent entries and uniformly distributed on $[0,1]$ has all real eigenvalues.
Let $Q_n$ be the same probability, for a standard normal distribution.
I've found, empirically (comments in this unanswered MSE question), that $P_n$ behaves quite similarly to $Q_{n-1}$ (at least for the small values of $n$ I tried).
$$\begin{array}{c}
n & P_n & Q_{n}& \\
1 & 1 & 1\\
2 & 1 & 0.7071\\
3 & 0.708 & 0.3536\\
4 & 0.346 & 0.1250\\
5 & 0.117 & 0.0313\\
6 & 0.028 & 0.0007 \end{array}$$
Values of $P_n$ are approximate, empirical, from my simulations. Values of $Q_n=2^{-n(n-1)/4}$, from
"The Probability that a Random Real Gaussian Matrix has
k Real Eigenvalues, Related Distributions,
and the Circular Law", A. Edelman, Journal of Multivariate Analysis, 60, 203-232 (1997)
I'd like to find out an expression for $P_n$, and/or some argument that helps to explain the approximation $P_n \approx Q_{n-1}$
Real eigenvalues of non-Gaussian random matrices and their products finds that the probability that all eigenvalues are real is larger for distributions with large weight
at the origin and that decay slowly away from the origin. The dependence on the distribution vanishes quickly for larger $n$, see section 8 of How Many Eigenvalues of a Random Matrix are Real?
|
2025-03-21T14:48:32.071524
| 2020-09-19T22:58:01 |
372117
|
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"Dmitri Panov",
"Sascha",
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|
Stack Exchange
|
Packing in uniform domains
Given $N$ points $X:=(x_i)_{i \in \{1,..,N\}}$, we now define a score function $S:X \rightarrow \mathbb{N}$ that is $S(X)= \sum_{i=1}^N S(x_i)$ where the score of $S(x_i)$ is
$$S(x_i) = 2* \vert \{x_j; \vert x_i-x_j \vert \in [1,2]\} \vert+ \vert \{x_j; \vert x_i-x_j \vert \in [2,3]\} \vert$$
where $\vert \bullet \vert$ denotes the cardinality of the set. Moreover, we require that for all $i\neq j$ we have $\vert x_i-x_j \vert \ge \frac{1}{2}.$
Question: Is it true that any configuration of $N$ points with maximal possible score is in a domain of diameter $c\sqrt{N}$ for some fixed c?
Do you assume that your configuration of $N$ points belongs to $\mathbb Z^2$, or you consider all possible configurations of points in $\mathbb R^2$?
@DmitriPanov I allow for any configurations of points.
My comment about the integer lattice was only intended to suggest that spacing of order 1 would in general lead to a configuration of size $\sqrt{N}.$
Thanks! And just to double check, $[1,2]$ means the interval $[1,2]$ (not the union of $1$ and $2$)
@DmitriPanov that's right.
I can write an answer if you replace $\frac{1}{2}$ by $1$ or something like $1-\varepsilon$, i.e., $|x_i-x_j|\ge 1-\varepsilon$. Let me know if you are interested. I am not sure the solution will work for $\frac{1}{2}$.
@DmitriPanov sure, I would be very interested in seeing an answer to that case :)
What follows below is the new answer to the modified question, where we assume in addition $|x_i-x_j|\ge 1$ (probably one can ask $|x_i-x_j|\ge 1-\varepsilon$ for sufficiently small $\varepsilon$). I want to propose a positive solution of this problem modulo the following guess, which, I hope, is correct.
Guess. Consider the equilateral triangular lattice $E$ with distance $1$ between neighbouring points. Then there are exactly $18$ points on distance at most $2$ from a given one, and $36$ on distance at most $3$. So the score of each point is $54=2*18+(36-18)$. I guess, that for any set $X$ such that any two points are on distance at least $1$, in the punctured $2$-neighbourhood of any point $x\in X$ there are at most $18$ points of $X$. I guess that the same holds for points on distance at most $3$. If this is true then we have the following corollary: for any set $X$ satisfying $|x_i-x_j|\ge 1$ we have $S(x_i)\le 54 $.
So from now on we assume that either the guess is correct, or we are working with a set $X$ such that each point of this set has score at most $54$.
I'll prove that under such condition the constant $c$ exists.
Proof. Note first of all that we can always construct a set $X$ with $N$ points, such that the score of $X$ is $54N-10^{10}\sqrt{N}$. Such a set can be given by intersecting $E$ with a disk of appropriate radius. (one can take a smaller constant than $10^{10}$, but it doesn't matter).
Assume by contradiction, that we constructed a set $X$ maximising the score and such that its diameter is more than $10^{10^{10}}\sqrt{N}$. Take the union of disks of radius $3$ around all points of $X$, and denote this set by $U_3$. It is easy to see that $U_3$ is connected. Indeed, if it is not, we can parallel translate its connected component by pushing one to another and increase this way the score of $X$. So, since the diameter of $X$ is at least $10^{10^{10}}\sqrt{N}$, the perimeter of the exterior boundary of $U_3$ is at least $10^{10^{10}}\sqrt{N}$. We will say that a point of $X$ contributes to the exterior boundary of $U_3$ if it is on distance $3$ from it. It is easy to see that the number of points of $X$ contributing to the exterior boundary is at least $10^{(10^{10}-2)}\sqrt{N}$ (because the length of a radius $3$ circle is $<100$). The final observation is that any point $x$ of $X$ that contributes to the boundary has score less than $54$. This is because the disk of radius $3$ around $x$ has a large sub-region, where points of $X$ can not lie (indeed, take a point $y\in \partial U_3$ on distance $3$ from $x$, then no point on distance less than $3$ from $y$ lies in $X$). Finally, taking into account the guess and the fact that the score of $X$ has to be at least $54N-10^{10}\sqrt{N}$, we get a contradiction.
Old answer
Let's consider two variations of this question. In both cases the answer is yes. In the first case $X$ is any subset of $\mathbb R^2$ in the second it is a subset of $\mathbb Z^2$.
1 We assume first that $X$ is any subset of $\mathbb R^2$. In such case the set with maximal possible score has diameter at most $6$. Let me prove this. Let's first construct a set with score approximatively $\frac{5}{3}N^2$. To do this we put $N/6$ points in each vertice of the regular hexagon with side of length $1$.
Now, suppose we have a set with maximal score and suppose its diameter is more than $6$. We will construct a set with larger score which will give us a contradiction.
So, suppose $X$ has two points $x_i, x_j$ such that $|x_i-x_j|>6$. Let's take two disks of radius $3$ around both points. One of them contains at most $N/2$ points, which means $S(x_i)$ or $S(x_j)$ is at most $N$. Without loss of generality assume $S(x_i)\le N$. On the other hand, we know that $S(X)\ge \frac{5 N^2}{3}$. So, there is a point $x_k$ such that $S(x_k)\ge\frac{5}{3}N$. Move $x_i$ at the place of $x_k$, this will increase the score $S(X)$. Contradiction.
2 What follows is just a sketch of proof. We assume $X\subset \mathbb Z^2$. In such case each point $x_i$ contributes at most $2*8+20=36$ to the sum $S(X)$. Indeed, there are $8$ integer points on distance at most $2$ from a given one, and $20$ on distance in $[2,3]$. From this one can deduce the answer applying the isomperietric inequality to the set that is the union of $2\times 2$ squares with centres at points of $X$. I can give more details, if you want.
interesting, thank you, indeed I was interested mostly in Case 1. Do you think one can include a penalization saying: All points, are at least a distance $1/2$ away from one another and then still get the $\sqrt{N}$ scaling? Cause now it seems there is nothing in my question that would make the volume grow...
I'll need to think about this. This might be harder. I think this depends on whether positioning points in a lattice (for, example Eisensteins one) gives you the best possible score for one vertex. So some variant of the problem (if you fiddle with the intervals [1, 2] and [2, 3] might have a solution using isoperimetric inequality
|
2025-03-21T14:48:32.071970
| 2020-09-19T23:05:53 |
372118
|
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|
Stack Exchange
|
A question related to the union-closed sets conjecture
Let $f(n)$ denote the maximum possible cardinality of a collection $\mathcal F$ of nonempty sets which is closed under unions ($X,Y\in\mathcal F\implies X\cup Y\in\mathcal F$) and is such that no element belongs to more than $n$ members of $\mathcal F$.
It follows from the union-closed sets conjecture that $f(n)\le2n-1$. (That's $2n-1$ instead of $2n$ because I'm omitting the empty set, which seems irrelevant to me.)
However, the upper bound $2n-1$ does not seem to be the best possible in all cases. I may have figured these things wrong, but it seems to me that $f(3)=4$ and $f(5)=8$ and $f(6)=9$. (Of course $f(n)\ge2n-1$ if $n=2^k$.)
Question. Is there a conjectured exact value for $f(n)$?
|
2025-03-21T14:48:32.072064
| 2020-09-20T01:36:07 |
372120
|
{
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"Bill Johnson",
"Dongyang Chen",
"https://mathoverflow.net/users/2554",
"https://mathoverflow.net/users/41619"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/372120"
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|
Stack Exchange
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$C[0,1]$ fails the property (K)
Recall that a Banach space $X$ has the property (K) if every $w^{*}$-convergent sequence in $X^{*}$ admits a convex block subsequence which converges with respect to the Mackey topology. The property (K) was invented by S. Kwapien to provide an alternative approach to some results of N. Kalton and A. Pelczynski on subspaces of $L_{1}[0,1]$. A. Pelczynski noted that $L_{1}(\mu)$ ($\mu$ is a finite measure) has the property (K). Further, Schur spaces, Grothendieck spaces and strongly weakly compactly generated spaces enjoy the property (K). It is known that $c_{0}$ and $C[0,1]$ fail the property (K). But it seems not easy to check that $C[0,1]$ fails the property (K).
Question. How to check that $C[0,1]$ fails the property (K)?
It would help if you would give references when you mention relatively obscure concepts. Where did Kwapien introduce property (K). I thought it was first defined in my paper with Figiel and Pelczynski.$$ $$ Figiel, Tadeusz; Johnson, William B.; Pełczyński, Aleksander Some approximation properties of Banach spaces and Banach lattices. Israel J. Math. 183 (2011), 199--231.
Oh, we defined a property called property (k) that is (probably) different from property (K). Kalton and Pelczynski defined property (K) somewhat differently; namely, where the convex block subsequence $y_n^$ satisfies $y_n^(z_n) \to 0$ for every weakly null sequence $(z_n)$ in $X$. Is that equivalent to your definition of property (K)?
Kalton, N. J.(1-MO); Pełczyński, A.(PL-PAN)
Kernels of surjections from ℒ1-spaces with an application to Sidon sets.
Math. Ann. 309 (1997), no. 1, 135--158.
A. Aviles and J. Rodriguez (Convex combinations of weak*-convergent sequences and the Mackey topology, Mediterr J. Math. 2016) mentioned that property (K) was invented by Kwapien, but did not give the reference.
In your paper with Figiel and Pelczynski, a weakening of property (K), called property (k), was introduced and proved that a Banach space $X$ would fail property (k) if $X$ contains a complemented copy of $c_{0}$. This answers my question.
Yes. Kalton and Pelczynski's definition of property (K) is equivalent to my definition.
Your property (K) uses the topology of uniform convergence on weakly compact sets, while the Kalton Pelczynski version of property (K) uses the topology of uniform convergence on weakly convergent sequences. How do you see that these two definitions are equivalent? Aviles and Rodriguez do not address this in their paper.
I give a detailed proof of the equivalence of my property (K) and the Kalton-Pelczynski version of property (K) in the following.
Let $(x_{n}^{*})_{n}$ be a weak*-null sequence in $X^{*}$. The following are equivalent:
(1)$\sup\limits_{x\in K}|\langle x^{*}_{n},x\rangle|\rightarrow 0$ for each weakly compact subset $K$ in $X$;
(2)$|\langle x^{*}_{n},x_{n}\rangle|\rightarrow 0$ for each weakly null sequence $(x_{n})_{n}$ in $X$.
Indeed, if (1) is false, there exist a subsequence $(x^{*}_{k_{n}})_{n}$ of $(x^{*}_{n})_{n}$, a sequence $(x_{n})_{n}$ in $K$ and $\epsilon_{0}>0$ so that $|\langle x^{*}_{k_{n}},x_{n}\rangle|>\epsilon_{0}$ for all $n$. Since $K$ is weakly compact, there is a subsequence $(x_{n_{m}})_{m}$ of $(x_{n})_{n}$ that converges weakly to $x\in K$. Let us define a weakly null sequence $(z_{n})_{n}$ in $X$ by $z_{k_{n_{m}}}=x_{n_{m}}-x$ and $z_{n}=0$ otherwise. By (2), $\langle x^{*}_{n},z_{n}\rangle\rightarrow 0$. Note that $\langle x^{*}_{n},x\rangle\rightarrow 0$. This implies that $\langle x^{*}_{k_{n_{m}}},x_{n_{m}}\rangle\rightarrow 0$, a contradiction.
It follows from the above fact that my property (K) is equivalent to the Kalton-Pelczynski version of property (K).
Thanks, Dongyang.
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2025-03-21T14:48:32.072297
| 2020-09-20T03:55:46 |
372124
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Stack Exchange
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Algebraic morphisms of affine varieties in positive characteristic
Let $\Omega$ be a completion of an algebraic closure of $\mathbb F_q\left(\left(\frac1T\right)\right)$ for the valuation $-\deg$.
Consider two matrices $M_1,M_2$ in $\mathcal M_2(\Omega)$ that are $\Omega$-linearly independent. Denote by $\varphi$ the morphism of affine algebraic varieties $\varphi:\Omega^2\to\Omega^2$ defined by $\varphi\begin{pmatrix}z_1\\z_2\end{pmatrix}=M_1\begin{pmatrix}z_1\\z_2\end{pmatrix}+M_2\begin{pmatrix}z^q_1\\z^q_2\end{pmatrix}$. Does $\varphi(\Omega^2)$ contain a non empty open disk (for the topology induced by $\deg$) centered at $(0,0)$?
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2025-03-21T14:48:32.072363
| 2020-09-20T04:03:26 |
372126
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Stack Exchange
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A problem on polynomial operators
Let $p(z)=\sum_{k=1}^na_kz^k$ be a polynomial of degree $n.$ Then we have these two results;
If $p(z)\neq 0 $ in $\{|z|<1\}$ then
$$
np(z)+(\alpha-z)p'(z)\neq 0\label{1}\tag{1}
$$ for all $z,\alpha$ in $\{|z|<1\}.$
If $p(z)\neq 0 $ in $\{|z|<1\}$ then
$$n\left(1-(\frac{|a_0|-|a_n|}{n(|a_0|+|a_n|)})\right)p(z)+(\alpha-z)p'(z)\neq 0 \label{2}\tag{2}
$$ for all $z,\alpha$ in $\{|z|<1\}$.
Let us generalise these two properties. Suppose $\gamma$ is any complex number with
$$
P(z)=\sum_{k=1}^na_kz^k\neq \gamma
$$
for all $z$ in $\{|z|<1\}$. Then we can easily check applying \eqref{1} to $P(z)-\gamma$ that
$$
nP(z)+(\alpha-z)P'(z)\neq n\gamma
\label{3}\tag{3}$$ for all $z,\alpha$ in $\{|z|<1\}$.
I have been trying this property on \eqref{2}. Here we have problem of coefficients!
Now is it true that
$$
n\left(1-(\frac{|a_0|-|a_n|}{n(|a_0|+|a_n|)})\right)P(z)+(\alpha-z)P'(z)\neq n\left(1-(\frac{|a_0|-|a_n|}{n(|a_0|+|a_n|)})\right)\gamma
$$ for all $z,\alpha$ in $\{|z|<1\}$ ?
[It is true if $a_0$ is replaced by $a_0-\gamma$].
If not for the general case, is it true if we have additional hypothesis that $P(z)\neq 0$ in $|z|<1$ ?
\eqref{3} implies |$P(z)+(\alpha-z)\frac{P'(z)}{n}|\leq \max_{|z|=1}|P(z)|$ on $|z|=1.$ Is it true that |$P(z)+(\alpha-z)\frac{P'(z)}{n\left(1-(\frac{|a_0|-|a_n|}{n(|a_0|+|a_n|)})\right)}|\leq \max_{|z|=1}|P(z)|$ on $|z|=1?$ (with/without the hypothesis $P(z)\neq 0$ in $|z|<1$)
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2025-03-21T14:48:32.072456
| 2020-09-20T04:27:05 |
372127
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Stack Exchange
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How can we interpret Groebner basis in a special case?
Let consider a free Lie algebra generated by $X$ with a set of relations $S$ such that the degree of leading monomial of relations in $S$ are greater than or equal to $2$. Let assume that we compute all possible compositions between relations in $S$, and we find a non-trivial composition, therefore, we add the non-trivial composition to $S$ and obtain a new set and call it the completion of $S$. In this case, can we claim that the our Grobner-Shirshov basis never include the relations with leading term of degree $1$, even our steps to get an $S$-reduced GSB be infinite?
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2025-03-21T14:48:32.072558
| 2020-09-20T04:33:01 |
372128
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Stack Exchange
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What is the cotangent complex good for?
The cotangent complex seems to be a pretty fundamental object in algebraic geometry, but if it's treated in Hartshorne then I missed it. It seems to be even more important in derived algebraic geometry, so I think I need to slow down and zoom out a bit. When first learning about object $X$, it's nice to have in mind some concrete applications of $X$ to structure one's thinking.
Question: Why study the cotangent complex? What problems is it intended to solve?
(Bonus points if there is something interesting to say about extending to the derived setting.)
I have the sense that the cotangent complex is such a fundamental object that it may be difficult to isolate its importance -- much like trying to articulate the significance of something like cohomology. In that case, it might be more appropriate to ask something like "what kinds of questions does the cotangent complex allow one to ask?".
EDIT: The answers so far are great, but I imagine there are a great many more examples which could be given (the more down-to-earth the better!). As suggested in the comments, it's probably appropriate to say a bit more about where I'm coming from.
I suppose the main ideas I have in my head right now are:
The cotangent complex generalizes the Kahler differentials.
The cotangent complex controls deformation theory.
This leaves me with a few difficulties:
I'm not used to thinking of differential forms primarily as "things that control deformations". So it might be helpful to simply illustrate the use of the cotangent complex by describing some deformation problem and its solution in the smooth case using differntials -- it would then seem natural to want to generalize this situation to the non-smooth case.
I'm not even sure why I should be interested in deformation theory as such. So it might be helpful to simply see an example of a problem which arises outside the context of deformation theory itself, see how it can be rephrased deformation-theoretically, and then see how its solution uses the cotangent complex. Bonus points if the story is geometric enough to see why the role of the cotangent complex here is really a generalization of the role of differentials.
First of all, make sure you understand the cotangent sheaf -- differential forms -- whose SpecSym is the tangent bundle. The cotangent complex degenerates to that in the smooth case, but is still very nicely behaved in case of mild singularities (e.g. local complete intersections). It was originally designed by Grothendieck et al in the guise of the "virtual tangent bundle" (its induced K-theory class) to formulate GRR formulas in the singular case.
-1: it is good to do some preliminary research before asking a question of this kind. The original source is Illusie's thesis, whose title (!) already gives you the first idea what the cotangent complex is good for.
Answers to this might partially answer your question: https://mathoverflow.net/q/2607
If $X$ is a smooth variety, then the first order deformations are parameterized by $H^1(X,T_X)$, where $T$ is the tangent sheaf. If $X$ is singular, then what? Ans: Use the cotangent complex. By the way, Hartshorne's AG book does discuss $T^1$ in the exercises I believe.
From a "topological" point of view, one of the utilities of the cotangent complex is (a refinement of) the following Hurewicz theorem (used in the proof of the representability theorem): a map R -> R' of simplicial commutative (in fact, this also works for connective E_oo-) rings is an equivalence iff it's an equivalence on pi_0 and L_{R'/R} = 0. (The idea is: the cotangent complex measures infinitesimal (derived) deformations, and the higher homotopy groups of a SCR are "fuzzy deformations" away from pi_0, so L_{R'/R} = 0 tells you that the Whitehead towers of R and R' are the same.)
@PiotrAchinger Fair enough. I guess part of my issue is not knowing what problems are meant to be solved by studying deformations as such.
Deformations of $X$ tell you what the tangent space of $X$ is in the moduli space of varieties. So e.g. if you have a moduli space and you want to show it is smooth, then you are interested in deformations.
Apparently it's good for asking questions on MathOverflow about it and getting internet points! :-)
Maybe this isn't an answer to what it's good for, but I've heard it said that $L_X$ is the derived functor (in a sense that is possible to make precise for non-Abelian source categories such as in this case) of $X\mapsto \Omega^1_X$.
fwiw the cotangent complex is indispesnable for the recognition principle in motivic homotopy theory
Here is an example mentioned in passing by user ali's answer, but I think it is cute (and powerful) enough to be worth fleshing out the details.
Lifting from characteristic $p$ to characteristic zero
In short, studying a geometric object (say, a scheme) $X$ in characteristic $p$ often involves lifting it to characteristic zero. For example, if $X$ is a smooth projective variety over $\mathbf{F}_p$, we may try to find a (flat) lift $\mathcal{X}$ over the $p$-adic numbers $\mathbf{Z}_p$. Now, $\mathbf{Z}_p$ embeds into $\mathbf{C}$ (in some completely noncanonical way), and we can apply powerful methods such as Hodge theory to the complex manifold underlying $\mathcal{X}_\mathbf{C}$.
Now, recall that
$$ \mathbf{Z}_p = \varprojlim_n \mathbf{Z}/p^{n+1}. $$
Thus lifting $X_0=X$ over $\mathbf{Z}_p$ involves finding compatible liftings $X_n$ over $\mathbf{Z}/p^{n+1}$ for all $n$. The system $\mathfrak{X} = \{X_n\}$ (or its inductive limit in locally ringed spaces) is a "$p$-adic formal scheme," and the next step involves checking that it is algebraizable, i.e. that it comes from an actual scheme $\mathcal{X}/\mathbf{Z}_p$ by the obvious "formal completion" functor.
Now the first step, finding the successive liftings $\{X_n\}$, is completely controlled by deformation theory. In our situation, it says the following:
If $X_0$ is a scheme over $\mathbf{F}_p$, and $X_n$ is a flat lifting of $X_0$ over $\mathbf{Z}/p^{n+1}$, there there exists an obstruction class
$$ {\rm obs}(X_n, \mathbf{Z}/p^{n+2}) \in {\rm Ext}^2(\mathbf{L}_{X_0/\mathbf{F}_p}, \mathcal{O}_{X_0}) = {\rm Hom}_{D(X_0)}(\mathbf{L}_{X_0/\mathbf{F}_p}, \mathcal{O}_{X_0}[2]), $$
which vanishes if and only if there exists a flat lifting $X_{n+1}$ of $X_n$ over $\mathbf{Z}/p^{n+2}$. It is functorial in the sense that for $f_n\colon X_n\to Y_n$ lifting $f_0\colon X_0\to Y_0$ we have a commutative square
$$\require{AMScd} \begin{CD}
\mathbf{L}_{Y_0/\mathbf{F}_p} @>>> \mathcal{O}_{Y_0}[2]\\ @VVV @VVV\\
Rf_{0, *}\mathbf{L}_{X_0/\mathbf{F}_p} @>>> Rf_{0, *}\mathcal{O}_{X_0}[2]
\end{CD}$$
In case the obstruction class vanishes, the set of isomorphism classes of such liftings $X_{n+1}$ is in a natural way a torsor under
$$ {\rm Ext}^1(\mathbf{L}_{X_0/\mathbf{F}_p}, \mathcal{O}_{X_0}) = {\rm Hom}_{D(X_0)}(\mathbf{L}_{X_0/\mathbf{F}_p}, \mathcal{O}_{X_0}[1]). $$
The group of automorphisms of any lifting $X_{n+1}$ restricting to the identity on $X_n$ is naturally isomorphic to
$$ {\rm Hom}(\mathbf{L}_{X_0/\mathbf{F}_p}, \mathcal{O}_{X_0}). $$
There is a similar story for lifting morphisms $f_0\colon X_0\to Y_0$.
So if you can show that ${\rm Ext}^2(\mathbf{L}_{X_0/\mathbf{F}_p}, \mathcal{O}_{X_0})$ vanishes, then you know that $X_0$ admits a formal $p$-adic lifting $\mathfrak{X}$. For example, if $X_0$ is a K3 surface, then this group can be identified with the space of global vector fields on $X_0$, and its vanishing is a difficult theorem due to Rudakov and Shafarevich. (And the fact that there is an algebraizable formal lifting, i.e. that an ample line bundle can be lifted to all $X_n$'s for a good choice of $\mathfrak{X}$, was shown later by Deligne.)
Perfect schemes and Witt vectors
Recall that for every perfect field $k$ of characteristic $p>0$ there exists a unique complete discrete valuation ring $W(k)$ (its ring of Witt vectors) with residue field $k$ whose maximal ideal is generated by $p$. It is a functor of $k$, and we have $W(k) \simeq k^{\mathbf{N}}$ as functors into sets. The addition and multiplication laws on $k^{\mathbf{N}}$ obtained this way are given by complicated universal formulas, e.g.
$$ (x_0, x_1, \ldots) + (y_0, y_1, \ldots) = (x_0 + y_0, x_1 + y_1 - \sum_{0<i<p} \frac 1 p \binom p i x_0^i y_0^{p-i}, \ldots). $$
We define $W_n(k) = W(k)/p^n$ and call these Witt vectors of length $n$.
For example, $W(\mathbf{F}_p) = \mathbf{Z}_p$, $W_n(\mathbf{F}_p) = \mathbf{Z}/p^n$.
In fact, the above can be defined for any ring $R$. If $R$ is a perfect $\mathbf{F}_p$-algebra, meaning that its Frobenius
$$ F_R \colon R\to R, \quad F_R(x) = x^p $$
is an isomorphism, then $W(R)$ is a flat lifting of $R$ over $W(\mathbf{F}_p) = \mathbf{Z}_p$.
Here is a beautiful argument (I think due to Bhargav Bhatt) employing the cotangent complex to show the existence of Witt vectors for perfect rings (or schemes) without using any strange-looking universal formulas for addition and multiplication.
Theorem. Let $X$ be a perfect $\mathbf{F}_p$-scheme. There exists a unique up to unique isomorphism formal $p$-adic lifting $\mathfrak{X} = \{X_n\}$ of $X_0=X$. Moreover, every morphism $f\colon X\to Y$ admits a unique lifting $\mathfrak{X}\to \mathfrak{Y}$.
The above implies that $\mathfrak{X}$ is a functor of $X$, denoted $W(X)$. It is not difficult to prove that it indeed coincides with the Witt vectors.
Proof. Consider the cotangent complex $\mathbf{L}_{X_0/\mathbf{F}_p}$ and the map
$$ F_X^* \colon \mathbf{L}_{X_0/\mathbf{F}_p}\to F_{X, *} \mathbf{L}_{X_0/\mathbf{F}_p} $$
induced by the absolute Frobenius $F_X\colon X\to X$. Since $F_X$ is an isomorphism, the map $F_X^*$ is an isomorphism too. The complex $\mathbf{L}_{X_0/\mathbf{F}_p}$ is defined by locally resolving $\mathcal{O}_X$ by free $\mathbf{F}_p$-algebras and considering their Kaehler differentials. And $F_A$ acts as zero on $\Omega^1_{A/\mathbf{F}_p}$ for every $\mathbf{F}_p$-algebra $A$:
$$ F_A^*(dx) = dF_A(x) = dx^p = px^{p-1} dx = 0. $$
Therefore the map $F_X^*$ above is the zero map. Since it is also an isomorphism, we conclude that $\mathbf{L}_{X_0/\mathbf{F}_p} = 0$!
Now by deformation theory, the obstructions to lifting lie in the zero group (and hence the successive liftings exist), the isomorphism classes of different successive liftings are permuted by the zero group (and hence the liftings are unique), and their automorphism groups are trivial (so the liftings are unique up to a unique isomorphism). Similarly, one handles the lifting of morphisms. $\square$
@TimCampion Thanks for the edit (and the hint in the changelog)!
One major application is in Artin's representablity theorem. Existence of a cotangent complex (what Artin calls existence of an obstruction theory) allows you to linearize the question of the existence of a smooth atlas. It's really powerful. I'd really suggest looking through DAG XIV if you're interested in this application (the derived version is in DAG-0 (Lurie's thesis)) and a simplified version is in one of the appendices of HAG-II by Toën and Vezzosi.
Representability was used to stunning effect in Toën's 2011 paper showing that every derived fppf-algebraic n-stack is admits a smooth atlas and therefore is a derived Artin algebraic n-stack, and vice-versa (showing that all Artin derived n-stacks are fppf sheaves).
I recently had to make use of these theorems to prove fppf descent for étale n-sheaves on spectral DM stacks in something I've been working on recently. The key point in all of this is showing the existence of a cotangent complex.
This is not down a stacky and geometric alley many people like, but I think it maybe useful to show how the (algebraic) cotangent complex can be used to organize and solve a very natural question of comparing cotangent cohomology theories as the classical HKR theorem does.
In joint work with R. Campos, we managed to use the cotangent complex (which is defined functorially for algebras over say algebraic operads) to solve the following problem:
Suppose that $f:P\longrightarrow Q$ is a morphism of (algebraic) operads and consider the induced morphism on cotangent complexes $\mathbb L_{P,A} \longrightarrow \mathbb L_{Q,A}$. If $A$ is a smooth $Q$-algebra (meaning the functor $X\to \operatorname{Der}_Q(X,M)$ is exact),
when can we find a functor $F$ that produces an HKR-type quasi-isomorphism of complexes
$ \mathbb L_{P,A} \longrightarrow F(\mathbb L_{Q,A})?$
It turns out that if one considers the category of left dg-$P$-modules, then such functors $F$ are more-or-less in correspondence with resolutions of $Q$ as a left $P$-module through $f$, and we managed to show that
If $f$ is left Koszul (meaning, we can choose a left resolution $(P\circ F,d)$ that is diagonally pure) then the generators $F$ of the resolution solve the problem above: for every smooth $Q$-algebra $A$ there is a map of complexes $\mathbb L_{P,A} \longrightarrow F(\mathbb L_{Q,A})$ that is a quasi-isomorphism.
To see how this is an HKR theorem, observe that for the projection $A\to C$ of the associative operad onto the commutative one, we can find a diagonally pure resolution given by $(A\circ F,d)$ where $F =\mathsf{Lie}^¡$ has weight degree equal to arity. Indeed, you get that the homology of this is $C$ because $A = C\circ \mathsf{Lie} $ and $\mathsf{Lie} \circ \mathsf{Lie}^¡ \simeq k$.
This recovers the classical HKR theorem, since of course $F(V) = S^c(V)$ and so we get that $HH_*(A)$ is the free cocommutative algebra over $\Omega_A^1$ (since $A$ is smooth, this is all that is left in homology of $\mathbb L_{C,A}$).
If you want to see the relation between differential forms and deformation theory you can look at the part B of Illusie article in FGA explained which shows that why the obstruction to the problem of lifting a morphism or a scheme to an infinitesimal neighborhood lies in the cohomology of a sheaf related to sheaf of differentials. It also discuss some classic application of this fact about fundamental group of schemes and similarly this is very important fact when you want to prove the base change theorems about étale cohomology.
One of the main motivation of cotangent complexes was studying deformations of $p$-divisible groups and proving Grothendieck–Messing theorem. This is important because for example you have a criterion for smoothness based on deformations which is in particular useful when you defined your scheme by it's functor of points. You can use Grothendieck–Messing theory to prove smoothness of Shimura varieties (and their integral models) among other things.
The cotangent complex is also useful in perfectoid geometry because you often want to lift a morphism or a scheme from residue field of a complete local ring and deformation theory and cotangent complex are the main tool for this kind of problems.and in the setting of prefectoid rings the cotangent complex (or at least its derived p-adic completion) vanish because of the surjection of the Frobenius. it is an easy exercise once you build the machinery of cotangent complexes but it is very important because you don't have any obstruction to lift morphisms.
Let me try to answer the questions in your edit. For the relationship to deformation theory, it is geometrically simpler to think about the tangent complex.
Let $X$ be a smooth algebraic variety. Then $H^0(T_X)$, the space of global vector fields on $X$, governs infinitesimal automorphisms of $X$. This is just the usual relationship between vector fields and one parameter subgroups from differential topology. From this fact, we can see why $H^1(T_X)$ governs deformations.
Algebraically, write $D = {\rm Spec}~ k[\epsilon]/\epsilon^2$. A deformation of $X$ is a (flat) variety $\tilde X \to D$ together with an isomorphism $X \to \tilde X \times_D {*}$. Deformations of $X$ form a groupoid in the obvious way. Further, for every open subset $U$ of $X$ we can consider the groupoid of deformations of $U$. This gives us a sheaf of groupoids on $X$, $U \mapsto Def(U)$.
Now the key facts are:
If $U$ is affine, then every deformation of $U$ is isomorphic to $U \times D$.
The automorphism group of the deformation $U \times D$ is canonically isomorphic to $H^0(T_U)$.
The second fact is an algebraic restatement of the fact from before. The first one uses smoothness of $X$. Granting these two facts, it follows formally that the sheaf of groupoids is the one associated to the sheaf of groups $H^0(T_U)$. So in particular the set of connected components is $H^1(T_X)$.
Now, since we are working in a homological/homotopical setup, it is intuitive that to carry out a similar story for a singular scheme, we should resolve it by smooth schemes, and work from there. In other words, if $X$ is singular, then the automorphism group of $X \times D$ is not necessarily governed by $H^0(T_X)$ any more, but we know we can compute its automorphism group in the category of derived schemes by resolving $X$ by smooth schemes. You can try to think about what happens with the simplest singularity $\mathbb V(xy)$.
As for applications of deformation theory-- algebraic geometers want to classify varieties. Moduli spaces/stacks are very useful, and deformation theory precisely tells you the local structure of moduli space. For instance, the moduli space of genus $g$ curves is smooth and $3g-3$ dimenisonal. Why? Because $H^1(T_C) = H^0(\Omega_C^{\otimes 2})$ has dimension $3g-3$ always by Riemann Roch. The cotangent complex comes in when you want to compactify the moduli-space, the boundary of the compactification will consist of singular curves. You can look at Deligne-Mumford's paper to see how deformation theory gets used here.
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2025-03-21T14:48:32.073780
| 2020-09-20T05:34:04 |
372129
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"AspiringMat",
"Brendan McKay",
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"https://mathoverflow.net/users/9025",
"მამუკა ჯიბლაძე"
],
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/372129"
}
|
Stack Exchange
|
Expected height of a poset?
I am interested in any known results/empirical studies done on the average height of a poset with $N$ elements. Obviously this would depend on how that poset relation was randomly defined, however, at this point, I'll take any reasonable result on the topic regardless of how the poset was formed.
In my case $N$ is a very large number (billions of elements), so I am interested whether the height $h \ll N$ asymptotically.
In Asymptotic Enumeration of Partial Orders on a Finite Set (1975), Kleitman and Rothschild showed that almost all partial orders on an $n$-element set have a simple description: they have three "layers" $L_1$, $L_2$, and $L_3$ of incomparable elements, of size $n/4$, $n/2$, and $n/4$ respectively. Each element of $L_1$ is covered by about half the elements of $L_2$. Likewise for $L_2$ and $L_3$, and in the reverse direction.
So almost all finite posets have height $3$.
(If you don't have the time/effort to read Kleitman and Rothschild's paper, I came across this reference via G. Brightwell, Linear extensions of random orders, Discrete Math. 125 (1994) pp. 87–96. If memory serves, Brightwell's paper presents a good summary of this one.)
Thank you, this is exactly what I need! I definitely have both the time and the willingness to read the paper. Thank you!
Thank you, this was a very heavy but insightful paper. The result is really interesting.
Astonishing result
The statistics for small sizes (as far as can be computed) don't look like the final shape. For example, the most popular number of levels for 16 points is 5. This suggests that the convergence to the final shape is quite slow.
|
2025-03-21T14:48:32.073947
| 2020-09-20T08:05:20 |
372134
|
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"Stanley Yao Xiao",
"Wojowu",
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"url": "https://mathoverflow.net/questions/372134"
}
|
Stack Exchange
|
Frey's elliptic curve and perfect numbers?
Let $E_n:y^2=x(x-\sigma(n)/2)(x+\sigma(n)/n)$ be a Frey-elliptic curve, where $\sigma$ denotes the sum of divisors of the natural number $n$.
If $n$ is a perfect number ($\sigma(n)=2n$) then the $j$-invariant is given by:
$$j(n) = \frac{16^2(n^2+2n+4)^3}{(2n)^2(n+2)^2}$$
If $n$ is a perfect number, then the points $(n:0:1),(-2:0:1)$ are integral points of $E_n$.
What properties does this curve have, if $n$ is an (odd) perfect number?
I am not sure what you are expecting. If $n$ is a perfect number then this curve has three rational 2-torsion points, and that’s about as much as anyone can say. Remember that the original Frey curve is useful for Fermat because if a non-trivial counterexample to FLT exists, then the curve is not modular, which would contradict the modularity theorem. I don’t think this curve can be shown to not be modular.
@StanleyYaoXiao Doesn't it have three rational 2-torsion points for any $n>0$? For perfect numbers those are moreover integral, but it is not an equivalenr condition.
One thing to note is that this elliptic curve has discriminant $(\sigma(n)/2\cdot\sigma(n)/n\cdot\sigma(n)(1/2+1/n))^2$, so that this curve's properties rely on the factorization of not only $n$ and $\sigma(n)$, but also $n+2$. I doubt perfectness of $n$ will be enough to guarantee the curve to have interesting properties.
|
2025-03-21T14:48:32.074080
| 2020-09-20T08:33:37 |
372135
|
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"Wlod AA",
"Wojowu",
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|
Stack Exchange
|
A detail oriented prime conjecture
Conjecture For arbitrary integers $\ 0 \le k \le m\ $ there exists
integer $\ n\ge m\ $ such that for every natural number $\ s\ $ at
least one of the numbers $\ p(x)+s\ (\text{where}\ k\le x\le n)\ $ is not prime.
Here, $\ p(0)=2, p(1)=3,\ldots\ $ is the strictly increasing sequence of all prime numbers.
Let $q=p(k)$. Using Dirichlet's theorem on primes in arithmetic progressions, there is $n\geq m$ large enough so that for any $a$ not divisible by $q$, there is some $k\leq x\leq n$ such that $p(x)\equiv a\pmod q$. Also taking $a=0$ and $x=k$, we see this is true for all residue classes mod $q$.
Take any natural $s$ (which presumably also requires $s>0$ for you). Pick some $k\leq x\leq n$ such that $p(x)\equiv -s\pmod q$ (which exists by construction). Then $p(x)+s$ is divisible by $q$ and greater than $q$.
Simple and nice. ### (Frankly speaking, somehow, I made a huge "typo". I am leaving this question as it is but you may try my new "typo"-free version -- see a new post).
@WlodAA Do you mean including $k=0$? I don't see how that makes any difference really. My solution should still work.
there was ASLO a huge typo, see my new post. That other typo, 0 < or = k, was trivial.
Ah sorry, I see you made a new post. I thought you just edited that one. I will take a look at the new one.
|
2025-03-21T14:48:32.074215
| 2020-09-20T09:07:07 |
372139
|
{
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"Wlod AA",
"Yaakov Baruch",
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"user44191"
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|
Stack Exchange
|
A typo-free detail oriented prime conjecture
I've grossly overstated things in my two posts before the last one. Thank you for providing references that have returned me to reality.
Conjecture For arbitrary integers $\ 0 \le k \le m\ $ there exists
integer $\ n\ge m\ $ such that for every natural number $\ s\ $ at
least one of the numbers
$$\ p(x+s+1)-p(x+s)\ \ne\ p(x+1)-p(x) $$
where $\ k\le x < n$.
Here, $\ p(0)=2, p(1)=3,\ldots\ $ is the strictly increasing sequence of all prime numbers.
An assembler-like equivalent formulation:
Conjecture'
$$ \forall_{m\in\mathbb Z_{\ge 0}}\,\forall_{k\in 0..m}\,
\exists_{n\in\mathbb Z_{\ge m}}\,\forall_{s\in\mathbb N}\,
\exists_{x\in k..n\!-\!1} $$
$$ p(x+s+1)-p(x+s)\ \ne\ p(x+1)-p(x) $$
Here (Perl notation),
$$\ u..v\ :=\ \{x\in\mathbb Z:\ u\le x\le v\} $$
I don't understand the statement, in particular the part "at least one of the numbers" followed by a formula... I would welcome a clarification, perhaps using less natural language and more quantifiers.
@YaakovBaruch or anybody, if you would like too, I'd provide an Image-Processing/Bourbaki-like terminology for my answer below (it was my first choice but I don't mean to write here purely for my own sake)..
Thank you! It's not that I'm so much of Bourbaki style fan myself, but simpy I didn't understand that "at least one of the numbers , for $k\le x <n$" meant "for at least one $x$ such that $k\le x <n$, "...
Correct me if I'm wrong. Isn't it true that $\forall_{m\in\mathbb Z_{\ge 0}},\forall_{k\in 0..m}, \exists_{n\in\mathbb Z_{\ge m}} :\forall_{s\in\mathbb N}, \exists_{x\in k..n-1} : P(x,s)$ is equivalent to $\forall_{a\in \mathbb N}, \exists_{b(a)\in\mathbb Z_{\ge a}} :\forall_{s\in\mathbb N}, \exists_{x\in a..b(a)} : P(x,s)$?
@YaakovBaruch, u'r right! ### I had the feeling that the initial segment was not important but, how nice! you took a step back and saw it clearly. Indeed, formally, by setting a=k=m we get your equivalence. (There is often a bit of psychology or inertia left beside the straight logic).
Like in the previous post, let $q=p(k)$ and let $n\geq m$ be such that primes $p(x),k\leq x<n$ cover all residue classes mod $q$.
Suppose this $n$ doesn't work. This means that $p(x+s+1)-p(x+s)=p(x+1)-p(x)$ for all $k\leq x<n$. Adding up a bunch of such equalities we get $p(x+s)-p(k+s)=p(x)-p(k)$ for all $k\leq x\leq n$. As $s>0$, we have $p(k+s)>q$ so it is indivisible by $q$. There is some $k<x<n$ such that $p(x)\equiv -p(k+s)\pmod{q}$, and we get $p(x+s)=p(k+s)+p(x)-p(k)\equiv 0\pmod q$, which is impossible as $p(x+s)>q$, hence a contradiction.
This is so nice! You've applied a simple-sounding but powerful Dirichlet Theorem twice (in the previous thread too), and this time this application is even more surprising since the profile definition seems to describe complicated objects, not linear. (On a personal note, I am happy that my intuition was justified -- I just got carried away far too far in my -2 and -3 note, by now already deleted).
In effect, Wojowu, you have provided a large class of lonely profiles.
@WlodAA This should be equivalent to "inadmissible" tuples, to my understanding, essentially the "exceptions" to the prime tuples conjecture.
"This" ? (Let's avoid pronouns -- my general principle for the "Art of Communication" and for poetry).
@WlodAA The profiles that cover all residue classes mod $q$. I should note that they may not actually be fully lonely - $q$ might "slot in" at different points. Wiki gives the example of $(0, 2, 8, 14, 26)$ (the "profile" being the consecutive differences), where $3, 5, 11, 17, 29$ and $5, 7, 13, 19, 31$ both fit.
@user44191, you've provided non-consecutive primes.
@user44191, the primes from your first sequence are $\ne$ 1 mod 3.
@WlodAA True - I guess I was aiming at a somewhat more general concept, not necessarily requiring consecutiveness.
@user44191, actually, consecutiveness and extending the sequence assures a full representation of residua mod q. Otherwise, it's not important, meaning that a full set of residua mod q makes the prime sequence lonely, i.e. not shiftable. ### I feel that the consecutiveness is important in general or else we get global-statistical problems about primes rather than local. The consecutiveness may spur new breakthroughs.
NOTATION:
$u_0\ \ldots\ u_n\,\ $ and $\ v_0\ \ldots\ v_n\,\ $ are arbitrary finite strictly increasing sequences of non-negative integers, of the same sequence's length $\ n+1$.
$p(0)=2, p(1)=3, \ldots\ $ is the strictly increasing sequence of all
(consecutive) primes.
$ p(u_0)\ \ldots\ p(u_n)\,\ $ and $\,\ p(v_0)\ \ldots\ p(v_n)\ $ are
arbitrary finite strictly increasing sequences of primes (not necessarily
consecutive), where the length of the two sequences is the same, namely
$\ n+1.$
The goal: We will see that under certain additional assumptions, the
above two finite sequences are identical; in particular when one of them is assured to exist as an extension of a different shorter sequence.
Wojowu has posted his answer in the Q&A style. He deserves a more rounded presentation. Let me be the first one to do it. It'll be but a presentation. The results are exclusively due to Wojowu (Wojciech Wawrów).
Theorem 1 Assuming our NOTATION, if there exists a prime $\ q\ $
such that for every integer $\ x\ $ there exists at least one integer (index)
$\ a(x)\ $ such that
$$ p(u_{a(x)})\ \equiv x\ \mod q $$
and if there exists an integer $\ s\in\mathbb Z\ $ such that
$$\forall_{t=0}^n\quad p(v_t) = p(u_{s+t}) $$
and
$$ \forall_{t=1}^n\quad p(v_t)-p(v_{t-1})\ =\ p(u_t)-p(u_{t-1}) $$
then $\ s=0$.
Remark The last assumption can be written equivalently as
$$ \forall_{t=1}^n\quad p(v_t)-p(u_t)\ =\ p(v_{t-1})-p(u_{t-1}) $$
Proof (of Theorem 1) Let the assumptions of the theorem hold. By the above Remark,
for every integer $\ x\ $ there exists at least one integer (index)
$\ b(x)\ $ such that
$$ p(v_{b(x)})\ \equiv x\ \mod q $$
The terms in each prime sequence that are $\ 0 \mod q\ $
are simply equal to $\ q.\ $ If integer $\ s\ $ were different from $\ 0\ $
then the same prime $\ q\ $ would appear twice -- $\ s\ $ places apart -- in the increasing sequence of all primes; that would be a contradiction.
End of Proof
More NOTATION: Given integers $\ 0\le m\le n,\ $ prime sequence
$ p(u_0)\ \ldots\ p(u_n)\,\ $ is called to be the minimal $n$-extension
of a strictly increasing sequence $ p(u_0)\ \ldots\ p(u_m)\,\
\Leftarrow:\Rightarrow$
$$ \forall_{k=m+1}^n\quad u_k=u_m+k-m, $$
Theorem 2 For every strictly increasing finite prime sequence $ p(u_0)\ \ldots\ p(u_m)\,\ $ there exists a minimal extension $ p(u_0)\ \ldots\ p(u_n)\,\ $ which is lonely, meaning that it is the only prime sequence $ p(v_0)\ \ldots\ p(v_n)\,\ $ such that there exists an integer $\ s\in\mathbb Z\ $ such that
$$\forall_{t=0}^n\quad p(v_t) = p(u_{s+t}) $$
and
$$ \forall_{t=1}^n\quad p(v_t)-p(v_{t-1})\ =\ p(u_t)-p(u_{t-1}). $$
Proof For $n$ large enough all residue classes \mod p(u_0) appear
in $ p(u_0)\ \ldots\ p(u_n)\,\ $ (by Dirichlet Theorem). This, by Theorem 1,
makes sequence $ p(u_0)\ \ldots\ p(u_n)\,\ $ lonely. End of Proof
|
2025-03-21T14:48:32.074637
| 2020-09-20T12:58:39 |
372145
|
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"Will Sawin",
"asv",
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"https://mathoverflow.net/users/18060"
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|
Stack Exchange
|
Asymptotic estimate of the number of points of variety over finite field
EDIT: Let $X$ be a geometrically irreducible $n$-dimensional variety over finite field $\mathbb{F}_{q_0}$. Let $\mathbb{F}_q$ denote any finite extension of $\mathbb{F}_{q_0}$.
It is known (e.g. follows from the Weil conjectures) that
$$\frac{|X(\mathbb{F}_q)|}{q^n}\to 1 \mbox{ as } q\to\infty.$$
I am wondering if there is an elementary proof of this fact.
You want $X$ to be geometrically irreducible here.
This is a theorem of Lang and Weil, proven well before the Weil conjectures. It relies only on Weil's proof of the Riemann hypothesis for curves.
The Riemann hypothesis for curves was given an elementary proof by Bombieri and Stepanov.
Combining these should give the elementary proof you seek.
Many thanks. Fixed the missing condition. The Lang-Weil estimate seems to be much stronger than what I asked. Also the Riemann hypothesis for curves is not very elementary for my taste. In fact I do not have an elementary answer to my question even for the case of a curve. Is Riemann hypothesis necessary in the latter case?
@makt For $X$ of dimension $1$, one can prove the weaker estimate by analytic number theory methods a la the usual proof of the prime number theorem over the integers (see Proposition 5.13 of Rosen's Number Theory in Function Fields). Maybe there is an analogue of the Erdős-Selberg elementary proof using Selberg's symmetry formula as well.
|
2025-03-21T14:48:32.074884
| 2020-09-20T13:44:40 |
372151
|
{
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"authors": [
"Max Alekseyev",
"fedja",
"gondolf",
"https://mathoverflow.net/users/1131",
"https://mathoverflow.net/users/4987",
"https://mathoverflow.net/users/7076"
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"url": "https://mathoverflow.net/questions/372151"
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|
Stack Exchange
|
An optimization problem about number series
Given $m>0$, we want to minimize
$$
\sum_{k=1}^r a_k \log b_k
$$
for arbitrary increasing number series $a_k\geq 1$ and $b_k\geq 1$ satisfies
$$
\sum_{k=1}^{\infty} \frac{1}{a_k}=1
$$
and $r$ satisfies
$$
a_rb_r\geq m.
$$
It should be a function of $m$. It seems $\log m$ is not reachable.
And what prevents you from having $b_1=0$?
What are variables and what are parameters?
@fedja Thank you very much!
@Alekseyev $m$ is the parameter, the rest are variables.
Should it be $\sum_{k=1}^{r} \frac{1}{a_k}<1$ or why do you care about $a_k$ for $k>r$?
Also, why not just take $r=b_1=1$ and $a_1=m$ ?
@Alekseyev $r$ is determined by $a_k$ and $b_k$. $b_k$ is increasing, therefore $b_1$ can not be $1$.
|
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