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2025-03-21T14:48:30.092522
| 2020-03-18T23:31:08 |
355220
|
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|
Stack Exchange
|
Generating all pentominoes by cutting and pasting
Is it possible to place the twelve pentominoes around a circle in such a way that if two of the pentominoes find themselves next to each other, it is because one of the two can be obtained from the other by cutting out one of its component squares (thus obtaining a tetramino) and glueing it elsewhere?
This question probably belongs on math.stackexchange.com and not here. In any case, NB that removing a square from an $n$-omino need not leave an $(n - 1)$-omino; in general it will leave some collection of $k_i$-ominos, with $\sum k_i = n - 1$. It is not clear to me from the wording of the question whether you mean to allow maneuvers for which the intermediate step is not an $(n - 1)$-omino; this makes a difference in whether, e.g., the $V$ and $W$ and pentominos can be adjacent in the circle (where $V$ and $W$ refer to Conway's notation: https://en.wikipedia.org/wiki/Pentomino#History ).
Intermediate step must be a tetramino.
Yes. Ugly representation below (* is the cell to be moved to . to obtain the next pentamino)
.
####* -> ###* -> ### -> ### -> ### -> *## -> ## -> ## -> ##. -> ### -> * -> ####. -> #####
. #. #* # . * #. ## ## #* # # ###. *
. * . .* #. #* * #
And Hexominoes?
@BernardoRecamánSantos, absent some special insight you answer these questions by constructing the graph where pairs of n-ominoes are joined by an edge if you can move between them in the way you describe, then looking for a Hamilton cycle. There is lots of off the shelf software (e.g. sage) available for the second step, but you probably have to do the first step yourself.
For pentominoes, there are 4080 inequivalent ways to do it. (Reversing the direction of a cycle doesn't count as a separate solution.) Answering the comment to Mikhail's answer, here is a solution for hexominoes:
I verified that such cycles exist for polyominoes with up to 13 cells. I have assumed that the intermediate step must be connected, as explained in the comments to the original question. Without this requirement, there are 16800 solutions for pentominoes.
(Edit) And here is a cycle for the 108 heptominoes (I curled it up a bit to save some space):
There are now OEIS sequences A367123 and A367436 for the number of Hamiltonian cycles, without and with the requirement that the intermediate step must be connected, respectively.
|
2025-03-21T14:48:30.092832
| 2020-03-19T00:01:21 |
355221
|
{
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"Ben McKay",
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|
Stack Exchange
|
What does a curve with unbounded acceleration tells about its shape?
Let $(M,g)$ be a compact Riemannian manifold and assume that $c : \mathbb{R} \to M$ is a smooth curve. I am work on a research problem where
$$\left\|\dfrac{\nabla}{dt}c'\right\|\to \infty.$$
What can I conclude about the shape of the curve?
Alternatively, given an extra condition on the problem, if the acceleration blows up, then for a smooth vector field $\nu$ along the curve, there is $t_0 > 0$ such that for all $t>t_0$,
$$g(\nabla_{\nu}\nu,\dfrac{\nabla}{dt} c') < 0,$$
does this imply something obvious?
Clearly you can reparameterize any noncompact curve to get its acceleration to be unbounded, so all you can say is that it is not defined on a compact interval. Since your curve is defined on the real number line, I don't see anything else.
The second condition, since it only involves a vector field $v$ defined along the curve, can surely be satisfied by one tangent to the curve, reducing the problem to $M=\mathbb{R}$, where it is clearly satisfied for any curve, with some choice of $v$ growing rapidly and pointing the other direction.
|
2025-03-21T14:48:30.092937
| 2020-03-19T00:23:01 |
355222
|
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"Hannes Jakob",
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|
Stack Exchange
|
Cardinal numbers and the Bolzano-Weierstrass theorem
Let $\kappa$ be a cardinal number, define $\textsf{M}(\kappa)$ and $\textsf{BW}(\kappa)$ as follows:
$\textsf{M}(\kappa)$ : For every sequence $(f_{n}:\kappa\to \mathbb{R})_{n\in\mathbb{N}}$ of real-valued function defined on $\kappa$ there is a subsequence $(f_{n}:\kappa\to \mathbb{R})_{n\in A}$ such that for each $\alpha\in\kappa$ the sequence $(f_{n}(\alpha))_{n\in A}$ is eventually monotonic.
$\textsf{BW}(\kappa)$ : For every sequence $(f_{n}:\kappa\to \mathbb{R})_{n\in\mathbb{N}}$ of pointwise bounded real-valued functions defined on $\kappa$ there is a subsequence $(f_{n}:\kappa\to \mathbb{R})_{n\in A}$ which is pointwise convergent.
Proposition: For each cardinal number $\kappa$, we have that $\textsf{M}(\kappa) \Longrightarrow \textsf{BW}(\kappa)$
Proof: Let $(f_{n}:\kappa\to \mathbb{R})_{n\in\mathbb{N}}$ be a sequence of pointwise bounded real-valued functions defined on $\kappa$ (i.e., for each $n\in\mathbb{N}$,
the set $\{f_{n}(\alpha):\alpha\in\kappa\}$ is bounded on $\mathbb{R}$). As $\textsf{M}(\kappa)$ holds, there exists a subsequence $(f_{n}:\kappa\to \mathbb{R})_{n\in A}$ such that for each $\alpha\in\kappa$ the sequence $(f_{n}(\alpha))_{n\in A}$ is eventually monotonic. Then, for each $\alpha\in\kappa$, there is an infinite subset $A_{\alpha}$ of $A$ such that $(f_{n}:\kappa\to \mathbb{R})_{n\in A_{\alpha}}$ is monotonic, consider $A^{\prime}=\bigcap_{\alpha<\kappa}A_{\alpha}$.
Can I conclude that $A^{\prime}$ is infinite and that $(f_{n}(\alpha))_{n\in A^{\prime}}$ is pointwise convergent (in fact, only I need that $(f_{n}(\alpha))_{n\in A^{\prime}}$ is bounded)? In case my reasoning is wrong, can someone give me an idea of how to fix it?
Thanks a lot
It seems unlikely that $A’$ would be infinite.
But since you only need pointwise convergence, does the statement not follow, because, or any $\alpha\in\kappa$, the sequence $(f_n(\alpha))_{n\in A}$ is eventually monotonic and therefore convergent (since it is also bounded)?
I think your notion of "pointwise bounded" is wrong -- don't you want ${f_n(\alpha) ,:, n \in \mathbb N}$ to be bounded for any given $\alpha$? If you fix $n$ and let $\alpha$ vary, as in your definition, then the constant functions $f_n(\alpha) = n$ satisfy your hypotheses, and BW$(\kappa)$ is false for every $\kappa > 0$.
Also, unless I'm misunderstanding your definition, I think $\mathsf{BW}(\kappa)$ is true if and only if $\kappa < \mathfrak{s}$. This follows from Theorem 3.2 in Andreas Blass' article in the Handbook of Set Theory (link: http://www.math.lsa.umich.edu/~ablass/hbk.pdf).
One more thing: I think $\mathsf{M}(\kappa)$ is true if and only if $\kappa < \mathrm{min}{\mathfrak{b},\mathfrak{s}}$. This follows from Theorem 3.5 in Andreas' article.
|
2025-03-21T14:48:30.093112
| 2020-03-19T02:16:52 |
355226
|
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|
Stack Exchange
|
Serre's theorem on global generations on stacks
Let $X$ be a quasi-projective scheme, the followings are quite useful.
Every coherent sheaf is globally generated after tensoring with a suitable line bundle.
Every coherent sheaf has trivial higher cohomology groups after tensoring with a suitable line bundle.
Every coherent sheaf is a quotient of a finite rank locally free sheaf.
Question
Are there any analogous discussions for Deligne- Mumford/Artin Stacks. My primary interests would be on the stacks of principal bundles/Higgs bundles on a curve.
Thank you in advance.
About 1 & 2, I doubt there are sensible results without some hypothesis like the existence an ample family of line bundles that makes the stack actually a scheme, in fact a so-called divisorial scheme.
As for 3, there is a paper that settles the issue, namely a quasi-compact and quasi-separated algebraic stack has affine stabilizer groups at closed points and satisfies the resolution property if and only if it is the quotient stack of a quasi-affine scheme by an action of $\mathop{GL}(n)$ for some $n$.
It is:
Gross, Philipp:
Tensor generators on schemes and stacks.
Algebr. Geom. 4 (2017), no. 4, 501–522.
ArXiv version: https://arxiv.org/abs/1306.5418
Tame Artin stacks (in the sense of Abramovich, Olsson and Vistoli, https://math.berkeley.edu/~molsson/tame.pdf) with quasi-projective moduli spaces will have property 2: the line bundle is the pullback of an ample line bundle on the moduli space.
As to property 1, a line bundle will not be enough, but you can get a version for quotient tame Artin stacks using a generating sheaf (in the sense of Olsson and Starr, https://math.berkeley.edu/~molsson/quot2a.pdf).
|
2025-03-21T14:48:30.093246
| 2020-03-19T02:29:28 |
355227
|
{
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"authors": [
"Mark Lewko",
"Sam Hopkins",
"Wlod AA",
"domotorp",
"https://mathoverflow.net/users/110389",
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|
Stack Exchange
|
How can we find n points on a plane so that as many pairs of points as possible have the same distance?
There are $n$ points on the plane, and we need to maximize the number of pairs of points which have the same Euclidean distance.
See work related to Erdos' Unit Distance Problem: https://en.wikipedia.org/wiki/Unit_distance_graph#Counting_unit_distances
$P(0)=P(1)=0; P(2)=1; P(3)=3; P(4)=5; P(5)=7; \ldots$
$P(6)=9; P(7)=12; \ldots$
I assume "unordered pairs of different points".
The subsection of the Wiki page @MarkLewko linked to changed its name to https://en.wikipedia.org/wiki/Unit_distance_graph#Number_of_edges
The number is tabulated at OEIS. It seems that it's only known up to $n=14$ (and some scattered larger values). Links are given there to some papers on the topic. Evidently, no one knows how to do it for general $n$.
Also discussed on math.stackexchange.
Hey, I am glad, thank you. I knew also P(8)=14 but didn't want to risk it.
There aren't any values bigger than 14 for which it would be known, see the table at the end of http://www1.pub.informatik.uni-wuerzburg.de/eurocg2020/data/uploads/papers/eurocg20_paper_85.pdf.
|
2025-03-21T14:48:30.093365
| 2020-03-19T02:36:33 |
355228
|
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"KhashF",
"Willie Wong",
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|
Stack Exchange
|
Obstruction to the existence of a globally defined integrating factor
Let $U$ be an open subset of $\Bbb{R}^n$ and take $\omega$ to be a nowhere-vanishing smooth $1$-form on $U$. The Frobenius Theorem implies that, near each point of $U$, $\omega$ may be written as $g\,{\rm{d}}f$ for suitable locally defined smooth functions $f$ and $g$ iff $\omega\wedge{\rm{d}}\omega=0$. Here is my question(s): For such a $1$-form $\omega$ on $U$ (never vanishing and $\omega\wedge{\rm{d}}\omega=0$) what is the obstruction to the existence of global smooth functions $f$ and $g$ on $U$ with $\omega=g\,{\rm{d}}f$? Can that be formulated as the vanishing of any homotopical invariant of $U$? What is a good non-example in which such a global presentation of $\omega$ fails to exist?
For non-examples, just take any closed but non-exact $\omega$ on $U$ that fails to be simply connected?
@WillieWong Well, in that case $\omega$ is not in the form of ${\rm{d}}f$, but why not a multiple of such a thing? For instance, $\omega:=-\frac{y}{x^2+y^2}{\rm{d}}x+\frac{x}{x^2+y^2}{\rm{d}}y$ is famously closed but not exact on $U:=\Bbb{R}^2-{(0,0)}$. But it is a multiple of ${\rm{d}}\left(\frac{y^2}{x^2+y^2}\right)$.
In your "counter example", your function $g$ is not smooth. (It is singular whenever $x = 0$ or $y = 0$.)
(More precisely, your function $g = \frac{x^2 + y^2}{2xy}$.)
To expand on my comment: suppose you have $U$ not simply connected, and $\omega$ closed but not exact, and suppose there exists a closed loop $\gamma: [0,1]\to U$ such that $\omega(\dot{\gamma})$ is signed. (This is in particular the case with the "example" in your comment, where you can take $\gamma$ to be any circle centered at the origin.)
Suppose $\omega= g df$ for smooth $g$ and $f$, then you must have
$$ g \nabla_{\dot\gamma} f $$
is signed. This implies both $g$ and $\nabla_{\dot\gamma} f$ are signed along $\gamma$, but this is absurd, since integrating from $0$ to $1$ you have
$$ \int_0^1 \nabla_{\dot\gamma(s)} f(\gamma(s)) ~ds = f(\gamma(1)) - f(\gamma(0)) = 0 $$
since $\gamma$ is a closed curve.
Simply-connectedness is not enough, however, to ensure that the integrating factor can be globalized. To see this, recall that Frobenius theorem states that the nonvanishing one-form $\omega$ satisfies $\omega\wedge d\omega = 0$ IFF its kernel is the tangent bundle of a regular foliation.
If $\omega = g ~df$ for some function $f$, then necessarily $f$ will be constant on the leaves of this foliation. Hence a counterexample will be found if you have a regular foliation of a simply connected compact manifold. (As on a compact manifold the function $f$ must attain a maximum and $df = 0$ there, contradicting the assumption that $\omega$ is non-vanishing.)
One such example is given by the Reeb foliation of the 3-sphere.
If we want to examine domains in Euclidean space: embed $\mathbb{S}^3 \hookrightarrow \mathbb{R}^4$ and slightly thicken it radially by $\epsilon$. Define a foliation on this 4 dimensional (simply-connected) domain by extending the Reeb foliation trivially in the radial direction. The lifted one-form $\tilde{\omega}$ is the pull-back of the Reeb $\omega$ from $\mathbb{S}^3$ by radial projection, and hence $\tilde{\omega} \wedge d \tilde{\omega} = 0$. Any $\tilde{f}$ that realizes $\tilde{\omega} = \tilde{g} d\tilde{f}$ will be constant on the foliation, and hence factors through some $f$ on $\mathbb{S}^3$, and the argument above shows that this contradicts the fact that $\omega$ is non-vanishing.
You seem to be saying that if the vector field corresponding to $\omega$ has a closed integral curve $\gamma(t)$, then $\omega$ cannot be written as $g,{\rm{d}f}$: Since $\omega$ is non-zero everywhere, $g$ should be either always positive or always negative. Now $f$ is strictly monotonic along $\gamma(t)$ (a Lyapunov function) due to the fact that
$\frac{\rm{d}}{\rm{d}t}f(\gamma(t))=g(\gamma(t)),\big|\nabla f(\gamma(t))\big|^2$
does not change sign. This cannot be the case since $\gamma$ is closed.
Thanks, it works. But here we assume something extra about the $1$-form $\omega$: There is a closed integral curve which of course prevents $\omega$ from being exact. I was looking for a topological property of the domain $U$. For instance, if $U$ is simply connected can we always write a non-vanishing $\omega$ with $\omega\wedge{\rm{d}}\omega=0$ as $g,{\rm{d}}f$?
"Closed integral curve" is not a requirement; that $\omega(\dot{\gamma})$ is signed is a weaker condition. // I don't think simply connected is enough; the Reeb foliation of $\mathbb{S}^3$ should be a counterexample. (The three-sphere is compact so there cannot be global $f$ with non-vanshing $df$.)
Can you please elaborate on what do you mean by $\omega(\dot{\gamma})$ being ``signed''?
$\dot\gamma$ is a vector field along $\gamma$. You pair it against $\omega$. You get a scalar. Nothing about this requires $\omega$ having a vector field corresponding to it (not using any Riemannian structures at all). In the case where you do have a Riemannian structure, you can imagine the corresponding vector field being perturbed to be spiralling (ever so slightly) instead of having closed integral curves.The circles are no longer integral curves but $\omega(\dot{\gamma})$ can still be signed.
To be more explicit: on an annulus, using the polar coordinates, instead of using $d\theta$ as your $\omega$, use $d\theta + \epsilon dr$ where $\epsilon$ is small.
Thanks, now I understand. As for your mention of Reeb foliation, I am looking at a domain in a Euclidean space, not a compact manifold. For instance, if $U$ is a simply connected (contractible) domain in $\Bbb{R}^2$, any $1$-form $\omega$ satisfies $\omega\wedge{\rm{d}}\omega=0$. One needs to work harder to find a nowhere-vanishing $\omega$ and a Jordan curve $\gamma$ in $U$ with $\omega(\dot{\gamma})$ signed since $\int_{\gamma}\omega=0$ if $\omega$ is closed (and hence exact).
What if you take a slightly thickened copy of $\mathbb{S}^3 \subset \mathbb{R}^4$ and extend the Reeb foliation trivially in the radial direction? If the leaves were defined as level sets of a function the function projects to a function on $\mathbb{S}^3$ and the same argument goes through, no?
I think that may work. Would you mind writing it as an answer? I think it would be interesting as it indicates that there are $1$-forms $\omega$ on simply connected domains of Euclidean spaces that satisfy the integrability condition $\omega\wedge{\rm{d}}\omega=0$ but lack a globally defined integrating factor. So the problem of formulating conditions on a domain that guarantee the existence of a globally defined integrating factor for any integrable $1$-form on the domain seems to be subtle.
|
2025-03-21T14:48:30.093775
| 2020-03-19T07:55:50 |
355234
|
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"Gerry Myerson",
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|
Stack Exchange
|
Maximal families of equal length intervals consist of equilateral triangles
My question is a follow up to How to find n points on a plane so that as many pair of points as possible have the same distance? -- see the conjecture at the bottom of this post.
Let $\ n\ $ be a positive integer. Let $\ e(A)\ $ be the maximal cardinality
of set $\ P\subseteq\binom A2\ $ such that distance $\ d(x\ y)\ $ is constant
over all $\ \{x\ y\}\in P.$
Let $\ D_n\in\binom{\Bbb R^2}n\ $ be such that
$$ \forall_{A\in\binom{\Bbb R^2}n}\quad e(D_n)\ge e(A) $$
Set $\ \Delta\subseteq\binom{D_n}2\ $ such that distance $\ d(x\ y)\ $ is constant
over all $\ \{x\ y\}\in \Delta\ $ and $\ |\Delta|=e(D_n)\ $ is called critical.
CONJECTURE: Let $\ n>2.\ $ Let $\ \Delta\in\binom{D_n}2\ $ be critical.
Then for every $\ \{x\ y\}\in\Delta\ $ there exists $\ z\in D_n\ $ such that
$\ \{x\ z\}\,\text{and}\, \{y\ z\}\,\in\,\Delta.$
I am sure that the following weaker version is indeed true. Namely, only
cardinalities $\ e(D_n)\ $ are defined uniquely but not sets $\ D_n.\ $ Thus I am confident that the conjecture is true for at least one $\ D_n\ $ (for every $n$)
but -- who knows -- the conjecture fails for some $\ D_n.$
Have a look at the diagrams at the end of http://www1.pub.informatik.uni-wuerzburg.de/eurocg2020/data/uploads/papers/eurocg20_paper_85.pdf and see whether this fails for $n=13$.
Following up on Gerry Myerson's suggestion, the following graph from Ágoston and Pálvölgyi's Improved constant factor for the unit distance
problem may be a counterexample.
This is the $n=13$ example illustrating the maximum number (30) of unit distances among 13 points in the plane. Schade's 1993 thesis establishes that this is the unique maximal "unit distance graph" for 13 points. Looking at the 3 neighbors of the lefthand marked vertex $x$ and the 5 neighbors of the righthand marked vertex $y$, you can see that there is no vertex $z$ adjacent to both $x$ and $y$.
|
2025-03-21T14:48:30.093930
| 2020-03-19T08:27:01 |
355235
|
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"Mare",
"Vladimir Dotsenko",
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|
Stack Exchange
|
Commutator of finite global dimension algebras
Let $A=KQ/I$ be a finite dimensional quiver algebra of finite global dimension.
Is it true that the dimension of $A/[A,A]$ is equal to the number of simples of $A$?
Here $[A,A]$ is the vector space generated by all elements of the form $ab-ba$.
Note that it is known that in general for such $A$ that the dimension of $A/([A,A]+rad(A))$ is equal to the number of simple $A$-modules. Thus the question should be equivalent to asking whether we have $rad(A) \subseteq [A,A]$ in case $A$ has finite global dimension.
I am maybe missing something, but what if your quiver has one vertex, and your algebra $A$ is the algebra of polynomials? What does your suggested result want to say then?
@VladimirDotsenko A quiver algebra is finite dimensional for me. I add that assumption, thanks.
Yes. See the result of Section 2.5 of a wonderful paper of Bernhard Keller :
https://webusers.imj-prg.fr/~bernhard.keller/publ/ilc.pdf
(and the references therein).
|
2025-03-21T14:48:30.094030
| 2020-03-19T09:13:44 |
355238
|
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|
Stack Exchange
|
What is the jacobian of an image lookup function?
I posted this question on Robotics Stack Exchange (link) but thought it could be relevant here as well.
I'm trying to solve a computer vision problem whereby I wish to use Levenberg–Marquardt non-linear optimization to solve the following equation:
$\left(\begin{array}{c}\mathbf{p}_{\min } \\ \mathbf{n}_{\min }\end{array}\right)=\underset{\mathbf{p}, \mathbf{n}}{\operatorname{Argmin}} \underbrace{\sum_{i=1}^{N}\left(I_{k+1}\left(\mathbf{x}_{i}\right)-I_{k}\left(\mathbf{H}_{k+1}(\mathbf{p}, \mathbf{n})\left(\mathbf{x}_{i}\right)\right)\right)^{2}}_{Q(\mathbf{p}, \mathbf{n})}$
Whereby $\mathbf{x}_i$ stands for the $x,y$ coordinates of a provided image, $I$ is a function which represents the grayscale value of the $x,y$ coordinate in the image. For instance, $I(r,c)$ refers to the pixel value at row $r$ and column $c$ in the image, which has fixed dimension e.g $1024 \times 768$. $H$ is the function, or $3\times 3$ homography matrix constructed from $\mathbf{p}$ and $\mathbf{n}$, converting vector $\mathbf{x}_i$ (coordinates $x_i, y_i$) into $\mathbf{x}_i'$ (coordinates $x_i', y_i')$.
Because gradient-related optimization requires the calculation of the Jacobian, can someone tell me how do I formulate the partial deriviatives of the Image function I? I understand that the size of the Jacobian is $n \times 2n$, where $n$ is the number of points to be used in the optimization.
The author of the paper did stated the following, but I did not quite get it.
To numerically compute the optimization, the loss function is further modified by introducing a linearization of the image function $I$ as we do not know directly in which way the image function itself changes when the normal vector $\mathbf{n}$ is modified. Thus, we use the first order term of its Taylor expansion.
Welcome to MO. As it is, little chances that anyone would be able to answer your question. Too many details are "hidden" under image processing jargon, which is not the usual language here. We do not know what $I$ and $H$ are explicitly, for example. Try rephrasing the question so that any analyst/applied mathematician can read it
thank you @AmirSagiv, I have added more details about I and H .
@goh it would help if you a) gave a link to the paper and b) typed out equations, rather than included images. I have only managed to do the second of these, Google doesn't return hits that match the quoted text exactly. I corrected an ungrammatical typo to read "I understand that the size of the Jacobian is $n\times 2n$", but I'm not 100% confident that's what you meant; it's the only thing that makes sense to me in the context, but I don't know what you know about this stuff.
I also added a link to the other question, and you should include a link to here in your Robotics.SE question.
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2025-03-21T14:48:30.094244
| 2020-03-19T09:37:38 |
355239
|
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|
Stack Exchange
|
Given a cubic fourfold is it possible to extend an isometry on the primitive cohomology to a complete marking?
Let $X$ be a (smooth) cubic fourfold and $F$ its Fano variety of lines.
It is known that $F$ is an hyperkähler fourfold. Let $(L,u)$ be an abstract lattice with a distinguished element $u$ isomorphic to $(H^4(X,\mathbb{Z}),h^2)$ , where $h^2$ is the class of a hyperplane section of $X$. We recall that a complete marking of $X$ is an isometry $H^4(X,\mathbb{Z}) \to L$ which sends $h^2$ to $u$. Moreover, let $(L',u')$ an abstract lattice, with a distinguished element $u'$, isomorphic to the pair $(H^2(F,\mathbb{Z}),g)$, where $g$ is again the class of a hyperplane section of $F$. It is known that the Abel-Jacobi map induces an isometry
$$
H^4(X,\mathbb{Z})_{pr} \to H^2(F,\mathbb{Z})_{pr}(-1)
$$
and also we may identify the primitive part of $L'$ (i.e. the orthogonal complement of $u'$ in $L'$) with the lattice $L_0(-1)$, where $L_0$ is the orthogonal complement of $u$ in $L$. Now, let $(F,\psi)$ be a complete marking of $F$, i.e. an isometry
$$
\psi \colon H^2(F,\mathbb{Z}) \to L' \text{ s.t. } g \mapsto u'
$$
which clearly induces an isometry
$$
\phi_0 \colon H^4(X,\mathbb{Z})_{pr} \to L_0.
$$
My questions are:
Is it possible to extend $\phi_0$ to a complete marking $\phi$ of $X$, i.e. an isometry $\phi \colon H^4(X,\mathbb{Z}) \to L$, with $\phi(h^2)=u$?
How is justified the existence of a complete marking for a cubic fourfold and for the variety of lines of a cubic fourfold?
If we start from a complete marking of a cubic fourfold, is it possible to "extend" this to a complete marking for its variety of lines?
|
2025-03-21T14:48:30.094370
| 2020-03-19T10:31:06 |
355240
|
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|
Stack Exchange
|
Is an automorphic form of $\operatorname{GL}_{n}(\mathbb{A}_{\mathbb{Q}})$ determined by its L-function?
To an automorphic representation $\pi$ of $\operatorname{GL}_{n}(\mathbb{A}_{\mathbb{Q}})$ one can associate its L-function $s\mapsto L_{\pi}(s)$.
Is the map $\pi\mapsto L_{\pi}$ bijective?
Edit March 26th, 2021: can such a bijection be seen as a fully faithful fonctor between Tannakian categories $\mathcal{C}$ whose objects are automorphic representations and $\mathcal{D}$ whose objects are L-functions suggesting every automorphism of a given object of the latter arises from an intertwinning operator? If so, can such an automorphism $\varphi$ be written as $\varphi_{\sigma}:L_{\pi}\mapsto\sigma\circ L_{\pi}\circ\sigma^{-1}=L_{\pi}\mapsto{L_{\sigma'\circ\pi\circ\sigma'^{-1}}}$ where $\sigma$ is a field automorphism of $\mathbb{C}$ and $\sigma'$ the corresponding intertwinning operator?
We are dealing with irreducible representations so it is generated by the $GL_n(A_Q)$-translates of one "vector" (which concretely is a left-$GL_n(Q)$-invariant function $GL_n(A_Q)\to C$), an automorphic form, that you can choose with some concrete desirable properties, translate, or make abstract by decomposing the representation into local parts.
I think it's true at least for cuspidal automorphic reps. If $L(s,\pi) = L(s,\pi')$, there should be some way to conclude that the local components $L(s,\pi_p) = L(s,\pi'_p)$ for almost all primes $p$. At most of the primes $\pi_p$ is unramified and determined by its local L-function. Then strong multiplicity one gives you $\pi = \pi'$.
Thank you. This should be also true for primitive L-functions, from Lemma 4.2 in M. Ram Murty, Selberg conjectures and Artin L-functions (1994).
Certainly true for irreducible cuspidal representations by the multiplicity one theorem: https://en.wikipedia.org/wiki/Multiplicity-one_theorem
@GHfromMO can you conclude the local L-functions are equal if the global L-functions are?
@D_S: Yes, since the local $L$-functions are just the Euler factors of the global $L$-function. To put another way, if you restrict the Dirichlet coefficients of $L(s,\pi)$ to powers of $p$, you get $L(s,\pi_p)$.
@GHfromMO : just out of curiosity, what happens for automorphic form over $\mathrm{GL}_n(\Bbb A_K)$, when $K$ is a number field?
@GHfromMO Not quite: For number fields, you can't necessarily recover the local $L$-factors from the global $L$-function -- for example if $K$ is Galois over $\mathbb Q$, a Galois twist of $\pi$ will have the same $L$-function as $\pi$, but the local $L$-factors are permuted. This has been studied, for example by Harry Smit.
@Peter Scholze: are these permutations of local L-factors the analogue of permutations of roots of the minimal polynomial of an algebraic number? If yes, can one deduce from it that any "symmetry" of an L-function is an element of some Galois group?
It's not a symmetry of any given $L$-function -- rather, two distinct $L$-functions happen to agree (by virtue of the local $L$-factors being permuted around). But I think the whole phenomenon is even more subtle than this. One way to think about it is that the $L$-function only knows about the automorphic induction from $K$ to $\mathbb Q$.
But aren't they the same as functions from $C$ to itself?
@PeterScholze: Thank you, I agree of course. I delete my comment regarding number fields.
|
2025-03-21T14:48:30.094609
| 2020-03-19T11:25:37 |
355242
|
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|
Stack Exchange
|
Complex cobordism and Chern numbers
Let $L$ be the Lazard's universal ring, and $R=\mathbb{Z}[b_1,b_2,\cdots,b_n,\cdots]$, regarded as a graded ring with the degree of $b_i$ equal to $2i$. Let $\theta: L\rightarrow R$ be the homomorphism carrying the universal formal group law $\mu^L$ to the formal group law
$$\mu^R(x_1,x_2)=\exp(\log(x_1)+\log(x_2)),$$
where the power series
$$\exp(x)=x+\sum_{i\geq 1}b_ix^{i+1},$$
and $\log(x)$ its inverse, denoted as
$$\log(x)=x+\sum_{i\geq 1}m_ix^{i+1}.$$
Let $MU$ be the complex cobordism spectrum, and by Quillen's theorem we have the following commutative diagram
$\require{AMScd}$
\begin{CD}
L @>\theta>> R\\
@V \cong V V @VV \cong V\\
\pi_*(MU) @>>h> H_*(MU;\mathbb{Z})
\end{CD}
where $h$ is the Hurewicz homomorphism.
In Section 9, Part II of
Adams, J. F., Stable homotopy and generalised homology, Chicago Lectures in Mathematics. Chicago - London: The University of Chicago Press. X, 373 p. 3.00 (1974). ZBL0309.55016.**
it is stated that the class $[\mathbb{C} P^n]\in\pi_*(MU)$ is sent to $(n+1)m_n\in H_*(MU;\mathbb{Z})$ by $h$, and it is indicated there that the argument is a Chern number computation, but I am not seeing the argument.**
I would greatly appreciate your help if you could sketch the proof or point out a reference containing a proof. Thank you!
I think the idea to make this a characteristic class computation is that the image of $[\mathbb{C}P^n]$ under the Hurewicz homomorphism is the same as pushing forward the fundamental class into $BU$ via classifying the tangent bundle, then using the Thom isomorphism to get a homology class in $MU$.
@Connor Malin Thank you for your helpful comment.
It is a key result that the composite
$$ MU_* \xrightarrow{h} H_*(MU;\mathbb Z) \xrightarrow[\sim]{\Phi^{\vee}}H_*(BU;\mathbb Z),$$
where $\Phi^{\vee}$ is the dual of the Thom isomorphism $\Phi$, agrees with evaluating on normal Chern numbers.
In other words, $\langle \Phi(c), h([M])\rangle = \bar c(M)$ for all $c \in H^*(BU)$ and for all $[M] \in MU_*$.
(A reference in the real case is the diagram on page 228 of A concise course in algebraic topology by J.P.May. The complex case is identical.)
So the assertion is just a Chern number calculation:
$h([\mathbb CP^n]) = (n+1)m_n$ if and only if, for all $c \in H^{2n}(BU;\mathbb Z)$, $$\langle \Phi(c), (n+1)m_n\rangle = \bar c(\mathbb CP^n).$$
Thank you for your answer and reference! They are very helpful.
In the first displayed line, should it be $\pi_(MU)$ instead of $MU_$?
@MichaelAlbanese The latter is standard notation for the former.
Good to know. Thanks.
|
2025-03-21T14:48:30.094932
| 2020-03-19T11:53:15 |
355243
|
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|
Stack Exchange
|
Full-rank Hadamard product given a certain structure
Let us assume that we have a full-rank randomly chosen $k\times (m\cdot l)$ matrix, $\boldsymbol{H}$, with
$l \leq k \leq (m\cdot l)$ and no specific structure (e.g., a realization of an IID complex Gaussian random matrix), and a full-rank $(m-r)\times m$ matrix, $\boldsymbol{X}$, which has at least one non-zero element per column and its last $(m-r)$ columns conform a non-singular matrix. Let us also assume that
\begin{equation}
k\cdot (m-r)\geq m\cdot l.
\end{equation}
We have noticed that
\begin{equation}
\mathrm{rank}\left( (\boldsymbol{X}^H\boldsymbol{X} \otimes \boldsymbol{1}_{l\times l}) \odot \boldsymbol{H}^H\boldsymbol{H} \right) = m\cdot l.
\end{equation}
is in general fulfilled (we have tested for $\boldsymbol{X}$ being realizations of an IID complex Gaussian random matrix). Here $(.)^H$ denotes conjugate transpose, $\otimes$ denotes Kronecker product, $\boldsymbol{1}_{l\times l}$ is the $l\times l$ all-ones matrix, and $\odot$ denotes Hadamard product.
We would like to know if there exists an $\boldsymbol{X}$, given the previous restrictions (i.e., full-rank and at least one non-zero element per column, and last $(m-r)$ columns linearly independent), for having
\begin{equation}
\mathrm{rank}\left( (\boldsymbol{X}^H\boldsymbol{X} \otimes \boldsymbol{1}_{l\times l}) \odot \boldsymbol{H}^H\boldsymbol{H} \right) < m\cdot l.
\end{equation}
In case there's no $\boldsymbol{X}$ fulfilling this, how would it be proved?
|
2025-03-21T14:48:30.095039
| 2020-03-19T12:41:58 |
355245
|
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|
Stack Exchange
|
Commutator of translation invariant operators on $L^2(\mathbb{R})$
I have a question concerning the commutator of translation invariant operators on $L^2(\mathbb{R})$.
Recall that $S:L^2(\mathbb{R})\to L^2(\mathbb{R})$ is translation invariant if $Su_t=u_tS$ for all $t\in\mathbb{R}$ where $(u_tf)(x)=f(x+t)$. The space of such operators is well-studied due for example to the work of Hörmander. The question is the following: Let
$$
\mathcal{A}:=\{S:L^2(\mathbb{R})\to L^2(\mathbb{R}):\ S\ \text{is translation-invariant}\},
$$
what is $\mathcal{A}'$ (the commutant of $\mathcal{A}$ inside $L^2(\mathbb{R})$)? Is there any explicit description of this space, and if so, do you have some literature sources for such results? Thank you very much in advance for your help.
Don't you mean the commutant in $B(L^2(R))$ rather than in $L^2(R)$?
@YCor Sorry of course I mean $B(L^2(\mathbb{R}))$.
Go to the Fourier transform picture. Then these become the operators that commute with multiplication by exponentials, which means they commute with all multiplication operators, which means you have just described the operators which become multiplication operators when you conjugate by the Fourier transform.
(I.e., $\mathcal{A}' = \mathcal{A}$.)
Thank you for the clarification. Is there a way to see that $\mathcal{A}'=\mathcal{A}$ without using the Fourier transform? In fact one has that if I apply the Fourier transform to $\mathcal{A}$ this space becomes $L^{\infty}(\mathbb{R})$. How can one use this?
"Is there a way to see ...?" --- in principle, yes, but I think you'd have to build up a lot of machinery that's immediately available for multiplication operators. "How can one use this?" --- sorry, I don't understand, use it for what?
|
2025-03-21T14:48:30.095194
| 2020-03-19T14:47:37 |
355248
|
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|
Stack Exchange
|
Integral transformation, Laplace-like
Is the following integral transformation of $f$ known (for suitable $f$ and $s\in\mathbb{C}$)?
$$
\int_1^\infty f(t) \frac{e^{-ts}}{1-e^{-ts}}dt
$$
It resembles somewhat the Laplace transformation.
What about properties, references …?
Expand the ratio of exponentials as a geometric series in exponentials. This is called a Lambert series. Interchange sum and integral.
Thanks a lot! Nice and simple idea.
Also looks like it will have something to do with Bernoulli polynomials.
For what it worth, in the terms of divergent integrals, your transform can be rewritten as
$$\operatorname{reg} \int_1^\infty f(t)e^{-t s \omega _-}dt$$
Looks like some kind of analog of Fourier transform, if you ask me...
fixed the typo, the previous form was of course, meaningless
|
2025-03-21T14:48:30.095287
| 2020-03-19T15:58:54 |
355250
|
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|
Stack Exchange
|
A characterisation of weakly symmetric algebras
A finite dimensional algebra $A$ over a field $K$ is called a Frobenius algebra in case there exists a $K$-linear map $f: A \rightarrow K$ such that $ker(f)$ contains no non-zero right ideal of $A$ (such an $f$ is called trace of $A$). $A$ is called symmetric in case additionally $f(xy)=f(yx)$ for all $x,y \in A$.
$A$ is called weakly-symmetric in case every indecomposable projective module $P$ has the property that $top(P)=soc(P)$.
Symmetric implies weakly symmetric, which implies being Frobenius. But the other directions are not true in general.
Question: Is there a characterisation when a Frobenius algebra $A$ is weakly-symmetric using the trace map $f$?
Some experiments suggest that maybe in case $A$ is weakly-symmetric we have that $f(xy)=s_{x,y} f(yx)$ for some non-zero $s_{x,y} \in K$ for all $x,y$. Is this true?
In case this is true this would prove an old guess by me that being weakly symmetric over the field with two elements implies that the algebra is symmetric (since in this case the values of $f$ are either 0 or 1).
Note that there is the following connection between $f$ and the socles $soc(P)$ of the indecomposable projective modules $P$ in case $A=KQ/I$ is a quiver algebra:
$soc(e_i A)=<a_i>$ is 1-dimensional and spanned by a maximal path $a_i$ (I guess it can always be choosen to be a path, or could it be that it is not possible and it must be a sum of paths with coefficients?) and then $f(a_i)=1$ and $f(r)=0$ for all $r$ in a basis of $A$ that contains the elements $a_i$.
So at least for quiver algebras, the problem is very combinatorial.
My guess is that it would be equivalent to some condition on Nakayama automorphism, for instance, that the Nakayama automorphism acts trivially on the set of simple modules by twisting them.
|
2025-03-21T14:48:30.095429
| 2020-03-19T16:11:30 |
355251
|
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|
Stack Exchange
|
Characterizing the Haagerup property of finite von Neumann algebras via unbounded derivations
A correspondence $_{N} H_{N}$ is a Hilbert space with two normal commuting left/right representations of $N$ in $B(H)$. The following characterization of property (T) for finite von Neumann algebras was proved in [Pe].
Theorem ([Pe,Theorem 3.2.]) Let $N$ be a factor and $N_0 \subset N$ weakly-$\ast$ dense, unital $\ast$-subalgebra containing a
non-$\Gamma$-set*. Then, the following are equivalent:
$N$ has property (T).
Let $H = _{N}H_N$ be any correspondence. Any a priori unbounded closable derivation $\delta: D(\delta) \subset N \to H$ defined over a
domain $N_0 \subset D(\delta)$ is inner.
The intuition in the theorem above is that derivations $\delta$ over a correspondence are the
von Neumann analogue of $1$-cocycles over a unitary representation. The characterization in point 2 is thus the analogue of the characterization of property (T) groups as those whose $1$-cocyles are all inner, which is due to Delorme/Guichardet.
There is a notion of Haagerup property for factors that was introduced in [Ch].
Question: Is there a characterization of the Haagerup approximation property for finite von Neumann algebras in terms of unbounded derivations?
A group has the Haagerup approximation property iff it admits a metrically proper $1$-cocycle $\beta: G \to H$, ie a cocycle such that $\| \beta(g_n) \| \to \infty$ as $g_n$ escapes from every compact subset of $G$. What is the analogue of properness for cocycles?
A characterization of Haagerup property for von Neumann algebras in terms of quantum Markov semigroups has appeared in [JM]. I guess the natural thing to do would be to use the theory developed by Cipriani/Sauvageot to obtain a derivation from the Markovian semigroup and try to characterize which derivations correspond to semigroups of compact operators. Is this characterization known already in the literature?
A guess would be those derivations such that, for every unitary $u \in U(N_0)$ such that $u^k \to 0$ weakly, $\| \delta(u^k) \| \to \infty$.
[*] $N_0$ contains a non-$\Gamma$ set if there is a finite $F \subset N_0$ and a $K > 0$ such that for all $\xi \in L^2(N , \tau )$, we have $\| \xi - \tau(\xi) \|_2^2 \leq K \, \sum_{x \in F} \|[\xi,x]\|_2^2$, i.e.: a set implementing the failure of property $\Gamma$.
[JM] Jolissaint, Paul; Martin, Florian, Finite von Neumann algebras with Haagerup property and (L^2)-compact semigroups, Bull. Belg. Math. Soc. - Simon Stevin 11, No. 1, 35-48 (2004). ZBL1068.46037.
[Ch] Choda, Marie, Group factors of the Haagerup type, Proc. Japan Acad., Ser. A 59, 174-177 (1983). ZBL0523.46038.
[Pe] Peterson, Jesse, A 1-cohomology characterization of property (T) in von Neumann algebras, Pac. J. Math. 243, No. 1, 181-199 (2009). ZBL1178.22010.
For separable von Neumann algebras the Haagerup property for $N$ is equivalent to the existence of a real closable derivation $\delta$ such that $\delta^*\delta$ has compact resolvents in $\mathcal B(L^2(N))$. This is true even in the non-tracial case. See Theorem 7.7 in:
Martijn Caspers, Adam Skalski, The Haagerup Approximation Property for von Neumann Algebras via Quantum Markov Semigroups and Dirichlet Forms, Commun. Math. Phys. 336, 1637–1664 (2015)
Thanks for the reference! Let me think a little bit whether this can be expressed in terms of unitaries before accepting it as an answer.
Finding natural equivalent conditions to the Haagerup property based on the behavior of a derivation on unitaries might be difficult. This seems likely to be connected to the still open problem of whether the Haagerup property is equivalent to the compact approximation property. See Definition 4.13 from the paper: Anantharaman-Delaroche, Amenable correspondences and approximation properties for von Neumann algebras, Pacific J. of Math. 171, no. 2, 309-341 (1995).
|
2025-03-21T14:48:30.095682
| 2020-03-19T16:13:43 |
355252
|
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|
Stack Exchange
|
An inequality in harmonic analysis with the BMO flavour
I am asking myself this question (which seems to be a natural generalization of Remark 4.4 of these lecture notes).
Question. Let $I_s, s \in \mathcal{S}$ be a collection of intervals included in
$[0,1]$ such that there exists $K>0$ such that for every
interval $J$ there holds $$\frac{1}{|J|} \sum_{I_s\subset J} |I_s|\leqslant K\,,$$ then is it true that the function $f:=\sum_{s\in\mathcal{S}} \mathbf1_{I_s}$ is in $\bigcap_{p \geqslant 1}L^p$? Is it also true that $|\{x \in [0,1]\,, f(x)\geqslant n\}| \lesssim e^{-cn}$ for some $c>0$?
I have the strong intuition that the assumption on the intervals implies that the function $f$ is not big on large measure sets.
A first guess would be that $f$ lies in $\mathit{BMO}$ (i.e., has bounded mean oscillation) but as pointed out by fedja it is not the case. If we take the family of intervals $[1/2,1/2+2^{-j}]$ we see that it defines a function which is roughly $x \mapsto \log (x-1/2) \mathbf{1}_{x>1/2}$ which is not in $\mathit{BMO}$ (see the averages on $[1/2-\delta, 1/2+\delta]$).
However, if we define $\alpha_n$ as the measure of the $x\in [0,1]$ such that $f(x)=n$ (i.e. $x$ lies in $n$ intervals $I_s$), I thing that we can prove an estimate of the form $\alpha_n \lesssim e^{-cn}$, and for that, it is sufficient to prove that for large $n$ there is a positive real $b<1$ such that $\alpha_{n+1} \leqslant b \alpha_n$.
In order to prove the last statement I tried to select the dyadics intervals $J$ on which $\frac{1}{|J|}\int_J f$ is large or not, and run pretty much the same argument as in the proof of the John-Nirenberg inequality [see Grafakos "Modern Fourier Analysis" page 124]. However the problem is that even if we have:
$$\frac{1}{|J|}\int_J f = \frac{1}{|J|}\sum_{\cup I_s \cap J} |I_s|\,,$$ the hypothesis does not say something about this quantity ... (if $I_s$ is not included in $J$ then this interval does not appear in the hypothesis ...).
Can someone give me a hand for proving this claim? I really believe that the proof should be an adaptation of the John-Nirenberg inequality.
Edit. I edited the question after comments.
Something is fishy: consider the collection of open intervals centered at $1/2$. Then the condition is almost void. Are you sure those are not dyadic intervals or that the condition is not imposed with $J$ being a union of 2 adjacent dyadic intervals?
Well, I realize that it is not clear. In fact in the lecture notes I just attached a 'tiling' is defined with dyadic intervals so all the intervals involved are in fact dyadic. However we can cahnge the question to make it more interesting by letting $J$ be any interval!
But then there is something I do not understand with the Remark 4.4 in the attached lecture notes. If any interval involved are dyadic then by Lebesgue differentiation theorem we get that $f$ is bounded and thus in any $L^p$ ... which is strange.
@fedja I edited the question so that the hypothesis concerns any interval $J$, thus ruling out your example (in your case we can take $J=[1/2-\eta,1/2+\eta]$...
The BMO understood literally is still out of question: Just consider the family $[0.5,0.5+2^{-j}]$ and look at a very short interval centered at $0.5$. So, you need to be a bit more inventive. Hint: replace each interval by a trapezoid function of comparable width; then the sum will be in BMO, indeed.
@fedja you are right. However with the family $[1/2,1/2+2^{-j}]$ we have $|f|_{L^p} < \infty$ for any finite $p$. So we should not expect $f$ to lie in BMO but I expect that $\alpha _n \lesssim e^{-cn}$ anyway. I could not find any counterexample to that estimate ... I really feel that the condition on any interval $J$ constrained the measure of sets on which $f$ is large, but I do not find any way of making that a rigorous statement.
|
2025-03-21T14:48:30.095952
| 2020-03-19T16:51:53 |
355256
|
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|
Stack Exchange
|
First Dirichlet eigenvalue below second Neumann eigenvalue?
Let $\Omega$ be a bounded domain in $\mathbb R^n $ with smooth boundary.
I was wondering if there exist any known conditions on $\Omega$ such that the 1st Dirichlet eigenvalue of the (positive) Laplacian is below the $2$nd Neumann eigenvalue, i.e.
$$\lambda_1^{\operatorname{Neumann}}(\Delta) \le \lambda_1^{\operatorname{Dirichlet}}(\Delta)$$
is clear but when do we also have
$$\lambda_1^{\operatorname{Dirichlet}}(\Delta) \le \lambda_2^{\operatorname{Neumann}}(\Delta)? $$
You have always for $n \ge 2$ that $\lambda_{n+1}^{Neumann} <\lambda_n^{Dirichlet}$. The easiest prrof is due to Filonov (st. Petersburgh Math. J. 2005).
|
2025-03-21T14:48:30.096027
| 2020-03-19T16:59:01 |
355257
|
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"Pol van Hoften",
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|
Stack Exchange
|
Quotienting $G(\mathbb{Q})_{+}$ by $G^{\text{sc}}(\mathbb{Q})$ and inner forms
Let $G/\mathbb{Q}$ be a connected reductive group, let $G^{\text{ad}}$ be the adjoint group, let $G^{\text{der}}$ be the derived group and let $\rho\colon G^{\text{sc}} \to G^{\text{der}}$ be the simply connected cover. Let $G^{\text{ad}}(\mathbb{R})^{0}$ be the identity component (in the real topology) of $G^{\text{ad}}(\mathbb{R})$, let $G(\mathbb{R})_{+}$ be the inverse image of $G^{\text{ad}}(\mathbb{R})^{0}$ under the natural map $G(\mathbb{R}) \to G^{\text{ad}}(\mathbb{R})$ and let $G(\mathbb{Q})_{+}$ be the intersection of $G(\mathbb{R})_{+}$ with $G(\mathbb{Q}$). Note that $\rho(G^{\text{sc}}(\mathbb{R})) \subset G(\mathbb{R})_{+}$ because $G^{\text{sc}}(\mathbb{R})$ is connected.
Question: Let $H/\mathbb{Q}$ be an inner form of $G$ and let the notation be as above, is there a `natural isomorphism' of abelian groups $H(\mathbb{Q})_{+}/H^{\text{sc}}(\mathbb{Q}) \simeq G(\mathbb{Q})_{+}/G^{\text{sc}}(\mathbb{Q})$?
When $G^{\text{sc}}=G^{\text{der}}$ this is true: Let $Z$ be the center of $G$, let $\nu:G \to D$ be the maximal abelian quotient of $G$, define $D(\mathbb{R})^{\dagger}:=\operatorname{Im}(Z(\mathbb{R}) \to D(\mathbb{R}))$ and let $D(\mathbb{Q})^{\dagger}=D(\mathbb{R})^{\dagger} \cap D(\mathbb{Q})$. Then Lemma 5.10 of https://www.jmilne.org/math/xnotes/svi.pdf shows that
\begin{align}
\nu(G(\mathbb{Q})_{+}) = D^{\dagger}(\mathbb{Q}).
\end{align}
Note that because $H$ is an inner form of $H$, we can identify its center with $Z$ and its maximal abelian quotient with $\mu:H \to D$. Applying the lemma again we see that
\begin{align}
\mu(H(\mathbb{Q})_{+})=D^{\dagger}(\mathbb{Q})
\end{align}
and we are done. I have tried to follow a similar strategy in the general case, by which I mean comparing to $Z(\mathbb{Q})/Z^{\text{sc}}(\mathbb{Q})$, but I havent been able to get it to work.
The answer is Yes. We denote $K(G)=G({\mathbb Q})_+/\rho G^{\rm sc}({\mathbb Q})$.
We compute $K(G)$; see the corollary below.
It is clear from the corollary that $K(G)$ is canonically isomorphic to $K(H)$.
We will use Section 3 of M. Borovoi, Abelian Galois cohomology of reductive groups. Memoirs of the AMS 132 (1998), No. 626,
although all necessary results can be found in Deligne's paper
Variétés de Shimura: interprétation modulaire, et techniques de construction de modèles canoniques,
Proc. Sympos. Pure Math. 33, Part 2, pp. 247–289.
We consider the crossed module $(G^{\rm sc}\to G)$ and the hypercohomology
$$H^0_{\rm ab}({\mathbb Q},G):=H^0({\mathbb Q},G^{\rm sc}\to G),$$
where $G$ is in degree 0;
see the Memoir. By definition $H^0_{\rm ab}({\mathbb Q},G)$ is a group.
We consider the abelian crossed module $(Z^{\rm sc}\to Z)$, where $Z=Z(G)$ and
$Z^{\rm sc}=Z(G^{\rm sc})$.
The morphism of crossed modules
$$(Z^{\rm sc}\to Z)\,\longrightarrow\,(G^{\rm sc}\to G)$$
is a quasi-isomorphism, and hence it induces a bijection on hypercohomology,
permitting us to identify $H^0_{\rm ab}({\mathbb Q},G)$ with the abelian group $H^0({\mathbb Q},Z^{\rm sc}\to Z)$.
We conclude that $H^0_{\rm ab}({\mathbb Q},G)$ is naturally an abelian group and that it does not change under inner twisting of $G$.
The short exact sequence
$$1\to(1\to G)\to (G^{\rm sc}\to G)\to (G^{\rm sc}\to 1)\to 1$$
(where $(G^{\rm sc}\to 1)$ is not a crossed module) induces a hypercohomology exact sequence
$$ G^{\rm sc}({\mathbb Q})\to G({\mathbb Q})\to H^0_{\rm ab}({\mathbb Q},G)\to H^1({\mathbb Q},G^{\rm sc}),$$
where
$${\rm ab}^0\colon G({\mathbb Q})\to H^0_{\rm ab}({\mathbb Q},G)$$
is the abelianization map.
This permits us to identify $G({\mathbb Q})/\rho G^{\rm sc}({\mathbb Q})$ with the kernel
$${\rm ker}[H^0_{\rm ab}({\mathbb Q},G)\to H^1({\mathbb Q},G^{\rm sc})]$$
(yes, this kernel is a subgroup of the abelian group $H^0_{\rm ab}({\mathbb Q},G)$ ).
This kernel might change under inner twisting of $G$, because $H^1({\mathbb Q},G^{\rm sc})$ changes under inner twisting.
By definition,
$G({\mathbb R})_+=Z({\mathbb R})\cdot\rho G^{\rm sc}({\mathbb R})$, and hence
$$G({\mathbb R})_+/\rho G^{\rm sc}({\mathbb R})={\rm ab}^0(Z({\mathbb R}))\subset
{\rm ker}[ H^0_{\rm ab}({\mathbb R},G)\to H^1({\mathbb R}, G^{\rm sc})].$$
We see that $K(G):=G({\mathbb Q})_+/\rho G^{\rm sc}({\mathbb Q})$ can be identified with the preimage of ${\rm ab}^0(Z({\mathbb R}))\subset H^0_{\rm ab}({\mathbb R},G)$ in ${\rm ker}[H^0_{\rm ab}({\mathbb Q},G)\to H^1({\mathbb Q},G^{\rm sc})]$.
Lemma. The preimage of ${\rm ab}^0(Z({\mathbb R}))\subset H^0_{\rm ab}({\mathbb R},G)$ in ${\rm ker}[H^0_{\rm ab}({\mathbb Q},G)\to H^1({\mathbb Q},G^{\rm sc})]$
coincides with the preimage of ${\rm ab}^0(Z({\mathbb R}))$ in $H^0_{\rm ab}({\mathbb Q},G)$.
Proof.
Let $\xi\in H^0_{\rm ab}({\mathbb Q},G)$ lie in the preimage of
$${\rm ab}^0(Z({\mathbb R}))\subset
{\rm ker}[ H^0_{\rm ab}({\mathbb R},G) \to H^1({\mathbb R}, G^{\rm sc})].$$
Then the image of $\xi$ in $H^1({\mathbb R},G^{\rm sc})$ is trivial, and therefore,
the image of $\xi$ in $H^1({\mathbb Q},G^{\rm sc})$ lies in the kernel of the localization map
$$ H^1({\mathbb Q}, G^{\rm sc})\to H^1({\mathbb R},G^{\rm sc}).$$
By the Hasse principle for simply connected groups, this kernel is trivial.
Thus the image of $\xi$ in $H^1({\mathbb Q},G^{\rm sc})$ is trivial, and hence $\xi$ lies in the preimage of ${\rm ab}^0(Z({\mathbb R}))$ in ${\rm ker}[H^0_{\rm ab}({\mathbb Q},G)\to H^1({\mathbb Q},G^{\rm sc})]$, as required.
Corollary. The abelianization map ${\rm ab}^0\colon G({\mathbb Q})\to H^0_{\rm ab}({\mathbb Q},G)$ with kernel $\rho G^{\rm sc}({\mathbb Q})$ induces a canonical isomorphism between the abelian groups $K(G):=G({\mathbb Q})_+/\rho G^{\rm sc}({\mathbb Q})$
and the preimage of ${\rm ab}^0(Z({\mathbb R}))\subset H^0_{\rm ab}({\mathbb R},G)$ in $H^0_{\rm ab}({\mathbb Q},G)$.
Thank you, this is exactly what I was looking for!
|
2025-03-21T14:48:30.096572
| 2020-03-19T19:35:31 |
355261
|
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|
Stack Exchange
|
Stability of weak solutions to quasilinear parabolic equations
Consider the quasilinear operator $A(x,t,\nabla u)$ satisfying $$A(x,t,\nabla u).\nabla u \geq C_0 |\nabla u|^p$$ and $$|A(x,t,\nabla u)| \leq C_1 |\nabla u|^{p-1}$$ where $1<p<\infty$. Note that there is no assumption made regarding $<A(x,t,\zeta)-A(x,t,\eta).\zeta-\eta>$, in particular, there is no comparison principle. Moreover, $A(x,t,\nabla u)$ is just measurable function.
Suppose $u \in C^0(0,T; L^2(\Omega)) \cap L^p(0,T; W^{1,p}(\Omega))$ is a local weak solution of $$u_t - \text{div}A(x,t,\nabla u)=0.$$
My question is the following: I need to find a sequence of nonlinearities $A_{\epsilon}(x,t,\zeta)$ satisfying the same structure conditions as $A(x,t,\zeta)$, but having the following properties:
1) $A_{\epsilon}$ is smooth.
2) Let $u_{\epsilon}$ be the unique weak solution of
$$
\begin{array}{l}
(u_{\epsilon})_t - \text{div} A_{\epsilon}(x,t,\nabla u_{\epsilon}) = 0\quad \text{on} \quad Q\\
u_{\epsilon} = u \quad \text{on} \quad \partial_p Q
\end{array}
$$
where $Q$ is a fixed cylinder and $\partial_p Q$ denotes the parabolic boundary of $Q$. Then $u_{\epsilon}$ converges to $u$ locally in $Q$.
Basically, how do I get a local smooth approximation to weak solutions of quasilinear parabolic equations?
|
2025-03-21T14:48:30.096680
| 2020-03-19T21:15:30 |
355264
|
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"Asvin",
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|
Stack Exchange
|
Torsion in the jacobian of a super elliptic curve
Let $y^n = f(x)$ define a smooth projective curve $C$ over some field $k$ with $\deg f \geq n$ and odd and with $f(x)$ having no repeated roots. Let $J$ be the Jacobian of $C$ and $J[n]$ it's (geometric) n-torsion. Then is it true that the points $x= x_i, y=0$ generate the group $J[n]$ for the roots $x_i$ of $f(x)$?
Is there a similar explicit description if $f$ happens to have repeated roots?
This article (http://math.mit.edu/~varul/torsion.pdf) seems relevant
@JacksonMorrow Thank you for the suggestion! All his papers seem relevant.
As mentioned by the previous answers, this cannot be true for $n \ge 3$ by size considerations. When you identify the points $(x_i, 0)$ of $C$ inside its jacobian $J$, you are implicitly using some base-point. I will assume that $n$ and $\deg f$ are coprime, so that there is exactly one rational point at infinity (which I will denote by $\infty$). So you are considering the subgroup of $J[n]$ generated by the divisor classes
$$
[(x_i, 0) - \infty] \in J[n].
$$
Let $\zeta_n$ be a primitive $n$th root of unity and use $\zeta$ to denote both (i) the automorphism of $C$ defined by $(x, y) \mapsto (x, \zeta_n y)$ and (ii) the corresponding automorphism this induces on $J$. Note that each of the aforementioned divisor classes $[(x_i, 0) - \infty]$ are fixed by $\zeta$ and hence
$$
[(x_i, 0) - \infty] \in J[1 - \zeta].
$$
When $f$ is separable, it turns out that the $[(x_i, 0) - \infty]$ generate the subgroup $J[1 - \zeta]$ of $J[n]$; for a proof, see Proposition 3.2 of
(1) Schaefer, Edward F. "Computing a Selmer group of a Jacobian using functions on the curve." Math. Ann 310 (1998): 447-471.
which is also available as arxiv:1507.08325. This proof might actually even work when $f$ is not assumed to be separable (I couldn't immediately see anything that breaks if the separability assumption is dropped).
When $n = p$ is prime, it turns out that $J[(1 - \zeta)^{p - 1}] = J[p]$; this is shown in Section 3 of (1). There has been some work on attempting to understand generators for/the field of definition of $J[(1 - \zeta)^{i}]$. Using my "division by $1 - \zeta$ formula" (arxiv:1810.07299), one can compute generators when $i \le 2$ and the field of definition for $i \le 3$ (this work does not assume that $n$ is prime, but when $n$ is not prime, it is not necessarily true anymore that $J[(1 - \zeta)^{i}] \subseteq J[n]$ for $i \le n - 1$). I also compute the field of definition of $J[(1 - \zeta)^{i}]$ for $i \le p$ in the specific case $y^p = x^q + 1$ (see arxiv:1910.14251, Theorem 3.6 5.1.6).
Additionally, the case $y^p = u^s (1 - u)$ for $1 \le s \le p - 2$ is studied in
(2) Greenberg, Ralph. "On the Jacobian variety of some algebraic curves." Compositio Mathematica 42.3 (1980): 345-359.
(3) Tzermias, Pavlos. "Explicit rational functions on Fermat curves and a theorem of Greenberg." Compositio Mathematica 122.3 (2000): 337-345.
Letting $J_{p, s}$ be the jacobian of $y^p = u^s (1 - u)$ and $K$ be the cyclotomic field $K = \mathbf{Q}(\zeta_p)$, Tzermias's Theorem 1 in (3) (which is a combination of work by Greenberg (2), Gross-Rohrlich, Kurihara) classifies exactly "how much" of the $J_{p, s}[p^\infty]$ and $J_{p, s}[\ell^\infty]$ for $\ell \neq p$ is defined over $K$; the answer is that $J_{p, s}[p^\infty](K) = J_{p, s}[(1 - \zeta)^3]$ and $J_{p, s}[\ell^\infty](K) = \{ 0 \}$ except in the special cases $(\ell, p, s) \in \{ (2, 7, 2), (2, 7, 4) \}$. In addition, Tzermias in (3) computes generators for $J_{p, s}[(1 - \zeta)^i]$ for $i \le 3$.
Not in general. $J[n]$ has $n^{2g}$ elements, where $g$ is the genus. If $n$ and $\deg f$ are relatively prime, then $g=(n-1)(\deg f-1)/2$. The points you ask for generate a subgroup of order $n^{\deg f-1}$ (the upper bound is easy, for equality see my preprint). So for $n\geq 3$ $J[n]$ has too many points for your claim to hold.
Do you know of any explicit description of the complementary $n^{n-1}$ order subgroup? Even for special cases?
You don't have enough points to generate the group if $n \ge 3$. Moreover, if $f(x)$ has rational roots, then all the points on the Jacobian you write down are rational, whereas $\mathbf{Q}(J[n])$ contains $\mathbf{Q}(\zeta_n) \ne \mathbf{Q}$ for $n > 2$.
|
2025-03-21T14:48:30.097065
| 2020-03-19T21:30:03 |
355265
|
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|
Stack Exchange
|
Can the dimension of Hom space between vector bundles on an algebraic curve predicted by Riemann-Roch type formula be the minimal possible?
Let us study vector bundles $E$ and $F$ on a smooth projective curve $C$. There is a Riemann-Roch type formula for the Euler characteristic $\chi(E,F)=dim\, Hom(E,F)-dim\, Ext^1(E,F)$ in terms of degrees $d(E),d(F)$, ranks $r(E),r(F)$ and the genus $g$ of $C$. Assume that this formula gives a non-positive number (this is equivalent to $\mu(F)-\mu(E)\le g-1$).
Is it true that for generic $E$ and $F$ with such ranks and degrees one has $Hom(E,F)=0$? Or is it true at least for some $E,F$?
(This is clearly true for line bundles, but I failed to answer the question for arbitrary vector bundles after several days of thinking)
I have found the answer, it is positive.
The statement is equivalent to a theorem of Hirschowitz, see Th. 1.2 in arXiv:alg-geom/9710019v2.
The paper you mention is published: B. Russo, M. Teixidor,
On a conjecture of Lange,
J. Algebraic Geom. 8 (1999), no. 3, 483–496.
|
2025-03-21T14:48:30.097170
| 2020-03-19T21:32:28 |
355267
|
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"Monroe Eskew",
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"url": "https://mathoverflow.net/questions/355267"
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|
Stack Exchange
|
What's the consistency strength of resemblance + global failure of the continuum hypothesis?
Let $T$ be a theory formalized in first order logic with equality and membership and the additional primitive constant symbol $W$, with the following axioms:
Extensionality: $\forall z (z \in x \leftrightarrow z \in y) \to x=y$
Class comprehension: if $\phi$ is a formula in which $x$ doesn't occur free, then all closures of: $$\exists x \forall y (y \in x \leftrightarrow \phi \land \exists z (y \in z) ) $$; are axioms.
Define: $x=V \iff \forall y (\exists z (y \in z) \to y \in x)$
Sub-world: $W \in V$
Resemblance: if $\varphi^X$ is a formula in prenex normal form, whose matrix is in $L(=,\in)$, and each variable in its prefix must appear as bounded either $\in X$ or $\subseteq X$, then: $$\varphi^V \leftrightarrow \varphi^W $$, is an axiom.
All axioms of "$\sf ZF + \text {CH fails everywhere}"$ relativized to $W$.
What's the consistency strength of $T$?
Why do you keep asking this question?
@MonroeEskew, that's the first time I asked this question. I never asked about the effect of resemblance on global failure of choice.
Zuhair, I think @Monroe means that a lot of your questions are quite similar: take a theory, vary some axiom scheme, ask for consistency strength. To a professional set theorist, these questions are kind of odd. We don't see the picture in your head that motivates this sort of research, and at some point it starts to become repetitive for no good reason. We know that you're a hobbyist, and I think it's great people are interested in this on a hobbyist level. But seeing how this is not what set theorists normally do, seeing a continuous flow of these questions is just... befuddling.
@AsafKaragila, I'm interested in axiomatics for their own sake, not for a motivation the lies prior to them (like capturing an informal notion, construction, etc...), the motivation of an axiomatic system is itself, so I only need to know its relative consistency to other axiomatics, and see what it can bear. That is hobby, but it doesn't mean that it is not serious. I want to investigate these various axiomatics, especially those that are connected to recombination of known axiomatic systems, since otherwise there is endless axiomatics. Thanks!
@AsafKaragila for example here in this question, I want to know how far can resemblance go! Can it add strength to something that is known to be stronger than it, which is failure of continuum hypothesis everywhere; can indescribability (what resemblance is about) on top of failure of CH everywhere also yield stronger results? or it becomes idle at such altitude? I expect such a question to be easy for a professional set theorist interested in large cardinals to answer.
Based on the fact that among your last 30 questions only 8 received an answer and you get +6 summing all of their up and downvotes, the other users of the website don't appear to be interested in such questions for their own sake. As Asaf was saying those theories appear arbitrary/unmotivated, usually set theorists start by studying a problem interesting by itself which then turns out to have high consistency strength, they don't sit around randomly modifying axioms until they stumble in an interesting one. Of course you can ask any question, but if you never get answers you should wonder why
It seems that you are just imagining different axiom systems and telling us about them, while putting the question, "What is the consistency strength?" at the end to justify these being mathoverflow questions. I think this is not an appropriate use of mathoverflow, because it is not really about posing and solving mathematical problems. Moreover, "What is the consistency strength?" is not in general a well-defined question because the set of all theories is not linearly ordered or well-founded in the relative consistency partial order.
https://cs.nyu.edu/pipermail/fom/1998-August/001929.html
@MonroeEskew, No! I'm asking about the consistency strength to find if it is consistent or not, and to see in which of the known systems it is equi-interpretable. For example here I'm asking specifically about whether resemblance would increase the strength of failure of CH everywhere when added on top of it? This is a question. My aim is not to show you an axiomatic system as you think, its not really a hard job to come up with such axiomatics, I'm presenting that problem axiomatically in order for it to be presented it in rigorous terms, and I think that's how it should be presented.
My bad, I was thinking about regular cardinals
|
2025-03-21T14:48:30.097485
| 2020-03-19T21:36:40 |
355269
|
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|
Stack Exchange
|
Closed points of a closed subscheme of $\mathbb{P}^n$ over the residue field and the fraction field of a valuation ring $R$
Let $(R, M)$ be a valuation ring with algebraically closed fraction field $k$. Let $L = R/M$ be the residue field of $R$; it follows that $L$ is algebraically closed. I would like to understand the following: Suppose $Z$ is an irreducible closed subscheme of $\mathbb{P}^n_k$ defined by the homogeneous ideal $A \subset k[X_0, ..., X_n]$. Let $B = \pi\{ A \cap R[X_0, ..., X_n] \}$ where $\pi$ is the quotient modulo $M$. Let $W \subset \mathbb{P}^n_L$ be defined by $B$. Suppose $[\bar{a_0}:...: \bar{a_n}]$ is a closed point of $W$. Then I would like to know how one can prove that there exit $b_0, ..., b_n \in R$ (not all in $M$) such that $b_j + M = \bar{a_j}$ for each $0 \leq j \leq n$ and $[b_0:...: b_n]$ is a closed point of $Z$.
I am asking this question because I would like to understand the proof of the following result.
In Mumford's 'Red book of schemes', Theorem 1 Chapter II Section 8, he proves:
For all closed subsets $Z \subset \mathbb{P}^n_k$, there is a unique closed subset
$W \subset \mathbb{P}^n_L$ such that
$$
\rho(Z(k)) = W(L),
$$
where $\rho: \mathbb{P}^n(k) \to \mathbb{P}^n(L)$.
I am having similar issues with the proof as in this MathSE question. If someone could provide an alternative reference for this result or an explanation for the statement in the first paragraph, it would be appreciated. Thank you
Let me first reformulate this question slightly. Set $S = \text{Spec}(R)$, denote $\eta \in S$ the generic point (with residue field $k$) and denote $s \in S$ the closed point (with residue field $L$). We may assume $Z$ is not only irreducible but also reduced as the question is about points. I assume that $A \subset k[X_0, \ldots, X_n]$ is exactly the set of homogeneous polynomials vanishing on $Z$. Thus we see that $A$ is a homogeneous prime ideal. The homogeneous prime ideal $A \cap R[X_0, \ldots, X_n]$ defines an integral closed subscheme $\overline{Z} \subset \mathbf{P}^n_R$ whose generic fibre is $Z$ and whose closed fibre is $W$. We are going to think of $\overline{Z}$ as a scheme over $S$. Picture:
$$
\overline{Z} \longrightarrow S
\quad\text{with}\quad
\overline{Z}_\eta = Z \quad\text{and}\quad \overline{Z}_s = W
$$
Denote $\xi \in Z = \overline{Z}_\eta$ the generic point. Clearly, $\xi$ is also the generic point of $\overline{Z}$. Denote
$w = [\overline{a}_0 : \ldots : \overline{a}_n]$ the closed point of $W = \overline{Z}_s$ given to us. Then we have a specialization
$$
\xi \leadsto w
$$
which maps to the specialization $\eta \leadsto s$ in $S$. What you are asking for is to show that there is a closed point $z$ of $Z = \overline{Z}_\eta$ and specializations $\xi \leadsto z \leadsto w$.
(Hints: Observe that a closed point of $Z$ always is a $k$-rational point because $k$ is algebraically closed and that a $k$-rational point of $\mathbf{P}^n_k$ is the same as an $R$-rational point of $\mathbf{P}^n_R$ by the valuative criterion for projective space, so this will give you the sequence $b_0, \ldots, b_n \in R$ not all in $M$ well defined up to a unit in $R$, which specializes to $[\overline{b}_0 : \ldots : \overline{b}_n]$ in the closed fibre.)
The existence of $z$ is a general fact about the topology of finite type morphisms of schemes! AFAIK this property in general was first discovered by Brian Osserman and Sam Payne in a paper entitled "Lifting tropical intersections". In the Stacks project you can find it here.
On the other hand, I am pretty sure that you don't need this result in order to prove the statement in Mumford's book. In fact, I just looked at his proof of this fact and it uses just a little bit of algebraic geometry (projections) and going down for finite morphisms to prove this particular case of the Osserman-Payne result (the proof is on the bottom of page 130 in my edition). So I urge you to stick to this case and read what Mumford wrote there.
|
2025-03-21T14:48:30.097760
| 2020-03-19T22:04:31 |
355271
|
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"Ali Taghavi",
"Salvatore Siciliano",
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"url": "https://mathoverflow.net/questions/355271"
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|
Stack Exchange
|
When a finite codimensional subalgebra contains a finite codimension ideal?
What is a classification of all algebras $A$ (purely algebraic algebras, Banach or $C^*$ algebras or Lie algebras) with the following property:
Every finite codimensional subalgebra $B$ of $A$ contains a subalgebra $I$ such that $I$ is an ideal in $A$ and $A/I$ is a finite dimensional algebra.
One can ask a similar question for "Rings" but consider finite quotient rather than finite codimensionality in algebra case.
This question is inspired by this post and its comments conversation
@MarkSapir The motivation for this question is the pure group theoric problem written in the attached link you find in this pist "every finit index subgroup contains a finit index normal sibgroup". By classification I mean some results as "An algebra has this property if ......" or " An algebra has this property if and only if it satisfies......". Or some examples or counter examples.
@MarkSapir I think you are collecting some (and not all ) sufficient conditions which looks trivial. A classification is a "iff" theorem. For example I think it is the case for commutative unital $C^*$ algebra. Do you have a counter example? Moreover what about Lie algebra case?
In my opinion, a possible classification of Lie algebras with the required property is hopeless. However, it is worth mentioning that examples of infinite-dimensional simple Lie algebras containing subalgebras of finite codimension were constructed by Amayo in a paper published in the Proc. Lond. Math. Soc. in 1976. On the other hand, by a Theorem of Kukin, if a restricted Lie algebra $L$ over a field of characteristic $p>0$ contains a restricted subalgebra of finite codimension, then $L$ also contains a restricted ideal of finite codimension,
@SalvatoreSiciliano Thank you very much for your very interesting comment. I was not aware of the concept "restricted Lie algebra"
|
2025-03-21T14:48:30.097921
| 2020-03-19T22:43:01 |
355274
|
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|
Stack Exchange
|
Minors of low rank skew-symmetric matrix
Let $A$ be an $n\times n$ skew-symmetric matrix of rank $r$.
Given subsets $X$ and $Y$ of row and column indices respectively, let $A_{X,Y}$ denote the submatrix of $A$ obtained by only keeping rows with indices in $X$ and columns with indices in $Y$.
Prove that for any subsets $X, Y\subseteq \{1, 2, \ldots, n\}$ each of size $r$, we have
$$\det A_{X,X} \cdot \det A_{Y,Y} = (-1)^r (\det A_{X,Y})^2.$$
I've heard that this theorem is due to Frobenius, but have not been able to track down a reference that proves this result.
Since no one else has posted a complete answer so far, let me give one.
Note that it is only complete in the sense of answering the OP's question;
several other questions arise that I cannot easily address.
In Section 1, I will prove the main result (Theorem 1), which is more general
than the OP's equality. In Section 2, I will derive the latter from the
former. In Sections 3 and 4, I will generalize the statement and ask further questions.
1. On size-$r$ minors of a rank-$r$ matrix
We fix a field $\mathbb{K}$. (In Section 3, we will generalize this to a
commutative ring.)
Let $\mathbb{N}=\left\{ 0,1,2,\ldots\right\} $. For each $n\in\mathbb{N}$,
let $\left[ n\right] $ denote the set $\left\{ 1,2,\ldots,n\right\} $.
If $A=\left( a_{i,j}\right) _{1\leq i\leq n,\ 1\leq j\leq m}\in
\mathbb{K}^{n\times m}$ is an $n\times m$-matrix (for some $n,m\in\mathbb{N}
$), and if $I\subseteq\left[ n\right] $ and $J\subseteq\left[ m\right] $
are arbitrary subsets, then $A_{I,J}$ will denote the submatrix $\left(
a_{i_{x},j_{y}}\right) _{1\leq x\leq p,\ 1\leq y\leq q}$ of $A$, where the
subsets $I$ and $J$ have been written as $I=\left\{ i_{1}<i_{2}<\cdots
<i_{p}\right\} $ and $J=\left\{ j_{1}<j_{2}<\cdots<j_{q}\right\} $. (Thus,
$A_{I,J}$ is the matrix obtained from $A$ upon removing all rows other than
the rows indexed by elements of $I$ and removing all columns other than the
columns indexed by elements of $J$.)
The crucial result is the following:
Theorem 1. Let $n,m,r\in\mathbb{N}$. Let $A\in\mathbb{K}^{n\times m}$ be a
matrix such that $\operatorname*{rank}A\leq r$. Let $X$ and $Y$ be subsets of
$\left[ n\right] $, and let $U$ and $V$ be subsets of $\left[ m\right] $;
assume that $\left\vert X\right\vert =\left\vert Y\right\vert =\left\vert
U\right\vert =\left\vert V\right\vert =r$. Then,
\begin{align*}
\det\left( A_{X,U}\right) \cdot\det\left( A_{Y,V}\right) =\det\left(
A_{X,V}\right) \cdot\det\left( A_{Y,U}\right) .
\end{align*}
Theorem 1 is surprisingly easy to prove using the following two basic facts:
Lemma 2. Let $n,m,r\in\mathbb{N}$. Let $A\in\mathbb{K}^{n\times m}$ be a
matrix such that $\operatorname*{rank}A\leq r$. Then, there exist two matrices
$B\in\mathbb{K}^{n\times r}$ and $C\in\mathbb{K}^{r\times m}$ such that $A=BC$.
Proof of Lemma 2. The particular case of Lemma 2 when $\operatorname*{rank}
A=r$ is a well-known result in elementary linear algebra, but since the
general case is rarely stated, let me sketch the proof (for the general case):
Let $\operatorname*{Col}A$ denote the span of the columns of $A$. This is a
$\mathbb{K}$-vector subspace of $\mathbb{K}^{n\times1}$. Its dimension is
$\dim\left( \operatorname*{Col}A\right) =\operatorname*{rank}A\leq r$. In
other words, the $\mathbb{K}$-vector space $\operatorname*{Col}A$ has
dimension $\leq r$. Thus, this $\mathbb{K}$-vector space has a basis with at
most $r$ elements. Therefore, this vector space can be generated by exactly
$r$ elements (just take a basis with at most $r$ elements, and insert the zero
vector enough times to get exactly $r$ generators). In other words, there
exist $r$ vectors $v_{1},v_{2},\ldots,v_{r}$ that span the vector space
$\operatorname*{Col}A$. Consider these $v_{1},v_{2},\ldots,v_{r}$.
Now, let $B\in\mathbb{K}^{n\times r}$ be the matrix whose $r$ columns are
$v_{1},v_{2},\ldots,v_{r}$. For each $i\in\left\{ 1,2,\ldots,m\right\} $, we
have
\begin{align*}
\left( \text{the }i\text{-th column of }A\right) & \in\operatorname*{Col}
A\qquad\left( \text{by the definition of }\operatorname*{Col}A\right) \\
& =\operatorname*{span}\left( v_{1},v_{2},\ldots,v_{r}\right)
\end{align*}
(since $v_{1},v_{2},\ldots,v_{r}$ span $\operatorname*{Col}A$); thus, there
exist scalars $c_{i,1},c_{i,2},\ldots,c_{i,r}\in\mathbb{K}$ such that
\begin{align*}
\left( \text{the }i\text{-th column of }A\right) =c_{i,1}v_{1}+c_{i,2}
v_{2}+\cdots+c_{i,r}v_{r}.
\end{align*}
Consider these scalars $c_{i,1},c_{i,2},\ldots,c_{i,r}$. Define the matrix
$C\in\mathbb{K}^{r\times m}$ by $C=\left( c_{j,i}\right) _{1\leq i\leq
r,\ 1\leq j\leq m}$. Then, it is easy to see that $A=BC$. This proves Lemma 2.
$\blacksquare$
Lemma 3. Let $n,m,r\in\mathbb{N}$. Let $B\in\mathbb{K}^{n\times r}$ and
$C\in\mathbb{K}^{r\times m}$ be two matrices. Let $X\subseteq\left[ n\right]
$ and $U\subseteq\left[ m\right] $ be two subsets. Then,
\begin{align*}
\left( BC\right) _{X,U}=B_{X,\left[ r\right] }C_{\left[ r\right] ,U}.
\end{align*}
Proof of Lemma 3. Straightforward entry-by-entry verification (using the
definition of matrix multiplication). $\blacksquare$
Proof of Theorem 1. Lemma 2 yields that there exist two matrices
$B\in\mathbb{K}^{n\times r}$ and $C\in\mathbb{K}^{r\times m}$ such that
$A=BC$. Consider these $B$ and $C$. Now, Lemma 3 yields $\left( BC\right)
_{X,U}=B_{X,\left[ r\right] }C_{\left[ r\right] ,U}$. In view of $A=BC$,
this rewrites as $A_{X,U}=B_{X,\left[ r\right] }C_{\left[ r\right] ,U}$.
But the matrix is $B_{X,\left[ r\right] }$ is square (since $\left\vert
X\right\vert =r=\left\vert \left[ r\right] \right\vert $), and so is the
matrix $C_{\left[ r\right] ,U}$ (since $\left\vert U\right\vert
=r=\left\vert \left[ r\right] \right\vert $); hence, $\det\left(
B_{X,\left[ r\right] }C_{\left[ r\right] ,U}\right) =\det\left(
B_{X,\left[ r\right] }\right) \cdot\det\left( C_{\left[ r\right]
,U}\right) $. In view of $A_{X,U}=B_{X,\left[ r\right] }C_{\left[
r\right] ,U}$, this rewrites as
\begin{equation}
\det\left( A_{X,U}\right) =\det\left( B_{X,\left[ r\right] }\right)
\cdot\det\left( C_{\left[ r\right] ,U}\right) .
\label{eq.darij1.pf.t1.AXU}
\tag{1}
\end{equation}
Similar reasoning shows that
\begin{align}
\det\left( A_{X,V}\right) & =\det\left( B_{X,\left[ r\right] }\right)
\cdot\det\left( C_{\left[ r\right] ,V}\right) ;
\label{eq.darij1.pf.t1.AXV}
\tag{2}
\\
\det\left( A_{Y,U}\right) & =\det\left( B_{Y,\left[ r\right] }\right)
\cdot\det\left( C_{\left[ r\right] ,U}\right) ;
\label{eq.darij1.pf.t1.AYU}
\tag{3}
\\
\det\left( A_{Y,V}\right) & =\det\left( B_{Y,\left[ r\right] }\right)
\cdot\det\left( C_{\left[ r\right] ,V}\right) .
\label{eq.darij1.pf.t1.AYV}
\tag{4}
\end{align}
Multiplying the equalities \eqref{eq.darij1.pf.t1.AXU} and
\eqref{eq.darij1.pf.t1.AYV}, we obtain
\begin{align*}
\det\left( A_{X,U}\right) \cdot\det\left( A_{Y,V}\right) & =\det\left(
B_{X,\left[ r\right] }\right) \cdot\det\left( C_{\left[ r\right]
,U}\right) \cdot\det\left( B_{Y,\left[ r\right] }\right) \cdot\det\left(
C_{\left[ r\right] ,V}\right) \\
& =\underbrace{\det\left( B_{X,\left[ r\right] }\right) \cdot\det\left(
C_{\left[ r\right] ,V}\right) }_{\substack{=\det\left( A_{X,V}\right)
\\\text{(by \eqref{eq.darij1.pf.t1.AXV})}}}\cdot\underbrace{\det\left(
B_{Y,\left[ r\right] }\right) \cdot\det\left( C_{\left[ r\right]
,U}\right) }_{\substack{=\det\left( A_{Y,U}\right) \\\text{(by
\eqref{eq.darij1.pf.t1.AYU})}}}\\
& =\det\left( A_{X,V}\right) \cdot\det\left( A_{Y,U}\right) .
\end{align*}
This proves Theorem 1. $\blacksquare$
2. The skew-symmetric case
Recall that a square matrix $A\in\mathbb{K}^{n\times n}$ is said to be
skew-symmetric if it satisfies $A^{T}=-A$. Now you claim:
Theorem 4. Let $A\in\mathbb{K}^{n\times n}$ be a skew-symmetric matrix.
Let $r\in\mathbb{N}$ be such that $\operatorname*{rank}A\leq r$. Let $X$ and
$Y$ be subsets of $\left[ n\right] $ such that $\left\vert X\right\vert
=\left\vert Y\right\vert =r$. Then,
\begin{align*}
\det\left( A_{X,X}\right) \cdot\det\left( A_{Y,Y}\right) =\left(
-1\right) ^{r}\cdot\left( \det\left( A_{X,Y}\right) \right) ^{2}.
\end{align*}
Proof of Theorem 4. The matrix $A$ is skew-symmetric; thus, $A^{T}=-A$.
Now,
\begin{align*}
\left( A_{X,Y}\right) ^{T} & =\left( A^{T}\right) _{Y,X}=\left(
-A\right) _{Y,X}\qquad\left( \text{since }A^{T}=-A\right) \\
& =-A_{Y,X}.
\end{align*}
Therefore, $\det\left( \left( A_{X,Y}\right) ^{T}\right) =\det\left(
-A_{Y,X}\right) =\left( -1\right) ^{r}\det\left( A_{Y,X}\right) $ (since
$A_{Y,X}$ is an $r\times r$-matrix). But the determinant of a matrix does not
change when the matrix is transposed. Hence,
\begin{align*}
\det\left( A_{X,Y}\right) =\det\left( \left( A_{X,Y}\right) ^{T}\right)
=\left( -1\right) ^{r}\det\left( A_{Y,X}\right) .
\end{align*}
Now, Theorem 1 (applied to $U=X$ and $V=Y$) yields
\begin{align*}
\det\left( A_{X,X}\right) \cdot\det\left( A_{Y,Y}\right) =\det\left(
A_{X,Y}\right) \cdot\det\left( A_{Y,X}\right) .
\end{align*}
Comparing this with
\begin{align*}
& \left( -1\right) ^{r}\cdot\left( \det\left( A_{X,Y}\right) \right)
^{2}\\
& =\left( -1\right) ^{r}\cdot\det\left( A_{X,Y}\right) \cdot
\underbrace{\det\left( A_{X,Y}\right) }_{=\left( -1\right) ^{r}\det\left(
A_{Y,X}\right) }\\
& =\left( -1\right) ^{r}\cdot\det\left( A_{X,Y}\right) \cdot\left(
-1\right) ^{r}\det\left( A_{Y,X}\right) \\
& =\underbrace{\left( -1\right) ^{r}\cdot\left( -1\right) ^{r}}_{=\left(
-1\right) ^{2r}=1}\cdot\det\left( A_{X,Y}\right) \cdot\det\left(
A_{Y,X}\right) =\det\left( A_{X,Y}\right) \cdot\det\left( A_{Y,X}\right)
,
\end{align*}
we obtain
\begin{align*}
\det\left( A_{X,X}\right) \cdot\det\left( A_{Y,Y}\right) =\left(
-1\right) ^{r}\cdot\left( \det\left( A_{X,Y}\right) \right) ^{2}.
\end{align*}
This proves Theorem 4. $\blacksquare$
Note that if $\mathbb{K}$ is a field of characteristic $\neq2$, then the rank
of any skew-symmetric matrix $A\in\mathbb{K}^{n\times n}$ is even. This is a
known fact, implicit in Keith Conrad, Bilinear
Forms
(combine Theorem 1.6 and formula (5.3) in this document).
3. Generalizing to commutative rings
Now, let us generalize our situation. Instead of requiring $\mathbb{K}$ to be
a field, we now merely assume that $\mathbb{K}$ be a commutative ring. Does
Theorem 1 still hold? The answer will depend on how we define
$\operatorname*{rank}A$. There is no "obviously right" definition of "rank" of
a matrix over a commutative ring, but there are several contenders. We look
specifically for a notion that generalizes the statement
"$\operatorname*{rank}A\leq r$". Two candidates are the following:
We say that a matrix $A\in\mathbb{K}^{n\times m}$ has strong rank $\leq r$
(for some $r\in\mathbb{N}$) if there exist two matrices $B\in\mathbb{K}
^{n\times r}$ and $C\in\mathbb{K}^{r\times m}$ such that $A=BC$.
We say that a matrix $A\in\mathbb{K}^{n\times m}$ has weak rank $\leq r$
(for some $r\in\mathbb{N}$) if every two $\left( r+1\right) $-element
subsets $X\subseteq\left[ n\right] $ and $Y\subseteq\left[ m\right] $
satisfy $\det\left( A_{X,Y}\right) =0$.
It is easy to see (using the Cauchy--Binet theorem) that every matrix that has
strong rank $\leq r$ must also have weak rank $\leq r$. The converse is not
true (in fact, for $n=2$ and $r=1$, it boils down to the pre-pre-Schreier condition).
Our above proof of Theorem 1 obviously proves the following generalization:
Theorem 5. Let $n,m,r\in\mathbb{N}$. Let $A\in\mathbb{K}^{n\times m}$ be a
matrix that has strong rank $\leq r$. Let $X$ and $Y$ be subsets of $\left[
n\right] $, and let $U$ and $V$ be subsets of $\left[ m\right] $; assume
that $\left\vert X\right\vert =\left\vert Y\right\vert =\left\vert
U\right\vert =\left\vert V\right\vert =r$. Then,
\begin{align*}
\det\left( A_{X,U}\right) \cdot\det\left( A_{Y,V}\right) =\det\left(
A_{X,V}\right) \cdot\det\left( A_{Y,U}\right) .
\end{align*}
But we also have the following more general fact:
Theorem 6. Let $n,m,r\in\mathbb{N}$. Let $A\in\mathbb{K}^{n\times m}$ be a
matrix that has weak rank $\leq r$. Let $X$ and $Y$ be subsets of $\left[
n\right] $, and let $U$ and $V$ be subsets of $\left[ m\right] $; assume
that $\left\vert X\right\vert =\left\vert Y\right\vert =\left\vert
U\right\vert =\left\vert V\right\vert =r$. Then,
\begin{align*}
\det\left( A_{X,U}\right) \cdot\det\left( A_{Y,V}\right) =\det\left(
A_{X,V}\right) \cdot\det\left( A_{Y,U}\right) .
\end{align*}
The only way I can currently prove Theorem 6 is by reducing it to Theorem 5
using the "meta-theorem" saying that every polynomial identity in the entries
of a matrix that holds (universally, i.e., over every $\mathbb{K}$) for all
matrices having strong rank $\leq r$ must also hold for all matrices having
weak rank $\leq r$. This "meta-theorem" follows from some standard monomial
theory, specifically the result that the quotient of the coordinate ring of
$\mathbb{K}^{n\times m}$ modulo the ideal generated by all $\left(
r+1\right) \times\left( r+1\right) $-minors can be embedded into the
coordinate ring of $\mathbb{K}^{n\times r}\times\mathbb{K}^{r\times m}$ via
the $A=BC$ substitution (with $B\in\mathbb{K}^{n\times r}$ and $C\in
\mathbb{K}^{r\times m}$). The latter fact is proved implicitly in Section 5 of
Richard G. Swan, On the straightening law for minors of a matrix, https://arxiv.org/abs/1605.06696v1.
Question 7. Is there a "natural" proof of Theorem 6?
4. Pfaffians
If we are already looking at skew-symmetric matrices, it isn't a long stretch
to move on to alternating matrices any more.
Recall that a square matrix $A\in\mathbb{K}^{n\times n}$ is said to be
alternating if it is skew-symmetric (i.e., satisfies $A^{T}=-A$) and all its
diagonal entries are $0$. If the field $\mathbb{K}$ has characteristic $\neq
2$, then every alternating matrix is skew-symmetric; but this is not true in
characteristic $2$. Any alternating matrix $A\in\mathbb{K}^{n\times n}$ has a
well-defined Pfaffian
$\operatorname*{Pf}A\in\mathbb{K}$ satisfying $\det A=\left(
\operatorname*{Pf}A\right) ^{2}$. Note that $\operatorname*{Pf}A=0$ if $n$ is
odd. Properties of Pfaffians come up in the literature all the time; there is
a MathOverflow question collecting references on
them.
(It is probably worth adding the recent preprint arXiv:2008.04247v1.)
It is now not too hard to show the following:
Theorem 8. Let $\mathbb{K}$ be a field. Let $n,r\in\mathbb{N}$. Let
$A\in\mathbb{K}^{n\times n}$ be an alternating matrix such that
$\operatorname*{rank}A\leq r$. Let $X$ and $Y$ be subsets of $\left[
n\right] $ such that $\left\vert X\right\vert =\left\vert Y\right\vert =r$.
Then,
\begin{align*}
\operatorname*{Pf}\left( A_{X,X}\right) \cdot\operatorname*{Pf}\left(
A_{Y,Y}\right) =\det\left( A_{X,Y}\right) .
\end{align*}
Indeed, this relies on the following analogue of Lemma 2:
Lemma 9. Let $\mathbb{K}$ be a field. Let $n,r\in\mathbb{N}$. Let
$A\in\mathbb{K}^{n\times n}$ be an alternating matrix such that
$\operatorname*{rank}A\leq r$. Then, there exist a matrix $S\in\mathbb{K}
^{r\times n}$ and an alternating matrix $B\in\mathbb{K}^{r\times r}$ such that
$A=S^{T}BS$.
Proof of Lemma 9 (rough outline). We will use the results of the
(expository) paper Keith Conrad, Bilinear
Forms.
We translate the claim into the language of alternating bilinear forms. Set
$V=\mathbb{K}^{n\times1}$, and let $f:V\times V\rightarrow\mathbb{K}$ be the
alternating bilinear form corresponding to $A$. (Thus, explicitly, it is the
form $\left( x,y\right) \mapsto x^{T}Ay$.) Let $R$ be the subspace $\left\{
v\in V\ \mid\ f\left( v,w\right) =0\text{ for all }w\in V\right\}
=\operatorname*{Ker}A$ of $V$. (This is often called the radical of $f$.)
Then, $f$ descends to an alternating bilinear form $\overline{f}$ on the
quotient space $V/R$. Choose any basis $\left( w_{1},w_{2},\ldots
,w_{g}\right) $ for the latter quotient space; then,
\begin{align*}
g=\dim\left( V/R\right) =\underbrace{\dim V}_{=n}-\dim\underbrace{R}
_{=\operatorname*{Ker}A}=n-\dim\left( \operatorname*{Ker}A\right)
=\operatorname*{rank}A\leq r.
\end{align*}
Let $B_{0}\in\mathbb{K}^{g\times g}$ be the alternating matrix that represents
the bilinear form $\overline{f}$ on this basis $\left( w_{1},w_{2}
,\ldots,w_{g}\right) $. Let $S_{0}\in\mathbb{K}^{g\times n}$ be the matrix
that represents the projection map $V\rightarrow V/R$ with respect to the
standard basis of $V=\mathbb{K}^{n\times1}$ and the basis $\left( w_{1}
,w_{2},\ldots,w_{g}\right) $ of $V/R$. Then, it is easy to see that
$A=S_{0}^{T}B_{0}S_{0}$. By adding $r-g$ zero rows and columns to $B_{0}$ and
$r-g$ zero rows to $S_{0}$, we obtain two matrices $B\in\mathbb{K}^{r\times
r}$ and $S\in\mathbb{K}^{r\times n}$ such that $B$ is alternating and such
that $A=S^{T}BS$. This proves Lemma 9. $\blacksquare$
[Remark: Lemma 9 can be made a lot stronger if we assume
$\operatorname*{rank}A=r$. Indeed, in this case, we can say that $r$ is even,
and that $A$ can be written in the form $A=S^{T}JS$, where $S\in
\mathbb{K}^{r\times n}$ is some matrix, and where $J$ is the "standard"
skew-symmetric $r\times r$-matrix (i.e., the block matrix $\left(
\begin{array}
[c]{cc}
0 & I_{r/2}\\
-I_{r/2} & 0
\end{array}
\right) $). This follows from the fact that the alternating bilinear form
$\overline{f}$ in the above proof of Lemma 9 is non-degenerate and thus has a
symplectic basis. (And this basis has size $g=r$, because
$\operatorname*{rank}A=r$.)]
Lemma 10. Let $r\in\mathbb{N}$. Let $B\in\mathbb{K}^{r\times r}$ be an
alternating matrix. Let $K\in\mathbb{K}^{r\times r}$ be any matrix. Then,
\begin{align*}
\operatorname*{Pf}\left( K^{T}BK\right) =\det K\cdot\operatorname*{Pf}B.
\end{align*}
For proofs of Lemma 10, see (e.g.) Teorema 8.6.3 (1) in Marco Manetti,
Algebra lineare, per matematici,
2020-09-13.
A more general fact -- which relates to Lemma 10 in the same way as the
Cauchy-Binet formula relates to the classical $\det\left( UV\right) =\det
U\cdot\det V$ -- is Ishikawa/Wakayama's "Minor Summation Formula" as stated in
Masao Ishikawa, Masato Wakayama, Minor Summation Formula of Pfaffians,
Linear and Multilinear Algebra 39 (1995), pp. 285--305.
Proof of Theorem 8 (rough outline). Lemma 9 yields that there exist a matrix
$S\in\mathbb{K}^{r\times n}$ and an alternating matrix $B\in\mathbb{K}
^{r\times r}$ such that $A=S^{T}BS$. Consider these $S$ and $B$.
From $A=S^{T}BS$, it is easy to see that
\begin{equation}
A_{X,Y}=\underbrace{\left( S^{T}\right) _{X,\left[ r\right] }}_{=\left(
S_{\left[ r\right] ,X}\right) ^{T}}BS_{\left[ r\right] ,Y}=\left(
S_{\left[ r\right] ,X}\right) ^{T}BS_{\left[ r\right] ,Y}.
\label{eq.darij1.pf.t8.4}
\tag{5}
\end{equation}
Hence,
\begin{align}
\det\left( A_{X,Y}\right) & =\det\left( \left( S_{\left[ r\right]
,X}\right) ^{T}BS_{\left[ r\right] ,Y}\right) =\underbrace{\det\left(
\left( S_{\left[ r\right] ,X}\right) ^{T}\right) }_{=\det\left(
S_{\left[ r\right] ,X}\right) }\cdot\det B\cdot\det\left( S_{\left[
r\right] ,Y}\right) \nonumber\\
& =\det\left( S_{\left[ r\right] ,X}\right) \cdot\det B\cdot\det\left(
S_{\left[ r\right] ,Y}\right) .
\label{eq.darij1.pf.t8.5}
\tag{6}
\end{align}
Similarly to \eqref{eq.darij1.pf.t8.4}, we also obtain
\begin{align*}
A_{X,X}=\left( S_{\left[ r\right] ,X}\right) ^{T}BS_{\left[ r\right]
,X},
\end{align*}
so that
\begin{align*}
\operatorname*{Pf}\left( A_{X,X}\right) =\operatorname*{Pf}\left( \left(
S_{\left[ r\right] ,X}\right) ^{T}BS_{\left[ r\right] ,X}\right)
=\det\left( S_{\left[ r\right] ,X}\right) \cdot\operatorname*{Pf}B
\end{align*}
(by Lemma 10). Likewise,
\begin{align*}
\operatorname*{Pf}\left( A_{Y,Y}\right) =\det\left( S_{\left[ r\right]
,Y}\right) \cdot\operatorname*{Pf}B.
\end{align*}
Multiplying the preceding two equalities and recalling that $\left(
\operatorname*{Pf}B\right) ^{2}=\det B$, we obtain
\begin{align*}
\operatorname*{Pf}\left( A_{X,X}\right) \cdot\operatorname*{Pf}\left(
A_{Y,Y}\right) =\det\left( S_{\left[ r\right] ,X}\right) \cdot\det\left(
S_{\left[ r\right] ,Y}\right) \cdot\det B=\det\left( A_{X,Y}\right)
\end{align*}
(by \eqref{eq.darij1.pf.t8.5}). This proves Theorem 8. $\blacksquare$
Note that squaring the equality in Theorem 8 yields the claim of Theorem 4
(albeit only in the case when $A$ is alternating).
Question 9. Can Theorem 8 be generalized to commutative rings $\mathbb{K}$
? This is trickier than one might expect, as the notions of "weak rank" and
"strong rank" probably need to be updated.
A partial solution: Assume $X$ and $Y$ partition $\{1,\dots,n\}$. Hence $A$ is in the block form
$$
A=\begin{bmatrix}
B_{r\times r}&C_{r\times r}\\
-^{\rm{T}}C_{r\times r}&D_{r\times r}
\end{bmatrix}
$$
where $B$ and $D$ are skew-symmetric. One needs to show that if the rank of the matrix above is $r$, then $\det(B)\det(D)=(-1)^r\left(\det(C)\right)^2$. The dimension of the null space must be $r$. A column vector
$\begin{bmatrix}
v_{r\times 1}\\
w_{r\times 1}
\end{bmatrix}$
is killed by $A$ iff
\begin{equation*}
\begin{cases}
Bv+Cw=\mathbf{0}\\
-^{\rm{T}}Cv+Dw=\mathbf{0}
\end{cases}.
\end{equation*}
If $B$ is invertible, we can solve the first equation for $v$ to obtain $v=-B^{-1}Cw$. Substituting in the second equation yields $(^{\rm{T}}CB^{-1}C+D)w=0$. Thus the dimension of the null space of $A$ is the same as that of the $r\times r$ matrix $^{\rm{T}}CB^{-1}C+D$. Hence this matrix must be zero. We conclude that $^{\rm{T}}CB^{-1}C=-D$. Taking determinants of both sides we obtain $\det(^{\rm{T}}CB^{-1}C)=(-1)^r\det(D)$, and hence $\det(B)\det(D)=(-1)^r\left(\det(C)\right)^2$.
If $B$ is singular, it suffices to argue that $C$ is singular as well. Aiming for a contradiction, if $C$ is invertible one may solve the first equation for $v$ to obtain $w=-C^{-1}Bv$. Substituting in the second equation, a similar argument shows $^{\rm{T}}C+DC^{-1}B=\mathbf{O}_{r\times r}$. This is a contradiction since $^{\rm{T}}C$ is invertible while $B$ and hence $DC^{-1}B$ are singular.
|
2025-03-21T14:48:30.099084
| 2020-03-20T01:33:31 |
355277
|
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],
"authors": [
"Pol van Hoften",
"https://mathoverflow.net/users/56856"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/355277"
}
|
Stack Exchange
|
Kottwitz global gerbes
I've been trying to understand global gerbes constructed by Kottwitz in $B(G)$ for all local and global fields (arXiv:1401.5728). Scholze explained in $p$-adic geometry (arXiv:1712.03708) why I would be interested in these. However, I find Kottwitz article fairly hard to read. In particular, it is hard for me to relate the notation in Scholze's article to Kottwiz's notation. Does anyone know a nice reference for Kottwitz's construction of global gerbes, or would care to sketch it for me?
I don't think there is any other reference. But it might be useful to try to understand $B(G)$ for a connected reductive group $G/\mathbb{Q}p$ in terms of morphisms from the the Dieudonné gerb $\mathfrak{D} \to \mathfrak{G}{G}$, where $\mathfrak{G}_{G}$ is the neutral $\mathbb{Q}_p^{\text{ur}}/\mathbb{Q}_p$ gerb associated to $G$. This is explained on page 109 and 110 of Reiman's book 'the semi-simple zeta function of quaternionic Shimura varieties'. See also the remark on page 86 of Langlands-Rapoport.
|
2025-03-21T14:48:30.099192
| 2020-03-20T01:58:47 |
355279
|
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"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"authors": [
"Hannes",
"JustWannaKnow",
"https://mathoverflow.net/users/150264",
"https://mathoverflow.net/users/85906"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:627305",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/355279"
}
|
Stack Exchange
|
Functional derivatives on Banach spaces
Physicists often use functional integrals and I'm trying to make sense of it in more precise terms. As you can see here, the functional derivative in Physics is defined in terms of Taylor expansions. Let me elaborate.
[The Physicist point of view] Let $f$ be a functional defined on some convenient function space and let $\varphi, \eta$ be fixed functions on this space. We expand $f$ in Taylor:
\begin{eqnarray}
f(\varphi + t\eta) = f(\varphi) + \frac{df}{dt}(\varphi + t\eta)\bigg{|}_{t=0}t + \frac{1}{2}\frac{d^{2}f}{dt^{2}}(\varphi+t\eta)\bigg{|}_{t=0}t^{2} +\cdots + \frac{1}{n!}\frac{d^{n}f}{dt^{n}}(\varphi+t\eta)\bigg{|}_{t=0}t^{n} +\mathcal{o}(t^{n+1}) \tag{1}\label{1}
\end{eqnarray}
where $\frac{d^{k}f}{dt^{k}}(\varphi+t\eta)\bigg{|}_{t=0}$ denotes the $k$-th Gâteaux derivative of $f$ at $\varphi$ evaluated at $\eta$. Thus, the $k$-th functional derivative of $f$ at $\eta$ is the function $\frac{\delta^{k} f}{\delta \varphi(x_{1})\cdots\delta\varphi(x_{k})}$ satisfying the equality:
\begin{eqnarray}
\frac{d^{k}f}{dt^{k}}(\varphi+t\eta)\bigg{|}_{t=0} =\int dx_{1}\cdots dx_{k} \frac{\delta^{k}f}{\delta \varphi(x_{1})\cdots\delta\varphi(x_{k})}\eta(x_{1})\cdots \eta(x_{k}) \tag{2}\label{2}
\end{eqnarray}
[The Mathematician point of view] Let $E, F$ be Banach spaces. A continuous bilinear functional ${\langle \cdot\,, \cdot \rangle }: E \times F \to \mathbb{R}$ is called $E$-non-degenerate if $\langle x,y\rangle = 0$ for all $y \in F$ implies $x=0$ (Similarly for $F$-non-degenerate). Equivalently, the two maps of $E$ to $F^{*}$ and $F$ to $E^{*}$ defined by $x \mapsto \langle x, \cdot \rangle$ and $y \mapsto \langle \cdot\,, y\rangle$, respectivelly, are one-to-one. If they are isomorphisms, $\langle \cdot\,, \cdot \rangle$ is called $E$ or $F$-strongly non-degenerate. We say that $E$ and $F$ are in duality if there is a non-degenerate bilinear functional $\langle \cdot\,, \cdot \rangle: E\times F \to \mathbb{R}$, also called a pairing of $E$ with $F$. If the functional is strongly non-degenerate, we say the duality is strong.
Consider the following definition (from this book).
Definition: Let $E$ and $F$ be normed spaces and $\langle \cdot, \cdot \rangle$ a $E$-non-degenerate pairing. Let $f: F \to \mathbb{R}$ be Fréchet differentiable at the point $\varphi \in F$ (denote this derivative as $Df(\varphi)$). The functional derivative $\delta f/\delta \varphi$ of $f$ with respect to $\varphi$ is the unique element in $E$, if it exists, such that:
\begin{eqnarray}
Df(\varphi)(\eta) = \left\langle \frac{\delta f}{\delta \varphi}, \eta\right\rangle\quad\forall\gamma \in F. \tag{3}\label{3}
\end{eqnarray}
Now, take $E=F=C(\Omega)$ to be a space of functions defined on a region $\Omega \subset \mathbb{R}^{n}$, which is Banach, and take the pairing $\langle \cdot, \cdot \rangle : C(\Omega)\times C(\Omega) \to \mathbb{R}$ given by:
\begin{eqnarray}
\langle f,g\rangle := \int_{\Omega}f(x)g(x)dx \tag{4}\label{4}
\end{eqnarray}
If $f$ is Fréchet differentiable at $\varphi$, then it is also Gâteaux differentiable at $\varphi$ and the following identity holds:
\begin{eqnarray}
Df(\varphi)(\eta) = \frac{df}{dt}(\varphi+t\eta)\bigg{|}_{t=0} \tag{5}\label{5}
\end{eqnarray}
Thus, the above definition together with the pairing \eqref{4} and \eqref{5} implies that the functional derivative of $f$ at $\varphi$ is the element $\delta f/\delta\varphi$ satisfying:
\begin{eqnarray}
\frac{df}{dt}(\varphi+t\eta)\bigg{|}_{t=0} = \int \frac{\delta f}{\delta \varphi}\eta(x)dx \tag{6}\label{6}
\end{eqnarray}
Note that \eqref{6} is exactly the physicist definition \eqref{2} for $k=1$. Now, my question is how to extend the mathematician's definition to consider higher order derivatives. If $f$ has, say, $k$ Fréchet derivatives at $\varphi$, then it has $k$ Gâteaux derivatives at this point. But now, the $k$-th Fréchet derivatives is a $k$-linear map, so I wonder if I should extend the definition by considering not pairings but $k$ linear maps instead, and then demand that these $k$-linear maps satisfy something like:
\begin{eqnarray}
D(\varphi_{1},\ldots,\varphi_{k})(\eta) = \left\langle \frac{\delta^{k}f}{\delta \varphi^{k}},\eta,\ldots,\eta\right\rangle \nonumber
\end{eqnarray}
where, now, $\langle \cdot, \cdots, \cdot \rangle$ is a $k$-linear non-degenerate map. Another possible approach is to use the same pairings and define high order derivatives as successive applications of the first derivative (I don't know how to do it though) and then prove a representation theorem when $E=F=C(\Omega)$, i.e. prove that if we take $E=F=C(\Omega)$ and use the pairing \eqref{4} then this $k$-th functional derivative becomes \eqref{2}. I'm really lost at this point, and I'd appreciate any help or tips on how to proceed.
EDIT: A nice discussion in my previous question led me to some clarifications and possible directions. First, suppose that $f$ is twice Fréchet differentiable at $\varphi \in E$. Then, there exists a bounded bilinear functional $D^{2}f[\varphi]$ satisfying
\begin{eqnarray}
\lim_{\eta \to 0}\frac{Df[\varphi+\eta](\gamma)-Df[\varphi)](\gamma)-D^{2}f[\varphi](\eta,\gamma)}{\Vert\eta\Vert} = 0. \tag{7}\label{7}
\end{eqnarray}
But, using \eqref{3}, we also have
$$
\begin{split}
Df[\varphi+\eta](\gamma)-Df[\varphi](\gamma) &= \left\langle \frac{\delta f}{\delta(\varphi+\eta)},\gamma\right\rangle - \left\langle\frac{\delta f}{\delta \varphi},\gamma\right\rangle \\
&=\left\langle\frac{\delta f}{\delta(\varphi+\eta)}-\frac{\delta f}{\delta \varphi},\gamma\right\rangle = \langle \mathcal{L}[\varphi](\eta),\gamma\rangle
\end{split}
$$ for some linear operator $\mathcal{L}[\varphi]:E\mapsto E$. If we take $E=F=C(\Omega)$ as I mentioned before, it seems that the physicist's result is obtained by taking
$$
\begin{eqnarray}
\mathcal{L}[\varphi](\eta) := \int \frac{\delta^{2}f}{\delta \varphi^{2}} (x,y)\beta(x) dx \tag{8}\label{8}
\end{eqnarray}
$$
where, now, $\delta^{2}f/\delta\varphi^{2} = \delta^{2}f/\delta\varphi^{2}(x,y)$ is a function on $C(\Omega\times\Omega)$ and this would be our second order functional derivative of $f$. But I still have doubts about that. Why taking \eqref{8} as my linear map? It seems very arbitrary.
The question seems to be an extension of the previous unanswered question by the OP.
Yes! This one is more elaborated.
Premise: almost (if not) all derivations below are kept at a formal level, i.e. (apart from the notes, almost) no discussion of the hypotheses needed to make the result rigorous are given. This is because the question asks to show a way to extend a particular notion of derivative of functionals in order to include differentiability of order higher than one, not when this way of doing things is rigorously justified.
The problem. When trying to figure out how a general notion applies to narrower scope, I think that the best (or perhaps the simpler) approach is to start from the very basic, primitive general concept which originates it. In this case, the originating concept is that of functional derivative: a map $f: F\to G$ is between two (real) Banach spaces $F$ and $G$ is Gâteaux differentiable if its functional derivative exists and is linear respect to the increment, i.e.
$$
\bigg{[}\frac{\mathrm{d}}{\mathrm{d}\varepsilon}f[\varphi+\varepsilon \eta]\bigg{]}_{\varepsilon = 0} \triangleq D f[\varphi](\eta)\in\mathscr{L\!i\!n}(F, G) \quad\forall \varphi,\eta\in F\label{A}\tag{A}
$$
where $\varepsilon$ is a real parameter and $\mathscr{L\!i\!n}(F, G)$ is the vector space of all linear maps from $G$ to $F$ (including unbounded, and thus possibly discontinuous, ones).
If the functional derivative is also continuous respect to the topology of $G$ i.e. it belongs to $\mathscr{L}(F, G)\subsetneq\mathscr{L\!i\!n}(F, G)$, then it is a Fréchet derivative, thus $f$ is said to be Fréchet differentiable.
Let's define a (possibly multilinear) higher order functional derivative analogous to \eqref{A}: the classical idea seems to have been first explicitly stated by Fantappiè in [1], §5, pp. 513-514 (see also [2], chapter VI, §(25÷27), pp. 70-78). If $\varepsilon_1,\ldots\varepsilon_k$ are $k\in\Bbb N\setminus\{0\}$ real parameters, for all $\varphi,\eta_i,\ldots,\eta_k \in F$, the $k$-th order functional derivative is recursively defined as follows:
$$
\begin{split}
D f[\varphi](\eta_1)&=\bigg{[}\frac{\partial}{\partial\varepsilon_1}f[\varphi+\varepsilon_1 \eta_1]\bigg{]}_{\varepsilon_1 = 0}\\
D^2 f[\varphi](\eta_1,\eta_2)&=\bigg{[}\frac{\partial^2}{\partial\varepsilon_2\partial\varepsilon_1}f[\varphi+\varepsilon_1 \eta_1+\varepsilon_2 \eta_2]\bigg{]}_{\varepsilon_1, \varepsilon_2= 0} \\
&\qquad=\bigg{[}\frac{\partial}{\partial\varepsilon_2}Df[\varphi+\varepsilon_2 \eta_2](\eta_1)\bigg{]}_{\varepsilon_2 = 0}\\
\vdots\;\qquad &\:\,\vdots\quad\quad\qquad\qquad\vdots\\
D^k f[\varphi](\eta_1,\ldots,\eta_k) &=\left[\frac{\partial^k}{\partial\varepsilon_k\cdots \partial\varepsilon_k}f\bigg[\varphi+\sum_{i=1}^k\varepsilon_i\eta_i\bigg]\right]_{\varepsilon_1,\ldots,\varepsilon_k= 0}\\
&\qquad=\bigg{[}\frac{\partial}{\partial\varepsilon_k}D^{n-1}f[\varphi+\varepsilon_k \eta_k](\eta_1,\ldots,\eta_{k-1}) \bigg{]}_{\varepsilon_k= 0}
\end{split}\label{B}\tag{B}
$$
Be it noted that $D^kf[\phi](\eta_1,\ldots,\eta_k)$, $k\ge1$ is a $k$-linear functional (Fantappiè, working with analytic functionals i.e. $ F=\mathscr{O}(\Bbb C)$ and $G=\Bbb C$ is also able to prove \eqref{A}, i.e. he does not need to assume that the functional derivative is linear, since the structure of functionals guarantees this fact, as shown in [1], §2, pp. 510-511 or [2], §25, pp. 73-74).
Now applying the hierarchy \eqref{B} to equation \eqref{3} in the definition given by Abraham, Marsden and Ratiu, for all $\varphi,\eta_1,\ldots,\eta_k \in F$ we have
$$
\begin{split}
D^2 f[\varphi](\eta_1,\eta_2)&=\left[\frac{\partial}{\partial\varepsilon_2}\left\langle \frac{\delta f}{\delta \varphi}[\varphi+\varepsilon_2\eta_2], \eta_1\right\rangle\right]_{\varepsilon_2=0}\\
&\qquad= \left[\left\langle \frac{\partial}{\partial\varepsilon_2}\frac{\delta f}{\delta \varphi}[\varphi+\varepsilon_2\eta_2], \eta_1\right\rangle\right]_{\varepsilon_2=0}\\
&\qquad= \left\langle \frac{\delta^2 f}{\delta \varphi^2}[\varphi](\eta_2), \eta_1\right\rangle\simeq \left\langle \frac{\delta^2 f}{\delta \varphi^2}[\varphi], \eta_1, \eta_2\right\rangle_{\!\!1+2}\\
\vdots\;\qquad &\:\,\vdots\quad\quad\qquad\qquad\vdots\\
D^k f[\varphi](\eta_1,\ldots,\eta_k)&=\left[\frac{\partial}{\partial\varepsilon_k}\left\langle \frac{\delta^{k-1} f}{\delta \varphi^{k-1}}[\varphi+\varepsilon_k\eta_k], \eta_1,\ldots,\eta_{k-1} \right\rangle_{\!\!1+(k-1)}\right]_{\varepsilon_k=0}\\
&\qquad = \left\langle \frac{\delta^k f}{\delta \varphi^k}[\varphi](\eta_k), \eta_1,\ldots, \eta_{k-1}\right\rangle_{\!\!1+(k-1)} \!\!\simeq \left\langle \frac{\delta^k f}{\delta \varphi^k}[\varphi], \eta_1,\ldots, \eta_k\right\rangle_{\!\!1+k}
\end{split}\label{C}\tag{C}
$$
where $\langle{ \cdot\,, \ldots,\, \cdot}\rangle_{1+k}:E\times\big(\times^k_{i=1}F\big)\to\Bbb R$ is a $1+k$-linear mapping for all $k\ge 1$: thus if this $(1+k)$-linear mapping is definable, the $k$-th order functional derivative at $\varphi\in F$ of a functional $f: F\to\Bbb R$, $k\ge 1$ is the unique element $\delta^k f/\delta\varphi^k\in E$, if it exists, verifying the last of the equalities \eqref{C}.
Notes
As stressed in the premise, the development above is by no means a proof of the existence (and uniqueness) of a $(1+k)$-linear mapping and an element in $E$ such that \eqref{C} holds. However, assuming that the hypotheses guarantee the existence of such objects, the formal step at the right side of each equality in \eqref{C} is justifiable in the same standard way used to prove isomorphisms theorems of the following kind
$$
\mathscr{L}\big(F, \mathscr{L}(F,\Bbb R)\big)\simeq \mathscr{L}\big(\times_{i=1}^{2}F,\Bbb R\big)\triangleq \mathscr{L}_2(F,\Bbb R)
$$
Earlier work. The key information in the formalism expressed by \eqref{B} is that it emphasizes the intrinsic multilinearity of higher order functional derivatives. In general function spaces, the concept of product of functions may not be defined therefore it couldn't be possible to define polynomials: therefore the straight notation
$$
\bigg{[}\frac{\mathrm{d}^k}{\mathrm{d}\varepsilon^k}f[\varphi+\varepsilon \eta]\bigg{]}_{\varepsilon = 0} \triangleq D^k f[\varphi](\eta^k)
$$
could be misleading and overshadowing instead of shed light on the problems at hand. Nevertheless it is possible define abstract multilinear mappings and work with them. And while I have not being able to find earlier references for the notation \eqref{B} respect to the works of Fantappiè, Volterra is surely aware of the necessity of working in a multilinear setting: his development of higher order functional derivative (remember that he works essentially in a Banach space setting) leads exactly to formula \eqref{2} ([3], §2.7, p. 102, formula (5) or [3], §29, p. 25) and to an extension of Taylor's theorem but his approach, while being essentially the same is much more involved, and requires several restriction on the functional $f$ ([2] §2.5, p.99, or [3], §27, pp. 22-24).
Life outside Banach spaces. As I briefly said in the main body of the question, formulas \eqref{B} were developed by Fantappiè to work with locally analytic functionals i.e. functionals defined on germs of holomorphic functions $ F=\mathscr{O}(\Bbb C)$ and $G=\Bbb C$. However, a more general and modern approach is developed by prof. Michor and described in the monograph linked in his answer: his approach is substantially based on the use of smooth/analytic curves with values in bornological spaces, i.e. on the use of functions of one real variable $c:\Bbb R\to B$ with $c\in C^\infty(\Bbb R, B)$ or $c\in C^\omega(\Bbb R, B)$, used as arguments of the functionals to be analyzed. This is truly an evolution of the classical methods by Volterra and Fantappiè, as it allows to give a precise meaning to straightforward expressions like the following one
$$
\bigg{[}\frac{\mathrm{d}^k}{\mathrm{d}t^k}f[c(t)]\bigg{]}_{t = 0} \triangleq D^k f[c_0](c_1,\ldots,c_n)\quad c_0,c_1,\ldots,c_k\in B
$$
for very general functionals.
References
[1] Fantappiè, Luigi, "La derivazione delle funzionali analitiche [Derivation of analytic functionals]" (Italian), Atti della Accademia Nazionale dei Lincei, Rendiconti, VI Serie, vol. 1, 1° semestre, pp. 509-514 (1925), JFM 51.0314.03.
[2] Fantappiè, Luigi, Teoría de los funcionales analíticos y sus aplicaciones. Curso de conferencias desarrollado en el Instituto de Matemáticas ""Jorge Juan"" de Madrid y en el Seminario Matemático de Barcelona en el año académico 1942-1943, recopiladas por R. Rodríguez Vidal (Catalan), Barcelona: Seminario Matemático de Barcelona [Imprenta-Escuela de la Casa Provincial de Caridad], pp. 174 (1943), MR0014598.
[3] Volterra, Vito, "Sulle funzioni che dipendono da altre funzioni [On functions which depend on other functions]" (in Italian), Atti della Reale Accademia dei Lincei, Rendiconti (4) III, No. 2, 97-105, 141-146, 153-158 (1887), JFM 19.0408.01.
[4] Volterra, Vito, Theory of functionals and of integral and integro-differential equations. Dover edition with a preface by Griffith C. Evans, a biography of Vito Volterra and a bibliography of his published works by Sir Edmund Whittaker. Unabridged republ. of the first English transl, New York: Dover Publications, Inc. pp. 39+XVI+226 (1959), MR0100765, Zbl 0086.10402.
The `functional derivative' $\frac{\delta f}{\delta \varphi}$ in your sense is the gradient of the derivative $d f(\varphi)\in L(E,\mathbb R)$ with respect to duality $\langle\quad,\quad\rangle$ which you are specifying. It need not exist in $F$ since $F$ might be smaller then the dual of $E$. $\frac{\delta^2 f}{\delta \varphi^2}$ then would be a second order gradient with respect to an extension of $\langle\quad,\quad\rangle$, whose existence is also not sure, but $d^2f(\varphi)$ exist as a bounded bilinear map $E\times E\to \mathbb R$. In your example with $E=C(\Omega)$ the second derivative or Hessian is, in general, if it exists, a measure on $\omega\times \Omega$, and not a function.
See here for a concise setting of all this.
Added:
How to extend $\langle\quad,\quad\rangle$? Since $d^2f(\varphi): E\times E\to \mathbb R$ is symmetric bilinear bounded, it linearizes to the projective tensor product as $E\hat\otimes E\to \mathbb R$. So it lies in the dual $L(E,E')$ of the projective tensor product and is symmetric.
It might lie in a subspace of $L(E,E')$, for example in the subspace of compact operators, which is $E'\hat{\hat\otimes} E'$ (under the assumption of the approximation property, or in $F \hat{\hat\otimes}F$ (this would be one extension of $\langle\quad,\quad\rangle$), depending on the properties of the functional.
The easiest way for you would be to let $F=E'$ and to let $\langle\quad,\quad\rangle$ be just the duality, and to use full dual spaces all around. Of course you have symmetry.
Thanks for your answer! Can I clarify some points? First, does your comment imply that my second order derivative should be a measure such as $\eta(\omega\times \Omega):=\int_{\Omega}f(\omega, \omega')\beta(\omega')d\omega'$ so that $\omega \to \eta({\omega}\times \Omega)$ is the 'function' to be considered in (8)? Second, how exactly should I extend $\langle \cdot, \cdot \rangle$. Does $\eta$ defined above extend my pairing (4), for instance?
Not a great idea to use normed spaces here. In physics, one typically deals with a functional on a space of smooth functions. Via polarization, the higher derivatives become symmetric continuous multilinear maps on that space of smooth functions.
Then via the Schwartz Kernel Theorem the latter become continuous linear maps, i.e., Schwartz distributions. In other words, what physicists write as
$$
\frac{\delta^k f}{\delta\phi(x_1)\cdots\delta\phi(x_k)}
$$
is a distributional kernel. For a reference that explores the corresponding theory, see "Properties of field functionals and characterization of local functionals" by Brouder, Dang, Laurent-Gengoux and Rejzner.
Thanks for your answer and reference! Just to clarify, it seems that my notion of functional derivative should be this so-called Bastiani derivative and representation (2) will follow from Schwartz Kernel Theorem?
|
2025-03-21T14:48:30.100184
| 2020-03-20T05:09:29 |
355283
|
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|
Stack Exchange
|
Girth and diameter of a graph with minimum degree at least 3
The problem is motivated by generalizing Moore graphs, graphs with maximum possible girth ($2\text{diam}+1$) given the diameter.
Question. Does there exist a graph $G$ with $\text{g}(G)-\text{diam}(G)>8$ and minimum degree at least $3$, where $\text{g}(G)$ and $\text{diam}(G)$ are the girth and diameter of $G$, respectively?
Remark. There's an infinite class of graphs with $\text{g}(G)-\text{diam}(G)=8$: the (point,line) incidence graphs of generalized octagons have $\text{g}(G)=16$ and $\text{diam}(G)=8$.
Bonus points (+200 bounty) for a proof that $\text{g}-\text{diam}$ is unbounded for graphs with minimum degree at least $3$.
Have you checked any of the cubic cages in here for whether they give you examples?
Maybe relevant: https://mathoverflow.net/q/145045
The graphs of girth 19, 20 and 22 have diameters 14, 15 and 16 respectively. The graph of girth 21 is missing.
The recent preprint "A randomized construction of high girth regular graphs" by Linial and Simkin suggests in its closing remarks that the answer (at least, to your bonus question) is unknown:
The possible relation between a graph’s girth and its diameter is particularly intriguing. It follows from [12] and Moore’s bound that
$$2 \ge \mathrm{lim\ sup} \frac{\mathrm{girth}(G)}{\mathrm{diam}(G)} \ge 1,$$
where the lim sup ranges over all graphs where all vertex degrees are $\ge 3$. Nothing better seems to be known at the moment.
Even more remarkably, we do not know whether
$$\mathrm{sup}(\mathrm{girth}(G) − \mathrm{diam}(G))$$
is finite or not. The sup is over all $G$ in which all vertex degrees are $\ge 3$.
|
2025-03-21T14:48:30.100327
| 2020-03-20T07:23:44 |
355286
|
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|
Stack Exchange
|
Polynomial algebra and its special ideals
Consider a polynomial algebra $A=\mathbb{K}[x_1,\ldots,x_n]$ and its ideal $I$, such that $A/I=\mathbb{K}[y_1,\ldots, y_k]$. Is it true that there exist new polynomial generators $z_1,\ldots,z_n$ (in a sense that $A=\mathbb{K}[z_1,\ldots,z_n]$) such that $I=(z_1,\ldots,z_{n-k})$?
It's unclear in general what "such $A/I=K[y_1,\dots,y_k]$ means: does it mean that $A/I$ is generated by $k$ elements, or that it's freely generated by $k$ elements (i.e., is isomorphic to a polynomial algebra over $k$ indeterminates). Given the context it certainly means freely generated. Also "new variables" should be "elements".
Variables are elements that are part of a set of polynomial generators.
It means that $y_i$ are algebraically independent. By variables I mean a part of independent generators of polynomial algebra.
No, you don't mean that: $x^2$ and $y^2$ are algebraically independent but they are not "variables"...
See if my edit is OK
You are right. In my comment I used that if some independent elements generate algebra, then they must be its variables
The general situation (independent of the characteristic of the base field) is, it is true when $n\geq k+2$ in your notation and not known in general otherwise and as @BugsBunny pointed out, false in positive characteristic.
This is called the Embedding Problem. I believe that it is false in positive characteristics and unknown in characteristic zero, except a few small cases. See Kraft's review
for the extent of my knowledge of the state of the problem. The counterexample in characteristic $p$ is from this paper:
$$
K[x,y]/(y^{p^2}-x-x^{2p})
$$
|
2025-03-21T14:48:30.100573
| 2020-03-20T11:15:00 |
355294
|
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|
Stack Exchange
|
Under Ramanujan conjecture, is primitivity equivalent to cuspidality and irreducibility?
Lemma 4.2 in M. Ram Murty, Selberg conjectures and Artin L-functions(1994), states that under Ramanujan conjecture, an irreducible cuspidal automorphic representation of $\operatorname{GL}_{n}(\mathbb{A}_{\mathbb{Q}})$ gives rise to a primitive L-function.
Is the converse true? That is, assuming Ramanujan conjecture, if a degree $n$ primitive L-function comes from an automorphic representation of $\operatorname{GL}_{n}(\mathbb{A}_{\mathbb{Q}})$, is this representation necessarily irreducible and cuspidal?
Yes, and this is true without the Ramanujan conjecture (but see also Peter Humphries' comment below).
If $\pi$ is not irreducible, say $\pi=\pi_1\oplus\pi_2$, then $L(s,\pi)=L(s,\pi_1)L(s,\pi_2)$.
If $\pi$ is not cuspidal, then by Langlands' theory of Eisenstein series, there is a nontrivial partition $n=\sum n_j$ and cuspidal irreducible representations $\pi_j$ of $\mathrm{GL}_{n_j}(\mathbb{A}_\mathbb{Q})$ such that $\pi$ is parabolically induced from $\prod\pi_j$ on $\prod\mathrm{GL}_{n_j}(\mathbb{A}_\mathbb{Q})$ as a Levi subgroup of $\mathrm{GL}_n(\mathbb{A}_\mathbb{Q})$. Then, $L(s,\pi)=\prod L(s,\pi_j)$.
It's not true without the Ramanujan conjecture if by "primitive $L$-function" you mean "primitive element of the Selberg class whose Dirichlet series coefficients satisfy $a(n) \ll_{\varepsilon} n^{\varepsilon}$", which is what I imagine the OP meant.
@PeterHumphries: I guess I thought of primitivity within automorphic $L$-functions or some extended Selberg class. But you are absolutely right.
Many thanks to the both of you.
|
2025-03-21T14:48:30.100705
| 2020-03-20T11:21:15 |
355295
|
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|
Stack Exchange
|
Is there an upper bound on the number of points in point cloud for which we compute the persistent homology?
I am interested in the topic of persistent homology (topological data analysis). According to what I read, there is some roadblock in the analysis of "big data" using persistent homology as it is considered rather computationally intensive. Computation of the simplicial complexes of a large point cloud often relies on extracting a sample. For example we use a witness complex.
My main question is, what is the current maximum or upper bound on the number of points in the point cloud for computing persistent homology?
Or in other words, what is the "world-record" in the number of points allowed by state-of-the-art persistent homology algorithms? Is it in the range of thousands of points, or millions of points?
It depends very much on the type of simplicial complex you're using. If you have points in 3d then doing Cech/Delaunay complex is feasible with millions of points. If you have high dimensional data, complexes will generally blow up in size and millions of points will be too much. Finding ways to decrease the size of complexes, possibly using approximations (eg witness complexes), is an active research area. It is hard to formulate a precise answer because it depends too much on how nice the data is and how much error is tolerated.
Possibly of interest to you is this paper, A roadmap for the computation
of persistent homology by Otter, Porter, Tillmann, Grindrod, and Harrington. They compare different pieces of software for computing persistent homology, and list off the maximum sizes of simplicial complexes that each software could handle under their constraints.
It looks like in the paper that they were mostly working with either $2$- or $3$-dimensional data sets. Anyways, to answer your question, we can compute the persistent homology of data sets consisting of at least $10^9$ points. The above table is from this presentation by Nina Otter.
|
2025-03-21T14:48:30.100855
| 2020-03-20T12:05:46 |
355297
|
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|
Stack Exchange
|
NCF, P-points, weak P-points, and cardinalities
The post is a bit long, but all the questions are similar or concern the same topic.
Let $\omega^*=\beta\omega\setminus\omega$. A well-known topological definition of a P-point (on $\omega$) is as follows: a point $x\in\omega^*$ is a P-point if every intersection of countably many open neighborhoods of $x$ contains an open neighborhood of $x$. Similarly, a point $x\in\omega^*$ is a weak P-point if $x$ is not in the closure of any countable subset not containing $x$. We have also an equivalent definition of P-points in terms of functions $\omega\to\omega$: an ultrafilter $x\in\omega^*$ is a P-point if for every function $f\colon\omega\to\omega$ there is $A\in x$ such that $f\restriction A$ is either constant or finite-to-one. Since every P-point is a weak P-point, my first question is as follows:
Question 1. Does there exist a similar characterization of weak P-points (i.e. in terms of functions $\omega\to\omega$)?
Kunen proved in ZFC that we always have $2^{\mathfrak{c}}$ weak P-points and at least $\mathfrak{c}$ incomparable in the sense of Rudin-Keisler weak P-points. My next questions concern the total number of (incomparable/incompatible) P-points provided that there is at least one P-point.
Question 2. Assume that a P-point exists. (a) Does there exist another (non-isomorphic or incomparable) P-point? (b) Do there exist $2^{\mathfrak{c}}$ different P-points? (c) Do there exist $\mathfrak{c}$ incomparable P-points?
(It is easy to see that there exist $\mathfrak{c}$ isomorphic P-points.)
Question 3. Assume that there exist $2^{\mathfrak{c}}$ many P-points. (a) Do we have then $2^{\mathfrak{c}}$ (or at least $\mathfrak{c})$ incomparable P-points? (b) Do there exist $2^{\mathfrak{c}}$ (or at least $\mathfrak{c}$, or even $2$) incompatible P-points?
Let us say that two ultrafilters $U,V\in\omega^*$ are near coherent if there exists a finite-to-one function $f\colon\omega\to\omega$ such that $f(U)=f(V)$. The near coherence is an equivalence relation, so we can count the number of the equivalence classes. The Near Coherence of Filters principle (NCF in short) states that there exists only one equivalence class. Blass and Shelah constructed a model of set theory in which the NCF holds (it is now also known to hold in the Miller model). On the other hand, Banakh and Blass proved that either we have finitely many equivalence classes or $2^{\mathfrak{c}}$ (the latter holds e.g. in each model where $\mathfrak{u}\ge\mathfrak{d}$, so e.g. under CH). The next question is in the same spirit as Questions 2 and 3.
Question 4. Assume that we have $2^{\mathfrak{c}}$ many near coherence classes. Does there exist $\mathfrak{c}$ (or $2^{\mathfrak{c}}$) many incompatible in the sense of Rudin-Keisler ordering weak P-points?
It is believed (but not proved so far) that there exists a model with exactly $2$ near coherence classes and for every $n>2$ there is no model with $n$ classes. Also, the NCF implies the existence of a P-point. Thus, my next (and last) question is the following.
Question 5. Assume there are exactly 2 classes of near coherence. Does there exist any P-point? If yes, then are there two that are not compatible?
Thank you very much for the help!
An answer to question 2: Shelah constructed a model with exactly one $P$-point (up to isomorphism). In fact, the one $P$-point is a selective ultrafilter. You can find the construction in section XVIII.4 of Proper and Improper Forcing.
An answer to question 5: If there are exactly $2$ classes of near coherence, then $\mathfrak{u} < \mathfrak{d}$. (You say this in your post, just before Question 4.) In other words, there is a non-principal ultrafilter generated by fewer than $\mathfrak{d}$ sets. Ketonen proved that any ultrafilter generated by fewer than $\mathfrak{d}$ sets is a (non-selective) $P$-point. See
J. Ketonen, "On the existence of $P$-points in the Stone-Cech compactification of the integers," Fundamenta Mathematicae 92 (1976), pp. 91-94. (link)
Thanks, Will! I was pretty sure that I had heard about this result, but couldn't find it in the literature, so started to doubt it...
Note that $\mathfrak{u}=\mathfrak{d}=\aleph_1$ in this model, so you have $2^{\mathfrak{c}}$ near coherence classes as well. You didn't ask about this specifically, but it seemed relevant to the spirit of your question.
In question 4 I'm asking about weak P-points.
For the second part of question 5, let me point out that finitely many equivalence classes implies there are non-isomorphic $P$-points. We already know it gives us $P$-points that are not selective. But an ultrafilter is RK-minimal iff it is selective, so there are $P$-points with ultrafilters strictly RK-below them. But anything RK-below a $P$-point is also a $P$-point.
Will, concerning your last comment, we have two P-points RK-below a P-point. This means that they're RB-below (Rudin-Blass) a P-point, which is equivalent to be RB-above an ultrafilter (Laflamme--Zhu), which then again must be a P-point. This means that they're compatible.
|
2025-03-21T14:48:30.101142
| 2020-03-20T14:27:09 |
355308
|
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|
Stack Exchange
|
Lax pair of an integrable non-linear PDE
The following is a fourth-order non-linear PDE that passes the Painleve integrability test
$$\left(1+x^{2}\right)^{2}u_{xxxx} + 8x\left(1+x^{2}\right)u_{xxx} + 4\left(1+3x^{2}\right)u_{xx}+ t\left(2xuu_{xx} + \left(1+x^{2}\right)\left(uu_{xx}\right)_{x} - 4\left(1+3x^{2}\right)u_{xxt} - 4x\left(1+x^{2}\right)u_{xxxt}\right)=0,$$
where $u=u(x,t)$. The leading-order behaviour of this PDE is of the form $u=A_{0}\left(x-x_{0}\right)^{-1}$, where $A_{0}=3\left(1+x_{0}^{2}\right)t^{-1}$. I am looking for a Lax pair $[\mathcal{L},\mathcal{M}]$ which satisfies the equation
$$\dot{\mathcal{L}}+[\mathcal{L},\mathcal{M}]=0,$$
but not able to find the same. I suspect that this may be due to the fact that although few systems may pass the Painleve test, they need not be integrable. Any help/comment(s) about the problem and/or the integrability of the PDE is appreciated. Also, are there any other (stronger) methods using which I can explicitly try and probe for the integrability of the PDE although it passed the Painleve test?
You could try using the Wahlquist-Estabrook prolongation structure technique, per H.D. Wahlquist and F.B. Estabrook, J. Math. Phys 16 (1975) 1-7 (covering the Korteweg-deVries equation), & F.B. Estabrook and H.D. Wahlquist J. Math. Phys. 17 (1976) 1293-1297 (covering the nonlinear Schrödinger equation).
A non-trivial prolongation structure would be a signal that your equation is indeed integrable. To construct a Lax pair, you need to find an explicit representation of the prolongation structure: the paper by R. Dodd and A. Fordy Proc Roy. Soc. Lond. A385 (1983) 389-429 provides a method of doing this.
Since your PDE includes both independent variables in the coefficients, you may also wish to review the note by B.A. Kupershmidt, Nonautonomous form of the theory of Lax equations, Lettere al Nuovo Cimento 33 (1982) 103–107.
Thanks for the answer and the references! I'll check them out.
|
2025-03-21T14:48:30.101301
| 2020-03-20T14:32:25 |
355309
|
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|
Stack Exchange
|
Is the number of commutation classes of reduced words of the longest element of $S_n$ even for $n\geq 3$?
Observably, the number of primitive sorting networks on $n$ elements (or the number of commutation classes of reduced words of the longest element of $S_n$) is even for $3\leq n\leq 15$. These are all the values for which it has been computed so far. Is there an abstract way to see that it is always even?
Sequence is at http://oeis.org/A006245
Isn't it true for $n=2$ as well? Anyways, maybe you can show that a word $s_{i_1}s_{i_2}\cdots s_{i_k}$ and its reverse $s_{i_k}s_{i_{k-1}}\cdots s_{i_1}$ or twisted reverse $s_{n-i_k}s_{n-i_{k-1}}\cdots s_{n-i_1}$ never belong to the same commutation class.
@SamHopkins For $n=2$ the value is $1$. The reverse doesn't work because sometimes words are palindromic.
@SamHopkins Twisted reverse or conjugating the generators by the element might work. Have to think about that.
Ah sorry yes I got $n=2$ and $3$ mixed up. Good point about palindromes.
is there an oeis entry which is relevant here?
@PerAlexandersson I posted a link.
On the linked OEIS entry I find the following:
Also the number of mappings X:{{1..n} choose 3}->{+,-} such that for any four indices a < b < c < d, the sequence X(a,b,c), X(a,b,d), X(a,c,d), X(b,c,d) changes its sign at most once (see Felsner-Weil and Balko-Fulek-Kynčl reference). - Manfred Scheucher, Oct 20 2019
There is an obvious fixed-point free involution on the set of such maps given by switching "+" and "-", assuming that $n \geq 3$ so that $\binom n 3 > 0$.
|
2025-03-21T14:48:30.101435
| 2020-03-20T15:15:18 |
355314
|
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"url": "https://mathoverflow.net/questions/355314"
}
|
Stack Exchange
|
Hilbert class field tower
Let $K$ is a number field,and $H_{K}^{i},i=1,2,\cdots$ be its Hilbert class field tower,suppose it is finite,and let $L=H_{K}^{n}$ is the top of the tower. Must $L$ be galois over $K$?
Each $H_K^i$ is Galois over $K$. Let $\sigma$ be any automorphism of the algebraic closure of $K$ fixing $K$. Inductively show that $\sigma(H_K^{i+1})$ is the maximal unramified abelian extension of $\sigma(H_K^i)=H_K^i$, so equals $H_K^{i+1}$.
Thank you very much !
More generally (and basically for the same reason), if $L/K$ is Galois and $\mathfrak{m}$ is a Galois stable modulus of $L$, then the ray class field of $L$ with modulus $\mathfrak{m}$ is Galois over $K$.
Posting as an answer to get out of the unanswered list.
Each field $H^i_K$ in the class field tower of $K$ is Galois over $K$. Indeed, let $\sigma$ be any automorphism of the algebraic closure of $K$ fixing $K$. Inductively we can show that $\sigma(H^i_K)=H^i_K$. Indeed, this is clear for $i=0$ (setting $H^0_K=K$) and assuming the equality for $i$, $\sigma(H^{i+1}_K)$ is the maximal unramified abelian extension of $\sigma(H^i_K)=H^i_K$, so equals $H^{i+1}_K$.
Therefore the union of the entire class field tower, be it finite or not, is Galois over $K$ as well.
|
2025-03-21T14:48:30.101541
| 2020-03-20T15:27:41 |
355315
|
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"Moishe Kohan",
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"url": "https://mathoverflow.net/questions/355315"
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|
Stack Exchange
|
p-adic analogue of classification of irreducible Riemannian symmetric spaces?
For Riemannian symmetric spaces, we have a nice result that the simply-connected ones are products of irreducible symmetric spaces, of which we have a list of ten infinite families and some exceptional ones (see e.g. wikipedia). In Paul Garrett's answer to this MSE question, after giving the non-compact forms of the infinite families of irreducible Riemannian symmetric spaces, he says
Note that in all cases the three $\mathbb{R}$-algebras $\mathbb{R}$,$\mathbb{C}$,$\mathbb{H}$ play roles. In fact, a completely parallel thing happens for classical groups over $p$-adic fields and in other cases, as in A. Weil's "Algebras with involutions and classical groups", Indian (not Indiana) J. Math. 1960.
I flipped through Weil's paper but wasn't enlightened. What I didn't find and am looking for in this question is an explicit list of p-adic homogeneous spaces $G/K$, where $K$ is presumably the fixed points of some involution, which is the analogue in whatever suitable sense of the irreducible Riemannian symmetric spaces; additionally I would like an explanation of what this 'suitable sense' is. Any help is appreciated!
At least for the analogue of symmetric spaces of noncompact type, the answer is probably "Euclidean building", as developed by Bruhat and Tits in the late 60's. The problem with $G/K$ is that it's just a discrete space. But the point is that it's (close to be) the 1-skeleton of a nice complex.
To complement Yve's comment: For compact symmetric spaces an analogue would be Tits buildings. Their classification (in the compact case) is well-known and can be found in any book on buildings, e.g. Brown or Ronan.
@YCor: are there pairs $(G,K)$ associated to these Euclidean buildings or to compact buildings, or are they something more complicated? (sorry, I know almost nothing about buildings). The fact that $G/K$ is a discrete space isn't a problem for my purposes. I'm really after a list of pairs of groups $(G,K)$, as I'm coming from a random matrix theory viewpoint. Can such a list be extracted from the classification of buildings? It looked like Ronan's treatment of the classification is using an axiomatic approach to buildings and not referencing p-adic groups too much. Thanks!
In the Riemannian case, the point is that you can define on something purely local (Riemannian symmetric space $X$ of non-positive sectional curvature) which essentially characterizes the pair $(G,K)$ (say, taking the isometry group of the universal covering and one point stabilizer). In the $p$-adic case, $G/K$ being discrete, one needs something new, and the new feature is the building structure. (Here I'm only addressing your sentence "the fact that $G/K$ is discrete isn't a problem for my purposes". I don't know what a good reference for Bruhat-Tits buildings is.)
|
2025-03-21T14:48:30.101755
| 2020-03-20T15:50:36 |
355318
|
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"Denis Serre",
"Lucia",
"Sylvain JULIEN",
"https://mathoverflow.net/users/13625",
"https://mathoverflow.net/users/38624",
"https://mathoverflow.net/users/8799"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/355318"
}
|
Stack Exchange
|
Inequality in Iwaniec-Kowalski
I am reading about Dirichlet polynomials in the book Analytic Number Theory by the said authors. Can anyone justify the following inequality? Assume that $a(n),b(m)$ are sequences of non-negative numbers such that $a(n)=0$ for all large enough $n$ and $b(m)=0$ for all large enough $m$. How does one prove $$ \sum_{\substack{ n,m , n_1, m_1 \in \mathbb N \\ n m =n_1 m_1 }}a(n)b(m)a(n_1) b(m_1) \leq \sum_n \sum_m a(n)^2 b(m)^2 \tau(nm )?$$ Here $\tau$ is the function that counts the number of positive integer divisors. This is after equation (9.7) in page 231.
There are infinitely many divisor functions. Would you give more detail about $\tau$ ? By the way, $\tau$ is often used for Ramanujan's function, which is not a divisor function.
Don't feel offended. Wikipedia page https://en.wikipedia.org/wiki/Divisor_function speaks of functions $\sigma_x(n)$ parametrized by a real $x\ge0$. Apparently, yours is $\sigma_0$. I am confined at home and don't have the book next to me.
Just use $a(n)b(m)a(n_1)b(m_1) \le (a(n)^2b(m)^2 + a(n_1)^2 b(m_1)^2)/2$, and note that given $n$, $m$, there are at most $\tau(nm)$ choices for $n_1$, $m_1$.
@Denis: in case you or your confined colleagues need references in this book, I have it at home, so feel free to ask. It would be nice to create some shared file among members of this site to know who owns which book, so that we can help one another.
|
2025-03-21T14:48:30.102142
| 2020-03-20T17:16:03 |
355322
|
{
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"authors": [
"Igor Khavkine",
"https://mathoverflow.net/users/2622"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/355322"
}
|
Stack Exchange
|
On various versions of the harmonic oscillator
The standard $n$-dimensional harmonic oscillator is the operator
$
\mathcal H=\frac{1}{2}\sum_{1\le j\le n}(D_j^2+x_j^2), \text{ $D_j=-i\partial_{x_j}$},
$
and its spectral decomposition is
$$
\mathcal H=\sum_{k\ge 0}(\frac{1}{2}+k)\mathbb P_{k,n}, \quad \mathbb P_{k,n}=\sum_{
\alpha\in \mathbb N^n,\alpha_1+\dots+\alpha_n=k}
\mathbb P_{\alpha_1}\otimes\dots\otimes\mathbb P_{\alpha_n},
$$
where $\mathbb P_{\alpha_j}$ stands for the orthogonal projection onto the one-dimensional Hermite function with level $\alpha_j$. Now let us consider for $µ=(µ_1,\dots, µ_n)$ with $\mu_j>0$,
the operator
$$
\mathcal H_µ=\frac{1}{2}\sum_{1\le j\le n}µ_j(D_j^2+x_j^2).
$$
With the notation $\vert µ\vert=\sum_{1\le j\le n} µ_j$ and $µ\cdot \alpha=\sum_{1\le j\le n}µ_j\alpha_j$, we have
$$
\mathcal H_µ=\sum_{\alpha\in \mathbb N^n}(\frac{\vert µ\vert}2+µ\cdot\alpha)
(\mathbb P_{\alpha_1}\otimes\dots\otimes\mathbb P_{\alpha_n}),
$$
so that the eigenspaces are the same as for $\mathcal H$ but the arithmetic properties of $µ$ make possible that all eigenvalues $(\frac{\vert µ\vert}2+µ\cdot\alpha)$ are simple. What are the typical examples and standard references on this topic?
The first Hamiltonian describes the isotropic harmonic oscillator, while the one with the non-trivial $\mu$ would be an anisotropic harmonic oscillator. Not sure what you mean by "this topic". I'm not sure there's much more to it than just giving names to the different cases.
|
2025-03-21T14:48:30.102262
| 2020-03-20T17:34:11 |
355324
|
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"authors": [
"Simon Henry",
"fosco",
"https://mathoverflow.net/users/22131",
"https://mathoverflow.net/users/7952"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/355324"
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|
Stack Exchange
|
On the subcategory of product-preserving functors
A well-known theorem of Freyd and Kelly says that the subcategory of functors $F : A \to Set$ that commute with a prescribed class of limits is reflective inside the whole $[A,Set]$.
The proof of this statement relies on an application of the small object argument: for a functor $F : A \to Set$ the following conditions are equivalente:
$F$ preserves binary products;
$F \in \{g\}^\perp$, where $g : A(X,\_\ )\sqcup A(Y,\_\ ) \to A(X\times Y; \_\ )$ is (induced by) the image under Yoneda of the terminal cone $X \leftarrow X \times Y \to Y$.
Of course, the same equivalent conditions apply to more general shapes of diagrams! $F$ preserves all limits of diagrams of shape $\Gamma \subset \sf Cat$ if and only if $F$ is right orthogonal to the set $\{\text{colim } A(\gamma,\_\ ) \to A(\lim\gamma, \_\ ) \mid \gamma \in\Gamma\}$.
Now, the SOA entails that there is a chain of endofunctors $R_\alpha : [A,Set] \to [A,Set]$ that "eventually" fall into $[A,Set]_\times$, or $[A,Set]_\Gamma$: this means that a sufficiently large transfinite composition $R^\kappa : [A,Set]\to [A,Set]_\times$ fall into product-preserving, or $\Gamma$-continuous, functors.
I have no problem with this theorem, apart its complete lack of explicitness:
How can I get a more concrete description of what is the universal product-preserving functor $\tilde F$ associated to $F$ under the reflection $[A,Set]_\times \leftrightarrows [A,Set]$?
I think the small object argument is the most explicit you can get in full generality. But I disagree with you that it 'completely lack explicitness': it presents $\widetilde{F}$ as a colimit of an explicitly constructed $\omega$-chain. Of course there are special cases where you can get a better description, for example, if this localization happen to be left exact, or if you have a good understanding of the category $[A,Set]_\times$, as you question boils down to computing colimits in it.
Thank you. My question stems from the particular case when $A$ is a Lawvere theory; is it possible to say something more in that particular case, when $A$ is such a concrete object?
|
2025-03-21T14:48:30.102431
| 2020-03-20T17:36:42 |
355325
|
{
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"authors": [
"Mare",
"Todd Leason ",
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"https://mathoverflow.net/users/14094",
"https://mathoverflow.net/users/17734",
"https://mathoverflow.net/users/61949"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/355325"
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|
Stack Exchange
|
Local submodules of finite rings
Let $A$ be a finite local ring $R$, such as a group algebra of $p$-groups over a finite field of characteristic $p$.
Question 1: Is it possible, using GAP, to obtain the poset of all submodules of $R$ of the form $uR$ up to equality where $u$ is an elements of $R$ (that are exactly the local submodules of $R$, that is the submodules with simple top)?
Question 2: Let $G_n$ be an elementary abelian $p$-group of order $p^n$ and $K$ the field with $p$ elements. What is the number of such submodules of the form $uR$ in this case when $R=KG$?
For $p=2$, the sequence starts with 2,5,23 for $n=1,2,3$. I wonder what the next term is?
Do you assume local rings are commutative?
@YCor No, but it is an interesting special case.
"such as a group algebra over a finite field": Should that mean that all group algebras over a finite field are local or should it mean that there are group algebras over finite fields that are local? Note that a group algebra kG over a field of char. p and G finite is local iff G is a p-group.
@ToddLeason thanks I added that.
The following lines of code should work for a finite ring where all the commands used are defined (here there is an example of a group algebra):
G := SmallGroup( 8, 5 );
KG := GroupRing( GF(2), G );
rightideals := [ ];
rightideals := Unique( List( Elements( KG ), u -> RightIdeal( KG, [ u ] ) ) );
n := Length( rightideals );
relations := List( [ 1..n ], i -> [ i ] );
for i in [ 1..n - 1 ] do
for j in [ i + 1..n ] do
I := rightideals[ i ];
J := rightideals[ j ];
if Intersection( I, J ) = I then
Add( relations[ i ], j );
elif Intersection( I, J ) = J then
Add( relations[ j ], i );
fi;
od;
od;
_P := [ 1..n ];
D := Domain( _P );
_P := Elements( D );
rel := [];
for r in relations do
for i in [ 2..Length( r ) ] do
Add(rel, DirectProductElement( [ Elements( D )[Position( _P, r[ 1 ] ) ], Elements( D )[ Position( _P, r[ i ] ) ] ] ) );
od;
od;
P := BinaryRelationByElements( D, rel );
P := ReflexiveClosureBinaryRelation( P );
P := TransitiveClosureBinaryRelation( P );
H := HasseDiagramBinaryRelation( P );
UnderlyingRelation( H );
However, it might take a while for big finite rings.
|
2025-03-21T14:48:30.102705
| 2020-03-20T18:28:57 |
355328
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/355328"
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|
Stack Exchange
|
On the global dimension of an endomorphism algebra
Let $G_n$ be the elementary abelian 2-group with $2^n$ elements and $R=R_n:=KG$ the group algebra over the field with 2 elements.
Let $M_n$ be the direct sum of all non-projective modules of the form $uR$ for some element $u \in R$ up to isomorphism. Let $B_n:=End_{R_n}(M_n)$.
Question: For which $n$ does $B_n$ have finite global dimension?
Interestingly, $B_n$ has Cartan determinant equal to -1 for $n=3$ and has finite global dimension for $n=1$ and $n=2$. So in case $B_3$ has finite global dimension, this would give a counterexample to the Cartan determinant conjecture. Sadly $B_3$ has vector space dimension 813 and thus even with a computer it looks impossible to understand the algebra $B_3$ directly, but maybe there is a trick?
In my answer to does-this-algebra-have-finite-global-dimension-human-vs-computer I recalled how one can go about computing a projective resolution of the simple modules of an endomorphism ring of a module. There is a missing piece of code in QPA to do this in general, but now I have made an attempt to rectify this. The code is not made public yet, but hopefully soon it will. Using this new code it seems that for one simple $B_3$-module, the third and the fourth syzygy of this particular simple $B_3$-module has a common isomorphic direct summand. Hence the projective dimension of this particular simple $B_3$-module is infinite and consequently the global dimension of $B_3$ is also infinite. I am stressing that the calculations only indicating this, I don't have a proof of this.
Addition April 17, 2020: Here is a copy of the GAP-session computing the above (using the newly uploaded additions to QPA as of April 17th, 2020):
gap> G := ElementaryAbelianGroup( 8 );
<pc group of size 8 with 3 generators>
gap> A := GroupRing( GF( 2 ), G );
<algebra-with-one over GF(2), with 3 generators>
gap> B := AlgebraAsQuiverAlgebra( A )[ 1 ];
<GF(2)[<quiver with 1 vertices and 3 arrows>]/<two-sided ideal in <GF(2)[<quiver with 1 vertices and 3 arrows>]>, (6 generators)>>
gap> elements := Elements( RadicalOfAlgebra( B ) );;
gap> mods := List( elements, m -> RightAlgebraModule( B, \*, RightIdeal( B, [ m ] ) ) );;
gap> Mods := List( mods, m -> RightAlgebraModuleToPathAlgebraMatModule( m ) );
[ <[ 0 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 2 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>,
<[ 4 ]>, <[ 2 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 2 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>,
<[ 4 ]>, <[ 2 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 2 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>,
<[ 4 ]>, <[ 4 ]>, <[ 2 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 2 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>,
<[ 4 ]>, <[ 4 ]>, <[ 1 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 2 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>,
<[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 2 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 2 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>,
<[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 2 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 2 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>,
<[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 2 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 2 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>,
<[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]> ]
gap> n := Length( Mods );
128
gap>
gap> non_isos := [ ];
[ ]
gap> isos := [ ];
[ ]
gap> multiplicities := [ ];
[ ]
gap> for i in [ 2..n - 1 ] do
> if not i in isos then
> Add( non_isos, i );
> testset := [ i + 1..n ];
> SubtractSet( testset, isos );
> num := 0;
> for j in testset do
> if IsomorphicModules( Mods[ i ], Mods[ j ] ) then
> num := num + 1;
> Add( isos, j );
> fi;
> od;
> Add( multiplicities, num + 1 );
> fi;
> od;
gap>
gap> basiclist := List( non_isos, i -> Mods[ i ] );
[ <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 2 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 4 ]>, <[ 2 ]>, <[ 4 ]>, <[ 4 ]>,
<[ 2 ]>, <[ 2 ]>, <[ 4 ]>, <[ 2 ]>, <[ 2 ]>, <[ 2 ]>, <[ 1 ]> ]
gap> Length( basiclist );
22
gap> test := ProjectiveResolutionOfSimpleModuleOverEndo( basiclist, 1, 4 );
[ "projdim > 4", [ <[ 3 ]>, <[ 10 ]> ] ]
gap> U := test[ 2 ][ 1 ];
<[ 3 ]>
gap> V := test[ 2 ][2];
<[ 10 ]>
gap> decomp := DecomposeModule( V );
[ <[ 3 ]>, <[ 7 ]> ]
gap> IsomorphicModules( U, decomp[ 1 ] );
true
|
2025-03-21T14:48:30.102922
| 2020-03-20T19:03:30 |
355329
|
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"Devlin Mallory",
"Evgeny Shinder",
"https://mathoverflow.net/users/111491",
"https://mathoverflow.net/users/90515"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/355329"
}
|
Stack Exchange
|
induced morphism through normalization
Hello i have a little question.Let $f:X\rightarrow Y$ where $X$ is the normalization of $Y$ (projective algebraic variety). Is $$ Hom(\mathbb{P}^{1},X) \rightarrow Hom(\mathbb{P}^{1},Y)$$ finite and surjective ?
I know that $X \rightarrow Y$ is finite and surjective by the fact that $X$ is normalization and $ Hom(\mathbb{P}^{1},X) \rightarrow Hom(\mathbb{P}^{1},Y)$ is surjective by the universal property of normalization. Anyone could help me with this thank u.
I think the map is not surjective in general. That is, if the image of $\mathbf{P}^1$ lands in the singular locus of $Y$, there is no reason why the map can be lifted to $X$. One can find an example for surfaces, and I think one such example is a surface $x^2 u = y^2 v$ in coordinates $[x,y,z] \in \mathbf{P}^2$, $[u,v] \in \mathbf{P}^1$. Its singular locus is isomorphic to a projective line, and taking normalization yields a connected $2:1$ cover on the exceptional locus, hence there is no lifting for the identity map on the exceptional locus.
As to why your argument using the universal property of normalization doesn't go through, see the answer at https://mathoverflow.net/questions/46/what-is-the-universal-property-of-normalization: the issue is that the universal property only allows you to lift maps $Z\to X$ up to $Y$ if they send associated points of $Z$ to those of $X$; if $\dim X\geq 2$, this won't be the case for maps $\mathbb P^1\to X$.
|
2025-03-21T14:48:30.103036
| 2020-03-20T19:12:39 |
355331
|
{
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"authors": [
"Evgeny Shinder",
"https://mathoverflow.net/users/111491"
],
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"sort": "votes",
"url": "https://mathoverflow.net/questions/355331"
}
|
Stack Exchange
|
The Picard scheme of an ordinary singular curve
Let $k$ be an algebraically closed field, $C$ a proper reduced connected scheme over $k$ of dimension 1, whose singularity is at worse ordinary, $\pi : \tilde{C} \to C$ the normalization of $C$ and $D_i$ be the irreducible components of $\tilde{C}$.
And write the structure morphism $f:C \to k, g : \tilde{C} \to k$.
Then, since "group algebraic spaces of locally of finite type" over a filed are schemes, its Picard functor is representable by a scheme, and denote its identity component by $J$.
Since $C$ is a curve, $J$ is smooth, so is an algebraic group.
How can I show that $J$ is semi abelian?
Here is what I have tried:
This is written in ch.9.2, proposition 10 in Bosch et. al.'s Neron Models.
First, write $r = (\text{the number of irreducible components of } C)$, $C_{\text{sing}} = \{ x_i \}_{i = 1, \dots, N}$, $ \pi^{-1}(x_i) = \{ x_{ij} \}_{j = 1, \dots, n_i}$ and $M$ the rank of $H^1(\Gamma, \mathbb{Z})$, where $\Gamma$ is the graph associated with $C$.
We have a short exact (S) $ 1 \to \mathscr{O}_C^* \to \pi_*\mathscr{O}_\tilde{C}^* \to \mathscr{Q} \to 1 $, where $\mathscr{Q} = \oplus_i k_{x_i}^{n_i - 1}$, where $k_x$ is the skyscraper sheaf at $x \in C$ associated with $k$.
Then the author says that we have the long exact
$$ 1 \to f_* \mathbb{G}_{m, C} \to f_* \pi_* \mathbb{G}_{m, \tilde{C}} \to f_* \mathscr{Q} \to R^1 f_* \mathbb{G}_{m, C} \to R^1 (f_* \pi_*) \mathbb{G}_{m, \tilde{C}} \to 1 $$
in the big etale topology on $\operatorname{Spec}k$.
I don't understand why.
(And I don't know what is $\mathscr{Q}$ as an etale sheaf.
A quasi-coherent sheaf of module (in the usual sense) induces the big etale sheaf.
But now $\mathscr{Q}$ is not a quasi-coherent module.)
I think there exists the "big etale version" of (S), $ 1 \to \mathbb{G}_{m, C} \to \pi_* \mathbb{G}_{m, \tilde{C}} \to \mathscr{C} \to 1 $ on the big etale topology of $C$, where
$\mathscr{C} = \oplus_i i_{x_i *} \mathbb{G}_{m, k}^{n_i - 1}$, $i_x$ is the canonical morphism $\{x\} \to C$.
If so, then since $R^1 f_* \mathbb{G}_{m, C} = \operatorname{Pic}_C, (R^1 f_*) \circ \pi_* = R^1 g_*, R^1f_* \mathscr{C} = 0$ and $f_* \mathscr{C} = \mathbb{G}_{m, k}^M$, we have
$$ 0 \to \mathbb{G}_m \to \mathbb{G}_m^r \to \mathbb{G}_m^M \to \operatorname{Pic}_{C/k} \to \operatorname{Pic}_{\tilde{C}/k} \to 0$$
on the big etale topology on $\operatorname{Spec}k$, hence ok.
How can I show it?
I could show it for the standard Neron n-gon (over a general scheme $S$).
(For the definition of Neron polygon, see the paper of Deligne-Rapoport.)
But I could not show it for general stable curves.
Any other proof will be appreciated, but if possible, I want online-available references.
Thank you very much!
Are you asking why Jacobian of a nodal curve is an extension of an abelian variety (in fact, the Jacobian of its normalization) by a torus? This is a result of Oort 1962, https://link.springer.com/article/10.1007%2FBF01440949, Prop. 2.3.
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2025-03-21T14:48:30.103223
| 2020-03-20T20:16:31 |
355335
|
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"Ihnatus",
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|
Stack Exchange
|
A distribution of maximum of sums if add to the minimal
Consider a vector of $n$ integer variables with initial values of 0. Each step we take random $w_i\thicksim NB(q, l)$ (independent randon values with the same negative binomial distribution) and add it to the variable with the minimal current value, even cases go to the leftmost of the equal variables. What is the CDF of the maximal value in the vector at a given step?
This is what stochastic simulation is for.
But it looks like a common optimization case, may there be an analytic expression?
|
2025-03-21T14:48:30.103289
| 2020-03-20T20:21:46 |
355336
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/355336"
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|
Stack Exchange
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galois deformation ring with type is union of irreducible components
Notation:
$K$ finite extension of $\mathbb{Q}_p$, $G_K$ absolute Galois group of $K$,
$E$ is finite extension of $\mathbb{Q}_p$ (coefficient field), $O_E$ is ring of integer in $E$.
In this paper of Toby Gee, (after lemma 2.1.3) for a given complete noetherian local $O_E$-algebra $A^\circ$ with finite residue field, and rank $n$ free $A^\circ$-module $V_{A^\circ}$ with $G_K$-action, there is a quotient $A^\tau$ of $A=A^\circ[1/p]$ such that a homomorphism $A\xrightarrow{} B$ where $B$ is a finite $E$-algebra, it factors through $A^\tau$ if and only if the Weil–Deligne representation associated to $V_B=V_{A^\circ}\otimes B$ has type $\tau$ (i.e. $\tau$ is representation of $I_K$ on $E$-vector space with open kernel, and $I_K$-action on the Weil-Deligne representation associated to $V_B$ is isomorphic to $\tau\otimes B$)
He adds that $A^\tau$ is union of irreducible components of $A$, because of the fact that the cohomology of finite group of positive degree with coefficients in characteristic 0 vanishes.
I want to understand this part, and I also tried to look at this paper of Kisin, which Toby Gee was following closely. He proved analogous result in theorem 2.7.6, and refer subsection 2.7 to argue that the trace of elements in $I_K$ is locally constant on $\operatorname{Spec}A$.
I interpret this as Zariski locally (or étale locally?) the trace of $I_K$ is valued in some finite extension of $E$ and so the action of $I_K$ comes from a representation on $E'$-vector space.
What I don't understand about subsection 2.7 in Kisin's paper is the existence of $B'$ (we can just add necessary coefficients from $\overline{K_0}$ and it will ensure étale but why local?) and how it is used in the context of theorem 2.7.6 (does $B'$ provide étale covering of $A$? even then it is not a component in $\operatorname{Spec}A$) and why the cohomology of finite group is mentioned here.
At last, theorem 2.7.6 is using complete noetherian local $O_E$-algebra, but subsection 2.7 is using finite local $E$-algebra. I don't think this finiteness will really matter, but I would be glad if one can make this point clear.
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2025-03-21T14:48:30.103443
| 2020-03-20T20:38:23 |
355337
|
{
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"Fabian Meumertzheim",
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"url": "https://mathoverflow.net/questions/355337"
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|
Stack Exchange
|
Is having no rational point always witnessed over a place?
Let $K$ be a finitely generated extension of $\mathbb{Q}$ of transcendence degree at least 1.
Recall that a valuation ring of $K/\mathbb{Q}$ is a sub-$\mathbb{Q}$-algebra $V\subset K$ such that for every $x\in K$ we have $x\in V$ or $x^{-1}\in V$. A place of $K/\mathbb{Q}$ is a $\mathbb{Q}$-algebra homomorphism $p\colon V\to L$, where $V$ is a valuation ring of $K/\mathbb{Q}$, $L$ is a field and $p(x^{-1})=0$ for any $x\in K, x\not\in V$. The image of $p$ is an extension of $\mathbb{Q}$, which is called the residue field of $p$.
If $X/\mathbb{Q}$ is a projective variety with a $K$-rational point and $L$ is the residue field of a place of $K/\mathbb{Q}$, then it is easy to see that $X$ admits an $L$-rational point.
My question: Assume now that we are given a projective variety $Y/\mathbb{Q}$ that does not admit a $K$-rational point. Is there always a place $p$ of $K/\mathbb{Q}$ with residue field $L$, where $\operatorname{trdeg}_{\mathbb{Q}} L < \operatorname{trdeg}_{\mathbb{Q}} K$, such that $Y$ also does not admit an $L$-rational point?
If $K=L(t)$ for $t$ trascendental over $L$, $L$ is a residue field but also $L$ embeds into $K$, so any $L$-rational point can be made a $K$-rational point?
@GTA Thanks for your comment! If I understood you correctly, this does seem to resolve my question in the special case where K can be written as a purely transcendental extension of an extension of $\mathbb{Q}$. But if we choose $K$ to be e.g. $\mathbb{Q}(t, \sqrt{t^3-t})$, this trick does not seem to apply anymore. Do you maybe have an idea how to handle the general case?
The trick will work as long as there exists a residue field that embeds into $K$. Say that $K$ is the function field of a curve $C$ over a number field $k$. Usually $C$ is geometrically connected, which means that $k$ is algebraically closed in $K$. This implies that any residue field embedding into $K$ must be equal to $k$. If $C$ has a $k$-rational place, then the answer to your question is yes. But in general the $k$-rational places correspond to the $k$-rational points of the normalisation of $C$, and this set can be empty.
I see now that this trick will only work in rather special situations. Do you know whether there is anything else one can do?
|
2025-03-21T14:48:30.103604
| 2020-03-20T20:54:13 |
355338
|
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"url": "https://mathoverflow.net/questions/355338"
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|
Stack Exchange
|
Definition of functions in the induced space from parabolic induction
Let $P$ be a parabolic subgroup of a connected, reductive group $G$ over a $p$-adic field. Let $M$ be a Levi subgroup of $P$, and let $N$ be the unipotent radical of $P$. If $(\pi,V)$ is a smooth, irreducible representation of $M$, extend $\pi$ to a representation of $P$ by making it trivial on $N$, and let $\sigma = \operatorname{Ind}_P^G \pi$, the smooth representation of $G$ obtained by parabolic induction.
By definition, a function $f: G \rightarrow V$ lies in the space of $\sigma$ if the following conditions are met:
$f$ is locally constant.
$f(mng) = \pi(m)f(g)$ for all $m \in M, n \in N, g \in G$.
There exists an open compact subgroup $K$ of $G$, depending on $f$, such that $f(gk) = f(g)$ for all $g \in G$ and $k \in K$.
Is the third condition redundant in this definition? I know in the general case for smooth induction in totally disconnected groups, it is necessary, but I have thought that since $P \backslash G$ is compact, there should be some way to show the third condition from the first two. I haven't been able to do this. I have seen some authors leave out the third condition in the definition of parabolic induction.
Let $H$ be any subgroup such that $H\backslash G$ is compact. Let $K$ be an open subgroup. Then there are $x_1,...x_n$ such that $G=Hx_1 K \cup...\cup Hx_nK.$ Suppose $f$ satisfies points 1 and 2. Replace $K$ with a smaller $K'$ so that $f(x_i k)=f(x_i)$ for all $i$ and for all $k \in K'.$ Given $a \in G,$ there are $h \in H, i,$ and $k' \in K'$ such that $a=hx_i k'$. Let $k \in K'$. Then $f(ak)=f(h x_i k' k)= \pi(h)f(x_i k' k)=\pi(h)f(x_i k')=f(h x_i k')=f(a).$
So $f$ satisfies point 3.
|
2025-03-21T14:48:30.103740
| 2020-03-20T22:07:39 |
355341
|
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"Ali Taghavi",
"Anton Petrunin",
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|
Stack Exchange
|
Riemannian manifolds which admit a smooth free $\mathbb{Z}/3\mathbb{Z}$ action but do not admit an equilateral triangle action
A free action of $\mathbb{Z}/3\mathbb{Z}$ on a Riemannian manifold $(M, g)$ is called an equilateral action if for every $x\in M$ all three points of orbit of $x$ have the same distance from each other.
What is an example of a Riemannian manifold which does not admit such an action but it already admit a smooth free action by cyclic group of order $3$?
Are you assuming compactness or completeness? If you don't I think you can find uninteresting counter examples of the form $M=\Bbb{R}^2\setminus{0,p, Ap, A^2p}$ where $A$ is a matrix that is conjugate to a rotation by $2\pi/3$, $p$ is any nonzero point and the Riemannian structure is induced from that of $\Bbb{R}^2$. I would expect you can make this into a complete example by tampering with the metric. So maybe you should assume compactness?
@OlivierBégassat I think you mean we do not remove $A^2p$, yes?otherwise the rotation irself give the required action right?
In this case what is the free action which already exist?
$A$ is only conjugate to a rotation (i.e. satisfies $A^3=I_2$) but it's not an isometry. The Z/3Z action is that of $A$ on $M$. I'm removing two orbits from the action of $A$ on $\Bbb{R}^2$.
@OlivierBégassat for sure it works for the plane minus 4 points, none 3 of which form an equilateral triangle.If $f$ is a homeomorphism, it acts on the end compactification, which is a 2-sphere. If it fixes all end points, then this is a homeo of order 3 with exactly 5 fixed points and this doesn't exist. So it permutes 3 and fixes the last two. A limit argument shows that the point at infinity is fixed, and that the three moved ones form an equilateral triangle.
@YCor I don't understand. The three points here don't form an equilateral triangle, and neither will the orbit of a point close to $p$.
I guess you want a manifold with a free action of $\mathbb{Z}_3$ such that there is no equilateral action for ANY Riemanninan metric. (Otherwise the question has no sense.)
@OlivierBégassat Precisely, I'm giving a proof of your assertion (that there's no "equilateral action") for the plane minus 4 points, none of which form an equilateral triangle.
@AntonPetrunin for any metric space an equilateral action makes sense (a free continuous action of $C_3$ in which every orbit is equilateral). So asking for a Riemannian manifold with no such action is perfectly meaningful. While your modified question is quite trivial: if there's a free $C_3$-action, you can average to get an invariant Riemannian metric.
@YCor Right,. ${}$
@YCor I thought that $|a(x)-x|=const$.
Here's a compact example without boundary (a 2-torus).
Choose a topological disc $D$ on the two-torus $T$, and choose a Riemannian metric so that $D$ has very close Gromov-Hausdorff distance (say $\le 1$) to a segment of length $20$, while $T\smallsetminus D$ has diameter $\le 1$ (so metrically $D$ is predominant, while all the homotopic part lies in $T\smallsetminus D$. Also assume (just to fix ideas) that there exists $x_0\in D$ such that $d(x_0,D)=20$ (so $x_0$ is the "opposite tip" of the segment). Again to fix ideas, suppose that $x_0$ is part a geodesic segment $(x_t)_{0\le t\le 10}$ such that $\sup_{x\in T}\inf_{t\in [0,20]}d(x,x_t)\le 1$.
(On the picture, the complement of $D$ is the little left tip, including the handle, and $x_0$ is at the right tip.)
Then $T$ (with this metric) has no large equilateral triangle: indeed if $r$ is the size of an equilateral triangle, such a triangle would be 1-close to a "triangle" $\{x_{t_1},x_{t_2},x_{t_3}\}$, with $t_1\le t_2\le t_3$ and $||t_i-t_j|-a\le 2$ for all $i\neq j$. So $t_3-t_1\le a+2$, $t_3-t_2,t_2-t_1\ge a-2$, hence $t_3-t_1\ge 2a-4$, so $2a-4\le a+2$, i.e., $a\le 6$.
Hence, if $f$ is an "equilateral" self-homeomorphism, we have $d(f(x),x)\le 6$ for all $x$. Let $B$ be the open $7$-ball around $x_0$. Then the open subset $U=B\cup f(B)\cup f^{-1}(B)$ is $f$-invariant, and contained in the $13$-ball around $x_0$. The connected component $U'\subset U$ of $x_0$ contains the $f$-orbit of $x_0$.
Let $T'$ be the quotient $T/\langle f\rangle$, so $T\to T'$ is a connected covering of degree 3. Since $\pi_1(U',x_0)\to \pi_1(T,x_0)$ is trivial, the covering is trivial in restriction to $U'$. We get a contradiction, since $U'$ is connected and contains as fiber the orbit of $x_0$.
I think you mean $a$ for the size of the equilateral triangle, and $x_0$ satisfies $d(x_0,T\setminus D)=20$. Very nice answer by the way, this is also how I read the OP's question.
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2025-03-21T14:48:30.104027
| 2020-03-20T22:17:24 |
355343
|
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|
Stack Exchange
|
Intersection of primes in a regular local ring
Suppose $R$ is a regular local ring of dimension at least 3, and suppose $P_1, P_2$ are dimension 1 primes. Does there necessarily exist a dimension 2 prime $Q$ contained in both?
In other words, is every pair of "little curves" contained in a "little surface"?
If this is false, will it hold if the ring is complete?
I am particularly interested in the mixed characteristic case.
Can you first show that $P_1\cap P_2\neq 0$?
Yes, since their product $P_1 P_2$ will be nonzero.
I prove below that the answer is positive in the following special case: $R$ is the localization of a finitely generated polynomial ring over a field $k$. By Cohen's structure theorem this will also cover the case that $R$ is finite dimensional, complete and equicharacteristic.
For the proof it suffices to consider the case that $k$ is algebraically closed and $R$ is the localization of $k[x_1, \ldots, x_n]$ at the maximal ideal generated by $x_1, \ldots, x_n$. let $C_1, C_2$ be two irreducible curves on $k^n$ passing through the origin. It suffices to show that there is an irreducible surface $S$ containing both $C_1$ and $C_2$.
Claim: there is a curve $C$ and non-constant (and therefore dominant) maps $\phi_j: C \to C_j$, $j = 1, 2$.
Proof of the claim: Take any polynomial $f$ which is non-constant on both $C_j$. The fields $k(C_j)$ of rational functions on $C_j$ are finite extensions of $k(f)$, so that there is a common finite extension $L$ of $k(C_1)$ and $k(C_2)$ (see this for a neat trick). Now define $C$ to be the (unique) nonsingular curve such that $k(C) = L$.
It is easy to construct $S$ with $C, \phi_1, \phi_2$ from the claim: take the closure of the image of $\phi: C \times k \to k^n$ given by $\phi(x, t) := t\phi_1(x) + (1-t)\phi_2(x)$.
Thanks for that -- I am actually mostly interested in the mixed characteristic case though! I will clarify this in the question.
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2025-03-21T14:48:30.104179
| 2020-03-20T22:19:26 |
355344
|
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"Dmitri Pavlov",
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|
Stack Exchange
|
What is a morphism of ∞-sites?
Recall that a morphism of sites
is a covering-flat functor
that preserves covering families.
Morphisms of sites can be identified with those
geometric morphisms of induced toposes
for which the inverse image functor preserves representables.
If both sites have finite limits, then covering-flat functors are precisely the functors that preserve finite limits.
What is the analogue of this story for ∞-sites?
Again, we want to identify morphisms of ∞-sites with those
geometric morphisms of induced (nonhypercompleted) ∞-toposes
for which the inverse image functor preserves representables.
Thus, a morphism of ∞-sites should be a functor between the underlying ∞-categories
that preserves covering families and satisfies the appropriate ∞-analogue of being a covering-flat functor.
What is a covering-flat functor between ∞-sites?
If both ∞-sites admit finite ∞-limits, then covering-flat functors could be defined
as finite ∞-limit preserving functors.
However, I am interested in ∞-sites that do not have finite ∞-limits,
such as the site of smooth manifolds or the site of cartesian spaces.
I havn't really thought about it, but you might get two different answer depending if you use site to represente hypercomplete $\infty$-topos or topological localization of presheaves categories...
@SimonHenry: The toposes in this question are not hypercompleted, although an answer in the hypercompleted case would also be interesting.
the appendix of this paper develops some of this theory (but maybe not in the generality you wanted?): https://arxiv.org/abs/1803.01804
@DylanWilson: Definition A.10 of morphisms of ∞-sites in this paper assumes the existence of pullbacks, i.e., finite ∞-limits. Also, Proposition A.13 constructs geometric morphisms of ∞-toposes under a rather strong additional assumption of a covering lifting property.
@DmitriPavlov that's why I said it might not be the generality you want. but not all pullbacks are assumed, only pullbacks along covers. so the site of smooth manifolds is okay here, right? anyway, I agree the theory in that appendix is probably not quite what you want, and only mention it because I thought it may still be helpful.
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2025-03-21T14:48:30.104340
| 2020-03-21T00:09:45 |
355346
|
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|
Stack Exchange
|
Dualizing complex description in Stacks project
The question is closely related to this one this one (more precisely the reference the comment by AGl earner) and is aimed to understand the proof of Lemma 20.2 from notes from Stacks notes from Stacks on dualizing complex. Lemma 20.2 states
Lemma 20.2. Let $(A,m, \kappa=A/m)$ be a Noetherian local ring with normalized dualizing
complex $\omega^{\bullet}_A$
and dualizing module $\omega_A:= H^{- \dim A}\omega^{\bullet}_A$. The following are equivalent:
(1) $A$ is Cohen-Macaulay,
(2) $\omega^{\bullet}_A$ is concentrated in a single degree, and
(3) $\omega^{\bullet}_A= \omega_A[\dim(A)]$. (i.e. a complex that is concentrated in degree $\dim(A)$ and is zero in other degrees)
In this case $\omega$ is a maximal Cohen-Macaulay module. The Proof refers to Lemma 16.7:
Lemma 16.7.
Let $(A,m, \kappa=A/m)$ be a Noetherian local ring with normalized dualizing complex $\omega^{\bullet}_A$. Let $M$ be a finite $A$-module. Here $M$ is considered as complex (or a class of this complex in $D(A)$) with $M[0]=M$ and $M[i]=0$ for $i \neq 0$. The following are equivalent
(1) $M$ is Cohen-Macaulay,
(2) $Ext^i _A(M,\omega^{\bullet}_A)$ is nonzero for a single $i$,
(3) $Ext^{-i} _A(M,\omega^{\bullet}_A)$
is zero for $i \neq \dim(Supp(M))$.
I not understand why 16.7 imply the $(1) \Rightarrow (3)$ in 20.2: why does it imply that $\omega^{\bullet}_A= \omega_A[\dim(A)]$? I guess that we apply 16.7 with $M=A$.
Then by definition for two complexes $X^{\bullet}, Y^{\bullet}$ Ext is defined by
$$Ext_A ^i(X^{\bullet}, Y^{\bullet}):= Hom_{D(A)}(X^{\bullet}, Y^{\bullet}[i])= Hom_{D(A)}(X^{\bullet}[-i], Y^{\bullet})$$
Moreover $Ext_A ^i(X^{\bullet}, Y^{\bullet})= H^i (RHom_{D(A)}(X^{\bullet}, Y^{\bullet}))$. If $Y^{\bullet} \to I^{\bullet}$ is an injective resolution, then we can calculate it by $Ext_A ^i(X^{\bullet}, Y^{\bullet})=Hom_{K(A)}(X^{\bullet}, I^{\bullet}[i])$ and similary with aprojective resolution of $X^{\bullet}$.
$A$ has most natural projective resolution $I^{\bullet} _A \to A$ by $... \to 0 \to A \xrightarrow{\text{id}} A $.
thus we obtain $Ext^i _A(A,\omega^{\bullet}_A) = Hom_{D(A)}(A,\omega^{\bullet}_A[i])= Hom_{K(A)}(I^{\bullet} _A, \omega^{\bullet}_A[i])=Hom_A(A, \omega^{\bullet}_A[i]_0)= (\omega^{\bullet}_A)_i$.
now comes the part which I not understand:
Why $\omega^{\bullet}_{-d}= H^{-d}(\omega^{\bullet}) =
\omega_A[d]_{-d}= \omega_A$ (recall definition of
$\omega_A$ above)?
Since we have observed that $Ext^i _A(A,\omega^{\bullet}_A) =
\omega^{\bullet}_i$ we have "only" to show that
$Ext^{-d} _A(A,\omega^{\bullet}_A)= H^{-d}(\omega^{\bullet})$.
Why is it true?
If we continue as before then $Ext^{-d} _A(A,\omega^{\bullet}_A)
= H^{-d}(Hom_{D(A)}(A, \omega^{\bullet}_A)=
H^{-d}(Hom_A(A, \omega^{\bullet}_A)=
= H^{-d}( \omega^{\bullet}_0)= \omega^{\bullet}_0$. This is nonsense.
|
2025-03-21T14:48:30.104621
| 2020-03-21T07:19:30 |
355353
|
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"Manfred Weis",
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|
Stack Exchange
|
Algorithms for heaviest edge-disjoint cycle collection contained in graph's set of edges
given a biconnected symmetric graph with weighted edges,
what is the algorithmic complexity of determining a set of pairwise edge-disjoint cycles with maximal sum of edge weights if there are no other constraints besides edge-disjointness of the cycles and maximal weightsum of their edges?
Determing such a set of cycles is a stepping stone in an algorithm for determining a heaviest euler tour in complete symmetric graphs with $n=2k$ vertices (which isn't eulerian), which in turn would yield an improved heuristic for the non-eulerian windy postman problem.
The problem is NP-hard, even in the unweighted case (all weights equal to $1$).
Indeed, given a graph $G$ and an integer $k$, deciding if $G$ contains an Eulerian subgraph with at least $k$ edges is NP-complete.
However, the problem is fixed parameter tractable (FPT) with respect to the parameter $k$. See this paper of Fomin and Golovach, and the references therein.
In the cited paper I read that it is concerned with packing a graph with "a maximal set of disjoint cycles". My question doesn't ask for finding a set of maximal cardinality, but for a set whose elements have the highest weightsum; that difference may imply a different algorithmic complexity, depending on the weights.
You are correct. I edited my answer accordingly to the correct reference.
|
2025-03-21T14:48:30.104733
| 2020-03-21T07:22:20 |
355354
|
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"Dominic van der Zypen",
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"url": "https://mathoverflow.net/questions/355354"
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|
Stack Exchange
|
Minimal generating set for $S_\omega$
If $G$ is a group and $S\subseteq G$, let $\langle S \rangle$ be the intersection of all subgroups of $G$ containing $S$.
Let $S_\omega$ denote the group of all bijections $f:\omega\to\omega$ with composition.
Is there $M\subseteq S_\omega$ such that $\langle M \rangle =S_\omega$, but for all $m\in M$ we have $\langle M\setminus\{m\} \rangle\neq S_\omega$?
No.
Indeed, F. Galvin proved in 1995 that every countable subset of $S_\omega$ is contained in a finitely generated subgroup (and also $S_\kappa$ for every infinite $\kappa$). By contradiction suppose $M$ exists. Let $I$ be an infinite countable subset of $M$, so $I\subset \langle F\rangle$ for some finite $F$, and hence there exists a finite subset $J$ of $M$ such that $F\subset \langle J\rangle$. Hence, for $g\in I\smallsetminus J$, we have $g\in\langle J\rangle$, and therefore $M\smallsetminus\{g\}$ generates $S_\omega$.
More generally no such $M$ exists in any uncountable group $G$ with uncountable cofinality. Indeed fix $I\subset M$, write finite subsets $I_0\subset I_1\subset I_2...$ with union $I$, and define $M_n=(M\smallsetminus I)\cup I_n$. Then $G =\bigcup\langle M_n\rangle$ (increasing union). By definition of uncountable cofinality, $G=\langle M_n\rangle$ for some $n$: contradiction. Thus this shows that for every generating subset $M$, there exists $M'\subset M$ with $M\smallsetminus M'$ infinite, such that $M'$ generates $G$.
This also shows some stronger consequence: $G$ is not "infinitely independently generated": there is no sequence $(S_n)_{n\in\omega}$ of subsets of $G$ such that $\bigcup_k S_k$ generates $G$, but $\big\langle\bigcup_{k\neq n}S_k\big\rangle\neq G$ for every $n$. The latter condition has the advantage of being purely intrinsic to the poset of subgroups of $G$.
Brilliant - thanks @YCor!
Just about the digression at the end: the group $\mathbf{Q}$ has no minimal generating subset, but yet is infinitely independently generated (because it admits a quotient that is an infinite direct sum).
|
2025-03-21T14:48:30.104879
| 2020-03-21T07:26:03 |
355355
|
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"Gerald Edgar",
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|
Stack Exchange
|
For $\mathcal{L}^1$-a.e. $t\in\mathbf R$, is $n-1$ the Hausdorff dimension of level sets of a locally Lipschitz function $f:\mathbf R^n\to\mathbf R$?
Let $f:\mathbf R^n\to\mathbf R$ be a locally Lipchitz function. Denote $\mathrm H^n$ the $n$-dimensional Hausdorff measure. We know that for any $\mathrm H^n$-measurable subset $A\subset\mathbf R^n$, for $\mathcal{L}^1$-a.e. $t\in\mathbf R$, $A\cap f^{-1}（t)$ is $\mathrm H^{n-1}$-measurable.
Thanks to Gerald Edgar's comment, I should ask: For $\mathcal{L}^1$-a.e. $t\in\mathbf R$, $f^{-1}(t)$ is empty or has Hausdorff dimension $n-1$?
A constant function is certainly locally Lipschitz. Its level sets are either dimension $n$ or empty. So not dimension $n-1$. Maybe add "non-constant"? But even then you can have the set $t$ with $f^{-1}(t) = \varnothing$ of positive measure. So maybe: for almost all $t$, level set $f^{-1}(t)$ is empty or has dimension $n-1$?
Since $f$ is Lip, then we can apply the Co-area formula to obtain
$$\DeclareMathOperator{\Dm}{d\!}
\int |Df|=\int\limits_{-\infty}^{+\infty}\!\!{\mathrm H}^{n-1}\!\!\left\{\left(x\mid f=t\right)\right\}\Dm t,
$$ thus the conclusion holds.
|
2025-03-21T14:48:30.104979
| 2020-03-21T08:07:28 |
355357
|
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|
Stack Exchange
|
Terminology for transforming a directed acyclic graph into a tree
I am looking for the term of converting a directed acyclic graph (DAG) into a tree by traversing its topologically ordered nodes and copying the subtrees of the nodes with in-degree $> 1$.
Such a transformation appears in many questions in this site, such as here and here, and across the web (e.g here).
The term arboreal decomposition appears here (p. 818, Algorithm 1) and here (p. 145/177) seems strongly connected to what I am looking for, but not exactly the same. However, I could not find the exact term of this graph operation.
More specifically, I am currently writing a paper and I need
1. A reference that this transformation is well-defined
2. What's the name of this transformation?
3. Is there some graph theory terminology could be used to depict this operation on a DAG with exactly one node with in-degree $> 1$, better than depicting the graph explicitly? (to make my paper more elegant)
How does this compare with the McCammond expansion (see Section 2 in Jon McCammond, John Rhodes, Benjamin Steinberg, Geometric Semigroup Theory, arXiv:1104.2301v1 and §2.5 in John Rhodes, Anne Schilling, Unified theory for finite Markov chains, arXiv:1711.10689v3)?
I'm accustomed to calling this process "unraveling" the dag. Unfortunately, I don't remember where I learned this terminology and I'm unsure how common it is.
See p. 8 of this paper: https://arxiv.org/pdf/1904.02309.pdf. The paper is in the context of network theory, and discusses a procedure called TENN for expanding networks to rooted trees.
Did you every find an answer to this question?
This is a directed version of the "universal covering graph": https://en.wikipedia.org/wiki/Covering_graph#Universal_cover. Nodes in the expanded tree correspond to paths in the graph starting with a root node. Unfortunately, I can't find a reference for a directed version of this concept; people seem to care more about the undirected version, which typically gives you an infinite graph. But imposing direction seems like an obvious generalization to me.
|
2025-03-21T14:48:30.105165
| 2020-03-21T08:44:08 |
355359
|
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|
Stack Exchange
|
A variant of Lambert function
How to express the solution of $x^{x+1}=a$ using Lambert function? I know that the standard Lambert function can be used to describe the solution of $x^x=a$. I wonder if $x^{x+1}=a$ can be addressed similarly.
Sorry, my answer is wrong. As it was pointed out by "Simply Beautiful Art" $x\ne W(z)$ but $x=e^{W(z)}.$
It is known that $$W'(z)=\frac{W(z)}{z(1+W(z))}.$$
Let $z=\log t$. Then $x=W(z)$ is a root of the equation $x^x=t$, in particular $x\log x=\log t=z.$ It means that $$W'(z)=\frac{x}{(x+1)\log t}=\frac{1}{(x+1)\log x}=\frac{1}{\log x^{x+1}},\quad x^{x+1}=e^{\frac{1}{W'(z)}}.$$
So solution of the equation $x^{x+1}=a$ is $x=W(z)$, where $z$ is defined by $W'(z)=1/\log a.$
What about if $x$ is in the denominator, i.e., how to solve $\left(\frac{a}{x}\right)^{x+1}=b$? Thank you
@lchen It is not clear.
Ｉmean to solve the equation $\left(\frac{a}{x}\right)^{x+1}=b$.
@lchen I don't know.
Thank you all the same. I thought that it was related to Lambert function but could not figure out the solution.
@lchen If you'll replace $x$ by $-x$ and $a$ by $-a$, then you'll get $\left(\frac{x}{a}\right)^{x-1}=b.$
Huh I've never thought to use the derivative of the Lambert W function, interesting approach.
Correct me if I'm wrong, but if $z=\ln(t)$ and $x=W(z)$, then $z=xe^x$ and $t=e^z=e^{xe^x}\ne x^x$.
Assuming I haven't made a mistake, using this approach but modified with $\ln(x)=W(z)$ will give us $$W'(z)=\frac1{x(\ln(x)+1)}$$which is reduces to$$\frac e{ex\ln(ex)}$$which is invertible using the Lambert W function. It does go to show, however, that the inverse of $W'$ is solvable in terms of $W$ and is given by$$(W')^{-1}(z)=x\ln(x),~x=\frac e{W^{-1}(\ln(z)+1)}$$
@SimplyBeautifulArt You are write, my answer is wrong.
Thank you both.
|
2025-03-21T14:48:30.105298
| 2020-03-21T14:58:45 |
355376
|
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"Federico Poloni",
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|
Stack Exchange
|
Estimates on norm Hessian Matrix
Let $u:\Omega \rightarrow \mathbb{R}$ a twice differential function, with $\Omega$ a subset of $\mathbb{R}^n$.
Suppose that we have the following:
$$D^2u\geq - \dfrac{(1+K^2)^{1/2}}{\epsilon}I$$
where the inequality is in the sense of quadratic form, so it means
$$
D^2u+\dfrac{(1+K^2)^{1/2}}{\epsilon}I\text{ is positive semidefinite.}
$$
Now if $A$ is a matrix, let's define its norm of as
$$
|A|=\sqrt{\sum a_{ij}^2}.
$$
I want to show that $$\epsilon|D^2u|\geq 1.$$
My attempts:
First, I thought I could considerate different definitions of the "norm" of a matrix, since all norms on the space of matrices are equivalent.
Then I tried to use the fact that all the eigenvalues of $D^2u$ belong to the set $\big[-\epsilon^{-1}{(1+K^2)^{1/2}},+\infty\big)$.
Last, I tried with fact that the sum of the elements of the Hessian is greater than $n{\epsilon^{-1}}{(1+K^2)^{1/2}}$.
Nothing useful arose.
$D^2u=0$ satisfies that inequality, and does not have norm greater than $1/\epsilon$.
Anyway, once you fix that statement: for a symmetric matrix, $\rho(A) = |A| \leq |A|$, where $\rho(A)$ is the spectral radius (largest modulus of eigenvalue), $|A|$ is the Euclidean induced norm, and $|A|$ is the norm you defined (Frobenius norm). Hopefully this is enough to get what you need.
|
2025-03-21T14:48:30.105418
| 2020-03-21T15:22:34 |
355377
|
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|
Stack Exchange
|
Is there a rectangular tiling based on the Padovan sequence?
I'm thinking of developing a rectangular tiling based on the Padovan sequence (think of the Fibonacci mosaic). It seems like something that should exist, but I can't find anything in the literature. Do you know of something like this and where I can find it?
I can't think of "the Fibonacci mosaic" since I've never heard of it. Reference? Link?
"Padovan tiling" turns up on top of page 7 of https://people.csail.mit.edu/ddeford/DeFord_Enumerating_Distinct.pdf but with no illustration.
@GerryMyerson See the two top tilings here: https://en.wikipedia.org/wiki/Fibonacci_number. I'm using the term mosaic because different size tiles are used. And thanks for the link, but it's not quite what I'm after.
OK. You wouldn't take a tiling with equilateral triangles, would you? https://en.wikipedia.org/wiki/Padovan_sequence
No. That one is very well known. I've done that one and the plastic pentagon gnomon as well.
I think your question is quite vague. It is not hard to build a sequence of plane filling rectangles in which some side lengths or areas resemble the Padovan sequence, but it is absolutely not clear whether these would satisfy you.
Can you please define rectangular tiling?
The Fibonacci numbers count number of ways to tile a $1\times n$-strip with rectangles of size $1 \times 1$ and $1\times 2$.
This is easy to see from the recurrence.
Now, the Padovan sequence count number of ways to tile a $1\times n$ rectangle with rectangles of size $1\times 2$ and $1\times 3$.
One needs to be a bit careful with initial conditions, so that there is one way to tile the $1\times 1$-rectangle.
By rectangular tiling I mean a whorled figure where the tiles circle around and increase in size according to some sequence or numbers raised to successive powers. See https://en.wikipedia.org/wiki/Fibonacci_number for an example with the Fibonacci sequence. Of course, the Padovan sequence would require rectangles rather than squares.
@CyeWaldman Which figure do you mean? (There are 10.) "Rectangular tiling" suggests the third one whose caption begins "Thirteen $(F_7)$ ways" showing tilings of a $1 \times 6$ board with $1\times1$ and $1\times 2$ tiles, in which case Per's answer seems appropriate.
@BrianHopkins Thanks for looking. I'm talking about the first and second figures on that web page. The point being to create a tessellated figure using only the integers in the sequence.
@CyeWaldman So you want a tiling of the plane using rectangles whose side lengths are sequential Padovan numbers, hopefully in some sort of spiral.
@BrianHopkins Well, that's exactly what I'm talking about and I've developed it already. I'm merely trying to find out if it's been done before. I didn't want to go posting it anywhere without verifying that it hasn't been done before. I apologize if I've caused any misrepresentations.
|
2025-03-21T14:48:30.105625
| 2020-03-21T15:55:26 |
355379
|
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|
Stack Exchange
|
Containment of $c_0$ in projective tensor products
Let $X$ and $Y$ be Banach spaces and denote by $X\hat{\otimes}_\pi Y$ the projective tensor product.
Question:
If $X\hat{\otimes}_\pi Y$ contains an isomorphic copy of $c_0$, must then $X$ or $Y$ contain an isomorphic copy of $c_0$ also?
The answer is no. Bourgain and Pisier have given a counterexample (A construction of $\mathcal{L}_\infty$-spaces and related Banach spaces. Bol. Soc. Bras. Mat. 14, No. 2, 109-123 (1983). See Zbl 0586.46011 https://zbmath.org/?q=an%3A0586.46011 ).
Dear Prof. Werner, thank you very much for such a quick reply and pointing me to such an interesting paper!
|
2025-03-21T14:48:30.105702
| 2020-03-21T16:05:22 |
355381
|
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"url": "https://mathoverflow.net/questions/355381"
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|
Stack Exchange
|
Is this Siegel's formula correct?
In the paper Zum Beweise des Starkschen Satzes Siegel considers the function
$$L_q(s)=\sum_{n=1}^{\infty}\left(\frac{q}{n}\right)n^{-s},$$
where $q$ is a discriminant of a quadratic number field and the character is the Kronecker symbol. Then he writes that "according to Dirichlet" we have, in case $G>0$,
$$L_G(1)=2G^{-1/2}h_G\log \varepsilon_G,$$
where $h_G$ is the corresponding class number and $\varepsilon_G$ the fundamental unit.
However, according to the book Zetafunktionen und quadratische Körper the formula reads
$$h_G=\frac{G^{1/2}}{\log \varepsilon_G}L_G(1).$$
Which one is correct? Am I making some mistake?
you mean where does the factor of two in the first formula come from?
@CarloBeenakker Yes, that is the problem.
Dirichlet's class number formula is quoted with this factor of two in several other sources, for example here.
Maybe it would be wiser to read Dirichlet's proof itself.
@CarloBeenakker Thanks for the reference.
Siegel is always right.
Dirichlet proved his class number formula for quadratic forms; in particular he was working with class numbers $h^+$ in the strict sense, and his unit $\varepsilon$ was the fundamental solution of the Pell equation $t^2 - Du^2 = 1$, not the fundamental unit $\eta$ of the corresponding number field. The relation
$$ \eta^{2h} = \varepsilon^{h^+} $$
encodes the two cases
$\varepsilon = \eta$, $h^+ = 2h$
$\varepsilon = \eta^2$, $h^+ = h$.
Thank you for your answer!
|
2025-03-21T14:48:30.105842
| 2020-03-21T17:08:14 |
355384
|
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|
Stack Exchange
|
How to understand the interface of the consistency strength hierarchy, reverse mathematics, and proof-theoretic ordinal analysis?
I am aware of three major "hierarchies" of mathematical theories, but I don't know how to relate these hierarchies to one another. Here are the hierarchies I have in mind:
Consistency strength. My understanding is that here one considers (recursively-enumerable?) theories $T$ of arithmetic (or which interpret the language of first-order arithmetic) of sufficient strength to fix some scheme for the syntax of first-order languages. One (partially) orders these theories by saying that $T > T'$ if $T$ proves the consistency of $T'$ (when $T'$ is coded up according to the aforementioned syntactic scheme).
Reverse mathematics. My understanding is that here one considers theories $T$ of second-order arithmetic (or which interpret the language of second-order arithmetic) and (partially) orders them directly by their implications.
Proof-theoretic ordinal analysis. My understanding is that here one considers theories $T$ of arithmetic (or which interpret the language of first-order arithmetic) and orders them by their proof-theoretic ordinal, i.e. the supremum of all (countable) ordinals $\alpha$ such that there exists a relation $R \subseteq \mathbb N \times \mathbb N$ definable in $T$ such that $T$ proves that $R$ is a well-order (although the sense in which $T$ can even express that $R$ is a well-order if $T$ is first-order is something that I don't quite understand), and $R$ is (externally) isomorphic to $\alpha$.
Questions:
Where do the domains of applicability of these hierachies overlap?
For instance, it seems that the "strongest" theories (like ZFC+ large cardinals) are usually studied in terms of consistency strength as opposed to reverse math or proof theory. I have the impression that proof-theoretic ordinals are most commonly used for relatively weak theories, and that reverse math lies somewhere in the middle. But I'm not even sure where to look for the overlaps in these domains, partly because the sorts of theories considered in each hierarchy are slightly different.
Where there domains overlap, how do these hierarchies relate?
In general, I imagine there are no direct implications saying that any of these partial orders refines any of the others (even where their domains of applicability coincide). But I imagine that there are some general tendencies -- a stronger theory in one hierarchy should probably typically be stronger in another as well.
Should I really be thinking of these 3 hierarchies as "comparable" in the sense that they give some notion of "strength" of a theory? And are there other hierarchies I should have in mind in this regard as well?
Sorry this is a bit disjointed - there's a lot of stuff here. I hope this helps though.
All of these notions are applicable in all contexts - or at least, all sufficiently rich contexts (we probably want at least to interpret $PRA$). That said, once we get to reasonably strong theories (basically anything above $\Pi^1_2$-$CA_0$) we don't know how to calculate proof-theoretic ordinals, so in practice ordinal analysis doesn't reach (anywhere close to) ZFC-type theories.
Of course, ordinal analysis provides more than just a hierarchy - it assigns a "value" to each theory, independent of what other theories we're considering. The implication and consistency hierarchies don't do this, or at least not directly (see here for some pushback on this claim), so it's not surprising that calculating proof-theoretic ordinals is harder than comparing consistency strengths - although it may be surprising (it was to me) that it's that much harder.
At this point it's a good idea to get to defining proof-theoretic ordinals properly. There isn't a single definition here, and some definitions (most?) have an element of subjectivity (they require a preexisting notion of "natural ordinal notation"). My favorite definition - which is fully formal - is the following:
Set $PTO(T)$ to be the smallest $\alpha$ such that there is no (index for a) primitive recursive well-ordering isomorphic to $\alpha$ which $T$ proves is well-ordered.
This definition makes sense for theories in sufficiently rich languages (e.g. second-order arithmetic and set theory). This isn't too big an issue (e.g. $RCA_0$ is conservative over $I\Sigma_1$ and $ACA_0$ is conservative over $PA$), but we can whip up versions for first-order arithmetic by talking about provable induction schemes (e.g. "$T$ proves $\Sigma_1$-induction along (that notation for) $\alpha$"). Here we have another level of flexibility, namely how much induction along the notation we want; if I recall correctly at this point $\Sigma^0_1$ induction is the standard choice.
But there are many other kinds of proof-theoretic ordinal, and beyond this one notion I really have no relevant competency.
The key point is that we need to talk about only primitive recursive relations. For example, in $\Pi^1_1$-$CA_0$ we can define a canonical relation on $\omega$ of ("true") ordertype $\omega_1^{CK}$, and in $ZFC$ we can go galactically beyond that. The issue is that these aren't really "concrete," and if we're thinking of ordinal analysis as a tool for Gentzen-style consistency proofs we really want to work lower down.
That said, we could look at very different ways of assigning ordinals to theories - see e.g. here.
Implication strength of course behaves quite differently from consistency strength/ordinal analysis. First, differences in language are a bit more significant here, and we have to talk about interpretations/conservative extensions. More importantly, while there are definitely large chunks of well-foundedness (for example: $I\Sigma_n/B\Sigma_n$; the Big Five (+ higher $\Pi^1_k$-$CA_0$s); $KP\omega+\Sigma_n$-Replacement; large chunks of the large cardinal hierarchy) there are important nonimplications amongst the natural theories (plenty in the context of choice fragments over $ZF$, and in reverse mathematics probably the most important - sociologically speaking - being $WKL_0\perp RT^2_2$).
That said, I know of relatively few situations where we have a strict inequality in consistency strength and no corresponding strict implication (or incompatibility: over ZFC, a measurable doesn't imply V=L but it does resolve it). Some occur in the large cardinal hierarchy, and by Montalban/Shore we know that the $n$-$\Pi^0_3$-determinacy hierarchy has lots of incomparabilities with the $\Pi^1_n$-$CA_0$ hierarchy, but it does seem to be pretty rare.
The three contexts mentioned - first-order arithmetic, second-order arithmetic, and set theory - do "glue together" reasonably well in terms of modified implication (= folding in appropriate interpretations to deal with language differences). E.g. $RCA_0$ is conservative over $I\Sigma_1$, and every model of $ATR_0$ is the set of reals of some model of $KP\omega$ (the converse fails though!). The real gulf comes in when we try to get from weak set theories (like $KP\omega$, $Z$, etc. - see here) to ZFC and its ilk. This gap is gigantic, and I know very little about it.
As a trivial observation, implication strength is also much broader than consistency strength and ordinal analysis. The division between "classifying members/subclasses of a given class" and "comparing different axiom systems" is pretty subjective - there's no reason we couldn't think of either as the other.
|
2025-03-21T14:48:30.106319
| 2020-03-21T17:15:25 |
355385
|
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|
Stack Exchange
|
Interpolation of product spaces
Suppose that $X_{\theta}$ is an interpolation space between the Banach spaces $X_0$ and $X_1$. Let $\mathcal{B}$ be another Banach space.
Is it true that $X_{\theta}\times\mathcal{B}$ is an interpolation space between $X_{0}\times\mathcal{B}$ and $X_{1}\times\mathcal{B}$?
Thank you for any suggestion.
Yes, interpolation on product spaces works componentwise, so $$\Bigl(\prod_{i=1}^n X_i,\prod_{i=1}^n Y_i\Bigr) = \prod_{i=1}^n (X_i,Y_i)$$ for any interpolation functor $(\cdot,\cdot)$ even with equal norms for a fixed choice of $\ell_p$ norm on the product spaces. This follows from the restriction/corestriction theorem in Section 1.2.4 in the bible book of Triebel (or alternatively from interpolation for complemented subspaces in Section 1.17.1): One shows that the $j$-th component of the interpolation space of products is exactly $(X_j,Y_j)$.
For this, let $R_j$ be the mapping which extracts the $j$-th component of an $n$-tuple, and let $E_j$ insert the $j$-th component in an $n$-vector of zeroes. Then $E_jR_j$ is a linear continuous projection and $E_jR_j\prod_{i=1}^n X_i$ is a complemented subspace which is isometrically isomorphic to $X_j$; analogously for $Y_j$. Hence, by the mentioned theorem(s),
$$E_jR_j\Bigl(\prod_{i=1}^n X_i,\prod_{i=1}^n Y_i\Bigr) =(X_j,Y_j)$$
and this again isometrically.
Triebel, Hans, Interpolation theory. Function spaces. Differential operators, Berlin: Deutscher Verlag des Wissenschaften. 528 p. M 87.50 (1978). ZBL0387.46033.
|
2025-03-21T14:48:30.106563
| 2020-03-21T18:15:06 |
355388
|
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"url": "https://mathoverflow.net/questions/355388"
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|
Stack Exchange
|
Representing modus ponens in a Polish propositional logic with NAND as the only connective
In a base for propositional logic using the Polish connective $\uparrow$ for not both, J. Nicod isolated one axiom as sufficient:
$\uparrow\uparrow p\uparrow q r\uparrow\uparrow t\uparrow tt\uparrow\uparrow s q\uparrow\uparrow p s\uparrow ps$
J. Nicod used the somewhat odd inference rule $r$ if $\uparrow p\uparrow q r$ and $p$. May we use $\uparrow p\uparrow r r$ and $p$ instead of Nicod's rule, or is there some deeper reason for including $q$?
About the title: It seems to me that the rule "infer $r$ from $\uparrow p\uparrow rr$ and $p$" (i.e., in Nicod's rule replace $q$ with $r$ rather than with $p$) is intuitively closer to modus ponens than the rule you asked about.
@AndreasBlass Thank you for identifying that typo. I will edit.
You could really make the format of the second part readable form.
This weakening of Nicod's inference rule is too weak. The 4-element counter-model above is one on which (a) your rule holds, (b) Nicod's single axioms holds, but (c) the formula s(s(X,X),X) fails to be a theorem (even though it is a classical theorem), where s(•,•) is the sheffer stroke.
Here is the code (3-lines of TPTP format) I used to find this model with mace4.
fof(mpweak,axiom, ![X,Z]: ((t(s(X,s(Z,Z))) & t(X)) => t(Z))).
fof(nicod,axiom, ![X,Y,Z,U,V]: t(s(s(X,s(Y,Z)),s(s(V,s(V,V)),s(s(U,Y),s(s(X,U),s(X,U))))))).
fof(luka1,conjecture,![X]: t(s(s(X,X),X))).
For a discussion of alternative axiomatizations, I suggest the following article and the references therein.
https://projecteuclid.org/euclid.ndjfl/1093958259
Axiomatization of propositional calculus with Sheffer functors.
Thomas W. Scharle
Notre Dame J. Formal Logic
Volume 6, Number 3 (1965), 209-217.
Welcome to Mathoverflow! I don't understand how the countermodel works, but maybe others can contribute?
As far as I can see, the Lukasiewicz axiom in the answer differs from the Nicod axiom in the question in that U replaces both $s$ and $t$. Does the Nicod axiom hold in your countermodel?
Right, sorry, thanks Andreas. I have fixed this above. Now a 4-element model is required. Note that t(•) is the theorem-hood predicate, and c1 = 0 is the instance of s(s(X,X)X) that fails to be a theorem. i.e., t(s(s(0,0),0)) = 0.
note, also that Nicod's stronger rule also works with many other known axioms, including Lukaiewicz's. so, even if somehow this rule had worked with his axiom (in syntactically idiosyncratic way) it would still in this sense have been less generally useful than his stronger rule.
I accepted this answer because of the very useful literature reference in an edit. Scharle's article settles the question in the negative, but it does not exclude that there are other single axioms in the single connective $\uparrow$ for not both ,,, and ... which are adequate for classical propositional logic jointly with normal detachment as the only inference rule.
yes, i suspect there is a single axiom for the weaker rule. i'll be working on this now.
Note: Scharle's two-basis contains a typo. Here is a sound and complete 2-axiom system, when combined with your weaker detachment rule (which Scharle calls D3): (1) DpDpp, (2) DDpDqrDDDsqDDpsDpsDDsqDDpsDps
or, in TPTP format: (1) fof(scharle1,axiom,![X]: t(s(X,s(X,X)))). (2) fof(scharle2,axiom,![X,Y,U,R]: t(s(s(X,s(Y,R)),s(s(s(U,Y),s(s(X,U),s(X,U))),s(s(U,Y),s(s(X,U),s(X,U))))))).
@BrandenFitelson Thank you very much for this important information.
@BrandenFitelson Did you find another single axiom which combines with the weaker rule?
|
2025-03-21T14:48:30.106819
| 2020-03-21T19:05:59 |
355391
|
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|
Stack Exchange
|
English translation Gaston Darboux paper works
I'm looking for an English translation of Gaston Darboux's thesis:
Sur les équations aux dérivées partielles du second ordre
and
Mémoire sur la théorie des fonctions discontinues
Does anyone know where can I find it? I wasn't able to google it, so it might be very nice to provide me any suggestions to find it.
I doubt translations were ever published of these papers (19, 53), which incidentally, were not his thesis. You can likely get a passable machine translation of the pdfs on numdam.
@FrancoisZiegler I have always been in mind that these papers is important part of modern math analysis, so translations should be sth obvious. Are u really sure that there is no any published translations?
Well, absence of evidence is not evidence of absence, but I guess any translations would be listed in such histories as Hawkins or Pier if published before them.
|
2025-03-21T14:48:30.106916
| 2020-03-21T19:08:14 |
355392
|
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|
Stack Exchange
|
Good reference on the algebraic geometry of non-associative rings
I am looking for a good reference about the algebraic geometry of non-associative rings. I am in particular interested in derivation algebras.
Preferrably an online resource or a book that is available online, since every library near me is in lockdown right now.
As mentioned in the answer by user6976, there is the idea of development of algebraic geometry to (essentialy) any general algebraic system. This is carried out(following Plotkin's work) by E. Daniyarova, A. Miasnikov, V. Remeslennikov and co-authors in a series of papers.
A more recent survey (2016) of this area, called Universal Algebraic Geometry, of which (not necessarily associative) algebras is naturally of central interest, is Artem N. Shevlyakov's Lectures notes in universal algebraic geometry.
This survey is very easy to read, and has a very good bibliography ponting out to more specific topics.
|
2025-03-21T14:48:30.107013
| 2020-03-21T19:37:52 |
355394
|
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"Alexander Schmeding",
"Glen M Wilson",
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"url": "https://mathoverflow.net/questions/355394"
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|
Stack Exchange
|
The set of embeddings is open in the strong Whitney topology
In Hirsch's book "Differential Topology," he claims in Chapter 2, Theorem 1.4 that the set of $C^1$-embeddings is open in the strong Whitney topology $C^1(M, N)$ where $M$ and $N$ are $C^1$ manifolds. My question is: How do you see that the set $\mathcal{N}_1$ is open? See below for the setup and description of $\mathcal{N}_1$.
$M$ and $N$ are $C^1$ manifolds and we have an embedding $f: M \to N$. We equip $M$ with a locally finite atlas $\{\phi_i, U_i\}$ of $M$ which admits compact sets $K_i \subseteq U_i$ whose interiors cover $M$ too. We find a cover $\{\psi_i, V_i\}$ of $N$ so that $f(U_i) \subseteq V_i$. A lemma says we can find numbers $\epsilon_i > 0$ so that any $C^1$ function $g: M \to N$ that is in the neighborhood
$$\mathcal{N}_0 = \{ g : M \to N \,\vert\, (\forall i) (\forall x \in K_i)(g(K_i) \subseteq V_i)\text{ and }\, \lvert D^k(\psi_i f\phi_i^{-1} (x) - D^k(\psi_i f \phi_i^{-1}(x) \rvert < \epsilon_i \text{ for } k = 0,1 \}
$$
satisfies $g\vert_{\mathrm{Int}(K_i)}$ is an embedding.
Since $f$ is an embedding, we can find open sets $A_i$, and $B_i$ in $N$ so that $f(K_i)\subseteq A_i$ and $f(M\setminus U_i) \subseteq B_i$ and $A_i\cap B_i = \emptyset$. The claim is that there is an open set $\mathcal{N}_1$ in $C^0(M,N)$ about $f$ such that if $g \in \mathcal{N}_1$ then $g(K_i) \subseteq A_i$ and $g(M\setminus U_i) \subseteq B_i$.
The result is certainly true if $M$ is compact, so I believe it is possible to extend it to the case where $M$ is not compact using the strong Whitney topology. I just can't get it to work.
A proof of this is also contained in Michor: Manifolds of mappings. There Proposition 5.3 establishes that the $C^r$-embeddings are $WO^1$ open in the space $C^r (M,N)$ (no restriction on the manifolds $M,N$). Thus if you set $r=1$ this implies that they are open with respect to the strong Whitney topology, as the $WO^1$-topology in Michor is coarser then the strong $C^1$-Whitney topology. For definitions and a comparison of the topologies see the book, it can be downloaded for free here: https://www.mat.univie.ac.at/~michor/manifolds_of_differentiable_mappings.pdf.
Going a bit in the details concerning the proof (answering to the comment by the OP, you have to decide whether this is an easy way to see that it is open):
Michor defines the set
$$B = \{g \in C^1(X,Y) : d(g(x),f(x)) <\varepsilon (x), d_1 (j_1 f (x),j_1 g (x))< \delta (x), g(K_\alpha) \subseteq A_\alpha , g(X\setminus U_\alpha ) \subseteq B_\alpha \text{for all }x \in X, \text{ for all } \alpha \}.$$
For the first two conditions $d(g(x),f(x)) <\varepsilon (x), d_1 (j_1 f (x),j_1 g (x))< \delta (x)$ recall that $d,d_1$ are metrics adapted to the jet-bundles and $\varepsilon, \delta$ are continuous functions. Thus these conditions are open in $WO^1$ by the characterisation of the $WO^1$-topology in Michor 4.4.2.
I claim that the conditions $g(K_\alpha) \subseteq A_\alpha , g(X\setminus U_\alpha ) \subseteq B_\alpha$ are open conditions in the $WO^1$-topology. To see this recall that $K_\alpha$ is compact, the $A_\alpha, B_\alpha$ and $U_\alpha$ are open with $K_\alpha \subseteq U_\alpha$ and the $U_\alpha$ form a locally finite-family. Now $WO^1$ is induced by an embedding from the so called $LO$-topology (recorded as 4.6 in Michor) and for the $LO$-topology, 3.7 Lemma in Michor states that the above conditions are open.
As pointed out by the OP the second point is still missing details. Since the family of closed sets $X \setminus U_\alpha$ will in general not be locally finite. We can not use for this condition as it is written the cited result. The additional information making this possible is that we have an embedding f to construct a family of neighborhoods which is locally finite and implements a condition similar to the one asked in Michors book. A writeup on how to do this can be found her https://www.math.uni-hamburg.de/home/latschev/lehre/ws17/embeddings.pdf
Thus in conclusion the set B is open in $WO^1$. I admit it is not pretty and easy, but that is the shortest explanation I could come up with in limited time.
Thanks for the reference. I did take a look at Michor's book, but I was still confused by it. I don't see why in the proof of Proposition 5.3 on page 45 he gets $\mathfrak{B}$ is open immediately in the $WO^1$ topology. Do you have a good way of seeing this?
Hei, I ammended my answer with details concerning the openness of the set B (as asked for in your comment.
Thanks for the additional details Alexander! Your approach is certainly more systematic, which is what I was looking for. I am still not convinced by the second bullet point though, because the family of sets $X \setminus U_{\alpha}$ is not locally finite in general. E.g., we could cover $\mathbb{R}$ with intervals $U_i = (i-1,i+1)$. Then the $U_i$'s are locally finite yet $\mathbb{R} \setminus U_i$ is not.
Hei Glen, yes you are indeed correct I overlooked this (was late yesterday). This needs to be remedied and it is not apparent from what Michor does. The point should be that one can find a locally finite family of closed sets which still implements a similar condition. The main issue is to get injectivity in the end. The additional information needed seems to be that the map f for which we construct a neighborhood is an embedding. Fortunately, I found a writeup which addresses these problems: https://www.math.uni-hamburg.de/home/latschev/lehre/ws17/embeddings.pdf
I think the following works, but it is ad hoc. I would have suspected Hirsch had something more general in mind.
Because there is a tubular neighborhood of the embedding $f$ in $N$, we can work in an open subset of the zero section of the normal bundle $N_f(M)$. Because of this, I'll identify the points $f(x)$ with $x$ in the normal bundle. The open sets $A_i$ and $B_i$ can be taken to be of the form $\pi^{-1}(A'_i)$ and $\pi^{-1}(B'_i)$ where $A'_i$ and $B'_i$ are open subsets in $M$, with $M\setminus U_i \subseteq B'_i$ and $K_i \subseteq A'_i$.
We want to find $\epsilon$'s about each $x \in X$ so that all points in $B(x; \epsilon)$ in the normal bundle stay in $A_i$, whenever $x \in K_i$, and stay in $B_j$ whenever $x \in M \setminus B'_j$.
First step: If $x \in K_i$, there are only finitely many $K_j$ that intersect $K_i$ (if there were infinitely many, we can take a sequence of elements of $K_i$ that lie in some other $K_j$, this has a convergent subsequence with limit in $K_i$, but then this limit point lies in infinitely many other $K_j$, which is prohibited by local finiteness of the cover). So we can now measure the distance to any $x \in K_i$ to $N \setminus A_i$ to get $\epsilon_i$. We require that $x$ can't move more than $\epsilon_i$ then, but we must impose this condition for every $j$ that $x \in K_j$. We just showed there are only finitely many $j$ though, so there is a smallest positive of $\epsilon$ for each $x$.
Second step: Suppose now that $ x \in K_j$ and $x \in X \setminus U_i$. We can then measure the distance from $x$ to $N\setminus B_i$, call it $\epsilon_{i,j}(x)$ which is positive. We require that for a fixed $x$, for every $j$ and $i$ for which $x \in (X \setminus U_i) \cap K_j$ that we do not move from $x$ more than $\epsilon_{i,j}(x)$. The worry is that there is no positive $\epsilon(x)$ that is smaller than all of the $\epsilon_{i,j}(x)$. Let us check if this can happen.
We know that each $x$ lies in only finitely many $K_j$, but $x$ can lie in infinitely many of the sets $X \setminus U_i$. So for each $j$, consider if it is possible that there is a sequence of $i_k$ so that $x \in \setminus U_{i_k}$ with the distance $\epsilon_{j,i_k} \to 0$ as $k \to \infty$. If this were the case, then there would be a sequence of points $z_{i_k}\in N \setminus B_{i_k}$ in the normal bundle, lying over points $y_{i_k} \in M \setminus B'_{i_k}$ so that the distance $d(x, y_{i_k}) \to 0$ as $k \to \infty$. But then, as $x \in K_j \subseteq \mathrm{Int}(U_j)$, there are infinitely many values of $k$ for which $y_{i_k} \in U_j$. That is, local finiteness at $x \in K_j$ is contradicted. We therefore conclude that each $x \in X$ admits an $\epsilon(x)$ for which all points $z \in B(x; \epsilon(x))$ lie in all possible $N \setminus B_i$.
To finish up the argument, we just need to pick the smallest value of the $\epsilon$'s determined above on each compact set $K_j$. This then gives an open set $\mathcal{N}_1$ in the strong Whitney topology $C^0_S(M, N)$ for which all $g$ in this set satisfy $g(K_j) \subseteq A_j$ and $g(M \setminus U_j) \subseteq B_j$ for all $j$.
|
2025-03-21T14:48:30.107900
| 2020-03-23T17:26:28 |
355542
|
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|
Stack Exchange
|
A quotient stack $[X/G]$ and irreducibility of $X$
Let $k$ be an algebraically closed field.
Let $X$ be a variety over $k$ and $G$ be a algebraic group over $k$ acting on $X$.
Question
If the quotient stack $[X/G]$ is irreducible, then is $X$ also irreducible ?
Any comment welcome !
Edit: the original question is from line 8 on page 208 of
https://books.google.co.jp/books?id=jcrcPtr8aFMC&pg=PA208&lpg=PA208&dq=irreducible+quotient+stack&source=bl&ots=UxLrr9puIN&sig=ACfU3U2wQ9htPKj1_PYkWJQLS-Fpp26Klg&hl=ja&sa=X&ved=2ahUKEwiRot3li7HoAhU5yIsBHRZSB8wQ6AEwAnoECAgQAQ#v=onepage&q=irreducible%20quotient%20stack&f=false
What definition of variety does this book use? Usually they are assumed integral, hence irreducible. If not, you can just quotient a disjoint union of two copies of a space by an action of $\mathbb Z/2$.
Do you want to assume $G$ is connected?
In the book you refer to the group $G$ is connected and $[X/G]$ is smooth, so $X$ is smooth and irreducible.
Thank you for the comments ! @abx, Could you give me a detailed reason for that?
And I'm sorry I didn't make any assumptions in detail.
Pull back $X\rightarrow [X/G]$ by a surjective morphism$P\rightarrow [X/G]$ with $P$ irreducible. The pull back is a principal $G$-bundle over $P$, hence irreducible.
Thank you very much !
Is there one more question you want? How can I get irreducible $ P $ and surjection$ P → [X / G]$ ?
Stacks Project, Lemma 104.3.1.
Thank you many times.
@abx i'm slightly confused: doesn't that lemma merely assure that you can find P irreducible and with dense image in [X/G]? Anyway, it seems to me you can just argue using the fact that X->[X/G] is open with irreducible fibers (being a G-torsor), as you would if [X/G] were a scheme.
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2025-03-21T14:48:30.108059
| 2020-03-23T18:05:01 |
355545
|
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|
Stack Exchange
|
Examples of groups admitting a proper $1$-cocyle for a bounded representation
A representation $\pi: G \to B(H)$ of a group $G$ on a Hilbert space $H$ is called bounded iff $\sup_{g \in G} \| \pi(g) \|_{B(H)} = C < \infty$. A $1$-cocycle with respect to the representation $\pi$ is a map $\beta: G \to H$ such that
$$
\beta(g \, h) = \beta(g) + \pi(g) \, \beta(h).
$$
In the recent preprint
Nishikawa, Shintaro,
$\mathrm{Sp}(n,1)$ admits a proper $1$-cocycle for a uniformly
bounded representation,
https://arxiv.org/abs/2003.03769.
It is claimed that $\mathrm{Sp}(1,n)$, although does not have the Haagerup approximation property, possess a metrically proper $1$-cocyle with respect to a bounded representation on a Hilbert space.
Question: Are there other known examples of groups without the Haagerup property (or with property $\mathrm{(T)}$) which admit a proper
$1$-cocycler with respect to a bounded representation?
You automatically get their direct products (and direct product of their lattices), and subgroups of these direct products. Besides, I guess hyperbolic groups are natural candidates.
|
2025-03-21T14:48:30.108165
| 2020-03-23T18:08:56 |
355546
|
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|
Stack Exchange
|
Ergodicity of induced system
Suppose $(X,\mathcal{F},\mu,T)$ is an ergodic measure preserving dynamical system.
Let $Y\subset X$ be such that $\mu(Y)>0$ and suppose there is an integrable function $R:Y\to \mathbb{N}$ such that $T^{R(y)}(y)\in Y$.
Then we can define a function $F:Y\to Y$ by $F=T^R$ and consider the induced system $(Y,\mathcal{F}\cap Y, \mu|_Y,F)$.
Can we say that $F$ is ergodic?
When $R(x)=\inf\{n\ge 1 : T^n(x)\in Y\}$, this induced system is very well studied and the answer to my question is yes (see for example http://www.weizmann.ac.il/math/sarigo/sites/math.sarigo/files/uploads/ergodicnotes.pdf, Theorem 1.7). However, the given proof does not extend to the general return times I am considering above.
Thanks :)
In general $F$ is not ergodic. A very simple example can be constructed as follows: let $X=\mathbb{Z}_3=\{0,1,2\}$ and $\mu =1/3(\delta_0+\delta_1+\delta_2)$ and $T(x):=x+1$. This is an ergodic system. Let us define $Y:=\{0,1\}$ and $R\equiv 3$. Since $F=T^3=id$, it is not ergodic.
Is there any counter example with $T$ totally ergodic?
Isn't $T$ totally ergodic in Alejandro's counter example?
@Joël, Doesn't totally ergodic mean that all iterates are ergodic? That certainly is not the case if $T^3$ is identity.
Totally ergodic means that there exists only one $T$-invariant measure (it is a property of the space, the $\sigma$-algebra, and $T$, but not of the measure $\mu$)
@Joël I think that's unique ergodicity: https://en.wikipedia.org/wiki/Ergodicity#Unique_ergodicity
I don't think that this system is even automatically measure-preserving (unlike the traditional induced map, which as you note is measure-preserving and ergodic).
Just take something silly like $(X, T)$ an irrational circle rotation with Haar measure, $Y$ the left half $[0, 1/2)$, and $R$ the first return time of a point in $Y$ to $[0, 1/4)$ (this always exists by minimality).
Then $T^R(y) \in [0, 1/4))$ for all $y \in [0, 1/2)$, so, for instance,
$\mu|_Y([0, 1/4)) = 1/2$, but $\mu|_Y\big((T^R)^{-1}([0, 1/4))\big) = \mu|_Y(Y) = 1$. (I am here normalizing $\mu|_Y$ to make it a probability measure.)
I know my example is silly in that it's not even surjective on $Y$, but this isn't the problem; you could make a slightly trickier example where $R$ is the first return time to $[0, 1/4)$ for $y \in [0, 3/8)$ and the first return time to $[1/4, 1/2)$ for $y \in [3/8, 1/2)$, and then $T^R$ is surjective on $Y$, but still not measure-preserving.
|
2025-03-21T14:48:30.108355
| 2020-03-23T18:17:04 |
355548
|
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|
Stack Exchange
|
Semidirect product of metaplectic group and Heisenberg group
I know that Symplectic group has an action on Heisenberg group.
I am wondering how to extend this to non-trivial two fold metaplectic covering?
Thanks in advance!
Well, trivially, by projecting to the symplectic group …? Do you have some other action in mind? If so, why? Do you know that it exists, or just hope? Also, over what field?
@LSpice, Oh, the action uses the projection map from the metaplectic group to symplectic group. Since many people use the semi-direct product of metaplectic group and Heisenberg group, I just wondered people’s convention. Thank you!
Why close this beautiful question with even more beautiful answer?
@BugsBunny Because there was no effort from OP (even after LSpice's comment) to be more precise, e.g., symplectic group over which field, in what dimension, Heisenberg group in which sense?
I presume you are looking for a faithful action of $Mp_{2n}$ on something related to the Heisenberg group $H_{2n+1}$. This is well-known as Weil Representation.
In the modern language, consider $Mp_{2n}$ acting on $H_{2n+1}$ by automorphisms. This action has a kernel. Now consider the action on the category of unitary representations of $H_{2n+1}$ by twisting representations by automorphisms. This categorical representation still has the same kernel. Finally, choose a skeleton of the category of unitary representations of $H_{2n+1}$. The modern interpretation of all this Weil and Stone–von-Neumann business is that $Mp_{2n}$ acts on the skeleton and this yields a faithful categorical representation of $Mp_{2n}$.
Thank you very much! I learned much from your comments!
|
2025-03-21T14:48:30.108478
| 2020-03-23T18:54:18 |
355549
|
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|
Stack Exchange
|
When does the PDE $F\cdot \nabla u+Gu+H=0$ admit a global solution $u:\mathbb R^3\to\mathbb R$?
Let $F:\mathbb R^3\to \mathbb R^3$ and $G,H:\mathbb R^3\to\mathbb R$ be some given smooth maps. In order for the PDE $F\cdot \nabla u+Gu+H=0$ to admit a global solution $u:\mathbb R^3\to\mathbb R$, what are the restrictions needed on $F,G,H$?
Typically this type of quasilinear PDE can be solved by method of characteristics. But here we have no boundary condition so we gain some freedom in finding solutions.
The method of characteristic reduces the problem to solving a family of ODEs. To obtain a global solution of $u$, we need all the ODE to have a global solution, and these solutions should patch together smoothly. What condition on $F,G,H$ do we need to achieve this?
If $F, G, H$ are given and smooth, then all the ODEs will have global solutions. It seems to me that any issue would be only coming from patching.
|
2025-03-21T14:48:30.108556
| 2020-03-23T20:18:04 |
355552
|
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|
Stack Exchange
|
Global Arthur packet consist of only globally generic representations?
I would like to ask very stupid two questions to experts.
I am wondering whether every globally generic automorphic representation of unitary groups are contained some global Arthur packet associated to some A-parameter.
Conversely, I am also wondering whether global generic A-packet consists only globally generic automorphic representations.
If these two questions are right, then these consequences are the work of Mok and Kaletha, Shin, White, Minguez?
Thank you very much if you share your knowledge.
Yes. 2) No, at least for L-packets there should be exactly 1 generic member of each tempered packet (on the quasi-split forms). This is Shahidi's conjecture, and is not quite known even for unitary groups. (And KMSW is neither complete nor unconditional yet)
|
2025-03-21T14:48:30.108635
| 2020-03-23T21:51:57 |
355553
|
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|
Stack Exchange
|
Descent of projective bundles
A problem studied in GIT is the descending of vector bundles (or more in general coherent sheaves) to quotients.
It is a result of Kempf that whenever we have a vector bundle over a quasiprojective scheme defined over $\mathbb{C}$ (or any algebraically closed field of characteristic zero)
$$\pi:E\to X$$
and a reductive group $G$ acting linearly on $E$ and $X$ with the property that
$$\pi(g\cdot v)=g\cdot \pi(v)$$
Let $X//G$ be the GIT quotient, then $E$ descends to a vector bundle
$$\bar{E}\to X//G$$
if for any $x\in X^{ss}$ (closed point) $Stab(x)$ acts trivially on $E_x$.
My problem is the following: i am interested to know if there is a similar condition which ensures that a projective bundle descends to the quotient.
Thank you in advance.
You can get a condition by combining the result cited in the question with the observation that projective $G$-bundle on $X$ is the same as a $\widetilde{G}$-bundle on $X$, where $\widetilde{G}$ is a central extension of $G$ by $\mathbf{G}_m$.
|
2025-03-21T14:48:30.108731
| 2020-03-23T21:56:01 |
355554
|
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"Christian Remling",
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|
Stack Exchange
|
Polar decomposition of the Volterra integral operator
Repost of this Math.SE question due to a lack of answers (No one was able to help me find the closed form of $U_T$ and $|T|$ after two bounties). I also searched extensively online but couldn't find anything. A reference for the solution suffices.
Define the Volterra integral operator (or numerical differentiation operator)
$$
T: L^2([0,1]) \to L^2([0,1]), \
f(x) \mapsto \int_{0}^{x} f(y) dy.
$$
Find its polar decomposition $T = U_T | T |$.
First I tried to find $|T| := (T^* T)^{\frac{1}{2}}$.
We have
$$
T^*: L^2([0,1]) \to L^2([0,1]), \
f(x) \mapsto \int_{x}^{1} f(y) dy.
$$
and therefore
$$
T^* T: L^2([0,1]) \to L^2([0,1]), \
f(x) \mapsto \int_{x}^{1} \int_{0}^{y} f(z) dz dy.
$$
Now I got stuck when finding $\sqrt{T^* T}$. I found out that $T^2 \ne T^* T$ and $(T^*)^2 \ne T^* T$.
Can somebody please give me a hint?
I have also tried writing out $T^* T$ more explicitly: denote by $F$ the anti-derivative of $f$ and by $\mathscr{F}$ the antiderivative of $F$.
The we have
$$
(T^* T f)(x)
= \int_{x}^{1} F(y) - F(0) dy
= \int_{x}^{1} F(y) dy - (1 - x) F(0)
= \mathscr{F}(1) - \mathscr{F}(x) - (1 - x)F(0).
$$
Edit.
I also know that $T^* T$ is compact and self-adjoint and
$$
(T^{n + 1} f)
= \frac{1}{n!} \int_0^x (x - y)^n f(y) dy.
$$
holds.
Maybe this can help use find a closed form for $(I - T^* T)^n$?
Edit 2.
I know that
$$
| T |
= \sum_{n \in \mathbb N} \sigma_n \langle \cdot, v_n \rangle v_n,
$$
where $\sigma_n := \frac{1}{\pi\left(n - \frac{1}{2}\right)}$ are the singular values of $T^* T$ and $(v_n)_{n \in \mathbb N} \subset L^2([0,1])$ a orthonormal basis of $\overline{\text{ran}(T)}$ which fulfils
$$
T^* T x
= \sum_{n \in \mathbb N} \sigma_n^2 \langle x, v_n \rangle, v_n.
$$
The eigenfunctions of $T^* T$ are $f_n(x) = C \cdot \cos(\sigma_n^{-1} x)$ for some $C \in \mathbb{R}$.
Partial answer from Math.SE
We have $$\sqrt{T^* T} = I-\sum^{\infty}_{n=1}a_n(I-T^*T)^n,$$ where the $a_n$ are determined by $$1-\sqrt{1-x}=\sum^{\infty}_{n=1}a_nx^n.$$
This yields $\sum_{n = 0}^{\infty} a_n = 1$. The first $a_{2n}$ are $$\left(\frac{1}{2}, \frac{1}{8}, \frac{1}{16}, \frac{5}{128}, \frac{7}{256}, \frac{21}{1024}, \frac{33}{2048}, \frac{429}{32768}\right),$$ the odd terms are zero.
The numerators, omitting the zeros, are A098597 on OEIS and the denominators (omitting zeros again) are given by $2^{\ell}$, $\ell \in \{1,3,4,7,8,10,11,15\}$, which is this in OEIS.
@ChristianRemling The reason I hope there exists an explicit answer is that this exercise is taken from this problem set from LMU Munich wherein it states "Important: the full solution here is to produce the explicit "closed" form of the operators
$U_T$ and $|T|$. Giving the canonical decomposition of $|V|$ is not enough, show that in fact $|T|$ is an integral operator and $U_T$ is a unitary operator (with an explicit, "easy" form)."
I see. In that case, we need a kernel $K(x,t)$ with $\overline{K(t,x)}=K(x,t)$ and $\int_0^1 K(x,s)K(s,t), ds = 1-\max { x,t}$ (since this latter expression is the kernel of $T^*T$), and you would then hope to get the right idea by just staring at this long enough (?).
@ChristianRemling I know a kernel for $T$ is $K(x,t) := \mathbb{1}_{t \le x}(x,t)$, as in $$(T f)(x) = \int_0^1 K(x,t) f(t) dt.$$ Futhermore, $K(t,x)$ is a kernel for $T^$: $$(T^ f)(x) = \int_0^1 K(t,x) f(t) dt.$$ This is consistent with the kernel-free expression I found for $T^* T$ above:
$$
(T^* T f)(x)
= T^*\left(\int_0^1 K(x,s) f(s) ds \right)
= \int_0^1 K(t,x) \int_0^1 K(x,s) f(s) ds dt
$$ Am I on the right track? Because I think that $\overline{K(t,x)} = K(t,x) \ne K(x,t)$ and $$\int_0^1 K(x,s) K(set) ds = \int_x^t dx = t - x \ne 1 - \max(x,t).$$
So shouldn't the kernel of $T^* T$ be $\tilde{K}(x,t) := \int_0^1 K(t,x) K(x,s) ds$, as $$(T^* T f)x = \int_0^1 \tilde{K}(x,t) f(t) dt?$$
I think my formula $\widetilde{K}(s,t)=1-\max{ t,s }$ is fine. I simply changed the order of integration in $(T^*Tf)(x)=\int_x^1 dt\int_0^t ds, f(s)$ to obtain this.
But of course $\widetilde{K}(x,t) = \int_0^1 K(s,x)K(s,t), ds$ must give the same answer, as you pointed out.
Let us continue this discussion in chat.
|
2025-03-21T14:48:30.109114
| 2020-03-23T22:27:54 |
355557
|
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|
Stack Exchange
|
About the distribution of Fibonacci numbers that are primes
Let's consider the Fibonacci sequence, that is the sequence of naturals defined by:
$F_1=F_2=1$
$F_{n+1}=F_{n}+F_{n-1}$
It is an open problem whether the sequence contains an infinite number of primes.
It is known that, after $\,F_3=2\,$ and $\,F_4=3$, the only possible primes are those of the form:
$F_{6m-1}=F_{3m-1}^2+F_{3m}^2$
$F_{6m+1}=F_{3m}^2+F_{3m+1}^2$
One could expect that the number of primes of the form $\,F_{6m-1}\,$ was similar to the number of primes of the form $\,F_{6m+1}$, but brute force shows that:
up to $\,m=10$, there are $\,6\,$ primes of the form $\,F_{6m-1}\,$ and only $\,3\,$ of the form $\,F_{6m+1}$;
up to $\,m=100$, there are $\,14\,$ primes of the form $\,F_{6m-1}\,$ and only $\,5\,$ of the form $\,F_{6m+1}$;
up to $\,m=1000$, there are $\,15\,$ primes of the form $\,F_{6m-1}\,$ and only $\,7\,$ of the form $\,F_{6m+1}$;
up to $\,m=5522$, there are $\,17\,$ primes of the form $\,F_{6m-1}\,$ and only $\,10\,$ of the form $\,F_{6m+1}$.
How can this asymmetric behavior be explained?
Thanks.
$m=1000$ is a very small number in asymptotics, even given the size of $F_m$. Could there be some sort of Chebyshev bias going on?
I don't know ... the computation is much time-expensive ... my workstation is working on $,m=10000,$ ...
It seems appropriate to gather a numerically significant amount of data before asking for explanation of possible coincidences. (For that matter, one could also ask why the asymmetric behaviour needs an explanation—the only justification given for the symmetry was "One could expect" it.)
Well, in the context of natural numbers the same simmetry of distribution would be a consequence of Dirichlet's theorem on arithmetic progressions ...
There is already a lot of fine bias (i.e. asymmetry) in the distribution of all prime numbers in residue classes. I suggest that you study the relevant literature, e.g. https://en.wikipedia.org/wiki/Chebyshev%27s_bias and https://arxiv.org/abs/1603.03720
The sequence of n's such that $F_n$ is prime is in OEIS: https://oeis.org/A001605 which also has references.
|
2025-03-21T14:48:30.109304
| 2020-03-23T22:59:31 |
355558
|
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"Dylan Thurston",
"Inácio ",
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|
Stack Exchange
|
Automorphism group of formally real Jordan algebras of hermitian matrices
It is well known that the automorphism group of exceptional Jordan algebra $\mathcal{h}_{3}(\mathbb{O})$ is the exceptional Lie group $F_{4}$. I am trying to understand the automorphism group of the Jordan Algebra of Hermitian matrices $h_{3}(\mathbb{F})$, $\mathbb{F} = \mathbb{R}$, $\mathbb{C}$, $\mathbb{H}$. My suspicion is that at least the connected components are $\mathrm{SO}(3)$, $\mathrm{SU}(3)$ and $\mathrm{Sp}(3)$, respectively. My calculations, using the matrix model of projective spaces, result in
$$\mathbb{R}\mathrm{P}^{2} \simeq G_{\mathbb{R}} / \mathrm{SO}(2)$$
$$\mathbb{C}\mathrm{P}^{2} \simeq G_{\mathbb{C}} / \mathrm{SU}(2)$$
$$\mathbb{H}\mathrm{P}^{2} \simeq G_{\mathbb{H}} / \mathrm{Sp}(2)$$
where $G_{\mathbb{F}}$ are the desired groups, based on theorem 14.99 of Spinors and Calibration (F. Harvey). Of course, the above candidates lies in the desired groups by an action of the form $g(A) = gA\overline{g}^{\mathrm{t}}$. I am trying to compare with:
$$ \mathbb{R}\mathrm{P}^{2} \simeq \mathrm{O}(3) / \mathrm{O}(2) \times Z_{2}$$
$$\mathbb{C}\mathrm{P}^{2} \simeq \mathrm{U}(3) / \mathrm{U}(2) \times U(1)$$
$$\mathbb{H}\mathrm{P}^{2} \simeq \mathrm{Sp}(3) / \mathrm{Sp}(2) \times \mathrm{Sp}(1)$$
References are welcome.
I also need this result, and have been fairly frustrated looking for references: almost all references talk about the Lie algebra and not the Lie group of symmetries. An answer can be extracted from Michael Orlitzky's example, but it's hard for me to believe that this was not written down before April 2023.
Maybe much more elementary than what you are looking for, but I answered this specific question (at least for connected components and including the non-compact case) in my master's thesis: http://hdl.handle.net/1843/36039 (Theorems A.1 and A.2 in Appendix A). The proof, which is quite simple and has a geometric approach, is an adaptation of F. Harvey's proof for the exceptional case in the book Spinors and Calibrations. My mistake not having included this comment here before.
It is worth to remark the following. My original motivation was to show that the Jordan algebra structure completely replaces the vector structure (compact and non-compact case), which allows us to construct the octonionic (projective or hyperbolic) plane.
Thanks, I will look at that (and the other references you give)! Unfortunately for my application I need the other component for $h_3(\mathbb{C})$ too, which seems to have gone little-noticed in the literature.
In a simple Euclidean Jordan algebra, the Jordan-automorphism group of
the algebra is the subgroup of the cone of squares' automorphism group
that fixes the Jordan unit element. This is written down, for example,
in Theorem 8 of Gowda, but the result goes all the way back to
Vinberg, even for non-simple EJAs.
For $\mathbb{R}$ and $\mathbb{C}$, that's enough to squeeze the result
out of Theorem 2 in Schneider. He gives you the operators that
preserve the cones of squares (the PSD cones), namely,
$$
\operatorname{Aut}\left(\mathcal{H}^{n}_{+}\left(\mathbb{R}\right)\right)
=
\left\lbrace X \mapsto U^{T}XU \ \middle|\ U \in \mathbb{GL}_{n}\left(\mathbb{R}\right) \right\rbrace
$$
and
$$
\begin{aligned}
\operatorname{Aut}\left(\mathcal{H}^{n}_{+}\left(\mathbb{C}\right)\right)
&=
\left\lbrace X \mapsto U^{*}XU \ \middle|\ U \in \mathbb{GL}_{n}\left(\mathbb{C}\right) \right\rbrace\\
&\cup
\left\lbrace X \mapsto U^{*}\overline{X}U \ \middle|\ U \in \mathbb{GL}_{n}\left(\mathbb{C}\right) \right\rbrace.
\end{aligned}
$$
Imposing the additional condition that these preserve the identity matrix,
$$
\begin{aligned}
\operatorname{JAut}\left(\mathcal{H}^{n}\left(\mathbb{R}\right)\right)
&=
\left\lbrace X \mapsto U^{T}XU \ \middle|\ U \in \mathbb{R}^{n \times n}, U^{T}U = I \right\rbrace,\\
\operatorname{JAut}\left(\mathcal{H}^{n}\left(\mathbb{C}\right)\right)
&=
\left\lbrace X \mapsto U^{*}XU \ \middle|\ U \in \mathbb{C}^{n \times n}, U^{*}U = I \right\rbrace\\
&\cup
\left\lbrace X \mapsto U^{*}\overline{X}U \ \middle|\ U \in \mathbb{C}^{n \times n}, U^{*}U = I \right\rbrace.
\end{aligned}
$$
An analogous result based on the same sort of inertia theorem holds
over the quaternions, $\mathbb{H}$. The book by Rodman contains
enough of the spectral theory to show that,
$$
\operatorname{Aut}\left(\mathcal{H}^{n}_{+}\left(\mathbb{H}\right)\right)
=
\left\lbrace X \mapsto U^{*}XU \ \middle|\ U \in \mathbb{GL}_{n}\left(\mathbb{H}\right) \right\rbrace,
$$
from which it follows that
$$
\operatorname{JAut}\left(\mathcal{H}^{n}\left(\mathbb{H}\right)\right)
=
\left\lbrace X \mapsto U^{*}XU \ \middle|\ U \in \mathbb{H}^{n \times n}, U^{*}U = I \right\rbrace.
$$
I've just posted a preprint where I work out all of these
details. This method has the advantage that you wind up knowing the
cone automorphisms as well, but there is a more direct approach that
can be used to verify the results. Theorem 6.5 in Huang characterizes the matrix Jordan-automorphism groups over $\mathbb{R}$, $\mathbb{C}$, and
$\mathbb{H}$, assuming that you know the linear automorphism groups of
$\mathbb{R}$, $\mathbb{C}$, and $\mathbb{H}$ as algebras over
$\mathbb{R}$. Here again Rodman comes to the rescue with the
prerequisite result for $\mathbb{H}$. In any case, we eventually
deduce the same Jordan-automorphism groups that we do using the
stabilizer subgroup approach.
So to specialize to $n=3$ translate into the notation for Lie groups that the question used, your answer is that $G_{\mathbb{R}} = PO(3) = SO(3)$, $G_{\mathbb{C}} = PU(3) \rtimes \mathbb{Z}/2$, and $G_{\mathbb{H}} = PSp(3)$?
To expand on Michael Orlitzky's answer, unpacking the notation the claim is that, with $h_n(R)$ the $n\times n$ Jordan algebra of Hermitian matrices is
$$ \begin{aligned} \mathrm{Aut}(h_n(\mathbb{R})) &= PO(n) (= SO(n) \text{ if $n$ is odd})\\
\mathrm{Aut}(h_n(\mathbb{C})) &= PU(n) \rtimes \mathbb{Z}/2\mathbb{Z} = (SU(n) / \mathbb{Z}/n\mathbb{Z}) \rtimes \mathbb{Z}/2\mathbb{Z}\\
\mathrm{Aut}(h_n(\mathbb{H})) &= PSp(n) \end{aligned} $$
The projectivizations come from the way the action is defined, the center automatically acts trivially. The most interesting case is for $h_n(\mathbb{C})$, where the group is not connected; the $\mathbb{Z}/2\mathbb{Z}$ acts by the automorphism of the Dynkin diagram (as it exchanges the defining representation and its dual/conjugate representation).
This doesn't look compatible with the expectation in the original question that $\mathbb{C}P^{n-1} = \mathrm{Aut}(h_n(\mathbb{C}))/SU(n-1)$, I would need to see more of the argument to see what might be going wrong.
Update: For the cases of $h_n(\mathbb{R})$ and $h_n(\mathbb{H})$, this is proved by Kalisch (Theorem 6):
Kalisch, G. K., On special Jordan algebras, Trans. Am. Math. Soc. 61, 482-494 (1947). ZBL0032.25003.
The paper also says something about the $h_n(\mathbb{C})$ case, but I don't think it computes the automorphism group. This is also treated by Jacobson:
Jacobson, Nathan, Isomorphisms of Jordan rings, Am. J. Math. 70, 317-326 (1948). ZBL0039.02801.
Jacobson, Nathan, Some groups of transformations defined by Jordan algebras. I, J. Reine Angew. Math. 201, 178-195 (1959). ZBL0084.03601.
The 1948 Jacobson paper is rather inexplicit (referring to automorphisms of the matrix algebra that commute with the involution), and one could easily miss the central extension in the $h_n(\mathbb{C})$ case (which is the $A_{II}$ case in his notation). The 1959 Jacobson paper is way more general and more explicit, perhaps too general (and doesn't state succinctly the result above).
A modern reference is Yokota: https://arxiv.org/pdf/0902.0431.pdf. Some discussion about this is contained in Lie Groups in the Foundations of Geometry (p.165) by Freudenthal. I think this latter has a more geometric approach.
As far as I can tell, Yokota is almost exclusively interested in the cases that give exceptional groups, right? Does that paper consider the other cases in your original question?
On further investigation, Theorem 3 of the 1947 Jacobson paper appears to be wrong, it omits the non-identity component of automorphisms of $h_n(\mathbb{C})$ (coming from an anti-automorphism of the matrix algebra rather than an automorphism).
Yes, Yokota considers. Check Proposition 2.11.1 and 2.12.1. More specifically, he obtains the complete automorphism group in the complex case. It seems to me that the embeddings of the classical cases, although expected, require some algebraic technicalities to be described.
Just a naive question: do you have an explicit expression for these actions? I would expect the other connected component of $G_{\mathbb{C}}$ to have a correspondence with the anti-holomorphic isometries of the complex planes through this action.
Which sort of explicit expression do you want? Michael Berlitzky had it fairly explicitly in his answer. Yokota has a description of the Z/2 action on SU(3) before Prop. 2.12.1, but I think he has a typo: it should be \epsilon(A) = (A^T)^(-1) (transpose inverse, combining two anti-automorphisms)
|
2025-03-21T14:48:30.109841
| 2020-03-24T00:08:06 |
355563
|
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}
|
Stack Exchange
|
Weyl group stabilizer of semisimple element in adjoint group
Let $G$ be semisimple group over $\mathbb{C}$ of adjoint type. Let $T$ be a maximal torus, $s\in T$ semisimple element. Let $W$ be a Weyl group and $W(s)$ be a stabilizer of $s$ in $W$. I am interested in question: weather $W(s)$ is generated by reflections?
Let $\tilde{G}$ denotes simply connected cover of $G$, $\tilde{s}\in \tilde{G}$ denotes some lift of $s$. Then the centralizer $C_{\tilde G} (\tilde s)$ is connected and reductive. The $W(\tilde s)$ is Weyl group of $C_{\tilde G} (\tilde s)$ and generated by reflections.
The group $W(s)$ is genereted by $W(\tilde s)$ and also some elements in $W$ which permutes different lifts of $s$ to $\tilde{G}$. For $C_{\tilde G} (\tilde s)$ these elements are 'diagram automorphisms'
Question. It there any classification of $s$ such that $W(s)$ differs from $W(\tilde s)$?
This looks more or less straightforward (at least for classical types) but I couldn't find reference.
Another question is about affine analogue. Let $T^a=T\times \mathbb{C}^*$ be a torus of affine group, there is an action of affine Weyl group $W^a$ on $\hat{T}$. For any $s^a \in T^a$ let $W^a(s^a)$ denotes its stabilizer in $W^a$.
Question. Is there any reference about $W^a(s^a)$? I am interested about the same question: weather $W^a(s^a)$ is generated by reflections?
Recall the $W^a=W \ltimes Q^\vee$, where $Q^\vee$ is a coroot lattice. Since $Q^\vee$ acts on $T^a$ freely the natural map $W^a(s^a) \rightarrow W$ is injective, so the this question should be very close to the non affine case. But I could not find a reference.
Let $G=PGL_3$ and let $s$ be the diagonal matrix with eigenvalues given by 3 distinct roots of 1 of order 3. Then $W(s)={\mathbb Z}/3$, so it is not generated by reflections.
Yes, of course. My question is there any classification of such $s$.
For such $s$ it is necessary that $W(s)$ is larger then $W(\tilde{s})$. And for example for $PGL_N$ this means that multiset of eigenvalues of $\tilde{s}$ is invariant under multiplication on some $N$-th root of unity.
So I am asking for similar classification for other types.
|
2025-03-21T14:48:30.109996
| 2020-03-24T00:18:24 |
355564
|
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|
Stack Exchange
|
$p$-adic Harish-Chandra character of a stable virtual character
Let $F$ be a $p$-adic field and let $G$ be a reductive group over $F$. Associated to an irreducible admissible representation of $\pi$ of $G(F)$, we have a distribution character $\Theta_{\pi}$ defined by $\Theta_{\pi}(f)= \mathrm{tr}( f \mid \pi)$. Then a theorem of Harish-Chandra says there is a locally constant function $f_{\pi}$ on $G(F)_{sr}$, the strongly regular semisimple elements of $G(F)$, such that $\Theta_{\pi}(f)=\int_{G(F)_{sr}} f(g)f_{\pi}(g)dg$.
My question is as follows. What is a proof or reference for a proof of the following statement: If $a_1 \pi_1 + ... a_n \pi_n$ is a virtual representation such that the associated trace distribution is stable, then $a_1f_{\pi_1} + ... +a_n f_{\pi_n}$ is constant on stable conjugacy classes.
|
2025-03-21T14:48:30.110081
| 2020-03-24T00:31:59 |
355565
|
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"Wlod AA",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/355565"
}
|
Stack Exchange
|
Finding all squares in a generalized Fibonacci type sequence
I asked this two weeks ago at https://math.stackexchange.com/questions/3571697/squares-in-a-second-order-integer-recursive-sequence
Most of the related question shown to me while composing are about primes, not squares...
Given a sequence $x_n$ as in https://oeis.org/A001075
$$ 1, 2, 7, 26, 97, 362, 1351, $$
such that
$$ x_{n+2} = 4 x_{n+1} - x_n $$
These are the $x$ values in $x^2-3y^2 = 1$
Can we find, and prove, all squares in the sequence and all double squares?
I see that Cohn did this for the Fibonacci and Lucas numbers in the 1960's.
For this sequence, it seems $1$ is the only square and $2$ is the only doubled square.
alright, an answer at the recent Squares in Lucas sequences does give a reference Petho
Is this sequence somehow special? (and why?)
CW. this turns out to be a question from May 2020 in a Hungarian magazine for high school students; a bit deflating. The deadline was June 10th and an answer became available online; they did squares but did not bother with twice squares.
|
2025-03-21T14:48:30.110186
| 2020-03-24T00:54:27 |
355567
|
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"url": "https://mathoverflow.net/questions/355567"
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|
Stack Exchange
|
Isometries on the unit sphere
Suppose that $X$ and $Y$ are two Banach spaces, $S_{X}$ and $S_{Y}$ their unit spheres, and $f$ an onto isometry between $S_X$ and $S_Y$. Does it follow that $X$ and $Y$ are isometric?
I believe this is open when stated in full-generality, being sometimes referred to as Tingley's problem. I do not know the details here, but perhaps having this name to search for may help you out; for instance, the answer is apparently positive if $X$ or $Y$ is one of the classical sequence spaces
(I am assuming you are taking real scalars everywhere)
As of November 2018, the problem seems to be still open https://arxiv.org/pdf/1804.10674.pdf
Is there a reason why the insistence on real scalars? Doesn't the problem make sense for complex too?
Also, my question seems to be a tad weaker than Tingley's problem, no? $X$ and $Y$ may still be isometric, even if $f$ does not have an extension.
@Markus: There are many examples of complex Banach spaces that are not isomorphic as complex Banach spaces but are isometrically isomorphic as real Banach spaces. See MR0818448.
|
2025-03-21T14:48:30.110298
| 2020-03-24T01:22:32 |
355568
|
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"lcv"
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|
Stack Exchange
|
Convergence rate of Toda/Morse flow
Let $A(t), A_0$ be a $n\times n$ hermitian complex matrices and consider the following matrix flow
\begin{align}
\frac{dA}{dt} &= \left [ C\circ A , A \right ] \\
A(0) &= A_0 \ .
\end{align}
Here $\circ$ stands for Hadamard (pointwise) multiplication and the (antisymmetric) matrix $C$ has the form
$$
C=\left(\begin{array}{ccccc}
0 & -1 & -1 & \cdots & -1\\
1 & 0 & -1 & & \vdots\\
1 & 1 & 0 & \ddots & -1\\
\vdots & & \ddots & \ddots & -1\\
1 & 1 & \cdots & 1 & 0
\end{array}\right).
$$
I am not 100% sure of the nomenclature Toda vs Morse flow, in any case the flow above or slight variation thereof, has also been considered in a few occasions on MO, for example
here.
It can be shown that the flow is isospectral and that as $t\to\infty$ , $A(t)$ converges to a diagonal matrix with the eigenvalues of $A_0$ ordered from lowest to largest.
My question is the following:
Is anything known about the convergence rate of $\mathrm{diag}(A(t))$ to the eigenvalues of $A_0$? Can we show that the convergence is bounded by some exponential (and in this case can we estimate the rates)?
If the question above is too complicated I would be equally interested in the convergence rate of the flow with the following modified $C$:
$$
C_1=\left(\begin{array}{ccccc}
0 & -1 & -1 & \cdots & -1\\
1 & 0 & 0 & 0 & \vdots\\
1 & 0 & 0 & \ddots & 0\\
\vdots & & \ddots & \ddots & 0\\
1 & 0 & \cdots & 0 & 0
\end{array}\right).
$$
In case of the flow with $C_1$, it can be shown that the matrix element $A_{1,1}(t)$ converges to the lowest eigenvalue of $A_0$, $\lambda_1$.
For the $C_1$-flow, can it be shown that $A_{1,1}(t)$ converges exponentially fast to $\lambda_1$?
It is usually said that the Toda flow is a sort of continuous version of Lanczos diagonalization algorithm. This similarity is even more marked in case of the $C_1$ flow. In fact in the latter case one can show that the "diagonalization" amounts to a (continuous) series of rank-2 unitary rotation (reminiscent of Lanczos algorithm). Since the convergence of the Lanczos algorithm is exponential in the number of iteration,
I formulate the conjecture that the convergence of the above flows is also exponential. Is this true? Can it be proven? I couldn't find anything in the literature linked to the MO answers.
IN general, the strategy is to show that the correct diagonal matrix is the only stable fixed point, and hence the convergence is exponential. Are you looking to prove more than this ?
@PiyushGrover that is exactly what I would like to prove or see a reference for
It's easy to see that a diagonal $A$ is a fixed point, but how can you show from there that the convergence is exponential? (How do you prove the "hence" in your sentence?)
There are several versions of Toda flow, for some there are results by Tomei et.al. that it gives continous version of QR algorithm. Hence the convergene is the same as for QR algorithm which is quite well-known.
@AlexanderChervov thank you can you provide a reference?
https://www.google.com/search?q=toda+and+qr+algorithm&oq=toda+and+qr+algorithm&aqs=chrome..69i57.18020j0j1&sourceid=chrome&ie=UTF-8
@AlexanderChervov thank you I had seen that reference but missed that part. Will look into it now
|
2025-03-21T14:48:30.110548
| 2020-03-24T02:40:14 |
355571
|
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|
Stack Exchange
|
Fano $3$-folds with one singularity
Is there a classification of complex Fano $3$-folds, with Picard rank $1$ and a single cyclic quotient singularity of type $\frac{1}{2}(1,1,1)$?
This should be a bounded family by a result of Borisov.
I'm not sure if a nice clean statement exists, but this paper of Takagi at least covers some of the classification you hope for https://projecteuclid.org/download/pdf_1/euclid.nmj/1114649295
Is it a fact that such Fano 3-folds are not smoothable?
Yes, they are not smoothable. Indeed the singularity 1/2(1,1,1) is rigid (i.e. it doesn't deform) - see Schlessinger's "Rigidity of quotient singularities", Invent. math., vol. 14, pages 17-26 (1971).
|
2025-03-21T14:48:30.110632
| 2020-03-24T04:26:53 |
355574
|
{
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"M. Winter",
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|
Stack Exchange
|
Name for specific cycles in graphs
Is there an established name for cycles $C\subseteq G(V,E)$ with the property that
$$\lbrace u,v\rbrace\subseteq C\cap V\implies\mathrm{dist}_{|C}(u,v)\le \mathrm{dist}_{|G}(u,v)$$
I would be tempted to call them facets because vertices and edges that constitute to the boundary of a facet of a polyhedron are prototypical examples of such cycles.
@MarkSapir the term convex subgraph exists but has a slightly different meaning: all shortest paths from $G$ must me in $C$.
You are looking for the following:
Definition. A subgraph $H\subseteq G$ is called isometric if $\mathrm{dist}_H(u,v)=\mathrm{dist}_G(u,v)$ for all $u,v\in V(H)$.
So your cycles could be called isometric cycles.
Note that not all facets of a polyhedron are induced in this sense.
Consider the a $2n$-gonal pyramid for some $n\ge 3$.
Two antipodal vertices of the $2n$-gonal face $C$ have distance $n$ along $C$, but only distance two in the pyramid (via the apex vertex).
|
2025-03-21T14:48:30.110743
| 2020-03-24T07:43:56 |
355579
|
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|
Stack Exchange
|
Rational Peano curves
An rr function (i.e. rational rational function) is a quotient
$$ \frac fg\,:\, \Bbb Q\ \to\ \Bbb Q\cup\{\infty\} $$
such that $\ f,g\,\in\,\Bbb Z[X],\ $ where $\ g\ne 0.$
QUESTION Do there exist rr functions $\ \phi\ \psi\ $
such that set
$$ \{(\phi(x)\ \ \psi(x))\,:\, x\in\Bbb Q\}\ \subseteq
\ (\Bbb Q\cup\{\infty\})^2 $$
is dense in a non-empty open subset of
$\ \Bbb Q^2\,?$
I'd guess -- NO.
Of course, I assume the standard topology in $\ \Bbb Q^2\ $ induced by the euclidean topology in $\ \Bbb R^2.$
Even with $\phi ,\psi$ in $\mathbb{C}(X)$, your map extends to an algebraic map $\mathbb{P}^1\rightarrow \mathbb{P}^2$, whose image is an algebraic curve, certainly not dense in $\mathbb{P}^2$.
@abx, not dense even in an open subset of the plane? I guess I'll remove my question (since it's too well known). (?)
Is $f\ g$ just the product of $f$ and $g$? The space confuses me.
@Wojowu, $\ (f,\ g)\ $ is an ordered pair. I type operations explicitly, e.g. $\ f\cdot g\ $ and never as $\ fg\ $ To me $\ sin\ $ is not a product but sinus function (product would be $\ s\cdot i\cdot n$).
Since the Chinese and Gutenberg, we hardly ever need "," in mathematical formulas; comas are eyesores.
Thanks for clarification.
comas may be eyesores, but commas are sights for sore eyes. I'm putting 'em in.
Remember elimination theory from algebraic geometry. You just use a resultant to eliminate the parameter. See my notes on Concrete Algebra (on github) for explicit algorithm. So the image is an algebraic curve.
If we put $\cdot$ and $\circ$ in any multiplication and in any composition of linear operators, the result would be quite more heavy but not more clear --- Good notations only make explicit what is needed for clarity (like commas)
@GerryMyerson "comas [...]. I'm putting 'em in."
Your action is unethical. Shame on you and on everybody who imposes on others. DO NOT IMPOSE is the one and only ethical commandment. Decent people have to follow it or else they are indecent. If people followed "do not impose" commandment we would have a happy and universally prosperous society. ***** Of course, I am right about avoiding commas and having operations explicitly. You guys see a tree here and there but not the forest.
@PietroMajer, in 1977, I have written about the translation covariant transformations (mathematics which assisted my computer architectures). With multilevel usage of comp. $\ \circ.\ $ hiding composition symbols would be a disaster. However, I've invented notation that eased the pain and aided understanding. The mathematical notation is mostly an evolutionary creature, it begs for a consistent clean version. A minor example: $\ (a,b)\ (a,b]\ldots\ $ could be my $\ (a;b)\ (a;b]\ldots.\ $ As it is now, $\ (a,b)\ $ may stand for an open interval but also for an ordered pair.
I personally dislike the semicolon in mathematical notations, but, yes, that's a reasonable solution. Another could be the old $]a,b[$ for open intervals etc. Or maybe, in the spirit of the ceiling/floor functions, $\lfloor a,b \rfloor$, and $\lceil a,b \rceil$ for closed intervals, and so on. But the real problem about notations is to reach a general consensus, which seems still to come. Today $(a,b)$ may denote an ordered pair, an open interval, an inner product, and a greater common divisor!
Actually this notation for intervals that I invented right now is not bad at all
$\lfloor a,b \rfloor$
$\lfloor a,b \rceil$
$\lceil a, b \rfloor$
$\lceil a, b \rceil$
@PietroMajer, an analogy is fine but exact reproducing of a symbol is wrong. "Your" intervals, at first glance look like ceiling and floor which decreases the readability. Just imagine a text on number theory, ouch!
In case anyone is confused, when OP writes $(\phi(x)\ \ \psi(x))$, he means what most of us would write as $(\phi(x),\psi(x))$.
@WlodAA I see your point... so, you see, the main problem is really to reach a consensus ;)
@PietroMajer, "the main problem is really to reach a consensus ;)" ***** Unfortunately. But AoA says, don't vote. Instead -- THINK! It should be the profoundness as opposed to consensus. When 100 professors spoke in an open letter against Einstein's Relativity, Einstein said: if just one of them were right it would be enough.
@WlodAA I see that you have created the ([tag:peano-curves]) tag. It might be useful to create also the tag-wiki or at least the tag-excerpt. It might help other users to use the tag correctly. (This is probably not a problem here, since the tag name seems to be descriptive enough.) Another reason is that the tags used on only one question are automatically deleted after certain time unless they have a tag-wiki.
Use the resultant to eliminate the variable $X$. Since the resultant is computed over the rationals, the resultant is a rational coefficient polynomial in the two variables of the plane, satisfied on the image of the parameterized curve. See my (undergraduate!) lecture notes Concrete Algebra on github for complete (and elementary) details.
This sounds good. I'll check your link. (+1 for now). U'r more or less translating parametrization into equation (roughly speaking).
|
2025-03-21T14:48:30.111271
| 2020-03-24T09:13:52 |
355587
|
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/355587"
}
|
Stack Exchange
|
Resolution of an inequality on integers
I’m trying to resolve respect to $k$ the following inequality,
$$
k\left(\log k +\log \log k-\alpha+O\left(\frac{\log \log k}{\log k}\right)\right)\geq x,
$$
in order to obtain, under the condition $k\leq x$, that
$$
\ k\geq \dfrac{x}{\log x}\left(1+\dfrac{\alpha+o(1)}{\log x}\right)
$$
is this possible?
What is the context of this? Is this coming from an explicit version of the Prime Number Theorem?
I want to get a explicit upper bound of Prime counting function
That was my guess. Have you looked at the various papers by Dusart? His paper "Explicit estimates of some functions over primes" and others may have bounds which do what you want. http://www.unilim.fr/pages_perso/pierre.dusart/Publications.html Has a list of a variety of his papers many which have bounds which may be relevant.
I know Dusart’s work, but I want to get some one via an elementary method.
In that case, how are you getting your bound on p_k to start with? I don't know of any elementary proof of the PNT which gives an error term.
$\newcommand\ka{\kappa}$
Let $a:=\alpha$. Let $f$ be a function such that
$$f(k)=k\Big(\ln k +\ln\ln k-a+O\Big(\frac{\ln\ln k}{\ln k}\Big)\Big)
=k\big(\ln k +\ln\ln k-a+o(1)\big)$$
as $k\to\infty$.
For any real $b$ and $x>0$, let
$$\ka:=\ka_b(x):=\frac x{\ln x}\Big(1+\frac b{\ln x}\Big).$$
We have to show that
for any real $b\in(-\infty,a)$, if $x>0$ is large enough and $f(k)\ge x$, then $k\ge\ka_b(x)$.
Take any real $b\in(-\infty,a)$ and then take any $c\in(b,a)$. Let
$$f_c(k):=k\big(\ln k +\ln\ln k-c\big).$$
Let $x\to\infty$. Then
$$\ln\ka=\ln x-\ln\ln x+o(1)\sim\ln x,$$
$$\ln\ln\ka=\ln\ln x+o(1),$$
$$\ln\ka+\ln\ln\ka=\ln x+o(1),$$
$$f_c(\ka_b(x))=\frac x{\ln x}\Big(1+\frac b{\ln x}\Big)\big(\ln x-c+o(1)\big) \\
=\frac x{\ln x}\Big(1+\frac b{\ln x}\Big)\Big(1-\frac{c-o(1)}{\ln x}\Big)\ln x \\
=x\Big(1+\frac{b-c-o(1)}{\ln x}\Big)<x$$
for large enough $x>0$.
Now suppose that $f(k)\ge x$. Then $x\to\infty$ implies $k\to\infty$, because $f$ is bounded on any bounded subset of the set of all natural numbers. Therefore, for all large enough $x>0$ we have $f(k)\le f_c(k)$ and, because $f_c(k)$ is increasing in large enough $k$, we also have $f_c(k)\le f_c(\ka_b(x))$ if $k<\ka_b(x)$, so that
$$f(k)\le f_c(k)\le f_c(\ka_b(x))<x,$$
which means that indeed, if $x>0$ is large enough and $f(k)\ge x$, then $k\ge\ka_b(x)$.
It’s not classic but it works normally, thanks.
The problem is I must now have an effective version.
|
2025-03-21T14:48:30.111459
| 2020-03-24T09:15:34 |
355588
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/355588"
}
|
Stack Exchange
|
Continuous function covers in connected $T_2$-spaces
If $X$ is a topological space, we let $\text{End}(X)$ be the collection of continuous functions $f:X\to X$. We say that $f,g\in \text{End}(X)$ meet if there is $x\in X$ with $f(x) = g(x)$. We say that $D\subseteq \text{End}(X)$ is a cover for $\text{End}(X)$ if for every $f\in \text{End}(X)$ there is $g\in D$ such that $f$ and $g$ meet.
For $\text{End}(\mathbb{R})$, a boring example of a cover is the collection of all constant functions. A more interesting example is the following countable cover: for $k\in\mathbb{Z}$ let $f_k$ to be defined by $x \mapsto x+k$ for all $x\in\mathbb{R}$. Let $c_0$ be the constant $0$ function. Then $$\{f_k:k\in \mathbb{Z}\}\cup\{c_0\}$$ is a cover for $\text{End}(\mathbb{R})$.
This motivates the following question: If $\kappa$ is an infinite cardinal and $X$ is a connected $T_2$-space with $|X|=2^{\kappa}$, does $\text{End}(X)$ have a cover of cardinality $\kappa$?
The answer is strong YES for connected spaces admitting a non-constant continuous function and NO in the opposite case.
If $X$ is a connected topological space that admits a non-constant continuous function
$\gamma:X\to\mathbb R$, then the family
$$\{f\in C(X):f(X)=\{q\}\subset\mathbb Q\}\cup\{q+\gamma:q\in\mathbb Q\}$$is a countable cover for $C(X)$.
If $X$ is a connected space admitting no non-constant continuous function, then the family $C(X)$ consists of constant functions, has cardinality of continuum and each covering of $C(X)$ has cardinality of continuum.
Remark. Examples of connected Hausdorff spaces without non-constant continuous functions are well-known, take for example, the Golomb space. Or even a singleton for the trivial case.
Added in Edit. It turned out that I answered another question mixing $\mathrm{End}(X)$ with $C(X,\mathbb R)$. But for $\mathrm{End}(X)$ also there exists a counterexample: the Cook continuum $K$. It has countable weight but each continuous self-map $K\to K$ is constant. So each cover for $\mathrm{End}(K)$ has cardinality $\mathfrak c>\omega=w(K)$.
Nonetheless, the question remains open for Peano continua: Has the set $\mathrm{End}(X)$ a countable cover for each Peano continuum $X$ without the fixed point property? The answer is not known even for compact connected Lie groups, for example the multiplicative group $S^3$ of quaternions of unit norm.
Let us observe that for $n\le 2$ the $n$-dimensional sphere $S^n$ has a countable cover. More precisely, for $n\in\{0,2\}$ the family $\{\mathrm{id},-\mathrm{id}\}$ is a 2-element cover of $S^n$. For $n=1$, fix any countable dense set $Y\subset S^1$ and observe that the family $\{\mathrm{id}\}\cup\{f\in\mathrm{End}(S^n):\exists y\in Y\;f(S^n)=\{y\}\}$ is a countable cover of $\mathrm{End}(S^n)$. So, the first unclear case is the 3-dimensional sphere $S^3$.
|
2025-03-21T14:48:30.111682
| 2020-03-24T10:27:23 |
355595
|
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"https://mathoverflow.net/users/33741",
"leo monsaingeon"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/355595"
}
|
Stack Exchange
|
Second derivative estimates for a subsolution of linear elliptic equation
Definition. Let $u:\Omega \rightarrow \mathbb{R} $. A function $u$ is called semiconvex if $u=v+w$ for some $v\in C^{1,1}(\Omega)$ and a convex function $w$.
Note. Saying that $u$ is semiconvex is equivalent to say that there exists a $\lambda$ such that the function
$$
z(x)=u(x)+\dfrac{|x|^2}{2\lambda}\text{ is convex}.$$
Consider the elliptic operator of the form $$Lu=a^{ij}D_{ij}u+b^iD_iu$$ and let $L$ be uniformly elliptic.
I want to prove the following statement:
Let $u$ be semiconvex in $\Omega$ and suppose $Lu+f\geq0$ almost everywhere in $\Omega$ for some $f\in L^{n}(\Omega)$. We then have the following:
There exists a positive number $p$, depending only on $n$ and on the constant that define the uniformly ellipticity such that, for any subdomain $\Omega' \subset \subset \Omega$,
$$ \bigg(\int_{\Omega'} |D^2u|^p\bigg)^{1/p}\leq C(||f||_{L^n},n)$$
where $|D^2u|$ is a norm on the space of matrices.
Can somebody please help me?
Are you sure this should be expected for subsolutions? The estimate you're asking for is a typical elliptic regularity statement, but as far as I'm aware of you can only control derivatives for solutions, not just sub/super-solutions (as in DeGiorgi-Nash-Moser)
|
2025-03-21T14:48:30.111797
| 2020-03-24T11:15:40 |
355600
|
{
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"Gerry Myerson",
"Sylvain JULIEN",
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}
|
Stack Exchange
|
Perfect squares between certain divisors of a number
Let $n$ be a positive integer. We will call a divisor $d(<\sqrt{n})$ of $n$ special if there exists no perfect squares between $d$ and $\frac{n}{d}$. Prove that $n$ can have at-most one special divisor.
My progress: I boiled down the problem to the following:
Suppose $k^2\le a,b,c,d\le (k+1)^2$, then $ab=cd\implies \{a,b\}=\{c,d\}$. But I can't seem to prove this.
Arriving here isn't difficult so I am omitting any further details(one more reason being I am not sure if I am on the correct path).
This MO website is for questions of math research. It's not clear to me that this question qualifies. How did you come across it? (Also, you are overloading the symbol $n$.)
@Gerry Myerson My math professor says that this problem came up while he was trying to solve some open problem. I didn't bother asking him the details though. I would be very grateful if you can help me solve this problem. Or even any good idea would do.
Simulposted to m.se, https://math.stackexchange.com/questions/3592817/perfect-squares-between-certain-divisors-of-a-number without notice to either site. That's an abuse.
Doesn't $d$ being special imply $\sqrt{n/d}-\sqrt{d}<1$?
This follows from a result I have asked about a few years ago, namely:
For any $n\in\Bbb N$ there is at most one divisor of $n$ in the interval $[\sqrt{n},\sqrt{n}+\sqrt[4]{n}]$.
I claim that if $d<\sqrt{n}$ is special, then $n/d\in[\sqrt{n},\sqrt{n}+\sqrt[4]{n}]$. Indeed, if this were not the case, we would have $n/d>\sqrt{n}+\sqrt[4]{n}$ and
$$d<\frac{n}{\sqrt{n}+\sqrt[4]{n}}=\frac{n-\sqrt{n}}{\sqrt{n}+\sqrt[4]{n}}+\frac{\sqrt{n}}{\sqrt{n}+\sqrt[4]{n}}<\sqrt{n}-\sqrt[4]{n}+1.$$
Letting $x=\sqrt[4]{n}$, it remains to show that for any $x>2$ there is a perfect square between $x^2-x+1$ and $x^2+x$. By monotonicity, it is enough to show that if $x^2-x+1=k^2$ for some $k\in\mathbb N$, then $x^2+x\geq (k+1)^2$. The first equality resolves to $x=\frac{1}{2}(\sqrt{4k^2-3}+1)$, and we have
$$x^2+x=x^2-x+1+2x-1=k^2+\sqrt{4k^2-3}<k^2+2k<(k+1)^2.$$
I'm sure the last part of the argument can be carried out more cleanly, but it checks out.
Thanks for answering this question, so that I can stop racking my brain on it!
Here is an elementary approach. We want to rule out finding distinct $a,b,c,d$ in short interval (specifically $[n^2+1,n^2+2n]$) with $ad=bc.$ We may assume that $a$ is the smallest and that $b<c$. Then $a<b<c<d.$ I will show that $a+2\sqrt{a}+1\leq d$ so that if $n^2 \leq a$ then $(n+1)^2 \leq d.$
Claim: There are integers $u<v$ and $x<y$ with $$a,b,c,d=ux,uy,vx,vy.$$
Proof: Let $u=\gcd(a,b)$ so $a=ux$ and $b=uy$ with $\gcd(x,y)=1.$ Then $uxd=uyc$ so $xd=yc$ and thus (since $x$ and $y$ are co-prime) there is $v$ with $c=xv$ and $d=yv.$
ASIDE We might as well use $v=u+1$ and $y=x+1$ since $$a,b',c',d'=ux,u(x+1),(u+1)x,(u+1)(x+1)$$ give $ad'=b'c'$ with $a <b',c',d'\leq d.$ I'll comment a bit more about this at the end.
So we want to show
if $ad=bc$ with $n^2<a<b<c<d$ then $d>(n+1)^2.$
From the claim above, $n^2<a=ux$ and $d\geq (u+1)(x+1)=ux+u+x+1.$ But given $ux=a>n^2,$ we know $u+x\geq 2\sqrt{a}>2n.$ Thus $ux+u+x+1>n^2+2n+1$ as desired.
Consider this problem: Given $a$, find $a<b<c<d$ with $d$ minimal such that $ad=bc.$ The work above shows that the solution is to have $$a,b,c,d=ux,u(x+1),(u+1)x,(u+1)(x+1)$$ with $|u-x|$ minimal and that $d>a+2\sqrt{a}+1.$ If we allow $b=c$ then $$n^2\cdot (n+1)^2=(n^2+n)\cdot (n^2+n).$$ If we want $b < c$ then $(n^2-n)\cdot(n^2+n)=(n^2-1)\cdot(n^2).$ With $d-a\approx 2\sqrt{a}.$
If $ad=bc$ then $abcd$ is a perfect square. However this property is weaker and there are solutions such as $$a,b,c,d=2\cdot 120^2,3\cdot 98^2,30\cdot 31^2,5\cdot 76^2=$$ $$28800, 28812, 28830, 28880$$ with $d<a+\frac12\sqrt{a}$ and all four factors between $28561=169^2$ and $28900=170^2.$
Only a partial answer for now. Let $d_{-1}$ the largest divisor of $n$ below $\sqrt{n}$, $d_{-(k+1)}$ the largest divisor of $n$ below $d_{-k}$ and $d_{k}:=n/d_{-k}$. Let $r_{k}:=d_{k}-\sqrt{n}$ and $l_{k}:=\sqrt{n}-d_{-k}$.
Define the $k$-th "square root divisor span" of $n$ as $s_{k}(n):=\sqrt{d_{k}}-\sqrt{d_{-k}}$. The sequence $(s_{k}(n))_k$ is strictly increasing, and its general term equals $\sqrt{\sqrt{n}+r_{k}}-\sqrt{\sqrt{n}-l_{k}}$ which is greater or equal than $\sqrt{\sqrt{n}+k}-\sqrt{\sqrt{n}-k}$.
I think the condition in my comment, namely $s_{k}(n)<1$, is fulfilled only when $\max(l_{k},r_{k})<n^{1/4}$, so for $m(n)=O(1)$ values of $k$, with $\displaystyle{\lim_{n\to\infty}m(n)=0}$.
|
2025-03-21T14:48:30.112086
| 2020-03-24T13:35:02 |
355607
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"Yaakov Baruch",
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"valle"
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|
Stack Exchange
|
When does $\min_x \max_yf(x,y) = \min_y \max_x f(x,y)$ hold for a real function $f(x,y)$?
Let $f(x,y)$ be a real function of the variables $x,y$ (which can be real vectors). Under what conditions do we have the following equality:
$$\min_x \max_yf(x,y) = \min_y \max_x f(x,y)$$
For example, this equality is true if $f(x,y) = xy$ and $x,y$ are real scalars.
Note that this is not the same as Von Neumann's minimax theorem (https://en.wikipedia.org/wiki/Minimax_theorem), because here the role of the variables is exchanged (e.g., $x$ is minimized on the left-hand side, but it is maximized on the right-hand side).
Though I do not know if convexity/concavity of $f(x,y)$ with respect to either arguments plays a role here (like it does for Von Neumann's minimax), I am using the convex-related tags here since that's the context where I've seen related questions. Similarly I am tagging game-theory, though I'm not sure it's directly applicable.
I also expect that the a condition for this equality to be true is that the saddle-points of both sides of the equation be attained at points where the gradient of $f$ vanishes (see my answer below).
Also posted here: https://math.stackexchange.com/q/3592868/10063
Why the close vote?
It's a slightly strange question, since for example the equality wouldn't even hold for $ax+by$ in $[0,1]\times [0,1]$ (unlike the Minimax Theorem). Any particular reason why you are interested in this?
@YaakovBaruch I have a variational formulation of a certain statistical mechanics problem, where I derive an optimization like one side of this equation, but where it would make a lot of sense physically that the optimization were like the other side. I need to understand when both sides are equal, because if they are not it would be potentially interesting. Numerically I find they typically are equal (but this could be because I am only looking at the points where the derivative are zero).
In your example $ax+by$, the saddle occurs at the boundary of the domain, where the gradient of $f$ need not be zero. I do not expect my equality to hold in that case.
Well, I would maybe consider in this case editing some of the rationale into the question itself. I think it deserves more visibility than it has received so far, perhaps due to it being misunderstood as some sort of playing around with Minimax variants?
Also, what are the domains for $x$ and $y$ that you are most interested in?
@YaakovBaruch The domains of $x,y$ are $\mathbb R^n$ (with different dimensions possibly). In my current application they are unbounded.
Here is a tentative proof, under some assumptions. Would like to see some additional arguments (or counterexamples) to make this clearer.
Assumptions:
We suppose that $f(x,y)$ is concave in $y$ for all $x$.
That the equation $\partial f/\partial y=0$ has a unique solution $x$ for every $y$.
That the solution to both saddle-point optimizations is attained at a point where $\partial f/\partial x = \partial f/\partial y = 0$.
Proof:
Then finding $\max_y f(x,y)$ is equivalent to $\partial f/\partial y = 0$. It follows that the original problem is equivalent to a constrained minimization over all variables:
$$\begin{aligned}
\min_x \max_y f(x,y) &= \min_{x,y} f(x,y)
\quad \left(\text{subject to }\frac{\partial f}{\partial y}=0\right) \\
&= \min_y \min_x f(x,y)
\quad \left(\text{subject to }\frac{\partial f}{\partial y}=0\right)
\end{aligned}$$
where we simply changed the order of the minimizations. Since there is a unique $x$ that makes $\partial f/\partial y=0$ for any $y$, the inner minimization on $x$ is trivial, and can be formally changed to a maximization:
$$\begin{aligned}
\min_x \max_y f(x,y) &= \min_y \max_x f(x,y)
\quad \left(\text{subject to }\frac{\partial f}{\partial y}=0\right)
\end{aligned}$$
Finally, if the unconstrained version of the right-hand side optimization is solved at a point where $\partial f/\partial y=0$ (this assumption 3 above), then the constrain can be ignored:
$$\min_y \max_x f(x,y) = \min_y \max_x f(x,y)
\quad \left(\text{subject to }\frac{\partial f}{\partial y}=0\right)$$
In this case, we obtain:
$$\min_x \max_y f(x,y) = \min_y \max_x f(x,y)$$
as desired.
|
2025-03-21T14:48:30.112343
| 2020-03-24T15:00:04 |
355612
|
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"Dylan Wilson",
"Maxime Ramzi",
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|
Stack Exchange
|
$\mathbb Z \otimes_\mathbb S \mathbb Z$ is concentrated in degree $0$ : mistake in the argument
I'm not sure this is research level so if this is not appropriate, feel free to move the question to StackExchange. However, I post it here since my "fake proof" is based on a (recent) paper and I'm not sure it would be fit for SE.
Everything is done $\infty$-categorically.
I'm using the paper A simple universal property of Thom ring spectra, by Antonin-Camarena and Barthel (see here for the arXiv version), and specifically using proposition 4.9 and corollary 5.9 of that paper.
For clarity, let me quote them here :
Prop 4.9 : if $f: S^{k+1}\to BGL_1(R)$ is a based map and $\overline f : \Omega^n\Sigma^n S^{k+1}\to BGL_1(R)$ is the corresponding $n$-fold loop map, then for any $A\in Alg_R^{\mathbb E_n}$, there is an equivalence of spaces $Map_{Alg_{R}^{\mathbb E_n}}(M\overline f,A) \simeq \Omega^{\infty +k+1}A$, if $A$ has characteristic $\chi(f)$, otherwise the space is empty.
Where $M\overline f$ denotes the Thom ring spectrum associated to $\overline f$, that is, the colimit of $\Omega^n\Sigma^nS^{k+1} \to BGL_1(R)\to Mod_R$
And
Corollary 5.9 : There is a map $\Omega^2\Sigma^2S^1\to BGL_1S^0$ whose associated Thom spectrum is equivalent to $H\mathbb Z$ as $\mathbb E_2$-ring spectra.
I changed the phrasing : in the paper it was written $\Omega^2S^3$, but I wrote it that way to emphasize the connection between the two results, and to emphasize the fact that a map is obtained as a $2$-fold loop map from $S^1$. So corollary 5.9 puts us in the situation of proposition 4.9 with $k+1=1$.
We get
Let $A\in Alg_\mathbb S^{\mathbb E_2}$, then if that space is nonempty, there is an equivalence of spaces $Map_{Alg_\mathbb S^{\mathbb E_2}}(H\mathbb Z,A) \simeq \Omega^{\infty +1}A$
My problem is the following : consider $K= H\mathbb Z\otimes_{\mathbb S}H\mathbb Z$. This is an $\mathbb E_\infty$-algebra (which is the coproduct of $H\mathbb Z$ with itself as $\mathbb E_\infty$-algebras), and in particular there are two inclusion map $H\mathbb Z\overset{in_i}\to K$, $i=0,1$ that are $\mathbb E_\infty$-maps.
Forgetting some structure, they are also $\mathbb E_2$-maps. But note that $Map_{Alg^{\mathbb E_2}}(H\mathbb Z, K) \simeq \Omega^{\infty +1}K$, so its $\pi_0$ is $\pi_1K = H_1(H\mathbb Z; \mathbb Z) = 0$ (indeed, one can easily prove that $H_{n+1}(K(\mathbb Z,n);\mathbb Z) = 0$ for all $n$, for instance by providing a cell-structure with no $n+1$-cells).
It follows that $Map_{Alg^{\mathbb E_2}}(H\mathbb Z, K)$ is connected, so $in_0\simeq in_1$ as $\mathbb E_2$-maps.
What this means is that $in_0\otimes in_1 : H\mathbb Z\otimes H\mathbb Z\to K\otimes K$ (tensor products will be over $\mathbb S$, unless explicitly stated) is equivalent to $i\otimes i$ as $\mathbb E_2$-maps, or even as maps of spectra, where $i : H\mathbb Z\to K$ is any $\mathbb E_2$-map (for instance $in_0$ or $in_1$)
But the identity of $K$ factors as $H\mathbb Z\otimes H\mathbb Z\overset{in_0\otimes in_1}\to K\otimes K\to K$ as a map of spectra, and so, as a map of spectra, it factors as $H\mathbb Z\otimes H\mathbb Z\overset{i\otimes i}\to K\otimes K\to K$.
But now this latter map can be upgraded to a map of $\mathbb E_\infty$-algebras : indeed if we chose $i=in_0$ for instance, we can see this latter map as $H\mathbb Z\otimes H\mathbb Z \to H\mathbb Z \overset{in_0}\to K$ where the first map is obtained from the coproduct property, applied to the identity two times.
This means that $id_K$ factors, as a map of spectra, through $H\mathbb Z$. Looking at homotopy groups, this is absurd (it would mean $\pi_iK = 0$ for $i\neq 0$, which is known to be false)
Of course there must be a mistake somewhere but I don't see where it is, so my question is :
Where is the mistake ?
the first mistake is a typo in the paper: $\mathbb{Z}$ is a Thom spectrum on $\Omega^2(S^3\langle 3\rangle)$ not $\Omega^2S^3$, so proposition 4.9 does not apply.
the next question one might ask is "ok, then what goes wrong with the same argument for $\mathbb{F}_p$"? and the answer is that $\pi_1(\mathbb{F}_p \wedge \mathbb{F}_p) \ne 0$. So assuming everything else you wrote is correct (I didn't check) you would be showing that $\tau_0$ detects the difference between the left and right unit as $\mathbb{E}_2$-maps, which would be a fun fact
@DylanWilson : ah that's what that was ! it did feel weird when reading the appendix, but since they referred to some other paper for how to globalize their maps (from $\mathbb Z_p$ to $\mathbb Z$), I figured that the globalization process somehow made this go from $S^3\langle 3\rangle$ to $S^3$ (although I should have known that it wasn't reasonable). Thanks ! And yeah; I had figured that out for $\mathbb F_p$ - I guess what you call $\tau_0$ is the generator of the $\pi_1$ ?
@DylanWilson : and I guess your comment completely settles the question (unless someone comes along and finds another mistake), so if you could write it as an answer, I could accept it and close the question
|
2025-03-21T14:48:30.112676
| 2020-03-24T15:45:01 |
355614
|
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}
|
Stack Exchange
|
Convergence of semi convex functions
Definition. Let $u:\Omega \rightarrow \mathbb{R} $. A function $u$ is called semiconvex if $u=v+w$ for some $v\in C^{1,1}(\Omega)$ and a convex function $w$.
Note. Saying that $u$ is semiconvex is equivalent to say that there exists a $\lambda$ such that the function
$$
z(x)=u(x)+\dfrac{|x|^2}{2\lambda}\text{ is convex}.$$
It's known from the Alexandrov Theorem that a semi-convex function is twice differentiable a.e., that is
for almost everywhere $x_0$ there exists a unique $(p,A)$ such that
$$ u(x)=u(x_0)+<p,x-x_0>+<A(x-x_0),x-x_0> + o(|x-x_0|^2) $$
Let $u_k = u\star \eta_k $ the mollification of $u$. I would show some kind of convergence of the second and first derivatives of $u$, that is $D_{ij}u_k$ and $D_iu_k$.
Can I say that $D_{ij}u_k-D_{ij}u$ for a.e. $x\in\Omega$?
Can be shown the uniformly convergence?
Can somebody please help me?
You get almost everywhere convergence for both first and second derivatives. In general, no uniform convergence can be expected. Take for example $\Omega=(-1,1)\subset\mathbb{R}$ and $u(x)=|x|$. Then $Du=\frac{x}{|x|}$ is discontinuous and $D^2u=2\delta_0$ in the sense of measures. On the other hand, both $Du_k$ and $D^2u_k$ are continuous functions, hence they can't converge to anything discontinous.
Now to the proof. Let me assume that $u$ is convex: if not, just add $|x|^2/2\lambda$. Let me also assume that $\eta$ is supported in $B_1(0)$ and define $\eta_k(x)=k^n\eta(kx)$; I expect what follows to be true also when $\eta$ doesn't have compact support, but I haven't checked. Since $u$ is locally lipschitz, $Du\in L^\infty_{loc}(\Omega)$. In particular, for every $x$,
$$ Du_k(x)=\int_{B_{1/k}}Du(y)\eta_k(x-y)\,dy.$$
Now
$$ |Du(x)-Du_k(x)|\le\int_{B_{1/k}}|Du(x)-Du(y)|\eta_k(x-y)\,dy\le \frac{C||\eta||_{\infty}}{|B_{1/k}|}\int_{B_{1/k}}|Du(x)-Du(y)|\,dy $$
which goes to $0$ as $k\to\infty$ whenever $x$ is a Lebesgue point for $Du$, that is almost everyhwere.
The proof for the second derivative goes more or less the same way. Indeed, by convexity, $D^2u$ is a positive-semidefinite Radon measure whose components are
$$(D^2u)_{i,j}=D_{ij}u\mathcal{L}^n+\Gamma_{ij}$$
for some positive-semidefinite Radon measure $\Gamma=\{\Gamma_{ij}\}$ which is singular with respect to $\mathcal{L}^n$ and $D_{ij}u\in L^1(\Omega)$ is as in Aleksandrov's theorem. Therefore, as above, we get
$$ |D_{ij}u(x)-D_{ij}u_k(x)|\le\int_{B_{1/k}}|D_{ij}u(x)-D_{ij}u(y)|\eta_k(x-y)\,d\mathcal{L}^n(y)+\int_{B_{1/k}}\eta_k(x-y)\,d\Gamma_{ij}(y) $$
which tends to $0$ as $k\to\infty$ for every $x$ which is a Lebesgue point for $D^2u$ and such that
$$\lim_{\rho\searrow0}\frac{|\Gamma|(B_\rho(x))}{\mathcal{L}^n(B_\rho(x))}=0, $$
which happens at $\mathcal{L}^n$-a.e. $x\in\Omega$.
I hope I haven't made any stupid mistake and that my answer was useful!
Thank you very much @Carlo Gasparetto. Now, do you think there is a $L^p$ convergence if one assumes a bounded domain $\Omega$?
@GiovanniFebbraro If you are still interested in the $L^p$ convergence, ask it as a question and notify me about that so I can answer it.
Piotr it is not a question yet? How can I do?
|
2025-03-21T14:48:30.112989
| 2020-03-24T16:13:23 |
355615
|
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|
Stack Exchange
|
fast V representation update of polytope
Say that I have both the V and the H representation of a (possibly unbounded) polytope $P$. I want to append a some rows to the H representation, how can I quickly update the V representation to incorporate the additional constraints?
I case it matters, the additional rows are just intersections with orthants, that is, restricting values to be of a particular sign.
The number of vertices could grow exponentially with the number of hyperplanes, no?
Yes, that's true. In fact, it is the essence of why it is important to have an updating formula. I'm hoping for some clean and simple specialization of something like the "Double Description Method" described at Lemma 9.1 here: https://inf.ethz.ch/personal/fukudak/lect/pclect/notes2015/PolyComp2015.pdf
|
2025-03-21T14:48:30.113076
| 2020-03-24T16:53:24 |
355617
|
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|
Stack Exchange
|
Is there a common notation to indicate the final form of a simplified definition?
I'm trying to become better with using proper terminologies and standard notation when taking notes, which lead me to think:
Similar to the indication of a completed proof by use of the Q.E.D. mark, or "∎", is there a standard method as to indicate the end of a simplified definition?
What do you mean by "simplified definition"?
QED "quod erat demonstrandum", literally meaning "what was to be shown", doesn't fit, but I see no problem with the box sign. An alternative is (using 'theoremstyle" in latex) to use italics in the definition environment. I guess anyway that the most standard marking is the blank line, which is automatic after definition environment.
I concur with @Carl-FredrikNybergBrodda: it would help to know just what you mean here. If you have a definition and then it gets simplified, perhaps in several steps to equivalent but shorter forms, then I would do this: Don’t give the definition first, but rather prove the lemma that the various descriptions of the phenomenon are equivalent, and then define: A (blah) that satisfies any one of the conditions of the above Lemma will be said to be (wudge).
Following Euclid, you could use QEF (quod erat faciendum – which had to be done). Euclid used the Greek version of this (ὅπερ ἔδει ποιῆσαι) to close propositions that were not proofs of theorems, but constructions of geometric objects.
What if I'm not presenting the constructions? I could be interpreting the use of QEF wrong; I'm questioning whether or not I'm able to use QEF due the sake of my simplification being for ease of notation, despite the unsimplified definition being usable.
|
2025-03-21T14:48:30.113242
| 2020-03-24T16:57:09 |
355618
|
{
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|
Stack Exchange
|
How did Hilbert prove the Nullstellensatz?
All of the many proofs of the Nullstellensatz I have seen use results from long after Hilbert’s time: Zariski’s lemma, Noether normalization, the Rabinowitch trick, model theory, etc. How did Hilbert’s original proof go?
Duplicate of this stackexchange post.
That MSE link gives proof written in German.. Can one ask for translation to English of Hilbert's proof of Nullstellensatz?
an English transation is available here
Some discussion of the proof of the Nullstellensatz taken from https://webusers.imj-prg.fr/~michael.harris/theology.pdf pg. 3, footnote 3 which cites 'Algorithms in invariant theory' by Bernd Sturmfels and pg. 7, footnote 7 which states, "These are notably differential methods related to the Lie-Klein theory of continuous groups
which later served Hilbert as a framework for invariant theory; and irrational invariants closely
tied to his discovery and use of the Nullstellensatz."
Of course either of these would be survey in nature, I reckon. But if diving into Hilbert's paper is undesired...
I should add that that paper is 'Theology and Its Discontents: The Origin Myth of Modern Mathematics' by Colin McLarty and (as the title suggests) about the basis theorem and not the Nullstellensatz. But it came to mind and is a fun and interesting paper.
Here is a scan of the relevant pages from the english translation of Hilbert's 1893 paper.
For reference, these are pp. 234 and 235 of the book.
As noted in math.stackexchange, the theorem is in part 3 of this paper: Hilbert, D. (1893). Über die vollen Invariantensysteme. Math. Ann. 42, pp. 313–373.
From a quick overview, I think the idea is to first prove the theorem when the set of common zeros is finite, using elimination theory. Everything is done with homogeneous polynomials.
Now to find time to read it carefully.
I copied the relevant pages from the english translation of Hilbert's 1893 paper. If there is an interest, drop me a line and I can email these.
I would definitely like a cooy but I don’t know how to “drop you a line”.
@Fernando --- I've posted the scan in the answer box, hoping distribution of these few pages qualifies as "fair use".
|
2025-03-21T14:48:30.113697
| 2020-03-24T17:17:27 |
355621
|
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}
|
Stack Exchange
|
Sard's theorem and Cantor set
Sard's famous theorem asserts that
Theorem. The set of critical values of a smooth function from a manifold to another has Lebesgue measure $0$.
I am asking for the curiosity that is it possible to find such a function whose set of critical values is
Cantor set or
Any other uncountably infinite set?
You might have looked at the first version of Sard's Theorem, which is more heavily cited. He published a follow-up paper where he gave an upper bound on the Hausdorff dimension.
It is not hard to construct a smooth function $f$ on $\mathbb R$ such that $f \ge 0$ with $f(x) = 0$ if and only if $x$ is in the Cantor set $E$. If $F$ is an antiderivative of $f$, the critical values of $F$ will be an uncountably infinite perfect set.
By a theorem of Whitney (easy in this $1$-dimensional case), any compact subset $K$ in the interval $I$ is the set of zeroes of a
smooth ($C^\infty$) nonnegative function $f$. As Robert said, take a primitive $F$. Provided that $K$ has no interior point, the critical values $F(K)$ of $F$ are homeomorphic with $K$.
The paper "Hausdoff Measures and the Morse-Sard Theorem" by Moreira gives in section 4 a general construction of near-counterexamples involving generalized Cantor sets.
Reference
Carlos Gustavo T. de A. Moreira, "Hausdorff measures and the Morse-Sard theorem" (English), Publicacions Matemàtiques 45, No. 1, 149-162 (2001), MR1829581, Zbl 0995.58007.
|
2025-03-21T14:48:30.113833
| 2020-03-24T17:17:57 |
355622
|
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|
Stack Exchange
|
Minimum of $\mathrm{rank}\left( \boldsymbol{W} \boldsymbol{H} \right)$, with $\boldsymbol{W}$ block diagonal
Let us assume that we have a full-rank $(n\cdot l)\times k$ matrix, $\boldsymbol{H}$, with no specific structure (e.g., a realization of a Gaussian i.i.d. random matrix), and an $m\times (n\cdot l)$ full-rank block diagonal matrix $\boldsymbol{W}$ such that
\begin{equation}
\boldsymbol{W}= \begin{bmatrix}
\boldsymbol{W}_1 & \boldsymbol{0}_{m_1\times l} & \dots & \boldsymbol{0}_{m_1\times l}\\
\boldsymbol{0}_{m_2\times l} & \boldsymbol{W}_2 & \dots & \boldsymbol{0}_{m_2\times l}\\
\vdots & \vdots & \ddots & \vdots \\
\boldsymbol{0}_{m_n\times l} & \boldsymbol{0}_{m_n\times l} & \dots & \boldsymbol{W}_n
\end{bmatrix},
\end{equation}
where $\boldsymbol{W}_i$ are $m_i \times l$ matrices with $\sum_{i=1}^nm_i=m$, $1\leq m_i \leq l$, and $\boldsymbol{0}_{m_i\times l}$ denotes the $m_i\times l$ all-zero matrix. A new $\boldsymbol{W}$ can be selected for every possible $\boldsymbol{H}$
What is the minimum rank of the matrix product $\boldsymbol{WH}$?
In the case where $m_i = 1,\; \forall i\in\{1,2,\dots,n \}$, we have been able to roughly demonstrate that
\begin{equation}
\mathrm{rank}\left( \boldsymbol{W} \boldsymbol{H} \right)>\frac{m\cdot (k-l)}{k},
\end{equation}
but we still lack an elegant proof.
I think for a "generic" choice of a full rank matrix $H$, ${\rm{rank}}(WH)$ coincides with $\min{{\rm{rank}}{W},{\rm{rank}}{H}}$. Indeed, assuming that $m\leq k$ so that ${\rm{rank}}(H)=\min{nl,k}$ is not smaller than ${\rm{rank}}(W)=m$, the condition ${\rm{rank}}(WH)<m$ could be described as the vanishing of certain $m\times m$ minor. So ${\rm{rank}}(WH)=m$ for any $H$ except those belonging to a closed nowhere dense subset of positive codimension.
That's true, but we are interested in the case where $\boldsymbol{W}$ can be constructed for each $\boldsymbol{H}$. We have edited the post accordingly to clarify this fact. Sorry for the misunderstanding.
|
2025-03-21T14:48:30.113981
| 2020-03-24T17:19:55 |
355623
|
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|
Stack Exchange
|
How does this calculation of Siegel make sense?
I am reading Siegel's paper Zum Beweise des Starkschen Satzes. Let $K$ be an imaginary quadratic field with $d_K=-p$, $p=4k+3$ a prime, and such that $h_K=1$.
Let $f=4m+1$ be a prime inert in $K$, and consider the order $\mathcal{O}=\mathbb Z+f\mathcal O_K$ with conductor $f$. Let $$\omega=\frac{fp+\sqrt{-p}}{2}.$$
Siegel defines the lattices
$$\mathfrak c_k=[f,k-\omega]\quad\text{for } -\frac{f-1}{2}\leq k\leq \frac{f-1}{2}$$
and
$$\mathfrak{c}_\infty =[1,-f\omega].$$
These are proper fractional ideals of $\mathcal O$, mutually non-homothetic. By a well-known formula relating the class numbers $h_K$ and $h(\mathcal O)$ we know that $h(\mathcal O)=f+1$, so the ideals above represent all ideal classes of $\mathcal O$. A little calculation shows that $N(\mathfrak{c}_k)=N(\mathfrak{c}_\infty)=1$. Indeed
$$N(\mathfrak c_k)=\frac{N([f^2,fk-f\omega])}{f^2}=\frac{1}{f^2}\frac{\text{disc}([f^2,fk-f\omega])^{1/2}}{\text{disc}(\mathcal O)^{1/2}}=\frac{1}{f^2}\frac{\begin{vmatrix}f^2 & fk-f\omega \\ f^2 & fk -f\overline \omega\end{vmatrix}}{\begin{vmatrix}1 & -f\omega \\ 1& -f\overline \omega\end{vmatrix}}=1.$$
Then Siegel proceeds to calculate the values of the character defined by
$$\chi(\mathfrak a)=\left(\frac{f}{N(\mathfrak a)}\right)=\left(\frac{fd_K}{N(\mathfrak a)}\right).$$
This does not make sense to me, because the norms are equal to $1$, but Siegel gets different values. See the referenced paper, beginning of the section 2., p. 183.
Update
We have $\mathfrak c_k \not \subset \mathcal O$ but $f\mathfrak c_k\subset \mathcal O$, so we can use the relation $N(f)N(\mathfrak c_k)=N(f\mathfrak c_k)$:
$$ N(\mathfrak c_k) =\frac{N(f\mathfrak c_k)}{N(f)} = \frac{N(f\mathfrak c_k)}{f^2} .$$ To compute $N(f\mathfrak c_k)$ we use the following fact: if $M\subset L$ are free modules of the same rank $n$, $(e_i)$ and $(u_i)$ bases for $L,M$ respectively, $u_i=\sum c_{ij}e_j$, then $(L:M)=\lvert \det(c_{ij})\rvert.$ Therefore
$$\mathcal O=[1,-f\omega],\qquad f\mathfrak c_k=[f^2,fk-f\omega],\qquad N(f\mathfrak c_k)=\begin{vmatrix}f^2 & 0\\ fk &1\end{vmatrix}=f^2.$$
Consequently, $N(\mathfrak c_k)=1$.
Update 2
The $\mathfrak c_k$ are not ideals of $\mathcal O_K$. Let $m$ be a rational integer. We prove that if $m\omega \mathfrak c_k\subset \mathfrak c_k$ then $m$ is a multiple of $f$.
Suppose that $m\omega \mathfrak c_k\subset \mathfrak c_k$ and that $(f,m)=1$. Then $$m\omega(k-\omega)=xf+y(k-\omega),\qquad x,y\in \mathbb Z,$$
$$(m\omega-y)(k-\omega)\in f\mathcal O_K.$$
But $f$ was assumed to be inert in $K$, so $f\mathcal O_K$ is a prime ideal, and $\omega\not\in \mathcal O$. Therefore $\omega\equiv y/m$ modulo $f\mathcal O_K$, because $m$ is invertible. On the other hand, from the definition of $\omega$ we have $4\omega^2\equiv p$ modulo $f\mathcal O_K$. Therefore $(p|f)=1$. But since $f=4m+1$ is inert $-1=(-p|f)=(p|f)$, a contradiction.
This shows that $\mathfrak c_k$ is a proper ideal of $\mathcal O$.
Your ideals are integral ideals, and their norms seem to be $f$, not $1$.
@FranzLemmermeyer We have $\mathfrak c_k\subset \mathcal O_K$, but $\mathfrak c_k\not\subset \mathcal O $. Could you please show how you calculated the norm?
The norm of an ideal with basis $[m, a+n\omega]$ is $mn$.
@FranzLemmermeyer Then what is wrong with my computation of the norm (see above)?
I don't understand what you are doing - what references are you using?
Anyway, using the definition of the norm of an ideal as the cardinality of its residue class ring, use the element $k - \omega$ to reduce any algebraic integer to an ordinary integer; using the element $f$ of the ideal we find that the residue class ring is simply ${\mathbb Z}/f{\mathbb Z}$, so the norm if the ideal is $f$.
@FranzLemmermeyer But $\mathfrak c_K \not \subset \mathcal O$, so what do you mean by the residue class ring in this case? One cannot use $k-\omega$ to reduce elements of $\mathcal O$, because $k-\omega \not \in \mathcal O$. I have updated the question to give a more detailed computation of the norm.
These are ideals in ${\mathcal O}_k$, and of course we are talking about the ordinary norms of these ordinary ideals. You seem to want to compute the norm of ideals in a ring to which they do not belong.
@FranzLemmermeyer That is not true. We actually have $\mathcal O =\lbrace \alpha \in K\colon \alpha \mathfrak c_k\subset c_k\rbrace$.
@FranzLemmermeyer Please see Update 2 above.
You're right. But they are ${\mathcal O}_k$-modules, and these have norm $f$. Siegel's "Ringideal" is defined as in Hecke's lecture "Analysis und Zahlentheorie" (edited by Roquette), p. 125ff.
@FranzLemmermeyer How they can be $\mathcal O_K$-modules?
${\mathbb Z}-modules$ in ${\mathcal O}_k$ of course - sorry. I'll go through the stuff carefully until tomorrow.
In the definition on p.182 using the Kronecker symbol, $\mathfrak{p}$ and $\mathfrak{c}$ are meant to be ideals of $\mathcal{O}_k$, and their norms are meant to be $[\mathcal{O}_k:\mathfrak{p}]$ and $[\mathcal{O}_k:\mathfrak{c}]$. So I think you need to consider the product $\mathfrak{c}_k\mathcal{O_k}$, which is a fractional ideal of $\mathcal{O_k}$: write it as a quotient of two ideals of $\mathcal{O}_k$, and then apply the definition on p.182 using the Kronecker symbol.
Here's what I found out so far.
Let $K$ be a complex quadratic number field with discriminant $\Delta < -4$. The ring class group modulo $f$ is a special case of a ray class group: Two ideals (coprime to $f$, as everything below) are equivalent in the ring class group modulo $f$ if $\alpha {\mathfrak a} = \beta {\mathfrak b}$ for elements $\alpha, \beta \in {\mathcal O}_K$ congruent to a rational integer modulo $f$. The different classes can be represented by ideals in ${\mathcal O}_K$ (as I just did), as ${\mathbb Z}$-modules, or as ideals in the order ${\mathcal O}_f$. There are a lot of isomorphisms floating around, and the underlying sets of these objects are, in general, not the same.
Let me give an example. Consider $K = {\mathbb Q}(\sqrt{-7})$ and $f = 5$. The formula for the number of ring classes (see Cox, Primes of the form $x^2 + ny^2$ or, better yet, Cohn's Advanced number theory) gives $h = 6$. The corresponding ring classes are represented by the ideals $(1)$ (the principal class) and the ideals $(k+\alpha)$ for $k = 0, 1, \ldots, 4$, where $\alpha = \frac{1 + \sqrt{-7}}2$. This does not contain the number theoretic information we are interested in.
We therefore consider the ${\mathbb Z}$-modules $M_k = [5, k-\omega]$ and $M_\infty = [1, -5\omega]$, where $\omega= \frac{35 + \sqrt{-7}}2$. To these modules $M_k = [\alpha, \beta]$ we associate quadratic forms $Q_k = N(\alpha x + \beta y)$. Here's what we get:
$$ \begin{array}{c|cc}
k & Q_k & \text{reduced form} \\ \hline
1 & 25x^2 - 165xy + 274y^2 & (4, -1, 11) \\
2 & 25x^2 - 155xy + 242y^2 & (2, 1, 22) \\
3 & 25x^2 - 145xy + 212y^2 & (2, -1, 22) \\
4 & 25x^2 - 135xy + 184y^2 & (4, 1, 11) \\
5 & 25x^2 - 125xy + 158y^2 & (7, 7, 8) \\
\infty & x^2 - 175xy + 7700y^2 & (1, 1, 44)
\end{array} $$
These are the six form classes of binary quadratic forms with discriminant $-5^2 \cdot 7$. These form classes contain all the information we need for computing class fields using complex multiplication.
The only nontrivial quadratic character $\chi$ on the ring class group is the one with values $-1$ on the nonsquare classes. Since the forms $Q_1$, $Q_4$ and $Q_\infty$ obviously represent squares, we have $\chi(Q_1) = \chi(Q_4) = \chi(Q_\infty) = 1$
and $\chi(Q_2) = \chi(Q_3) = \chi(Q_5) = -1$.
We can also attach ideals in the rings ${\mathcal O}_f$ representing the six equivalence classes by simply associating the ideal $(a, \frac{b - f\sqrt{\Delta}}2)$ to the form $(a, b, c)$. I have not yet checked how the evaluation of the genus character works using these ring ideals.
Thank you for your answer. However, it is still not clear to me what is wrong with my calculation of the norm. According to the question above, every ideal class of $\text{Cl}(\mathcal O_f)$ contains a fractional ideal of norm equal to $1$. But by this question/answer this happens if and only if the class number $h_f$ is odd. But $h_f=f+1$ above is even.
Neither do I. For the moment, I think that Siegel is a little bit sloppy with his language.
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2025-03-21T14:48:30.114479
| 2020-03-24T18:03:20 |
355624
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"authors": [
"Oeyvind Solberg",
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Stack Exchange
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Comparing zeroth Hochschild homology and cohomology for algebras
Let $A$ be a finite dimensional algebra and $[A,A]$ its commutator and $Z(A)$ its center.
Question 1: Do we have $dim(A/[A,A]) \geq dim(Z(A))$?
Note that A/[A,A] is the zeroth Hochschild homology of $A$ and $Z(A)$ the zeroth Hochschild cohomology.
Question 2: In case $A$ is a Frobenius algebra, do we have equality $dim(A/[A,A]) = dim(Z(A))$ iff $A$ is a symmetric algebra (meaning $A \cong D(A)$ as bidmodules)?
Note that in case $A$ is symmetric we have $dim(A/[A,A]) =dim DHom_{A^e}(A,D(A))=dim Hom_{A^e}(A,A)= dim(Z(A))$ , where $A^e$ is the enveloping algebra of $A$. So one direction is clear, and at least for $A$ weakly symmetric, the computer gives good evidence that the other direction is also true. This would also have very practical use as testing whether an algebra is symmetric takes very long at the moment with qpa while the truth of the above criterion would give a much quicker test.
Question 2 is not true for general algebras. For example we also have $dim(A/[A,A]) = dim(Z(A))$ for the representation-finite blocks of Schur algebras $A$.
(this is a new formulation of a recent question)
In the repository for QPA1 there is, what we believe, a faster implementation of the 'IsSymmetricAlgebra' as of yesterday. Download the new version, and see if there is any improvement for you in your examples.
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