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2025-03-21T14:48:30.491144
| 2020-05-04T16:56:30 |
359351
|
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"Ben McKay",
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|
Stack Exchange
|
Inner product between gradients of two functions
It is well known that the orthogonality of two functions can be measured by their inner product. check https://en.wikipedia.org/wiki/Orthogonal_functions In discrete case, the inner product is basically $\sum_i f(x_i)g(x_i)$
But have people studied the inner product between their 1st order derivatives? In particular, in the discrete case, it would be $\sum_i \left<\nabla f(x_i), \nabla g(x_i)\right>$, where $\left<\vec a, \vec b\right>$ denotes the inner product between vector $\vec a$ and $\vec b$. Does it measure anything of intuitive sense?
The name Sobolev comes to mind: https://en.wikipedia.org/wiki/Sobolev_space
|
2025-03-21T14:48:30.491232
| 2020-05-04T17:46:28 |
359355
|
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"Jochen Wengenroth",
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"https://mathoverflow.net/users/37855",
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"thomashennecke"
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}
|
Stack Exchange
|
How to find eigenvalues following Axler?
Preparing my Linear Algebra lecture I like the determinant free approach of Axler because the proof that operators $T$ on an $n$-dimensional complex vector space have eigenvalues is so simple:
Fix any non-zero vector $x$, observe that $x,T(x),\ldots,T^n(x)$ are linearly dependent to get a non-trivial linear combination $\sum\limits_{k=0}^nc_k T^k(x)=p(T)=0$, factorize the polynomial, and conclude that at least one factor $\lambda_k-T$ is not injective just because compositions of injective maps are injective.
The existence of a single eigenvalue is enough to prove the spectral theorem for normal operators by induction.
However, I also try to mention the algorithmic aspects, and to make this proof an algorithm you really have to find all zeros of $p$. This is in contrast to the usual determinant approach where you only need one zero of the characteristic polynomial to get an eigenvalue.
Assume you have an oracle telling you one zero of a complex polynomial each time you ask. Is there a determinant free argument similar to Axler's which would lead to an eigenvalue?
Of course, without asking the oracle $n$ times, but if one quesion isn't enough perhaps at least considerably less than $n$ calls.
Without using the oracle you can find the minimal polynomial of $T$ by looking for the first linear dependence among $I, T, T^2,\dotsc,T^r$ (minimal $r$). Alternatively find the first linear dependence among $x, T(x),\dotsc, T^r(x)$, for a fixed (arbitrary) nonzero $x$. Either way, the resulting polynomial has the property that all roots are eigenvalues. So, only one oracle call is needed at that point. (In practice it seems to me that the oracle goes the other way: to find a root of a polynomial, you look for an eigenvalue of the companion matrix.)
Minimality is a great concept. Shame on me that I missed that.
@ZachTeitler Of course, I would like to accept your comment as an answer.
Using minimality can help.
Without using the polynomial-root oracle you can find the minimal polynomial of $T$ by looking for the first linear dependence among $I, T, T^2, \dotsc, T^r$ (for the minimal $r$—starting with just $I,T$ and increasing $r$ until you find a linear dependence). Alternatively, find the first linear dependence among $x, T(x), \dotsc, T^r(x)$ for a fixed (arbitrary) nonzero $x$. Either way, the resulting polynomial will have the property that all its roots are eigenvalues of $T$. So, only one call to the oracle is needed at that point.
(In practice, it seems to me that it probably goes the other way: to find a root of a polynomial, one looks for an eigenvalue of the companion matrix. But pedagogically it makes sense.)
I would actually go one step further and DEFINE eigenvalues as the zeros of the minimal polynomial, which I would introduce just as above asap. One can then easily prove the "if and only if"-theorem connecting this to the usual definition involving eigenvectors. In this way, one focuses on the central problem of finding zeros, instead of sweeping it under the rug. Consequently, one should move forward to the Frobenius normal form from there and can discuss the Jordan Form as special case when the minimal polynomial factors completely.
This ties in neatly with computations over the rational numbers and their finite extensions that can actually be performed in practice. I think Axler falls short of his own title by relegating the minimal polynomial to the back of the book without using it in this context, even though it is natural to introduce it while talking about eigenvalues from the start. Here, I refer to the second edition, I do not know the later ones.
Section 1.5 of Householders "The theory of matrices in numerical analysis" contains a proof of the minimal polynomial having degree at most $n$. I learned an explicit constructive variant of it from Harald Löwe, TU Braunschweig.
|
2025-03-21T14:48:30.491524
| 2020-05-04T18:39:10 |
359359
|
{
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"authors": [
"Sergei Akbarov",
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|
Stack Exchange
|
"Constructive" proof that compact sets $K\subseteq L_1(\mu\times\nu)$ are contained in products $S\widehat{\otimes} T$
A.Defant and K.Floret in chapter 7 of their Tensor Norms and Operator Ideals prove the equality
$$
L_1(\mu\times\nu)\cong L_1(\mu)\widehat{\otimes}L_1(\nu)
$$
for measures $\mu$ and $\nu$. At the same time, A.P.Robertson and W.Robertson in chapter VII of their Topological Vector Spaces write that
if $E$ and $F$ are metrizable locally convex spaces, then each compact set $K\subseteq E\widehat{\otimes} F$ is contained in the closed absolutely convex hull of a sequence $x_n\otimes y_n$, where $x_n\to 0$ and $y_n\to 0$.
Together this implies, that
for each compact set $K\subseteq L_1(\mu\times\nu)$ there are compact sets $S\subseteq L_1(\mu)$ and $T\subseteq L_1(\nu)$ such that
$$
K\subseteq S\widehat{\otimes} T,
$$
where $S\widehat{\otimes} T$ means closed absolutely convex hull of the set $\{x\otimes y;\ x\in S, \ y\in T\}$ in $L_1(\mu\times\nu)$.
I think, that this result can be proved directly, inside the theory of the Banach space $L_1(\mu)$ (and without references to the theory of topological vector spaces, in particular, to Robertsons' proposition). Is that true?
I am asking because I am trying to study the properties of the spaces of universally integrable functions, where I suspect the same must be true (but with a much more complicated topology, which is not metrizable, and therefore one can't apply Robertsons' lemma).
P.S. I consider the case where $\mu$ and $\nu$ are (positive) Radon measures on compact topological spaces.
@erz, I think, you can shed some light on this.
|
2025-03-21T14:48:30.491679
| 2020-05-04T19:07:23 |
359361
|
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"0xbadf00d",
"Aleksei Kulikov",
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|
Stack Exchange
|
Determine the singular values of a compact operator in terms of the eigenvalues of an alternating tensor product of operators
Let $H$ be a $\mathbb R$-Hilbert space, $A\in\mathfrak L(H)$ be compact and $$|A|:=\sqrt{A^\ast A}$$ denote the square-root of $A$. By definition, the $k$th largest singular value $\sigma_k(A)$ of $A$ is equal to the $k$th largest eigenvalue $\lambda_k(|A|)$ of $A$.
Now, if $S,T:H\to H$ are linear, let $S\otimes T$ denote the linearization of $$H^2\ni(x,y)\mapsto Sx\otimes Ty\tag1$$ and $S^{\otimes 2}:=S\otimes S$. I would like to endow $H^{\otimes2}$ with an inner product so that $S\otimes T$ is bounded, whenever $S,T$ are bounded, and for its completion $H^{\hat\otimes2}$ the operator $S\otimes T$ has a unique bounded linear extension $S\hat\otimes T$.
Moreover, it should hold that the largest singular value $\sigma_1(A^{\hat\otimes k})$ of $A^{\hat\otimes k}$ is given by $\prod_{i=1}^k\sigma_i(A)$. Note that this is a property which is fulfilled by the exterior product $A^{\wedge k}$.
It's trivial to see that if $S:H\to H$ is linear, $\lambda_i\in\mathbb R$ and $e_i\in\mathcal N(\lambda_i-S)$, then $$S^{\otimes k}\underbrace{\left(\bigotimes_{i=1}^ke_i\right)}_{=:\:e}=\underbrace{\prod_{i=1}^k\lambda_i}_{=:\:\lambda}\bigotimes_{i=1}^ke_i\tag2$$ and hence $e\in\mathcal N(\lambda-S^{\otimes k})$.
The problem with this is that the largest eigenvalue of $|A|^{\otimes k}$ is clearly $\lambda_1(|A|)^k$; not $\prod_{i=1}^k\lambda_i(|A|)$ as I would like. The crucial difference to what I remarked about the exterior product before is that the exterior product consists of alternating multilinear forms (and hence $e_1\wedge\cdots\wedge e_1$ cannot be an eigenvector of $A^{\wedge k}$).
So, we would need some kind of alternating tensor product. Is such a construction possible?
Remark: Note that the natural choice for the inner product on $H^{\otimes2}$ is given by the Hilbert-Schmidt tensor product; maybe we can consider some kind of closed subspace of alternating tensors of it.
EDIT: What I'm trying to find is an analogue of Proposition 3.2.7 in Random Dynamical Systems:
The proposition you quote is only stated in finite dimensions. Is it important to you to have a generalization that works in infinite dimensions, or is your question more about the algebraic construction (full tensor powers of a vector spaces versus its exterior powers)?
If $\sigma_1(B) = 0$ then $B = 0$. Now pick your favorite $2\times 2$ matrix $A$ with $\sigma_1(A) = 1$, $\sigma_2(A) = 0$ and get a counterexample ($A\otimes A \neq 0$, but you want $\sigma_1(A\otimes A) = 1*0 = 0$).
Thank you for your answer. Please consider the exterior/wedge product of your example and see my edit of the question for clarification.
@0xbadf00d perhaps I do not understand what you are asking for, sorry. If you want the norm on $H\otimes H$ so that your conditions on the singular values are satisfied, then I gave an example showing that is not possible. If on the other hand you are OK with using some other space, then why $\bigwedge^2 H$ is not suitable for you?
$\wedge^2H$ is suitable, but I wondered whether the same construction is possible using tensor products. It should be possible to construct an "alternating" tensor product as the subspace of $H^{\otimes2}$ spanned by the thensor $(x\otimes y-x\otimes y\rangle/2$.
I've found something in the book Ryan, but cannot manage to understand his notation: https://math.stackexchange.com/q/3660077/47771.
@0xbadf00d I am having trouble understanding your comments. If one actually wants to construct the wedge product then you do it precisely by taking a tensor power and applying antisymmetrisation. But the question in the body of your post asks for a tensor power that behaves like the wedge product. Clearly Aleksei's answer addresses the question of the tensor power, showing it does not behave like the wedge product. If you want to antisymmetrise then you just get back the wedge product, which seems to be covered by the proposition you quote
|
2025-03-21T14:48:30.491977
| 2020-05-04T19:21:39 |
359362
|
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"Iosif Pinelis",
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|
Stack Exchange
|
Proof of Hitting-time theorem in branching processes
I want to understand theorem 5.21 (page 224) in this link and here is where I don't understand: $$
\{W = t\} = \{t \text{ is the first ladder index in }R_1, \dots, R_t\},$$
i.e. $\{R_t = 1, R_1 < 1, \dots, R_{t-1} < 1\}$.
And if anyone can offer easier ways to prove it, it would be great to hear your ideas.
In the proof of Theorem 5.21 (page 224) in the notes at your link, we find the definitions
$R_i:=1-S_i$, with $R_0:=0$. By Lemma 5.17 and the three-line display on page 221 of those notes,
$$W=\tau_0=\inf\{t\ge0\colon S_t=0\},$$
whence
$$W=\tau_0=\inf\{t\ge0\colon R_t=1\}.$$
Thus,
$$W=t\iff R_1<1,\dots,R_{t-1}<1,R_t = 1,$$
as claimed.
Can you also explain how $\tau_0 = \inf{t \geq 0 : S_t = 0}$ works only regarding branching process? The notes explains in regard to random walk and it is hard to relate with branching process for me.
Or would it be better to ask another question? Thank you in advance!
In case that my question was not clear enough. I want to know what $S_n$ means in branching process where $S_n = 1 + \sum_{i=1}^{n}(X_i - 1)$ where $X_i$ is an offspring process and why $\tau_0 = \inf{t \geq 0 : S_t = 0}$. Thank you!
@MathislikeFriday : This additional question should indeed be posted separately.
|
2025-03-21T14:48:30.492099
| 2020-05-04T19:30:40 |
359363
|
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|
Stack Exchange
|
Max-norm projection of a Hermitian matrix onto the set of positive semidefinite matrices
For a given Hermitian matrix $A$ (i.e. complex matrix with $A_{ij}^{\ast}=A_{ji}$) find its max-norm projection onto the set of complex positive semi-definite matrices:
$$\Pi(A)=\mathrm{argmin}_{M\succeq0}\|A-M\|_{\infty}.$$
Here $\|A\|_{\infty}=\mathrm{max}_{ij}|A_{ij}|$ is the entry-wise max-norm. This problem has exact solution for the Frobenius and spectral matrix norm (see for example here). Is there a closed-form solution for the max-norm? Is there an efficient algorithm to calculate the projection and what is the computational cost (for $n\times n$ complex matrix)?
Partial answer regarding an algorithm to find projection, not a closed form.
The problem is of the following convex SDP:
$$
\min_{t,M} ~~~~~~~t\\
\mbox{subject to}\\
\hspace{5cm} |A_{i,j}-M_{i,j}|\leq t, ~\forall ~i,j\\
\hspace{3cm}M\succeq 0.
$$
This can be solved using CVXPY (or a software alike). You might want to check Boyd's Convex Optimization for the exact details of the computational complexity of interior point methods.
|
2025-03-21T14:48:30.492198
| 2020-05-04T19:38:45 |
359364
|
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|
Stack Exchange
|
intersection of left and right orthogonals of a module
$\DeclareMathOperator\Ext{Ext}$Let $(R, m)$ be a commutative Noetherian Gorenstein local ring and let $M$ be an $R$-module. Let
${^\perp M}$ and $M^\perp$ be respectively the left and the right orthogonal classes of $M$ defined by vanishing of the $\Ext$ functor; namely, $M^\perp$ consists of those $R$-modules $X$ such that $\Ext^1_R(M, X)=0$, and ${^\perp M}$ is defined dually.
Does anybody have particular information on the (homological) properties of ${^\perp M} \cap M^\perp$?
Any reference to relevant papers or books is highly appreciated even in case of particular assumptions on $M$.
|
2025-03-21T14:48:30.492277
| 2020-05-04T21:07:17 |
359368
|
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"Arrow",
"LSpice",
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|
Stack Exchange
|
Adjunctions capturing duality between ideals and saturated monoids in a commutative ring?
Let $R$ be a commutative ring. A saturated monoid in $R$ is a multiplicative submonoid $S\subset R$ which is closed under divisors, i.e $xy\in S\implies x\in S$. This is the converse of the analogous axiom for ideals $x\in I\implies xy\in I$. So saturated monoids and ideals have a sort of duality, part of which interchanges addition and multiplication, and part of which reverses the direction of implication.
The only elements we are given suggest a pair of operations between ideals and saturated monoids
$$\begin{gathered}\mathrm{Ideal}(R)\rightleftarrows\mathrm{SatMon}(R), \\ I\mapsto (1+I)_{\text{sat}}\text{ and } S \mapsto \left\langle S-1 \right\rangle. \end{gathered}$$
Taking the poset structures induced from the powerset of $R$, given by inclusions, both of the above operations are poset morphisms (take inclusions to inclusions). However, they are not adjoint functors:
$$\begin{gathered} \left\langle S-1 \right\rangle ⊂ I\iff S-1 \subset I\iff S\subset 1+I \\ S\supset(1+I)_\text{sat} \iff S\supset 1+I \end{gathered}$$
This is a strange situation of functors $\mathsf C\rightleftarrows \mathsf D$ say $F,G$ and yet bijections $$\mathsf D(FA,B)\cong \mathsf C(GB,A)= \mathsf C^\text{op}(A,GB).$$
Question 1. Do the operations above underlie some adjunction I'm missing?
If not, then:
Question 2. Is the formal duality between ideals and saturated monoids captured by some other adjunction?
If not, then:
Question 3. Is there more structure to the abstract setting I described above, with interesting category theory?
Added remark. Between prime ideals and prime saturated monoids (prime here means $0\notin S$ and $x+y\in S\implies x\in S\vee y\in S$), taking set-complements is actually a bijection. This perfect duality is really a consequence of the fact the added primeness assumptions make the definitions perfectly dual.
I certainly see how the duality between ideals and saturated monoids reverses the direction of implication, but in what sense does it interchange addition and multiplication?
An ideal is an additive subgroup, as opposed to a multiplicative submonoid.
I fail to see how the "strange situation" you describe actually occurs in your case: the sets $\mathrm{Ideal}(R)(\langle S-1 \rangle, I), \mathrm{SatMon}(R)((1+I)_{\mathrm{sat}}, S)$ will not be bijective for every pair $I, S$. If $I \subsetneq J$ are two ideals and $S=(1+J)_{\mathrm{sat}}$, then $\mathrm{SatMon}(R)((1+I)_{\mathrm{sat}}, S)=\{\subseteq\}$ while $\mathrm{Ideal}(R)(\langle S-1 \rangle, I)=\emptyset$ since $\langle S-1\rangle \supseteq J \supsetneq I.$
To describe what adjoint situations actually do occur here, I would suggest adding an intermediate step to make the situation a bit clearer:
Consider first rather $\mathrm{Ideal}(R)\overset{F}{\underset{G}\leftrightarrows}\mathrm{Mon}(R)$ where $F: S \mapsto \langle S-1 \rangle$ and $G: I \mapsto 1+I$. Then your first line of iffs shows that $F \dashv G$.
Now consider separately $\mathrm{Mon}(R)\overset{U}{\underset{\mathrm{sat}}\leftrightarrows}\mathrm{SatMon}(R)$ where $U$ is the forgetful functor. When you replace in your second line of iffs "$1+I$" by some general multiplicative monoid $S'$, the statement still holds. That is, given a monoid $S'$ and a saturated monoid $S,$ we have $S'_\mathrm{sat}\subseteq S$ iff $S' \subseteq S$. Thus, $\mathrm{sat}\dashv U$.
So the adjunction-like thing that you have can be described as the pair given by a composition of a left adjoint with a right adjoint, and the composition of their respective adjoints in the opposite order.
I don't think there are adjunctions like that because the variance of $F$ and $G$ goes the wrong way. The functors are both covariant, but the "strange situation" looks like an adjunction of contravariant functors. I think that's supposed to be the point of the question (but this is not made terribly clear).
@R.vanDobbendeBruyn I am not sure, however, that the "strange situation" as described actually occurs in the discussed case (does it? note that the two lines of "iffs" are not connected). So instead, I tried to answer question 1 by providing the adjunctions that do appear in the context. I edited the answer to make this clearer.
|
2025-03-21T14:48:30.492566
| 2020-05-04T22:50:43 |
359377
|
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|
Stack Exchange
|
Metrics on the space of distributions in terms of p.d.fs
If two probability distributions (on the same measure space) are s.t they have p.d.fs and the $L^1$ distance between the p.d.f.s is large, then is there a choice of a ``nice" metric $d_{\rm smooth}$ and a positive integer $p$ s.t the two distributions are far in the Wasserstein $p-$norm in $d_{\rm smooth}$?
If we use the trivial metric $d(x,y) = {\bf 1}(x \neq y)$ then the ${\rm W}_{1,d} = 0.5 {\rm TV}$ and TV is the $L^1$ distance between the p.d.fs when they exist. But suppose that I want to use a smoother metric than this.
Secondly if two probability distributions are such that their inverse CDFs are close (say in the sup norm) then are the two distributions close in some Wasserstein norm?
Note that one can create examples where the two inverse CDFs are close in the sup norm but their TV distance is large hence their $1-$Wasserstein norm in the trivial metric is large.
At least in the one-dimensional case the Lp distance between quantile functions happens to be exactely the Wasserstein-p metirc for the corresponding measures.
Any reference you could give for this?
It's a fairly standard result which you will find in every book on optimal transport, e.g. Santambrogio "Optimal Transport for Applied Mathematicians" just to name one.
Not really standard to beginners in the field :D Can you kindly point to the theorem number?
|
2025-03-21T14:48:30.492701
| 2020-05-04T23:03:48 |
359378
|
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|
Stack Exchange
|
Basis pursuit algorithms for exponentially large matrices?
Are there any efficient algorithms/heuristics for basis pursuit for exponentially large matrices?
That is
$$\begin{array}{ll} \underset{x \in \Bbb R^n}{\text{minimize}} & \lVert x \rVert_0\\ \text{subject to} & Ax = y\end{array}$$
where $A$ is an exponentially large matrix, i.e., $A \in \mathbb{R}^{m \times n}$ where $n = \Theta(2^m)$.
Is $\lVert x\rVert_0$ the number of non-$0$ entries of $x$? If so, then it's positive, so taking its argument seems pointless. Also, what is $y$?
(Also also, you somehow switched your delimiters, typing \rVert x\lVert instead of \lVert x\rVert. I have edited to correct.)
There is a minimal number of measurements required for uniform sparse recovery, i.e.:
$$
m \asymp s \log(n)
$$
where $m$ is the number of measurements (the number of rows of your matrix) and $n$ is the ambient dimension of the signal (the number of columns of the matrix).
Adding the condition $n \equiv 2^m$ simply gives the condition
$$
m \geq Csm,
$$
Or $s$ is simply a constant independent of the dimension of the matrix. I don't think that's a useful think to look at.
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2025-03-21T14:48:30.492810
| 2020-05-05T01:27:33 |
359383
|
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|
Stack Exchange
|
Analogous $H^1$-space for pseudo inner products
Perhaps this is a naive question but I could not find anything related to this.
Imagine we are on a bounded and regular open subset $\Omega$ of $\mathbb{R}^3_1$, i.e, $\mathbb{R}^4$ is considered with a Lorentzian inner product, let's say $ds^2 = -dx_1^2 + \sum_{i=1}^3dx_i^2.$
In this setting, is it possible to define analogously $W^{1,2}(\Omega)$ with inner product
$$-\int_{\Omega}(u')^2 + \int_{\Omega}|\nabla u|^2?$$
If it is possible, is there analogous Poincaré inequality and Kondrakov immersion theorem's?
What would be the point? This is not even a norm, as you can easily take any function $u(t,x)=u_1(t)+u_2(x)$ with $\int |u'|^2-\int |\nabla u|^2=0$ but of course $u\not\equiv cst$. I mean, sure, this particular difference can arise somewhere as just a particular functional (I don't have any example, though), but if you want to construct a "theory" based on this idea then it would lack all the proper analytic-functional properties, wouldn't il? (perhaps I misunderstood the question)
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2025-03-21T14:48:30.493038
| 2020-05-05T01:40:24 |
359384
|
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|
Stack Exchange
|
Spaces that are comparable with their compacts
This is an outgrowth of this question.
For a (metrizable) space $X$ consider the following increasingly strong properties:
(i) For every compact $K\subset X$ there is a map $f:X\to X$ such that $K\subset f(X)$ and $f(X)$ is relatively compact in $X$.
(ii) For every compact $K\subset X$ there is a map $f:X\to X$ which is the identity on $K$ and $f(X)$ is relatively compact in $X$.
(iii) For every compact $K\subset X$ there is a compact $L$ that contains $K$ and which is a retract of $X$ (this property was in the focus of the original question).
Obviously every compact space satisfies (iii).
As was pointed out in the comments to the linked question, among examples of spaces that does not satisfy (i) are so-called connected punctiform spaces, i.e. such that have no connected compact subsets other than singletons. If $X$ is such a space, and $K$ is a set of two points, then $\overline{f(X)}$ has to be connected, compact and contain these two points, which gives a contradiction.
Again, as was pointed out in the comments, connected punctiform spaces can be analytically as good as polish, so the problem has to have a geometric flavor. This is supported by the fact (as was pointed in the original question) that every closed convex subset of a locally convex space satisfies (iii). In my opinion the next step is the class of ANR's.
Does every ANR satisfy any of the properties (i), (ii) or (iii)? How about open sets in locally convex spaces?
I would like to present another example of a space that does not satisfy (i). Let $Y$ be a continuum constructed in Mackowiak - Singular arc-like continua. It has a lot of cool features, what is relevant here is that it is embeddible into a plane, hereditarily decomposable and if $L\subset Y$ is a subcontinuum, and $g:L\to Y$, then either $g(L)=L$ and $g$ is the identity, or $g(L)$ is a singleton.
Let $X=Y\backslash\{y\}$, for arbitrary $y$. Using the properties of $Y$ one can show that the only non-constant continuous map $f:X\to X$ is the identity. Now arguing as above, we can see that $X$ does not satisfy (ii). This example is valuable, because $Y$ satisfies (iii) being compact, and yet removing a single point destroys even (i). Moreover, since $Y$ is compact and embeddible into the plane, $X$ is embeddible into a punctured plane as a closed set. One can show that the latter satisfies (iii), while $X$ doesn't satisfy (i). Hence, (i)-(iii) is not stable with respect to taking closed or open subsets.
The question about ANR's seems to reduce to the question about locally finite simplicial complexes. Namely, is it true that any finite subcomplex $K$ of a locally finite simplicial complex $X$ is contained in a finite subcomplex $R$ of $X$ such that $R$ is a retract of $X$?
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2025-03-21T14:48:30.493251
| 2020-05-05T01:40:33 |
359385
|
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|
Stack Exchange
|
On the Absolute Value of the Riemann Zeta Function on the Critical Line
Is there any sort of (closed form preferably, though if not, it's fine) function for $|\zeta(\frac12+it)|$ where $\zeta$ is the Riemann zeta function? Anything is welcome, so I can take it from there. Upper bounds, lower bounds and asymptotics are also accepted.
@zz7948 I'm not sure how that helps.
@zz7948 Yeah, but that was for zeta, not for the absolute value of it.
It is not known whether, for $t>0$ large, $|\zeta(1/2+it)|$ is smaller than $t^{1/7}$, while it is conjectured that $1/7$ can be replaced by any positive exponent. This should answer your question. For more information, consult Titchmarsh's book on the Riemann zeta function, there is a whole chapter on this topic. Another good starting point is the relevant Wikipedia entry.
Are there any unconditional results?
@QuoteDave: There are hundreds of papers on the subject. As a starter, read Titchmarsh's chapter and the Wikipedia entry (in my post). See also the recent work of Soundararajan, Harper etc. on the arXiv. This is one the hot topics of analytic number theory.
Though when they say "$c_{k,j}$ as some constant", what kind of constants are they talking about?
@QuoteDave: That asymptotic formula is only conjectured. As the Wikipedia page says, a conjectured formula for $c_{k,j}$ was given by Keating & Snaith (2000). The best unconditional upper bounds appear in the table in the Wikipedia page. Please study the literature. It is vast and easy to find.
No, I am really only interested in $k=1$, which has been proven, and the Wikipedia page never gave any values of $c$.
@QuoteDave: That conjectured bound is a moment bound (which implies pointwise bounds though, see Titchmarsh's book). Pointwise bounds of the form $|\zeta(1/2+it)|<c(1+|t|)^{1/6}$ were proven with explicit constants $c>0$. I don't have time to look up this part of the literature, just use Google, MathSciNet etc. I gave you pointers. At any rate, it is a more substantial question how the $1/6$ can be reduced, so analytic number theorists are mainly occupied with that. The current record is by Fields medalist Bourgain from 2017 as recorded in the Wikipedia page. Discussion is closed on my part.
|
2025-03-21T14:48:30.493442
| 2020-05-05T02:00:53 |
359386
|
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|
Stack Exchange
|
Residual and continuous spectra of $L^2( G(k) \backslash G(\mathbb A) ; \omega)$, and cuspidal automorphic data
Let $G$ be a connected, reductive group over a number field $k$. Let $\mathbb A$ be the ring of adeles of $k$, $\omega$ be unitary character of $Z_G(\mathbb A)/Z_G(k)$, and $V = L^2(G(k) \backslash G(\mathbb A); \omega)$ the space of functions $f: G(\mathbb A) \rightarrow \mathbb C$, modulo functions vanishing almost everywhere, such that:
$f(\gamma zg) = \omega(z) f(g)$ for all $\gamma \in G(k), z \in Z_G(\mathbb A), g \in G(\mathbb A)$.
$\int\limits_{Z_G(\mathbb A) G(k) \backslash G(\mathbb A)} |f(g)|^2dg < \infty.$
Then $V$ is a Hilbert space with inner product $\langle f_1, f_2 \rangle = \int\limits_{Z_G(\mathbb A) G(k) \backslash G(\mathbb A)} f_1(g) \overline{f_2(g)} dg$, and $G(\mathbb A)$ acts continuously on $V$ by right translation.
Let $V_{\textrm{disc}}$ be the closure of the subspace linearly spanned by closed, irreducible $G(\mathbb A)$-invariant subspaces. A function $f \in V$ is called cuspidal if $$\int\limits_{N(k) \backslash N(\mathbb A)} f(ng)dn = 0$$ for all unipotent radicals $N$ of all proper parabolic subgroups of $G$ and almost all $g \in G(\mathbb A)$. The set $V_{\textrm{cusp}}$ of cuspidal functions in $V$ is a closed subspace of $V_{\textrm{disc}}$.
We let $V_{\textrm{res}}$ be the orthogonal complement of $V_{\textrm{cusp}}$ in $V_{\textrm{disc}}$, and $V_{\textrm{cont}}$ the orthogonal complement of $V_{\textrm{disc}}$ in $V$. Then $V$ decomposes into a direct sum of subrepresentations:
$$V = V_{\textrm{disc}} \oplus V_{\textrm{cont}} = V_{\textrm{cusp}} \oplus V_{\textrm{res}} \oplus V_{\textrm{cont}}. \tag{1}$$
I have the following questions:
$V_{\textrm{res}}$ is said to consist of "residues of Eisenstein series." What does this mean precisely? I understand that an Eisenstein series can be associated to a function in the induced space $\operatorname{Ind}_{P(\mathbb A)}^{G(\mathbb A)} \sigma$, where $P$ is a parabolic subgroup of $G$, and $\sigma$ is a cuspidal automorphic representation of $P/N(\mathbb A)$, where $N$ is the unipotent radical of $N$.
$V_{\textrm{cont}}$ is said to be "spanned by Eisenstein series." What does this mean exactly?
There is apparently an orthogonal direct sum decomposition of $V$:
$$V = \bigoplus\limits_{(P,\sigma)} V_{P,\sigma} \tag{2}$$
where $(P,\sigma)$ are equivalence classes of pairs, where $P$ is a parabolic subgroup of $G$, $M$ is a Levi subgroup of $P$, and $\sigma$ is a cuspidal representation of $M(\mathbb A)$. These equivalence classes are called "cuspidal automorphic data." If I understand correctly, the sum of the spaces $V_{P,\sigma}$ with $P = G$ constitute the space $V_{\textrm{cusp}}$. Whatever this decomposition means, is supposed to tell us that understanding all the cuspidal representations of $G$ and its Levi subgroups is the same as understanding $V$ itself.
What is the connection between the decompositions (1) and (2)? If cuspidal representations of Levi subgroups of $G$ show up in $V$ somehow, how do we determine whether they appear in $V_{\textrm{cont}}$ or in $V_{\textrm{res}}$?
What have you read? Do you understand the situation for GL(2)? If not, maybe you could start with looking at Gelbart-Jacquet's Corvallis article and, for instance, Gelbart's lectures on the trace formula.
Not as well as I should. I've been reading Gelbart's lectures
|
2025-03-21T14:48:30.493661
| 2020-05-05T02:35:15 |
359389
|
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|
Stack Exchange
|
Locally connectedness and accessibility in $\mathbb{C}$
Suppose $\Omega$ to be a bounded area in the complex plane $\mathbb{C}$ with a locally connected boundary $\partial\Omega$, then every point of $\partial\Omega$ is accessible from Ω.Here accessibility of a point $\omega_0\in\Omega$ means that there exists a continuous mapping $\alpha:[0,1)\to\Omega$ such that $ \lim\limits_{t\to1^-}\alpha(t)=\omega_0$.
Actually this proposition is part of the proof of the Caratheodory Theorem of comformal mapping which states: If $f:B_1(0)\to\Omega$ is a comformal mapping and $\Omega$ is a bounded simply connected area, then the following two are equivalent:
(1)$f$ can be extended to a continuous surjective map $\overline{B_1(0)}\to\overline{\Omega}$
(2)$\partial\Omega$ is locally connected
Of course, if we assume that the Caratheodory Theorem of comformal mapping has already been proven, then the proposition I asked above can also be solved by applying the Theorem. However, as I have said above, the proposition is used in the proof of Caratheodory Theorem of comformal mapping (more exactly, this proposition is used to prove surjectivity), so I am looking for a topological proof or any proof without using the Caratheodory Theorem. I have already searched for some resourses of the proof of the Caratheodory Theorem, but most of them either deal with the case that $\partial\Omega$ is a Jordan curve, or simply omit the proof that $f$ is a surjective map.
Any hint or help is highly appreciated.
This is Theorem 14.4. in Newman - elements of the topology of plane sets of points
I don't think the Theorem 14.4 in Newman book is talking about the same thing as I posted. It states that if $D$ is locally connected at $a$, where $a$ is a boundary point of $D$, then $a$ is accessible from $D$. However, in my post, I do not assume that $a\in D$, and it's not hard to give a counterexample if we replace the requirement that $\partial D$ is locally connected everywhere with it is locally connected at a point $a\in \partial D$ @erz
ok, i see what you mean, sorry for a wrong reference. Check out Rempe-Gillen - On prime ends and local connectivity, I think he discusses something along these lines there
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2025-03-21T14:48:30.493859
| 2020-05-05T02:42:13 |
359390
|
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"Anthony Quas",
"Anton Petrunin",
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"მამუკა ჯიბლაძე"
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|
Stack Exchange
|
Is it possible to connect every compact set?
Let $X$ be a "nice" space: metrizable, connected, locally path connected perhaps. Let $K\subset X$ be a compact set.
Is there a always a compact connected $L\subset X$ such that $K\subset L$?
This is true if we assume local compactness: cover $K$ with a finite number of connected relatively compact open sets, take their closure, and then join with arcs. However, without local compactness I don't know what to do.
Isn’t the long line a counterexample?
@AnthonyQuas The long line not metrizable, or are you referring to something else?
Sorry. Missed that part.
Choose a sequence $\varepsilon_n\to 0$ and a $\varepsilon_n$-net $N_n$ for each $n$.
Assume $N_0$ is a one-point set.
For each point in $x\in N_k$ choose a closest point in $y\in N_{k-1}$ and connect $x$ to $y$ by a curve.
Note that we can assume that diameter of the curve is at most $\delta_k$ for a fixed sequence $\delta_k\to 0$.
Consider the union $K'$ of all these curves with $K$; observe that $K'$ is compact and path connected.
Why is it possible to connect $x$ to $y$? Are you assuming path connectedness of $X$? I think the assumption is connected but only locally path connected?
@მამუკაჯიბლაძე X is assumed to be locally path connected (perhaps), likely one may do the same with a weaker assumption.
@მამუკაჯიბლაძე local path connected + connected implies path connected: the path components are open and disjoint, and therefore there is just one such path component
Do I understand correctly that the control of the diameters of the curves comes from the fact that any locally path connected metrizable space admits a metric such that every ball of diameter less than some fixed number is path connected? Or is there a simpler way?
@erz, yes, it can be done this way, but this theorem is hard (and I do not know its proof). Instead one may directly apply existence of arbitrary small path connected neighborhood.
I think understand how to do that, but I decided to present the theorem I referred to (which is not hard by the way), since I was going to post a complement to your answer with a more detailed explanation.
This is meant to fill in some of the details outlined by Anton Petrunin's answer, and also to refine the statement slightly. Recall that a compact connected Hausdorff space is called a continuum.
We will call a topological space $X$ continuum-wise connected if every $x,y\in X$ can be joined by a continuum, i.e. there is a continuum $K\subset X$ that contains both $x$ and $y$. We will call $X$ locally continuum-wise connected if for every $x\in X$ and open neighborhood $U$ of $x$ there is an open neighborhood $V$ of $x$ such that every $y\in V$ can be joined by a continuum within $U$. It is easy to see that continuum-components of locally continuum-connected space are open and disjoint, and so a connected locally continuum-connected is continuum-connected.
Proposition. A metrizable space $X$ is locally continuum-wise connected if and only if there is a metric $\rho$ on $X$ compatible with the topology and such that every open ball of radius less than $1$ is continuum-connected.
This is analogous to Theorem IV.7.1 in Newman - Elements of the topology of plane sets of points. There it is stated for (locally) connected metrizable spaces, but works also for any (locally) set-wise connected metrizable spaces, for an appropriate collection of connected sets (e.g. separable, bounded, arcs).
Proof. Sufficiency is clear. Let us prove necessity. Choose an arbitrary metric $d$ on $X$ bounded by $1$. For $x,y\in X$ declare $\rho(x,y)$ to be the infimum of diameters of the continuums that join $x$ and $y$ (if $x$ and $y$ are not joined by any continuum put $\rho(x,y)=1$). It is easy to see that $\rho$ is a metric, and moreover $d\le\rho$. Furthermore, if $x_n\to x$, since $X$ is locally continuum-wise connected, $x_n$ and $x$ can be joined by arbitrarily small continuums, and so $\rho(x_n,x)\to x$. Thus, $\rho$ is equivalent to $d$, and so is compatible with the topology of $X$.
It is left to show that every ball of radius less than $1$ is continuum-wise connected. Let $x\in X$ and let $R<1$. Assume that $y\in B_{\rho}(x,R)$, i.e. $\rho(x,y)=r<R<1$. By definition of $\rho$, there is a continuum $K$ with $d$-diameter at most $\frac{r+R}{2}$ that joins $x$ and $y$. Every point $z\in K$ is joined with $x$ by $K$, and so $\rho(x,z)=\frac{r+R}{2}<R$. Hence, $K\subset B_{\rho}(x,R)$, and so $y$ is joined by $x$ by a continuum in $B_{\rho}(x,R)$. $\square$
Corollary. A metrizable space $X$ is locally continuum-wise connected if and only if every point has a base of open continuum-wise connected neighborhoods.
Now, having these characterizations we can answer the original question.
Theorem. Let $X$ be a connected and locally continuum-wise connected metrizable space. Then for every compact $K\subset X$ there is a continuum $L\subset X$ that contains $K$.
Before proving the theorem, let us prove the following characterization of compactness.
Lemma Let $Y$ be a metric space for which there is a compact $K\subset Y$ such that for every $\varepsilon>0$ there is a compact $N$ such that $K$ is an $\varepsilon$-net of $Y\backslash N$. Then $Y$ is compact.
Proof. It is clear that $Y$ is completely bounded. We only need to prove completeness. Let $\{y_m\}\subset Y$ be a Cauchy sequence. It is enough to find a convergence subsequence. For every $k$ let $N_k$ be compact and such that $K$ is $\frac{1}{k}$-net for $Y\backslash N_k$. We may assume that $N_k\subset N_{k+1}$.
If an infinite subsequence of $\{y_m\}$ was contained in $N_k$, for some $k$, then there would be a convergent subsequence due to compactness of $N_k$. Hence, we can choose a subsequence $\{z_m\}$ such that $z_m\not\in N_m$. Since $K$ is an $\frac{1}{m}$-net for $Y\backslash N_m$, there is $x_m\in K$ with $\rho(x_m,z_m)<\frac{1}{m}$. Since there is a subsequence of $\{x_{m_k}\}$ that converge to $x\in K$, so does $\{z_{m_k}\}$. $\square$
Proof of the theorem. Using the proposition, we can metrize $X$ with a metric such that open balls of radius less than $1$ are continuum-wise connected.
For natural $n$, let $K_n\subset K$ be a finite $\frac{1}{2^n}$-net of $K$. For every $x\in K_{n+1}$ there is $y\in K_{n}$ such that $\rho(x,y)<\frac{1}{2^n}$. Since $B(y, \frac{1}{2^n})$ is continuum-wise connected, there is a continuum $L^n_{x}\subset B(y, \frac{1}{2^n})$. Then for any $m>n$ and $x\in K_m$ and $z\in L_x$ there $y\in K_{n}$ such that $\rho(z,y)<\frac{1}{2^{n-1}}$.
Let $z\in K$ and for $x\in K_1$ let $L^1_x$ be a continuum that joins $x$ with $z$.
Observe by induction that $M_n=\bigcup_{i\le n, x\in K_n} L_{x}^i$ is a continuum, and so $M= \bigcup M_k$ is connected. Since $M$ contains an $\frac{1}{2^n}$-net of $K$, for every $n$, it follows that $K\subset \overline{M}$. Hence, $M\subset M\cup K\subset \overline{M}$ from where $Y=M\cup K$ is connected.
Finally, since $K_n\subset K$ is a $\frac{1}{2^{n-1}}$-net for $K\cup \bigcup_{k>n} M_k\supset Y\backslash M_n$, for every $n$, $Y$ is compact due to the Lemma.$\square$
Remark. I also would like to present a nice example that bof gave in the comments (now deleted), that at least local connectedness is required: Consider the following modification of the topologist's sine curve $X=\{(t,\sin \frac{1}{t}), 0<t\le 1\}\cup\{(0,0\}$, which is connected and moreover is a polish space. However the compact set $\{(x,y)\in X, y=0\}$ cannot be connected by a continuum. Note that for a completely metrizable space local connectedness is equivalent to local path-connectedness.
The usual term you will find in the literature is "continuum-wise connected". I have also seen "semi-continuum" or "semicontinuum" to refer to such a space.
@D.S.Lipham thank you! now fixed
Here is another answer, based again on Anton Petrunin's idea, but obtaining a slightly different result.
Theorem. Let $X$ be a connected and locally path-connected completely metrizable space. Then for every compact $K\subset X$ there is a Peano continuum $L\subset X$ that contains $K$.
In order to prove this result we need a version of the proposition from my previous answer.
Proposition. A locally path-connected completely metrizable space $X$ supports a complete metric $\rho$ on $X$ compatible with the topology and such that every open ball of radius less than $1$ is path-connected.
Proof. Let $d$ be a complete metric on $X$ bounded by $1$. Apply the same construction as in my previous answer (but with paths instead of continuums) and obtain $\rho$. Since $\rho\ge d$ are equivalent, and the latter is complete it is easy to see that the former is also complete (a $\rho$-Cauchy sequence is a $d$-Cauchy sequence, hence it is $d$-convergence, and so $\rho$-convergent).$\square$
Proof of the Theorem. We will construct a convergent sequence of paths $\varphi_n:[0,1]\to X$ such that the image of $\gamma_n$ contains a $\frac{1}{2^n}$-net of $K$.
For natural $n$, let $K_n\subset X$ be a finite $\frac{1}{2^n}$-net of $K$. Using connectedness one can choose them so that $K_n\cap K_m=\varnothing$. Moreover, let $K_1=\{x_0,...,x_n\}$.
Let $\gamma_1:[0,1]\to X$ be a continuous path such that $\gamma_1|_{[\frac{2i}{2n+1},\frac{2i+1}{2n+1}]}\equiv x_i$, $i=0,...,n$ (on the intermediate segments $\gamma_1$ joins $x_i$ with $x_{i+1}$, which is possible since $X$ is path connected).
For $0\le a<b\le 1$ and $x,y\in X$ with $\rho(x,y)<r<1$ let $\gamma:[a,b]\to X$ be a a continuous loop such that $\gamma(a)=\gamma(b)=x$, $\gamma|_{[\frac{2a+b}{3},\frac{a+2b}{3}]}\equiv y$, and the image of $\gamma$ is contained in $B(x,r)$ (which is possible since open balls of radius less than $1$ are path connected).
Now assume that $\gamma_n$ is constructed so that its image contains $K_n$ and for every $x\in K_n$ there are $a<b$ such that $[c,d]\subset \gamma^{-1}_n(x)$. Let $x_1,...,x_m\in K_{n+1}$ be such that $\rho(x_i,x)<\frac{1}{2^n}$. Re-define $\gamma_n$ on $[c,d]$ to be a series of loops defined above from $x$ to $x_1$ and back, then from $x$ to $x_2$ and back, and so on.
Applying the same construction to all elements of $K_n$ (simultaneously) we get $\gamma_{n+1}$ such that for every $y\in K_{n+1}$ there are $a<b$ such that $[a,b]\subset \gamma^{-1}_{n+1}(y)$. Moreover, if $\gamma_{n+1}(t)\ne \gamma_n(t)$, it follows that $\gamma_{n+1}(t)\in B(\gamma_n(t),\frac{1}{2^n})$, from where $\rho(\gamma_{n+1},\gamma_n)\le \frac{1}{2^n}$.
Note that the image of $\gamma_{n+1}$ contains the image of $\gamma_n$. Moreover, from construction and the fact that $K_m$'s are disjoint that if $x\in K_n$, then $\gamma_{n+1}^{-1}(x)\ne \varnothing$ and if $t\in \gamma_{n+1}^{-1}(x)$, then $\gamma_m(t)=x$, for all $m>n$.
It follows that $\{\gamma_n\}$ is a Cauchy sequence of maps from $[0,1]$ into a complete space $X$. Consequently, it uniformly converges to a map $\gamma:[0,1]\to X$. From the previous paragraph it follows that the image of $\gamma$ contains every $K_n$, and since it is compact, it contains $\overline{\bigcup K_n}\supset K$. $\square$
|
2025-03-21T14:48:30.494578
| 2020-05-05T02:49:48 |
359391
|
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|
Stack Exchange
|
Injective continuous operators between Banach spaces
Suppose $X$ and $Y$ are two infinite dimensional Banach spaces. What can we say about the set of all injective continuous linear operators between $X$ and $Y$? Is it always nonempty?
What if $X$ is a non-separable Hilbert space, and $Y$ is a separable one?
Or more generally, if the Hamel dimension of $X$ is larger than that of $Y$. Then there will be no injective linear operators at all, never mind continuous.
The answers here show that your second question fails but I think that you have defined an interesting preorder on the family of Banach spaces: $X$ is dominated by $Y$ if such an injection exists. Call the corresponding equivalence relation “weak equivalence”. There is a number of easy pickings (stability properties, etc) that one can quickly deal with but also many interesting questions, in particular, on weak equivalence of specific spaces. I would start with: when is $C(K)$ dominated by (equivalent to) $C(L)$? There is a plethora of such questions for the classical Banach spaces.
Of course the dimension is an obvious obstacle, but even if the space have the same cardinality of Hamel bases the answer is no. For example in the paper
A. Avilés, P. Koszmider,
A Banach space in which every injective operator is surjective.
Bull. Lond. Math. Soc. 45 (2013), no. 5, 1065–1074
the authors constructed an infinitely dimensional Banach space $X$ such that if $T:X\to X$ is bounded and injective, then $T(X)=X$. Therefore if $Y$ is a subspace of $X$, then one cannot find an injective operator $T:X\to Y$.
Piotr Hajłasz' answer nails the problem, however, let me point out that there are easier examples of such pairs of spaces among spaces that have the same density.
Suppose that $X$ fails to have a strictly convex renorming. Thus, there is no injective operator $T$ from $X$ into any space $Y$ that is strictly convex, as if it were $\|x\|^\prime = \|x\| + \|Tx\|$ would be a strictly convex norm on $X$.
Spaces that do not have a strictly convex norm include
$X = \ell_\infty^c(\Gamma)$, the space of all bounded scalar-valued functions on an uncountable set $\Gamma$ that have at most countable support (Day);
$X = \ell_\infty / c_0$ (Bourgain).
In the latter case, you may even take $Y= \ell_\infty$.
Is it absolute that there is an example where the density character of both spaces is $\aleph_1$? Are there examples where both spaces are reflexive and have the same density character?
@BillJohnson, for reflexive it is even easier: take $X = \ell_p(\Gamma)$ and $Y=\ell_q(\Gamma)$ for $q>p$. By Pitt's theorem, every operator from $X$ to $Y$ is compact, hence it has separable range. But an injection will be a homeomorphic embedding of the weakly compact unit balls; a contradiction.
Duh, now. :) (Actually, I was thinking about not having an injective operator in either direction when I asked that stupid question.)
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2025-03-21T14:48:30.494828
| 2020-05-05T03:33:46 |
359393
|
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|
Stack Exchange
|
Central limit theorem versus entropy in dynamical systems context
A dynamical system $(S^1,T, \mu)$, $T_* \mu=\mu$, $T$ ergodic, $S^1$ is circle. Assume it has central limit theorem.
Want to know the relation between its measure-theoretic entropy $h_{\mu}(T)$ and the central limit theorem? Is there a good reference for it? Thanks!
For the measure-theoretic entropy $h_{\mu}(T)$, see https://en.wikipedia.org/wiki/Measure-preserving_dynamical_system#Measure-theoretic_entropy
For the central limit theorem, means: for any smooth observable $\phi$ on $S^1$ with $\int \phi d\mu=0$, $\frac{\sum_{i \le n}\phi \circ T^i}{\sqrt{n}} \stackrel{d}{\longrightarrow} N(0,\sigma^2_{\phi})$ holds where $\sigma^2_{\phi} \ge 0$ is a constant.
As I see it the normal distribution arises when minimizing the differential entropy subject to a given expectation and variance, when we don't know the expectation and variance what we minimize is the relative/cross entropy (KL divergence) and we get a normal distribution with 0 expectation estimator if unbiased and variance estimator given by the Fisher information...
...then in CLT S converges to $N(\mu, \sigma)$ while the mle converges to $N(0,I(\mu,\sigma)^{-1})$ so I think some kind of similar play should be play between the measure entropy and the normal here
If the CLT as you stated it holds for $\phi$, it cannot hold for $2\phi$, right?
@Kostya I edited my question.
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2025-03-21T14:48:30.494968
| 2020-05-05T04:54:33 |
359397
|
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|
Stack Exchange
|
A graded version of the Koethe conjecture
The most straightforward graded version of the Koethe conjecture would seem to say that the result of summing of two graded-nil ideals produces a graded-nil ideal. Here, graded-nil means having all homogeneous elements nilpotent. The case of rings graded by the natural numbers interests me most, but I'd like to learn whatever I can.
Does this occur in the literature? I haven't spotted it myself in papers and books on radicals of graded rings, but among other things, I could have easily missed an equivalent formulation.
Does the original Koethe conjecture imply this more general version (which trivially contains the original by giving every element degree 0)?
Or is this more general version actually known to be false?
|
2025-03-21T14:48:30.495166
| 2020-05-05T06:08:37 |
359399
|
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|
Stack Exchange
|
Need help to check a quote from Hecke's "Lectures on Dirichlet Series, Modular Functions and Quadratic Forms"
I'd like to check the accuracy of a reference to Hecke's 1938 "Lectures on Dirichlet Series, Modular Functions and Quadratic Forms" implicit in eq. 4.8 on p. 50 of J. G. Leo's dissertation, which is here:
http://halfaya.org/ucla/research/thesis.pdf
The Hecke theorem in question describes an analogue of the normalized weight 4 Eisenstein series for the full modular group that is itself modular with weight 4 under the transformations of the $m^{th}$ Hecke group $G(2\cos (\pi/m))$. (I'll typeset Leo's rather intricate eq.4.8 if necessary, but I'm hoping to avoid that, since people can find it with a few mouse-clicks.) Leo seems to say that his equation 4.8 is an immediate consequence of something in Hecke's book, but doesn't give page numbers, and the book is hard to find under the present no-libraries circumstances. I want to implement Leo's eq. 4.8 in a computer algebra system; right now my code is not generating plausible output. So, while I debug, I think it's sensible to rule out a typographical error or other mistake in Leo's quote. If someone with access to Hecke's book can reply with a yes or a no as to Leo's accuracy, it would help me, and if that person could provide me with the relevant page numbers in Hecke's book, it would help me even more, because I could then look for a copy of those few pages legally, under fair-use.Thank you.
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2025-03-21T14:48:30.495331
| 2020-05-05T06:32:14 |
359401
|
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|
Stack Exchange
|
Can learning Riemann surfaces be more beneficial than numerical analysis for an analyst?
I am in master program of mathematics, specialized in PDE and numerical analysis. Now I am trying to decide which classes to take for next semester. Of course I want to become an expert in my field, but I am also interested in Geometry. I learned some differential geometry in my bachelor class, and I want to take the next tier, which is Riemann surfaces in my university program. But I am not very sure this will be the right choice, since if I don't take this, I will have some time to take more numerical analysis courses which directly connect to my area. But in same time I feel like I can self-study some numerical methods in the future. So now, the question is:
Can taking a Riemann surface course be more beneficial (which is a very vague concept) over taking more numerical analysis course?
If I want to do research in my future career (i.e. proceed to PhD) will there be any topics in analysis related to Riemann surfaces?
Analysis is a very broad topic... If you want to have anything to do with functions of complex variables, then knowing the bases of Riemann surfaces is a must.
I recall several speakers at the last ICM in Brazil saying that their research in applied mathematics drew ideas from the study of the dynamics of geodesic flow on Riemann surfaces, i.e. constant curvature surfaces. Perhaps not the usual material in a course on Riemann surfaces, but related by the close connection between dynamics and the Laplacian operator.
@BenMcKay Would you remember any names of those people ? Just curious to see applications of the topic
The question as titled is trivial. (If you're ever wondering whether it will be beneficial to learn any new piece of math, then the answer is 'yes'.) The more focussed questions in the body are much better. Maybe re-phrase the title as "How can learning Riemann surfaces …"?
@LSpice I agree, I tried to rephrase.
The topic of Riemann surfaces is beautiful, and connected to so many things. They can be looked at from so many different perspectives. Do not get fooled by how easy they may seem. Some of the deepest questions in Mathematics have to do with them (the Riemann hypothesis for one). I know this does not answer your question, but I wanted to write it as a comment.
@Malkoun Thanks, maybe I just needed some motivations.
I should add that they are also the easiest examples of moduli spaces. Many times, you are looking at the set of all solutions, whether algebraic or analytic, of some equations (algebraic or differential). Often the general solution depends on some parameters. In case, there is only one complex parameter, and the general solution depends holomorphically on it, then the natural habitat for such a parameter is a Riemann surface.
@Malkoun Thank you so much for the detailed comment. Can you recommend any introductory book with regarding the topic you stated?
It is a tough question to answer in general. It depends on one's taste and background. I would for instance recommend Otto Forster's book "Lectures on Riemann surfaces". It is a good one.
Riemann surfaces connect to many areas of geometry, dynamics, analysis and physics.
My wife used to study "Minimal Surfaces". Is there a connection between "Minimal Surfaces" and "Riemann Surfaces" ?
Not really, although one can have Riemann's minimal surface and Riemannian surfaces :). You can think of a minimal surface as something which always wants to minimize its surface area at every point (like a soap bubble). A Riemann surface is a one-dimensional complex manifold, but complex manifolds are closer to objects called algebraic varieties than they are to the objects which you think of when you say 'surface', hence many of the applications in algebraic geometry.
As another application, in string theory (if you are interested in learning a bit more about that) the quantum field theory which one studies lives on a Riemann surface.
You might enjoy the edited volume Computational Approach to Riemann Surfaces, Eds Bobenko & Klein, Springer 2011. I've left this as a comment instead of an answer, as it may not address the question in the title. That said, given your interests in PDEs and numerical analysis, it could be worth taking a look at. If you'd like me to convert this comment into an answer, let me know.
Painleve - It came to appear that, between two truths of the real domain, the easiest path quite often passes through the complex domain.
Hadamard - It has been written that the shortest and best way between two truths of the real domain often passes through the imaginary one."
Read first "The Magic Wand Theorem of A. Eskin and M. Mirzakhani" by Anton Zorich for motivation on Riemann sufaces.
And, "A Singular Mathematical Promenade" by Etienne Ghys (particularly p. 87-93 and glance at the Wikipedia article on monodromy).
More prosaically:
To understand integral transform solutions (Mellin, Laplace, Fourier) to pdes, you need to understand poles and branch cuts of complex functions.
Solutions to Laplace's equation in two dimensions are called harmonic functions that are the real and complex components of a complex function and give mutually orthogonal contour lines on the complex plane and Riemann sphere.
From "THE THEOREM OF RIEMANN-ROCH AND ABEL’s THEOREM" by Siu:
The theorem of Riemann-Roch and Abel’s theorem could be interpreted as answering the question: for which configuration of charges, dipoles, or multipoles on a compact Riemann surface of genus ≥ 1 would the flux functions (whose level curves are the flux lines and which are the harmonic conjugates of the electrostatic potential functions) in the case of the theorem of Riemann-Roch, or their exponentiation after multiplication by 2πi in the case of Abel’s theorem, be single-valued on the Riemann surface so that the flux lines are closed curves?
At a more basic level for numerical analysis, in understanding convergence of real power series and, therefore, series solns. to pdes, you need to understand singularities (poles and branch cuts in the complex domain) and these involve Riemann surfaces.
Same for Newton (finite difference) and sinc function (Nyquist-Shannon) interpolations of sequences of real/complex numbers and their numerical analytic continuations and for asymptotic series a la Poincare. (Norlund, Poincare, and Berry wrote well on these topics.)
Helps in understanding convolutions, Dirac Delta functions and their derivatives, and, therefore, fractional calculus and operational calculus.
Necessary in understanding basic string theories.
The list is endless. Without such knowledge, you live (perhaps blissfully) in Abbott's Flatland.
The examples really suggest that you may be imposing a gratuitous, restrictive dichotomy--there is plenty of synergy between the study of numerical analysis and Riemann surfaces and both provide paths to other intriguing areas of the grand, evolving tapestry of mathematics, engineering, and science. (Of course, if you are looking where the money is in America, well I suggest a medical degree or starting a munitions factory.)
Electromagnetics involves solving Maxwell's equations (pdes), of course.
Haha, I liked you mentioned the Flatland. Of course I understand the importance of the Riemann surfaces in PDEs, but what hesitates me is 'I don't know taking a course is beneficial compare to take other analysis course'. I guess you are saying knowing the Riemann surfaces is (almost) mendatory, but do you think I have to take a course? No way of self-study when I needed?
@JingeonAn Most math professors, engineers, and scientists are specialists and go through the motions when teaching. Take courses under the rare motivated, conscientious generalist and/or study good textbooks and articles on your own and use Q&A sites and MathCad to check your analysis symbolically or numerically.
Thank you for the great answer!
Examples: The two papers by McMullen "Advanced Complex Analysis" and "Riemann surfaces, dynamics, and geometry" and the books "Visual complex analysis" by Needham and "On Riemann's Theory of Algebraic Functions and Their Integrals" by Klein.
For basic numerical analysis, I'd guess understanding the role of Riemann surfaces in analytic continuation of finite difference and sinc (cardinal series) interpolation and summability of series and in understanding asymptotic series is a good start at least. Norlund, Poincare, and Berry wrote well on these topics.
Thank you so much. Do you have any book recommendation for self-study on the subject?
I no longer have Klein's book, but because I have a background in mathematical physics and he was one of the last great mathemages (all keenly interested in physics), it's worth looking at. Klein very much promoted visual insight--see comments in my response to https://mathoverflow.net/questions/32479/what-are-some-mathematical-sculptures/252321#252321
Check out the refs I've already given. Also https://math.stackexchange.com/questions/262677/what-was-klein-working-on-when-he-replaces-his-riemann-surface-by-a-metallic-su and https://mathoverflow.net/questions/313254/references-for-riemann-surfaces
@Copeland Thank you so much!
Speak of the devil. Klein and Riemann surfaces are disucussed in the recent AMS article "Higgs Bundles—Recent Applications" by Laura Schaposnik (https://arxiv.org/abs/1909.10543).
and https://www.ams.org/journals/notices/202005/rnoti-p625.pdf
See also "Divergent series: taming the tails" by Berry and Howls for a glimpse at the interplay of numerical analysis, pdes, and Riemann sheets.
In support of my claim of interconnectedness, see my answer and the Papadopoulos and Elizalde refs in the comments in https://mathoverflow.net/questions/127601/does-the-derivative-of-log-have-a-dirac-delta-term/364456#364456. (Riemann was a master at numerical computations as well as an amazing theoretician.)
For superb visualizations and conceptual analysis starting from square one, see "Exploring Visualization Methods for Complex Variables" by Andrew J. Hanson and Ji-Ping Sha.
Nice motivating forward and mix of math and history in "Uniformization of Riemann Surfaces:
Revisiting a hundred-year-old theorem" by Henri Paul de Saint-Gervais, et al.
Let me answer the question in the title. The answer is definitely yes.
Just to mention an important topic, the proof of Uniformization Theorem for Riemann surfaces requires to construct at least one holomorphic or meromorphic form with prescribed singularies. All known proofs use some Analysis, and none of them is simple.
In fact, you will be led to study deep properties of elliptic operators on the surface (aka "Hodge Theory"), and this will surely boost your analysis skills.
This is very clear and motivating. Thank you!
|
2025-03-21T14:48:30.496164
| 2020-05-05T09:04:03 |
359409
|
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|
Stack Exchange
|
Can a polytope with vertex-transitive edge graph or face lattice be made vertex-transitive?
Let $P\subset\Bbb R^d$ be a convex, full-dimensional polytope (convex hull of finitely many points, affine hull is the whole space), $G_P$ its edge graph and $\mathcal F_P$ its face lattice. Any of these is called vertex-transitive if the respective symmetry group (isometries, graph isomorphisms or lattice isomorphisms) acts tansitively on its vertices.
Are there any known results on the following questions?
Questions:
If $G_P$ is vertex-transitive, is then $\mathcal F_P$ also vertex-transitive?
If $G_P$ is vertex-transitive, is there a vertex-transitive polytope $Q$ with $G_P$ as edge graph, maybe in the same dimension $d$, maybe even combinatorially equivalent to $P$ (i.e. with isomorphic face lattice)?
If $\mathcal F_P$ is vertex-transitive, is there a vertex-transitive polytope $Q$ with $\mathcal F_P$ as its face-lattice?
The answer is always No if we not already know that $G_P$ or $\mathcal F_P$ belong to an actual polytope, or if we replace vertex-transitivity by a weaker form of symmetry (e.g. central symmetry).
On the other hand, for $d\le 3$ all answers are Yes.
2-neighborly polytopes (https://en.wikipedia.org/wiki/Neighborly_polytope) have vertex-transitive edge graphs, but $\mathcal{F}_P$ is rarely vertex transitive.
@RichardStanley Are there any easily accessible examples for this? As far as I know the cyclic polytopes can be realized vertex-transitively.
When $d$ is odd and the number of vertices is at least $d+3$, then no realization of the cyclic polytope is vertex transitive. See Theorem 8.3 of http://www.math.uni-magdeburg.de/~kaibel/ALT/Downloads/cyclaut.pdf. Shemer (https://link.springer.com/article/10.1007/BF02761235) showed in 1982 that there are super-exponentially many neighborly polytopes (improved by Padrol in https://arxiv.org/abs/1202.2810). I suspect that most of them are not vertex transitive.
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2025-03-21T14:48:30.496342
| 2020-05-05T10:18:59 |
359414
|
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Stack Exchange
|
A question about dense sets
Suppose that $A$ is a given subset of $I=[0,1],\ $ and
$ \left\{ x_j = \frac{j}{m} \right\}_{j=0}^{m}\ $ is the $m$-partition of $I$, and $\nu(m)$ is the number of
$\ [x_{i-1},x_{i}]\ $ such that $\ [x_{i-1},x_{i}]\cap A \neq \emptyset\ $ for intervals $\ [x_{i-1},x_{i}]\ $ belonging to the $m$-partition. Also assume that $M>0$ is such that
$$ \forall_{m>M}\quad \nu(m) \ge \ln(m) $$
Does it follow that $A$ is dense on some open non-empty subinterval of $I$?
Remark (editorial by wlod): the rest of the post is unclear to editor wlod. Once the things get clear, this remark should be removed. ### This remark is here to avoid a prolonged (if momentary) discussion in comments. ### Also, Wlod followed by Andreas have provided counter-examples, and -- to make it better -- they need not the complication of constant $M$. Wlod stated the result for general functions $f$ in place of $\ f:=\ln$. Then Andreas had it for a wider class of functions $f$ (but Wlod's proof handles it too).
An example:
Let $A = \{\frac{1}{j}\}_{j=1}^{\infty}$. By some calculation we can get the value of $n$:
$$1,7,23,70,211,649,$$
when $m$ is $1,10,100,1000,10000,100000$,respectively.
Notes: I want to emphasize that you cannot construct $A$ with $m$ related to the precondition, because $m$ is given and changing to $\infty$ after $A$ has been supposed. Even if you want to construct $A$ with some variables, please notice that when $A$ is constructed, your variables are fixed and cannot follow the change of $m$.
What if you construct the set $A$ as follows? For each $m$, pick just enough points as close to 0 as possible so that the desired inequality holds for $m$. Then I think for every interval $(a,b)$, where $a>0$, $A \cap (a,b)$ is finite.
This is basically what the answer below does.
@MonroeEskew. $m$ is given and changes after $A$ has been constructed. You cannot do anything with the constructed $A$ while $m$ changes. It's like you pick up a real number first, and you can find certain real numbers greater than the number you pick up. But you cannot find certain real numbers greater than an arbitrary real number.
$A$ is a countably infinite set. I construct it once.
@MonroeEskew Please give me the $\sup$ of $A$ you have constructed.
Can OP provide a formal formulation, using $\ \forall\ $ and $\ \exists\ $, to make the question clearer to some of us (to me)?
I'll construct an $A$ that satisfies the conditions of the question but is not dense in any interval. I'll use the notation in the question and abbreviate the $i$-th interval of the $m$-th partition as $J(m,i)$. There are two requirements that I need to satisfy:
(1) For every $m$, $A$ contains points from at least $\ln m$ of the $m$ intervals $J(m,i)$.
(2) For all $m$ and all $i$, $J(m,i)$ has a nontrivial subinterval $K(m,i)$ disjoint from $A$.
Note that requirement (2) ensures that $A$ is not dense in any nontrivial interval, because any such interval includes $J(m,i)$ for some $m$ and some $i$.
I'll build $A$ by an inductive process. At step number $m$, I'll choose some finitely many points to put into $A$ in order to satisfy (1) for the current $m$, and I'll choose the $m$ subintervals $K(m,i)$ to satisfy (2) for the current $m$ and all $i\leq m$. Of course I need to make sure that the points I put into $A$ for the sake of (1) are not in any previously chosen excluded interval $K$ and that the intervals $K$ that I exclude for the sake of (2) don't contain any of the points previously put into $A$.
Fix, for the rest of the construction, a sequence of positive real numbers $r(m)$ small enough so that $\sum_{m=1}^\infty mr(m)<\frac12$. The subintervals $K(m,i)$ chosen at step $m$ will each have length $\leq r(m)$.
Now here's how to do step $m$. The previous steps have put (altogether) finitely many points into $A$. So I can choose, inside each of the $m$ intervals $J(m,i)$, a subinterval containing none of the points already put into $A$. Shrinking those subintervals if necessary, I can arrange that each has length $\leq r(m)$. Take those shrunken subintervals as $K(m,i)$.
What is the total length of all the intervals $K(q,i)$ chosen so far, i.e., with $q\leq m$ and $i\leq q$? Well, at step $q$, we chose $q$ intervals of length at most $r(q)$, so the total length is at most $\sum_{q=1}^m q\cdot r(q)$, which is less than $\frac12$ by our choice of the $r$ sequence. So those intervals, though they might intersect all of the current $J(m,i)$'s, cannot completely cover more than $\frac m2$ of them. In each of the $J(m,i)$ that are not completely covered, which is at least $\frac m2$ of them, we can choose a point and put it into $A$. Do so; this satisfies requirement (1) with room to spare ($\lceil\frac m2\rceil$ instead of $\ln m$). This completes the construction of the counterexample.
Remark: The "room to spare" that I pointed out at the end of the construction can be amplified. For any fixed $\epsilon$, we can get $A$ to meet all but $\epsilon m$ of the intervals $J(m,i)$ in the $m$-th partition. Just choose the $r(m)$'s a little smaller.1
At the beginning of your answer, could you reformulate the OP's question in a formal way (using nothing but $\ \forall, $ and $\ \exists\ $, etc. ? I don't find the present formulation of the OP's question clear.
@WlodAA I understood the problem to be proving or refuting the statement: If a subset $A$ of the unit interval has the property that, for each $m$, at least $\ln m$ of the $m$ intervals $[\frac{i-1}m,\frac im]$ contain points from $A$, then $A$ must be dense in some nontrivial subinterval.
Isn't my simple answer providing a counter-example?
Thus, say, $\ J(m\ i):=\left[\frac{i-1}m;\frac im\right]$ ?
And, say, $\ \emptyset\ne K(m\ i):=(a_{m\ i};b_{m\ i})\subset J(m\ i)\ $ for certain $\ a_{m\ i}\ $ and $\ b_{m\ i}\ $ (where $\ a_{m\ i}<b_{m\ i})\ $) ?
@WlodAA I have changed the way to state my question by your suggestion. Now the question statement is using the form similar to $\varepsilon$-$\delta$ of the definition of limit.
@WlodAA I think I misunderstood your answer originally. It now looks correct to me (assuming, of course, that the question means what I think it means). So the only advantage in my more complicated answer is that it works even for $f(m)=(1-\varepsilon)m$ rather than $o(m)$.
@AndreasBlass, $\ (1-\epsilon)\cdot m\ $ is nice! (However, my proof works also for this tougher case too).
Let the number of the $m$-th partition intervals which intersect $\ A\ $ be $\ \ge f(m).\ $ Then
Theorem There exists $\ A\ $ which is not dense in any non-trivial interval whenever
$$ \lim_{m=\infty} \frac {f(m)}m\ =\ 0 $$
Proof The required $\ A\ $ can be given as follows:
$$ A\ :=\\
\ \left\{ \frac {2\cdot k-1}{2\cdot m}:
\ m\in\Bbb N\ \ \text{and}
\ \ k\in\Bbb Z\ \ \text{and}\ \ 0<k\le f(m)\,\right\} $$
End of Proof
By “it is possible” you mean that a counterexample exists, whereas the answer asked “is it possible to prove”. Could you edit so as to remove this confusion? (Generally speaking, it's best to avoid “yes” or “or” or variants thereupon.)
Have you noted that $A$ has been supposed at the beginning of the question? This indicates $A$ does not depend on variables given behind, like $m$ and $n$.
@Gro-Tsen, I hope that I followed your observation and suggestion.
If you want to unrelate your $m$ to the $m$ in the precondition, you have to take the limit for the $m$ in your theorem. But this will render your $A={0}$.
@WlodAA Let's get this straight. We can get $\sup A = \frac{2f(m)-1}{2m}$ from your constructed $A$. If the limit is taken, $\sup A = 0$ so that $A = {0}$ or $A=\emptyset$.
OP, please make the formulation of your Question straight. Your comments here, under my question, are not straight at all. If you still stand by them then, please, clarify them in a straightforward manner - please, make things explicit.
@Watheophy, have you noted that given function $\ f\ $ (e.g. $\ f:=\log),\ $ my $\ A\ $ is constant, it doesn't depend on any variables like $\ n\ $ or $\ m$?
Let us continue this discussion in chat.
|
2025-03-21T14:48:30.497255
| 2020-05-05T10:20:59 |
359415
|
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|
Stack Exchange
|
Characteristic function of comonotone gammas
Suppose we have two random variables $X$ and $Y$ both distributed as gamma random variables with parameters $\alpha_X,\beta_{X},\alpha_{Y},\beta_Y$, with characteristic functions given by:
$$\phi_{X}(t) = \left(1 - \frac{it}{\beta_X}\right)^{-\alpha_X} \;\;\text{ and }\;\;\phi_{Y}(t) = \left(1 - \frac{it}{\beta_Y}\right)^{-\alpha_Y}$$
Furthermore, suppose that $X$ and $Y$ are comonotone : the joint c.d.f is then given, via the upper frechet-hoffding bound, by :
$$F_{X,Y}(x,y) = \min(F_X(x),F_Y(y))$$
Note that this random vector has no density.
Can we obtain a formulation for the joint characteristic function ? Same question for more than $2$ comonotone gammas, say $X_i$ with parameters $\alpha_i,\beta_i$ for $i \in \{1,...,n\}$.
If you are asking for a closed-form expression, this is quite impossible, at least with the current state of the art. Indeed, for any cdf $F$ and its generalized inverse defined by
$$F^{-1}(u):=\inf\{x\in\mathbb R\colon F(x)\ge u\}
=\min\{x\in\mathbb R\colon F(x)\ge u\}$$
for $u\in(0,1)$, we have
$$x\ge F^{-1}(u)\iff F(x)\ge u$$
for all $x\in\mathbb R$ and $u\in(0,1)$.
Letting now $F:=F_X$ and $G:=F_Y$, we see that the joint distribution of comonotone $X$ and $Y$ with cdf's $F$ and $G$ is the same as the joint distribution of $F^{-1}(U)$ and $G^{-1}(U)$, where $U$ is a random variable uniformly distributed on the interval $(0,1)$.
Therefore, the joint characteristic function (c.f.) $f$ of $X$ and $Y$ is given by
$$f(s,t)=\int_0^1\exp\{i(s F^{-1}(u)+t G^{-1}(u))\}\,du$$
for real $s,t$.
This integral cannot be taken in closed form when $X$ and $Y$ have arbitrary gamma distributions.
This cannot be done even in the simplest case when $\alpha_X=2$, $\beta_X=1$, $\alpha_Y=1$, $\beta_Y=1$. Indeed, then $F(x)=1-(x+1)e^{-x}$ and $G(x)=1-e^{-x}$ for $x>0$, whence for real $s,t$
$$f(s,t)=\int_0^\infty\exp\{i(s x+t G^{-1}(F(x)))\}\,dF(x) \\
=\int_0^\infty \exp\{i(s+t)x-x\}\,\frac{x\,dx}{(x+1)^{it}}.$$
Mathematica cannot do anything with this integral, returning just a trivially identical expression:
Thanks for all the work; Except you made a mistake at the end : the $i$ is in factor of everything, not just (s+t)x, so you obtain an exponent of $i(s+t-1)x$ if i'm right. Can mathematica solve it with this correction?
@lrnv : I think everything is correct as written. $-x$ in the exponent comes from $dF(x)=e^{-x},dx$.
Ho yeah you are right. Here comes down all my hopes ;( Thanks anyway !
@lrnv : Oops! I did confuse $F$ and $G$ at some point ($G$ must be the simpler one, as compared to $F$, to make $G^{-1}$ explicit). This is now fixed, and the integral has become only slightly more complicated, with the integrand now containing the extra factor $x$.
Hahaha it got worse :)
|
2025-03-21T14:48:30.497485
| 2020-05-05T10:30:39 |
359416
|
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|
Stack Exchange
|
Game on a square grid
Not research level, comments are welcome.
Consider the following game:
The board is the vertices of an $n$ by $n$ square grid.
Two players take moves in turns.
A move is picking two vertices and drawing a straight line between them.
If the line intersects another line or passes through a third vertex,
the game ends and the player who made the move loses the game. Two or more lines are allowed to end at the same vertex.
Is there winning strategy depending on $n$?
Partial result:
We believe if we take the board to be the vertices of regular polygon,
the first player always wins, even if they don't have any skills
except finding a non-losing move if it exists.
On a hexagon, the first player can win, but the strategy is non-trivial as they must make sure an odd number of interior edges will be drawn. If the second player manages to draw lines from 1 to 3 and from 4 to 6, they'll win instead.
@Glorfindel Confusion is possible, but we believe in a $n$ regular polygon there are $n-3$ non-intersecting diagonals and $n$ sides.
Ah, I missed a long diagonal from 1 to 4 (or 3 to 6) is still possible in this case. However, for higher $n$ it still might be non-trivial.
Doesn't a symmetry argument make this question rather simple?
If $n$ is odd, the second player (blue) can always mirror the first player's (red) moves (reflected through, or rotated by $\pi$ radians around the origin of the grid, i.e. its center vertex), so the first player loses.
If $n$ is even, the first player can draw one of the diagonals of the central square (orange; this move can't be mirrored) and can then mirror all of the second player's moves, so the second player loses.
Many thanks! Are there boards for which this game is hard? This is not rigorous, but what if the board is random $n$ points in the plane?
Perhaps it depends on the number of edges and of possible edge crossings? (Note that unlike the square grid, you can draw a line between all points if they're randomly distributed; the square grid excludes some because there's another vertex in the middle.)
your "reflection through the origin" is actually rotation by $\pi$ radians, right? So your argument would work for any board which is preserved by such a rotation.
@NickGill that's correct.
For 3 by 3 or larger odd board do you have a game where the first player wins when the second player plays weak, but doesn't make obviously bad moves? We have conjecture that this impossible.
It could very well be the nature of the game that it doesn't matter which moves you make (and hence there are no 'weak' moves, and not much of a game either :-)).
I think not all moves allow your symmetry. What if some player takes the main diagonal? The symmetry move will be crossing.
@joro You can't take the entire main diagonal because of the rule if the line intersects another line or passes through a third vertex .... Taking one segment will allow the other player to copy that move (unless it's the middle segment, but that's the opening move in the n is even case).
FYI I asked about generalization of the game: https://mathoverflow.net/questions/414574/game-on-a-square-gripd-part-ii
|
2025-03-21T14:48:30.497907
| 2020-05-05T10:40:36 |
359419
|
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"Liviu Nicolaescu",
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|
Stack Exchange
|
When is the Morse equivalence local?
Let $f:X \to \mathbb{R}$ be a Morse function on some compact submanifold $X \subset \mathbb{R}^n$, and assume that $p \in X$ is not a critical point of $f$. For some $\epsilon > 0$ let $D_\epsilon(p)$ denote the Euclidean disk of radius $\epsilon$ around $p$. I'd like to claim that there are some small $\epsilon > 0$ and $\delta > 0$ so that we have a deformation-retraction (or at least a homotopy equivalence)
$$ \rho:D_\epsilon(p) \cap \{f(p) \leq f \leq f(p)+\delta\} \stackrel{\simeq}{\longrightarrow} D_\epsilon(p) \cap {\{f = f(p)\}}.$$
Let's call the codomain $A$ and the domain $B$. My questions are: (a) can we impose conditions on $f$ which make $\rho: B \to A$ exist, (b) and is there a precise reference for this?
The Hope: The intuition is simply that even if the interval $[f(p),f(p)+\delta]$ contains a billion critical values of $f$, so long as none of the critical points involved are in our $\epsilon$-ball, the handle attachments will be far away from $p$ and therefore the gradient vector field of $f$ should take the $(f(p)+\delta)$-sublevelset to the $f(p)$-sublevelset without any serious incidents en route.
The Problem: Of course, there is no reason for the gradients $-\nabla f$ to point into $D_\epsilon(p)$ along the bounding upper hemisphere $$H^+ = \partial D_\epsilon(p) \cap (B-A),$$ which means that the gradient flow might be pushing points outside $D_\epsilon(p)$ laterally into $B$ rather than flowing down to $A$. I suspect that the following should suffice: if for every point $x$ in $H^+$, the gradient $-\nabla_xf$ does not lie in the tangent space $T_xH^+$, then the desired map $\rho:B \to A$ is furnished by flowing along $-\nabla f$. I could certainly try to write all of this down, but it seems like overkill and it's hard to believe that it hasn't been done before (one might expect to see it in Nicolaescu's nice Morse theory textbook for instance).
If $p_0$ is not a critical point of $f$ then the implicit function theorem states that, there exists local coordinates $(x^1,\dotsc, x^n)$, defined in an open neighborhood $U$ of $p_0$ in $\newcommand{\bR}{\mathbb{R}}$ $\bR^n$ such that, in these coordinates we have ($m=\dim X$)
$$
x^i(p_0)=0,\;\;\forall i,
$$
$$
X=\{ x^{n-m+1}=\cdots =x^n=0\},
$$
$$
f(x^1,\dotsc, x^n)=f(0,\dotsc,0)+x^m.
$$
If you now define the (non-Euclidean) box $\newcommand{\ve}{{\varepsilon}}$
$$
B=\big\{ |x^i|< \ve;\;\;i=1,\dotsc, m\big\}.
$$
In this neighborhood, that is not an Euclidean ball, the deformation you seek is obvious.
To deal with the region $D_{\ve}(p)$ consider the smooth function $\DeclareMathOperator{\Hess}{Hess}$
$$
g:X\to\bR,\;\;g(x)=\Vert x-p\Vert^2,
$$
where $\Vert-\Vert$ is the standard Euclidean norm on $\bR^n$. The Hessian of $g$ at $p$, viewed a symmetric bilinear form $T_pX\times T_pX\to\bR$ is positive definite. $\newcommand{\pa}{\partial}$
Choose local coordinates $(x^1,\dotsc, x^m)$ on $X$ near $p$ as above. In these coordinates the vector field $\pa_{x^m}$ is a gradient like vector field for $f$.
For $\ve>0$ sufficiently small the region $R_{\ve}=\{g\leq \ve\}$ is strictly convex in the above coordinates $x^i$ because the second fundamental forms along the boundary $\pa R_{\ve}$ are positive definite being small perturbations of $\Hess$.
This reduces the problem to the following situation. Suppose that $R_\ve$ is a compact, convex neighborhood of the origin in $\bR^m$ with smooth boundary. Then for $\delta>0$ sufficiently small we have a deformation retract
$$
R_\ve \cap \{ 0\leq x^m\leq \delta\}\to R_{\ve}\cap\{x^m=0\}.
$$
I think this is clear.
Thank you for this answer (and your book, from which I learned Morse theory!) One question: I don't see how you get $$f(x^1,\ldots,x^n) = \text{stuff}$$ when $f$ is only defined on $X$ consisting of the first $m$ coordinates. Should that $n$ be an $m$, or are you using an extension? And similarly, I'm not sure what the $b$ is in the index set of your box.
You are right. Use an extension of $f$. Indeed, $b$ was a typo. I meant $m$ so $B$ is a box living inside $X$. If you are interested specifically in the intersection of $X$ with a Euclidean ball then things get a bit more technical. See the new addition to my answer.
Ah, wonderful. Thank you for the update, this is precisely what I was hoping for!
I don't see how you could exclude that the gradient lies in the tangent space near the equator.
Seems to me that you need to use the diffeomorphism provided by the Morse lemma applied to $f$ to straighten the level set of $f$ into a hyperplane locally. After that, work in the "straightened domain" and apply the Morse lemma to the distance to (the image of) $p$ to "straighten" the distorted ball into a round ball. Of course some care is needed to avoid distorting the hyperplane. Composing the two diffeomorphisms you're in a situation where an orthogonal projection gives the desired retraction.
It seems that an extension of the Morse Lemma to vector valued functions should be feasible and would solve this more directly but I don't know references on this.
EDIT: the Morse lemma for vector functions approach is probably too restrictive, as discussed in
Modification of Morse lemma with two functions
The first approach above probably can be made to work though.
The space $B-A$ excludes the equator, and in general I'd expect the limiting gradient line to actually lie on the tangent at the equator (even in a hyperplane with parallel downward flow). This is fine as long as off the equator we don't have any points of tangency.
yes but close to the equator higher order terms would in general make a tangency possible I think
|
2025-03-21T14:48:30.498383
| 2020-05-05T11:51:52 |
359426
|
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|
Stack Exchange
|
Intersection of subgroup of a free group with the lower central series
If I have a subgroup $S$ of a free group $\mathcal{F}_m$, what can I say about the behaviour of the descending sequence of subgroups
$\left< S, \Gamma_c(\mathcal{F}_m) \right>$ (where $\Gamma_c(\mathcal{F}_m)$ is the $c$-th term of the lower central series)?
In particular, does this sequence always converge to $S$ or not?
(Below is what I've tried so far)
If after some point $S$ contains $\Gamma_c$, then this is obviously true, but that's only the case if the corresponding f.g. group $\mathcal{F}_m/S$ is nilpotent, so there's still a lot to go. If after some point $\Gamma_c \cap S =\{1\}$ then it's also true, but that would mean $S$ is nilpotent which is impossible since $S$ is free as a subgroup of a free group.
I've tried to use the diamond isomorphism theorem to use that $S \Gamma_c/S \cong \Gamma_c/S \cap \Gamma_c$, but it doesn't seem clear to me that my sequence converging to $S$ has much of a connection with $S \Gamma_c/S$ converging to the trivial group - indeed it's not entirely clear what "converging" means in the context of these quotients.
Thanks a lot!
(By the way: an answer only in the case where S is finitely generated is fine for my purposes)
In general you can't say much. For example, if $m \ge 2$ and $S \lhd \mathcal{F}_m$ is such that $\mathcal{F}_m/S \cong A_5$, then $\langle S,\Gamma_c \rangle = \mathcal{F}_m$ for all $c$.
Terms of the lower central series define a basis of neighborhoods of $1$, which gives rise to a pro-nilpotent topology on $\mathcal{F}_m$. Your sequence of subgroups converges to $S$ iff $S$ is closed in this topology.
Thanks a lot for your very complete answer!
|
2025-03-21T14:48:30.498544
| 2020-05-05T11:55:30 |
359427
|
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|
Stack Exchange
|
weak convergence of positive currents vs. $L^1$ convergence of normalized potentials
I have run into the following statement in the literature (e.g. here, p.5, after Theorem 1.1): that weak convergence of positive $(1,1)$-currents on a complex manifold is equivalent to $L^1$ (I presume $L^1_{loc}$ in the non-compact case) convergence of their $dd^c$-potentials which are normialized (I presume, having a fixed mean). I tried to search the book of Demailly for it but couldn't quite find such a statement. What is the precise statement and how can it be proved?
Let $X$ be a compact complex manifold and let $T_n=\theta+dd^c \varphi_n$ be a sequence of positive currents where $\theta$ is a fixed smooth $(1,1)$-form on $X$. Assume that $\varphi_n$ are normalized such that $\sup_X \varphi_n=0$ (one could have an analogous statement choosing the normalization $\int_X \varphi_n dV=0$ for some fixed volume form $dV$ on $X$).
If $\varphi_n$ converges weakly to $\varphi$ in $L^1$, then we get immediately $dd^c \varphi_n \to dd^c \varphi$ in the sense of currents, hence $T_n \to \theta+dd^c \varphi$.
Conversely, assume that there exists a positive current $T:=\theta+dd^c \varphi$ such that $T_n \to T$ and $\sup_X \varphi=0$. Given that the functions $\varphi_n$ are $\theta$-psh and sup-normalized, the set $\{\varphi_n,n\in \mathbb N\}$ is precompact in $L^1$. Let $\psi=\lim \varphi_{\sigma(n)}$ be any cluster value. It satisfies $dd^c \psi=\lim dd^c \varphi_{\sigma(n)}=\lim T_{\sigma(n)}-\theta=T-\theta=dd^c \varphi$. Hence $\psi-\varphi$ is pluriharmonic on the compact manifold $X$ and it is constant. By the normalization choice, $\psi=\varphi$. Therefore, $(\varphi_n)$ converges to $\varphi$.
Side remark. I assumed that $\theta$ does not depend on $n$ for simplicity. One can adapt the statements to a slightly more general setting if necessary.
why is the set of $\varphi_n$ precomact in $L^1$?
you can look e.g. at Demailly's book, Chapter I, Proposition 4.21 for the local version of the statement. For the global version, the observation is that there exists $C>0$ depending only on $X$ and $\theta$ such that for any $\theta$-psh function $\varphi$ on $X$ and $dV$ s.t. $\left| \sup_X \varphi - \int_X \varphi dV \right| \le C$. And this, in turn, follows from the existence of the Green kernel.
I see; thanks very much, this is the ingredient I was not aware of!
|
2025-03-21T14:48:30.498772
| 2020-05-05T13:49:04 |
359437
|
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|
Stack Exchange
|
Relationship between fans and root data
A (split) reductive linear algebraic group is equivalently described by combinatorial information called a root datum.
A toric variety is described by combinatorial information called a fan.
Both correspondences use the character lattice.
The reference:
http://u.cs.biu.ac.il/~margolis/Linear%20Algebraic%20Monoids/Renner-%20Lin.%20Alg.%20Monoids.pdf
says that spherical varieties are a nice class of objects that include
all my favorite spaces (e.g. symmetric spaces, toric varieties) . And, moreover, that a spherical variety is equivalent to combinatorial information called a colored fan. Is there any way of recovering a root datum from a colored fan? Or is a reductive group actually given as part of the data of a colored fan?
Are fans/ Toric varieties and root data/ reductive groups both special cases of a larger pattern (for example, colored fans/ spherical varieties)?
Isn't there a pretty big difference between these two in that the algebraic groups people study from root datum are affine varieties, whereas the toric varieties are usually projective varieties?
Yes. I agree it’s a long shot. Also, Sam I did lumina with you back in the day. My older sister is Lauren Teixeira; she was in Trojan Women with you when you were Poseidon. Hi I haven’t talked to you in a decade
Oh my god, I recognized the last name but didn't think that connection was possible. Way off topic, but I think saw an interview your sister did with a Chinese stunt drinker, and it was incredible.
Haha ya that guy is wild but his wife made him stop the stunt drinking because she was worried about his health so now he just makes motivational videos for his fans, he’s actually a totally sweetheart. And ya the one thing that was giving me hope about affine and projective things somehow having the same classification is the example of symmetric spaces where the spaces of compact and noncompact type are as different as possible but have the same classification by duality
Title of the reference, mostly recoverable from the URL but just in case of link rot: Rex - Linear algebraic monoids.
(1) A (connected) reductive group $G$ over an algebraically closed field $k$ is described by a combinatorial object called the based root datum ${\rm BRD}(G)$.
(2) A spherical homogeneous space $Y=G/H$ is a homogeneous space on which a Borel subgroup $B$ of $G$ acts with an open Zariski-dense orbit. It is described (uniquely at least in characteristic 0) by its homogeneous combinatorial invariants. These combinatorial invariants constitute an additional structure on ${\rm BRD}(G)$.
(3) A spherical embedding $G/H\hookrightarrow Y^e$ is a normal $G$-variety $Y^e$ containing a spherical homogeneous space $G/H$ as an open dense $G$-orbit. It is described by its colored fan, which is an additional structure on the homogeneous combinatorial invariants.
By spherical varieties one means spherical homogeneous spaces and spherical embeddings.
Therefore, I think that the based root datum of $G$ should be regarded as a part of data describing the $G$-variety $Y^e$.
In the case when $G=T$ is a torus, we take $H=1$, and then the spherical embeddings of $G/H=T$ are the same as the toric varieties for $T$, and the corresponding colored fans are just fans.
Reference: Nicolas Perrin, On the geometry of spherical varieties.
Not an answer, but: you can construct a fan from a root system. Let $R$ be a root system in an Euclidean space, and let $\Lambda_R$ be the root lattice with dual lattice $\Lambda_R^\vee$. The fan $\Sigma$ in $\Lambda_R^\vee$ associated to $R$ consists of the Weyl chambers of $R$ and all their faces. For instance, if $R=A_1$, then the associated toric variety is $\mathbf{P}^1$. I don't know how to determine when a fan comes from a root system, but I'm guessing someone here does.
These are called "permutohedral varieties," I believe.
Well in Type A they are called permutohedral varieties, at least; in other types maybe they would be called $W$-permutohedral varieties.
|
2025-03-21T14:48:30.499101
| 2020-05-05T14:26:06 |
359442
|
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|
Stack Exchange
|
What conditions are sufficient for two points to be independent in the Mordell-Weil group?
Consider a finite field $\mathbb{F}_{q}$ and an elliptic surface
$$
\mathcal{E}\!: y^2 + a_1(t)xy + a_3(t)y = x^3 + a_2(t)x^2 + a_4(t)x + a_6(t),
$$
where $a_i(t) \in \mathbb{F}_{q}[t]$. I am mainly interested in the case when $\mathcal{E}$ is a rational surface and $\mathrm{char}(\mathbb{F}_q) > 3$. Also, let
$$
P_0 = \big( x_0(t), y_0(t) \big),\qquad P_1 = \big( x_1(t), y_1(t) \big)
$$
be two sections of $\mathcal{E}$ of infinite order in the Mordell-Weil group $\mathrm{MW}(\mathcal{E})$. What conditions are sufficient for $P_0$, $P_1$ to be independent in $\mathrm{MW}(\mathcal{E})$?
If we had one element and wanted to detect wether it was torsion, we could specialize to various t, compute the orders and if for some set of specializations, the orders turned out to be coprime, then we know the original point was non torsion. For two points, we can quotient out by one point and test wether the other point is torsion. I don't know how practical this is.
Nonzero height pairing determinant is one. Otherwise if you had to calculate a Selmer group anyway on your way to determine the rank, then you could be lucky that they are independent in there. Just like over number fields, I don't hink specialisations help.
To add to Chris's answer, the height pairing reduces to intersection theory in the elliptic surface world. If you understand the geometry of the special fibers well, this should not be difficult at all to compute for a rational surface, and the determinant is an if-and-only-if condition so once you compute it you are done.
Could you please clarify your comment. Points $P_0, P_1$ are independent if and only if $\mathrm{det} = 1$?
det non-zero. The intersection pairing is a naive height which has to be modified slightly to get the canonical height which yields a non-degenerated bilinear form. magma has it all implemented : https://magma.maths.usyd.edu.au/magma/handbook/elliptic_curves_over_function_fields
Thank you very much for Magma link!
|
2025-03-21T14:48:30.499298
| 2020-05-05T15:00:25 |
359447
|
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|
Stack Exchange
|
what is the number of paths returning to 0 on the hexagonal lattice
I am looking for an estimation of the number of paths of length $n$ going from 0 to 0 on the hexagonal (or honeycomb) lattice.
I can find plenty on references on self avoiding paths, but I am looking into every paths. Is it considerably more difficult? Has anyone a reference?
It should be easier: you are just looking for a diagonal entry on a power of a certain infinite adjacency matrix.
For any finite graph $G$, if $A$ is its adjacency matrix, then $(A^n)_{ij}$ counts the number of walks from $i$ to $j$ of length $n$. So take the adjacency matrix for the lattice up to, say, $n/2$, and compute its powers.
This is OEIS A002898 (http://oeis.org/A002898).
This is answered by Ian Agol here, with the reference "All Roads Lead to Rome-Even in the Honeycomb World", Brani Vidakovic, Amer. Statist. 48 (1994) no. 3, 234-236.
An exact formula is
$$ p(n) = \sum_{k=0}^m \binom{2k}{k} \binom{m}{k}^2$$
if $n= 2m$ is even, and $0$ otherwise.
This is sequence A002893 on OEIS.
According to OEIS, the number of paths is asymptotic to
$$ p(n) \sim \frac{1}{2\pi n} 3^{n + 3/2}$$
when $n$ is even,
which agrees with the estimate given by shurtados.
In the above reference, Vidakovic proves that $p(n) \geq C \cdot {3^n}/{n}$ for some constant $C$.
If you want a rough answer, it is something of the order of $\frac{3^n}{n}$. This random paths are easier than self avoiding walks, you can think of these paths in this way: If you consider the even steps, these paths describe a random walk in a triangular lattice, which is a bit easier to describe. Each step $X_i$ is given by adding a sixth root of unity $\rho^{j}$, $j= 1,2,\dots, 6$. And we want to understand when $S_n = X_1 + X_2 + \dots X_n$ is equal to zero. After $n$ steps what you have $S_n = A_n 1 + B_n\rho + C_n \rho^2 = (A_n - C_n)1 + (B_n + C_n)\rho$ (here I'm using the fact that $\rho^2 = \rho -1$). You want to estimate the probability that $A_n - C_n = 0$ and $B_n + C_n = 0$.
I think the heuristic is that $A_n, B_n, C_n$ behave like standard random walks in the line (This is also true for $A_n - C_n$, and $B_n + C_n$) and so the probability of $A_n - C_n = 0$ or that $B_n - C_n = 0$ is of the order of $\frac{1}{\sqrt{n}}$, if these events were independent this gives you $(\frac{1}{\sqrt{n}})^2 = \frac{1}{n}$.
Observe also that this answer is the same as in the square lattice, and it seems this estimate should hold for more general tilings. See also Random walk on a Penrose tiling.
|
2025-03-21T14:48:30.499509
| 2020-05-05T15:41:28 |
359450
|
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|
Stack Exchange
|
Images of measurable function
My question is as follows. Consider an $L^\infty$ function $f:\mathbb{R}^n\times\mathbb{R}^n\to\mathbb{R}^n$ such that, for almost all $y$, the function $f({\cdot}, y)$ is continuous.
$\DeclareMathOperator\conv{conv}$Let $\conv D$ stand for the closed convex hull of a set $D$ and let $B_r(x)\subset\mathbb{R}^n$ be the open ball
with radius $r>0$ and with center at $x$.
I want to write
$$\bigcap_{r,r'>0}\bigcap_N\conv f(B_r(x_0),B_{r'}(y_0)\setminus N)=\bigcap_{r'>0}\bigcap_N\conv f(x_0,B_{r'}(y_0)\backslash N).$$
Here $\bigcap_N$ stands for the intersection over all measure-$0$ sets $N\subset\mathbb{R}^n$.
I feel that there must be some uniform continuity conditions on $f$ for this equality to be valid but I do not understand what to do.
This is stated in a way that makes it look like a homework question. Is it one? If so, then it doesn't belong here.
How does it arise? Do you know that it is true in any particular, non-convex case?
Oleg: the software on this website seems to think you are two different people. Can you comment using the same computer or browser window as you used to write the question?
|
2025-03-21T14:48:30.499624
| 2020-05-05T15:56:12 |
359454
|
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"url": "https://mathoverflow.net/questions/359454"
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|
Stack Exchange
|
Preprojective algebra of finite dimensional algebras
The preprojective algebra of a module $M$ over a finite dimensional algebra $A$ is
defined as $P_M:= \bigoplus\limits_{n=0}^{\infty}{Hom_A(M, \tau^{-n}(M))}$ with the canonical multiplication.
Question 1: Is there an easy way to obtain quiver and relations of $P_M$ in case it is finite dimensional? Are explicit quiver and relations known when $M=A$ and $A$ is an acyclic Nakayama algebra?
Question 2: Can $P_M$ be obtained in qpa in some way in case it is finite dimensional, at least in some special cases like Nakayama algebras?
For Question 2: In QPA one can do the following using the latest uploaded extensions of QPA:
gap> A := NakayamaAlgebra( GF(2), [ 3, 2, 1 ] );
<GF(2)[<quiver with 3 vertices and 2 arrows>]>
gap> M := DirectSumOfQPAModules(IndecProjectiveModules(A));
<[ 1, 2, 3 ]>
gap> B := PreprojectiveAlgebra( M, 3 );;
gap> C := B[1];
<GF(2)[<quiver with 3 vertices and 4 arrows>]/
<two-sided ideal in <GF(2)[<quiver with 3 vertices and 4 arrows>]>, (5 generators)>>
gap> Display( AdjacencyMatrixOfQuiver( QuiverOfPathAlgebra( C ) ) );
[ [ 0, 1, 0 ],
[ 1, 0, 1 ],
[ 0, 1, 0 ] ]
The command PreprojectiveAlgebra computes the preprojective algebra of the module $M$ if it is finite dimensional and degree $n$ is zero and it is given as a quotient of a path algebra.
The QPA-team.
Thank you very much. So degree $n$ means that $\tau^{-n}(M)=0$ or does it mean that $Hom_A(M,\tau^{-n}(M)=0$? Also I noted that I put $\tau^n$ instead of $\tau^{-n}$ in the question, I will edit that.
It means $\operatorname{Hom}_A(M, \tau^{-n}(M)) = 0$. Should be zero from degree $n$ on.
|
2025-03-21T14:48:30.499751
| 2020-05-05T16:08:23 |
359456
|
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|
Stack Exchange
|
Convergence of a spectral sequence of a double complex
In Weibel's book, a spectral sequence $E^r_{p,q}$ is said to weakly converge to a graded object $H_{\ast}$ if for every $n$ there exists a filtration $\dots \subset F_{r}H_{n} \subset F_{r-1}H_{\ast} \subset \dots H_{n}$ such that $E^{\infty}_{p,q} \simeq \text{gr}_{p}H_{p+q}$. Moreover, a filtration is called exhaustive if $\cup_r F_{r}H_n = H_n$, and it is called Hausdorff if $\cap_r F_r H_n = 0 $. A spectral sequence which weakly converges and that is both exhaustive and Hausdorff is said to approach $H_{\ast}$.
Later on in the book the filtrations for a double complex are introduced, and the two spectral sequences that arise from these filtrations are studied. For example, the filtration by columns of a double complex is defined as
$$
(\tau^{\leq n}C_{\ast,\ast})_{p,q} =
\left\{
\begin{array}{lr}
C_{p,q} & p \leq n\\
0 & p > n
\end{array}
\right.
$$
If we consider the filtration by columns of a double complex $C_{p,q}$ which is zero in the fourth quadrant, we get a spectral sequence that weakly converges to $H_{\ast}(\text{Tot}^{\tiny\prod}C_{\ast, \ast})$. What I don't understand is why the spectral sequence doesn't approach $H_{\ast}(\text{Tot}^{\tiny\prod}C_{\ast,\ast})$. In the book it is said that the filtration on the total complex is exhaustive, but to me it seems it is also Hausdorff. Am I getting something wrong?
I think that even if it is not true in full generality (strange categories might be badly behaved), it is nevertheless true for categories of modules or categories of sheaves, where we have an explicit construction for direct sums and direct products.
The issue here, I believe, is the failure of inverse limits to be exact. As a result, even if the filtration on the total complex is Hausdorff, the filtration on homology might not be. This is discussed at length in Boardman's "Conditionally convergent spectral sequences."
Yeah, searching among old questions I found an answer that said exactly the same thing, thank you.
And that implies it doesn't work even for modules or sheaves of modules, right?
Just to give a short example: A homology class lies in $F_rH_$, if it has a represetative in $F_rC_$ Now look at the complex of vector spaces over $\mathbb{F}2$ given by $0\rightarrow F[t]\rightarrow F[t]\rightarrow 0$ ($F[t]$ denotes the polynomial ring) and let the differential be given by multiplication with $1+t$. This complex has a filtration given by $F{-r}C_* =t^rC_$. Now the Kokernel of the differential is just an 1-dimensional vector space, and the nontrivial class is represented given by $[1]=[t]=[t^2]=\ldots$ and hence that class lives in $\bigcap_r F_rH_$.
|
2025-03-21T14:48:30.499967
| 2020-05-05T16:28:05 |
359460
|
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|
Stack Exchange
|
Does this series, related to the Hasse/Ser series for $\zeta(s)$, converge for all $s \in \mathbb{C}$?
I have asked this question at math stack exchange, however it did not get any traction. Still curious about the answer though.
Numerical evidence suggests that:
$$\lim_{N \to +\infty} \sum_{n=1}^N\frac{1}{n}\sum_{k=0}^n(-1)^k\ {n \choose k}\ \frac{1}{(k+1)^{s}}=s$$
or equivalently
$$\lim_{N \to +\infty} H\left( N \right)+\sum _{n=1}^{N} \left( {
\frac {1}{n}\sum _{k=1}^{n}{\left( -1 \right) ^{k}{n\choose k}\frac { 1
}{ \left( k+1 \right) ^{s}}}} \right) = s$$
with $H(N)$ = the $N$-th Harmonic Number.
Convergence is quite slow, but clearly goes faster for negative $s$. Also, the computations for non-integer values of $s$ require high accuracy settings (I have used Maple, pari/gp and ARB).
However, according to Mathematica the series diverges by the "Harmonic series test", although when taking $s$ as an integer, it agrees on convergence.
Does this series converge for $s \in \mathbb{C}$ ?
Some numerical results below:
s=0.5
0.497702121, N = 100
0.499804053, N = 1000
0.499905919, N = 2000
s=-3.1415926535897932385
-3.14160222, N = 100
-3.14159284, N = 1000
-3.14159272, N = 2000
s=2.3-2.1i
2.45310498 - 1.94063637i, N = 100
2.33501943 - 2.09308517i, N = 1000
2.31996958 - 2.09923503i, N = 2000
We have to show that
$$l(s):=\sum_{n=1}^\infty\frac1n\,S_n(s)=s,$$
where
$$S_n(s):=\sum_{k=0}^n(-1)^k\binom nk\frac1{(k+1)^s} \\
=\sum_{k=0}^n(-1)^k\binom nk\int_0^\infty du\,u^{s-1}e^{-(k+1)u}/\Gamma(s) \\
=\int_0^\infty du\,u^{s-1}e^{-u}\sum_{k=0}^n(-e^{-u})^k\binom nk/\Gamma(s) \\
=\int_0^\infty du\,u^{s-1}e^{-u}(1-e^{-u})^n/\Gamma(s).$$
So,
$$l(s)=\int_0^\infty du\,u^{s-1}e^{-u}\sum_{n=1}^\infty\frac1n\,(1-e^{-u})^n/\Gamma(s) \\
=\int_0^\infty du\,u^s e^{-u}/\Gamma(s)
=\Gamma(s+1)/\Gamma(s)
=s,$$
as desired.
very nice! So, the Mathematica divergence test wasn't correct after all.
|
2025-03-21T14:48:30.500228
| 2020-05-05T16:47:51 |
359463
|
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|
Stack Exchange
|
Bifurcation points in parametric Hammerstein Integral equation
I am a theoretical physicist working on integrable systems, hence I preemptively ask forgiveness for the eventual lack of mathematical rigor.
My question concerns the properties of a particular example of a Hammerstein equation depending non-trivially on a parameter. The specific property I am interested in is the existence of multiple solution branches and bifurcation points. In order to be as clear as possible and establish what I already know, I am going to present some background information, define the branching points and some elementary property and then present the specific case I am dealing with. Then I will pose a few questions that I would like to have an answer to. Also any reference to relevant literature is very much welcome.
Background
In the course of my research, I often have to deal with non-linear integral equations of the following type:
\begin{equation} \varepsilon(\theta) = r\cosh\theta - \intop_{\mathbb R} d\theta' \Phi\left(\theta - \theta'\right)\log\left[1+e^{-\varepsilon(\theta')}\right] \;. \tag{1}\end{equation}
In integrable systems literature an equation like (1) is known as Thermodynamic Bethe Ansatz (TBA) equation. The usual way to deal with such equations is to solve them numerically, often by Neumann series method. While this approach is generally efficient and yields sufficient informations, it is sometime desirable to have some degree of analytic control on certain features of TBA equations. In my case the relevant feature is the existence of bifurcation points in the parametric dependence of the solution(s) $\varepsilon^\ast(\theta)$ on the parameter $r$. I will be more concrete below; what I wish to stress here is that the existence of such a point is of physical relevance and any analytic information on it would be precious. This is also due to the fact that in the vicinity of said bifurcation points, usual numerical routines suffer from heavy slowing down, loss of accuracy and overall non-convergence: any analytical foothold would be precious. Similarly, any information concerning the existence and position of bifurcation points, obtainable directly from the properties of the kernel $\Phi$ is of huge importance as it allows to avoid a case by case study and makes contact with the general properties of the underlying physical system -- which are reflected in the kernel $\Phi$.
Bifurcation points
Going through some references on integral equations, I realized that equation (1) is an example of a parametric Hammerstein equation on the real line, whose general form is
\begin{equation}
u(x) = \intop_{\mathbb R} dt K(x,t)\Psi\left(t,u(t);\lambda\right)\;,\qquad x,t\in\mathbb R\;,\quad \lambda\in\mathbb C\;. \tag{2}
\end{equation}
As is the case whenever non-linearity is involved, the equation (2) might display some peculiar behaviour, one of which is the existence of bifurcation points.
Bifurcation point (adapted from [1])
A value $\lambda_c$ is a bifurcation point for the equation (2) if $\forall \epsilon>0\;,\; \exists \lambda\in\left(\lambda_c-\epsilon,\lambda_c+\epsilon\right)$ such that there exist two solutions $u(x) = u(x;\lambda)$ and $\tilde{u}(x) = \tilde{u}(x;\lambda)$ satisfying $\left\vert u(x) - \tilde{u}(x)\right\vert<\epsilon$.
Here $\lambda_c\in\mathbb R$, but it should be easy to generalize to the complex plane. In [1], pag. 836, I found these two nice little theorems
Theorem 1
Let $\lambda_c$ be a bifurcation point and $u(x) = u(x;\lambda)$ a solution of (2). Then $\nu = 1$ is a characteristic value of the kernel $L(x,t;\lambda_\ast) = K(x,t)\frac{d}{dy}\Psi(t,y;\lambda_\ast)\Big\vert_{y=u(x;\lambda_\ast)}$.
and
Theorem 2
Let $\nu(\lambda) = 1$ be a characteristic value of the kernel $L(x,t;\lambda) = K(x,t)\frac{d}{dy}\Psi(t,y;\lambda)\Big\vert_{y=u(x;\lambda)}$ defined in the above Theorem and let its multiplicity be equal to $1$. Then the solution(s) $\lambda_\ast$ to the equation $\nu(\lambda) = 1$ is a(are) bifurcation point(s).
Thus the question about the existence of bifurcation points reduces to the analysis of the spectrum of the linearized kernel $L(x,t;\lambda)$. While this is indeed a simplification, one still needs to have control over one solution $u(x;\lambda)$, which is often achievable only at a numerical level.
The specific case
In order to make things more concrete let me display the specific equation I am working with:
\begin{equation}
\varepsilon(\theta) = r\cosh\theta - \intop_{\mathbb R} \frac{d\theta'}{2\pi} \sum_{\sigma,\sigma' = \pm 1}\frac{1}{\cosh\left[\theta - \theta' + \sigma \omega + i \sigma' \gamma\right]}\log\left[1+e^{-\varepsilon(\theta')}\right]\;, \quad \omega\in\mathbb R\;,\quad \gamma\in\left[0,\frac{\pi}{2}\right)\;. \tag{3}
\end{equation}
Numerical analysis convinces me that there exist a value $r=r_c\in\mathbb R_{>0}$ which has the features of a bifurcation point. More specifically I see the following behaviour for $r\gtrsim r_c$:
\begin{equation}
\varepsilon(\theta) \underset{r\searrow r_c}{\sim} A(\theta) + B(\theta)\sqrt{r_c-r}+\mathcal O(r_c-r)\;. \tag{4}
\end{equation}
If I insert expression (4) into (3) and compare the dependence on $r$, I obtain the following equations (here $\Phi$ is the kernel of (3))
\begin{equation}
A(\theta) = r_c \cosh\theta - \intop_{\mathbb R}dt \Phi\left(\theta - \theta'\right) \log\left[1+e^{-A(\theta')}\right]\;, \tag{5}
\end{equation}
\begin{equation}
B(\theta) = \intop_{\mathbb R}dt \Phi\left(\theta - \theta'\right) \frac{B(\theta')}{1+e^{A(\theta')}}\;. \tag{6}
\end{equation}
This is exactly what I would expect from the Theorems mentioned above: equation (6) is the linearized equation with unit eigenvalue, evaluated on a solution $A(\theta)$ of (3) at the (supposed) bifurcation point $r=r_c$.
My questions
After this, admittedly lengthy, introduction, I think I have a reasonable grasp of the general situation. I have however a few questions that I hope you could help me answer (or at least shed some light on). Although I am currently working on the specific example (3), I would be very interested in an answer to the questions below for the more general case (1) -- possibly also an extension to systems of equations of type (1).
First of all I would like to have a workable way to determine the existence of bifurcation points, given the properties of the kernel $\Phi$. I ask this because, while most of TBA equations considered in the literature do not possess such a point -- e.g. sinh-Gordon TBA [2], there appear to exist a large class of kernels $\Phi$ for which the physics derived from the TBA equation share many properties with the model considered above(*). It would be important to understand if the mechanism at play is indeed the bifurcation and which conditions must be imposed on $\Phi$ in order to (not) have it. As mentioned above, the analytic properties of the kernel $\Phi$ reflect the physical properties of the underlying system, hence relating the existence of bifurcation points to specific properties of the kernel will allow for a possibly very interesting physical interpretation of the phenomenon.
I would like to understand the structure of the second solution.
In the specific case above, numerically I have access only to one of the solutions to (3). The behaviour (4) suggests then that the second solution might be the analytical continuation around the branching point of the first one, i.e.
\begin{equation}
\tilde{\varepsilon}(\theta) = A(\theta) - B(\theta)\sqrt{r_c-r} + \mathcal O(r_c-r)\;.
\end{equation}
However I performed some simple numerical analysis and found that $\tilde{\varepsilon}$ does not appear to satisfy the equation (3). It might be that I am doing something wrong with this "analytic continuation" or that there are some numerical errors that I did not consider.(**) I would like to be sure that $\tilde{\epsilon}$ is a solution to (3).
I would like to obtain an analytic handle on $r_c$ as a function of the external parameters $\omega$ and $\gamma$.
As far as I can understand, as long as I have a solution to (5), (6) is an equation fixing the value of $r_c$. One might try to use the method of Fredholm determinant to compute the eigenvalues of the kernel $\Phi(\theta - \theta')/\left(1+e^{A(\theta')}\right)$, however as long as I cannot obtain an analytic expression for $A(\theta)$ I am going nowhere. Numerical analysis is also tricky, since the standard numerical procedures stop working when approaching $r_c$ -- I guess since the Fredholm determinant vanishes.
Given the probable impossibility of an analytic answer, I would content myself with some approximate or general result, e.g. a bound on the position of $r_c$ or a condition on the existence of such a bifurcation point.
Comments
(*) Without going too deep in the subject, it has to do with the fact that any function computed from the TBA equation, amongst which are the energy and the entropy of the underlying physical system, display a square root behaviour as functions of $r$. The temptation to assert that this is a consequence of a bifurcation phenomenon is great, but I would like to make sure that no other phenomenon can be the cause of this behaviour.
(**) What I did was to fit the expression (4) -- actually up to terms $\mathcal O(r_c-r)^{10}$ -- against the numerical data close to $r_c$, for various values of $\theta\in\mathbb R$. This gave me a numerical approximation of $A(\theta)$ and $B(\theta)$ -- and the subsequent terms of the expansion. I then reversed the signs in front of the square root terms and computed $$\frac{1}{\cosh(\theta)}\left[\tilde{\varepsilon}(\theta) + \intop_{\mathbb R}d\theta'\Phi(\theta - \theta')\log\left[1+e^{-\tilde{\varepsilon}(\theta')}\right]\right]\;.$$ What I find is that it is not equal to $r$.
References
[1] A. D. Polyanin and A. V. Manzhirov, Handbook of integral equations, CRC Press, Boca Raton, 1998
[2] Al. Zamolodchikov On the thermodynamic Bethe ansatz equation in the sinh-Gordon model J. Phys. A: Math. Gen. (2006) 39 12863; arXiv: https://arxiv.org/pdf/hep-th/0005181.pdf.
|
2025-03-21T14:48:30.500844
| 2020-05-05T18:58:01 |
359476
|
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"Gerhard Paseman",
"Nik Weaver",
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|
Stack Exchange
|
Decreasing sequences in a finitely generated closure algebra
I am interested in finitely generated closure algebras (as a special case of Heyting algebras), and in decreasing sequences of elements within such an algebra that have no lower bound.
Call two decreasing sequences equivalent if any member of one sequence includes some member of the other sequence as a subset, and vice versa (informally, they have the same limit point in some completion).
After factoring out equivalence, can there be an uncountable number of such decreasing sequences?
Aren't all finitely generated Boolean algebras actually finite? Have I misunderstood?
The free one generated Heyting algebra is infinite. Gerhard "You Missed The Word 'Heyting'" Paseman, 2020.05.05.
@GerhardPaseman I had never heard of "closure algebras" but the definition I found says they are Boolean algebras with a closure operator. You missed the phrase `as a special case'.
Touche. And if closed set system is meant instead? Gerhard "Assumes There Is A Question" Paseman, 2020.05.05.
Wouldn't any finitely generated closed set system also be finite?
This is not an answer, but a long comment.
Nik asks:
Wouldn't any finitely generated closed set system also be finite?
Let me describe an infinite $1$-generated closure algebra.
Start with the Boolean algebra ${\mathcal P}(\omega)$ and take as the closed subsets of a closure algebra structure on ${\mathcal P}(\omega)$ the sets $\emptyset$ and the final segments $[n,\infty)$ (including the whole set $\omega = [0,\infty)$). The infinite $1$-generated subalgebra I will describe is generated by the set $X = \{1,3,5,\ldots\}$ = odds. We are allowed to generate with the Boolean operations and the closure operation $C$. We can get these things:
$\omega = \{0,1,2,\ldots\}= [0,\infty)$.
$X = \{1, 3, 5, \ldots\}$.
$C(X) = \{1,2,3,\ldots\} = [1,\infty)$.
$C(X)-X = \{2,4,6,\ldots\}$.
$C(C(X)-X) = \{2,3,4,\ldots\} = [2,\infty)$.
$C(C(X)-X) - (C(X)-X) = \{3,5,7,\ldots\}$.
$C(C(C(X)-X) - (C(X)-X)) = \{3,4,5,\ldots\} = [3,\infty)$.
Etc. (If you look at every other set on this list you see that all nonempty
closed sets appear.)
|
2025-03-21T14:48:30.501030
| 2020-05-05T19:14:05 |
359477
|
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|
Stack Exchange
|
Does forgetting colimits preserve colimits?
For each regular cardinal $\kappa$ let $\operatorname{Cat}_{\kappa}$ be the $(2,1)$-category of small categories with $\kappa$-small colimits, and functors that preserve those colimits. For each pair of regular cardinals $\tau > \kappa$ there is an evident forgetful functor $U_{\tau, \kappa}: \operatorname{Cat}_{\tau} \rightarrow \operatorname{Cat}_\kappa$. It can be seen that $U_{\tau,\kappa}$ preserves $\kappa$-small coproducts, as these can be computed in terms of $\kappa$-small products, which are preserved by virtue of the fact that $U_{\tau,\kappa}$ is a right adjoint.
Question: Does $U_{\tau,\kappa}$ preserve all $\kappa$-small colimits?
Feel free to assume $\kappa$ to be uncountable or even bigger if it helps. I'm ultimately interested in the $\infty$-categorical version of this question so answers in that context are welcome as well.
What sort of colimits? All weak 2-colimits? Or just those that can be built from coproducts coinserters and coequifers? Etc.
@DavidRoberts I had in mind all weak 2-colimits. I also want to consider these as (weak) $(2,1)$-categories, and since $U_{\tau, \kappa}$ already preserves $\kappa$ small coproducts, the question really reduces to whether it also preserves weak coequalizers.
Yeah, that's what I was thinking, but it might be easier to break that down from weak coequalisers to other specific colimits that generate all weak 2-colimits.
As you may know, if you work with 2-categories rather than (2,1)-categories, then the argument showing that the forgetful functor preserves $\kappa$-small coproducts also shows that it preserves Kleisli objects and $\kappa$-small lax colimits, since these are also absolute and can be computed in terms of limits. This doesn't answer your question however. I would be surprised if the answer is yes.
@MikeShulman That's interesting. So I guess your claim is that Kleisli objects in $\operatorname{Cat}\kappa$ are computed in terms of Eilenberg-Moore objects, which are preserved by the forgetful functor since they are weighted limits, and then an arbitrary $\kappa$-small lax colimit in $\operatorname{Cat}\kappa$ can be written as the Kleisli object of a certain monad on a $\kappa$-small coproduct, so it is also preserved.
I agree a positive answer to the original question would be surprising. There are different weakenings one could imagine that might have a better chance of being true: for instance, given a $\kappa$-small diagram $F: \mathcal{I} \rightarrow\operatorname{Cat}\tau$ one could ask whether the map $\operatorname{colim} U{\tau,\kappa} F \rightarrow U_{\tau,\kappa} \operatorname{colim} F$ is necessarily surjective/fully faithful/a localization.
Yes; see for instance section 15 of https://arxiv.org/abs/1301.3191.
When $\kappa = 0$ (so that $Cat_\kappa = Cat$), the answer is no for any infinite $\tau$. For instance, consider the inclusion $f : \ast \to (\to)$ into the arrow category of its initial object, which preserves all colimits. The pushout of $f$ against itself in $Cat$ is the walking span, which does not have finite colimits. So it differs from the pushout computed in $Cat_\tau$ for any infinite $\tau$.
Here is an example showing that $U_{\tau, \kappa}$ does not preserve pushouts for any pair of regular cardinals $\tau > \kappa$. Let $C = \tau$, where here $\tau$ is thought of as an ordinal (i.e. the smallest ordinal of cardinality $\tau$) and $C$ is regarded as a category where there is an arrow $x \rightarrow y$ if and only if $x \leq y$. Using the ordinal identity $2 \cdot \tau = \tau$ we may construct full subcategories $C_0$ and $C_1$ of $C$ where $C_i$ contains those objects that correspond to elements of $2 \cdot \tau$ of the form $(x, i)$. The inclusions $C_i \rightarrow C$ admit left adjoints, so we have a span $C_0 \leftarrow C \rightarrow C_1$ inside $\operatorname{Cat}_{\tau}$. Let $D_\tau$ be its pushout, and let $D_\kappa$ be the pushout of its image under $U_{\tau, \kappa}$. We will show that $D_\tau \neq D_\kappa$.
The object $D_\tau$ in $\operatorname{Cat}_{\tau}$ is defined by the following property: for each $E$ in $\operatorname{Cat}_{\tau}$, the category of $\tau$-small colimit preserving functors $D_\tau \rightarrow E$ is equivalent to the category of $\tau$-small colimit preserving functors $C \rightarrow E$ which factor through the localizations $C \rightarrow C_i$. The latter is the same as the category of $\tau$-small colimit preserving functors which invert the arrow $x \rightarrow x+1$ for every $x$ in $C$. A functor has this property if and only if it is the left Kan extension of its restriction along $\emptyset \rightarrow C$. It follows that $D_\tau$ is the initial object of $\operatorname{Cat}_\tau$, so that $D_\tau$ is the singleton category.
Arguing as above, to show that $D_\kappa \neq D_\tau$ it now suffices to show that there is an object $E$ in $\operatorname{Cat}_\kappa$ and a non-constant $\kappa$-small colimit preserving functor $F: C \rightarrow E$ which inverts the arrow $x \rightarrow x + 1$ for every $x$ in $C$. We may take $E = 2$ to be the walking arrow, and $F$ to be the functor sending $x$ to $0$ if and only if $x < \kappa$.
Here's an example to play around with, with $\kappa = \omega$ and $\tau = \omega_1$ and $Vect = Vect_k$ for $k$ a finite field, look at the pushout of $Vect \leftarrow Set \to Vect$, where both functors are the free functor. The question becomes whether $$Ind(Vect_{\omega_1} \times_{Set_{\omega_1}} Vect_{\omega_1}) \to Ind(Vect_{\omega_1}) \times_{Ind(Vect_{\omega_1})} Ind(Vect_{\omega_1})$$
is an equivalence. I think maybe it's not essentially surjective: the domain consists of Ind-systems of countable-sets-equipped-with-two-vector-space-structures, whereas the codomain consists of ind-countable-sets-equipped-with-two-ind-countable-vector-space-structures. If we restrict to $\omega$-indexed ind-systems, it already looks problematic to try to rectify the latter to the former.
(EDIT: The following argument is problematic --see the comments below.) Maybe the answer is actually yes?
Translating across the equivalence $Ind_\kappa : Cat_\kappa {}^\to_\leftarrow Pr^L_\kappa : (-)_\kappa$ (and similarly for $\tau$), we are asking whether the functor $\Phi: \mathcal K \mapsto Ind_\kappa(U^\tau_\kappa(\mathcal K_\tau)) : Pr^L_\tau \to Pr^L_\kappa$ preserves pushouts. Since $\Pr^L_\kappa$ and $Pr^L_\tau$ are closed under colimits in $Pr^L$, we may compute our pushouts there.
So consider a span $B \xleftarrow F A \xrightarrow G C$ in $Pr^L_\tau$, and let $F^\ast, G^\ast$ be the right adjoints of $F, G$ respectively. The pushout of this span in $Pr^L$ is the pullback $B \times_A C$ computed in $Cat$ using $F^\ast, G^\ast$. For some regular cardinal $\rho \geq \tau$ which is large enough that $F^\ast, G^\ast$ preserve $\rho$-compact objects, consider the fully faithful inclusion $Y^\ast : A \to A' = Ind_\tau(A_\rho)$, with left adjoint $Y$. Then the pullback is equally $B \times_{A'} C$, since $Y^\ast : A \to A'$ is fully faithful. Moreover, $(FY)^\ast$ and $(GY)^\ast$ preserve $\tau$-compact objects.
Now, the pushout of $\Phi B \xleftarrow{\Phi F} \Phi A \xrightarrow{\Phi G} \Phi C$ in $Pr^L$ is the pullback in $Cat$ $\Phi B \times_{\Phi A} \Phi C$, computed using the right adjoints $(\Phi F)^\ast, (\Phi G)^\ast$. Since $\Phi$ is a left 2-adjoint, it preserves localizations. Since $Y$ is a localization, so is $\Phi Y$, and thus $(\Phi Y)^\ast$ is fully faithful. So this pullback is equally $\Phi B \times_{\Phi A'} \Phi C$.
In other words, we have reduced to the case where $F^\ast$ and $G^\ast$ preserve $\tau$-compact objects. Note that $(-)_\tau$ is in fact 2-functorial in all functors preserving $\tau$-compact objects, and $U^\tau_\kappa$ and $Ind_\kappa$ are 2-functorial in all functors (in the sense that they carry functors to functors and natural transformations to natural transformations coherently). Moreover, these functors commute with pullbacks as computed in $Cat$ -- $(-)_\tau$ and $U^\tau_\kappa$ preserve all limits in this sense while $Ind_\kappa$ preserves $\kappa$-small limits as a functor $Cat \to Cat$. So $\Phi$, as the composite of these functors, likewise preserves the relevant $Cat$-pullback as desired.
The proof I know of $\operatorname{Ind}\kappa$ preserving pullbacks of categories seems to require at least some preservation of colimits by the functors involved. For instance HTT proposition <IP_ADDRESS> begins with a cospan in the image of $\operatorname{Ind}\kappa$ but only shows that the pullback is in the image of $\operatorname{Ind}_\tau$ for some $\tau \gg \kappa$.
It seems to me that $\operatorname{Ind}\kappa$ in fact does not preserve all pullbacks. For example, let $C = \kappa$ (thought of as an ordinal and in particular as a category) and let $C_0$, $C_1$ be disjoint cofinal full subcategories of $C$. Then the pullback $C_0 \times_C C_1$ is not preserved by $\operatorname{Ind}\kappa$.
@G.Stefanich Yeah, I guess you're right.
|
2025-03-21T14:48:30.501597
| 2020-05-05T19:30:09 |
359479
|
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|
Stack Exchange
|
Less fundamental applications of Zeta regularization:
As we all know, zeta regularization is used in Quantum field theory and calculations regarding the Casimir effect.
Are there less fundamental applications of zeta function regularization? By "less fundamental" I mean
it 'naturally' pops up in more of an artificially / purely mathematically ideal constructed scenario.
Thanks!
Zeta-function regularization of the determinant of the Laplacian, for example on a torus, might qualify as a "purely mathematical" application.
See, for example, On functional determinants of Laplacians in polygons and simplicial complexes or Zeta functions and regularized determinants on projective spaces.
Zeta function regularization computes the asymptotics of smoothed sums.
https://terrytao.wordpress.com/2010/04/10/the-euler-maclaurin-formula-bernoulli-numbers-the-zeta-function-and-real-variable-analytic-continuation/
Also, the regularized determinant of the Laplacian is related to the Ray-Singer analytic torsion, which is equal to the Reidemeister torsion:
https://en.wikipedia.org/wiki/Analytic_torsion
|
2025-03-21T14:48:30.501725
| 2020-05-05T19:39:30 |
359481
|
{
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"authors": [
"Dmitry Ezhov",
"Gerry Myerson",
"Jeremy Rouse",
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/359481"
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|
Stack Exchange
|
Natural number solutions for equations of the form $\frac{a^2}{a^2-1} \cdot \frac{b^2}{b^2-1} = \frac{c^2}{c^2-1}$
Consider the equation $$\frac{a^2}{a^2-1} \cdot \frac{b^2}{b^2-1} = \frac{c^2}{c^2-1}.$$
Of course, there are solutions to this like $(a,b,c) = (9,8,6)$.
Is there any known approximation for the number of solutions $(a,b,c)$, when $2 \leq a,b,c \leq k$ for some $k \geq 2.$
More generally, consider the equation $$\frac{a_1^2}{a_1^2-1} \cdot \frac{a_2^2}{a_2^2-1} \cdot \ldots \cdot \frac{a_n^2}{a_n^2-1} = \frac{b_1^2}{b_1^2-1} \cdot \frac{b_2^2}{b_2^2-1}\cdot \ldots \cdot \frac{b_m^2}{b_m^2-1}$$
for some natural numbers $n,m \geq 1$. Similarly to the above question, I ask myself if there is any known approximation to the number of solutions $(a_1,\ldots,a_n,b_1,\ldots,b_m)$, with natural numbers $2 \leq a_1, \ldots, a_n, b_1, \ldots, b_m \leq k$ for some $k \geq 2$. Of course, for $n = m$, all $2n$-tuples are solutions, where $(a_1,\ldots,a_n)$ is just a permutation of $(b_1,\ldots,b_n)$.
"I wonder" means "I haven't put any work into this myself"?
What does like mean in "solutions ... like $(a,b,c)=(9,8,6)$"?
Let $a^2\to u+v$, $b^2\to u-v$ and $c$ as parameter. Then solving difference of squares $\Bigl(2 (u - c^2)\Bigr)^2 - \Bigl(2 v\Bigr)^2 = (2 c^2 - 1)^2 - 1$. (a,b,c)=(9,8,6),(26,15,13),(55,24,22),(99,35,33),(50,49,35),(161,48,46),(120,55,50),(244,63,61),(351,80,78),(485,99,97),...
@Dmitry do all $c$ of the form $n^2-3$ give solutions? Most of the solutions you give have $c$ of the form.
@Gerry yes, for $c$ in 2..1000 give all $n$ in 3..31
It seems worth noting that the equation in the title does have infinitely many solutions in positive integers, as for all $n$ it is satisfied by $$a={n(n^2-3)\over2},\ b=n^2-1,\ c=n^2-3.$$ The number of solutions of this form with $a\le k$ will be on the order of $\root3\of{2k}$, but Dmitry has found solutions not of this form.
Here's another infinite family. Let $x,y$ be positive integers such that $x^2-2y^2=\pm1$ – there are infinitely many such pairs. Let $a=x^2$, $b=2y^2$, $c=xy$, then a little algebra will show that $(a,b,c)$ satisfy the equation in the title.
E.g., $x=3$, $y=2$ leads to $(9,8,6)$, and $x=7$, $y=5$ yields $(49,50,35)$, two triples already found by Dmitry, while $x=17$, $y=12$ gets us $(289,288,204)$.
This infinite family is much thinner than the one in the other answer.
[I seem to have become disconnected from the account under which I posted the other answer.]
EDIT: A third infinite family. $$a=4n(n+1)(n^2+n-1),\ b=(2n+1)(2n^2+2n-1),\ c=2(2n+1)(n^2+n-1)$$
I've alerted the moderators to your "account multiplicity" issue.
Please follow the directions at https://mathoverflow.net/help/merging-accounts to merge accounts.
Above equation shown below, has solution:
$\frac{a^2}{a^2-1} \cdot \frac{b^2}{b^2-1} = \frac{c^2}{c^2-1}$
$a=9w(2p-1)(18p-7)$
$b=4w(72p^2-63p+14)$
$c=3w(72p^2-63p+14)$
Where, w=[1/(36p^2-7)]
For, $p=0$ we get:
$(a,b,c)=(9,8,6)$
For which values of $p$ does this lead to a solution in integers? Very few, I think. Your formula gives $b=8-28(7p-2)/(36p^2-7)$, and a linear over a quadratic can only be an integer for finitely many integers $p$.
There are infinitely many integer solutions with $3b=4c$. Let $n$ be odd, and $(8+3\sqrt7)^n=x+y\sqrt7$, $x,y$ integers. Then $a=3y$, $b=x$, $c=3x/4$ is a solution in integers with $3b=4x$. $n=1$ gives $(9,8,6)$; $n=3$ gives $(2295,2024,1518)$.
|
2025-03-21T14:48:30.501982
| 2020-05-05T20:14:38 |
359485
|
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|
Stack Exchange
|
Example of a bounded function whose mean-zero mollification diverges at a point
For a Schwartz function $\psi(x)=xe^{-x^2}$ define $\varphi(x):=\psi'(x)$ and consider a family of $L^1$-dilations of $\varphi$ given by:
$$
\varphi_t(x)=\frac{1}{t}\varphi(x/t), \qquad t>0.
$$
$\textbf{Question:}$ Is there a function $f\in L^\infty(\mathbb{R})$ such that
\begin{equation}\label{eq:1}
\liminf_{t\rightarrow 0^+} |\varphi_t\ast f(0)|>0,
\end{equation}
where $\ast$ denotes a convolution on $\mathbb{R}$ ?
If not, does exist $f\in L^\infty(\mathbb{R})$ satisfying a weaker condition
\begin{equation}
\int_0^1 |\varphi_t\ast f(0)|\, \frac{dt}{t}=\infty ?
\end{equation}
I editted the question after Aleksei's comment.
Doesn't $\chi_{[0, \infty)}$ work?
Aleksei, thank you, you're right. However, I had something different on my mind when asking the question. I wanted to state it in a relatively general form and unfortunately I made it trivial. Let me edit it and please have a look at the updated version :] My apologies!
It seems to me I can construct an example for the weaker question. If you are interested I can post it as an answer.
@AlekseiKulikov, please do. I'm excited to see it!
$\newcommand{\ph}{\varphi}$
$\newcommand{\eps}{\varepsilon}$
Let $a_k$ be a very fast-growing sequence of integers (I think $a_k = 2^{1000k}$ should be enough). Consider the function $f$ defined as
$$f(x) = \sum_{k = 1}^\infty \chi_{[a_k, 2a_k]}.$$
I claim that there is a constant $c > 0$ such that for $\frac{1}{200a_n} \le t \le \frac{1}{100a_n}$ we have $\ph_t(x)*f(0) \ge c$. This is clearly enough since integral of $\frac{1}{t}$ over each of these intervals is $\log(2)$. We have
$$\ph_t(x)*f(0) = \sum_{k = 1}^{n-1}\int_{ta_k}^{2ta_k}\ph(-x)dx +\int_{ta_n}^{2ta_n}\ph(-x)dx +\sum_{k = n+1}^\infty\int_{ta_k}^{2ta_k}\ph(-x)dx.$$
For the first sum we can bound it by the integral $\int_0^{2ta_{n-1}}|\ph(-x)|dx$, which is at most $2Cta_{k-1}$, where $C = \max\limits_{x\in \mathbb{R}} \ph(x)$ which we can make smaller than any $\eps > 0$ if $\frac{a_{n-1}}{a_n}$ is small enough.
For the third sum again we can bound it by $\int_{ta_{n+1}}^\infty |\ph(-x)|dx$. Since $\int_\mathbb{R} |\phi(s)|ds$ converges we can make it smaller than any $\eps$ as long as $ta_{n+1}$ is big enough, that is $\frac{a_{n+1}}{a_n}$ is big enough.
Finally for the middle term we have $x$ is always at least $\frac{1}{200}$ and at least $\frac{1}{50}$. For these $x$ we have that $\phi(-x)$ is at least some constant $p > 0$ (can be seen from the direct computation although the whole argument essentially works for any nonzero, continuous $L^1$ function).
In total we have
$$\ph_t(x)*f(0) \ge ta_np - 2\eps\ge \frac{p}{200} - 2\eps.$$
Choosing $\eps = \frac{p}{1000}$ finishes the story.
As I said in the beginning, exponential grows of $a_k$ is enough for this argument to work so we can actually replace $\frac{1}{t}$ by any decreasing positive weight $w(t)$ with $\int_0^1 w(t)dt = \infty$.
Brilliant! Thank you so much for sharing this beautiful argument, Aleksei!
|
2025-03-21T14:48:30.502339
| 2020-05-05T20:46:33 |
359487
|
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|
Stack Exchange
|
What Stanley-Reisner rings are $\mathbb{Q}$-Gorenstein?
Let $\Delta$ be a simplicial complex and let $R$ be the associated Stanley-Reisner ring. We can characterize when $R$ is Cohen-Macaulay or when $R$ is Gorenstein in terms of the topology of $\Delta$ (c.f. Stanley's book). Can we similarly characterize when $R$ is $\mathbb{Q}$-Gorenstein in terms of $\Delta$? I would be grateful for an answer even in the case when $\Delta$ is a traingulation of a manifold.
To define $\mathbb Q$-Gorensteiness you typically needs the ring to be normal domains (so we can talk about class group). But Stanley-Reisner rings are almost never normal domain.
Thanks for your answer! I was hoping reduced would be enough instead of domain, but I did not know that SR rings are rarely normal...
Perhaps there is hope, despite my original comment. Hartshorne in this paper developed a theory of "generalized divisors" that works on any scheme that is generically Gorenstein and $(S_2)$.
Going quickly through his definitions and early results, the set of "almost Cartier divisors" (basically locally principal in codimension $1$) form a group. For the canonical ideal $w_R$ to be in this group, you need $R$ to be Gorenstein in codimension $1$.
Okay, so $R=k[\Delta]$ needs to be $(S_2)$ and $(G_1)$ to start with. These conditions are well-studied for Stanley-Reisner ring and can be checked relatively easily.
The last thing we need is $\mathbb Q$-Gorensteiness. This would mean that the class of $w_R$ is torsion in this group of "almost Cartier divisors". Algebraically, it says that $(I^n)^{**}$ is principal for some $n>0$, where $I$ is a fractional ideal representing $w_R$, and $^*$ means $Hom(-,R)$.
There are formulas for canonical ideals of Stanley-Reisner ring, I believe. So perhaps you can get there with some efforts.
|
2025-03-21T14:48:30.502497
| 2020-05-05T20:59:36 |
359488
|
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|
Stack Exchange
|
Finitely additive, $\kappa$-additive atomless measures in ZFC
Under Martin's Axiom (and non-CH) the Lebesgue measure is $2^\omega$-additive in the sense that unions of fewer than continuum ($2^\omega$) many null sets are measureable and null. In ZFC we may however extend the Lebesgue measure to a finitely-additive measure on the power set of $[0,1]$ and still call it atomless.
Are there ZFC examples of finitely-additive measures that extend an atomless measure $\mu$ and are $\kappa$-additive in the above sense w.r.t $\mu$-null sets, where $\kappa > 2^\omega$?
I claim that a non-atomic measure $\mu$ can never be $<{2^\omega}^+$-additive. Then the same applies to any finitely-additve extension.
Let $(\Omega, \frak{A}, \mu)$ be a measure space and let us assume that $\mu$ is non-atomic. It follows that there exists $A \in \frak{A}$ such that $0 < \mu(A) < \infty$. I now want to partition $A$ into $2^\omega$ many null sets. Start by splitting $A$ into $A_0$ and $A_1$ both are sets in $\frak{A}$ of positive measure. This can be done, since $\mu$ is non-atomic. Assume that $A_s$ for $s\in 2^{<\omega}$ has been defined and partition it into $A_{s^\frown 0}$ and $A_{s^\frown 1}$. For every $x \in \,^{\omega}2$ define $A_x:= \bigcap_{n < \omega} A_{x \restriction n}$, which is the first limit step.
First note that $\mu(A_x)=\inf_{n < \omega} \mu(A_{x \restriction n})$ and that some $A_x$ may already have measure 0, while others may still have positive measure. If $A_x$ has measure 0, then we do not have to take care of it anymore. If $A_x$ has positive measure, we continue as before and split it into $A_{x^\frown 0}$ and $A_{x^\frown 1}$. This way we get a (transfinite) binary tree, such that some branches die out at limit stages.
I claim that every branch dies out at countable height, i.e. there is no $\omega_1$-branch. If not, there exists a $\subseteq$-decreasing sequence $(A_\alpha)_{\alpha < \omega_1}$ of length $\omega_1$ such that $\mu(A_\beta) < \mu(A_\alpha)$ if $\beta > \alpha$. But this is impossible, since there cannot exist an uncountable decreasing sequence in $\mathbb{R}$ (separability). Therefore, there are only $2^\omega$ branches and so $A=\bigcup_{\alpha < 2^\omega} A_\alpha$.
This and (much) more can be found in Jech's book in chapter 10.
Right, so it seems the right question is whether there is a ZFC example of an atomless $(2^\omega)$-additive measure?
Let me ask a follow-up question.
|
2025-03-21T14:48:30.503093
| 2020-05-05T22:01:53 |
359491
|
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|
Stack Exchange
|
Finitely generated uniformly amenable groups
Keller in "Amenable groups and varieties of groups" introduces uniformly amenable groups as groups such that
there is a function $a: ]0,1[ \times \mathbb{N} \to \mathbb{N}$ such that
for any finite subset $A \subset G$ and
for any $k \in ]0,1[$
there is a finite subset $U$ of $G$ such that
$|U|< a(k,|A|)$ and
$\forall a \in A$, $|U \cap aU| > k |U|$.
Wysoczánski "On uniformly amenable groups" gives an example of a amenable group which is not uniformly amenable
(it's $G = \oplus G_p$ where $G_p$ are the upper-triangular 3x3 matrices wit 1 on the diagonal and entries in $\mathbb{Z}/p\mathbb{Z}$).
Bożejko in "Uniformly amenable discrete groups" proves that groups of polynomial growth are uniformly amenable.
An example of a finitely generated [elementarily amenable] group which is not uniformly amenable is given in Lemma 3.24 of Ol'shanskii, Osin & Sapir's paper.
This is a consequence of a result that [non-virtually cyclic] uniformly amenable groups satisfy a law by Corollary 6.17 of Druţu & Sapir
Another example is mentioned by de Cornulier and Mann (the linked paper contains a few questions on group laws): it is mentionned that the Grigorchuk group does not satisfy a law. In particular it may not be uniformly amenable.
Question: Are there examples of finitely generated amenable but not uniformly amenable groups which satisfy a law?
Another question (but it would imply a positive answer to a question from de Cornulier and Mann, so I assume it is open) is: is there a group of intermediate growth which is uniformly amenable?
nilpotent $3\times 3$ matrices do not form a group
"non-uniformly amenable" means "not (uniformly amenable)" or "amenable but not uniformly amenable"?
sorry for the abusive language, I thought of the obvious nilpotent group in the 3x3 matrices and contracted it (into a nonsensical statement). Hopefully clarified both points...
But this group (the direct sum) seems to be uniformly amenable, since it is solvable, the author probably messed up something in his estimate. On the other hand such a direct sum but increasing both $p$ and the size of matrices, is non-uniformly-amenable.
The first f.g. elementary amenable groups satisfying no laws are due to BH Neumann in 1937 (subgroup of permutations of $\mathbf{Z}$ generated by infinite cycle $n\mapsto n+1$ and transposition $01$. That the ultrapower criterion implies that a group with no law is not uniformly amenable is immediate. So it should be attributed to Keller (1972).
@YCor Thanks for these informations (and correction of reference; should I edit?). I am right to extend BH Neumann's example into saying that for any infinite group $G$, $Sym_f(G) \rtimes G <Sym(G)$ has no law? (where $Sym_f(G)$ are the finitely supported permutations on the set $G$, $Sym(G)$ all permutations and $G$ is naturally identified with a subset of $Sym(G)$,since group multiplication of a given element is a permutation on $G$) The proof (probably) relies on the fact that the group of finitely supported permutations (of an infinite set) has no law.
Yes this is correct. In addition this group is finitely generated if $G$ is. One reason this group has no law is that it is dense in the group of all permutations of $G$, which contains a free subgroup.
Very nice way to see it. Do you have a reference for "solvable implies uniformly amenable"?
If $G$ satisfies some law $w$, then so do its ultrapowers. Hence if $G$ is solvable, so are its ultrapowers, which are therefore amenable. Hence $G$ is uniformly amenable.
|
2025-03-21T14:48:30.503355
| 2020-05-05T22:26:40 |
359494
|
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|
Stack Exchange
|
singularities of the theta divisor $\Theta$
By $\Theta_{sing}$ we denote the singularities of the theta
divisor $\Theta$ of the Jacobian variety $J(R)$ of a compact Riemann surface
R of genus $g\ge 4$. Then
$$
\text{dim} \Theta_{sing}=
\{\begin{array}{l}g-4\, \, \, \, \,\text{if $R$ is non-hyperelliptic,} \\g-3\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \text{if $R$ is hyperelliptic.}\end{array}
$$
This result is due to Andreotti and Mayer. For $ g = 2 $, we have that the divisor $\Theta$ is smooth. My question is in gelation at $ g = 3 $. What can you say when $ g = 3 $? Is the divisor $\Theta$ smooth too or $\Theta_{sing}\not=\emptyset$ ?
Would the following statement be correct?
Considering $g = 3$ ...
Marten's Theorem (Geometry of Algebraic Curves
Volume I, Arbarello, E., Cornalba, M., Griffiths, P., Harris, page 191) + (4.5) Corollary (from the same book, page 190) $\Longrightarrow $ dim$(W_{2})_{sing}=W^{1}_{2}=0$. So $\Theta$ is smooth for $g = 3$.
dim zero means it is a finite set of points, not an empty set. and recall that a point of this set is a line bundle of degree 2 with 2 sections, which does happen when R is hyperelliptic. Just take a double cover of the line branched at 8 points, say y^2 = x^8-1. So the same dimensionality result holds that you stated for g ≥ 4. Hence for g=3, Theta has a singularity if and only if R is hyperelliptic.
If you look at pages 201, 212 of A-M's paper, you will see the explanation for the reason the singular locus of theta has dimension g-3 in the hyperelliptic case, i.e. that the general such singular point has the form: the g(1,2) + g-3 general points. This holds for g ≥ 3, although they do not say that. I.e. this argument assumes that g-3 ≥ 0. http://www.numdam.org/item/ASNSP_1967_3_21_2_189_0/
|
2025-03-21T14:48:30.503515
| 2020-05-05T23:08:03 |
359497
|
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|
Stack Exchange
|
The numerical evaluations of special values of $L$-functions associated to modular forms
LMFDB has a database of lots of classical modular forms. Suppose we have a modular form $f$ listed in LMFDB, which usually gives the first 100 terms of the $q$-expansion. Are there any packages/programs that can compute the special values of the $L$-function $L(f,s)$ to a high precision. For example, compute the first 100 digits of the special value $L(f,s=2)$!
This is possible in Pari/GP and Magma, see the respective documentations.
Sage can do this too. I believe all three packages use the same algorithms, which are due to Tim Dokchitser -- see his paper "Computing special values of motivic L-functions", https://projecteuclid.org/euclid.em/1090350929
Yes, the starting point was TIm Dokchitser's algorithm, initially implemented in Pari/GP and now an improved version in Magma. The recent version of Pari/GP also has improved algorithms using ideas of Pascal Molin, and can be combined with the new modular forms package.
You may also find the following course of Henri Cohen useful: https://arxiv.org/abs/1809.10904
|
2025-03-21T14:48:30.503636
| 2020-05-05T23:55:15 |
359501
|
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|
Stack Exchange
|
Generalized Erdős multiplication table problem
Consider multiplication operation $$f(x_1,\dots, x_k)=\prod_{i=1}^kx_i$$ where $x_i\in\{1,\dots, n_i\}$ with $n_1,\dots, n_k\in\{1,\dots,\infty\}$.
What is the cardinality of the range?
At $k =2$ with $n_1=n_2$ this is the standard Erdos multiplication table problem whose estimates are in Distinct numbers in multiplication table.
this should be easy to look up. i know it's been done before
Standard name for the problem?
How about googling the title of this question?
This problem is solved for $k\leq 5$ and open for $k\geq 6$. See Koukoulopoulos's paper for more details.
|
2025-03-21T14:48:30.503718
| 2020-05-06T03:07:59 |
359507
|
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"Alex Petzke",
"Lennart Meier",
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|
Stack Exchange
|
Roadmap to homotopical group theory
I have been lurking here for a long time just enjoying the scenery from my beginner's viewpoint. I have a math.SE account but I think this question is appropriate here based on the nature of the subject and similar questions I have seen. It also seems that some of those qualified to answer are much more active here than on math.SE.
Question: What are some references that would supply me with the algebra, algebraic topology, category theory, etc. I need to get into the literature on homotopical group theory? If that is too broad then at least as a starting point, what would let me read Part III: Fusion and homotopy theory from Fusion systems in algebra and topology by Aschbacher, Kessar, and Oliver?
I have a general sense of what I need and I am aware of standard references for some of the background material, but there are many components to this and I would appreciate sources that may be especially relevant to the direction I want to go in (e.g. an algebraic topology text that has a good treatment of classifying spaces).
My background: I am not pursuing a career in mathematics but I have been teaching myself math for a number of years now. I have always connected most with group theory and topology and have recently become deeply fascinated by homotopical group theory (i.e. the study of a group via the homotopy theory of its classifying space). I would like to begin focusing my studies in that direction and could use some help with where to direct my efforts.
I have taken introductory courses in real analysis and algebra and have taught myself some linear algebra, graph theory, point-set topology, ring theory, module theory, and field theory, at least to the point of being able to work with the basics. Additionally, I have gone a bit further into group theory (e.g. group actions, Sylow theory, solvable and nilpotent groups, a tiny bit of fusion) and have done a little bit of algebraic topology (basics of homotopy and singular homology) from Rotman's text. I do have access to a university library, at least once the campus reopens.
Some specific concepts I find especially interesting relate to cellularization of classifying spaces (e.g. Flores and Scherer 2007, Cellularization of classifying spaces and fusion properties of finite groups) and homotopy group extensions (e.g. Broto and Levi 2002, On spaces of self-homotopy equivalences of p-completed classifying spaces of finite groups and homotopy group extensions).
I'm not an expert, but I enjoyed some years ago a course Dwyer gave in Copenhagen: http://web.math.ku.dk/~jg/homotopical2008/Dwyer.CopenhagenNotes.pdf
@LennartMeier Thanks. It looks like those notes could serve as a good guide for what I need to learn, at a minimum.
|
2025-03-21T14:48:30.503930
| 2020-05-06T03:16:31 |
359508
|
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|
Stack Exchange
|
Coherent cohomology of the generic fiber of Lubin-Tate space vs. of Lubin-Tate space considered rationally?
I am trying to compare the coherent cohomology of the generic fiber of Lubin-Tate space to the coherent cohomology of Lubin-Tate space considered rationally, and I am going in circles! I would be very appreciative of any insight or direction on the topic.
Let's establish a bit of background and notation. Fix a $h \in \mathbb{N}$. Let $k = \mathbb{F}_{p^h}$. Let $\mathrm{Art}_k$ be the category of Artinian local rings $R$ and their maximal ideals, such that $R/\mathfrak{m}$ is a $k$-algebra. Let $\mathrm{FGL}_{R}$ be the category of formal group laws over a ring $R \in \mathrm{Art}_k$. Fix $H \in \mathrm{FGL}_k$ to be a height $h$ formal group law. The groupoid $\mathrm{Def_H(R)}$ has as objects: $$\{G \in \mathrm{FGL}_{R}, \iota: H \xrightarrow{\simeq} G \otimes_R k\},$$ and as morphisms: isomorphisms $f$ of formal group laws over $R$ which reduces to the identity modulo $\mathfrak{m}$.
We define another groupoid $\mathrm{DefU_H(R)}$ which has as objects: $$\{G \in \mathrm{FGL}_{R}, a \in R^\times, \iota: H \xrightarrow{\simeq} G \otimes_R k\},$$ and as morphisms: $f \in \mathrm{Hom}((G, a, \iota), (F, b, \delta))$ are isomorphisms $f$ of formal group laws $G, F$ over $R$, such that $a = f'(0)b$.
There is an action of the group $J := \mathrm{Aut}_k(H)$ on $\mathrm{Def}_H(R)$ and $\mathrm{DefU}_H(R)$, which acts by each $j \in J$ taking $(G, \iota)$ to $(G, \iota \circ j),$ and $(G, a, \iota)$ to $(G, a, \iota \circ j)$ respectively. The functor $\mathrm{Def}_H(-)$ is represented by a formal scheme, $LT := \mathrm{Spf} \text{ } W(k)[[u_1, ..., u_{h-1}]],$ where $W(k)$ denotes the Witt vectors of $k$. The functor $\mathrm{DefU}_H(-)$ is represented by the formal scheme $LTU :=\mathrm{Spf} \text{ }W(k)[[u_1, ..., u_{h-1}]][u, u^{-1}]$.
We denote the adic space associated to $LT$ by $LT^{ad}$. The adic space $LT^{ad}$ lives over $\mathrm{Spa}(\mathbb{Z}_p, \mathbb{Z}_p)$. We may take the rigid generic fiber of the formal scheme $LT$, $$LT^\eta := LT^{ad} \times_{\mathrm{Spa}(\mathbb{Z}_p, \mathbb{Z}_p)} \mathrm{Spa}(\mathbb{Q}_p, \mathbb{Z}_p).$$
Here is my question: What is the relationship between these two continuous group cohomologies? Are they isomorphic?
$$H_{cts}^*(J,\Gamma(\mathcal{O}_{LT}, LT)) \otimes_{\mathbb{Z}} \mathbb{Q} \stackrel{?}{\simeq} H_{cts}^*(J, \Gamma(\mathcal{O}_{LT^\eta}, LT^{\eta})).$$
I have the same question for the case with units thrown in: how are these two related?
$$H_{cts}^*(J,\Gamma(\mathcal{O}_{LTU}, LTU)) \otimes_{\mathbb{Z}} \mathbb{Q} \stackrel{?}{\simeq} H_{cts}^*(J, \Gamma( \mathcal{O}_{LTU^\eta}, LTU^{\eta})).$$
Here is my sub-question: Is $\Gamma(\mathcal{O}_{LT^\eta}, LT^{\eta}) \simeq \Gamma(\mathcal{O}_{LT}, LT) \hat{\otimes}_{\mathbb{Z}_p} \mathbb{Q}_p$?
By the following argument, it seems that this is true. However, I have a strange feeling in my gut about it. I am almost sure I have missed something, but combing through it I can't figure out what.
Let us denote the ring $L := W(k)[[u_1, ..., u_{h-1}]] \simeq \Gamma(\mathcal{O}_{LT}, LT)$. One might expect $LT^{\eta}$ to be $\mathrm{Spa}(L \hat{\otimes}_{\mathbb{Z}_p} \mathbb{Q}_p, L)$, but $L$ isn't open in $L \hat{\otimes}_{\mathbb{Z}_p} \mathbb{Q}_p$. This is because, for example, $p^{-1}u_1^{\text{ }m} \to 0$, but this sequence never enters $L$. Thus, $(L \hat{\otimes}_{\mathbb{Z}_p} \mathbb{Q}_p, L)$ isn't a Huber pair. Instead, if I understand correctly, $$LT^{\eta} := \mathrm{colim}_m \text{ }\mathrm{Spa} \left(L \left\langle \frac{u_1^{\text{ }m}}{p}, ...,\frac{u_{h-1}^{\text{ }m}}{p} \right\rangle \otimes \mathbb{Q}_p, L\left\langle \frac{u_1^{\text{ }m}}{p}, ...,\frac{u_{h-1}^{\text{ }m}}{p} \right\rangle \right),$$
where $U^{\underline{s}}$ denotes $u_1^{s_1}\cdots u_{h-1}^{s_{h-1}}$, and
$$ L \left\langle \frac{u_1^{\text{ }m}}{p}, ...,\frac{u_{h-1}^{\text{ }m}}{p} \right\rangle := \left\{ \sum_{\underline{s} \in \mathbb{N}^{h-1}} c_{\underline{s}}U^{\underline{s}}; c_{\underline{s}} \in L^{h-1} \text{ } \Big| \text{ } \lim_{s_i \to \infty} |c_{s_i}p^{s_i/m}| = 0 \text{ },\forall i \in \{1, ..., h-1\}\right \}.$$
This $(LT^{\eta})_m$ defines an increasing family of neighborhoods of the origin. Since a power of $u_i$ is divisible by $p$, the $(p, u_1, ..., u_{h-1})$-adic topology agrees with the $p$-adic topology on each of these rings. It seems that the global sections $\Gamma(\mathcal{O}_{LT^\eta}, LT^{\eta})$ are the completion of the intersection of these rings as $m \to \infty$, which is $L \hat{\otimes} \mathbb{Q}_p$. Thus, $\Gamma(\mathcal{O}_{LT^\eta}, LT^{\eta}) \simeq L \hat{\otimes} _{\mathbb{Z}_p} \mathbb{Q}_p$ as claimed.
|
2025-03-21T14:48:30.504205
| 2020-05-06T03:49:18 |
359511
|
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}
|
Stack Exchange
|
Varieties corresponding to closure of a function field
Let $V$ be a variety over a field $k$, and let $k(V)$ be its function field.
Let $\overline{k(V)}$ be a separable algebraic closure of $k(V)$, and let $\overline{V}$ be a variety (in the birational equivalence class) whose function field is $\overline{k(V)}$.
Provided it makes sense to talk about $\overline{V}$ (what is its field of definition?), what is the geometrical relationship of $\overline{V}$ to $V$? Is it related to some "universal cover" of $V$?
The "what is its field of definition" seems to be the main issue, to start with.
The problem is that $\overline{k(V)}$ is not usually finitely generated over, say, $\overline{k}$ (think about $\mathbb{C}(x)$). So it wouldn't be the function field of a variety defined over $\overline{k}$.
|
2025-03-21T14:48:30.504295
| 2020-05-06T05:33:18 |
359513
|
{
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"authors": [
"Johannes Schürz",
"Kant",
"Monroe Eskew",
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"https://mathoverflow.net/users/134910",
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/359513"
}
|
Stack Exchange
|
Atomless, c-additive measures in ZFC
This is a follow-up question to this one.
Is there a ZFC example of an atomless measure that is $2^\omega$-additive, meaning, fewer than continuum many null sets have measurable union that is null?
Are you assuming the collection of measurable sets is a $\sigma$-algebra?
@MonroeEskew, standardly, measures are defined on $\sigma$-algebras.
I think what you are looking for are real valued measurable cardinals and their existence is equiconsistent to a measurable cardinal.
@JohannesSchürz, no, because I do not insist on the measure to be defined for all sets.
The answer is negative. Suppose $(X,\mu)$ is such a measure space. By the argument in the linked question, there is a partition of $X$ into continuum-many pairwise disjoint $\mu$-null sets. This is done by building a binary tree that splits a given node into two nodes of one half the measure. Each branch corresponds to a point in Cantor space. We induce an atomless $2^\omega$-additive measure $\nu$ on Cantor space via this correspondence.
However, it is consistent with ZFC that the continuum is singular. Under this hypothesis, any $2^\omega$-additive ideal is $(2^\omega)^+$-additive. But each singleton is $\nu$-null, and so the Cantor space is the union of $2^\omega$-many $\nu$-null sets, meaning the whole space has measure 0. But this means that $\mu(X)=0$.
It looks like even restricting to finitely additive measures will not help?
@Kant: What do you mean? That the null sets form a continuum-additive ideal but the measure is not countably additive on positive sets?
@Kant I think you‘re right. The binary tree construction will transfer the additivity of the null ideal to the induced measure on the reals. But of course we have to assume the measure is defined on a sigma algebra to make sure the intersections along branches are measurable.
@Kant: Yes, I think so. Take some regular cardinal $\kappa \geq 2^\omega$ and consider the nonstationary ideal on $\kappa$. Then build a complete binary tree of stationary sets with $\kappa$ at the top. Then just define the probability function on finite unions of nodes in the tree such that the immediate children of a node each have probability 1/2 of the parent. The algebra of sets is just the algebra generated by these stationary sets plus/minus a nonstationary set. And the measure of each nonstationary set is zero.
Thanks. So it seems I can been every cardinal by the additivity of the null ideal?
@Kant: Yes. But some may consider the algebra of sets I described to be not very rich.
|
2025-03-21T14:48:30.504507
| 2020-05-06T05:44:32 |
359514
|
{
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"authors": [
"GGT",
"Per Alexandersson",
"Richard Stanley",
"darij grinberg",
"https://mathoverflow.net/users/1056",
"https://mathoverflow.net/users/2530",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/359514"
}
|
Stack Exchange
|
Jack function in power symmetric basis
In Macdonald's book, the Jack symmetric function $J_{\lambda}(x_1,\ldots, x_n)$ for a partition $\lambda$
is defined by three properties (orthogonality, triangularity, and normalization). In the following paper (http://www-math.mit.edu/~rstan/pubs/pubfiles/73.pdf) its existence and uniqueness appear as Theorem 1.1.
The Jack symmetric functions can be seen as eigenfunctions for the operators
$$D(\alpha)= \alpha/2 \sum_{i=1}^{n}x_i^2\frac{\partial}{\partial_i^2}+\sum_{i\neq j}\frac{x_i^2}{x_i -x_j}\frac{\partial}{\partial x_i}$$
with the eigenvalues as given in Theorem 3.1 of the above paper.
Now a recent paper of Chapuy and Dolega
(https://arxiv.org/pdf/2004.07824.pdf)
defines
the following operator, which is given in the power symmetric basis,
$$D_{\alpha}^{'}= \alpha/2 \sum_{i,j\geq 1}ij p_{i+j}\frac{\partial^2}{\partial p_i \partial p_j}
+ 1/2 \sum_{i,j\geq 1}(i+j) p_{i}p_{j}\frac{\partial^2}{\partial p_{i+j}}+(\alpha -1)\sum_{i\geq 1}\frac{i(i -1)}{2}p_i\frac{\partial}{\partial p_i}$$
and it defines the Jack symmetric functions to be those function which are eigenfunctions to these operators; they give the eigenvalues in terms of $\alpha$, which takes a nice form. In the paper it appears in Proposition 5.1.
My question is how to derive this operator from Stanley's paper and to apply it on the Jack symmetric function. Do we need to express the Jack symmetric function in the power symmetric basis? I cannot do it in the case of a general partition. Also, the operator in Stanley's paper involves finitely many variables, but in the Chaupy and Dolenga paper it involves infinitely many. I hope someone can give me more details.
Are you claiming that the operators $D\left(\alpha\right)$ and $D'_\alpha$ are identical? In that case, this shouldn't be hard to prove. Indeed, both are differential operators of second order (i.e., operators $f$ satisfying $f\left(abc\right) - f\left(ab\right)c - f\left(ac\right)b - f\left(bc\right)a + f\left(a\right)bc + f\left(b\right)ac + f\left(c\right)ab = 0$ for all $a, b, c$ in the algebra), and thus their equality needs only to be checked on a generating set of the algebra and the pairwise products of generators from this generating set.
Regarding the number of variables, you can restrict to n variables, if you are only interested in Jack polynomials of degree n. This is a standard observation when dealing with symmetric functions.
I guess they are identical as per the new paper goes. How does one get the operator in power symmetric basis from the monomial basis is the jacobian change of coordinates. I can't get the transformation.
@RichardStanley Can you comment on this? I too am confused by equality (11) in your paper; it seems to be tailored to $n$ variables, but $\Lambda$ is a ring of symmetric functions in infinitely many variables.
@darijgrinberg Yow! The operator $D(\alpha)$ that I define should be on $\Lambda_n\otimes \mathbb{Q}(\alpha)$. (I use $\Lambda_n$ to denote $n$ variables and $\Lambda^n$ to denote degree $n$, same as Macdonald.)
This explanation can be found in Macdonald's book "Symmetric Functions and Hall Polynomials" by looking at Ex.VI.4.3.
Note that Stanley's Laplace-Beltrami operator $D(\alpha)$ depends on $n$ and acts on the algebra $\mathbb{Q}(\alpha)\otimes \Lambda^n$, where $\Lambda^n$ denotes the algebra of symmetric polynomials in $n$ variables $x_1,\dots,x_n$. Because of that I prefer to modify your notation and denote Stanley's operator by $D_n(\alpha)$. Let me introduce a modified version of this operator $D'_n(\alpha)$, which also acts on $\mathbb{Q}(\alpha)\otimes \Lambda^n$ and is defined by setting
$$ D'_n(\alpha)f := \big(D_n(\alpha)-(n-1)\deg(f)\big)f$$
for homogenous $f$ and extended by linearity.
Recall that the algebra $\Lambda$ of symmetric functions is defined as the projective limit of $\Lambda_n$ with respect to the morphism $\rho_n : \Lambda_{n+1}\to \Lambda_n$ which kills the last variable:
$$\rho_n(f)(x_1,\dots,x_n) := f(x_1,\dots,x_n,0).$$
It is easy to check that
$$\rho_n D'_{n+1}(\alpha) = D'_{n}(\alpha)\rho_{n},$$
so you can define an operator $D'_\alpha := \lim D'_{n}(\alpha)$ which acts on $\mathbb{Q}(\alpha)\otimes \Lambda$. Stanley's computations used in the proof of his Theorem 3.1 give you immediately an expression for $D'_\alpha$ as a differential operator in power-sums.
|
2025-03-21T14:48:30.504934
| 2020-05-06T07:16:01 |
359518
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"authors": [
"Laurent Moret-Bailly",
"dorebell",
"https://mathoverflow.net/users/56878",
"https://mathoverflow.net/users/7666"
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"url": "https://mathoverflow.net/questions/359518"
}
|
Stack Exchange
|
Why is the set of parabolic reductions of a G-torsor E bijective to the set of parabolic subgroups of Aut(E)?
Let $G$ be a reductive group scheme over some base $X$ and $P \subseteq G$ a parabolic subgroup. To a $P$-torsor $\mathscr{E}_P$, we may associate a $G$-torsor $\mathscr{E} = G \times^P \mathscr{E}_P$, which is $G \times \mathscr{E}_P$ mod the relation $(gp, s) \sim (g, ps)$, with $G$ acting by $g \cdot (h, s) = (gh, s)$. This gives a $P$-equivariant monomorphism $\mathscr{E}_P \hookrightarrow \mathscr{E}$ sending $s$ to $(1, s)$, and this gives us an inclusion of group schemes $\mathrm{Aut}(\mathscr{E}_P) \xrightarrow{\sim} \mathrm{Stab}(\mathscr{E}_P) \subseteq \mathrm{Aut}(\mathscr{E})$.
$\mathrm{Aut}(\mathscr{E}_P)$ is an inner form of $P$ (indeed, it is given by twisting $P$ by $\mathscr{E}_P$), and likewise $\mathrm{Aut}(\mathscr{E})$ is an inner form of $G$. Upon passing to an étale cover $\widetilde{X} \rightarrow X$ and choosing a trivialization $\mathscr{E}_P$ (which automatically gives a trivialization of $\mathscr{E}$ compatible with the map $\mathscr{E}_P \rightarrow \mathscr{E}$), we obtain an isomorphism $G|_{\widetilde{X}} \rightarrow \mathrm{Aut}(E)|_{\widetilde{X}}$ sending $P$ to $\mathrm{Aut}(\mathscr{E}_P)|_{\widetilde{X}}$.
This discussion defines a functor from the groupoid of $P$-torsors on $X$ to the groupoid of pairs $(\mathscr{E}, \mathscr{P})$ where $\mathscr{E}$ is a $G$-torsor on $X$ and $\mathscr{P}$ is a parabolic subgroup of $\mathrm{Aut}(\mathscr{E})$. A morphism $(\mathscr{E}, \mathscr{P}) \rightarrow (\mathscr{E}', \mathscr{P}')$ is an isomorphism of $G$-torsors $\varphi \colon \mathscr{E} \rightarrow \mathscr{E}'$ carrying $\mathscr{P}$ to $\mathscr{P}'$ under the induced map on automorphism groups (given by conjugation by $\varphi$).
Example 10.6.2 of the book Weil's conjecture for function fields by Gaitsgory-Lurie, it is claimed that this functor is an equivalence of categories. What's the quasi-inverse functor? I can't seem to find a proof anywhere.
If it helps, feel free to assume that $X$ is a curve over a finite field $k$ and that $G = G_0 \times_{\mathrm{Spec}(k)} X$ for an adjoint split reductive group $G_0$ over $\mathrm{Spec}(k)$.
As stated, the functor is not essentially surjective: clearly, $\mathscr{P}$ should be (at least) an inner form of $P$.
Thanks to Laurent Moret-Bailly for pointing out that I missed a crucial hypothesis! Now I can construct the quasi-inverse, which I'll record below in case some future person is confused by the same problem:
The hypothesis is that "$\mathscr{P} \subseteq \mathrm{Aut}(\mathscr{E})$ is an inner form of $P \subseteq G$". By this, I mean:
(*): There is an étale cover $\widetilde{X} \rightarrow X$ and a trivialization of $\widetilde{\mathscr{E}}$ such that the induced isomorphism $\widetilde{G} \rightarrow \mathrm{Aut}(\widetilde{\mathscr{E}})$ takes $\widetilde{P}$ to $\widetilde{\mathscr{P}}$.
(I'm using the notation $\widetilde{Y}$ for $Y|_{\widetilde{X}}$). Note that if we change the trivialization, $\widetilde{P}$ gets mapped to a conjugate of $\widetilde{\mathscr{P}}$.
Gaitsgory-Lurie give an equivalent formulation of this condition, which is more intrinsic to $\mathscr{P} \subseteq \mathrm{Aut}(\mathscr{E})$ and extends to general inner forms of $G$. Namely: that there is an étale cover $\widetilde{X}$ of $X$ and an isomorphism $\varphi \colon \widetilde{G} \rightarrow \mathrm{Aut}(\widetilde{\mathscr{E}})$ taking $\widetilde{P}$ to $\widetilde{\mathscr{P}}$ which "is compatible with the inner structure on $\mathrm{Aut}(\widetilde{\mathscr{E}})$". This compatibility means that if we use $\mathscr{E}$ to realize $\mathrm{Aut}(\mathscr{E})$ as an inner form of $G$, then (after passing to a further étale cover of $X$ if necessary), the resulting isomorphism $\widetilde{G} \rightarrow \mathrm{Aut}(\widetilde{\mathscr{E}})$ differs from $\Phi$ by an inner automorphism of $\widetilde{G}$.
Now, we define the quasi-inverse functor as follows:
Let $(\mathscr{E}, \mathscr{P})$ be as above, and define a subsheaf $\mathscr{E}_{\mathscr{P}}$ of $\mathscr{E}$ by $\mathscr{E}_{\mathscr{P}}(U) = \{s \in \mathscr{E}(U) \colon \forall V \rightarrow U, \forall \varphi \in \mathscr{P}(V), \varphi(s|_V) \in P(V) \cdot s|_V \}$. Note that for $s \in \mathscr{E}_\mathscr{P}(U)$, $\mathscr{P}|_U$ is the full stabilizer of $s$ mod $P$ in $\mathrm{Aut}(\mathscr{E}|_U)$ (as can be seen from (*) by passing to $\widetilde{U} \rightarrow U$ and using $s$ to trivialize $\mathrm{Aut}(\mathscr{E}|_{\widetilde{U}})$ and noting that the stabilizer of $s$ mod $P$ must be a conjugate of $\mathscr{P}$).
We claim that $\mathscr{E}_\mathscr{P}$ is a $P$-torsor, so our quasi-inverse functor is $(\mathscr{E}, \mathscr{P}) \mapsto \mathscr{E}_{\mathscr{P}}$. It's easy to see that it is a subsheaf of $\mathscr{E}$ preserved by the action of $P$, so it suffices to prove:
The action of $P$ on $\mathscr{E}_{\mathscr{P}}$ is transitive.
$\mathscr{E}_{\mathscr{P}}$ has sections over some étale cover of $X$.
To prove 1., suppose that $s, s' \in \mathscr{E}_{\mathscr{P}}(U)$ are two sections. Then there is a unique automorphism $\varphi \in \mathrm{Aut}(\mathscr{E}|_U)$ such that $s' = \varphi(s)$. It suffices to prove that $\varphi \in \mathscr{P}(U)$ (since then by definition of $\mathscr{E}_{\mathscr{P}}$, $s' = \varphi(s) \in P(U) \cdot s$). Since $\mathscr{P}$ is the stabilizer of $s$ mod $P$, $\varphi \mathscr{P} \varphi^{-1}$ is the stabilizer of $s'$ mod $P$, so we have $\varphi \mathscr{P} \varphi^{-1} = \mathscr{P}$. Since parabolic subgroups are self-normalizing, this implies that $\varphi \in \mathscr{P}$.
To prove 2., choose $\widetilde{X}$ and $s_0 \in \mathscr{E}(\widetilde{X})$ be as in (*). We claim that $s_0 \in \mathscr{E}_{\mathscr{P}}(\widetilde{X})$. Indeed, the resulting isomorphism $\widetilde{G} \rightarrow \mathrm{Aut}(\widetilde{\mathscr{E}})$ maps $g$ to the unique automorphism sending $s_0$ to $gs_0$, so the stabilizer of $s_0$ mod $\widetilde{P}$ is the image of $\widetilde{P}$ under this isomorphism, which is $\mathscr{P}$.
For what it's worth, the hypothesis that $P$ and $\mathscr{P}$ are parabolic is only used in the proof of point 1. above. In general, this conversation should carry over to the case that $H \subseteq G$ is a (connected, smooth) closed subgroup to show an equivalence between the category of $N_G(H)$-torsors and the category of pairs $(\mathscr{E}, \mathscr{H} \subseteq \mathrm{Aut}(\mathscr{E})$ where $\mathscr{H}$ is an "inner form of $H$" in the above sense.
|
2025-03-21T14:48:30.505363
| 2020-05-06T08:39:21 |
359522
|
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"url": "https://mathoverflow.net/questions/359522"
}
|
Stack Exchange
|
Independence in a sequential problem with observations getting added to buckets
Consider a sequence of random observations $(O(t))_{t\geq 1}$, with $O(t)=(D(t),J(t),Y(t))$. Denote $\mathcal{F}(t) := \sigma(O(1),\ldots,O(t))$, the filtration induced by the first $t$ observations.
$J(t)$ takes values in $\{1,\ldots,J\}$, for some integer $J \geq 2$, and should be understood as the index of a bucket to which we add the observation $Y(t)$. Denote $\widetilde{Y}(j,n)$ the value of the $n$-th observation $Y(t)$ added to the bucket $j$. More formally, letting $n(j,t) := \sum_{\tau=1}^t I(J(\tau)=j)$, the number of observations added to bucket $j$ up till time $t$, and letting $t(j,n) := \min\{t \geq 1: n(j,t) =n \}$, the time $t$ at which we added its $n$-th observation to bucket $j$, we have $\widetilde{Y}(j,n):=Y(t(j,n))$. Let $\widetilde{\mathcal{F}}(j,n):=\sigma(\widetilde{Y}(j,1),\ldots,\widetilde{Y}(j,n))$, the filtration induced by the first $n$ observations in bucket $j$.
We have the following information on the dependence structure of $(O(t))_{t\geq 1}$.
$(D(t))_{t\geq 1}$ is a sequence of independent Bernoulli random variables taking value 1 with probability $\delta \in (0,1)$.
We know that $J(t)|\mathcal{F}(t-1),D(t)=1 \sim \text{Uniform}(\{1,\ldots,J\})$, however, we don't know anything on the conditional distribution $J(t)|\mathcal{F}(t-1), D(t)=0$.
For any $t \geq 1$, $Y(t)|\mathcal{F}(t-1),D(t),J(t)=j \sim Y(t)|\widetilde{\mathcal{F}}(j,n(j,t-1)), J(t)=j$. That is, $Y(t)$ depends on its past only through $J(t)$ and the observations added before it to its bucket. Another way of writing it is $Y(t)|\mathcal{F}(t-1),D(t),J(t)=j \sim \widetilde{Y}(j,n(j,t)) | \widetilde{Y}(j,1),\ldots,\widetilde{Y}(j,n(j,t)-1)$.
Are $\widetilde{Y}(j,n)$ and $D(t)$ independent for any $j$, $n$ and $t$? How would you prove it?
Let for all $t\geq 1$,$\mathcal{F}^-(t):=\sigma(\mathcal{F}(t-1), D(t), J(t)).$
The hypothesis in the third bullet point can be rephrased as $Y(t)|\mathcal{F}^-(t) \stackrel{d}{=} \widetilde{Y}(j,n(j,t)) | \widetilde{\mathcal{F}}(j,n(j,t)-1)$.
I prove the claim by induction. Fix $j$ and $t$. I show by induction that for all $n \geq 1$, $D(t) {\perp\!\!\!\perp} \widetilde{\mathcal{F}}(j,n)$. I treat the base case at the end of the proof. Suppose that for some $n\geq1$, $D(t) {\perp\!\!\!\perp} \widetilde{\mathcal{F}}(j,n)$. Let us show that $D(t) {\perp\!\!\!\perp} \widetilde{\mathcal{F}}(j,n+1)$. It suffices to show that $D(t) {\perp\!\!\!\perp} \widetilde{Y}(j,n+1) | \widetilde{\mathcal{F}}(j,n)$. Observe that
\begin{align}
&P\left[\widetilde{Y}(j,n+1)=y,D(t)=d \middle| \widetilde{\mathcal{F}}(j,n)\right] \\
=& P\left[\widetilde{Y}(j,n+1)=y,D(t)=d, t(j,n+1)<t \middle| \widetilde{\mathcal{F}}(j,n)\right] \\
&+ P\left[\widetilde{Y}(j,n+1)=y, D(t)=d, t(j,n+1) \geq t \middle| \mathcal{F}(j,n)\right].
\end{align}
I start with the first term. I have that
\begin{align}
&P\left[\widetilde{Y}(j,n+1)=y,D(t)=d, t(j,n+1)<t \middle| \widetilde{\mathcal{F}}(j,n)\right] \\
=& P \left[D(t)=d\middle| \widetilde{Y}(j,n+1)=y,t(j,n+1)<t \middle| \widetilde{\mathcal{F}}(j,n)\right] P \left[ \widetilde{Y}(j,n+1)=y,t(j,n+1)<t , \widetilde{\mathcal{F}}(j,n)\right] \\
=& P \left[D(t)=d\right] P \left[ \widetilde{Y}(j,n+1)=y,t(j,n+1)<t \middle| \widetilde{\mathcal{F}}(j,n)\right]
\end{align}
since $\{\widetilde{Y}(j,n)=y, t(j,n+1) < t \} \cap \widetilde{\mathcal{F}}(j,n)$ is $\mathcal{F}(t-1)$ measurable and $D(t) {\perp\!\!\!\perp} \mathcal{F}(t-1)$. Moreover, observe that $\{t(j,n+1) < t\} \cap \widetilde{\mathcal{F}}(j,n)$ is $\mathcal{F}^-(t(j,n+1))$-measurable, and therefore,
\begin{align}
&P \left[ \widetilde{Y}(j,n+1)=y \middle| t(j,n+1)<t, \widetilde{\mathcal{F}}(j,n)\right] \\
=& E \left[ P \left[ Y(t(j,n+1))=y \middle| \mathcal{F}^-(t(j,n+1))\right] \middle| t(j,n+1)<t, \widetilde{\mathcal{F}}(j,n)\right] \\
=& E \left[ P \left[\widetilde{Y}(j,n+1) \middle| \widetilde{\mathcal{F}}(j,n) \right] \middle| t(j,n+1) < t, \widetilde{\mathcal{F}}(j,n) \right] \\
=& P \left[\widetilde{Y}(j,n+1) \middle| \widetilde{\mathcal{F}}(j,n)\right].
\end{align}
Therefore,
\begin{align}
&P\left[\widetilde{Y}(j,n+1)=y,D(t)=d, t(j,n+1)<t \middle| \widetilde{\mathcal{F}}(j,n)\right] \\
=& P\left[\widetilde{Y}(j,n+1) \middle| \widetilde{\mathcal{F}}(j,n) \right] P \left[D(t)=d\right] P \left[ t(j,n+1) < t \middle| \widetilde{\mathcal{F}}(j,n) \right]\\
=& P\left[\widetilde{Y}(j,n+1) \middle| \widetilde{\mathcal{F}}(j,n) \right] P \left[D(t)=d, t(j,n+1) < t \middle| \widetilde{\mathcal{F}}(j,n) \right],
\end{align}
since $D(t) {\perp\!\!\!\perp} \{ t(j,n+1) < t\}$, as $\{ t(j,n+1) < t\}$ is $\mathcal{F}(t-1)$-measurable, and $D(t) {\perp\!\!\!\perp} \widetilde{\mathcal{F}}(j,n)$ by induction hypothesis, which imply that $D(t) {\perp\!\!\!\perp} \{ t(j,n+1)<t\} | \widetilde{\mathcal{F}}(j,n)$.
I now turn to the second term. Observe that $\{D(t)=d, t(j,n+1) \geq t \} \cap \mathcal{F}(j,n)$ is $\mathcal{F}^-(t(j,n+1))$-measurable. Therefore,
\begin{align}
&P \left[ \widetilde{Y}(j,n+1)=y \middle| D(t)=d, t(j,n+1) \geq t, \widetilde{\mathcal{F}}(j,n) \right] \\
=& E \left[ P \left[ Y(t(j,n+1))=y \middle| \mathcal{F}^-(t(j,n+1)) \right] \middle| D(t)=d, t(j,n+1) \geq t, \widetilde{\mathcal{F}}(j,n) \right] \\
=& E \left[ P \left[ \widetilde{Y}(j,n+1)=y \middle| \widetilde{\mathcal{F}}(j,n) \right] \middle| D(t)=d, t(j,n+1) \geq t, \widetilde{\mathcal{F}}(j,n+1) \right] \\
=& P \left[ \widetilde{Y}(j,n+1)=y \middle| \widetilde{\mathcal{F}}(j,n) \right].
\end{align}
Therefore,
\begin{align}
&P\left[\widetilde{Y}(j,n+1)=y, D(t)=d, t(j,n+1) \geq t \middle| \widetilde{\mathcal{F}}(j,n)\right] \\
=& P \left[\widetilde{Y}(j,n+1)=y \middle| \widetilde{\mathcal{F}}(j,n) \right] P\left[D(t)=d, t(j,n+1) \geq t \middle| \widetilde{\mathcal{F}}(j,n) \right].
\end{align}
Therefore, adding up the identities for the two terms, I get
\begin{align}
P\left[\widetilde{Y}(j,n+1)=y, D(t)=d \middle| \widetilde{\mathcal{F}}(j,n)\right] = P \left[\widetilde{Y}(j,n+1)=y \middle| \widetilde{\mathcal{F}}(j,n) \right] P\left[D(t)=d \middle| \widetilde{\mathcal{F}}(j,n) \right].
\end{align}
I have thus shown that $\widetilde{Y}(j,n+1) {\perp\!\!\!\perp} D(t) | \widetilde{\mathcal{F}}(j,n)$, which implies that $D(t) {\perp\!\!\!\perp} \widetilde{\mathcal{F}}(j,n+1)$.
The base case can be treated with the same arguments.
|
2025-03-21T14:48:30.505674
| 2020-05-06T08:51:43 |
359523
|
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|
Stack Exchange
|
Derivative of the Bott-Chern forms
The Bott-Chern forms are constructed formally in Bismut's "Analytic Torsion and Holomorphic Determinant Bundle I" (page 74). This construction can be found as well in "Lectures on Arakelov Geometry" page 79.
These forms are used in Donaldson's "ANTI SELF-DUAL YANG-MILLS CONNECTIONS OVER COMPLEX ALGEBRAIC SURFACES AND STABLE VECTOR BUNDLES" to define a functional on the space of metrics on a holmorphic vector bundle $E\rightarrow X$. In that setup we consider the acyclic complex
$$\mathcal{E}: \ 0\rightarrow (E,h_1)\xrightarrow{Id} (E,h_2) \rightarrow 0 $$
and we define $\widetilde{ch}(h_1,h_2)=\widetilde{ch}(\mathcal{E})$.
Donaldson writes the properties of the Bott-Chern forms in that constext and states that if $h_t$ is a path of metrics we have the following identity:
$$\dfrac{\partial}{\partial t}\widetilde{ch}(h_t,h_1)= -Tr(h_t^{-1}h_t'exp(i/2\pi F_{h_t})).$$
(this exact identity is taken form Itoh and Nakajima's "Yang-Mills Connections and Einstein-Hermitian Metric" p450).
I wonder how to obtain that property from the aforementioned abstract construction.
In order to construct the Bott-Chern class $\widetilde{ch}(h_0,h_1)$ we deform the Id complex over $\mathbb{P}^1$
\begin{equation*}
0 \rightarrow (E,\tilde{h}) \rightarrow (E,h_1) \rightarrow 0.
\end{equation*}
so that $i_0^*\tilde{h}=h_0$ and $i^*_{\infty}\tilde{h}\simeq h_1$. We get then
\begin{equation*}
\widetilde{ch}(h_0,h_1) = \int_{\mathbb{P}^1}ch(\tilde{h})\log|z|^2
\end{equation*}
We now take a path of metric $h_t$ instead of $h_1$, with the corresponding $\tilde{h}_t=h_0(\tilde{g}_t\cdot,\cdot)$. Differentiating along that path we obtain
\begin{eqnarray*}
\dfrac{\partial}{\partial t}\widetilde{ch}(h_0,h_t) &=& \dfrac{\partial}{\partial t}\int_{\mathbb{P}^1}ch(\tilde{h}_t)\log|z|^2\\
&=& \int_{\mathbb{P}^1}\dfrac{\partial}{\partial t}ch(\tilde{h}_t)\log|z|^2\\
&=& -\int_{\mathbb{P}^1}Tr(\partial\bar{\partial}(\tilde{g}_t^{-1}\tilde{g}_t')exp(-\tilde{F}_t))\log|z|^2
\end{eqnarray*}
where $\tilde{g}_t\in\Omega^0(X\times \mathbb{P}^1,End(E))$ and $\tilde{F}_t=F_{\tilde{h}_t}\in \Omega^2(X\times \mathbb{P}^1,End(E))$. Now notice that $\tilde{F}_t$ hence $exp(-\tilde{F}_t)$ are $\partial$ and $\bar{\partial}$ closed so we have
\begin{equation*}
Tr(\partial\bar{\partial}(\tilde{g}_t^{-1}\tilde{g}_t')exp(-\tilde{F}_t)) = \partial\bar{\partial} Tr(\tilde{g}_t^{-1}\tilde{g}_t'exp(-\tilde{F}_t)).
\end{equation*}
We can then keep going
\begin{eqnarray*}
\dfrac{\partial}{\partial t}\widetilde{ch}(h_0,h_t) &=& -\int_{\mathbb{P}^1}Tr(\partial\bar{\partial}(\tilde{g}_t^{-1}\tilde{g}_t')exp(-\tilde{F}_t))\log|z|^2\\
&=& -\int_{\mathbb{P}^1}\partial\bar{\partial}(Tr\tilde{g}_t^{-1}\tilde{g}_t'exp(-\tilde{F}_t))\log|z|^2\\
&=& \int_{\mathbb{P}^1}Tr(\tilde{g}_t^{-1}\tilde{g}_t'exp(-\tilde{F}_t))\partial\bar{\partial}\log|z|^2\\
&=& \int_{\mathbb{P}^1}Tr(\tilde{g}_t^{-1}\tilde{g}_t'exp(-\tilde{F}_t))\partial_z\bar{\partial}_z\log|z|^2\\
&=& -2\pi i (\delta_0-\delta_{\infty})[Tr(\tilde{g}_t^{-1}\tilde{g}_t'exp(-\tilde{F}_t))].
\end{eqnarray*}
Recall that $(\tilde{h}_t)_{|0}=h_0$, so that $(\tilde{g}_t)_{|0}=Id$. We get therefore $Tr(\tilde{g}_t^{-1}\tilde{g}_t'exp(-\tilde{F}_t))_{|0}=0$. We obtain
\begin{eqnarray*}
-2\pi i (\delta_0-\delta_{\infty})[Tr(\tilde{g}_t^{-1}\tilde{g}_t'exp(-\tilde{F}_t))] &=& 2\pi i Tr(g_t^{-1}g_t'exp(-F_{h_t})).
\end{eqnarray*}
|
2025-03-21T14:48:30.505874
| 2020-05-06T09:00:29 |
359524
|
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"Federico Barbacovi",
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/359524"
}
|
Stack Exchange
|
Does direct image via proper map preserve coherence of unbounded complexes?
As for the title, I'm considering a proper map $f : X \rightarrow Y$ of Noetherian schemes and I'm trying to understand whether the direct image $Rf_{\ast} : D_{qc}(X) \rightarrow D_{qc}(Y)$ sends the subcategory $D_{coh}(X)$ to $D_{coh}(Y)$. Here the subscripts means we are taking complexes with quasi coherent/coherent cohomologies.
I know that the statement is true if we restrict to bounded below complexes. The way I thought I could prove this is using the spectral sequence
$$
R^pf_{\ast}(H^{q}(A)) \implies R^{p+q}f_{\ast}(A)
$$
for any $A \in D_{coh}(X)$. The problem is that this spectral sequence converges only weakly, which means that the filtration on $R^{p+q}f_{\ast}(A)$ whose graded pieces are the sheaves on the $E^{\infty}$ page may be infinite.
In particular, even if everything on the $E^{\infty}$ page is coherent (and this is the case because in page $2$ we have coherent sheaves concentrated in a vertical strip of finite width), we can't claim coherence of $R^{p+q}f_{\ast}(A)$.
Any help?
According to Hartshorne's Residues and Duality, ch. II, Proposition 2.2, this holds if $X$ has finite Krull dimension — this implies $R^pf_*(A)=0$ for $p$ greater than some fixed integer and any sheaf $A$, and one concludes with standard arguments.
I checked out the proposition, and it doesn't say much on the proof. I don't understand why from the fact that $R^pf_{\ast}(A)$ vanishing for $p$ big enough the result should follow. I was implicitly assuming it in what I wrote above, and I couldn't get to the end anyway. It seems to me that the fact that the filtration might be infinite and not stabilise to zero creates problems.
|
2025-03-21T14:48:30.506038
| 2020-05-06T09:47:10 |
359525
|
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"Giles Gardam",
"Maxime Ramzi",
"Philippe Gaucher",
"Robbie Lyman",
"https://mathoverflow.net/users/102343",
"https://mathoverflow.net/users/135175",
"https://mathoverflow.net/users/15728",
"https://mathoverflow.net/users/24447",
"https://mathoverflow.net/users/24563",
"qkqh"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/359525"
}
|
Stack Exchange
|
injectivity of pushout?
We have the following pushout diagram:
$$\begin{array}{ccc} \langle X, Y \rangle & \xrightarrow{\alpha} & \mathbb{Z}_a \ast \mathbb{Z}_b \ast \mathbb{Z}_c \ast \mathbb{Z}_d \\ \downarrow \scriptsize{\beta} && \downarrow \scriptsize{g} \\ \mathbb{Z}_e & \xrightarrow{f} & G\end{array}$$
Suppose that $\alpha$ is injective, and $\beta$ maps $X,Y$ to $1\in \mathbb{Z}_e$. ($\mathbb{Z}_n$ means the cyclic group of order $n$.)
I wonder that in which cases, $f$ is injective.
For any positive integers $a,b,c,d,e \geq 2$? Otherwise, only for relative prime or distinct prime $a,b,c,d,e$?
(I found that if $\phi$ is injective, then $\psi$ is injective in "adhesive" categories. The category of abelian groups is adhesive, and the category of groups isn't. I also wonder which subcategory of groups are adhesive. )
$$\begin{array}{ccc}A & \xrightarrow{\phi} & B \\ \downarrow && \downarrow \\ C & \xrightarrow{\psi} & D\end{array}$$.
In the category of groups, there is a counterexample:
http://math.stackexchange.com/questions/601463/a-monomorphism-of-groups-which-is-not-universal
I can change my diagram similarly to the above counterexample:
$$\begin{array}{ccc}\mathbb{Z} & \xrightarrow{\alpha'} & (\mathbb{Z}_a \ast \mathbb{Z}_b \ast \mathbb{Z}_c \ast \mathbb{Z}_d)/\langle X=Y \rangle \\ \downarrow \scriptsize{\beta'} && \downarrow \scriptsize{g'} \\ \mathbb{Z}_e & \xrightarrow{f} & G\end{array}$$
$\alpha'(1)=X$, any words in $X,Y$ are nontrivial in $\mathbb{Z}_a \ast \mathbb{Z}_b \ast \mathbb{Z}_c \ast \mathbb{Z}_d$, and $\beta'$ is surjective. Then, $\alpha'$ is also injective. For instance, fix $Y=1_a 1_b 1_c 1_d$ with a generator $1_n$ of $\mathbb{Z}_n$. May $f$ not be a monomorphism?
The pushout of a one-to-one map in the category of sets is one-to-one. In the category of groups, my guess is that you have to add the free compositions. So I would be surprised that $f$ is not one-to-one.
What has this got to do with topos theory ?
@MaximeRamzi I found that the above injectivity holds in any topos (toposes are adhesive). I cannot understand what topos is, but I think it may provide some answer. If topos theory has nothing to do with my question, sorry. Then, tell me. I will subtract it.
@PhilippeGaucher What does it mean to add the free compositions in the category of groups? Could you explain it in detail?
@qkqh I've learnt something today, I could not imagine that there would be a counterexample. In my mind, a group is a one-object category, and the pushout must contain all possible compositions.
The first exercise in Serre’s book Trees spells this out (I’ve corrected a missing “normal”):
Let $f_1\colon A \to G_1$ and $f_2\colon A \to G_2$ be two homomorphisms and let $G$ be their pushout. We define subgroups $A^n$, $G^n_1$ and $G^n_2$ of $A$, $G_1$ and $G_2$ recursively by the following conditions:
$A^1 = \{1\}, \qquad G^1_1=\{1\}, \qquad G^1_2 = \{1\}$
$A^n = $ normal subgroup of $A$ generated by $f^{-1}_1(G^{n-1}_1)$ and $f^{-1}_2(G^{n-1}_2)$
$G^n_i = $ normal subgroup of $G_i$ generated by $f_i(A^n)$.
Let $A^\infty$, $G^\infty_i$ be the unions of the $A^n$, $G^n_i$, respectively. Show that $f_i$ defines an injection $A/A^\infty \to G_i/G^\infty_i$ and that $G$ may be identified with the amalgam of $G_1/G^\infty_1$ and $G_2/G^\infty_2$ along $A/A^\infty$.
It follows (using the results of no. 1.2 in Trees) that the kernel of $A\to G$ is $A^\infty$ and that the kernel of $G_i \to G$ is $G_i^\infty$.
Thank you very much!! Applying the theory to my lower diagram ($f_1:=\alpha', f_2=\beta'$), $A^n=A^\infty=e\mathbb{Z}, G_1^n=G_1^\infty=\langle \langle X^e \rangle \rangle, G_2^n=G_2^\infty=1$ for any $n>1$, and so $A/A^\infty=\mathbb{Z}_e \to G_1/G_1^\infty=G$ is injective. Hence, $f$ is alway injective!! Wow! Could you tell me some references to study it?
I want to insert missing subindex "1" on the codomain of $f_1$ in your answer, but I couldn't. I should fix at least 6 characters.
I've found that it is a book. However, I cannot find the nomality condition that you've modified.
In the exercise as written Serre does not say that $A^n$ and $G^n_i$ should be the normal subgroups generated by the conditions above. I spent a full day confused about this; I seem to recall that one family of subgroups was likely normal, but couldn't see any reason for the other family to be normal.
also, I really love Trees, I think it's a great introduction to a subject I like a lot.
I remember getting stumped by this omission of normal as well. Serre does say dist. (for distingué) in the French original so it was lost in (the generally excellent) translation.
Yes, we need the normality condition only to ensure that $A/A^\infty, G_i/G_i^\infty$ become groups. But, sadly, my first comment was wrong; I do not know what $f_1^{-1}(\langle\langle X^e \rangle\rangle)$ is. (On the above counterexample, $\langle\langle h^n \rangle\rangle$ is the whole group.) That's what I've wanted to know: ${X^n} \cap \langle\langle X^e \rangle\rangle_{G_1}$. I'm back to square one.
|
2025-03-21T14:48:30.506388
| 2020-05-06T10:20:12 |
359527
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/359527"
}
|
Stack Exchange
|
Fourier transform of Green function and its derivative
Consider a real Sturm-Liouville operator $L$ on $[0,+\infty)$ and use the following notations : https://www.encyclopediaofmath.org/index.php/Titchmarsh-Weyl_m-function
Assume $a = 0$, $\alpha \in [0,\pi)$ is fixed, $w \equiv 1$ and let's say that we are in the limit-point case so that $m(\lambda)$ is unique and for $\beta \in [0,\pi)$ fixed,
$$m(\lambda) = \ell_{\infty}(\lambda) = \lim \limits_{b \to +\infty}\ell_b(\lambda)=\lim \limits_{b \to +\infty}-\frac{\cot\beta\,\phi(b,\lambda)+p(b)\phi'(b,\lambda)}{\cot\beta\,\psi(b,\lambda)+p(b)\psi'(b,\lambda)}$$
For $0 < b \leq +\infty$, consider the Green function on $[0,b]$
$$G_b(x,y,\lambda) := \begin{cases} \phi(x,\lambda)\chi_b(y,\lambda) \quad 0 \leq x \leq y \leq b\\
\chi_b(x,\lambda)\phi(y,\lambda) \quad b \geq x > y \geq 0 \\
0 \quad x,y \gt b
\end{cases}$$
In Sturm-Liouville and Dirac Operators from Levitan & Sargsjan, at page 57, it is said that the Fourier transform of $$\dfrac{dG}{dx}(x,y,z)$$ (where $G = G_{\infty}$) is exactly
$$\int_0^{+\infty}\dfrac{dG}{dx}(x,y,z)\phi(y,\lambda)dy = \dfrac{\phi'(x,\lambda)}{(z-\lambda)}$$
but at this point in the book (at this point, we have proved the limit-point and limit-circle theorem, an integral expansion theorem for arbitrary functions and an integral representation for the resolvant. For example, the surjectivity of the Fourier transform is not proved), it is unclear why this is true. Indeed, in the regular case, we have
$$\int_0^{b}\dfrac{dG_b}{dx}(x,y,z)\phi(y,\lambda_{n,b})dy = \dfrac{\phi'(x,\lambda_{n,b})}{(z-\lambda_{n,b})}$$
where the $\lambda_{n,b}$'s are the eigenvalues of the Sturm-Liouville operator $L$ restricted to $[0,b]$.
In the book, it is said that it suffices "to take the limit $b \to +\infty$", but I do not understand why this should work.
|
2025-03-21T14:48:30.506656
| 2020-05-06T10:31:56 |
359529
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/359529"
}
|
Stack Exchange
|
Is the cone $\Sigma(T)$ orthogonal to the singular support of a distribution?
Hello I am totally new to microlocal analysis and I have a question. Is the cone $\Sigma(T)$ orthogonal to the singular support of a distribution?
$$\xi \notin \Sigma(T) \iff \exists V\ conic\ neighbourhood\ of\ \xi;\ \forall \eta \in V, \ \forall N,\ |\widehat{T}(\eta)|\le C_N (1+|\eta|)^{-n}$$
|
2025-03-21T14:48:30.506719
| 2020-05-06T10:36:54 |
359530
|
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/359530"
}
|
Stack Exchange
|
Hopf algebras structure and quantum affine algebras
I'm looking for some information about the Hopf algebras structure and the quantum groups.
In particularly I was wondering if (and eventually where) is defined in the case of quantum affine algebras of twisted type in terms of Drinfeld generators.
Thank you for any suggestion.
|
2025-03-21T14:48:30.506769
| 2020-05-06T10:47:17 |
359531
|
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"authors": [
"Jeremy Rickard",
"Mare",
"Oeyvind Solberg",
"Pedro",
"https://mathoverflow.net/users/130741",
"https://mathoverflow.net/users/21326",
"https://mathoverflow.net/users/22989",
"https://mathoverflow.net/users/61949"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/359531"
}
|
Stack Exchange
|
Higher analogue of the Auslander-Bridger transpose
Let $A$ be an Artin algebra and $M$ a module with $Ext^i(M,A)=0$ for $i=1,...,n-2$.
Then in case $P_{n-1} \rightarrow ... \rightarrow P_0 \rightarrow M \rightarrow 0$ is the beginning of a minimal projective resolution of $M$, we get an exact sequence
$0 \rightarrow M^{*} \rightarrow P_0^{*} \rightarrow ... \rightarrow P_{n-1}^{*}$
Here $(-)^{*}=Hom_A(-,A)$.
In case $n=2$, the cokernel of the map $P_0^{*} \rightarrow P_1^{*}$ is the well known Auslander-Bridger transpose $Tr(M)$ of $M$.
Question: Is there a name or nice expression for the cokernel $P_{n-2}^{*} \rightarrow P_{n-1}^{*}$ for general $n \geq 3$?
Isn't your $\mathrm{Tr}(M)$ what people call the Auslander--Reiten transpose? Or am I missing something? (Perhaps just different nomenclature?)
@PedroTamaroff I think Tr is called Auslander-Bridger transpose (which exists for any noetherian ring) while $\tau$=DTr is called Auslander-Reiten translate for Artin algebras.
Ah! I see. Thanks! :)
You probably already know that the cokernel of $P_{n-2}^{} \rightarrow P_{n-1}^{}$ is $\text{Tr }\Omega^{n-2}M$. I don't know a name, but Auslander and Bridger gave it a symbol, $J_{n-2}M$.
@JeremyRickard Thanks, one can find that in their book. I would hope that somewhere in the literature there has appeared an official name for that and not just a symbol. (maybe in the theory of general noetherian rings, which I do not follow very closely)
If you apply the duality to $\operatorname{Tr}\Omega^{n-2}(M)$, then you get the $(n-1)$-Auslander-Reiten translation of $M$ (see https://www.sciencedirect.com/science/article/pii/S0001870806001721).
|
2025-03-21T14:48:30.506899
| 2020-05-06T11:13:21 |
359535
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/359535"
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|
Stack Exchange
|
Fastest way to calculate the eigenvalues of a product of two Toeplitz matrices
I have the following problem:
I need to find the fastest way to calculate the eigenvalues of a matrix that is the product of two Toeplitz matrices. $B = A U$.
The first is a regular Toeplitz matrix $A$, while the second is an upper triangular matrix $U$.
I thought to use the QR decomposition on $A$ in such a way that $A U = (Q R) U = Q (R U)$
Since the product of two upper triangular matrices is an upper triangular matrix I would still get a QR type factorisation.
Now I have two questions:
Is this the best way to calculate eigenvalues or are there more efficient ways since they are structured matrices?
If the proposed way is correct, which is the best algorithm to factorize the Toeplitz A matrix? Is it better to transform the Toeplitz matrix into a Hessenberg matrix?
Thank you very much
|
2025-03-21T14:48:30.506988
| 2020-05-06T11:20:09 |
359537
|
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"Stéphane Laurent",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/359537"
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|
Stack Exchange
|
Associativity property of the gyrobarycenter
I'm using Ungar's terminology and notations. In the open unit ball of $\mathbb{R}^n$, let $GB(A_1, \ldots, A_N; m_1, \ldots, m_N)$ be the gyrobarycenter of the points $(A_1, \ldots, A_N)$ with gyrobarycentric coordinates $(m_1, \ldots, m_N)$.
I numerically found that (for $n=2$, I have not checked for other values of $n$)
$$
GB(A_1,A_2,A_3;1,1,1) = GB\bigl(GB(A_1,A_2;1,1), A_3; 2\gamma_{(-A_1)\oplus A_2}, 1\bigr)
$$
and
$$
GB(A_1,A_2,A_3;2,2,1) = GB\bigl(GB(A_1,A_2;1,1), A_3; 4\gamma_{(-A_1)\oplus A_2}, 1\bigr).
$$
I also found that there is $p$ not depending on $A_3$ such that
$$
GB(A_1,A_2,A_3;2,1,1) = GB\bigl(GB(A_1,A_2;2,1), A_3; p, 1\bigr)
$$
but I didn't find the expression of $p$.
So it looks like there is an associativity property of the gyrobarycenter. What is this property? There is nothing about that in Ungar's books.
EDIT
The gyrobarycenter (I'm interested in the Möbius gyrovector space) is defined as
$$
GB(A_1, \ldots, A_N; m_1, \ldots, m_N) =
\frac{1}{2} \otimes
\frac{\sum m_k \gamma^2_{A_k} A_k}
{\sum m_k \Bigl(\gamma^2_{A_k}-\frac{1}{2}\Bigr)}.
$$
The $\gamma$ factor of a point $X$ is defined by
$$
\gamma_X = \frac{1}{\sqrt{1-\Vert X \Vert^2}}.
$$
The scalar multiplication $\otimes$ is defined at page 10 of this paper, and the gyroaddition $\oplus$ at page 9.
@MattF. Ok, done.
|
2025-03-21T14:48:30.507120
| 2020-05-06T11:57:18 |
359538
|
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"url": "https://mathoverflow.net/questions/359538"
}
|
Stack Exchange
|
Upper density of subsets of an amenable group
Let $G$ be an amenable group (so locally compact Hausdorff) and also assume it is second countable if needed. My question is that what are the standard ways (across literature) of defining the upper density for a subset of $G$?
If we are talking about $\mathbb{N}$, then given a Folner sequence $\mathcal{F}=\{F_n: n\in \omega\}$ (basically each $F_n$ is finite and for any $m\in \mathbb{N}$, $\lim_{n\to \infty} \frac{|(m+F_n)\Delta F_n|}{|F_n|}=0$), then the upper density associated with $\mathcal{F}$ can be $\bar{d}_\mathcal{F}(A)=\limsup_{n\to \infty} \frac{|A\cap F_n|}{|F_n|}$.
One way to define the density is to replace the counting measure by the Haar measure $\mu$ on $G$ and the Folner sequence in a more general sense ($F_n$ are compact sets now) and we can say $\bar{d}_\mathcal{F}(A)=\limsup_{n\to \infty} \frac{\mu^*(A\cap F_n)}{\mu(F_n)}$, where $\mu^*$ is the outer measure (I want to define density for any subset $A$).
Another way is to say $\bar{d}(A)=\sup\{\alpha: \text{for every finite }H\subset G \text{ and $\varepsilon>0$, there is a finite $K$ with }\frac{|hK\Delta K|}{|K|}<\varepsilon\; \forall h\in H \text{ and }\frac{|A\cap K|}{|K|}\geq \alpha\}$.
I'm not a group theorist nor topologist and hopefully this question is okay here. Thanks.
One difficulty is the dependence on the Følner sequence. For instance, take $X=\bigcup_n [2^n,2^n+3n]$, and $F_n=\bigcup_{k=0}^n[2^k+n,2^k+2n[$. Then $F_n$ has cardinal $n^2$ and boundary of cardinal $2n$, so is Følner. But $F_n\subset X$ while $X$ has natural density zero. Only forcing $F_n$ to cover won't be enough to fix the issue (change $F_n$ to $F_n\cup [-\sqrt{n},\sqrt{n}]$ if necessary). In $\mathbf{Z}^2$ one has less artificial examples with $F_n=[-n,n]\times [-n^2,n^2]$ and $X={(x,y):|x|\le |y|}$ which has density zero w.r.t this choice.
So maybe one way is to look over \emph{all} Folner sequences?
The problem with this is that you have disjoint ones, i.e. two Følner sequences and two disjoint subsets which have density 1 for one and zero for the other; in particular the density will never be independent of the Følner sequence apart from being in ${0,1}$. The notion "subset that has density 0 for every Følner sequence" is thus quite restrictive. Still it seems not void, i.e., not reduced to finite subsets (excludes $\bigcup [2^n,2^n+n]$, but seems to include $\bigcup{2^n}$).
By the way I forgot to say, but we can always consider $\sup m(X)$ when $m$ ranges over invariant means. Probably it means the same as the supremum over all Følner sequences of the upper densities wrt the Følner sequence. Nevertheless it's frustrating that it maps $\bigcup [2^{2^n},2^{2^n}+n]$ to 1.
More generally, in $\mathbb{Z}$ or $\mathbb{N}$, $\sup m(X)=1$ if and only if $X$ contains arbitrarily large intervals, which is equivalent to the notion of a "thick" set (as defined in my answer below).
In the case that $G$ is discrete, what you've defined last as $\bar{d}(A)$ is usually considered the upper (Banach) density of $A$. Moreover, in this case $\bar{d}(A)$ is the supremum of $\bar{d}_{\mathcal{F}}(A)$ over all Folner sequences $\mathcal{F}$ (or nets in the uncountable case).
Recall also that a discrete group $G$ is amenable if and only if it admits at least one Folner sequence (or net) if and only if it admits a left-invariant finitely additive probability measure on its subsets. So another equivalent formulation of $\bar{d}(A)$ in this case is:
$$
\bar{d}(A)=\sup\{\mu(A):\text{$\mu$ is a left-invariant finitely additive probability measure on $G$}\}.
$$
In the non-discrete case, all of the same facts are true if we use compact sets instead of finite sets. In particular, a locally compact group $G$ (with Haar measure $\eta$) is amenable if and only if for any compact $H\subseteq G$ and $\epsilon>0$, there is a Borel set $K\subseteq G$, with $0<\eta(K)<\infty$, such that $\eta(hK{\vartriangle} K)/\eta(K)<\epsilon$ for all $h\in H$.
In this case, if one defines $\bar{d}(A)$ analogously (for Borel $A$), then
\begin{align*}
\bar{d}(A) &= \sup\{\bar{d}_{\mathcal{F}}(A):\text{$\mathcal{F}$ is a Folner net for $G$}\}.\\
&= \sup\{\mu(A):\text{$\mu$ is a left-invariant finitely additive Borel probability measure on $G$}\}.
\end{align*}
A final comment is that in groups like $\mathbb{Z}$, there are other "canonical" upper densities, such as the upper asymptotic density:
$$
\bar{\delta}(A)=\limsup_n|A\cap [\text{-}n,n]|/(2n+1).
$$
But notice that this is just $\bar{d}_{\mathcal{F}}(A)$ for a particular choice of Folner sequence in $\mathbb{Z}$. So in more general groups $G$, one can work with upper densities defined by "special" Folner sequences too.
Edit: The discussion above I think motivates some further remarks about combinatorics. For simplicity, let $G$ be a countable discrete amenable group. Then the notion of "density $0$ with respect to every Folner sequence", i.e. $\bar{d}(A)=0$, is a useful notion of "sparse". For example Szemeredi's Theorem can be rephrased as saying that (in $\mathbb{Z}$) if $\bar{d}(A)>0$ then $A$ contains arbitrarily large finite arithmetic progressions).
For some easier facts, define the lower Banach density with respect to a Folner sequence $\mathcal{F}$ to be $\underline{d}_{\mathcal{F}}(A)=\liminf|A\cap F_n|/|F_n|$. The lower Banach density is then
$$
\underline{d}(A)=\inf\{\underline{d}_{\mathcal{F}}(A):\text{$\mathcal{F}$ a Folner sequence}\}
$$
Then some basic combinatorial facts are:
$\bar{d}(A)=1-\underline{d}(G\backslash A)$ for any $A$ (this works at the level of Folner sequences.
$\underline{d}(A)>0$ if and only if $G=FA$ for some finite $F\subseteq G$ (i.e., $A$ is syndetic).
$\bar{d}(A)=1$ if and only if for any finite $F\subseteq G$, there is some $g\in G$ such that $Fg\subseteq A$ (i.e., $A$ is thick).
A final combinatorial notion is that of a piecewise syndetic set, which is a set $A$ such that $FA$ is thick for some finite $F\subseteq G$. Then another fact that is used a lot is that if $A$ is piecewise syndetic then $\bar{d}(A)>0$ (but the converse fails in general).
We also have the following facts.
$\underline{d}(A)=\inf\{\mu(A):\text{$\mu$ is a left-invariant finitely additive probability measure on $G$}\}$.
For any set $A$ there are Folner sequences $\mathcal{E}$ and $\mathcal{F}$ such that $\bar{d}(A)=\bar{d}_{\mathcal{F}}(A)$ and $\underline{d}(A)=\underline{d}_{\mathcal{F}}(A)$.
In additional to Szemeredi's Theorem, other well-known results are:
(Jin) If $A,B\subseteq\mathbb{Z}$ are such that $\bar{d}(A),\bar{d}(B)>0$, then $A-B$ is piecewise syndetic. This is a more difficult variation of the easier fact that if $\bar{d}(A)>0$ then $A-A$ is syndetic.
(Moreira, Richter, Robertson) If $A\subseteq\mathbb{Z}$ is such that $\bar{d}(A)>0$ then $A$ contains $B+C$ for some infinite sets $B$ and $C$. This was originally conjectured by Erdos as a replacement for the failure of a density version of Hindman's Theorem.
so $\bar{d}(A)$ is only defined for $\eta$-measurable $A$'s?
Yes, probably. I have to admit that my familiarity with this largely focused in the discrete case (I'll add this for safety).
thanks for the info. I was thinking maybe we can do outer measure in the definition of $\bar{d}_\mathcal{F}$, then it will make sense for all subsets. But maybe this will yield too many paradoxical examples.
That's a good question. There may be some discussion of this in Paterson's book "Amenability". Unfortunately my copy is in my office, which I can't access at the moment. I see now that I didn't fully appreciate this part of your question. So perhaps someone will post a better answer.
I guess one thing is that, if your precise definition of $\bar{d}(A)$ is well-defined in the sense that for any finite $H$ and $\epsilon>0$ there is such a $K$, then $G$ is amenable as a discrete group. So $\bar{d}(A)$ is the upper Banach density with respect to the counting measure, and can be applied to all subsets of $G$.
yeah that definition completely ignores the Haar measure, which does not seem to do justice to many things.
Just to add something to the already well written answer, there's an account of the relation between Szemeredi's theorem and Furstenberg-Zimmer's theorem in the excellent book of Kerr and Li about ergodic theory, where they not only treat $\mathbb{Z}$, but general discrete groups acting on locally compact Hausdorff spaces.
So Thickly syndetic sets fit in this picture? $A$ is thick if for any finite $F \subset G$, $\cap_{f \in F} fA \neq \emptyset$. The notion "thickly syndetic" is defined by: $A$ is thickly syndetic if for any finite $F \subset G$, $\cap_{f \in F} fA$ is a syndetic set.
@ARG Yes this notion fits with the others. A set is thickly syndetic if and only if it's complement is not piecewise syndetic (somewhat analogous to how a set is thick if and only if its complement is not syndetic).
@GabeConant thanks! so if $A$ is such that $\overline{d}(A^{\mathsf{c}}) =0$, then $A$ is thickly syndetic (but the converse fails)?
@ARG Yes, that's correct.
There is a dual description of $\overline d$ which goes back to Banach and Følner (in the general countable case), see Theorem 6 of Granirer. For locally compact groups it should be more or less the same if you deal with the means defined on $L^\infty$ with respect to the Haar measure.
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2025-03-21T14:48:30.507691
| 2020-05-06T12:32:41 |
359542
|
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|
Stack Exchange
|
Analyticity of the semigroup generated by a time-changed Brownian motion
Let $d$ be an integer. We denote by $m$ the Lebesgue measure on $\mathbb{R}^d$. We define $BL(\mathbb{R}^d)$ by
\begin{align*}
BL(\mathbb{R}^d)=\{f \in L^2_{\rm loc}(\mathbb{R}^d,m) \mid |\nabla f|\in L^2(\mathbb{R}^d,m)\},
\end{align*}
where $L^2_{\rm loc}(\mathbb{R}^d,m)$ denotes the space of locally square integrable function on $\mathbb{R}^d$. For $f \in L^2_{\rm loc}(\mathbb{R}^d,m)$, we write $\nabla f$ for the distributional derivative. For $f,g \in BL(\mathbb{R}^d)$, we define $\mathcal{E}(f,g)=\int_{\mathbb{R}^d}\nabla f \cdot \nabla g\,dm$, where $\cdot$ denotes the standard inner product on $\mathbb{R}^d$.
We take a positive continuous function $V\colon \mathbb{R}^d \to \mathbb{R}$,which may be unbounded. We set
\begin{align*}
\mathcal{F}=\left\{f \in BL(\mathbb{R}^d) : \int_{\mathbb{R}^d}f^2\,Vdm<\infty \right\}
\end{align*}
We can show that $(\mathcal{E},\mathcal{F})$ becomes a Dirichlet form on $L^2(\mathbb{R}^d,V\,dm)$. Therefore, $(\mathcal{E},\mathcal{F})$ generates a strongly continuous contraction semigroup $\{T_t\}_{t>0}$ on $L^2(\mathbb{R}^d,V\,dm)$, which is extended to a contraction semigroup on $L^{\infty}(\mathbb{R}^d,V\,dm)$. The extension is still denoted as $\{T_t\}_{t>0}$. In fact, $\{T_t\}_{t>0}$ is identified with the semigroup of a time-changed Brownian motion, and we can show that $\{T_t\}_{t>0}$ is a strongly continuous contraction semigroup on $C_{0}(\mathbb{R}^d)$. Here, $C_{0}(\mathbb{R}^d)$ stands for the space of continuous functions vanishing at infinity.
My question
Can we show that $\{T_t\}_{t>0}$ is extended to an bounded analytic semigroup on $C_{0}(\mathbb{R}^d)$? The generator of $\{T_t\}_{t>0}$ is given by $\frac{1}{V}\Delta$, where $\Delta$ is the Laplacian on $\mathbb{R}^d$.
I am interested in whether analyticity of semigroups are stable under time change, which is one of the most fundamental transformations of stochastic processes.
If I understand correctly, the question is about analyticity of the semigroup generated by $\frac{1}{V} \Delta$ in spaces of continuous functions.
@GiorgioMetafune Thank you for your comment. That's right. Certainly, this is a question for the analyticity of $\Delta/V$.
I notew that in the functional analysis community it is also usual to denote this space with $C_0$, your $C_\infty$ might be a source of confusion. See https://en.wikipedia.org/wiki/Function_space#Functional_analysis
@AndrásBátkai Thank you for your comment. I would like to write $C_{0}$ in stead of $C_{\infty}$.
I find that the answer is no. Let us work on the half line $(0,\infty)$ with Dirichlet boundary conditions at $0$; however the problems come from $\infty$. Let $L=a(x)D^2$ where $a=1/V$ is supposed to be smooth, positive and $a(0)=1$ and consider the change of variable $s=\phi(x)=\int_0^x \frac{1}{\sqrt{a(t)}} dt$. If $u$ is a $C^2$-function, then $$a(x)u_{xx}=u_{ss}-D_x (\sqrt{a})u_s.$$ To be more precise, the change of variable above induces an isometry $T: C_0([0,\infty[) \to C_0([0,\ell]$, $\ell=\int_0^\infty\frac{1}{\sqrt{a(t)}} dt$ given by $Tu(s)=u(\phi^{-1} s)$ and such that $TLT^{-1}=M$, where $$M=D_{ss}-D_x(\sqrt{a})D_s.$$ This chanhe of variable simplifies the diffusion but adds a drift $D_x(\sqrt a)$ which, however, should be written in the variable $s$. Next we choose $a$ in such a way that $\ell=\infty$ and $D_x(\sqrt {a})=s$. Letting $b=\sqrt {a}$ this leads to the Cauchy problem $$b''=\frac{1}{b}, \quad b'(0)=0, \quad b(0)=1.$$
This equation can be solved almost explicitely by multiplying by $b'$ and using the initial values, thus leading to $$\int_0^{\sqrt{\log b(x)}} e^{t^2}dt =\frac{\sqrt {2}}{2} x.$$ However, one can see directly from the equation that $b$ is globally defined for $x \ge 0$, positive, increasing and convex. Finally
$$
\ell=\int_0^\infty \frac{1}{b}=\int_0^\infty b''=\lim_{x \to \infty}b'(x).$$
If $\ell <\infty$, then $b(x) \le 1+\ell x$ and again $1/b$ is not integrable near $\infty$. With this choice of $b=\sqrt{a}$ the operator $M=D_{ss}-sD_s$ is the Ornstein-Uhlenbeck in the half-line which is known not to be the generator of an analytic semigroup. By similarity, the same happens for $L$.
Hoping it is correct. I find very interesting the question and let me point out that it seems that the counterexample cannot be obtained by using powers: if $a(x)=x^\alpha$, then $D_x(\sqrt{a})\approx 1/s$ with the above notation and the semigroup is analytic (also the singularity for small $s$ can be treated by using Bessel functions). It is not very clear to me what is behind.
Thank you very much for your very kind reply. I feel there is some relationship between the analyticity of Laplacian with unbounded drift and that of the generator of a time-changed Brownian motion. Your answer is very interesting!
|
2025-03-21T14:48:30.508010
| 2020-05-06T13:24:10 |
359545
|
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|
Stack Exchange
|
Polynomial satisfying a functional equation
I am currently stuck with the following question:
Let $q$ be a polynomial of degree $n+1$ with distinct positive zeros $x_0, ... , x_n$. Find a polynomial $p \in P_n$ that satisfies the functional equation $q(x)p(x) + q(-x)p(-x) = 1$ for all $x$ in $\Re$. Is such a polynomial unique?
Would appreciate your kind explanations! Thank you!
You can solve the problem by looking at the smallest (or largest coefficient first) and then working your way up (or down). For example, by setting $x=0$ you get $p(0) = 1/2q(0)$. So you can try to work the coefficients of $p$ one by one. For an abstract argument, consider $a_i$ to be the coefficients of $p$ (you have $n+1$ of them). Then $F(x) = q(x)p(x)$ has degree $2n+1$. When looking at $F(x) + F(-x)$ all monomials of odd degree cancel. There remains $n+1$ even degree monomials. So $n+1$ linear constraints and $n+1$ variables. There is a solution (use distinct roots for uniqueness).
Nice exercise, though I doubt that it can be considered as a research question. Put $y=x^2$, and write
$q(x)= a(y)+xb(y)$. I claim that the polynomials $a$ and $b$ are coprime: if a non-constant polynomial $f$ divides $a$ and $b$, then $f(x^2)$ divides $q(x)$, but that implies that some of the roots $x_i$ of $q$ are negative. Note also that since the roots are positive $y$ does not divide $a$. Therefore
there exist polynomials $c,d$ such that $a(y)c(y)+yb(y)d(y)=\dfrac{1}{2} $. Put $p(x)=c(x^2)+xd(x^2)$; then $q(x)p(x)=\dfrac{1}{2}+ x g(x^2) $, with $g= ad+bc$, so $q$ answers the question.
As usual you can replace $c(y),d(y)$ by $c(y)-yb(y)e(y), d(y)+ a(y)e(y)$ for some polynomial $e$, so the solution is far from unique.
Thank you! I have understood!
Out of curiosity, are we able to use Lagrange interpolation formula for this question? @abx
I don't think so. For one thing, there are many solutions for $p$, with different zeroes.
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2025-03-21T14:48:30.508195
| 2020-05-06T14:00:28 |
359550
|
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"Dmitri Pavlov",
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|
Stack Exchange
|
Čech cocycles and monodromy
It is well known that over a topological space $X$ (and choosing an open cover $\mathfrak{U}$) every locally constant Cech cocycle $g$ on $\mathfrak{U}$ with coefficients in a group $G$ yields a $G$-covering space $X_g \rightarrow X$. As such, the monodromy action of this covering space gives a homomorphism from the fundamental group $\pi_1(X,x)$ on a point $x\in X$ to $G$.
I am trying to write this homomorphism explicitly in terms of the cocycle $g$. In his Bachelor Thesis, Lemma 5.5, M.P. Noordman claims that this can be done in the following way. You consider a loop $\sigma:[0,1]\rightarrow X$ and apply the Lebesgue number lemma to get a finite subcover $\{U_1,...,U_n\}$ of $\mathfrak{U}$ and a partition $t_0<t_1<...<t_n$ of the interval $[0,1]$ in such a way that $\sigma([t_{i-1},t_i])\subset U_i$. Now, you can define the homomorphism $f:\pi_1(X,x)\rightarrow G$ as
$$
f([\sigma])=g_{12} g_{23} g_{34} \cdots g_{(n-1) n}.
$$
However, it is not clear to me why this does not depend on the choice of the "Lebesgue subcover" $\{U_1,...,U_n\}$ or on the choice of the representative of the class $[\sigma]$.
For example, consider the case where $\mathfrak{U}=\{U,V,W\}$ consists on three open sets with $U\cap V \neq \varnothing$ and $U,V \subset W$. If we choose a path contained in $U\cup V$, we could choose the Lebesgue covering to be $\{U,V\}$, which would yield $f([\sigma])=g_{UV}$ or we could choose the covering to be simply $\{W\}$, which would yield $f([\sigma])=1$, and I do not see why these should coincide.
A Cech cocycle always produces a bundle which is trivializable on each open set in the cover, so its monodromy on any loop contained in the open set should be trivial.
Also, you need to choose the first and last open sets to be the same for this formula to be correct.
First, one can pull back the Čech cocycle to S^1 and work directly with S^1
instead of X.
Any two open covers have a common refinement,
so it suffices to show that the monodromy map does not change
under passing to refinements.
As already pointed out in the comments, the open cover must be cyclic: $U_0=U_n$.
By induction, it suffices to show invariance under adding a single new point
$t_{i-1}<t_{i-1/2}<t_i$ with $[t_{i-1},t_{i-1/2}]⊂σ^*U_{i-1/2}$ and $[t_{i-1/2},t_i]⊂σ^*U_i$.
But this invariance is precisely what the Čech cocycle condition says for the open
sets $U_{i-1}$, $U_i$, and $U_{i-1/2}$: $$g_{i-1,i}=g_{i-1/2,i}g_{i-1,i-1/2}.$$
Great answer, thank you! However, what is the point about pulling back to S^1?
@G.Gallego: After pulling back, similar arguments can be used to define holonomy of bundles with connection.
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2025-03-21T14:48:30.508409
| 2020-05-06T14:26:19 |
359553
|
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"Igor Belegradek",
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|
Stack Exchange
|
Extrinsic horizontal path lifting
As a follow up question to my previous question about the orthonormal frame bundle, I would like to understand a simple example explicitly.
Let $\mathbb{S}^2$ be written extrinsically as $$\mathbb{S}^2 = \{x\in\mathbb{R}^3|\|x\|=1\}$$ and let an arbitrary smooth path $w:[0,1]\to\mathbb{R}^2$ be given.
The ultimate goal is to lift $w$ to a path $\gamma:[0,1]\to\mathbb{S}^2$ which has the same "energy", i.e. $$ \langle\dot{w},\dot{w}\rangle_{\mathbb{R}^2} \stackrel{!}{=} g_\gamma(\dot{\gamma},\dot{\gamma}) $$ on $[0,1]$, where $g$ is the Riemannian metric on $\mathbb{S}^2$ (which, as written here, is induced by the Euclidean metric on $\mathbb{R}^3$.
I presume eventually there would be some choice of (arbitrary) initial conditions and a (1st oder?) ODE to solve in order to obtain a path $\gamma:[0,1]\to\mathbb{S}^2\subseteq\mathbb{R}^3$.
I tried to follow this in a systematic way according to the prescription of: 1) building an orthonormal frame bundle $O\mathbb{S}^2$ on top of $\mathbb{S}^2$, 2) lifting $w$ to a horizontal path $\tilde{\gamma}:[0,1]\to O\mathbb{S}^2$, and 3) projecting down from $O\mathbb{S}^2$ to $\mathbb{S}^2$. I tried to do all of this extrinsically without using charts, and that's where I got stuck (perhaps this is a pointless endeavor, but I thought one point of using the frame bundle is to work with global objects rather than within charts).
Question 1: Is there a better procedure to achieve this goal rather than follow the horizontal path lifting? Perhaps something more explicit in this particular setting.
Question 2: How to follow the horizontal path lifting procedure extrinsically in this case? Here's how I got stuck:
Define the orthonormal frame bundle extrinsically as $$ O\mathbb{S}^2 = \{ (x,A) \in \mathbb{R}^3\times\mathbb{R}^{9} | x\in\mathbb{S}^2 \land A \in O(3) \text{ s.t. }Ax=x\}\,. $$ In the case of the sphere it's easy to picture that the fiber is one dimensional ($\dim(O(2))=1$) and amounts to the angle by which to rotate a basis of a 2D tangent space to each point on the sphere.
Now we need to define the tangent bundle of this, $$TO\mathbb{S}^2 = \{ (x,A,v_x,v_a) \in \mathbb{R}^3\times\mathbb{R}^{9}\times\mathbb{R}^3\times\mathbb{R}^{9} | (x,A)\in O\mathbb{S}^2\land \langle x,v_x\rangle+\langle A,v_a\rangle=0\}\,.$$ and its horizontal sub-bundle $HO\mathbb{S}^2 = ???$, find the two vector fields $H_1,H_2$ that build a global frame for $HO\mathbb{S}^2$, I guess they are called the canonical horizontal vector fields. This is the step where I get stuck because as far as I know, to check that a curve $u:[0,1]\to O\mathbb{S}^2$ is horizontal, I need to verify the equation $$ \nabla_{\dot{x}} v = 0 $$ for all columns $v$ in $A$ which are not equal to $x$, where $(x,A)=u$. Here $\nabla$ is the covariant derivative, which I understand in this extrinsic description is just the gradient along a vector projected to the tangent space of the manifold. So if $P_x = I - x\otimes x^\ast$, then the covariant derivative of two vector fields $a,b$ equals $$(\nabla_a b)(x) = P_x (a_j \partial_j b)(x)\,.$$ Using this interpretation I find the equation for a horizontal curve to equal $$ P_x \dot{v}(x) = 0 $$ for any column $v$ in $A$ not equal to $x$. This stopped making sense to me.
How to find $H_1,H_2$ in this description? Is there any point to write them as elements of $TO\mathbb{S}^2 \subseteq \mathbb{R}^3\times\mathbb{R}^{9}\times\mathbb{R}^3\times\mathbb{R}^{9}$?
Solve the ODE $$\dot{\tilde{\gamma}} = \sum_{i=1}^2 H_i(\tilde{\gamma}) \dot{w_i}$$ for $\tilde{\gamma}$ and project $\tilde{\gamma}\mapsto\gamma$. Here the notation $H_i(\tilde{\gamma})$ means evaluate the vector field $H_i$ at the base point $\tilde{\gamma}$.
I haven't read the whole question but to talk about lifting you need a map $\mathbb S^2\to \mathbb R^2$. Which map are you using? Without a map the word "lift" has no meaning. In general if a map is a Riemannian submersion, then you can lift any path in the base to a horizontal path in the total space. Any Riemannian submersion of manifolds of the same dimension is a local isometry. Round sphere is not locally isometric to the Euclidean plane. What you call "energy" is actually the square of the speed. It depends on how you parametrize the curve. Any curve can be parametrize by unit speed.
@IgorBelegradek, thanks for the comment. I understand this depends on the parametrization, but I exactly want to capture that (i.e. not allow for any parametrization), but stick to a given one that whomever gave me $w$ chose. When I say lift (technically) I mean a lift of curves from the sphere to the orthonormal frame bundle. I think my usage of the word 'lift' to denote taking a curve in Euclidean space to the sphere was technically incorrect.
You need to edit the question so that it makes sense mathematically. Otherwise, it may get closed as "unclear what you are asking". It seems this is what you want: fix a unit speed parametrization of a curve in $\mathbb S^2$ and $\mathbb R^2$. Then any other parametrization differs from the fixed one by a self-map $\phi$, and then the new speed will be $|\phi^\prime|$. This is true both in $\mathbb S^2$ and $\mathbb R^2$. Am I missing something?
@IgorBelegradek, as far as I understood from the construction of the orthonormal frame bundle, there is a 1-1 correspondence between paths in $\mathbb{R}^2$ and paths in $\mathbb{S}^2$ in an isometric way, which doesn't require charts. The question is how to realize this 1-1 map explicitly, thinking extrinsically of the sphere embedded in $\mathbb{R}^3$, starting from a curve in $\mathbb{R}^2$. In your comment you fix two curves on the two spaces, but it's not clear (to me) how to relate the two
Maybe the chapter 'rolling without twisting or slipping' in Appendix B of Sharpe's Differential Geometry book might be an interesting read, it sounds similar to what you are trying to do, and can be applied extrinsically.
So I think I have an answer, but instead of using the ODE in Step 3, it uses a simpler equation that implies it: $$ \dot{w} = \tilde{\gamma}^{-1}\dot{\gamma}\,. $$
Here $w:[0,1]\to\mathbb{R}^2$ is a given curve, $\gamma:[0,1]\to\mathbb{S}^2$ is the unknown curve, and $\tilde{\gamma}$ is the horizontal curve in $O\mathbb{S}^2$ lifted from $\gamma$.
It turns out that it is after all quite easy to write the horizontal curve in $\tilde{\gamma}$ induced by a given $\gamma$ if one uses spherical coordinates (and later on one may switch back to Cartesian coordinates if need be). Then if $\theta,\varphi:[0,1]\to\mathbb{R}$ parametrizes the curve $\gamma$ in spherical coordinates, find $\psi:[0,1]\to\mathbb{R}$ out of the equation $$ \dot{\psi} = -\dot{\varphi}\cos(\theta)\,. \tag{H}$$
Then $\psi$ gives the angle of rotation compared with the standard orthonormal frame on $T_\gamma\mathbb{S}^2$ given by the (co-moving) orthonormal frame $\hat{\theta},\hat{\varphi}$.
Then for each $t\in[0,1]$, $\tilde{\gamma}(t)$ may be viewed as a map $$ \tilde{\gamma}(t):\mathbb{R}^2\to T_{\gamma(t)}\mathbb{S}^2 $$ which is in fact an isometric isomorphism by construction. In our case, parametrized by $\psi$, it is given by $$ \mathbb{R}^2\ni v\mapsto (R_\psi v)_1\hat{\theta}+(R_\psi v)_2\hat{\varphi} $$
where $$R_\psi=\begin{bmatrix}\cos(\psi) && -\sin(\psi) \\ \sin(\psi)&&\cos(\psi)\end{bmatrix}$$ is the $2\times 2$ rotation matrix associated with $\psi$. Hence there is an easy way to write the inverse map $$ \tilde{\gamma}(t)^{-1}:T_{\gamma(t)}\mathbb{S}^2\to\mathbb{R}^2 $$ which is given by $$ y_\theta \hat{\theta} + y_\varphi\hat{\varphi} \mapsto R_\psi^{-1}\begin{bmatrix}y_\theta \\ y_\varphi\end{bmatrix}\in\mathbb{R}^2\,. $$
Now $$\dot{\gamma} = \dot{\theta}\hat{\theta}+\sin(\theta)\dot{\varphi}\hat{\varphi}$$ and so this finally yields the following ODE to be solved for the unknowns $\theta,\varphi$: $$ \dot{w} = R_\psi^{-1}\begin{bmatrix}\dot{\theta} \\ \sin(\theta)\dot{\varphi}\end{bmatrix} $$ where $\psi$ is also a function of $\theta,\varphi$ via (H).
|
2025-03-21T14:48:30.509393
| 2020-05-06T15:26:33 |
359560
|
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|
Stack Exchange
|
Is there an infinite order $\mathbb{F}_{p}$-section for a certain elliptic surface $\mathcal{E}_n$?
Consider a natural number $n$, a finite field $\mathbb{F}_{p}$ (such that $p$ is prime, $p \equiv 1 \ (\mathrm{mod} \ 3)$, $p \equiv 3 \ (\mathrm{mod} \ 4)$, and $\sqrt[3]{2} \notin \mathbb{F}_p$), and the elliptic $\mathbb{F}_{p}$-surface
$$
\mathcal{E}_n\!: y^2 = x^3 + (t^{6n} + 1)^2.
$$
Here I asked about infinite order $\mathbb{F}_{p}$-sections of $\mathcal{E}_1$. Is there such a section of $\mathcal{E}_n$ at least for some natural $n$? In other words, is the Mordell-Weil group of $\mathcal{E}_n$ of positive rank for some $n$?
The answer seems to be negative, but I cannot prove this. I have only confirmed my conjecture by the computer algebra system Magma for small $p$ and $n$.
I would recommend to state your question clearly here rather than to refer to your previous question. Here your notations are different from the previous question.
You define an elliptic surface, but in order to talk about sections one needs a morphism (do you mean $(x,y,t)\mapsto t$ ?). Also, do you mean regular sections or rational sections? (I expect rational sections).
Yes, I mean the projection $(x,y,t) \mapsto t$ as an elliptic fibration and I would like to consider rational sections.
|
2025-03-21T14:48:30.509516
| 2020-05-06T16:00:13 |
359563
|
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}
|
Stack Exchange
|
Global splitting field for algebras
Let $A$ be a finite dimensional algebra.
A field $K$ is a splitting field for an indecomposable $A$-module $M$ in case the local algebra $End_A(M)/(rad(End_A(M))$ is 1-dimensional.
$K$ is called a global splitting field for an algebra $A$ in case every indecomposable $A$-module splits.
Question: Is there a concrete example of a representation-infinite algebra $A$ over a finite field that is a global splitting field for $A$?
Answer by Jeremy Rickard: No.
This motivates the follow up question:
Question: Is a field $k$ algebraically closed if and only if it is the global splitting field of a representation-infinite $k$-algebra $A$?
Let $k$ be a finite field and $A$ a representation-infinite finite dimensional $k$ algebra. By the second Brauer-Thrall conjecture, $\bar{k}\otimes_kA$ has infinitely many nonisomorphic indecomposable modules of some dimension, and so has some that are not defined over $k$ (i.e., not of the form $\bar{k}\otimes_kM$ for any $A$-module $M$). However, such a module is defined over some finite field extension of $k$: namely, the extension generated by its structure constants.
So we have a finite extension $K$ of $k$, and an indecomposable $K\otimes_kA$-module $N$ that is not of the form $K\otimes_kM$ for any $A$-module $M$.
Let $X$ be any indecomposable direct summand of the restriction $\text{Res}^{K\otimes_kA}_A(N)$.
Since $K\otimes_k\text{Res}^{K\otimes_kA}_A(N)$ is the direct sum of $|K:k|$ Galois conjugates of $N$, $K\otimes_kX$ is the direct sum of some of the Galois conjugates of $N$, and more than one, since $N$ is not defined over $k$.
Hence (using the fact that $K$ is a separable extension of $k$, so that extending scalars to $K$ commutes with taking radicals of algebras)
$$\dim_k\text{End}_A(X)/\text{rad}(\text{End}_A(X))
=\dim_K\text{End}_{K\otimes_kA}(K\otimes_kX)/\text{rad}(\text{End}_{K\otimes_kA}(K\otimes_kX))$$
is greater than $1$.
Thanks, I asked the question for general fields now (that was a question in brackets before). It seems your proof works for any perfect field (equivalently fields where any finite extension is seperable)?
@JeremyRickard What does your first sentence mean when A is specialized to A=k? I am not understanding something.
@NicholasKuhn The question has a hypothesis "representation-infinite". Sorry, I forgot to carry that into the answer: I'll fix that. The second Brauer-Thrall conjecture, now a theorem, states that, for a representation infinite algebra over an infinite field, there is some $d$ for which there are infinitely many nonisomorphic indecomposable modules of dimension $d$.
|
2025-03-21T14:48:30.509715
| 2020-05-06T16:10:40 |
359564
|
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|
Stack Exchange
|
How to solve a non-local self-consistent equation
I have been struggling lately with solving numerically an equation of the form:
$$ g(x\pm x_{0}) = F[ g(x) ] $$
where $g(x)$ is a matrix satisfying the condition $g(x\to\pm\infty)=0$. My question is on how does one solve this recursively. I am running over some predefined sets $S=\{x_{n}\}_{n=1}^{N}$ of arguments where the function is supposed to be evaluated, so ideally the algorithm should run over a loop that yields:
$$g(x_{1}\pm x_{0})$$
$$g(x_{2}\pm x_{0})$$
$$ \vdots$$
$$g(x_{N}\pm x_{0})$$
in this order with $x_{1}<x_{2}< ... <x_{N}$ for $x<0$ $\forall x\in S$ and $x_{1}>x_{2}> ... >x_{N}$ for $x>0$ $\forall x\in S$. Any help would be greatly appreciated thanks!
Edit: For clarity, consider the example:
$$A(x-x_{0})=B(x) + C(x)A(x)D(x)$$
and we know $A(x\to\infty)=0$ and B,C,D are matrices that also depend on $x$.
More of an example:
Let's say I start with a value $ x = -a<0$. My desired goal is to calculate a matrix product for that specific value of x that has the form
$$ M(x) = inv( D(x) + T(x)g_{1}(x - x_{0})C(x) + S(x)g_{2}(x+x_{0})P(x) ) $$ where all quantities are matrices but the only unknown ones are $g_{1},g_{2}$. Note that $g_{1},g_{2}$ represent two different matrices. Now, what we know about these matrices is that they satisfy $g_{1,2}(x \to \pm \infty) = 0$. And, additionally, the recursion relation of each of them ( $g_{1} $ and $g_{2}$) has the form written in the edit. More specifically:
$$g_{1}(x-x_{0}) = H( g_{1}(x) )$$
$$ g_{2}(x+x_{0}) = H(g_{2}(x))$$
and both need to be evaluated for that specific value of $x=-a$. My question is on how to do that, since eventually, I want to do it not only for $x=-a<0$, but for an entire grid of $\{x\}$ that includes both positive and negative values.
What is the meaning of $\pm$ in your $x\pm x_0$? Do the signs generate two different equations that have to be considered simultaneously? If not, choose $x+x_0$ for example and rewrite your equation as the recurrence $g(x) = F[g(x-x_0)]$. Then, assigning an arbitrary value to $g(x)$ on the interval $[a,a+x_0)$, the recurrence uniquely defines $g(x)$ for any $x>a+x_0$.
Thanks for the quick reply. The $\pm$ sign represent independent cases. For a fixed value of $x$, I want to solve for the case $x-x_{0}$ and $x+x_{0}$.
So I guess my suggestion was not what you were looking for? But then your "for clarity" edit writes exactly the same kind of equation as I wrote. There's still something not very clear about the question. What is known (the input to the problem) and what is unknown (the desired output) about $g(x)$?
Aha, the problem is that the recurrence relation leaves $g_1$ and $g_2$ highly underdetermined, unless their values are known on some interval of width $x_0$. It seems that you want to fix that underdeterminacy by the boundary conditions $g_{1,2}(x\to\pm \infty) = 0$. That might work, but it means that you need to solve the recurrence relation on the entire real line (or at least an infinite sub-grid) before being able to determine the value of $g_{1,2}(x)$ at any given $x$. So there is no algorithm that will give you the solution that involves only finitely many points ${x_n}_{n=1}^N$.
So a proposed solution was to start with the boundary condition as initial value, and then iterate the solution, but I am having serious difficulties trying to understand this and how this might work numerically. Because what I want to do eventually is to evaluate $g_{1,2}$ on my fixed set of points ${\x_{n}}$ to later carry out the matrix product, which has to be one to one, that is, for a fixed value of $x$ of my original grid ${x_{n}}$ I want to know $g_{1}(x-x_{0})$ in order to carry out the products.
Just to clarify, so far, I have not made any suggestion for how to handle your boundary value problem numerically. I can only mention that your boundary value problem seems to have much in common with singular boundary value problems for ODEs (which includes unbounded domains), and their numerical solution. I would recommend looking into that literature for inspiration. But there is no unique approach, and the approaches would strongly depend on the $B(x), C(x), D(x)$ coefficients.
|
2025-03-21T14:48:30.510029
| 2020-05-06T16:16:48 |
359566
|
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|
Stack Exchange
|
Diagonally similar to submatrix of orthogonal matrix
Given $A \in \mathbb{R}^{n \times n}$, with $0 < |\det(A)| < 1$. Does a diagonal matrix $D$ exist such that
$$
B = D^{-1} A D
$$
is the principal submatrix of an $(n+1)\times(n+1)$ orthogonal matrix, i.e.,
$$
X = \begin{bmatrix} B & b \\ c & d \end{bmatrix} \textrm{ with } X^T X = I.
$$
This is equivalent to $B$ having singular values all 1 except one, i.e.,
$$
\sigma(B) = [1, 1, \dots, 1, |\det(A)| ] = s^*.
$$
There is, of course, a non-diagonal $D$ which satisfies the problem for any $A$. For diagonal $D$, however, not all $A$ do have a solution, e.g., a diagonal $A$ with $\sigma(A) \neq s^*$. I'd be interested also in a numerical algorithm to find $D$ (if it exists).
|
2025-03-21T14:48:30.510118
| 2020-05-06T18:26:54 |
359574
|
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"John Rognes",
"Nicholas Kuhn",
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|
Stack Exchange
|
Are the AHSS and Adams spectral sequence the same when computing connective Morava K-theory of a space?
Let $k(n)$ be the $n$th connective Morava K-theory, with $k(n)_* = \mathbb F_p[v]$ where $|v| = 2p^n-2$. If $X$ is a space or a spectrum (assumed bounded below), one can compute $k(n)_*(X)$ using either the classical Adams spectral sequence or the even more classical Atiyah-Hirzebruch spectral sequence.
Both spectral sequences are spectral sequences of modules over $k(n)_*$. (In the ASS the bidegree of $v$ is $(1,2p^n-1)$.) Both spectral sequences start with $k(n)_* \otimes H_*(X;\mathbb F_p)$, at $E_1$ for the ASS, and $E_2$ for the AHSS. Both have first possible nontrivial differential given by the formula $ d(x) = vQ_n(x)$,
where $Q_n$ is the $n$th Milnor primitive in the Steenrod algebra (acting on homology by going down in degree by $2p^n-1$).
So it seems that these must really be the same spectral sequence. Is this true? Is this a fact in the literature? (I am a tad bothered by the fact that the AHSS arises from an increasing filtration of $k(n) \wedge X$ while the ASS arises from a decreasing filtration of $k(n) \wedge X$.)
Hi Nick, I think this is true. The AHSS for $E_* X$ can either use a cellular filtration of X (starting at $E_2$) or the Postnikov filtration of E (starting at $E_1$), and "shearing" the spectral sequence accounts for the increasing/decreasing difference -- I think this is in Appendix B of Greenlees-May's "Generalized Tate cohomology". The result should then follow from the fact that the Postnikov tower for k(n) is an Adams tower, because mutliplication by $v_n$ becomes null after smashing with $H$.
@TylerLawson: this is perfect! And that Appendix is well written and definitive. (And I am pretty sure Peter taught me this in grad school, though that was awhile ago!)
The comparison of the exact couples (and spectral sequences) derived from the cellular and Postnikov filtrations goes back to:
Maunder, C. R. F.
The spectral sequence of an extraordinary cohomology theory.
Proc. Cambridge Philos. Soc. 59 (1963), 567–574.
In terms of publication dates, the Adams spectral sequence (Comment. Math. Helv., 1958) predates the Atiyah-Hirzebruch spectral sequence (PSPM III, 1961). Is there an earlier reference for the AHSS?
@JohnRognes I've always thought of the paper you mention as the `source' of the AHSS. So I guess you are right, the Adams SS is curiously older than the AHSS. The Atiyah-Hirzebruch paper is also introducing us to generalized cohomology theories. Just a year later, in 1962, Adams is happily computing with the AHSS in his Vector Fields on Spheres paper!
@JohnRognes Nice to know that Maunder reference, and it is amusing to see that this 1962 paper starts by saying that it is `well known' that there is a spectral sequence [like the AHSS]. One wonders if the AHSS was already well known earlier (without a name) by folks working with stable (co)homotopy in the early 50's.
Tyler's comment answers my question. A bit more detail: the Postnikov tower of $k(n)$ is an Adams resolution, because the `bottom class' map $k(n) \rightarrow H\mathbb F_p$ is onto in mod $p$ cohomology; indeed $H^*(k(n);\mathbb F_p) = A_p//E(Q_n)$.
The appendix by Greenlees and May has the details that two spectral sequences converging to $\pi_*(Y \wedge X)$, one coming from filtering $Y$ by its Postnikov tower and the other by filtering $X$ by its skeleta, agree.
|
2025-03-21T14:48:30.510408
| 2020-05-06T18:31:54 |
359575
|
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|
Stack Exchange
|
Closed graph theorem for cones?
In the paper "A strong open mapping theorem for surjections from cones onto Banach spaces, Marcel de Jeu and Miek Masserschmidt, Adv. Math." it is proved (among other things) that if $X, Y$ are complex Banach spaces, $C\subset X$ is a closed cone and $T:C\mapsto X$ is a continuous additive posotively homogeneous and surjective map, then it is open.
As far as I can see, if one follows the proof of the closed graph theorem from the open mapping mutatis mutandis, one can arrive at a " closed graph theorem for cones". I.e. for $X,Y,C$ as before and $T:C\subset Y \mapsto X$ is additive, positively homegeneous and has closed graph (in the topology of $X \times Y$) then there exists $K>0$ such that $\Vert T c\Vert \leq K \Vert c \Vert, \,\,\, \forall c \in C $. Although I am bit suspicious because nothing like this is mentioned in the paper.
Am I missing something ?
But then is you have a projection from the graph in $X\times Y$ onto $C$, not the whole space, so you cannot conclude that it is invertible
@erz yes, of course you are right
|
2025-03-21T14:48:30.510527
| 2020-05-06T20:45:29 |
359583
|
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|
Stack Exchange
|
Closure theorem for weak limits of "foliation currents"
A "foliation current" in the sense of Ruelle-Sullivan (https://www.math.stonybrook.edu/~ebedford/PapersForM655/RS.pdf) is essentially a closed subset of a manifold foliated by equidimensional oriented submanifolds along with a transverse measure, so integration makes sense: just integrate along each leaf, then integrate via the transverse measure.
I was wondering (although I don't have much hope) if a weak-* limit of a sequence of foliation currents is again a foliation current? Perhaps we would need to put some dimensional assumptions to exclude bad formations of singularities, or expand our notion of foliation current to a more singular one.
|
2025-03-21T14:48:30.510627
| 2020-05-06T21:32:06 |
359585
|
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|
Stack Exchange
|
Is there a formula with one free variable in NBG that defines a class that does not exist?
This question concerns Godel's Theorem on existence of classes in Set Theory of von Neumann–Bernays–Gödel.
This theorem implies that for any formula $\varphi(x)$ with one free variable $x$ whose quantifiers are of the form $\forall x\in\mathbf U$ or $\exists x\in\mathbf U$ the class $\{x:\varphi(x)\}$ exists.
Question. Is there a formula $\varphi(x)$ with one free variable in the language of set theory for which the class $\{x:\varphi(x)\}$ does not exist?
Added in Edit. Reading the comments I realized that such a formula indeed exists. So, I reformulate my question to
Question'. Find a (relatively) short formula $\varphi(x)$ for which the existence of the class $\{x:\varphi(x)\}$ cannot be proved in NBG.
I need such a formula to stress that the restricting quantifiers to run over sets in Godel's theorem on existence of classes is essential. Desirably to have such a formula as simple as possible.
Which theorem of Godel do you have in mind? What is $\mathbf{U}$? With "the class does not exist" do you mean it provably (in NBG) doesn't exist, or one for which it is consistent that it doesn't exist? When saying "the language of set theory" do you admit quantifiers over all classes or just ones over sets?
If you admit quantifiers over all classes, then the answer is as follows: for provable nonexistence, the answer is probably no, as that would imply Morse-Kelley set theory is inconsistent. If you mean consistent nonexistence, the answer is positive, as it's consistent that the set of truths about $V$ (which is definable in the language of classes) doesn't exist, as its existence implies consistency of ZF, and hence NBG.
@Wojowu: "as its existence implies consistency of ZF, and hence NBG", is it understood to mean "as its existence would imply consistency of ZF, and hence of NBG, is provable in ZF"?
@Qfwfq Existence of the truth class literally implies consistency of ZF, so if the existence was provable, so would be consistency.
@Wojowu, I think that @Qfwfq was referring to the grammar, not the logic. That is, "its existence implies consistency of ZF" seems to imply that it does exist, and so we can deduce consistency of ZF; but the subjunctive mood ("its existence would imply") says that we don't know it exists, and explains what we could deduce if we did.
@LSpice "The existence of the truth class" is (formalizable as) a sentence in the language of NBG, and that sentence implies the (formalization of) the sentence "ZF is consistent".
@Wojowu For me $\mathbf U$ is the class of all sets, $\mathbf V$ is the class of well-founded sets, i.e., the union of the von Neumann cumulative hierarchy. The axiom of foundation (or else regularity) is equivalent to $\mathbf U=\mathbf V$. As I understood from your comment, a simple formula defining a non-existent class does not exist because it would imply the inconsistency of the Morse-Kelley Set Theory. This explanation is sufficient for me. Thank you.
@AndreasBlass So, the answer to my question in that the formula $\varphi(x)$ expressing that $x$ is a natural number encoding a theorem of ZF defines a class ${x\in\omega:\varphi(x)}$ whose existence cannot be proved. Right? But also cannot be disproved. And all such $\varphi$ should be of such form, i.e., encoding some paradox, like Berry paradox? Is there some short $\varphi(x)$ of this form? Because Russell uses a very short formula $x\notin x$ for his paradox.
No, the set of theorems of ZF does provably exist in weak fragments of Z. The truth set for $\langle V,{\in}\rangle$ cannot be proved to exist in NBG. That is, ${\phi:\langle V,{\in}\rangle\models\phi}$. Or possibly ${\langle\phi,x\rangle:\langle V,{\in}\rangle\models\phi(x)}$ (I’m not sure which of these Wojowu meant; neither class is provable to exist in NBG).
@EmilJeřábek Writing ${\phi:\langle V,\in\rangle \models \phi}$ did you have in mind ${n\in\omega:\langle V,\in\rangle \models \phi_n}$ where $\phi_n$ is the formula whose Godel number is $n$?
That’s a possibility, but when working in set theory, there is no particular reason to encode formulas by integers. You can directly work with their most natural syntactic definition, e.g. as strings over some alphabet, or as trees.
@EmilJeřábek But formulas are not elements of the universe $\mathbf V$ (which is built up from the empty set). So, you should encode them as sequences of natural numbers anyway.
No, I do not have to encode them as sequences of natural numbers. I can encode them using their most natural syntactic definition, e.g. as strings over some alphabet, or as trees.
@EmilJeřábek But anyway, you alphabet should be a subset of the universe. Because there is no such sets as $\wedge,\vee$ or $\forall$.
There is no such set as $\mathbb R$ either, until I define it. So just fix any sets you like and define them to be the basic symbols of the alphabet. This choice does not matter.
@EmilJeřábek There is a subtle difference between $\mathbb R$ and say $\forall$. Defining $\mathbb R$ you defined a concrete set which has properties of the real line. But defining $\forall$ you just choose any set to replace the symbol $\forall$ which is not the set. Exactly this I had in mind writing that natural numbers can be used for encoding symbols. But anyway, both of us understand what is going on. A good question though remains: find a short formula $\varphi$ for which the existence of the class ${x:\varphi(x)}$ cannot be proved in NBG.
No, it is exactly the same: you define a concrete set which has the properties of the symbol $\forall$. What are the properties of the symbol $\forall$? Answer: none save that it is distinct from the symbols $\land$, $\neg$, $=$, and $\in$.
@EmilJeřábek Ok. I agree. Thank you.
Concerning “simple” formulas: something equivalent to the existence of the truth set for $V$ can be expressed without syntax coding as follows. We consider coding of families of classes such that $X$ is a coded element of $Y$ if $X={x:\langle u,x\rangle\in Y}$ for some set $u$. Then (using the notation from https://mathoverflow.net/a/356979) you state “there exists a coded family of classes that contains $E$ and is closed under the operations $X\setminus Y$, ${\rm dom}(X)$, $X\times V$, $X^{-1}$, and circular permutation”. However, this is not a definition of a unique class by comprehension.
@EmilJeřábek Ok. Very good. So, this defines an indexed family of classes that contains all classes of constructible hierarchy. And its existence cannot be proved because of Russell and because of the consistency of $U=L$. Right?
No, this has nothing to do with $L$. I'm talking about operations on classes, not on sets. It gives an indexed family of classes that's closed under definability by first-order formulas without class quantifiers. It cannot be proved to exist in NBG because it implies the existence of a conventinal satisfaction predicate for $(V,\in)$, which, as already explained, implies the consistency of ZFC, which implies the consistency of NBG, which is not provable in NBG by Gödels's incompleteness theorem.
@EmilJeřábek Writing $L$ I had in mind not only constructible sets but also constructible classes (that is the minimal inner model of NBG). The same Godel' operations (with pairing applied only to sets) can be also applied to classes and produce "constructible classes" starting with the class $\mathbf E={\langle x,y\rangle:x\in y}$. So, the existence of non-constructible class (= non-definable) is not provable in NBG. Right?
But NBG does prove the existence of non-definable classes, such as the global well ordering of the universe. Furthermore, the argument I outlined applies all the same to any consistent extension of NBG that proves itself relatively consistent w.r.t. ZFC, disregarding whether it proves the existence of other nondefinable classes or not. It also applies to any consistent extension of NBG by sentences that only quantify over sets, such as $V\ne L$, or $V\ne HOD$.
The well-ordering of $L$ is definable in NBG, so NBG cannot prove that any well-ordering of the universe is non-definable.
I think the problem in the last two comments, about provable existence of non-definable classes, arises from the lack of associativity of the English language. @EmilJeřábek meant that (NBG proves $\exists X,\psi(X)$) for a certain formula $\psi$ such that, for no formula $\delta$ does NBG prove "there is a unique $X$ satisfying $\delta$ and this $X$ also satisfies $\psi$." Taras took instead the meaning that "For every formula $\delta$, NBG proves that, if there is a unique $X$ satisfying $\delta$ then that $X$ does not satisfy $\psi$."
@AndreasBlass From all the discussion above I realized that my innocent question was not so innocent. I just wanted after the theorem on the existence of classes in NBG (= Theorem of comprehension) in a textbook for students to add an exercise showing that the restriction of quantifiers to run only over sets in that theorem was indeed essential. But no simple formula (like $x\notin x$ in Russell's paradox) exists for this purpose because of Morse-Kelley. All such formulas need to encode that some proof or some construction does not exist and this not so simple as $x\notin x$.
I agree, because I know of no easy argument for why NBG isn't the same as MK, where "easy" means avoiding metamathematical facts like Gödel's incompleteness theorems. On the other hand, if one admits such facts, then you (though probably not your students) might enjoy Emil's, Joel's, and my answers at https://mathoverflow.net/questions/87238 .
|
2025-03-21T14:48:30.511424
| 2020-05-06T22:31:54 |
359586
|
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|
Stack Exchange
|
Genes mirror geography on a torus?
Disclaimer: this is an open-ended, imprecise question, asking for speculation in a topic that I know relatively little about (random matrix theory and principal component analysis). I originally asked it on MSE and have copied it over here. I wanted to air this curiosity, but if it is not appropriate for MO, flag it and let me know.
Two years ago, I participated in a directed reading group on the Marchenko-Pastur law, led by one of my favorite professors ever, Brent Nelson. To motivate the subject, he showed us a paper entitled "Genes Mirror Geography Within Europe." Since there is a paywall, I will briefly summarize this motivating result.
The researchers compiled genetic data from a group of Europeans, ensuring (whenever possible) that each person's grandparents were of the same nationality, so that an individual country of origin could reasonably be assigned. These genetic markers were compiled into a matrix $X$. Random matrix theory predicts a spectrum that the eigenvalues of $X^T\!X$ should closely fit to, if the genetic data were indeed random. However, two main outliers appeared, representing signals among the random noise. The eigenvectors corresponding to these eigenvalues gave two data points on each individual. When this data was plotted (color-coding individuals by their country of origin), the resulting scatter plot had a remarkable resemblance to a map of Europe (color-coded by country). I believe that the scale of the axes was adjusted to show this resemblance, but their angle was not (i.e. the eigenvectors represent orthogonal geographical directions). They say that this outcome "fits the theoretical expectation for models in which genetic similarity decays with distance in a two-dimensional habitat, as opposed to expectations for models involving discrete well-differentiated populations." They also note that the much larger outlying eigenvalue corresponds to a NNW direction, indicating that genetic variation is more strongly correlated with latitude than longitude (this makes sense to me, since the former has much more to do with climate).
Now Europe covers a small portion of the Earth's surface, so this may as well be happening on a plane instead of a sphere. There are several good reasons for this: most of the Earth's surface is water, where human's do not often reside; probably, more genetic and family-historical data is available to the researchers for Europeans than for other groups; and projects of colonialism, imperialism and enslavement have had their impact on geography, genetics and family history of people in many other parts of the world. But I am curious about what would happen if a similar analysis were performed on a population occupying a topologically different region.
From the researchers' quote about theoretical models, it seems reasonable that letting "genetic similarity [decay] with distance in [an $r$-dimensional] habitat" should produce $r$ principal components. In fact, I would be completely unsurprised if this were already common knowledge, which was just unknown to me. But the output (vectors of data-points) is inherently Euclidean, so I wonder what would happen if the populations were no longer simply evolving in a portion of Euclidean space. For example, if a species populated the entire surface of a planet, how would the geography of the spheroid be reflected in the principal component analysis? What about on a torus? Would there still be 2 principal components corresponding to the 2 dimensions of the surface, or would there be more? If there were 2 principal components, how would points of the surface be identified under this map? More broadly, since distance certainly plays a role, how would the Riemannian geometry of a surface be reflected in the data?
I would be interested to hear any guesses or speculation from those with a better understanding of principal component analysis or genetic variations than myself.
Not on a torus, but related: scientists have investigated the existence of "ring species" https://en.wikipedia.org/wiki/Ring_species
where genetic distance (at least as measured by ability of individuals to breed) does not reflect geometry/topology.
Thank you, that's interesting! So, to put this into topological terms, this gives a phenomenon that prompts speciation to occur more frequently on a non-simply connected region?
|
2025-03-21T14:48:30.511786
| 2020-05-06T22:41:00 |
359587
|
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|
Stack Exchange
|
Understanding moduli of shtukas of non-minuscule cocharacter
I have kind of a soft question. I've studied the basics of L. Lafforgue's proof of function field Langlands for GLn, and its use of the moduli of shtukas with two legs, and the cocharacters $[1,0,\ldots, 0]$ and $[0,0,\ldots,-1]$. I vaguely know that V. Lafforgue's approach to the general reductive group involves higher legs; does it also involve other cocharacters? For instance, when would moduli of shtukas, even just with one leg, with cocharacter $[2,0]$, say, be useful? Or are they easily described in terms of the moduli of shtukas with minuscule cocharacter?
This question is also partly motivated by the fact that locally, shtukas with one leg of cocharacter $[1,1,\ldots, 1, 0,0, \ldots, 0]$ are closely analogous to moduli of $p$-divisible groups via the Dieudonne functor. I'd like a similar intuition/idea for other dominant cocharacters.
For non-experts, maybe you could give some references? (I guess the intuition is likely to come only from experts, but still.)
a decent introduction to shtukas and l. lafforgue's work is laumon's https://arxiv.org/pdf/math/0003131.pdf. for more general cocharacters and groups, seciton 7 of v. lafforgue's survey https://arxiv.org/pdf/1803.03791.pdf might be helpful.
Thanks! Names and abstract links for your references: Laumon - La correspondance de Langlands sur les corps de fonctions; Lafforgue - Shtukas for reductive groups and Langlands correspondence for function fields.
Yes, in general you need to consider all cocharacters.
$\mathrm{GL}_n$ has the special property that the dominant coweights are all sums of minuscule cocharacters, i.e. ones of the form $[1,\cdots,1,0,\cdots,0]$ or $[0, \ldots, 0, -1,\ldots, -1]$. (For a general group, a dominant cocharacter is minuscule if its pairing with any positive root is at most $1$).
In terms of representations, this says that all irreducible algebraic representations can be built inside tensor products of exterior powers of the standard representation and its dual. A manifestation of this fact is the linear algebra theorem that a semisimple element of $\mathrm{GL}_n(\overline{\mathbf Q_\ell})$ is determined up to conjugation by the coefficients of its characteristic polynomial. It's true more generally for a reductive group $G$ over an algebraically closed field $k$ that a conjugacy class of semisimple elements $g \in G(k)$ is determined by the trace of $g$ acting on each irreducible representation. So if I want to construct a Galois representation into $G(\overline{\mathbf Q_\ell})$ (up to semisimplification), I need to know the traces on all irreps.
On the automorphic side, the image of the minuscule representations (representations with minuscule highest weight) under the Satake isomorphism $\mathrm {Rep}(\mathrm{GL}_n) \rightarrow \mathscr{H}$ generates the spherical Hecke algebra $\mathscr{H}$, so all Hecke eigenvalues of an eigenform are determined by these special elements.
Putting this together, to verify the (automorphic to Galois direction of the) Langlands conjecture for $\mathrm{GL}_n$, given a Hecke eigenform $f$, it suffices to construct a Galois representation such that the coefficients of the characteristic polynomial of Frobenius elements at each point are the eigenvalues of $f$ for the elements corresponding to minuscule characters.
For general groups $G$, the minuscule characters do not generate the dominant weights. Thus, in order to make the Langlands correspondence unique up to semisimplification, we must match the eigenvalues of $f$ under the Hecke operators corresponding to any representation $V$ of $\widehat{G}$ with traces of Frobenius elements acting on $V$ through $\widehat{G}$.
Loosely speaking, V. Lafforgue's approach to this is to construct a Galois representation for each $V$ by studying the cohomology of shtukas with legs bounded by the corresponding cocharacter of $G$. (There's the further complication that there are no shtukas with one leg and thus we have to consider tuples of representations/cocharacters and verify a ton of tricky compatibilities coming from adding or subtracting legs - this is one of the principal difficulties of his paper).-
On shtukas for $\mathrm{GL}_n$ for more general cocharacters:
We like minuscule cocharacters because the corresponding modifications of vector bundles are parametrized by the usual Grassmannian, which is a smooth projective variety. This means we don't need any perverse sheaves or intersection cohomology!
The fact that any dominant cocharacter of $\mathrm{GL}_n$ is a sum of minuscule ones lets us (locally) write an arbitrary modification of vector bundles as a sequence of modifications by minuscule cocharacters. If $\mu = \sum n_i \mu_i$ with $\mu_i$ minuscule, there is a resolution of the Schubert cell for $\mu$ in the affine Grassmannian by the "iterated affine Grassmannian" which parametrizes sequences of $\mu_i$ modifications. This space can be constructed as an iterated bundle of usual Grassmannians: in particular, it is smooth.
We can pull this back to an analogous resolution of the stack of shtukas with a leg bounded by $\mu$. One can prove that this map doesn't change the IC cohomology, so we don't get new Galois representations (I think this is right but I'm not 100% sure).
Aside: mixed characteristic
The fact that minuscule cocharacters generate everything for $\mathrm{GL}_n$ is essential to the proof of the local Langlands correspondence in that case! For similar reasons to above, this means that we can construct the required Galois representations by studying the cohomology of Lubin-Tate space.
$p$-divisible groups are "local shtukas with minuscule cocharacter" or "$p$-adic Hodge structures of weight $1$". The minuscule property is exactly what lets us parametrize these by an honest rigid-analytic space/formal scheme.
Recently, Scholze and Fargues have proposed an interpretation of more general shtukas in mixed characteristic, in the hopes that they could carry over Lafforgue's ideas to the $p$-adic setting. This uses Scholze's theory of diamonds, and involves a lot of very exotic geometry. (Meanwhile, local Langlands has been proved for many $p$-adic groups using different methods. The Scholze-Fargues program gives a very nice conceptual framework for the problem, but there are many fundamental foundational problems to be solved before it has any hope of working).
Also, I should mention that in the number field setting, Shimura varieties are some vague analogue of moduli of shtukas with minuscule legs - a minuscule cocharacter is part of the data defining the Shimura variety. This is a fundamental obstruction to "cohomologically" constructing a Langlands correspondence for general groups over number fields (never mind the fact that there's no hope of handling e.g. Maass forms with these sorts of algebro-geometric methods).
thanks very much for the detailed response! one more naive question: could you elaborate why there are "no shtukas with one leg"? in what setting is this true?
maybe you mean for GLn? i can see that, at least (reduce to the case of line bundles via determinants and then it's just basic facts about positivity), but it seems like there are shtukas with one leg for, say, PGLn. take for instance simply the natural ideal sheaf inclusion O(-p)^n -> O^n for some rational point p; this is a shtuka for PGLn, no?
For $\mathrm{GL}_n$, this should come from the fact that pulling back a vector bundle $\mathscr{E}$ on $X_S$ by $F = 1 \times \mathrm{Frob}_S$ doesn't change the degree, so an everywhere-defined morphism $F^*\mathscr{E} \rightarrow \mathscr{E}$ that only vanishes at one point must be an isomorphism. For general groups, I think you can just use the above argument via the Tannakian description of torsors.
as i said, i agree for GLn. i'm not sure i understand the second part, though i think i get why my example above doesn't work; do you have a reference?
I can't find a reference right now but the argument is: The category of $G$-torsors is the same thing as the category of faithful exact tensor functors from the category of representations of $G$ into vector bundles. So a $G$-shtuka is the same thing as a collection of $GL(V)$-shtukas for each representation of $V$ which are compatible with tensor products etc. Thus if I had a $G$-shtuka with one leg, all of the corresponding $\GL(V)$ shtukas would be trivial, and thus the original $G$-shtuka must be as well.
oh i see, very neat. i'm satisfied; thank you!
|
2025-03-21T14:48:30.512384
| 2020-05-06T22:41:20 |
359588
|
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|
Stack Exchange
|
Continuous functions on a compact $T_1$ space
Let $X$ be a compact $T_1$ topological space consisting of more than one point, and suppose that $X$ is locally compact (i.e. every point has a local base of compact neighbourhoods), second countable, and is a Baire space. Does $X$ admit a non-constant continuous function into the real numbers?
[In C$^*$-algebra terms, let $A$ be a non-primitive separable C$^*$-algebra whose primitive ideal space is compact and $T_1$. Does the multiplier algebra of $A$ have non-trivial centre?]
Why is the cofinite topology on $\Bbb N$ not a counterexample?
@Henno Brandsma. I don't think that it is a Baire space. The intersection of all the co-singleton sets is empty.
True, I see. Thx.
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2025-03-21T14:48:30.512475
| 2020-05-06T23:01:28 |
359590
|
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|
Stack Exchange
|
Convergence of free boundary minimal surfaces
I suspect the following statement is true:
Let $(M^3,g)$ be a compact and orientable Riemannian $3$-manifold with nonempty boundary. Let $\{\Sigma_n\}_{n \geq 1}$ be a sequence of compact and orientable free boundary minimal surfaces embedded in $M$. Suppose that there are positive constants $C_0$ and $C_1$ such that
(i) $\sup_{x \in \Sigma_n} \Vert A_{\Sigma_n}(x) \Vert^2 \leq C_0$ for every $n \geq 1$, where $A_{\Sigma_n}(x)$ denotes the second fundamental form of $\Sigma_n$ at the point $x$;
(ii) $\operatorname{Area}(\Sigma_n) \leq C_1$ for every $n \geq 1$.
Then there exists a subsequence $\{ \Sigma_{n_k} \}_{k \geq 1}$ of $\{ \Sigma_n \}_{n \geq 1}$ that converges smoothly and locally uniformly to an embedded free boundary minimal surface $\Sigma_{\infty} \subset M$.
Do you have a possible reference for this result?
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2025-03-21T14:48:30.512562
| 2020-05-06T23:42:09 |
359592
|
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"Emil Jeřábek",
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|
Stack Exchange
|
Double Diophantine approximation
Let $0 < \alpha < 1$. For any $n$ there is a closest lower Diophantine approximation $\max p / q \leq \alpha$ with integer $0 \leq p < q \leq n$. It can be found efficiently, e.g., with Stern-Brocot tree in $O(\log^2 n)$, assuming there is a constant-time oracle comparing any rational with $\alpha$. It is known that the best approximation satisfies $\alpha - p / q = O(n^{-2})$.
Let's introduce lower double Diophantine approximation $D_n(\alpha)$ as $\max p_1 / q_1 \cdot p_2 / q_2 \leq \alpha$ with integer $0 \leq p_i < q_i \leq n$ for $i = 1, 2$.
Q1. How fast can one find best lower double Diophantine approximation (assuming the constant-time comparison oracle)? Trivial solution is to sort all rationals $p / q$ with $p < q \leq n$ and use two-pointers search for $O(n^2 \log n)$ complexity. Can we do $o(n^2)$? $o(n)$?
Q2. Let $\Delta(\alpha, n)$ be $\alpha - D_n(\alpha)$. One can look for a bound $\Delta(\alpha, n) = O(n^{-c})$ uniform for all $\alpha$. A natural guess that $c = 4$ is sharp. Is it true, and how can one prove this?
Note that for a fixed $n$ we have $\max_{\alpha} \Delta(\alpha, n) = \Omega(n^{-2})$, attained at $\alpha = 1/n^2 - \varepsilon$.
What do you mean by $\alpha=1/n^2-\epsilon$? Surely $\alpha$ has to be a constant independent of $n$?
Ah, there are two different bounds we can speak of, I didn't clarify that enough.
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2025-03-21T14:48:30.512691
| 2020-05-07T01:16:47 |
359594
|
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|
Stack Exchange
|
Can information be extracted more precisely using more random trials?
Write $n$ iid draws of $(x,y)$ as $(x^n, y^n)$. Fix $R\in (0,H(x))$. What is the min of $n^{-1}H(y^n|f(x^n))$ over maps $f$ with range $\lbrace 1,\dots,\exp nR\}$, taking $n\to \infty$?
What is $H$? Any function of the data or something more specific? Also what do we know about the distribution of $(x, y)$?
$H$ is (conditional) Shannon entropy. $(x,y)$ is distributed over $\mathcal{X}\times \mathcal{Y}$, both sets finite.
My current thinking is to find $\min_s n^{-1}H(y^n|x^n \in s)$ over sets $s$ of $\exp(n(H(x)-R))$ sequences typical for $x^n$, taking $n\to\infty$. Then if one can design $f$ with fibers similar to the optimum $s$, the main question is solved. I would be surprised if this new min wasn't equal to the information bottleneck optimum, which would make design of $f$ straightforward.
$f$ is over $n$ vars jointly. The other case is already a single-letter problem whose answer can (in principle) be numerically estimated.
The characterization is given in terms of a so-called auxiliary random variable. It is as explicit of an answer as you'll get, unless you consider very special cases (like jointly Gaussian $X,Y$, or binary-valued $X,Y$). Namely, you have
$$
\lim_{n\to\infty}\min_{f: x^n \mapsto f(x^n)\in \{1,\dots,2^{nR}\}} \frac{1}{n}H(Y^n|f(X^n)) = \inf_{U : Y-X-U, I(X;U)\leq R} H(Y|U).
$$
In the expression on the right, we look over all random variables $U$, jointly distributed with $X,Y$, such that $Y$ and $U$ are conditionally independent given $X$, and $I(X;U)\leq R$, where $I$ denotes the usual mutual information.
This is a special case of the indirect source coding problem (or, more generally CEO problem) under logarithmic loss. The information bottleneck problem corresponds to the unconstrained optimization problem associated to that on the right (i.e., the IB functional is equivalent to the Lagrangian of the thing on the right).
Here is a quick sketch of the proof.
Claim (Lower Bound): For any $f: x^n \mapsto f(x^n) \in \{1,\dots, 2^{nR}\}$, there is a random variable $U$ satisfying $Y-X-U$, $I(X;U)\leq R$ and $\frac{1}{n}H(Y^n|f(X^n)) \geq H(Y|U)$.
Proof: By the chain rule, write
$$
\frac{1}{n}H(Y^n|f(X^n)) = \frac{1}{n}\sum_{i=1}^n H(Y_i|Y^{i-1},f(X^n))= \frac{1}{n}\sum_{i=1}^n H(Y_i|V_i),
$$
where we define $V_i := (Y^{i-1},f(X^n))$. Observe that, since $(X^n,Y^n)$ are iid draws of $(X,Y)$, we have $Y_i - X_i - V_i$ for each $i=1,\dots, n$. Now, on a common probability space with $(X,Y)$ construct a random variable $U$ as follows. Given $X=x$, draw $Q$ uniformly from $\{1,\dots, n\}$, and take $U = (V_Q,Q)$, where $V_Q|\{Q=i, X=x\} \sim p_{V_i|X_i}(\cdot | x)$. Evidently, we have $Y - X - U$, and by definition of $U$,
$$
\frac{1}{n}H(Y^n|f(X^n)) = \frac{1}{n}\sum_{i=1}^n H(Y_i|V_i) = H(Y|U),
$$
since the $Y_i$'s are iid. Now, for each $i=1,\dots, n$, we have $X_i - (X^{i-1},f(X^n)) - (Y^{i-1},f(X^n))$ (again, using the fact that $(X^n,Y^n)$ are iid draws of $X,Y$ together with the fact that $f$ is a function of the $x_i$'s), so properties of entropy and the data processing inequality give
$$
R \geq \frac{1}{n}I(X^n ; f(X^n)) = \frac{1}{n}\sum_{i=1}^n I(X_i ; X^{i-1},f(X^n)) \geq \frac{1}{n}\sum_{i=1}^n I(X_i ; V_i) = I(X;U).
$$
Claim (Upper Bound): If $Y-X-U$ satisfy $I(X;U)\leq R$, then for $n$ sufficiently large, there is a function $f: x^n \to f(x^n) \in \{1,\dots, 2^{nR}\}$ with $\frac{1}{n}(H(Y^n| f(X^n))+o(n)) = H(Y|U)$
Proof: Fix any random variable $U$ on a common probability space with $(X,Y)$ such that $Y-X-U$ and $I(X;U) < R$. Generate $2^{nR}$ sequences $U^n(1), \dots, U^n(2^{nR})$ independently, according to $p_U^{\otimes n}$. By standard typicality arguments, if we observe $X^n$, then with high probability there is an index $f(X^n)\in \{1,\dots, 2^{nR}\}$ such that $U^n(f(X^n))$ is jointly typical with $X^n$ (and therefore also $Y^n$ by the Markov Lemma). By standard properties of (conditional) typical sets, we have
$$
\frac{1}{n}H(Y^n| f(X^n)) \sim H(Y|U).
$$
Tom, thanks for the proof. I think this is actually subtly different than source coding under logarithmic loss. Choosing the best encoder here is minimizing a different objective: in this case (1) the "loss function" simultaneously depends on all $n$ source letters and the encoder word in some complex way. in the source coding problem (2) the loss function is an average of $n$ identical single-letter losses. Your proof shows that the minima of the two problems are the same.
Put another way: after conditioning on the encoding, the source letters are no longer necessarily independent. This question's objective function keeps track of this non-independence, the source coding problem does not.
|
2025-03-21T14:48:30.513019
| 2020-05-07T02:16:08 |
359597
|
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"Gerhard Paseman",
"The Number Theorist",
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}
|
Stack Exchange
|
Sumset of $k$-smooth numbers
Take the set $T(k,n)=M_1(k,n)$ of all $k$ smooth numbers less than $n$.
What is the cardinality of $$\{1,\dots,n\}\cap M_2(k,n)$$ where every integer in $M_2(k,n)$ is the sum of two integers in $M_1(k,n)$?
Instead of $M_2(k,n)$ how would the behavior be for $M_i(k,n)$ where every integer is sum of $i$ integers in $T(k,n)$?
It is essentially like $((\log n)^k)^i)$, as smooth numbers tend to thin out when n is like exp(k). Gerhard "Log Smooth Is Really Smooth" Paseman, 2020.05.06.
@GerhardPaseman I think that's an upper bound?
You could take a look at the Croot's paper ''On a combinatorial method for counting smooth numbers in sets of integers'', Journal of Number Theory 126 (2007) 237–253. In particular, Theorems 3 and 4 there could be useful to your situation.
|
2025-03-21T14:48:30.513128
| 2020-05-07T04:30:18 |
359604
|
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|
Stack Exchange
|
What is the motivation of the $L^p$ differentiability?
I was reading some papers and come up with the next definition :
A function is differentiable in the $L^p$ sense at $x$ if there
exists a real number $f'_p(x)$ such that $$\bigg(\frac{1}{h}∫_{-h}^{h}|f(x+s)−f(x)−f_p'(x)s|^pds\bigg)^{1/p}=o(h)$$
And he states that many $f_p'$ are equivalent to an ordinary derivative on an almost everywhere basis. I saw this kind of differentiability is studied in some papers, but I don't know where does it started, i.e. what was it's motivation to define such operation and function space.
Ash, J. Marshall, An (L^p) differentiable non-differentiable function, Real Anal. Exch. 30(2004-2005), No. 2, 747-754 (2005). ZBL1107.26010.
The name Sobolev comes to mind, as it often does. https://en.wikipedia.org/wiki/Sobolev_space
Several references I looked at cite Local properties of solutions of elliptic partial differential equations by Calderón/Zygmund (1961), but all the citations I've seen are for both first order $L^p$ derivatives and higher order $L^p$ derivatives, leaving open the possibility that the first order $L^p$ derivative might have appeared earlier. Some motivation can be found in Ash's 2014 paper Remarks on various generalized derivatives. The top of p. 9 (.pdf file page number) Ash writes: (continued)
"The study of generalized $L^p$ differentiation began to become important in the 1950s in connection with the study of finding $L^p$ solutions for partial differential equations." However, as far as I can tell, Ash never discusses anything more, such as when the notion first appeared (but perhaps wasn't specifically singled out, and instead was buried in a theorem hypothesis or in a technical comment) and when the notion (but perhaps not using the present term) was first specifically singled out with a definition.
@BenMcKay: I am not entirely convinced this is the Sobolev notion. The $ | \cdot |p$ displayed in the question is $L^p$ mean in $h$. More precisely the definition is requiring that $$ \left( \frac{1}{h} \int{-h}^h | f(x+s) - f(x) - f_p'(x) s|^p ds \right)^{1/p} = o(h) $$
For example, the function $f(x) = \sqrt{|x|}$ on $[-1,1]$ is $W^{1,p}$ for any $p \in [1,2)$. But it is not $L^p$ differentiable at $x = 0$ for any $p$: one can check that for any choice of $f'_p(0)$ necessarily the mean shown above is $O(\sqrt{h})$. // (@OP: you should probably edit this very important part of the definition into your question.)
@WillieWong I stand corrected. Very interesting.
I don't know the literature, but it maybe helpful to understand exactly what this notion of differentiability gains for us.
First, unlike the Sobolev notions, this does not handle functions which belong in Holder classes. For example, based on the definition the absolute value function is not $L^p$ differentiable for any $p$, but it is certainly (locally) in the Sobolev class $W^{1,p}$ for every $p\in [1,\infty]$.
In fact, whenever $f$ is a function such that the one sided derivatives from the left and from the right independently exist, but do not agree, such a function is not $L^p$ differentiable in the sense defined in the paper.
So what does $L^p$ differentiability gain for us? It gains when your function fails to be differentiable near $x$ due to it being oscillatory in a certain way. An example:
Let the set $A = \cup_{k = 10}^\infty [ 1/k - 2^{-k}, 1/k + 2^{-k}]$, and let $f$ be the indicator function of $A$. This function is clearly not differentiable at $x = 0$. However, since
$$ \int_{-h}^h |f(s)|^p ~ds \leq \sum_{k = \lfloor 1/h \rfloor }^\infty 2^{1-k} = 2^{2-\lfloor 1/h \rfloor } $$
we see that for any $p\in [1,\infty)$, $f$ is $L^p$ differentiable at the origin with derivative $0$.
As to why measuring things with respect to $L^p$ means can be useful: there is a big hint in the paper of Calderon and Zygmund, concerning elliptic PDEs. They wrote [emphases mine]:
It seems that the notion of differentiability which is most suited to the treatment of the problems that concern us, is not the classical one. It appears that it is more convenient to estimate the remainder of the Taylor series in the mean with various exponents. This type of differentiability is much more stable ...
If you are familiar with some harmonic analysis, what this is screaming out is that a lot of analytic estimates (Sobolev regularity for elliptic PDEs, singular integrals, etc.) work generally for functions measured on the $L^p$ scale, but often fail (just by a little) at $L^\infty$ (and sometimes $L^1$).
You can of course ask why did Calderon and Zygmund not use the Sobolev class: the key is that they want to understand pointwise estimates of differentiability. As mentioned above, Sobolev classes are not so sensitive to pointwise properties.
Wow. Very impressive and informative answer. Thank you very much!
By the way, I am taking Prof.Krieger's course in EPFL, and I saw you work with him! It is an honor to get an answer from you :)
You are studying for an MA there? How did you come across this $L^p$ differentiability business?
Yes. I was reading some papers from who I was interested in his work, Daniel Spector. Specifically, D. Spector, On a generalization of $L^p$-differentiability, Calc. Var. Partial Differential Equations 55
(2016), no. 3, 55:62.
@Jingeon An: FYI, I'm a bit embarrassed that I didn't think of $L^p$ derivatives and their variations when I initially wrote, and then later expanded, this answer, given the fact that my Ph.D. supervisor's Ph.D. thesis involved $L^p$ derivatives (and I was well aware of this, even 25+ years ago).
|
2025-03-21T14:48:30.513647
| 2020-05-07T06:23:59 |
359608
|
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|
Stack Exchange
|
Derived symmetric powers and determinants
Given a vector bundle $V$ (on a scheme $X$, say), I can form $Sym(V[1])$, the symmetric algebra (in the derived/graded sense) on the shift of $V$; in other words this is the Koszul complex of the zero section of $V$, also known as the free exterior algebra on $V$. This is in fact a graded algebra, and its maximal non-vanishing graded component is $Sym^n(V[1]) = det(V)[n]$, where $n$ is the rank of $V$.
Thus we naturally get the graded determinant of $V$.
Now let $E$ be a perfect complex. The expression $Sym(E[1])$ still makes sense. Similarly the graded determinant of $E$ makes sense (the morphism $Vect(X) \to Pic(D(X))$ sending $V$ to its graded determinant is additive and hence factors over $K$-theory; in more down to earth terms the graded determinant of $V_1 \to V_2 \to \dots \to V_n$ is $det(V_1) \otimes det(V_2)^{-1} \otimes \dots$, placed in degree $rank(V_1) - rank(V_2) + \dots$).
However there need not be a maximal degree component of $Sym(E[1])$, and even if there is, it need not have anything to do with the graded determinant of $E$ (as far as I can tell).
My question: Is there a good way of thinking about this? For example, is there an alternative generalization of $Sym(V[1])$ to perfect complexes which relates to graded determinants?
|
2025-03-21T14:48:30.513782
| 2020-05-07T07:42:19 |
359612
|
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"M. Winter",
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|
Stack Exchange
|
Maximal distance between $2d+1$ points on the $(d-1)$-sphere
If one arranges $2d$ points on the sphere $\mathbf S^{d-1}\subset\Bbb R^d$ at the vertices of the crosspolytope, then one can achieve a minimal spherical distance of $\pi/2$ between any two points, and this is best possible.
What if I want to arrange $2d+1$ points on $\mathbf S^{d-1}$ as far apart as possible from each other? What are the best known upper and lower bounds on the minimal distance between any two points in such an arrangement, and is there anything known about how these arrangements look like?
I finally came to do the computations on Yoav Kallus' comment. After quite some tedious work one finds a cubic polynomial:
$$p_d(x)\,=\,d(d-2)^2 x^3 - d^2 x^2 - dx + 1$$
which has exactly one zero in the interval $[0,1]$. This zero is the cosine of the maximal distance between points in Yoav's solution. In fact, comparing with this source, these values seem to agree with the best known solutions (and for $d\in\{2,3\}$ they are proven to be optimal). These solutions are cited from
T. Ericson and V. Zinoviev, "Codes on Euclidean spheres"
but I wasn't able to find the exact place where they are computed.
OK, it's late and I may be wrong but I think that you can obtain the $2d$ points by using any set of orthonormal basis vectors $\{v_1,\ldots,v_d\}$ and their negatives.
Now if $n$ is such that a Hadamard matrix exists, you could take the rows of $H$ an $n\times n$ Hadamard matrix in its $\pm 1$ formulation, and its negative. If you then remove the first column, that would leave you with pairwise inner product $\pm \frac{1}{n} \sqrt{\frac{n}{n-1}}$ or something of this ilk. You can then convert this to the angle $\theta.$ This means that your $d=n-1.$
Maybe more general constructions exist, and feel free to shoot down my answer.
Thanks for your answer. If I made no mistake, then this gives $2d+2$ points on $\mathbf S^{d-1}$ with minimal spherical distance $\arccos(1/d)$ (assuming that a Hadamard matrix of size $(d+1)\times(d+1)$ exists).
In the case $d=3$ one can achieve a better result for $2d+1=7$ points using the vertices of a bipyramid with pentagonal base. This yields a slightly larger minimal distance $2\pi/5>\arccos(1/3)$. In this case, the Hadamard construction gives the vertices of a cube.
Also for eight points in three dimensions the maximin distance is not obtained with a cube. If you have eight points on a cube, you can twist one of the faces in-plane, thus increasing the length of the lateral edges. Then you are free to move the twisted square and its opposite slightly closer (and slightly expand them) to get a greater overall maximin distance.
Turns out kodlu's idea works in all dimensions, regardless of the existence of any Hadamard matrices.
Consider all coordinate permutations of
$$(1,...,1,-d)\in\Bbb R^{d+1}\quad\text{and}\quad (-1,...,-1,d)\in\Bbb R^{d+1}.$$
This gives $2d+2$ points in $\Bbb R^{d+1}$, but all these lie in the $d$-dimensional subspace orthogonal to $(1,...,1)$.
The smallest angle between two of these is attained e.g. between $a=(1,...,1,-d)$ and $b=(-1,...,-1,d,-1)$, whose cosine turns out to be
$$\cos\measuredangle(a,b)=\frac{\langle a,b\rangle}{\|a\|\|b\|} = \frac{\overbrace{(-1)+\cdots+(-1)}^{d-1}+d+d}{\underbrace{1+\cdots+1}_d+d^2} = \frac{d+1}{d(d+1)}=\frac1d.$$
One way to describe this configuration is to consider on of the points as the "north pole", then you have a (d-1)-simplex on one lattitudinal cross section (the "tropic of cancer") another one, with the opposite orientation, on the "tropic of capricorn", and another point on the souther pole. The minimum distance is not realized inside each line of lattitude, but only between adjacent ones. Since we have one more point than needed, we can remove the south pole, and pull the tropics south a little bit. This should increase the minimal distance slightly.
|
2025-03-21T14:48:30.514113
| 2020-05-07T09:29:06 |
359615
|
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"Ingo Blechschmidt",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/359615"
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|
Stack Exchange
|
Embedding abelian categories into abelian sheaves
The Yoneda functor from an abelian category into sheaves of abelian groups is shown to be exact in The Stacks Project, Lemma 19.9.2. I like this proof because it is constructive and it doesn't use injective/projective objects (as for instance the FM Embedding Theorem does). Is there anyone which has in mind a published references with an analogous proof?
Welcome on MO, Anna! I too like that proof very much. It's the workhorse of my favourite approach to diagram chasing in abelian categories: use the internal language of the resulting sheaf topos. (If desired, the definition of internal language can be unrolled to eliminate all mentions of sheaves and even sets, working just fine in restricted predicative foundations, yielding a convenient interface to this little jewel by George Bergman.
Unfortunately I also don't know a precise reference. The construction has certainly long been known in the topos community; it is an instance of "sheaves with respect to the regular topology". Perhaps this helps in tracking down a reference.
Numbered references in the stacks project are not stable. Please use tags instead.
|
2025-03-21T14:48:30.514235
| 2020-05-07T09:34:48 |
359616
|
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|
Stack Exchange
|
Continuity/Lipschitz regularity of exponential map from $C_c$ to $\operatorname{Diff}_c$?
For finite-dimensional Lie algebras, see this for a nice example, the exponential map is smooth and in particular, it is locally-Lipschitz onto its image. However, things are different when moving to the infinite-dimensional setting, as discussed for example in the answer to this post.
Let $G=\operatorname{Diff}_c(M)$ the space of compactly supported diffeomorphisms on $M$, say $M$ is Riemannian manifold diffeomorphic to $\mathbb{R}^k$, and let $\mathfrak{g}=C_c^{\infty}(M,M)$ be the $C^{\infty}$-vector fields on $M$. Again making use of comments in the answer to this post the exponential map taking a vector field $V$ in $\mathfrak{g}$ to its flow $\Phi^V:x\to x_1^x$ where $t\to x_t$ is well-defined by
$$
\partial x_t^x = V(x_t^x), \, x_0^x=x
.
$$
Is this map continuous?
Moreover, when is it locally-Lipschitz, in the sense that, for every $\emptyset \subset K \subset M$ compact there exists some $L_K>0$ such that for every $U,V \in C^{\infty}_c(M,M)$
$$
\sup_{x \in K} d_M\left(
\Phi^V(x),\Phi^U(x)
\right)\leq L_K \sup_{x \in X} d_M\left(
V(x),U(x)
\right)
$$
where $d_M$ is the metric induced by the Riemannian metric on $M$?
There is no need to be so specific in the question, you can indeed answer this in general for $M$ a (paracompact, finite-dimensional) manifold.
It is well known, that in this setup $\mathrm{Diff}_c(M)$ is an infinite-dimensional Lie group with Lie algebra $\mathfrak{X}_c(M)$ (this was your space of $C^\infty_c(M,M)$.
Before we continue, it needs to be mentioned that there are at least two settings in which the Lie group statement makes sense and is true:
The convenient setting of global analysis (a la Kriegl and Michor, this is Theorem 43.1. in 1)
The Bastiani setting of calculus (again Michor, Theorem 11.11 in 2)
In both settings the exponential is of course the flow map zou mention (this is said explicitely in the convenient calculus source 1, but NOT in the Bastiani source 2.
How does this relate to your question? Well $\exp$ is convenient smooth according to 1, but we can not use that! Convenient smooth mappings are in general NOT continuous with respect to the original topology. Note that this phenomenon happens only outside of Frechet spaces and some other nice classes of spaces. However, the model space of compactly supported vector fields is not one of the nice spaces, so we can not deduce it from this result.
In the Bastiani setting I know several sources wich mention Bastiani smoothness of the exponential map (this is a consequence of regularity of the Lie group, here regularity means that the flow map for curves in the Lie algebra exists and is smooth, in your picture this means that you take the flow map of time dependent vector fields) but only feature proofs in more restricted settings (i.e. $M$ compact). To my knowledge a proof of the regularity of $\mathrm{Diff}(M)$ in the Bastiani setting for paracompact $M$ appeared for the first time in print in my thesis 3 (as a special case of the orbifold diffeomorphism group discussed there).
Regularity of a Lie group (in the Bastiani setting) guarantees smoothness (in Bastiani sense) of the exponential map. Now as Bastiani smooth implies continuity, this answers your first question affirmatively.
First of all the second question seems to be off, or at least contain some non-trivial identification since the vector fields map into the tangent bundle and are still comared using the metric induced by the Riemannian metric on $M$. Anyway, a closer analysis of the problem should show that whatever you want is probably a finite-dimensional problem anyway. Therefore it might be misleading to think about your second property as Lipschitzness of the exponential.
To see this, note that the source of the differentiability of the exponential map is the so called exponential law (for function spaces, has nothing to do with Lie group exponentials). Here this translates roughly to the insight that you do not need to solve the defining equation on an infinite-dimensional manifold, but can place yourself in a situation where you are solving a differential equation on the underlying manifold $M$ (see e.g. 4 for more information on this topic). This is usually highly technical but since your question fixes the compact set, it should be doable (I would have to think more closely about how to do this though).
|
2025-03-21T14:48:30.514571
| 2020-05-07T11:27:12 |
359623
|
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"Asaf Karagila",
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|
Stack Exchange
|
Implications of the existence of a pair of surjective functions, without Axiom of Choice
The classical Cantor-Schroder-Bernstein Theorem says that there exists a bijective function $X\leftrightarrow Y$ if and only if there exist injective functions $X\hookrightarrow Y$ and $Y\hookrightarrow X$.
It is known that without Axiom of Choice the injectivity cannot be replaced by the surjectivity in this result.
Nonetheless, without AC, we have the following simple
Fact. If for two sets $X,Y$ there are surjective functions $X\twoheadrightarrow Y$ and $Y\twoheadrightarrow X$, then there exists a bijective function $\mathcal P(X)\leftrightarrow \mathcal P(Y)$ between the corresponding power-sets.
Can the power-sets in this result be replaced by some simpler sets built over $X$, for example, $X^\omega$ or $X\times\omega$?
More precisely:
Question. Assume that for two sets $X,Y$ there exist surjective functions $X\twoheadrightarrow Y$ and $Y\twoheadrightarrow X$.
Is it true (in ZF) that the sets
(i) $X^\omega$ and $Y^\omega$ have the same cardinality?
(ii) $X\times\omega$ and $Y\times\omega$ have the same cardinality?
Note: it appears to be an open problem whether existence of surjections both way implies AC: https://mathoverflow.net/q/38771/30186
@Wojowu I guess you mean, whether (existence of surjection both ways implies bijection) implies AC?
Take $X=c$ and $Y=t$, the cardinal of $\mathbf{R}/\mathbf{Q}$ (here "cardinal" is set modulo bijection). So, if I'm correct there are injections $c\to t$, and surjections in both ways. Also there is a bijection $c\to c^\omega$, and an injection $t\to t^\omega$. If there was a bijection $c^\omega\to t^\omega$ we would deduce an injection $t\to c$, which is not a theorem in ZF+DC as far I as I know. The same seems to apply to $\times\omega$.
@YCor Yes, I phrased that my previous comment very poorly.
@YCor In the second line of your comment you wrote that there exist injections $c\to t$. Why? (I mean in ZF?) And why there exists a surjection $t\to c$? (also in ZF)?
First if there's an injection $a\to b$ and $a$ is nonempty then there's a surjection $b\to a$ (trivial). For the other part it's easy to explicitly map injectively $c$ into $\mathbf{R}$ so that the image is a subset that is algebraically independent over $\mathbf{Q}$ (see for instance Wagon's book). Hence the composite map $c\to\mathbf{R}/\mathbf{Q}$ is injective.
@YCor I also thought about linearly independent Cantor sets in $\mathbb R$ but is it indeed constructed in ZF without use of Kuratowski-Mycielski Theorem, which uses the Baireness of the hyperspace of non-empty compact sets, which requires the Dependent Choice? But even constructing some canonical tree (representing the linearly independent Cantor set) we shall need dependent choice to show that the boundary of this tree is equipotent with the real line. Or everything is so constructive there?
It's explicit and you can find it in Wagon's book on the Banach-Tarski paradox. In spirit it looks like: choose an injection $i:\mathbf{N}^2\to\mathbf{N}$ such that $i(n,m)\ge n!$ for every $n$. Then for $x\in [0,1[$ written in binary expansion $x=\sum_{n\in J} 2^{-n}$, map it to $\sum_{n\in J}\sum_m 2^{-i(n,m)}$. I'm not sure this one works, but some variant in the same spirit should.
@YCor I would be happy to accept your answer in case you write your cemments as a "legal" answer. By the way, I have another question, but maybe I ask too many questions: writing down the Kuratowski definition of an ordered pair, we can see that $X\times X\subset\mathcal P(\mathcal P(X\times X))$, so $|X\times X|\le|\mathcal P(\mathcal P(X))$ in ZF. What about the inequality $|X\times X|\le|\mathcal P(X)|$? Is it true in ZFC?
@TarasBanakh I guess this fitted in a comment, and you have in any case another answer now to accept.
No, and here is a counterexample.
Suppose that $|\Bbb R|<|[\Bbb R]^\omega|$, that is, there are more countable subsets of reals than reals. This is indeed possible, e.g. if all sets of Lebesgue measurable.
Since $\sf ZF$ proves there are bi-surjections (in fact, an injection from $\Bbb R$ into $[\Bbb R]^\omega$), this would be a counterexample.
Now, $|\Bbb R^\omega|=|\Bbb R\times\omega|=|\Bbb R|$, and even without carefully checking what is $([\Bbb R]^\omega)^\omega$ and $[\Bbb R]^\omega\times\omega$, the cardinality cannot decrease.
Thnk you for the answer. In fact, @YCor suggested a bit different counterexample exploiting linearly independent Cantors sets in the real line. Could I ask another question also related to (the absence of) AC: Is $|X\times X|\le|\mathcal P(X)|$ in ZF? I can only prove (trivially) that $|X\times X|\le\min{|\mathcal P(\mathcal P(X))|,|\mathcal P(X\times{0,1})|}$, but $X\times{0,1}$ need not be equipotent with $X$ in ZF.
In fact, I am interested in the inequality $\mathcal P^2(X\times X)\le \mathcal P^3(X)$. Is it true in ZF, or the 4-th iterate of the power-set operation is necessary? In its turn, I need this for evaluating the Hartogs' number $\aleph(x)$ of a set $x$. It is easy to see that $\aleph(x)\le|\mathcal P^4(x)|$. So the question is about $|\aleph(x)|\le|\mathcal P^3(x)|$ in ZF.
I have already found the answer to my "Hartogs" question in https://caicedoteaching.files.wordpress.com/2009/04/580-choiceless.pdf Indeed, $|\aleph(x)|\le|\mathcal P^3(x)|$.
To your first question, no. It is consistent that there is a set $X$ such that $|X^2|\nleq|\mathcal P(X)|$. About the 3 Hartogs, you can find a question I asked here many years ago related to this.
Yes, I found this your question from 2012 and also the answer of Caicedo that 3 cannot be lowered to 2.
|
2025-03-21T14:48:30.514986
| 2020-05-07T11:57:28 |
359625
|
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|
Stack Exchange
|
Finding the nearest quadratic Bézier curve
Given a set of three-dimensional quadratic Bézier curves.
I'm looking for some analytical solution to find the nearest curve to an arbitrary point in space.
Example
I already have a brute force solution with the calculation of the closest point on each curve and selection one with the minimum distance. But this algorithm requires a lot of computational resources.
If you put boxes around all of your curves, take the centre-points of the boxes, and order them, then calculate the distance to the curve with the smallest centre-point distance, you have a lower bound on the smallest distance. You can now exclude all boxes which have all of their corners more than this distance from the target point. Then continue in the same way.
Too long to comment.
The square of the distance from a point $P$ to a point on a quadratic Bezier curve (parametrized by $t$) is a quartic in $t$. So, choose a Bezier curve and determine this quartic polynomial function. The minima $t^*$ in the range $[0,1]$ can lie at one of the critical points (gradient equal to zero) or the boundaries $0$ or $1$. The points at which the gradient vanishes are all roots of the cubic polynomial obtained by differentiating the quartic. One can then use Cordano's formula (https://en.wikipedia.org/wiki/Cubic_equation) to find a closed form for the roots. Choose all the real roots within $[0,1]$ and evaluate the quartic at these points. Similarly, evaluate the quartic at $0$ and $1$. Finally, choose a point which yields the minimum value.
Do the above for each Bezier curve and choose the one with the lowest minimum value.
Hope this helps.
It sounds like you already have a way to calculate the distance from a point to a Bézier curve, and you’d just like to speed things up by avoiding doing this calculation for every curve in your collection.
The way to do this is as outlined in the comment from @J.J Green -— you need to enclose each curve in some simple shape for which distance calculations are easy. Then, for a given curve, if the distance to its enclosing shape is larger than your current minimum, you don’t need to bother calculating distance to that curve itself.
Suitable enclosing shapes For quadratic Bézier curves are spheres, axis-aligned bounding boxes, or triangles
|
2025-03-21T14:48:30.515199
| 2020-05-07T12:06:53 |
359626
|
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|
Stack Exchange
|
Is the center of an abelian rigid monoidal category, abelian?
Is the Drinfeld-Majid center of an abelian rigid monoidal category, abelian?
[stated in 1J of On the center of fusion categories" by Bruguières and Virelizier (link at Virelizier's page)]
In particular, I’m not seeing why any monomorphism in the center would have to be a kernel of a morphism? (I’m relatively happy with the other axioms holding, but if anyone has a reference where this is discuss explicitly, it’s appreciated )
I think the point is that in a rigid tensor category, the tensor product is exact. This allows you to equip the cokernel of $A \to A'$ with a natural half braiding induced by the ones on $A$ and $A'$.
I agree that Ker/Cokers have induced braidings, but does this imply that every monomorphism in the center is the kernel of another morphism?
Once you check that the centre is finitely complete and cocomplete and the forgetful functor is exact, this shows that $f \colon (A,\sigma) \to (A',\sigma')$ is a monomorphism if and only if it is the kernel of $(A',\sigma') \to \operatorname{coker} f$, because this is true in $\mathscr C$.
Am I wrong in saying that an exact functor between abelian categories preserves Monos because it preserves kernels? But here we don’t know yet that the center is abelian to use that argument.
Any left exact functor between finitely complete categories preserves monomorphisms, because $X \to Y$ is a monomorphism if and only if $\Delta_{X/Y} \colon X \to X \times_Y X$ is an isomorphism. See for example Exercise III.4.4 in Mac Lane.
Thank you! I can’t believe how easily one forgets these facts
My question got answered in the comments, so I thought best to write a small answer for it here:
If $C$ is a rigid monoidal category, then the forgetful functor $U: Z(C) \rightarrow C $ creates colimits and limits: any (co)limit of the underlying objects of a diagram in $Z(C)$ will have an induced braiding satisfying the nessecary conditions. This follows from $A\otimes - $ and $-\otimes A$ preserving (co)limites.
Hence, if $C$ is abelian, then the center is finitely complete and cocomplete and naturally automatically additive. The only thing remaining is to check if any mono is a kernel and any epi is a cokernel which I was stuck at, which just follows from the fact that in a finitely (co)complete category, monos (epis) are just pullbacks and pushforwards, so they are preserved by $U$. If $m:(A,\sigma)\rightarrow (B,\tau) $ is a monomorphism in the center, then $m$ is a mono in $C$ and is the kernel of $f:B\rightarrow coker(f)$ and as mentioned above $coker(f)$ has an induced braiding compatible with $\sigma $ and $\tau $ and $f$.
U should go the other way, right?
@NoahSnyder Indeed! Thank you:)
|
2025-03-21T14:48:30.515736
| 2020-05-07T12:22:36 |
359628
|
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|
Stack Exchange
|
Can Alice ever fare the worst in this variant of the truel game?
In the well known classic three way duel puzzle, 3 players Alice, Bob and Carol take turns to shoot each other until only one survives. In his/her turn, a player can either choose to shoot or pass$^{1}$.
In this variant, the players must choose their guns before the game starts. The three guns available have hit probabilities $g_1\lt g_2\lt g_3$. To add more drama, the last person to choose enjoys the privilege to determine the shooting order (i.e. who gets to shoot first, who's second and who's last). Alice is the first to choose her gun, Bob the second, Carol the last.
Can there exist $g_1, g_2, g_3$ such that Alice has the lowest surviving probability among the three, given that everyone act to maximize his/her own surviving probability? If no, is there a not-too-involved proof that explains why not? If yes, what is an example?
$^{1}$Technical note: to rule out the bug equilibrium where everyone just passes, we simply add a powerful enforcer who ensures that as soon as three consecutive passes occur, all players will be killed right on spot. This will justify the best and middle shooters' motive to always shoot instead of passing under any situation.
Update: after trying a few examples, I think a complete list of endgames would be necessary. How Carol determines the optimal shooting order for herself is relatively easy, the only ambiguity being the middle gun. Thus there're eight possibilities as shown below:
where I use w, m and b to denote the worst, the middle and the best guns respectively.
For each possibility, Alice faring the worst means
where m w b(underlined) is a shorthand notation to denote the surviving probability of the best gun b under the shooting order m w b. The colors are there to make verification easy, and are otherwise inessential. (Red for Alice, Blue for Bob and Green for Carol)
But how do we proceed from here? It seems we can add 8 more constraints similar to those (1)~(8) above for Bob and Carol to decide to arrive at each possibilities, and hopefully teasing out some obvious contradictions from these 16 constraints (such as (m w b $\gt$ m w b) && (m w b $\lt$ m w b)). But I'm not sure we can work up from there to Alice. And I'm not sure it's the right way to go.
I don't know, but the "powerful enforcer" has to be named Ted. https://www.imdb.com/title/tt0064100/
Do we assume that Carol chooses randomly between orderings that she is indifferent to? For example, if Carol has the b gun, then bmw and bwm are both optimal for her. Depending how Carol breaks ties, this could affect Alice and Bob's strategies.
@TonyHuynh If Carol has the b gun, bwm can be strictly better than bmw for her, because there're 1,2,3 for which w gun would optimally choose to shoot for bmw but to pass for bwm. If we adopt the reasonable assumption that for all equally optimal strategies Carol will choose the simplest one (the one requiring least amount of calculation), she would just choose to simply stick to bwm under all circumstances. –
|
2025-03-21T14:48:30.516139
| 2020-05-07T12:36:47 |
359632
|
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|
Stack Exchange
|
Can every point of Wilson's adelic Grassmannian be obtained by Krichever construction of solutions to KP equations?
Igor Krichever introduced an algebro-geometric construction of solutions of KP equations starting from an algebraic curve with some additional data.
George Wilson introduced the adelic Grassmannian, which is a subspace of the Sato Grassmannian parametrizing all solutions to KP equations.
Question: can every point of Wilson's adelic Grassmannian be obtained by Krichever's algebro-geometric construction?
Yes - the adelic Grassmannian precisely parametrizes "rational solutions of KP", which are the Krichever solutions attached to rank 1 torsion free sheaves on cuspidal genus 0 curves -- i.e. curves with $P^1$ as their bijective normalization (or subrings of the field of rational functions). Its adelic (or factorization) nature is explained by keeping track of the finite subset of $P^1$ which is the location of the cusps - i.e. this space parametrizes "Hecke modifications" (Backlund transformations) supported at given points of $P^1$. This is stated as Corollary 5.21 in https://arxiv.org/abs/math/0212094 though I think was known to Wilson, Berest, Mulase and others.
It might be worth clarifying that these solutions are not called rational for the evident relation to P^1 (ie for the spectral curve), but rather because they give Lax operators which are rational (meromorphic rather than formal), ie for the “differential” parameter. This “coincidence” is a form of Wilson’s bispectrality for these solutions (a kind of Fourier transform)
|
2025-03-21T14:48:30.516281
| 2020-05-07T14:14:50 |
359636
|
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|
Stack Exchange
|
Is Krichever's constuction "inverse" to finding the spectral curve?
There is Krichever's algebro-geometric construction of solutions to the KP equations starting from a curve X together with extra data.
There is a way to find the spectral curve given a point of the Sato Grassmannian parametrizing solutions of KP equations, which is discussed here
Question: if I apply Krichever construction to a curve X with extra data, getting a point in the Sato Grassmannian, and then find the spectral curve of this point as above, will I get back the curve X?
|
2025-03-21T14:48:30.516355
| 2020-05-07T14:32:08 |
359637
|
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|
Stack Exchange
|
Geometric meaning of colocalization of modules?
Let $A$ be a commutative ring and $S\subset A$ a subset. A localization of $A$ at $S$ is defined as a ring morphsim $A\to A[S^{-1}]$ which is initial with respect to inverting $S$. Similarly, a localization of an $A$-module $M$ at $S$ is an $A$-module morphism $M\to M[S^{-1}]$ initial with respect to $f\in A$ acting invertibly.
For modules, principal localization at an element $f\in A$ is a special case of universally inverting an endomorphism $f$ in a category. (This fails for rings because the action of $f$ on $A$ is not a ring morphism.) As always there are two universal constructions - initial and terminal. Localization of a module is the initial variant.
What about the terminal way to invert the action of a ring element on a module? The "colocalization" at $f\in A$ is an $A$-module morphism $R_f(M)\to M$ which is terminal in the category of $A$-module maps to $M$ on whose domain $f$ acts invertibly.
The colocalization may be constructed as the following sequential limit $$R_f(M)\cong\varprojlim(\cdots \overset{f}{\to}M\overset{f}{\to}M),$$so an element is a string $(m_0,m_1,\dots )$ satisfying $m_n=f(m_{n+1})$.
Viewing a $C^\infty(X)$-module as a $C^\infty$ vector bundle over $X$, an element of the colocalization at some $f\in C^\infty(X)$ is like a "section obtained by arbitrarily many multiplications by $f$". I don't understand how to think of this. Maybe as a section "coming from" a certain completion?
Question. What is the geometric meaning of colocalization of a module at an element of the ring? Some kind of "cosupport" (whatever that is)?
If you work with "derived colocalization" then examples can get surprisingly complicated. I don't have a geometric intuition, but I'll give you several examples, due to Dwyer and Greenlees, and maybe you'll see a pattern. Everything I say is derived, and can be written down as a right Bousfield localization.
local cohomology is a colocalization, whereas local homology is a localization
To be $R/I$-torsion ($I$ an ideal of $R$) is the same as being colocal for a certain colocalization.
To be $R/I$-complete is also a colocalization
Even just taking $R = \mathbb{Z}$ and $A = \mathbb{Z}/p$ is complicated (see section 3.1)
Colocalization picks out pieces of spectral sequences and towers, in many settings.
Hope this helps! For even more examples, see recent work of Barthel, Heard, and Valenzuela, e.g. here.
|
2025-03-21T14:48:30.516559
| 2020-05-07T15:11:20 |
359639
|
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|
Stack Exchange
|
Stochastic differential equations with correlated Brownian Motions
let's consider an sde of this kind:
\begin{equation} \label{eq:system}
\begin{cases}
dX_t=b(t,X_t,Y_t)dt+\sigma(t,X_t,Y_t)dW_t \\
X_0=x_0 \\
dY_t=B(t,X_t,Y_t)dt+C(t,X_t,Y_t)dW_t^2 \\
Y_0=y_0
\end{cases}
\end{equation}
where t $\in [0,T]$, $T$ is the time horizon , $x_0 \in \mathbb{R}^d$, $y_0 \in \mathbb{R}^l$, b, $\sigma$, B, C are Borel functions and
$(\Omega, \mathcal{F}, \mathcal{F}_t,W_t , \mathbb{P})_{t \geq 0}$ is a continuous standard $\mathbb{R}^d$-Brownian motion and $\{ W_t^2 \}_{t \geq 0}$ is a continuous standard $\mathbb{R}^l$-Brownian motion on the same filtration having instantaneous correlation $\rho \in (-1,1)$
\begin{align*}
\mathbb{E}[dW_t^i dW_t^{2,j}]=\rho dt && \forall i \leq d, j \leq l
\end{align*}
Do the classic theorem of existence and uniqueness of strong solution holds also for this system?
Do you have any reference?
Thank you all in advance!
When you say "correlated", do you mean they are still jointly Gaussian, i.e. they form a two-dimensional Brownian motion with covariance $\left(\begin{smallmatrix} 1 & \rho \ \rho & 1 \end{smallmatrix} \right)$? If yes then by a linear transformation you can rewrite the system as driven by a standard 2-D BM and everything is fine. If no then things get much harder and I don't know what can be said.
yes, i mean jointly gaussian. But the BM are not 1D but d-dimensional and l-dimensional respectively. Does it work also in this sense?
I missed that. In general it will work whenever they are a linear transformation of a standard BM, which would be true for any covariance between them. But are you sure the condition you state is possible? If I compute correctly, when $d=l=3$ and $\rho = 1/2$ the resulting covariance matrix is not positive definite.
As mentioned in the comments you simply need to do a transformation.
Here in "Modelling the Stochastic Correlation" pg.23 they give the concrete transformation
You replace your SDE by the above. Then as usual you need to have regularity for the coefficients (see What work has been done on SDE with diffusion coefficients of bounded variation in $\mathbb R^d$? for references).
One reference I found online is section 3.3 in
"American-Type Options: Stochastic Approximation Methods, ".
|
2025-03-21T14:48:30.516767
| 2020-05-07T15:16:51 |
359640
|
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|
Stack Exchange
|
Approach to solve a coupled system of PDE (heat transfer in cylindrical coordinates)
I have the following two PDEs, which describe steady-state coupled heat transport between an externally heated axisymmetric solid body (Eq. 1, $T(r,z)$) and a fluid (Eq. 2, $t(z)$) flowing inside it:
$$
\frac{\partial^2 T}{\partial r^2}+\frac{1}{r}\frac{\partial T}{\partial r}+\frac{\partial^2 T}{\partial z^2}=0 \tag1
$$
$$
\frac{\partial t}{\partial z}+\alpha(t-T)=0 \tag2
$$
Eq. (1) is defined in the domain $r\in[r_1,r_2]$ where $r_1$ and $r_2$ describe the inner and outer radii of the cylinder and $z\in[0,L]$ where $L$ is the length of the cylinder. The boundary conditions for Eq. (1) are:
$$
\frac{\partial T(r,0)}{\partial z}=\frac{\partial T(r,L)}{\partial z}=0 \tag3
$$
$$
\frac{\partial T(r_2,z)}{\partial r}=\gamma \tag4
$$
$$
\frac{\partial T(r_1,z)}{\partial r}=\beta(T(r_1,z)-t) \tag5
$$
For Eq. (2) it is known that $t(z=0)=t_{\mathrm{in}}$
$\alpha, \beta, \gamma, t_{\mathrm{in}}$ are known constants. It seems the solid and fluid temperatures are coupled through the boundary condition at $r=r_1$ (solid-fluid interface, Robin condition).
I know that I can express $(2)$ as:
$$
\frac{\partial t}{\partial z} + \alpha (t - T(r,z))=0 \Rightarrow t=e^{-\alpha z}\int e^{\alpha z} T(r,z) \mathrm{d}z \\ \Rightarrow t=\alpha e^{-\alpha z} \Bigg[\int_0^z e^{\alpha s}T(r,s)\mathrm{d}s+\frac{t_{in}}{\alpha}\Bigg]
$$
This can be substituted in $(5)$ to get a bc only in terms of $(T)$. But I have no idea how to handle such integral type condition.
Any suggestion on how to approach this problem analytically is appreciated.
|
2025-03-21T14:48:30.516906
| 2020-05-07T15:29:24 |
359642
|
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|
Stack Exchange
|
Moduli of smooth curves in $|\mathcal{O}_{\mathbb{P}^1\times\mathbb{P}^1}(2,2)| $ and their invariants
It is well known that any smooth curve
$C\in |\mathcal{O}_{\mathbf{P}^1\times\mathbf{P}^1}(2,2)| $ has geometric genus equal to 1, so its isomorphism class is determined by its $j$-invariant. Nevertheless we have that $\dim(\mathbf P(H^0(\mathcal{O}_{\mathbf{P}^1\times\mathbf{P}^1}(2,2))))=8$ and $\dim(\operatorname{Aut}(\mathbf P^1\times\mathbf P^1))=6$. So the dimension of the GIT quotient
$$
\mathbf P(H^0(\mathcal{O}_{\mathbf{P}^1\times\mathbf{P}^1}(2,2)))//\operatorname{Aut}(\mathbf P^1\times\mathbf P^1)
$$
is 2. If we assume that the base field has characteristic 0, then by Castelnuovo theorem the quotient above is a rational surface. So, at least for a dense subset, there should be two numbers $I_1,I_2$ parametrizing the orbits of the action of $\operatorname{Aut}(\mathbf P^1\times\mathbf P^1)$.
My questions are:
1) Are these numbers known (in terms of the coefficients of the defining polynomial of $C$)?
2) How are they related to the $j$-invariant?
Thanks in advance.
Geometrically we can see not just a $j$-invariant but also
a divisor class on Jac$(C)$, coming from the difference of the
degree-$2$ classes ${\mathcal O}(1,0)$ and ${\mathcal O}(0,1)$.
Conversely, given a genus-$1$ curve $C$ and two degree-$2$ divisors $D_1,D_2$
you can recover the embedding $\iota: C \to {\bf P}^1 \times {\bf P}^1$ from
the two degree-$2$ maps $f_i : C \to {\bf P}^1$ obtained from sections of $D_i$.
$$ $$
Somebody else will probably post a description or reference to the
invariant theory of $(2,2)$ forms before I have the chance to
work it out or look it up.
I added the tag "classical invariant theory" which is the one most relevant to the question, I think. That's because what is asked is to figure out some explicit generators for the ring of invariants of bihomogeneous forms $F(x,y)$ of bidegree $(2,2)$ in $x=(x_1,x_2)$ and $y=(y_1,y_2)$ under the group $SL_2\times SL_2$. I suspect one should be able to construct three explicit invariants $J_1,J_2,J_3$ and take for $I_1$ and $I_2$ suitable ratios of powers of these invariants to make them absolute.
There is a single (up to scale) invariant $J_1$ in degree two, and $J_2$ in degree three. I still have to think about $J_3$.
If $G(x)$ is a binary quartic in $x=(x_1,x_2)$, one can define a biform $F$ by polarizing twice, i.e., applying $(y\cdot\partial_x)^2$ to $G$. Pulling back a $SL_2\times SL_2$ invariant of the resulting biform $F(x,y)$ gives an $SL_2$ invariant of $G$. This could help in finding the connection to $j$ invariants of elliptic curves.
The ring of invariants is ${\bf C}[J_2,J_3,J_4]$ where
$J_k$ (each $k=2,3,4$) is a polynomial invariant of degree $k$.
The Jacobian of $C$ is $y^2 = x (x-J_2)^2 - 4 J_4 x + J_3^2$,
with a visible rational point $(x,y) = (0,J_3)$.
This must be classical but it's more fun to work it out than to
try to hunt it down in the literature. I'll post one approach soon.
@NoamD.Elkies: I guess it makes more sense to index the $J$'s by their degree...:)
[EDITED to exhibit $j$ as a rational function of $J_2,J_3,J_4$,
and to fix various local errors etc.]
The action of ${\rm SL_2} \times {\rm SL_2}$ on the $9$-dimensional space of
$(2,2)$ forms has a polynomial ring of invariants, with generators in degrees
$2,3,4$. If we write a general $(2,2)$ form $P(x_1,x_2;y_1,y_2)$ as
$(x_1^2, x_1 x_2, x_2^2) M_3 (y_1^2, y_1 y_2, y_2^2)^{\sf T}$ where
$M_3$ is the $3 \times 3$ matrix
$$
M_3 = \left( \begin{array}{ccc}
a_{00} & a_{01} & a_{02} \cr
a_{10} & a_{11} & a_{12} \cr
a_{20} & a_{21} & a_{22}
\end{array} \right) \; ,
$$
then $J_k$ ($k=2,3,4$) can be taken to be the $x^{4-k}$ coefficient of the
characteristic polynomial $\chi^{\phantom.}_{M_4}$ of the $4 \times 4$ matrix
$$
M_4 = \left( \begin{array}{cccc}
\frac12 a_{11} & -a_{10} & -a_{01} & 2 a_{00} \cr
a_{12} & -\frac12 a_{11} & -2 a_{02} & a_{01} \cr
a_{21} & -2 a_{20} & -\frac12 a_{11} & a_{10} \cr
2 a_{22} & -a_{21} & -a_{12} & \frac12 a_{11}
\end{array} \right) .
$$
This matrix is characterized by the identity
$$
P(x_1,x_2;y_1,y_2) =
(z_{11},z_{12},z_{21},z_{22}) M_4 (z_{22},-z_{21},-z_{12},z_{11})^{\sf T}
$$
where each $z_{ij} = x_i y_j$, together with the requirement that
$M_4$ has trace zero and becomes symmetric when its columns are listed
in reverse order and columns $2,3$ are multiplied by $-1$.
The invariants of degree $2$ and $3$ can also be written as
$$
J_2 = -\frac12 a_{11}^2 + 2(a_{01} a_{21} + a_{10} a_{12})
- 4 (a_{00} a_{22} + a_{20} a_{02}),
\quad
J_3 = -4 \det M_3;
$$
of course $J_4 = \det M_4$.
The (Jacobian of the) genus-$1$ curve $P=0$ is isomorphic with the
elliptic curve
$$
y^2 = x (x-J_2)^2 - 4 J_4 x + J_3^2.
$$
In particular this lets us compute the $j$-invariant of this curve
as a rational function of $J_2,J_3,J_4$:
$$
j = \frac{256 (J_2^2 + 12 J_4)^3}{16 J_2^4 J_4 - 4 J_2^3 J_3^2
- 128 J_2^2 J_4^2 + 144 J_2 J_3^2 J_4 + 256 J_4^3 - 27 J_3^4} \, .
$$
One way to obtain these results is as follows. First compute the
Hilbert series of the invariant ring. We find that it is
$1 / \bigl( (1-t^2) (1-t^3) (1-t^4) \bigr)$; this suggests
a polynomial ring of invariants with generators of degrees $2,3,4$,
and shows that if we find independent invariants $J_2,J_3,J_4$
of those degrees then ${\bf C}[J_2,J_3,J_4]$ is the full invariant ring.
Now use the basis $\{z_{ij}\}$ of the four-dimensional space,
call it $Z$, of sections of ${\cal O}(1,1)$; it is well-known that
$\{z_{ij}\}$ embeds ${\bf P}^1 \times {\bf P}^1$ into ${\bf P}^3$
as the quadric $z_{11} z_{22} = z_{12} z_{21}$,
identifying ${\rm SL_2} \times {\rm SL_2}$ with
the special orthogonal group ${\rm SO}(Q)$ where
$Q$ is the quadratic form $z_{11} z_{22} - z_{12} z_{21}$.
This identifies $P$ with some other quadratic form $\tilde P$
in the $z_{ij}$, determined uniquely modulo ${\bf C} Q$.
Now $Q$ is nondegenerate, so it identifies $Z$ with its dual $Z^*$,
and thus identifies quadratic forms on $Z$ with self-adjoint maps
$T: Z \to Z$, with $Q$ itself mapping to the identity map.
It is known that generically ${\rm SO}(q)$ orbits of such $T$
are determined by their spectrum, and thus by the characteristic polynomial
$\chi^{\phantom.}_T$. There is a unique translate $\tilde P + cQ$ of trace zero,
represented by the above matrix $M_4$. Hence the coefficients
$J_2,J_3,J_4$ of $\chi^{\phantom.}_{M_4}$ are invariant and independent,
as claimed.
To identify the elliptic curve, write $C$ as a double cover of
one of the ${\bf P}^1$'s by taking the discriminant of $P$ with respect to
the other ${\bf P}^1$, and then use classical formulas for the
Jacobian of a genus-$1$ curve $y^2 = {\rm quartic}$.
The formulas, though not pretty to look at, are short enough
to let us identify the coefficients with polynomials in $J_2,J_3,J_4$.
The resulting curve has a rational point whose $x$-coordinate is
a multiple of $J_2$; translating $x$ to put this point at $x=0$
yields the model $y^2 = x (x-J_2)^2 - 4 J_4 x + J_3^2$ exhibited above.
The visible rational point $(x,y) = (0,J_3)$
ought to correspond to the difference between the divisors
${\mathcal O(0,1)}_C$ and ${\mathcal O(1,0)}_C$,
but I haven't checked this.
Nice answer. Does my polarization argument in my comments above help at all with the $j$ invariant question? It seems to provide an embedding of the moduli space of elliptic curves inside the OP's moduli space, but I don't know if the two ways computing $j$ coincide on this locus.
I'm not sure.
(In any case the natural map goes the other way,
from the moduli of $(2,2)$ curves to moduli of elliptic curves.)
I did use the invariant theory of binary quartics
to write the elliptic curve in terms of $J_2,J_3,J_4$.
The $j$-invariant can be computed from my answer -- e.g. ellinit([0,-2J2,0,J2^2-4J4,J3^2]).j in gp --
though inevitably it's not very pretty.
@AbdelmalekAbdesselam The composition of your map with Noam's is quadratic - it sends a binary quartic $f$ to $ (d^2f /dx_1^1) (d^2 f/dx_2^2) - (d^2 f/dx_1 dx_2)^2$, I think - so it shouldn't preserve the $j$ invariant.
@WillSawin: my map is linear it sends the binary quartic $f(x_1,x_2)$ to $D^2 f$ where $D$ is the differential operator $y_1\frac{\partial}{\partial x_1}+y_2\frac{\partial}{\partial x_2}$. The result is a biform in $x$ and $y$ of bidegree $(2,2)$. What you wrote sounds more like the Hessian of $f$ which is another binary quartic. Sorry if my notations were confusing.
@AbdelmalekAbdesselam Sorry, I should have been more clear. There are actually two maps from biforms of bidegree $(2,2)$ to binary quartics which preserve the $j$ invariant. These are given by taking the discriminant in one of the two variables. They were mentioned as the "natural map" by Noam. To see what your map does to the $j$ invariant, we can compose it with either of them. Doing this gives the Hessian. Thus your map preserves $j$ if and only if the Hessian does. (But it could still help somehow even if it does not preserve $j$.)
@WillSawin: Sorry I misread and missed the part about composition.
|
2025-03-21T14:48:30.517542
| 2020-05-07T15:42:00 |
359643
|
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"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"authors": [
"Carlo Beenakker",
"Drew Brady",
"Iosif Pinelis",
"R W",
"Zhanxiong",
"https://mathoverflow.net/users/11260",
"https://mathoverflow.net/users/121486",
"https://mathoverflow.net/users/36721",
"https://mathoverflow.net/users/498712",
"https://mathoverflow.net/users/71233",
"https://mathoverflow.net/users/8588",
"user135520"
],
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"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:628875",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/359643"
}
|
Stack Exchange
|
Marginal density of uniform spherical distribution
Suppose that $X$ is distributed uniformly in the scaled $n$-sphere $\sqrt{n} \mathbf{S}^{n-1} \subset \mathbf{R}^n$. Then apparently the distribution of $(X_1, \dots, X_k)$, the first $k < n$ coordinates of $X$ has density $p(x_1, \dots, x_k)$ with respect to Lebesgue measure in $\mathbf{R}^k$, moreover if $r^2 = x_1^2 + \cdots + x_k^2$, then it is proportional to
$$
\left(1 - \frac{r^2}{n}\right)^{(n-k)/2 - 1}, \quad \text{if}~0 \leq r^2 \leq n,
$$
and otherwise is 0. I tried to compute this using the fact that $(X_1, \dots, X_k) \stackrel{\rm d}{=} \sqrt{n} (g_1, \dots, g_k)/\sqrt{g_1^2 + \cdots + g_n^2}$, when $g_i$ are iid standard normal variables, but it was somewhat unclear to me how to use even this representation to compute the density. Can anyone sketch the details for me?
$\newcommand{\R}{\mathbb{R}}
\newcommand{\x}{\mathbf{x}}
\newcommand{\X}{\mathbf{X}}$
This is to present a formalization of the answer by Carlo Beenakker, without explicit use of the delta function.
We are going to assume that $\X=(X_1,\dots,X_n)$ is uniformly distributed on the unit sphere $\mathbb S^{n-1}$, rather than on $\sqrt n\,\mathbb S^{n-1}$.
For each real $t\in(0,1)$, define the measures $\mu_t$ and $\nu_t$ over $\R^n$ by the conditions
\begin{equation*}
\int f\,d\mu_t=\int_{\R^n}d\x f(\x)1_{1-t<|\x|^2\le1}
\end{equation*}
and
\begin{equation*}
\int f\,d\nu_t=\frac{\int f\,d\mu_t}{\int d\mu_t}
\end{equation*}
for all (say) nonnegative continuous functions $f\colon\R^n\to\R$, where $|\cdot|$ denotes the Euclidean norm. Then $\nu_t$ is a probability measure converging (as $t\downarrow0$) to the Haar measure $h$ on the unit sphere in $\R^n$, in the sense that (say)
\begin{equation*}
\int f\,d\nu_t\to\int f\,dh
\end{equation*}
for all nonnegative continuous functions $f\colon\R^n\to\R$.
Take now any function $f\colon\R^n\to\R$ such that
\begin{equation*}
f(\x)=g(\x_{n-1})
\end{equation*}
for some nonnegative continuous function $g\colon\R^{n-1}\to\R$ and all $\x\in\R^n$, where $\x_j:=(x_1,\dots,x_j)$ for $\x=(x_1,\dots,x_n)\in\R^n$ and $j=1,\dots,n-1$.
Then
\begin{align*}
\int f\,d\mu_t&=\int_{\R^{n-1}}d\x_{n-1}\,g(\x_{n-1})\int_\R du\, 1_{1-t<|\x_{n-1}|^2+u^2\le1} \\
&=\int_{\R^{n-1}}d\x_{n-1}\,g(\x_{n-1})
(1+o(1))t\,(1-|\x_{n-1}|^2)^{-1/2}\,1_{|\x_{n-1}|<1}.
\end{align*}
Also,
\begin{equation*}
\int d\mu_t=\int_{\R^n}d\x\, 1_{1-t<|\x|^2\le1}\propto(1+o(1))t,
\end{equation*}
where $\propto$ means an equality up to a constant factor, depending only on $n,k$.
So,
\begin{equation*}
\int f\,dh=\lim_{t\downarrow0}
\int f\,d\nu_t\propto\int_{\R^{n-1}}d\x_{n-1}\,g(\x_{n-1})
(1-|\x_{n-1}|^2)^{-1/2}\,1_{|\x_{n-1}|<1}.
\end{equation*}
Thus, the joint pdf of $\X_{n-1}=(X_1,\dots,X_{n-1})$ is given by
\begin{equation*}
p_{n-1}(\x_{n-1})\propto(1-|\x_{n-1}|^2)^{-1/2}\,1_{|\x_{n-1}|<1}.
\end{equation*}
Now successively integrating $p_{n-1}(\x_{n-1})$ ($n-1-k$ times) in $x_{n-1},\dots,x_{k+1}$ and each time using the formula
\begin{equation*}
\int_0^{b^{1/2}}(b-u^2)^p du=c_p b^{p+1/2}
\end{equation*}
for real $b>0$ and $p>-1$ with $c_p:=\int_0^1(1-u^2)^pdu\in(0,\infty)$ (so that $1/2$ is added to the exponent $p$ after such an integration), we see that the joint pdf of $\X_k=(X_1,\dots,X_k)$ is given by
\begin{equation*}
p_k(\x_k)\propto(1-|\x_k|^2)^{(n-k)/2-1}\,1_{|\x_k|<1},
\end{equation*}
as desired.
This is a wonderfully simple calculation and was illuminating. Thank you for the follow up questions, your persistence, and both answers.
@DrewBrady : Thank you for your kind words. However, I feel grateful to Carlo for teaching me something of value, with patience.
Do you by chance know how the ratio$\frac{ \int^{b}{a} (n-x^{2})^{(n-2)/2}dx}{\int^{\sqrt{n}}{-\sqrt{n}} (n-x^{2})^{(n-2)/2}dx} = \frac{ \int^{b}{a} (1-x^{2}/n)^{(n-2)/2}dx}{\int^{\sqrt{n}}{-\sqrt{n}} (1-x^{2}/n)^{(n-2)/2}dx}$ is related to marginal density of the first coordinate?
McKean in his paper: "Geometry of Differential Space" makes use of it to prove that the marginal density of the first coordinate tends to a standard Gaussian. The paper can be found here: https://projecteuclid.org/journals/annals-of-probability/volume-1/issue-2/Geometry-of-Differential-Space/10.1214/aop/1176996973.full
@user135520 : Thank you for the reference to McKean. Yes. it is well known and easy to prove that the distribution of the square of a coordinate, say $U_1$, of a uniformly distributed unit random vector $U$ in $\mathbb R^n$ has the beta distribution with parameters $1/2,(n-1)/2$. The Gaussian approximation to the distribution of $U_1$ easily follows from the representation of $U$ in terms of $n$ iid standard Gaussians.
Oh, is that the same formula that's in the answer? there seems to be an extra $-1$ when I set $k=1$ above? Oh, wait, it must have to do with the square of coordinate.
With $X$ uniformly distributed over the unit $n$-sphere, the joint probability distribution of all $n$ elements of $X$ is a Dirac delta function,
$$P(X_1,X_2,\ldots X_n)\propto\delta\left(1-\sum_{j=1}^n X_j^2\right).\qquad\qquad(1)$$
Now you integrate out elements one by one, to obtain the marginal distribution $P_k$ of $k<n$ elements. The first integration gives
$$P_{n-1}(X_2,X_3,\ldots X_n)\propto\left(1-\sum_{j=2}^n X_j^2\right)^{-1/2}\theta\left(1-\sum_{j=2}^n X_j^2\right),\qquad(2)$$
with $\theta$ the unit step function. The second integration gives
$$P_{n-2}(X_3,\ldots X_n)\propto\theta\left(1-\sum_{j=3}^n X_j^2\right),$$
the third integration
$$P_{n-3}(X_4,\ldots X_n)\propto\left(1-\sum_{j=4}^n X_j^2\right)^{1/2}\theta\left(1-\sum_{j=4}^n X_j^2\right),$$
and so on. Each additional integration increases the power by 1/2,
$$P_{k}(X_{n-k+1},\ldots X_n)\propto\left(1-\sum_{j=n-k+1}^n X_j^2\right)^{(n-k)/2-1}\theta\left(1-\sum_{j=n-k+1}^n X_j^2\right).$$
This is the answer in the OP (without rescaling the radius of the $n$-sphere, so $r^2/n\mapsto r^2$).
As requested in the comments, a more detailed exposition of the various steps.
• **First step:** the delta function. Denote the surface measure on the unit $n$-sphere as $d\Omega$, and $\int d\Omega=A_n$ the surface area. Uniformity of a distribution on the unit $n$-sphere means uniformity with measure $d\Omega$. I maintain that the joint probability distribution of the components of the vector ${\mathbf X}=(X_1,X_2,\ldots X_n)$, uniformly distributed on the unit $n$-sphere, is given by Eq. (1) with normalization constant $2/A_n$. Let us check this by calculating the expectation value of an arbitrary function $f$ of ${\mathbf X}$. For that purpose I transform to hyperspherical coordinates $r,\phi_1,\phi_2,\ldots\phi_{n-1}$,
$$\mathbb{E}[f(\mathbf{X})]=\int dX_1\int dX_2\cdots\int dX_n \,f(X_1,X_2,\ldots X_n)P(X_1,X_2,\ldots X_n)$$
$$\qquad=\int_0^\infty r^{n-1} dr \int d\Omega\, f(r,\phi_1,\phi_2,\ldots\phi_{n-1})\frac{2}{A_n}\delta(1-r^2)$$
$$\qquad=\frac{1}{A_n}\int d\Omega\, f(r=1,\phi_1,\phi_2,\ldots\phi_{n-1}).$$
In the last step I used that $\int_0^\infty r^{n-1}\delta(1-r^2)\,dr=1/2$ for $n\geq 2$.
• **Second step:** integration of the delta function, to arrive at Eq. (2). From now on I will ignore the normalization constants, these can easily be recovered at the end. Let me abbreviate $\sum_{j=2}^n X_j^2=s_2$. The marginal distribution $P_1(X_2,X_3,\ldots X_n)$ is obtained by definition upon integration of $P(X_1,X_2,X_3,\ldots X_n)$ over $X_1$. I carry out this integration in cartesian coordinates, changing variables to $q=X_1^2$,
$$P_1(X_2,X_3,\ldots X_n)\propto \int_{-\infty}^\infty dX_1\delta(1-s_2-X_1^2),$$
$$\qquad=\int_0^\infty\delta(1-s_2-q)\frac{dq}{\sqrt q}=(1-s_2)^{-1/2}\theta(1-s_2).$$
• **Third and following steps:** The following steps, subsequent integrations of $X_2,X_3,\ldots$ are now immediate consequences of the integral
$$\int_0^a(a^2-x^2)^p\,dx=c_p a^{1+2p}.$$
Hm. What does it mean that this distribution "is a delta function"?
@ Carlo Beenakker - are you sure you understand what a joint distribution is?
certainly, let's work this out for $n=2$; we then have in polar coordinates $x_1=r\cos\phi$, $x_2=r\sin\phi$ and we would expect the uniform distribution on the unit circle to be $P(\phi)=\text{constant}$. So let's check that the delta function distribution indeed gives this: take $P(x_1,x_2)\propto\delta(1-x_1^2-x_2^2)$ and integrate to obtain $P(\phi)=\int_0^\infty P(r\cos\phi,r\sin\phi) rdr\propto\int_0^\infty \delta(1-r^2)rdr=\text{constant}$. The multivariate Gaussian is a convenient way to generate the random coordinates, but for this calculation the delta function is easier.
@ Carlo Beenakker - how does the RHS of your formula know that you are dealing with the uniform distribution on the sphere?
the delta function $\delta(1-|X|^2)$ is zero unless the norm $|X|$ of the vector is unity, and it is independent of the orientiation of the vector.
What if the distribution is concentrated just on $(1,0,\dots,0)$?
I'm afraid I don't understand what you're aiming at; the delta function $\delta(1-|X|^2)$ is by definition rotationally invariant, since it depends only on the norm of the vector $X$, so it cannot be concentrated on $(1,0,\ldots 0)$.
It is precisely this definition that I find misleading - the argument of your $\delta$-function is one-dimensional, and I fail to see how this one-dimensional delta function can be rotationally invariant.
the argument of the delta function is a scalar, true, but so is the inner product of two vectors; wouldn't you agree that the inner product is rotationally invariant?
There is a difference: the rotation invariance of the inner product can be formulated mathematically, whereas the rotation invariance of something which is only defined in 1 dimension can not - anyway a good illustration of the difference between physics and mathematics :)
Carlo, I'd really like to understand your solution, especially because it promises an improvement over the Gaussian approach. Alas, I don't understand almost any part of your answer. Especially hard for me to understand your step-by-step integration. Can you write down in detail the integrals for at least the first two steps and how you take them?
@IosifPinelis -- I have added the steps, please do follow up if anything remains unclear.
@CarloBeenakker : Your "First step; the delta function" simply means that $Ef(\mathbf X)=\int f,dh$, where $h$ is the Haar measure on the unit sphere; this is something I had understood. Alas, I still don't understand even your first integration (which you labeled "Second step"). As you said, $\int f,dh$ can be written in hyperspherical coordinates.
Previous comment continued: However, in your "Second step", you seem to write $\int f,dh$ in Cartesian coordinates, as $\int_{-\infty}^\infty\delta(1-s_2-X_1^2),dX_1$. Is that so? And if so, why $\int f,dh=\int_{-\infty}^\infty f,\delta(1-s_2-X_1^2),dX_1$? I think it could help if you would write all the rounds of integration in detail for $n=4$ and $k=2$, say.
@IosifPinelis --- just to make sure I understand your point; once you have the delta function probability distribution all integrals are over $\mathbb{R}$ for each coordinate $X_i$, the delta function ensures that the integrand only contributes on the surface of the unit $n$-sphere; so the marginal distribution of $X_2,X_3,\ldots X_n$ is obtained by integrating $P(X_1,X_2,\ldots X_n)=\delta (1-|\mathbf{X}|^2)$ over $X_1$ from $-\infty $ to $+\infty$. If we agree on this, I can work out that integral.
to put it differently, if we agree on this equation $$\mathbb{E}[f(\mathbf{X})]=\int dX_1\int dX_2\cdots\int dX_n ,f(X_1,X_2,\ldots X_n)\delta(1-|\mathbf{X}|^2)$$ then the marginal distribution $P(X'_2,X'_3,\cdots X'_n)$ is obtained by substituting $f\mapsto \delta(X_2-X'_2)\delta(X_3-X'_3)\cdots \delta (X_n-X'_n)$. If desired, the integrals might be carried out in hyperspherical coordinates, but that is not convenient, integration in cartesian coordinates gives the answer directly.
@CarloBeenakker : I don't think I made any point. In the first part of my latest comment, I presented my (hopefully correct) understanding of your "First step" -- just to write $Ef(\mathbf X)=\int f,dh$; I can think of no other formal meaning of your $\delta$ symbol. In the second part of that comment, I just asked whether you write $\int f,dh$ as $\int_{-\infty}^\infty f,\delta(1-s_2-X_1^2),dX_1$, and if so, why.
@CarloBeenakker : I can understand the displayed equation in you latest comment only as $Ef(\mathbf X)=\int f,dh$, where $h$ is Haar measure on the unit sphere. I don't understand how you treat this equation in (apparently) Cartesian coordinates.
these are two equivalent ways to calculate the surface average of a function $f(\mathbf{X})$ of a rank-$n$ vector $\mathbf{X}$: either as an $n$-fold integration in cartesian coordinates of the product $f(\mathbf{X})\delta(1-|\mathbf{X}|^2)$, or as an $n-1$-fold integration $\int fdh$ over the unit sphere; the two ways are equivalent, but the first way is more convenient.
Can you formally define your $\delta$ symbol?
well, I would define $\delta(x)=d\theta(x)/dx$ as the derivative of the unit step function; I don't really need many properties; the only property I use is that $\int_{-\infty}^\infty f(x) \delta(x-x_0)dx=f(x_0)$ for suitable $f(x)$.
OK, now we are back to this: Why is your iterated integral the same as $\int f,dh$?
I think I got it -- thank you. Conversations with physicists are not always easy, but usually fruitful. :-)
thank you, Iosif.
Let $G_1,\dots,G_n$ be iid standard normal random variables. Then the random vector
\begin{equation*}
(Y_1,\dots,Y_n):=\Big(\frac{G_1}{\sqrt{\sum_1^n G_j^2}},\dots,
\frac{G_n}{\sqrt{\sum_1^n G_j^2}}\Big)
\end{equation*}
is uniformly distributed on the unit sphere $\mathbb S^{n-1}$.
Let
\begin{equation*}
\begin{aligned}
Z_i&:=Y_i=\frac{G_i}{\sqrt{\sum_1^n G_j^2}}&\text{ if }i\le k,\\
Z_i&:=G_i&\text{ if }i> k.
\end{aligned}
\end{equation*}
We want to find the joint pdf of $(Z_1,\dots,Z_k)$, which is the same as the joint pdf of $(X_1,\dots,X_k)/\sqrt n$.
The vector $(Z_1,\dots,Z_n)$ is obtained from $(G_1,\dots,G_n)$ by the transformation given by
\begin{equation*}
\begin{aligned}
z_i&:=\frac{g_i}{\sqrt{\sum_1^n g_j^2}}&\text{ if }i\le k,\\
z_i&:=g_i&\text{ if }i> k.
\end{aligned}
\end{equation*}
The transformation inverse to this is given by
\begin{equation*}
\begin{aligned}
g_i&:=\sqrt{s_2}\frac{z_i}{\sqrt{1-s_1}}&\text{ if }i\le k,\\
g_i&:=z_i&\text{ if }i> k,
\end{aligned} \tag{1}
\end{equation*}
where
\begin{equation*}
s_1:=\sum_1^k z_j^2,\quad s_2:=\sum_{k+1}^n z_j^2.
\end{equation*}
The Jacobian determinant of the inverse transformation is
\begin{equation*}
J=\det(cM)=c^k\det M,
\end{equation*}
where
\begin{equation*}
c:=s_2^{1/2}(1-s_1)^{-3/2},\quad M:=(1-s_1)I_k+UU^T,
\end{equation*}
$I_k$ is the $k\times k$ identity matrix, and $U:=[z_1,\dots,z_k]^T$.
Write $U=|U|Qe_1$, where $|U|=\sqrt{s_1}$ is the Euclidean norm of $U$, $Q$ is some orthogonal matrix, and $e_1:=(1,0,\dots,0)$. Then it is clear that the matrix $M$ is similar to $N:=(1-s_1)I_k+|U|^2e_1e_1^T=(1-s_1)I_k+s_1e_1e_1^T$, whence
$\det M=\det N=(1-s_1)^{k-1}$. So,
\begin{equation*}
J=s_2^{k/2}(1-s_1)^{-k/2-1}. \tag{2}
\end{equation*}
Also, the joint pdf of $(G_1,\dots,G_n)$ is given by
\begin{equation*}
(2\pi)^{-n/2}\exp\Big\{-\frac12\sum_1^n g_j^2\Big\}.
\end{equation*}
So, in view of (1) and (2), the joint pdf of $(Z_1,\dots,Z_n)$ is given by
\begin{equation*}
f_n(z_1,\dots,z_n)
=(2\pi)^{-n/2}\exp\Big\{-\frac12\frac{s_2}{1-s_1}\Big\}s_2^{k/2}(1-s_1)^{-k/2-1}.
\end{equation*}
So, the joint pdf of $(Z_1,\dots,Z_k)$ is given by
\begin{align*}
f_k(z_1,\dots,z_k)&=\int_{\mathbb R^{n-k}}dz_{k+1}\dots dz_n\,f_n(z_1,\dots,z_n) \\
&=(2\pi)^{-n/2}(1-s_1)^{-k/2-1} \\
&\times \int_{\mathbb R^{n-k}}dz_{k+1}\dots dz_n\,s_2^{k/2}\,\exp\Big\{-\frac12\frac{s_2}{1-s_1}\Big\} \\
&\propto(1-s_1)^{(n-k)/2-1},
\end{align*}
because $s_2=\sum_{k+1}^n z_j^2$. So, we have the desired result.
More specifically,
\begin{align*} f_k(z_1,\dots,z_k)
=\frac{\Gamma(n/2)}{\pi^{k/2}\Gamma((n-k)/2)}(1-s_1)^{(n-k)/2-1}
\end{align*}
for $s_1=\sum_1^k z_j^2\in(0,1)$.
The density seems not correct. According to Exercise 1.32 in Aspects of Multivariate Statistical Theory by Robb J. Muirhead, the normalization factor should be $\frac{\Gamma(n/2)}{\color{red}{\pi^{k/2}}\Gamma((n - k)/2)}$.
@Zhanxiong : Thank you for your comment. The constant factor is now fixed.
If I followed your calculation, the density I can get is $\frac{\Gamma(n/2)}{(n - k)\Gamma((n - k)/2)\pi^{k/2}}(1 - s_1)^{(n - k)/2 - 1}$, still cannot match the published density. Can you elaborate your calculation?
By the way, can you check where the argument below is problematic: If let $\sigma(S_{n - 1}(a))$ denote surface area of the $n$-dim sphere, then the joint density of $(X_1, X_2, \ldots, X_n)$ is $1/\sigma(S_{n - 1}(1))$. Partition $(X_1, \ldots, X_n)$ into $(T_1, T_2)$, where $T_1 = (X_1, \ldots, X_k)$, then the marginal density of $T_1$ is $f_{T_1}(t_1) = \int_{t_2: t_1't_1 + t_2't_2 = 1} 1/\sigma(S_{n - 1}(1))dt_2 = \frac{\Gamma(n/2)}{2\pi^{n/2}} \sigma(S_{n - k - 1}(\sqrt{1 - t_1't_1}))$.
(continued): $=\frac{\Gamma(n/2)}{2\pi^{n/2}}\frac{2\pi^{(n - k)/2}}{\Gamma((n - k)/2)}(1 - t_1't_1)^{(n - k - 1)/2} = \frac{\Gamma(n/2)}{\pi^{k/2}\Gamma((n - k)/2)}(1 - t_1't_1)^{(n - k - 1)/2}$. This is $(1 - t_1't_1)^{-1/2}$ short from the correct answer. But I can't see where the logic went wrong.
@Zhanxiong : The normalizing factor is uniquely determined by the variable factor $(1-s_1)^{(n-k)/2-1}$. So, I don't know what you mean by "If I followed your calculation, the density I can get is [...]".
What I meant is if the normalizing factor is not determined by integrating $(1 - s_1)^{(n - k)/2 - 1}$, but directly evaluating the integral you derived (using polar system transformation), then I got an extra term $(n - k)^{-1}$.
@Zhanxiong : If you are somehow getting the right variable factor $(1-s_1)^{(n-k)/2-1}$ but a wrong normalizing factor, then you have probably made a mistake along the way. If the mistake is caused by following my derivation of $(1-s_1)^{(n-k)/2-1}$, then you should be able to find a mistake in my derivation.
Let us continue this discussion in chat.
@Zhanxiong : I do not engage in such chat.
$\newcommand{\R}{\mathbb{R}}
\newcommand{\x}{\mathbf{x}}
\newcommand{\X}{\mathbf{X}}$
Here is yet another solution, which is partly informal but I think not hard to completely formalize. Its advantage is a strong and hopefully convincing appeal to geometric intuition.
Again, we are going to assume that $(X_1,\dots,X_n)$ is uniformly distributed on the unit sphere $\mathbb S^{n-1}$, rather than on $\sqrt n\,\mathbb S^{n-1}$.
Let $\X_j:=(X_1,\dots,X_j)$ and $\x_j:=(x_1,\dots,x_j)$ for $\x=(x_1,\dots,x_n)\in\R^n$ and $j=1,\dots,n-1$. Let $|\cdot|$ denotes the Euclidean norm.
The main point is that the probability density $p_{n-1}(\x_{n-1})$ of $\X_{n-1}$ at a point $\x_{n-1}\in\R^{n-1}$ with $|\x_{n-1}|<1$ is proportional to the ratio $r_{n-1}(\x_{n-1}):=vol_{n-1}(dS)/vol_{n-1}(dA)$, where $vol_{n-1}$ is of course the $(n-1)$-volume, $dA$ is an infinitesimal neighborhood of the point $\x_{n-1}$ in $\R^{n-1}$ and $dS$ is the preimage of $dA$ under the projection of the upper hemisphere $\mathbb S^{n-1}_+:=\{\x\in\mathbb S^{n-1}\colon\x\cdot e_n\ge0\}$ onto the closed unit ball in $\R^{n-1}$; this projection is given by $\mathbb S^{n-1}_+\ni(\x_{n-1},u)\mapsto\x_{n-1}$; here $e_n:=(0,\dots,0,1)$ and $\cdot$ denotes the dot product.
But
\begin{equation}
r_{n-1}(\x_{n-1})=\frac{vol_{n-1}(dS)}{vol_{n-1}(dA)}=\frac1{\cos\phi},
\end{equation}
where $\phi$ is the angle between the hyperplane $\R^{n-1}\times\{0\}$ of $\R^n$ and the tangent hyperplane to $\mathbb S^{n-1}$ at the point $(\x_{n-1},\sqrt{1-|\x_{n-1}|^2})\in\mathbb S^{n-1}_+$; that is, $\phi$ is the angle between the corresponding normal vectors $e_n$ and $(\x_{n-1},\sqrt{1-|\x_{n-1}|^2})$ of these two hyperplanes.
Thus,
\begin{equation}
p_{n-1}(\x_{n-1})\propto r_{n-1}(\x_{n-1})\propto\frac1{\cos\phi}=(1-|\x_{n-1}|^2)^{-1/2},
\end{equation}
where $\propto$ means an equality up to a constant factor, depending only on $n,k$.
Now successively integrating $p_{n-1}(\x_{n-1})$ ($n-1-k$ times) in $x_{n-1},\dots,x_{k+1}$ and each time using the formula
\begin{equation*}
\int_0^{b^{1/2}}(b-u^2)^p du=c_p b^{p+1/2}
\end{equation*}
for real $b>0$ and $p>-1$ with $c_p:=\int_0^1(1-u^2)^pdu\in(0,\infty)$ (so that $1/2$ is added to the exponent $p$ after such an integration), we see that the joint pdf of $\X_k=(X_1,\dots,X_k)$ is given by
\begin{equation*}
p_k(\x_k)\propto(1-|\x_k|^2)^{(n-k)/2-1}\,1_{|\x_k|<1},
\end{equation*}
as desired.
Here is a picture, for $n=3$, showing the upper hemisphere $\mathbb S^{n-1}_+$; a small neighborhood of a point $\x_{n-1}$ in the projection of $\mathbb S^{n-1}_+$ to the horizontal plane $\R^{n-1}\times\{0\}$ of $\R^n$; the preimage of that neighborhood under that projection; and the normal vectors of the horizontal plane and the tangent plane to the sphere -- with $\phi$ being the angle between these two normal vectors.
Although great answers have already been provided, the one provided below is perhaps be the shortest possible answer to the question.
Let $x_1,\ldots,x_k \in \mathbb R$ such that $r_k^2:=\sum_{i=1}^k x_i^2 < 1$, and note that the marginal density of $(X_1,\ldots,X_k)$ at $(x_1,\ldots,x_k)$ equals $p_n(r_k^2)$, where
$$
p_n(t) \propto \int_0^\infty\ldots \int_0^\infty dx_{k+1}\ldots dx_n\delta(\sum_{i=1}^n x_i^2-t) \,\forall t \in \mathbb R.
$$
The Laplace transform w.r.t $t$ is given by
$$
\begin{split}
\hat{p}_n(s) = \int_0^\infty e^{-ts}p_n(t)dt &\propto \int_0^\infty e^{-s\sum_{i=1}^n x_i^2}dx_{k+1}\ldots dx_n \\
&= e^{-s\sum_{i=1}^k x_i^2}\int_0^\infty e^{-s\sum_{i=k+1}^n x_i^2}dx_{k+1}\ldots dx_n\\
&= e^{-sr_k^2}\left(\int_0^\infty e^{-sz^2}dz\right)^{(n-k)} \propto e^{-sr_k^2}s^{-(n-k)/2}.
\end{split}
$$
Evaluating (e.g via mathematica, etc.) the inverse Laplace transform of the last term at $t=r_k^2$, we deduce that
$p_n(r_k^2) \propto (1-t)^{(n-k)/2-1}\delta(t-r_k^2)\bigg\rvert_{t=r_k^2} = (1-r_k^2)^{(n-k)/2-1}$.
I enjoyed thinking about these answers and this is my attempt to put them into (nonrigorous) geometrical terms. Writing the joint density compositionally as
$$p(\mathbf{x}_k \mid |\mathbf{x}| = 1)p(\mathbf{x}_{n-k} \mid \mathbf{x}_k, |\mathbf{x}| = 1) = p(\mathbf{x} \mid |\mathbf{x}| = 1) \propto 1,$$
we want to solve for the first term on the left. But since our density is proportional to a constant, this is just
$$p(\mathbf{x}_k \mid |\mathbf{x}| = 1) \propto \frac{1}{p(\mathbf{x}_{n-k} \mid \mathbf{x}_k, |\mathbf{x}| = 1)}.$$
Accordingly, instead of performing our calculation by integrating out $X_{k+1} \dots X_{n}$, we can think about it in terms of the conditional density for sampling $\mathbf{X}_{n-k}$, given $\mathbf{X}_k$ and the norm constraint $|\mathbf{X}| = 1$, denoted in the denominator above.
I propose a two step procedure. First, draw a point uniformly from within the $n-k-1$ dimensional ball with radius $$r_{n-k} = \sqrt{1 - |\mathbf{x}_k|^2}.$$ Each such vector has density proportional to $$\left(1 - |\mathbf{x}_k|^2\right)^{-(n-k-1)/2}.$$
This corresponds to the second and higher integrations in the previous answers, whereas here we directly use the formula for the volume of a ball.
Next, the $(n-k)$th coordinate must satisfy $|\mathbf{x}| = 1$, which is achieved by any point on the circle with radius $r_{n-k}$, a set with measure proportional to $r_{n-k}$. Proving this is the first integration in the previous answers.
Putting these two steps together and taking the reciprocal gives
$$p(\mathbf{x}_k \mid |\mathbf{x}| = 1) \propto \left(1 - |\mathbf{x}_k|^2\right)^{(n-k)/2 - 1}.$$
|
2025-03-21T14:48:30.519065
| 2020-05-07T15:49:52 |
359644
|
{
"all_licenses": [
"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Gerhard Paseman",
"Wojowu",
"Zach Teitler",
"https://mathoverflow.net/users/30186",
"https://mathoverflow.net/users/3402",
"https://mathoverflow.net/users/88133"
],
"include_comments": true,
"license": "Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/",
"provenance": "stackexchange-dolma-0006.json.gz:628876",
"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/359644"
}
|
Stack Exchange
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Interesting things you learned while grading/marking?
What are some interesting mathematical things you have learned while grading (or marking, if you prefer) student work? For example, clever proofs that students came up with; nice counterexamples or insights; interesting new questions inspired while grading; even just something you looked up to find out if a student's work was valid. Answers can be things that students wrote, or inspired by something a student wrote, or just something we learned during the grading process in some way.
It is final exams time here, so if anyone can help cast a more positive light on the grading experience, it would be most welcome.
Please refrain from snarky answers about things we learned about our students and their state of knowledge. Please stick to mathematical things. On the other hand, puns about graded rings, marked points, etc., are allowed.
Ah, I just thought it would be a fun chance to make grading just a little more pleasant. Oh well. I guess I should have posted it in matheducators.SE instead.
Either that or just Math.SE I would say.
I learned to be less clever. I was a linear algebra TA for a professor who found one of my questions "beautiful! The students won't understand it!", and put it as question 1 with my name for attribution on the final. Then he assigned me the task of grading that problem. Gerhard "Of Course He Was Right" Paseman, 2020.05.07.
https://matheducators.stackexchange.com/q/18297/12530
|
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