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2025-03-21T14:48:31.352117
| 2020-06-24T07:26:28 |
363995
|
{
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"authors": [
"Dieter Kadelka",
"Federico Poloni",
"Ivan Meir",
"MathGuest",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/363995"
}
|
Stack Exchange
|
Solving equation of matrix valued functions
Given $n\times n$ matrices with entire functions entries (holomorphic on all of the complex plane $\mathbb{C}$)
$A(z)=[a_{ij}(z)],B(z)=[b_{ij}(z)]$,
i.e.,
$a_{ij}(z),b_{ij}(z)$ are entire functions for all $i,j=1,\dots,n$.
Can we find a form of a third $n\times n$ matrix $C(z)=[c_{ij}(z)]$ (in terms of $A$ and $B$, and under some conditions on $A$ and $B$ as needed) such that
$$A(z)A^{*}(v)+B(z)B^{*}(v)=C(z)C^{*}(v)$$ for all $z,v\in\mathbb{C}$?
Edit:
where $A^{*}(v)=\left(\overline{A(v)} \right)^{T}$ (the Hermitian transpose of $A(v)$).
I tried adding and subtracting $A(z)B^{*}(v)$ to the LHS but it doesn't work!
I don't understand. Is it true that $A^(v) = (A(\bar v))^T$ and not $A^(v) = (\bar A(\bar v))^T$?
@DieterKadelka I think it is meant to say $A^(v)=A(v)^$, simply the hermitian transpose of A. So if $A(v)=(a_{ij}(v))$ then $A^*(v)=(\bar a_{ji}(v))$
@ivan, that's true, thank you!
First take the case $n=1$. Let $A(z)=(a(z))$, $B(z)=(b(z))$, $a(z),b(z)$ entire functions. Then $A(z)A^*(z)+B(z)B^*(z)=|a(z)|^2+|b(z)|^2$. Therefore we need $C(z)=(c(z))$ with $c(z)$ an entire function s.t $A(z)A^*(z)+B(z)B^*(z)=C(z)C^*(z)=|C(z)|^2=|a(z)|^2+|b(z)|^2$ for all $z\in \mathbb {C}$.
Therefore $|C(z)|^2\geq |a(z)|^2$ and $|C(z)|^2\geq |b(z)|^2$ for all $z\in \mathbb {C}$.
Liouville's theorem says that no entire function dominates another unless they are related by a constant multiple. Hence we must have $a(z)=\lambda b(z)$for some $\lambda \in \mathbb {C}$ and then $c(z)=\sqrt{1+|\lambda|^2}b(z)$.
Thus for $n=1$ we require that $A$ and $B$ are scalar multiples of one another and this also satisfies the more general equation with $z$ and $w$.
I believe in general you may need $A(z)=T(z)M$ and $B(z)=T(z)N$ where $M, N \in \mathbb{C}_{n\times n}$, constant matrices and $T(z)$ is any $n \times n$ matrix with entire entries.
This works because then
$$A(z)A^*(w)+B(z)B^*(w)=T(z)(MM^*+NN^*)T^*(w)$$$$=(T(z)Q)(T(w)Q)^*=C(z)C(w)^*$$
where $C(z)=T(z)Q$ a matrix with entire entries and $QQ^*=MM^*+NN^*$. $Q$ exists because $MM^*+NN^*$ is hermitian and has a Cholesky Decomposition.
To prove this we need the following lemma:
Lemma: For any $n \times n$ matrix $A(z)$ with entire entries there exists a matrix $A'(z)$ also with entire entries such that $A'(z)A(z)=A(z)A'(z)=D(z)I$ where $D(z)$ is entire.
Proof: $A'$ is just the Adjugate Matrix of $A$ and $D(z)=\det(A).$ $\blacksquare$
Theorem:
If $A(z)$, $B(z)$ and $C(z)$, all $n \times n$ matrices with entire entries, satisfy $$A(z)A^*(z)+B(z)B^*(z)=C(z)C^*(z) \tag{1} $$ for all $z \in \mathbb {C}$ then $A=C(z)M$, $B=C(z)N$ for constant $n \times n$ matrices $N$ and $M$ with $MM^*+NN^*=I$.
Proof:
By the lemma there exists an entire matrix $C'(z)$ s.t $C'(z)C(z)=C(z)C'(z)=\det(C(z))$.
Multiply equation (1) on the left by $C'$ and the right by $C'^*$. This gives
$$a(z)a^*(z)+b(z)b^*(z)=|\det(C(z))|^2 I $$
where $a(z)=C'(z)A(z)$, $b(z)=C'(z)B(z)$, both matrices with entire entries.
If $a(z)=(a_{ij}(z))$, $b(z)=(b_{ij}(z))$ then the main diagonal of the LHS, $L(z)=(l_{ij})$ is given by $$l_{ii}=\sum_{j=1}^n(|a_{ij}|^2+|b_{ij}|^2)=|\det(C(z))|^2. \tag{2}$$
Hence for all the entire functions $a_{ij}$ and $b_{ij}$ we have $$|a_{ij}|\leq |\det(C(z))|.$$ and $$|b_{ij}|\leq |\det(C(z))|$$ where $\det(C(z))$ is also an entire function.
Again by Liouville's Theorem we must have $|a_{ij}|=\lambda_{ij} \det(C(z))$ and $|b_{ij}|=\mu_{ij} \det(C(z))$ for some $\lambda_{ij}, \mu_{ij} \in \mathbb {C}.$
This clearly implies that $a=\det(C(z)) M$ and $b=\det(C(z)) N$ for constant $n \times n$ matrices $M=(\lambda_{ij})$ and $N=(\mu_{ij})$.
Hence by the definitions of $a$ and $b$,
$C'(z)A(z)=\det(C(z)) M$ and $C'(z)B(z)=\det(C(z)) N$.
Note that if $\det(C(z))=0$ then by equation (2) $A(z)=0$, $B(z)=0$ and hence $C(z)=0$. Since then $A(z)=C(z)I$ and $B(z)=C(z)I$ we may take $M=N=I$.
We can therefore assume henceforth that $\det(C(z))\neq 0$.
Multiplying both equations on the left by $C(z)$ gives $\det(C(z))A(z)=\det(C(z)) C(z) M$ and $\det(C(z))B(z)=\det(C(z)) C(z) N.$
Hence we have, as $\det(C(z))\neq 0$,
$A(z)=C(z) M$ and $B(z)= C(z) N.$
Substituting back into equation (1) gives
$$A(z)A^*(z)+B(z)B^*(z)=C(z)(MM^*+NN^*)C^*(z)$$$$=C(z)C(z)^*$$
for all $z$, which implies $MM^*+NN^*=I$ since $C(z)$ is invertible. $\blacksquare$
the equation says that $A(z) A^{*}(w)+... $, but you used $A(z) A^{*} (z) ... $ in the first line!
I was taking the special case $z=w$ which obviously must hold if the equation holds for all $z$ and $w$.
Note that I check that the consequence of the that A is proportional to B also satisfies your more general equation.
Is it possible to prove that this condition on $A$ and $B$ is the only one?
@MathGuest Yes I think it probably is possible. Note that for special cases like diagonal $A$ and $B$ it follows from the $n=1$ case. I'll have a think about it :-)
If you take $C(z) = $ any 2x2 matrix-valued function, $A=C(z)\begin{bmatrix}0 & 1\ 0 & 0\end{bmatrix}$, $B=C(z)\begin{bmatrix}0 & 0 \ 1 & 0\end{bmatrix}$, the equality holds but $B$ is not a matrix multiple of $A$ nor viceversa.
@FedericoPoloni Thank you Federico that is a great example! I think it works because $AB^=BA^=0$ and so $(AA^+BB^)=(A+B)(A+B)^*$. I had originally thought that case was covered by my condition but obviously not! Food for thought :-)
In fact if $AB^=BA^$ then $(AA^+BB^)=(A+iB)(A+iB)^*$ so we have a decomposition in general in this case also which as I said I originally thought was covered. Your example shows this is not the case.
I've updated the conditions on $A$ and $B$ in my solution.
@IvanMeir, Thank you, brilliant work!
@MathGuest No worries, it was an interesting challenge - glad I could help!
@Ivan So the form of the matrices $M$ and $N$ depends basically on $\det(C(z) ) \neq 0$ for all $z\in \mathbb{C} $, but what if $\det(C(z) ) =0$ for some points and $\det(C(z) ) \neq 0$ for other points? Should the matrices $M$and $N$ be independent of those $z$ where the determinant is zero or nonzero?
|
2025-03-21T14:48:31.352749
| 2020-06-24T07:52:48 |
363996
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
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"C7X",
"https://mathoverflow.net/users/479330"
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}
|
Stack Exchange
|
Self-contained proof of WO of Buchholz's ordinal notation system
I would like a self-contained proof that the ordinal notation system defined by Buchholz in this paper is indeed well-ordered. Meaning, I would like a proof that does not rely on ordinals. Buchholz's system defines ordinals below $\Psi_0(\Omega_\omega)$.
The system is recursively defined as follows: A term is an expression of the form $\beta = \alpha_1 + \cdots + \alpha_k$, $k\ge 0$, where each $\alpha_i$ is a previously defined principal term. A principal term is of the form $D_n(\beta)$ for $n\in\mathbb N$, where $D_n$ is just a formal symbol, and $\beta$ is a previously defined term. The term with zero principal terms is denoted $0$.
(Each term can be thought of as an ordered, rooted tree whose edges are labeled with natural numbers. A principal term is then a tree in which the root has a single child.)
Recursively define a binary relation "$<$" on terms and principal terms as follows: Let $\beta = \alpha_1 + \cdots + \alpha_k$ and $\beta' = \alpha'_1 + \cdots + \alpha'_\ell$ be two terms. Then $\beta<\beta'$ if either $\beta'$ is an extension of $\beta$ (meaning $k<\ell$ and $\alpha_i=\alpha'_i$ for all $i\le k$) or the smallest index $i$ at which $\beta,\beta'$ differ satisfies $\alpha_i<\alpha'_i$. Let $\alpha=D_n(\beta)$ and $\alpha'=D_m(\beta')$ be two principal terms. Then $\alpha<\alpha'$ if either $n<m$, or else $n=m$ and $\beta<\beta'$.
It's easy to show that the relation "$<$" is a total order. However, it is not a well order. For example, $D_1(0)>D_0(0)+D_1(0) > D_0(0)+D_0(0)+D_1(0)> \cdots$. Another example: $D_1(0)>D_0(D_1(0))> D_0(D_0(D_1(0)))>\cdots$.
We now define a subset of terms which we call ordinal terms (OT). First we define, for each term $\beta$ and each $n$, a set $G_n(\beta)$ of subterms of $\beta$. Suppose $\beta = \alpha_1 + \cdots + \alpha_k$. Then we define $G_n(\beta) = G_n(\alpha_1) \cup \cdots \cup G_n(\alpha_k)$. For a principal term $\alpha = D_m(\beta)$, we define $G_n(\alpha)$ as $\emptyset$ if $m<n$, and as $\{\beta\}\cup G_n(\beta)$ otherwise.
(If we think of terms as trees, then given a tree $t$ and an integer $n$, we start at the root of $t$. We follow all possible paths towards the leaves, continuing as long as we meet edges labeled $n$ or larger. For each such edge $e$ we take the tree supported by $e$ without $e$ itself. Edges with labels smaller than $n$ serve as no-entry barriers.)
Now, a term $\beta = \alpha_1+\cdots+\alpha_k$ is an ordinal term if each $\alpha_i$ is a principal ordinal term and $\alpha_1\ge\alpha_2\ge\cdots\ge\alpha_k$. And $\alpha = D_n(\beta)$ is a principal ordinal term if $\beta$ is an ordinal term and $\beta'<\beta$ for each $\beta'\in G_n(\beta)$. (Note that here it says $G_n(\beta)$ and not $G_n(\alpha)$.)
So for example, $D_n(D_m(0))$ is an OT for all $n,m$. But $D_0(D_0(D_1(0))$ is not an OT.
Finally, the "relevant ordinal terms" are those OT's of the form $D_0(\cdots) + \cdots$. (Because $D_0$ corresponds to the collapsing function $\Psi_0$ that produces countable ordinals... nevermind.)
So the question is to give a direct proof that the set of relevant ordinal terms is WO under "$<$". That's the end of the question.
To illustrate what I mean by a "self-contained proof", here is a sketch of the proof that, if we limit the subscripts $n$ in $D_n$ to just $n=0$, then the set is WO. (In fact, these OT's correspond to the ordinals below $\varepsilon_0$.) Call the height of a term the height of the corresponding tree (i.e. length of the longest path from the root to a leaf). Note that if two OT's have different heights, then the one with the larger height will always be larger, and that the requirement regarding the sets $G_n$ in the definition of OT's is irrelevant, since it is always satisfied.
We prove by induction on the height $h$ that there is no infinite descending sequence of OT's of height $\le h$. Suppose this is true for height $\le h-1$, and let $\beta_1 > \beta_2 > \cdots$ be an infinite decreasing sequence in which $\beta_1$ has height $\le h$. Denote $\beta_i = \sum_j D_0(\alpha_{i,j}) \cdot m_{i,j}$ (grouping equal principal terms). Then the sequence $\alpha_{i,1}$ is nonincreasing, and it becomes constant at some point (by the induction assumption). Afterwards, the coefficient $m_{i,1}$ is nonincreasing and becomes constant at some later point. Afterwards, $\alpha_{i,2}$ is nonincreasing and becomes constant at some later point. And so on. We obtain an infinite decreasing sequence $\alpha_{i,1} > \alpha_{i,2} > \cdots$, which is a contradiction.
Do you consider this result to be finitistic enough? https://core.ac.uk/download/pdf/82190282.pdf (Okada, "Note on a Proof of the Extended Kirby-Paris Theorem on Labeled Finite Trees".) I believe this proves that assuming consistency of $(\Pi_1^1-CA)+BI$, Buchholz's hydra game always terminates by relating the way Buchholz hydras expand with proof reductions in $(\Pi_1^1-CA)+BI$, whose proofs are all finite.
|
2025-03-21T14:48:31.353084
| 2020-06-24T08:52:42 |
363999
|
{
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"authors": [
"ABIM",
"Jochen Wengenroth",
"https://mathoverflow.net/users/21051",
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"url": "https://mathoverflow.net/questions/363999"
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|
Stack Exchange
|
Creating an inverse system which "stratifies density"
Setting:
Let $X'$ be a dense subset of an infinite-dimensional Fréchet space $X$ and suppose that $(X_n')_{n \in \mathbb{N}}$ is a nested sequence of non-empty subsets of $X'$ satisfying
$$
\bigcup_{n \in \mathbb{N}} X_n'=X - N ,
$$
where $N$ is a nowhere dense subset of $X$.
Question:
Are there reasonable conditions on $\{X_n'\}$, such that, for every $n \in \mathbb{N}$, there are topologies $\{\tau_n\}$ on $X$ satisfying:
$1_X:(X,\tau_{n+1})\rightarrow (X,\tau_{n})$ is continuous but the inverse map is not,
$\varprojlim_n (X,\tau_n) = X$
$\bigcup_{i=1}^n X_i'$ is dense in $(X,\tau_n)$,
$\tau_0$ is non-trivial, in the sense that: $\tau_0\neq \{\emptyset,X\}$.
If $X$ is a proper Fréchet space (i.e., not isomorphic to a Banach space) you can choose an increasing sequence of semi-norms $p_n$ giving the Fréchet space topology of $X$ such that the $p_{n}$-topology $\tau_n$ is strictly finer than $\tau_{n-1}$. These semi-normed topologies satisfy you requirements (although non-trivial is not very specific).
Do you have a reference? For example, it's not clear to me how this would make $X_n'$ dense in the $p_n$-topology?
If it is dense in $X$ it remains dense with respect to the coarser topologies $\tau_n$.
But $\bigcup_n X_n'$ is dense in $X$ (and therefore the coarser topologies) but why is $X_n'$ dense (on it's own) in $\tau_n$ (or $\bigcup_{i=1}^n X_i'$ dense in $\tau_n$)?
I described the standard way to make a Fréchet space the projective Limit of (semi-) normed spaces. In general, the sets $X_n'$ won't be dense. For example, each $X_n'$ could be a finite dimensional subspace, if the $p_n$ are proper norms (and not only semi-norms) then they are closed. What do you want to do with construction of the $\tau_n$?
Well, for me $X=C(\mathbb{R},\mathbb{R})$, $\cup X_n'$ is a set of continuous functions and I want to describe the "approximation-theoretic impact" of adding $X_{n+1}'-X_n'$. In my setting, $X_n'$ are affine subspaces (or "linear manifolds" as some authors say). I guess, it would be more appropriate if $X$ is instead required to be a (strict) LB/LF-space?
|
2025-03-21T14:48:31.353257
| 2020-06-24T10:22:38 |
364007
|
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/364007"
}
|
Stack Exchange
|
Average value of a fractional part of a function
Let $f(x): \mathbb{R} \to \mathbb{R}_{\geq 0}$ be a smooth function. I am interested in estimating
sums of the form
$$
\sum_{ A < n \leq B } \{ f(n)\}
$$
where $\{ c \}$ denotes the fractional part of $c \in \mathbb{R}$. Trivial upper bound is $B-A$. Are there any known non-trivial upper bounds for 'nice' $f$? say for example something like a polynomial or $f(x) = \sqrt{x}$ or $B/x$?
You should look at Equidistribution Theory, in particular equidistribution modulo 1. Equidistribution can be checked by confirming that $$\lim_{n\rightarrow \infty}\sum_{i=1}^n e^{2 \pi i f(n)}=0.$$
Then you should have $$\sum_{A<n<B}\{f(n)\}=(B-A)/2+o(B-A)$$ which gives you an upper bound of $$|\sum_{A<n<B}\{f(n)\}|\leq|B-A|(1/2+\epsilon)$$ for all $\epsilon>0$ and large enough $B-A$.
In particular this will hold if $f$ is a polynomial with at least one coefficient other than the constant term irrational. Also $\{x^n\}$ is equidistributed for almost all $x\in \mathbb{R_{>1}}$
A sum of fractional parts (1967) contains some results that may be of interest.
If $A=0$ and $B$ is a prime $p$, and $f(x)$ is a polynomial with integer coefficients, which does not vanish identically modulo $p$, then
$$\sum_{n=0}^{p-1}\{f(n)/p\}=\tfrac{1}{2}p+{\cal O}(\sqrt{p}\log p).$$
|
2025-03-21T14:48:31.353362
| 2020-06-24T11:18:08 |
364015
|
{
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"authors": [
"Jason DeVito - on hiatus",
"Malkoun",
"Michael Albanese",
"https://mathoverflow.net/users/1708",
"https://mathoverflow.net/users/21564",
"https://mathoverflow.net/users/81645"
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|
Stack Exchange
|
Is there a smooth $W_{G_2}$-equivariant map from the flag manifold of $U(4)$ to that of $G_2$?
The Weyl group $W$ of $G_2$, is a group of order $12$ which is generated by the subgroup of permutations of $e_1$, $e_2$ and $e_3$ and by by the element $\tau$ which maps $(e_1,e_2,e_3)$ to $(-e_1,-e_2,-e_3)$.
We now define an action of $W$ on the flag manifold $U(4)/T^4$. A permutation of $e_1$, $e_2$ and $e_3$ acts on $g \in U(4)$ by permuting the corresponding first $3$ columns $c_1$, $c_2$ and $c_3$ of $g$. For example, $(123)$ places $c_1$ in the second column, places $c_2$ in the third column and places $c_3$ in the first column. This action descends to an action of the permutation subgroup of $W$ on the flag manifold $U(4)/T^4$.
Now choose a quaternionic structure $j$ on $\mathbb{C}^4$ which is compatible with the standard hermitian inner product $(-,-)$ on $\mathbb{C}^4$. What this means is that
$(jv,jw) = \overline{(v,w)}$ for any $v$, $w$ in $\mathbb{C}^4$.
Note that $j$ acts on $g \in U(4)$ by acting on each of the $4$ columns of $g$ simultaneously. This action descends to an action, say $\tilde{j}$, on $U(4)/T^4$, satisfying $\tilde{j}^2 = \operatorname{Id}$. Let the element $\tau$ of $W$ act on $U(4)/T^4$ by $\tilde{j}$.
One can verify that we have defined an action of $W$ on $U(4)/T^4$. Moreover, since $W$ is the Weyl group of $G_2$, it acts naturally on the corresponding flag manifold $G_2/T^2$.
My question can now be formulated. Does there exist a smooth $W$-equivariant map $f: U(4)/T^4 \to G_2/T^2$? I am hoping there is, though I have a feeling there isn't. Your help is kindly appreciated!
Edit 1: I just realized that both manifolds are $12$-dimensional. Are they diffeomorphic by any chance? Answer: they are not, see the comment by Michael Albanese below.
Unless I am mistaken, it follows from the long exact sequence in homotopy that $\pi_2(SU(4)/S(T^4)) \cong \pi_1(S(T^4)) \cong \mathbb{Z}^3$ while $\pi_2(G_2/T^2) \cong \pi_1(T^2) \cong \mathbb{Z}^2$. In particular, the manifolds are not diffeomorphic.
@MichaelAlbanese, yes your argument is correct. I did not know for instance that $\pi_2(G)$ is trivial for any Lie group (is the statement correct, or does it require some assumptions?). Indeed, here is a relevant discussion https://mathoverflow.net/questions/8957/homotopy-groups-of-lie-groups.
It is true for all Lie groups.
Your action on $SU(4)$ is not an action on $SU(4)$. For example, the permutation $(12)$ changes the determinant of an element of $SU(4)$ to $-1$. Are you also, say, multiplying the last column by $(-1)^{|\sigma|}$ for a permutation $\sigma$?
@JasonDeVito, I fixed it now. One way to do things is indeed as you suggested. I wrote it in an equivalent way, by defining the action of the permutation subgroup on an element of $U(4)$ rather than $SU(4)$. Thank you.
|
2025-03-21T14:48:31.353626
| 2020-06-24T11:55:31 |
364018
|
{
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"authors": [
"Alexandre Eremenko",
"Dieter Kadelka",
"Wojowu",
"YCor",
"https://mathoverflow.net/users/100904",
"https://mathoverflow.net/users/14094",
"https://mathoverflow.net/users/25510",
"https://mathoverflow.net/users/30186"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/364018"
}
|
Stack Exchange
|
Functions $f \geq 0$ on $\mathbb{R}$ which are of the form $f = |g|^2$ for some entire function $g$
I think the answer to this question must be well known. Is it possible to characterize those functions $f \colon \mathbb{R} \to \mathbb{R}_+$ which are of the form $f(x) = |g(x)|^2, x \in \mathbb{R},$ for some entire function $g \colon \mathbb{C} \to \mathbb{C}$. As a simple counterexample let $f(x) = e^{-1/x^2}$.
Edit: Alexandre Eremenkos answer allows to reformulate the question.
Which entire functions $f \colon \mathbb{C} \to \mathbb{C}$ are nonnegative on $\mathbb{R}$?
In his proof he has (essentially) given a characterization with the help of the Weierstrass factorization theorem. Are there other (more direct) characterizations? I know, it's vague.
$e^{-1/x^2}$ is certainly not of this form, since it vanishes too quickly around $x=0$.
It was meant as a counterexample. I've edited my post.
It is not clear what "chatacterize" means. One characterization is this: there are exactly those positive functions on the real line which are entire. If this characterization answers your question, I can write a proof.
This would answer my question partially. If you have a reference this would suffice. Of course a short sentence of the form "An entire function $f$, defined on $\mathbb{C}$ is not negative on $\mathbb{R}$ iff ..." would be better.
These $f$ are exactly those non-negative functions on the real line which are entire (=represented by their Taylor series on the whole real line). For example, $f(x)=(\arctan x)^2$ is not in your class since the Taylor series at $0$ has finite radius of convergence. Neither $f(x)=e^{-1/x^2}$ is
in your class since the Taylor series at zero does not converge to the function).
Proof. Suppose that $g$ is an entire function.
Define $g^*(z)=\overline{g(\overline{z})}$ which is also entire. Then
on the real line $f(z)=|g(z)|^2=g(z)g^*(z)$, so your function $f(x)$ is non-negative on the real line and entire (as a product of entire functions).
Conversely. Let $f$ be an entire function which is non-negative on the real line.
Then all real roots are of even multiplicities, and the rest are symmetric with respect to the real line. Let $X$ be the divisor in the plane which consists of those roots which lie in
the open upper half-plane with their multiplicities,
and real roots with half of their
multiplicities. We have the Weierstrass factorization $f=P e^h$
where $P$ is the canonical product, and $h$ is entire, both $P$ and $h$ real on the real line. Let $P_1$ be the canonical product over $X$,
then $P=P_1P_1^*$, and set $g=P_1e^{h/2}$. Then on the real line
$$|g(x)|^2=|P_1(x)|^2|e^{h(x)}|=P(x)e^{h(x)}=f(x).$$
Remark. If $f$ has infinitely many non-real zeros, then there are infinitely many different $g$'s which give such a representation: the zeros can be split between $P_1$ and $P_1^*$ in many ways: if $Y$ is the divisor of zeros of $f$, then any $X$ such that $Y=X+\overline{X}$ will do the job.
Remark 2. How to determine that a function of a real variable is in fact entire. A criterion is that $|f^{(n)}(x)|^{1/n}/n\to 0$ uniformly on compact subsets of the real line.
This follows from the Taylor formula with remainder combined with Stirling's formula.
What is the definition of an entire function $\mathbf{R}\to\mathbf{C}$? I'm only used to the definition of entire function $\mathbf{C}\to\mathbf{C}$.
I wrote in the first sentence in parentheses.
Ah indeed, sorry.
@Ivan Meir: Yes. With essentially the same proof. Weierstrass representation can be generalized to any simply connected neighborhood of the real line.
@Alexandre Eremenko: I shall accept your answer when there are no more incoming "direct characterizations".
After thinking a little bit about your answer I think its just what I wanted. Thank you.
|
2025-03-21T14:48:31.354083
| 2020-06-24T11:57:02 |
364019
|
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|
Stack Exchange
|
Concavity of entropy difference
Suppose that $\mathrm{A}$ is a $n\times n$ random matrix with a given distribution. Suppose that $\mathrm{U}$ is a diagonal unitary random matrix, defined as
\begin{align*}
\begin{bmatrix}
\exp(i\theta_1)&0&\cdots&0\\
0&\exp(i\theta_2)&\cdots&0\\
0&0&\ddots&0\\
0&0&\cdots&\exp(i\theta_n)
\end{bmatrix},
\end{align*}
where $\theta_i$ are i.i.d. Uniform random variable over $[0,2\pi]$, independent of $\mathrm{A}$, and $i$ is the imaginary number.
I need to show that the following function is concave w.r.t. the input distribution:
\begin{align*}
F(p(\mathbf{x}))\triangleq H(\mathrm{A}\mathbf{X})- H(\mathrm{A} \mathrm{U}\mathbf{X}),
\end{align*}
where $\mathbf{X}$ is a continuous random vector of size $n$, with probability distribution $p(\mathbf{x})$, and $H(\cdot)$ is the Shannon entropy. This means that we need to show that for any $0 \leq \lambda \leq 1$, $p_1(\mathbf{x})$ and $p_2(\mathbf{x})$
\begin{align*}
\lambda F(p_1(\mathbf{x}))+ (1-\lambda) F(p_2(\mathbf{x})) \leq F(p(\mathbf{x})),
\end{align*}
where $p(\mathbf{x})=\lambda p_1(\mathbf{x})+ (1-\lambda) p_2(\mathbf{x}) $.
P.S. Some extra assumptions on $\mathrm{A}$ might be needed.
I don't understand your (interesting) question. Please clarify your notation: What does "concave r.r.t. the input distribution" mean? concave w.r.t. $p$ (the notation $p(x)$in this context does not make much sense) or w,r,t, $X$ or ...?
The entropy function (and hence the defined $F$ function), is a function of the underlying probability distribution, and not $\mathbf{X}$. I wrote it down. Please let me know if I'm not clear.
Now this is clear!
What happens in the situation when all distributions are discrete (with the diagonal entries, for instance, taking iid values +1 and -1), and one deals with the usual entropy instead of the differential one?
Still I don't know the answer. I'll think about it.
Are the entries of $\mathrm{A}$ and $\mathbf{X}$ real-valued?
No, they can have complex values.
Is $\mathbf{X}$ independent of $\mathrm{A}$ and $\mathrm{U}$?
Yes, it's independent.
Without further assumptions, I think $F$ is not necessarily concave.
Let $\mathbf{X}_1\sim p_1$, $\mathbf{X}_2\sim p_2$ and $B\sim\textrm{Bernoulli}(\lambda)$ be independent, and let
\begin{align*}
\mathbf{X} &:=
\begin{cases}
\mathbf{X}_1 & \text{if $B=1$,} \\
\mathbf{X}_2 & \text{if $B=0$.}
\end{cases}
\end{align*}
Then, $\mathbf{X}\sim p=\lambda p_1 + (1-\lambda) p_2$.
In general, for two random variables $Z$ and $C$, where $Z$ is continuous and $C$ is discrete, we have
\begin{align*}
h(Z) + H(C\,|\,Z) &= H(C) + h(Z\,|\,C) \;,
\end{align*}
where $H(\cdot)$ denotes the ordinary (discrete) entropy and $h(\cdot)$ is the differential entropy.
It follows that
\begin{align*}
&
\overbrace{h(\mathrm{A}\mathbf{X}) - h(\mathrm{A}\mathrm{U}\mathbf{X})}^{F(p)}
+
\overbrace{H(B\,|\,\mathrm{A}\mathbf{X})
- H(B\,|\,\mathrm{A}\mathrm{U}\mathbf{X})}^{\displaystyle(\sharp)} \\
&=
h(\mathrm{A}\mathbf{X}\,|\,B)
- h(\mathrm{A}\mathrm{U}\mathbf{X}\,|\,B)
+ H(B) - H(B) \\
&=
\lambda\big(\underbrace{h(\mathrm{A}\mathbf{X}_1) - h(\mathrm{A}\mathrm{U}\mathbf{X}_1)}_{F(p_1)}\big)
+
(1-\lambda)\big(\underbrace{h(\mathrm{A}\mathbf{X}_2) - h(\mathrm{A}\mathrm{U}\mathbf{X}_2)}_{F(p_2)}\big)
\end{align*}
provided that $p_1$ and $p_2$ are absolutely continuous w.r.t. the three-dimensional Lebesgue and $\mathrm{A}$ is almost surely non-singular. (Otherwise, the differential entropies become $-\infty$ and $F$ would not be well-defined.)
Therefore, in order for $F$ to be concave, we must have
\begin{align*}
(\sharp) = H(B\,|\,\mathrm{A}\mathbf{X})
- H(B\,|\,\mathrm{A}\mathrm{U}\mathbf{X})
&\leq 0 \tag{?}
\end{align*}
whenever $p_1$ and $p_2$ are absolutely continuous and $\mathrm{A}$ is almost surely non-singular.
[Update: The original example was not valid because it disregarded the requirement that $p_1$ and $p_2$ have to be absolutely continuous and $\mathrm{A}$ non-singular. The following sketch is meant to circumvent that issue.]
Fix $0<\lambda<1$. Let
\begin{align*}
\hat{\mathrm{A}} &:=
\begin{bmatrix}
1 & 1/2 & 1/2 \\
0 & -1/2 & 1/2 \\
0 & -1/2 & 1/2
\end{bmatrix}
&
\hat{\mathbf{X}}_1 &:=
\begin{bmatrix}
1 \\ 0 \\ 0
\end{bmatrix}
&
\hat{\mathbf{X}}_2 &:=
\begin{bmatrix}
0 \\ 1 \\ 1
\end{bmatrix}
\end{align*}
Let $\mathrm{A}$ be a non-singular (deterministic or random) matrix which is very close to $\hat{\mathrm{A}}$, and let $\mathbf{X}_1=\hat{\mathbf{X}}+\sigma\mathbf{Z}_1$ and $\mathbf{X}_2=\hat{\mathbf{X}}+\sigma\mathbf{Z}_2$, where $\mathbf{Z}_1$ and $\mathbf{Z}_2$ are two independent standard normal vectors and $\sigma$ is very small. Assume that $\mathbf{Z}_1$, $\mathbf{Z}_2$, $\mathrm{U}$ and $\mathrm{A}$ are all independent.
Note that both $\mathrm{A}\mathbf{X}_1$ and $\mathrm{A}\mathbf{X}_2$ are highly concentrated around a vector very close to $\hat{\mathbf{X}}_1$. By chooseing $\mathrm{A}$ close enough to $\hat{\mathrm{A}}$, we can make sure that $\mathrm{A}\mathbf{X}_1$ and $\mathrm{A}\mathbf{X}_2$ are hardly distinguishable. Hence, $\mathrm{A}\mathbf{X}$ would hardly have any information about $B$, and as a result
\begin{align*}
H(B\,|\,\mathrm{A}\mathbf{X}) &\approx H(B) = H(\lambda) \;.
\end{align*}
On the other hand, $\mathrm{A}\mathrm{U}\mathbf{X}_1$ and $\mathrm{A}\mathrm{U}\mathbf{X}_2$ will be distinguishable, with $\mathrm{A}\mathrm{U}\mathbf{X}_1$ still being close to the linear span of $\hat{\mathbf{X}}_1$ and $\mathrm{A}\mathrm{U}\mathbf{X}_2$ typically far from it. In particular, $\mathrm{A}\mathrm{U}\mathbf{X}$ has significant information about $B$ and hence
\begin{align*}
H(B\,|\,\mathrm{A}\mathrm{U}\mathbf{X}) &\ll H(B) = H(\lambda) \;.
\end{align*}
Therefore, in this example, $(\sharp)>0$ contrary to the claim.
For deterministic invertible matrices $\mathrm{A}$, the result is correct. I guess, I should have stated the invertible condition for $\mathrm{A}$.
Is $F(p)$ really concave when $\mathrm{A}$ is deterministic and invertible? Isn't $F(p)$ continuous as a function of $\mathrm{A}$? If so, we can choose $\mathrm{A}$ invertible but very close to the matrix I used above, and the concavity inequality would still fail. Am I missing something?
I argue as following: Let $\mathbf{Y}=(|X_1|,|X_2|,\cdots,|X_n|)$ and $\mathbf{Z}=(\angle X_1,\angle X_2 ,\cdots,\angle X_n)$. We have $h(\mathrm{A}\mathbf{X})=h(\mathbf{X})+\log(|det(\mathrm{A})|)$. Hence $F(P)=h(\mathbf{X})-h(\mathrm{U} \mathbf{X})=h(\mathbf{Z}|\mathbf{Y})-2n\pi$, and we know that entropy is a concave function. Note that in the last step, I changed the coordinate from Cartersian to polar.
In fact, $F(p)$ is not continuous w.r.t $\mathrm{A}$, since $\log$ is not continuous around 0.
I am not used to differential entropy, so I don't get how you did your last step. Regarding the (dis-)continuity of $F(p)$, we are not applying $\log$ around $0$ in $h(\mathrm{A}\mathbf{X})$ as long as $\mathbf{X}$ is absolutely continuous, are we?
I think there is a bug in my example. The example does provide a case in which $(\sharp)>0$. However in this case, the distribution of the vectors $\mathrm{A}\mathbf{X}_1$, $\mathrm{A}\mathbf{X}_2$, etc. are not absolutely continuous with respect to the 3d Lebesgue, hence the differential entropies are $-\infty$ and $F(p)$ is not well-defined.
However, I still think that it can be fixed by an approximation argument. I'll try to fix it.
To see the discontinuity of $F(P)$, note that due to the relation $h(\mathrm{A}\mathbf{X})=h(\mathbf{X})+\log(|det(\mathrm{A})|)$, whenever $\mathrm{A}$ is deterministic and invertible, $F(P)=h(\mathbf{X})-h(U\mathbf{X})$ does not depend on $\mathrm{A}$, while when $det(\mathrm{A})=0$, this does not hold.
This means that we should have $H(B|\mathrm{A}\mathbf{X})-H(B|\mathrm{A}\mathrm{U}\mathbf{X})=H(B|\mathbf{X})-H(B|\mathrm{U}\mathbf{X})$. However, both terms in $H(B|\mathbf{X})-H(B|\mathrm{U}\mathbf{X})$ are very small, while as you have argued, the differnce is not small.
How do you prove the below claim?
\begin{align*}
h(Z) + H(C,|,Z) &= H(C) + h(Z,|,C) ;,
\end{align*}
The joint distribution of $Z$ and $C$ can be described by a function $f(z,c)$ where $f_Z(z):=\sum_c f(z,c)$ is the probability density function of $Z$ and $p_C(c):=\int_z f(z,c)\mathrm{d}z$ is the probability mass function of $C$. Now, the quantity $-\sum_c\int_z f(z,c)\log f(z,c)\mathrm{d}z$ can be decomposed in two ways, by either writing $f(z,c)=f_Z(z)p_C(c,|,Z=z)$ or $f(z,c)=p_C(c)f_Z(z,|,C=c)$, where $p_C(c,|,Z=z)$ is the conditional mass of $C$ given $Z$, and $f_Z(z,|,C=c)$ is the conditional density of $Z$ given $C$. These two decompositions give the two sides of the equality.
Thanks. How do you define $p_{C}(C|Z=z)$? Is it like $\frac{f(z,c)}{f_Z(z)}$?
Yes! More precisely, for a fixed $c$, $p_C(c,|,Z)=\mathbb{P}(C=c,|,Z)$ is the conditional probability of the event ${C=c}$ given a continuous random variable $Z$, so it is only almost surely defined. The ratio $f(Z,c)/f_Z(Z)$ is a version of that conditional probability.
|
2025-03-21T14:48:31.354647
| 2020-06-24T12:23:07 |
364020
|
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|
Stack Exchange
|
Density function approximation with respect to $L^1$ distance
Given iid samples $X_1,...,X_N$ drawn from some unknown distribution with not necessarily continuous density function $f(x)$ are there any theorems/papers where based on the data $X_1,...,X_N$ an estimator $f_N(x)$ of $f(x)$ is defined and the approximation rate
$$\int_{\mathbb{R}}|f(x)-f_N(x)|dx$$
is estimated?
From the description of the book Nonparametric Density Estimation: The L1 View by Devroye and Gyorfi:
The first systematic single-source examination of density estimates. It develops, from first principles, the ``natural'' theory for density estimation, L1, and shows why the classical L2 theory masks some fundamental properties of density estimates.
|
2025-03-21T14:48:31.354747
| 2020-06-24T12:47:18 |
364023
|
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|
Stack Exchange
|
Generalization of Weak Nullstellensatz?
I believe the following is standard, namely when $k = \bar{k}$ is algebraically closed there is a bijection between points and maximal ideals
\begin{eqnarray*}
k^n &\longrightarrow& \operatorname{Specm}(k[X_1, \ldots, X_n]) \\
x &\longrightarrow& \ker(\operatorname{ev}_x)
\end{eqnarray*}
where surjectivity follows from Zariski's Lemma. It seems like the following should also be true, by essentially the same argument, but I couldn't find a reference. For $k$ not algebraically closed there is a bijection
\begin{eqnarray*}
\bar{k}^n / \operatorname{Aut}(\bar{k} / k) &\longrightarrow \operatorname{Specm}(k[X_1, \ldots, X_n])
\end{eqnarray*}
Is there a canonical reference for this ?
This is often given as an exercise in scheme theory classes... I'll see if I can find a book where they solve it :).
The closest reference in literature I have encountered is in Mumford's Red book of varieties and schemes, II.4 Theorem 1. More precisely, a direct citation is as follows:
Let $X_0$ be a prescheme over $k_0$, let $X = X_0 \times_{k_0} k$, and let
$p: X \rightarrow X_0$ be the projection. Assume that $k$ is an algebraic closure of $k_0$. Then
$p$ is surjective and both open and closed (i.e., maps open/closed
sets to open/closed sets).
For all $x,y \in X$, $p(x) = p(y)$ if and only if $x = \sigma_X(y)$, some
$\sigma \in Gal(k/k_0 )$. In other words, for all $x \in X_0$ $p^{-1}(x)$ is an
orbit of $Gal(k/k_0)$. Moreover, $p^{-1}(x)$ is a finite set.
The theorem applied to $X_0=\mathrm{Spec}\,k_0[x_1, \dots, x_n]$ together with the standard weak Nullstellensatz for $X=\mathrm{Spec}\,k[x_1, \dots, x_n]$ should give your statement.
Just discovered, it's demonstrated in Bourbaki Commutative Algebra Chapter V Section 3.3 Proposition 2.
¿Chapter? the reference is incomplete...
thank you, corrected
Or see Proposition 2.4.6 in Bjorn Poonen's book Rational Points on Varieties (link). This is almost exactly the result you conjectured, just a bit more general:
Let $X$ be a $k$-variety. Then the map
$$\left\{\text{$\operatorname{Gal}_k$-orbits in $X(\overline{k})$}\right\}\rightarrow \left\{ \text{closed points of $X$} \right\}$$
given by mapping the orbit of $f \colon \operatorname{Spec} \overline{k} \to X$ to $f(\operatorname{Spec} \overline{k} )$ is a bijection.
|
2025-03-21T14:48:31.354954
| 2020-06-24T13:34:07 |
364028
|
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|
Stack Exchange
|
Quadratic extensions of cyclotomic numbers by absolute values of elements
Summary
I was wondering whether there is an explicit description of the extension $\mathbb{Q}(\zeta_n, |z|)/\mathbb{Q}$ obtained from a cyclotomic field $\mathbb{Q}(\zeta_n)$ by adjoining any finite number of absolute values $|z| := \sqrt{z\overline{z}} \in \mathbb{R}^+$ of elements $z \in \mathbb{Q}(\zeta_n)$. I am of course interested in the cases where the absolute value $|z|$ is not itself a cyclotomic number, i.e. where $\mathbb{Q}(\zeta_n, |z|)/\mathbb{Q}$ is not an abelian extension, but see below for an additional desideratum.
Example of non-cyclotomic absolute value
As an example of an absolute value of interest, consider $z:=1+(2+\sqrt{2})i \in \mathbb{Q}(\zeta_8)$:
$$
|z| = \sqrt{x\overline{x}} = \sqrt{7+4\sqrt{2}}
$$
The minimal polynomial for $|z|$ is:
$$
p(X) := X^4-14X^2+17
$$
Using the characterisation of Galois groups for biquadratic quartics (e.g. see Galois group of a biquadratic quartic), it is easy to check that the Galois group for $p(X)$ is $D_8$, so that $\mathbb{Q}(\zeta_n, |z|)/\mathbb{Q}$ is not abelian.
Questions
I would like an algorithmic way to construct and work with a basis (over $\mathbb{Q}$) for the extension $\mathbb{Q}(\zeta_n, |z|)/\mathbb{Q}$, where $z$ is a given cyclotomic number.
If a description of the basis above is made easier by assuming a that $|z|$ is not itself a cyclotomic number, then I would like to have an algorithmic way of determining when $|z|$ is a cyclotomic number for a given cyclotomic number $z$.
I would like an algorithmic way to construct and work with a basis (over $\mathbb{Q}$) for the extension $\mathbb{Q}(\zeta_n, |z_1|, ..., |z_n|)/\mathbb{Q}$, where $z_1, ..., z_n$ are given cyclotomic numbers (possibly under the additional assumption that none of $|z_1|, ..., |z_n|$ is cyclotomic, see Q2).
If at all possible, I would like the basis for the extension to extend the basis described in [Breuer 1997] for the cyclotomic fields, or equivalently the one described in [Zumbroich 1989].
References
[Breuer 1997] T. Breuer. Integral Bases for Subfields of Cyclotomic Fields. 1997.
[Zumbroich 1989] M. Zumbroich. Grundlagen einer Arithmetik in Kreisteilungskörpern und ihre Implementation in CAS. 1989.
You have already given an explicit description of the field extension. What do you want?
@FranzLemmermeyer thank you for your comment! I have edited my question for clarification (and expanded it in the process).
The degree of $|z|$ over $\mathbb{Q}(\zeta_N)$ is 1 or 2 since $z\bar{z}$ is cyclotomic. We are in the first case if and only if $z \bar{z}$ is a square in $\mathbb{Q}(\zeta_N)$, there are algorithms to decide this (roughly by reducing modulo prime ideals). In either case a rational basis is readily obtained.
Here is one possible way:
We have that the $\zeta_k$ corresponds to some matrices
$$
\mu({\zeta_k}) = \mathbf{Diag}\left[e^{k\frac{2 \pi i }{n}},.... \right]
$$
so that $\mu$ is a representation of $\mathbb{Q} (\zeta_n)$ as a group algebra of matrices $\mathbb{Q}[\mathbb{Z}/n\mathbb{Z}]$. In particular,
for any $\mathbb{Q} (\zeta_n)$ we have that
$$
\mu(x )= x_1 \mu({\zeta_1} )+ ... +x_n \mu({\zeta_n } ) = \sum_{k} x_k\mu({\zeta_k})
$$
where $x = x_1 \zeta_1 + ... +x_n \zeta_n $.
Notice that
$$
\mu({\overline{\zeta_k})} = \mathbf{Diag}\left[\overline{e^{k\frac{2 \pi i }{n}}},.... \right]
$$
just permutes around the indices of the coefficients $x_i$ by some permutation $\pi$ so that
$$
\mu({\overline{x}}) = \sum_{k} x_{\pi(k)}\mu({\zeta_{\pi(k)}})
$$
and thus the problem is reduced to finding the square root, $y$, of the matrix
$$
y^2 = \mu(x ) \mu({\overline{x}}) =\left(\sum_{j} x_{j}\mu({\zeta_{j}})\right) \left(\sum_{k} x_{\pi(k)}\mu({\zeta_{\pi(k)}})\right) = \sum_{j,k} x_j x_{\pi(k)} \mu({\zeta_{j+\pi(k)}}).
$$
There are closed-form solutions (symbolic algorithms) for the square root of a matrix (see https://en.wikipedia.org/wiki/Square_root_of_a_matrix#Solutions_in_closed_form) and you may check whether the system of equations
$$
\sum_{j,k} y_j y_{k} \mu({\zeta_{j+k}}) = \sum_{j,k} x_j x_{\pi(k)} \mu({\zeta_{j+\pi(k)}})
$$
has solutions for rational $y_i \in \mathbb{Q}$. If it does have a solution then your
$$
y = |x \overline {x}|
$$
is already in the cyclotomic field and no work needs to be done, otherwise, $y$ gives you a new basis element and you then use the minimal polynomial of $y$ to get the extra basis elements needed.
Edit:
Things can be simplified further:
The system of equations
$$
\sum_{j,k} y_j y_{k} \mu({\zeta_{j+k}}) = \sum_{j,k} x_j x_{\pi(k)} \mu({\zeta_{j+\pi(k)}})
$$
is actually a linear system; if you notice that once we use a symbolic method to solve for the square root of $\mu(x ) \mu({\overline{x}})$,
we actually only need to solve a linear system
$$
\sum_{i}y_i \left[\mu(\zeta_i)\right]^{j}_{j}= \left[\sqrt{\mu(x ) \mu({\overline{x}})}\right]^j_j
$$
where $\left[ m \right]^i_j$ is the $(i,j)^\mathrm{th}$ coefficient of the matrix $m$.
|
2025-03-21T14:48:31.355281
| 2020-06-24T13:34:48 |
364029
|
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|
Stack Exchange
|
Group structure for distributive lattices
On the (finite) Boolean lattice there is a group structure given by the symmetric difference and this group is an elementary abelian 2-group.
Question: Does there exist a natural group structure on general (finite) distributive lattices?
Other examples of a group structures for a given lattice would also be interesting.
Hard to see how there could be something like symmetric difference for order ideals...
Would you mind clarifying what you mean by "natural"? If you mean that the commutative monoid given by the join is an (abelian) group, then the answer is that there are no examples other than the lattice with 1 element (the trivial group $\mathbb{Z}/1\mathbb{Z}$) and the lattice with 2 elements (the group $\mathbb{Z}/2\mathbb{Z}$). Indeed, for any other bounded (distributive) lattice you can find an element $z \neq 0$ such that $1 \vee z = 1 = 1\vee 0$, which means that $\vee$ cannot be cancellative. Not specifically what you asked, but a consideration to keep in mind.
@StefanoGogioso: well already the example of the Boolean lattice and symmetric difference is different than the join operation.
@SamHopkins Thank you: I know. I was just trying to clarify that cases to do with extending the lattice operations have some limitations. I edited my comment to make that clear.
It occurs to me that orthomodular lattices might be an interesting generalistion for you to look at, because people have studied various notions of symmetric difference there in the years. For example, a partial (negative) result for a certain definition of symmetric difference in OMLs is given by this paper. where the author shows that the symmetric differences they define can only be associative or cancellative when the orthomodular lattices are actually boolean algebras.
Take any finite distributive lattice and add a chain at the top so that it has a prime number $p$ of elements. It doesn't seem reasonable that there is a "natural" action of a cyclic group of order $p$, even if there is only the trivial automorphism.
No:
If it's natural, it should be invariant under the automorphism group of the original lattice.
let $X$ be the free distributive lattice on 2 generators $x,y$: it has 6 elements, $$0\quad<\quad x\wedge y \quad<\quad \stackrel{x}{_y}\quad<\quad x\vee y\quad<\quad 1 $$
with $x,y$ not comparable. It has an automorphism exchanging $x$ and $y$, and fixing the other elements.
But no group of order 6 has no automorphism with this property (if $G$ is a finite group and an automorphism fixes $>|G|/2$ elements, it's identity, just because the set of fixed points is a subgroup).
Initial answer: no for arbitrary finite lattices (the following example is not distributive).
Consider the lattice of subgroups of the Klein group $C_2^2$. It has cardinal 5 (0, the whole plane and 3 lines), and the automorphism group (of order 6) has two fixed points and a 3-element orbit.
Now a group structure on 5 elements is cyclic of order 5 and its automorphism group is cyclic of order 4, so a group structure on 5 elements cannot be preserved by the original automorphism group of order 6.
This question has already been answered, but I will add a slightly different answer.
First, the question and YCor's answer both seem to assume that by a natural group structure we mean that the group is a reduct of the original structure. This means that the group operations are given as words in the original structure, as symmetric difference is given by the Boolean word $x\oplus y = (x\wedge \neg y)\vee (\neg x\vee y)$. I also make this assumption.
Any compatible relation on a structure is also a compatible relation of any reduct. (A relation $R\subseteq A^n$ is compatible on an algebraic structure $A$ if $R$ is a subalgebra of $A^n$.)
YCor's answer uses automorphisms to answer the question. The graph of an automorphism of $A$ is a compatible binary relation on $A$, so any automorphism of a distributive lattice will also be an automorphism of any reduct. Then YCor exhibits a nonidentity automorphism of a distributive lattice that fixes more than half the points, which is something that can't happen in a group.
But to me the more obvious relation to consider is the order relation $\leq$. Every nontrivial distributive lattice has a nondiscrete compatible order, while no nontrivial group has a nondiscrete compatible order. (Even nontrivial ordered groups do not have nondiscrete partial orders that are compatible with all group operations. Reason: $a\leq b$ for compatible $\leq$ implies $a^{-1}\leq b^{-1}$, which implies $a\cdot \underline{a^{-1}}\cdot b\leq a\cdot \underline{b^{-1}}\cdot b$, which implies $b\leq a$.)
No, I didn't assume that the group structure should be a reduct of the lattice structure. I assumed that it should be compatible with isomorphisms.
Actually what you're saying just applies to finite total orderings, right? you're saying that there is no "natural" way to make every finite total ordered set a group. But of course it depends on "natural". Using reducts is very restrictive, and quite clearly from the 2-element total order viewed as lattice, we can't make a group law as reduct. But still identifying every finite total order to an initial segment of $\omega$ (nonnegative integers) endows it with a cyclic group law in a natural (= choice-free) way.
No, I am not talking about finite total orderings. Any compatible partial ordering on any group is discrete. (A partial ordering $\leq$ is compatible with the operations of $A$ if it is a subalgebra of $A^2$.)
Finite total orderings are a subclass of finite distributive lattices. What I'm saying is that under your restrictive interpretation of "natural", it's already true that there is no "natural" group law on every finite total ordering.
|
2025-03-21T14:48:31.355641
| 2020-06-24T13:59:02 |
364032
|
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|
Stack Exchange
|
Projective objects for compact po-spaces
Let us consider the following definition: a compact po-space is a pair $(X,\leq)$ where $X$ is a compact space and $\leq$ is an order, closed on $X^2$. Then, we can consider the category $KPoSp$ whose objects are compact po-spaces and whose morphisms are increasing continuous functions.
Moreover, let us denote by $KHaus$ the category of compact Hausdorff spaces with continuous functions.
It is known that the projective objects in $KHaus$ are the compact Hausdorff spaces which are extremally disconnected, in the sense that the closure of an open set is also open. (I have it from section 7 of Stone duality and Gleason covers through de Vries duality).
I would like to know if there is somewhere a similar result for the category $KPoSp$ ?
Are the objects of $KPoSp$ assumed to be Hausdorff? If not, I'd think one should first ask what are the projective objects in the category of compact (not-necessarily-Hausdorff) spaces. Also, what exactly is meant by "projective"? Probably you mean "object which has the left lifting property with respect to certain maps", but which maps? Surjections? Epimorphisms? Quotient maps? Finally, is $\leq$ assumed to be a poset structure or just a preorder?
For the definition of "projective", I must admit I don't know the kind of map I'm looking for (surjective maps would be perfect, but I can be satisfied with epimorphisms). The objects of $KPoSp$ are indeed Hausdorff. This is a consequence of $\leq$ closed. Finally, yes we have a poset structure (the relation between compact po-spaces and Priestley spaces is similar to the one between compact Hausdorff and Stone spaces, if it can help.)
Presumably you want to first show every compact pospace is a quotient of a Priestley space and then look at injective distributive lattices. Did you look in Johnstone's stone spaces or the compendium of continuous lattices?
I suspect that surjections are not what you want -- they lead to "too few" projective objects (and epimorphisms will be, if anything, worse in this regard). For instance, for $P$ to have the left lifting property against the bijection $f: {0,1} \to {0 < 1}$ (where ${0,1}$ has the discrete ordering) is a very strong requirement. In particular, if $P$ is totally disconnected and has the left lifting property with respect to $f$, then I believe the order on $P$ must be discrete. So perhaps you want "projective" to mean the left lifting property with respect to quotient maps.
On the other hand, I'm not sure I'd expect there to be a well-behaved notion of "projective" even in the category $Pos$ of posets without a topology -- generally I expect "projective" to be well-behaved in "algebraic" categories (by which I mean roughly: monadic over $Set$ or monadic over $Set^S$ for some set $S$), which $Pos$ is not. Analogously, I don't think there's a good notion of "projective object" in the category $Cat$ of categories.
@BenjaminSteinberg . Yes, I would like something like that. I have yet to look in Johnstone or in the compendium.
@TimCampion, you are presumably right. Again, I'm not still not sure about the property I'm really looking for. From "Proximity frames and regularisation", it seems to me that we can have a result "similar" to the one I mentioned in my original post. The problem is now that I still have to determine what "similar" means. The projective property on compact Hausdorff may have an equivalent characterisation and this is this latter equivalent characterisation which will be nicely generalised. This why I wanted to know if someone already explored this problem.
|
2025-03-21T14:48:31.355989
| 2020-06-24T14:08:45 |
364034
|
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"Đào Thanh Oai"
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}
|
Stack Exchange
|
Stewart's formula in plane geometry
In his book "Plane geometry and its groups", H. Guggenheimer proves the Stewart's formula:
If $A$, $B$, and $C$ are collinear, then for any point $Ρ$ in the plane
$$
PA^2 BC + PB^2 CA + PC^2 AB + AB . BC . CA = 0.
$$
and states that Stewart's formula [...], in principle, solves all computational problems in plane geometry.
What does he mean ? How do we solve all computational problems in plane geometry using that formula?
I understand that, for example, formulas given by the Gram determinant of the inner product of vectors are fundamental in the sense that two sets of vectors can be deduced from each other using an isometry if and only if these Gram determinants are the same, and this statement can be interpreted in terms of Invariant Theory. But I fail to find such an interpretation to Stewart's formula. What is the justification to his bold statement?
It possibly just means that many problems reduce to finding the fifth member of the set ${PA,PB,PC,AB,AC}$ given that other four are known. Another interpretation is that Stewart's formula is universal in a philosophical sense: it is an equivalent substitute to Cartesian coordinates, which are universal in the usual sense.
can't you say the same about Pythagoras? (since Pythagoras --> law of cosines --> Stewart)
https://mathoverflow.net/questions/234184/relation-of-some-euclidean-geometry-theorems-and-more-conjecture-generalizations
|
2025-03-21T14:48:31.356115
| 2020-06-24T15:25:51 |
364036
|
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|
Stack Exchange
|
Combinatorics of multivariate Faà di Bruno formula
This question is a followup on this one on stackoverflow where i implement in python this issue. I am applying the Faà di Bruno formula to obtain the $\mathbf{i}$th derivatives of the function $f = ln \circ g$ in terms of the derivatives of $g$, a function with $n$ variables. This yield the following formula, denoting by $f^{(\mathbf i)}$ the $\mathbf i$th derivative of a function f
$$f^{(\mathbf i)}(t) = \sum\limits_{\pi \in \Pi(\mathbf{i})} (\lvert \pi \rvert -1) (-1)^{\lvert \pi \rvert -1} g(t)^{- \lvert \pi \rvert} \prod\limits_{B \in \pi} g^{(\mathbf{i}(B))}$$
where the notation follows this section of the wikipedia page, plus a small variation that recover the multiindex $\mathbf i(B)$ from the set of dimension indexes $B$.
The problem is that, since the exponential function has all it's derivatives equal, as soon as we want to derivate twice on the same variable, i obtain several $\pi$ inside $\Pi(\mathbf i)$ that are equivalent.
For exemple, set $\mathbf i = (2,1)$ in a two-dimensional problem. Then, the partitions inside $\Pi(\mathbf i)$ are:
$(0,0,1)$
$(1), (0,0)$
$(0), (0,1)$
$(0, 1), (0)$
$(0) (0) (1)$
There is here a partition that appears twice (without taking into account the order of blocks inside a partition), and which will have the same value into the sum.
This happends a lot (run my code from stackoverflow to find out). Is there a way i could 'factorise' the formula ?
Edit: For $n = 1$, in a one-dimensional problem, the formula indeed factorises to the Bell polynomials, and the number of occurences of similar partitions are the Bell numbers. Is there somewhere a multivariate equivalent ?
Edit: This problem is equivalent to finding all multiset-partitions of a multiset. The solution in term of multivariate partial Bell polynomials exists (see article in a comment), but the polynomials in question are still hard to implement.
Just a little comment: are you aware of this paper ? https://www.ams.org/journals/tran/1996-348-02/S0002-9947-96-01501-2/S0002-9947-96-01501-2.pdf
@user69642 This is exactly what i need, precisely the Corrolary 2.10. Now i have trouble computing the $p(\mathbf \nu, r)$ set
The Faa-di-Bruno formulae are really a clumsy ways of expressing the fact that (in the appropriate quotient algebra) the Taylor polynomial of a composition is the (formal) composition of the relevant Taylor polynomials. Working with formal Taylor polynomials should (I think) give you cleaner code and let you delegate most of the fiddly algebra/combinatorics to something like SymPy.
You should be able to get the relevant composition rule by doing a Google-search for "composition rule for Taylor polynimials" or similar, but you can find it Chapter 1 of Malgrange's Ideals of differentiable functions if you need.
Edit: I thought I'd provide the following rough sketch just to illustrate what I mean... and because I couldn't resist having a go!
import sympy
import math
def get_indeterminates(n_indeterminates):
return [sympy.Poly('X{}'.format(i)) for i in range(1, n_indeterminates+1)]
def monomial_from_multi_index(multi_index, indeterminates):
monomial = indeterminates[0]**0
if type(multi_index) is int:
multi_index = (multi_index,)
for x, power in zip(indeterminates, multi_index):
monomial *= x**power/math.factorial(power)
return monomial
def taylor_polynomial_from_partials(partials, indeterminates):
"""
'partials' should be a dictionary mapping integer tuples to floats
"""
Tf = 0
for k, fk in partials.items():
Tf += fk * monomial_from_multi_index(k, indeterminates)
return Tf
def get_composite_taylor_polynomial(derivatives_of_f, partials_of_g, n_variables, truncation_order):
x = get_indeterminates(n_variables)
y = get_indeterminates(1)
Tf = taylor_polynomial_from_partials(derivatives_of_f, y)
Tg = taylor_polynomial_from_partials(partials_of_g, x)
Tfog = sympy.polys.polytools.compose(Tf, Tg)
return
This looks a lot like what i want; Except i'm using mpmath for precision, as Sympy will only have 60 digits or so and factorials add up quite quickly. But the spirit is exactly what i want, i'll try it
|
2025-03-21T14:48:31.356404
| 2020-06-24T15:48:52 |
364038
|
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|
Stack Exchange
|
Constructive factorisation of null-homology map through acyclic complex
Let $f: C \rightarrow D$ be a maps of chain complexes on an idempotent complete additive category with all kernel or cokernel (or chain complexes on abelian category).
If $f$ induces a null map in homology. By using the derivaded category, there exist a zig-zag
$$\require{AMScd}
\begin{CD}
X' @>f'>>W @> >> Y' \\
@V \sim V V && @AA \sim A \\
X @>>f>Y @= Y
\end{CD}$$
where $W$ is acyclic and $X' \rightarrow X$ and $Y \rightarrow Y'$ are quasi-isomomorphisms.
So here is my question :
Is there a way to construct explicitly $X'$, $W$ and $Y'$ by using only $X$, $Y$, the boundaries and cycles of these complexes ?
Are you sure that a map that's zero on homology factors through the zero complex (up to homotopy)? Unless I'm mistaken this is equivalent to Freyd's generating hypothesis for your derived category and that's true only under regularity assumptions.
As @Denis says, it's not true, even for complexes of abelian groups. If it were true then $f$ would be zero in the derived category, but there are usually plenty of nonzero maps in a derived category that induce the null map on homology.
I think you're right.
Is it the same for endomorphism (ie. X=Y ) ?
@MoreauT Yup, e.g. for $\mathbb Z/2 \oplus \mathbb Z/2 [1]$ and the map induced by the extension $\mathbb Z/2 \to \mathbb Z/2[1]$
|
2025-03-21T14:48:31.356547
| 2020-06-24T16:34:23 |
364040
|
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|
Stack Exchange
|
3-colored triangulations of the sphere $S^2$, and Sperner's Lemma
I noticed something about colored triangulations of the topological sphere $S^2$ and have a question about this.
Observation. If you triangulate the sphere $S^2$ and color the vertices with three colors: then the number of 3-colored triangles is always even (or zero). In particular, there is no coloring with exactly one 3-colored triangle.
For a proof, view $S^2$ as two triangulated disks with matching coloring of the boundaries that are glued together. As their boundaries have the same number of color changes, we know from Sperner’s Lemma that their triangulations have the same number (mod 2) of 3-colored triangles. So the total number of 3-colored triangles is even or zero.
As an interesting corollary, we get the characterization: A triangulated sphere has zero 3-colored triangles iff all cycles of the triangulation have an even number of color changes.
I looked at the torus, the Klein bottle, and the projective plane, and I find that the observation is also true for them.
Edit: Just for contrast, adding an example below of a "soap bubble" surface, where the two soap bubbles share a common disk. This surface allows for triangulations with even and odd numbers of 3-colored triangles (but like the other surfaces I looked at, cannot have just one).
Question. I wonder whether this also follows from more general theorems about triangulations of surfaces, or about maximal planar graphs? I have consulted algebraic topology and graph theory texts, but could not find any results in that direction. Would you have a suggestion where else to look, or maybe a reference for that?
This looks related to Sperner's lemma: https://en.wikipedia.org/wiki/Sperner%27s_lemma
@JanKyncl you are definitely right. It is like Sperner’s Lemma “without boundary”. Or the way I looked at it, 2 x Sperner’s Lemma and then glued together. Just in case you are interested, see this https://mathoverflow.net/q/362025/156936
"Uneven" numbers are commonly called odd.
@VictorProtsak thanks a lot for your comment! I corrected the wording in the text now.
A counting proof shows that this observation is unrelated to the global topology.
Every edge is monochromatic or dichromatic. How many dichromatic edges are there? If each triangle tells you its number of dichromatic edges (either 0, 2, or 3), you can add these up and divide by two to get the total number of dichromatic edges (since every edge contributes to two triangles). So the number of trichromatic triangles must be even.
This proof works for $k$-dimensional manifolds when $k$ is even, since the number of $k$-colored $(k-1)$-simplices bounding any $k$-simplex must be 0, 2, or $k+1$.
Your corollary similarly transfers to higher even dimensions, at least for orientable manifolds, replacing "cycles of edges" with "hypersurfaces of $(k-1)$-simplices", and "even number of color changes" with "even number of $k$-colored $(k-1)$-simplices".
I'm not sure which direction you want to go in with this, but I think it falls pretty squarely in combinatorics and graph theory. What you (not incorrectly) call "triangulations of $S^2$" are often referred to as maximal planar graphs. My proof is very similar to the handshaking lemma. People have also considered coloring and planarity on tori and other manifolds.
thanks a lot for this helpful comment and great links. This is excellent
Just to close the loop on this: the double-counting argument in the answer of user Matt allows for a nice visual proof of the (2-dim.) Lemma of Sperner. Just want to capture it here, as it connects nicely with the triangulation of the sphere / the maximal planar graph in my OP question.
Start with a triangulated polygon in the plane, and label each vertex with one of 3 colors. The example just shows the boundary of such a triangulated, 3-colored polygon. Claim (Sperner’s Lemma): If the boundary has an odd number of color changes, then a 3-colored triangle exists in the polygon triangulation. In fact, more generally, an odd number of such 3-colored triangles exists.
Proof: Go to 3-dimensional space, and build a “tent” over the polygon like in the diagram: add a colored vertex, and add the edges between this additional vertex and the boundary vertices of the polygon. This way, we have effectively created a triangulation of the topological sphere $S^2$.
If the boundary of the polygon has an odd number of color changes, this gives an odd number of 3-colored triangles in the “tent” over the polygon. But from the double-counting argument in user Matt’s answer, we know an even number of 3-colored Sperner triangles must exist. Hence the polygon at the bottom must have an odd number of 3-colored triangles (at least one) in its triangulation, which completes the proof.
|
2025-03-21T14:48:31.356886
| 2020-06-24T16:51:21 |
364041
|
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|
Stack Exchange
|
When can an $\mathfrak{S}_n$-equivariant map be extended to an $\textrm{O}(n)$-equivariant map?
The symmetric group $\mathfrak{S}_n$ can be regarded as a subgroup of the orthogonal group $\textrm{O}(n)$ via the permutation matrices. Let $V$ be a finite dimensional $\textrm{O}(n)$-module and $\varphi: \mathbb{R}^n\to V$ an $\mathfrak{S}_n$-equivariant linear map where $\mathfrak{S}_n$ acts on $\mathbb{R}^n$ in the obvious way. Finally, let $d:\mathbb{R}^n\to\textrm{Sym}_2(\mathbb{R}^n)$ be the map that sends a vector to the corresponding diagonal matrix.
Are there criteria on $\varphi$ that ensure the existence of an $\textrm{O}(n)$-equivariant linear map $\Phi: \textrm{Sym}_2(\mathbb{R}^n)\to V$ such that $\Phi\circ d=\varphi$?
Note that such a map, if it exists, is unique since every real symmetric matrix is diagonalizable with orthogonal matrices.
Edit: This is to show that the map suggested by Aurel is in general not linear. Let $V$ be the representation of $\textrm{O}(n)$ where rotation about angle $t$ acts by $\begin{pmatrix}\cos(6t)&-\sin(6t)\\ \sin(6t)& \cos(6t)\end{pmatrix}$ and reflection at the $x$-axes by $\begin{pmatrix}1&0\\ 0& -1\end{pmatrix}$. Then one can check that the map $\varphi:\mathbb{R}^2\to V$ with $\varphi\binom{1}{1}=0$ and $\varphi\binom{1}{-1}=\binom{1}{0}$ satisfies the conditions described by Aurel. However, $V$ is not part of the decompostion of $\textrm{Sym}_2\mathbb{R}^2$ into irreducibles.
Do you really seek invariant map or do you actually want equivariant map?
Sorry, I mean equivariant.
Here's a proposed answer: The map extends iff the Casimir element in ${U}(\mathfrak{o}_n)$ scales the image of the map $\varphi$ by the right scalar.
If I have understood the problem correctly, the map $\Phi$ deterimines $\varphi = \Phi \circ d$, so the question amounts to classifying possible compositions $\Phi \circ d$, where $d$ is the "diagonal" map, and $\Phi$ is $O(n)$ equivariant. Clearly the image of $\varphi$ must be contained in the image of $\Phi$ which is isomorphic to a quotient of $\mathrm{Sym}_2(\mathbb{R}^n)$.
Let's begin by noting that as a representation of $\mathfrak{S}_n$, we have the decomposition $\mathbb{R}^n = \mathbf{1}_{\mathfrak{S}_n} \oplus U$, where $\mathbf{1}_{\mathfrak{S}_n}$ is the trivial representation (spanned by the "all ones" vector), and $U$ is the (irreducible) standard representation (consisting of "mean zero" vectors).
Similarly, as a representation of $O(n)$, the representation $\mathrm{Sym}_2(\mathbb{R}^n)$ (which we are interpreting as symmetric $n \times n$ matrices under conjugation) decomposes as $\mathbf{1}_{O(n)} \oplus W$, where $\mathbf{1}_{O(n)}$ is the trivial representation (spanned by the identity matrix), and $W$ is an irreducible representation which consists of symmetric matrices of trace zero.
It is not difficult to see that the map $d: \mathbf{1}_{\mathfrak{S}_n} \oplus U \to \mathbf{1}_{O(n)} \oplus W$ is of the form $f \oplus g$, with $f : \mathbf{1}_{\mathfrak{S}_n} \to \mathbf{1}_{O(n)}$ (it sends the all-ones vector to the identity matrix) and $g: U \to W$.
Now, $\Phi$ must map $\mathbf{1}_{O(n)}$ to an invariant vector (possibly zero), and it must map $W$ to either a copy of $W$ or zero. So, we must give an "intrinsic" description of the image of $g$, i.e. traceless diagonal matrices, inside $W$.
Note that $O(n)$ contains not only $\mathfrak{S}_n$, but the larger hyperoctahedral group $\mathfrak{H}_n = C_2 \wr \mathfrak{S}_n = C_2^n \rtimes \mathfrak{S}_n$, via signed permutation matrices (permutation matrices, but the nonzero entries can be $\pm 1$). The hyperoctahedral group contains a subgroup $C_2^n$ consisting of diagonal matrices with entries $\pm 1$. It is not difficult to check that a matrix that commutes with every element of $C_2^n$ (viewed as a diagonal matrix) must itself be diagonal, and conversely it is clear that every diagonal matrix commutes with $C_2^n$. Phrasing this in terms of the module structure (commuting with = fixed under conjugation by), the diagonal matrices are fixed by the action of $C_2^n$.
We are now in a position to give the characterisation. A map $\varphi: \mathbb{R}^n \to V$ may be written in the form $\Phi \circ d$ if and only if all of the following conditions are satisfied.
1. The $O(n)$-module generated by the image of $\varphi$ is a quotient of $\mathrm{Sym}_2(\mathbb{R}^n)$.
2. The image of the "all ones" vector is $O(n)$ invariant (possibly zero).
3. The image of $\varphi$ is fixed pointwise by $C_2^n$ (and therefore has a $\mathfrak{H}_n$ action that factors through $\mathfrak{S}_n$).
We've demonstrated that these conditions are necessary, so let's check that they are also sufficient. Note that by (2), given $\varphi$ as above, the restriction $\mathbf{1}_{\mathfrak{S}_n} \to V$ determines the restriction $\Phi: \mathbf{1}_{O(n)} \to V$. So it suffices to construct the restriction $\Phi: W \to V$. By (1), we are guaranteed to find such a map whose image agrees with the $O(n)$-module generated by $\varphi(U)$; we check that that (3) implies that $\varphi(U)$ coincides with the image of the (traceless) diagonal matrices in $W$ (because these form an irreducible representation of $\mathfrak{S}_n$, rescaling $\Phi$ is all that is needed to guarantee pointwise agreement of $\varphi$ and $\Phi \circ d$).
As a representation of $\mathfrak{H}_n$, $\mathrm{Sym}_2(\mathbb{R}^n)$ can be decomposed as follows. Recognise that
$$
\mathbb{R}^n = \mathrm{Ind}_{\mathfrak{H}_1 \otimes \mathfrak{H}_{n-1}}^{\mathfrak{H}_n}(\varepsilon \boxtimes \mathbf{1}_{\mathfrak{H}_{n-1}}),
$$
where $\varepsilon$ is the sign character of $\mathfrak{H}_1 = C_2$. Now, using a little Mackey theory (to write down the tensor square, and then extract they symmetric part), we see that
$$
\mathrm{Sym}_2(\mathbb{R}^n) =
\mathrm{Ind}_{\mathfrak{H}_1 \otimes \mathfrak{H}_{n-1}}^{\mathfrak{H}_n}(\mathrm{Sym}_2(\varepsilon) \boxtimes \mathbf{1}_{\mathfrak{H}_{n-1}})
\oplus
\mathrm{Ind}_{\mathfrak{H}_2 \otimes \mathfrak{H}_{n-2}}^{\mathfrak{H}_n}(\varepsilon \otimes \varepsilon \boxtimes \mathbf{1}_{\mathfrak{H}_{n-1}})
$$
where $\varepsilon \otimes \varepsilon = \varepsilon \otimes \varepsilon \otimes \mathbf{1}_{\mathfrak{S}_2}$ is a one-dimensional representation of $\mathfrak{H}_2$ where a signed matrix acts by $(-1)^m$, where $m$ is the number of $-1$'s in the signed matrix. Enthusiasts of wreath products will recognise both summands as being irreducible representations of the hyperoctahedral group. The key point here is that $\mathrm{Sym}_2(\varepsilon) = \mathbf{1}_{C_2}$, so the first summand simply becomes
$$
\mathrm{Ind}_{\mathfrak{H}_1 \otimes \mathfrak{H}_{n-1}}^{\mathfrak{H}_n}(\mathbf{1}_{\mathfrak{H}_1} \boxtimes \mathbf{1}_{\mathfrak{H}_{n-1}}),
$$
which is simply $\mathbb{R}^n$, viewed as a $\mathfrak{H}_n$ module that factors through the action of $\mathfrak{S}_n$. In particular, this summand corresponds to diagonal matrices in $\mathrm{Sym}_2(\mathbb{R}^n)$. (The other summand does not factor through the $\mathfrak{S}_n$ action.)
One final comment: If $n \geq 4$, as a representation of $\mathfrak{S}_n$,
$$
\mathrm{Sym}_2(\mathbb{R}^n) = S^{(n)} \oplus S^{(n)} \oplus S^{(n-1,1)} \oplus S^{(n-1,1)} \oplus S^{(n-2,2)}
$$
where $S^\lambda$ is a Specht module (irreducible representation) indexed by $\lambda$. The key point here is that $U = S^{(n-1,1)}$ appears with multiplicity 2. This means that trace zero symmetric matrices are not the unique subspace of $\mathrm{Sym}_2(\mathbb{R}^n)$ isomorphic to $U$. This means that condition (3) is not automatic. In particular, Nate's suggestion of understanding the interaction with the Casimir element will help pin down condition (1) (it helps identify the ambient $O(n)$ representation), but some extra information will be required to detect condition (3).
Thanks for that wonderful answer! A last question regarding the Casimir element: I understand that the irreducible components of a representation will be eigenspaces of the Casimir operator. But is it really clear that non-isomorphic irreducibles will correspond to different eigenvalues?
@Christopher The second point is always satisfied, no? The $\mathfrak{S}_n$-homomorphism $\varphi$ sends invariant vectors to invariant vectors. I have trouble following your argument. How do you use $\mathfrak{h}_n$-module description?
@Hans It's not clear and maybe it's even not true. Casimir operator acts on a highest weight module by a scalar that can be calculated from the highest weight. You'd need to check whether this scalar uniquely determines the representation $\mathrm{Sym}_2^0.$
Thanks @VítTuček. Regarding the second point: For example the 2x2 matrix $\binom{01}{10}$ is $\mathfrak{S}_2$ invariant but not $\textrm{O}(2)$ invariant. So (2) is not automatically true.
You have already answered each other's questions, but I just wanted to mention that condition (2) is necessary because the image of $\mathbf{1}_{\mathfrak{S}_n}$ is a priori a trivial representation of $\mathfrak{S}n$ only. But there are many irreducible representations of $O(n)$ (including $W$) that contain $\mathbf{1}{\mathfrak{S}_n}$ when restricted to $\mathfrak{S}n$. It is because $d$ maps $\mathbf{1}{\mathfrak{S}n}$ to $\mathbf{1}{O(n)}$ (and not a $\mathfrak{S}_n$ trivial contained in $W$) that condition (2) arises.
Nice idea Chris to use the hyperoctohedral group to pick out the right copy of of $U$. I realized a few hours after posting my comment about the Casimir that $W$ had 2 copies of $U$. One more thing about my Casimir comment: While it is not true that the action of the Casimir always distinguishes irreducible representations (you need the full center for that) if usually does distinguish at least the first few smallest dimensional representations of $\mathfrak{g}$, and in this case I believe it does although I haven't done the full calculation.
@ChristopherRyba Yeah, I missed that there is another trivial subrepresentation of the symmetric group in traceless matrices.
Not an answer, just too long for a comment.
All representations in sight can be decomposed as direct sums of irreducible representations. By the Schur lemma, mapping between irreducibles can be either zero, or multiple of identity. In this way the problem can be broken down into subproblems. (I am not sure it helps much, but at least you can see immediately, once you have the decomposition, whether there is any $O(n)$ map at all.)
The symmetric matrices decompose as $O(n)$-modules into traceless matrices and multiples of identity matrix. I.e.
$$A \mapsto (A - \frac{1}{n}(\mathrm{Tr}\, A)\; \mathrm{Id}) \oplus \frac{1}{n}(\mathrm{Tr}\, A ) \; \mathrm{Id}.$$
If $V$ is a tensor representation (i.e. subrepresentation of $\bigotimes^k\mathbb{R}^n$) then the decomposition into $O(n)$-modules works in a similar way. One substracts all possible traces over all possible pairs of indices and then decomposes resulting modules according to their $\mathfrak{S}_n$-symmetries in indices. (See Goodmann, Wallach for reference.)
Unfortunately, I am not much familiar with representations of finite groups so I don't know what the decomposition of the symmetric matrices looks like.
Edit: this answer overlooks the requirement that $\Phi$ be linear.
For $a$ a diagonal matrix, let $Z_{O(n)}(a)$ be its centraliser in $O(n)$, i.e. $\{k\in O(n) \mid kak^{-1} = a\}$. This is easy to describe: partition $\{1,\dots,n\}$ into subsets of size $n_1,\dots,n_m$ according to where the components of $a$ are equal; then $Z_{O(n)}(a)$ is a corresponding copy of $O(n_1)\times\dots\times O(n_m)$ (blockwise diagonal if the components of $a$ are sorted).
A necessary condition for the existence of $\Phi$ is that for all $a$, we must have $\varphi(a) \in V^{Z_{O(n)}(a)}$. Simply write down what the equivariance says.
Let's prove that this condition is also sufficient, so assume it is satisfied. Let $s$ be a symmetric matrix, and write it as $kak^{-1}$ where the entries of $a$ are nondecreasing and $k\in O(n)$. Define $\Phi(s) = k\cdot \varphi(a)$. This is well-defined: $a$ depends only on $s$ and the only possible decompositions of $s$ with this $a$ are the $kza(kz)^{-1}$ with $z\in Z_{O(n)}(a)$.
The map $\Phi$ is $O(n)$-equivariant by construction (we defined it on each orbit by taking a representative and applying the action).
Finally, $\Phi$ agrees with $\varphi$ on all diagonal matrices because of the assumption of permutation equivariance.
Thanks for the answer. Could you also explain why $\Phi$ is a linear map?
Oh, I missed that requirement! I'll have to check whether it is.
I have added a counterexample where $\Phi$ is not linear.
Yeah, that was likely. Thanks for the counterexample. I'll think about the correct problem.
|
2025-03-21T14:48:31.357641
| 2020-06-24T17:00:42 |
364043
|
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"url": "https://mathoverflow.net/questions/364043"
}
|
Stack Exchange
|
Image of morphism of locally of finite type has a closed point?
Let $S$ be finite dimensional locally Noetherian regular scheme. Let $f \colon X \rightarrow S$ be locally of finite type.
Then $f(X) \subset S $ contains a closed point of $S$?
If it does, I'd like to know the proof.
Consider the case that $S$ is the spectrum of a DVR, and $X$ the spectrum of its field of fractions.
|
2025-03-21T14:48:31.357715
| 2020-06-24T18:18:09 |
364048
|
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|
Stack Exchange
|
When is $-1$ in the image of a field norm?
Let $p >0$ be an odd prime and let $\mathbb{K} = \mathbb{Q}(\zeta) \subseteq \mathbb{C}$ with $\zeta$ a primitive $p$th root of unity. There is a unique subfield $\mathbb{Q} \subseteq \mathbb{F} \subseteq \mathbb{K}$ satisfying $[\mathbb{F}:\mathbb{Q}]=2$. Specifically $\mathbb{F} = \mathbb{Q}(\alpha)$ where $\alpha^2 = (-1)^{\frac{p-1}{2}}p$.
If $p \equiv -1 \pmod{4}$ then $[\mathbb{K}:\mathbb{F}] = (p-1)/2$ is odd so $\mathrm{N}_{\mathbb{K/F}}(-1) = (-1)^{[\mathbb{K}:\mathbb{F}]} = -1$, where $\mathrm{N}_{\mathbb{K/F}} : \mathbb{K} \to \mathbb{F}$ is the field norm.
When $p \equiv 1 \pmod{4}$ why is $-1$ not in the image of $\mathrm{N}_{\mathbb{K/F}}$?
This should be very elementary and for reasons coming from representation theory of finite groups I know this statement is true. However I'd like a straightforward number theory argument for this.
My basic idea was the following. We have $[\mathbb{K}:\mathbb{Q}] = p-1$ and by assumption $4 \mid p-1$ so there exists a unique subfield $\mathbb{Q} \subseteq\mathbb{E} \subseteq \mathbb{K}$ with $[\mathbb{E}:\mathbb{Q}] = 4$. By transitivity of norms, if $-1$ is in the image of $\mathrm{N}_{\mathbb{K/F}}$ then it's also in the image of $\mathrm{N}_{\mathbb{E/F}}$. Hence, it suffices to show it's not in the image of $\mathrm{N}_{\mathbb{E/F}}$. Here is where I got a bit stuck as I wasn't sure what the field $\mathbb{E}$ is exactly. One can get a basis by taking Galois sums of $\zeta$. There might also be a way to use the discriminent.
I have the same question in the local case. So assume $\mathbb{K} = \mathbb{Q}_{\ell}(\zeta)$ with $\ell > 0$ a prime and $\zeta \in \overline{\mathbb{Q}}_{\ell}$ a primitive $p$th root of unity. Looking in Serre's Local Fields the Galois group $\mathrm{Gal}(\mathbb{K}/\mathbb{Q}_{\ell})$ should still be cyclic. So assume $2$ divides $[\mathbb{K}:\mathbb{Q}_{\ell}]$ then there is a unique subfield $\mathbb{Q}_{\ell}\subseteq\mathbb{F} \subseteq \mathbb{K}$ with $[\mathbb{F}:\mathbb{Q}_{\ell}]=2$. The following should be true:
If $\ell \neq p$ then $-1$ is in the image of the norm map $\mathrm{N}_{\mathbb{K/F}}$. If $\ell = p$ then $-1$ is in the image of the norm map $\mathrm{N}_{\mathbb{K/F}}$ if and only if $p \equiv -1 \pmod{4}$.
By a block theory argument from finite groups I know the statement when $\ell \neq p$ is true. Hasse's Norm Theorem would then imply that the $\ell = p$ case agrees with the global case. However this all feels far too overblown. There should be an elementary number theory argument for all of this.
Apologies in advance if this is all too elementary. I am sure these answers and arguments are well known.
In your last highlighted item, you are saying in part that for a prime $p \equiv 1 \bmod 4$, $-1$ is not a norm from $\mathbf Q_p(\zeta_p)$ down to $\mathbf Q_p(\sqrt{p})$. But this extension has degree $(p-1)/2$, and if $\omega$ is a $(p-1)$th root of unity in $\mathbf Q_p$ then its norm for the field extension $\mathbf Q_p(\zeta_p)/\mathbf Q_p(\sqrt{p})$ is $\omega^{(p-1)/2}$, which is $\pm 1$ and is $-1$ for half the choices of $\omega$.
My previous argument has nothing to do with $p$ being $1 \bmod 4$. For an odd prime $p$, the field $K = \mathbf Q_p(\zeta_p)$ has cyclic Galois group $(\mathbf Z/p\mathbf Z)^\times$ over $\mathbf Q_p$, so $K$ contains a quadratic subfield $F/\mathbf Q_p$ inside $K$. If $\omega$ is a $(p-1)$th root of unity in $\mathbf Q_p$ then ${\rm N}{K/F}(\omega) = \omega^{(p-1)/2}$, which is $1$ for half the choices of $\omega$ and $-1$ for the other half. In particular, $-1 \in {\rm N}{K/F}(K^\times)$ for all odd $p$.
Let $a = p+b\sqrt{p}$, where $b$ is an even, positive integer such that $b^2+c^2=p$ for some integer $c$. The subfield $\mathbb{E}$ must be $\mathbb{F}(\sqrt{\pm a })$, where the sign is $+$ if $4 \mid b$ and $-$ otherwise. See page 2 of http://www.pphmj.com/abstract/1486.htm .Your conclusion is simple enough in the case $4 \nmid b$, but the case $4 \mid b$ seems harder to me.
Maybe I've overlooked something here, but: if $p = 1 \bmod 4$, isn't $\mathbb{F}$ totally real and $\mathbb{K}$ totally complex, and so any norm from $\mathbb{K}$ to $\mathbb{F}$ has to totally positive and thus can't be $-1$?
I think it's unfortunate that the OP is using the exact same notation for a cyclotomic and quadratic extension of $\mathbf Q$ as for a cyclotomic and quadratic extension of a local field, which makes it a bit confusing to keep straight which norm mapping is being discussed. A rational number may be in the image of the norm mapping for the local field extension without being in the image of the mapping for the number field extension.
@DavidLoeffler Ah, thanks, I was hoping to get some argument like that by passing to the smaller extension $\mathbb{E}$, so as to mimic the situation with $\mathbb{R}(\zeta)/\mathbb{R}$. I see now if you take $\mathbb{E} = \mathbb{Q}(\zeta+\zeta^{-1}) \subseteq \mathbb{K}$ to be the maximal real subfield then for any $x \in \mathbb{E}$ you have $\mathrm{N}{\mathbb{K/F}}(x) = \mathrm{N}{\mathbb{E/F}}(x)^2$ so it will be positive.
@KConrad Thanks for pointing that out about the $\ell = p$ case. I had mostly been thinking of the $\ell \neq p$ case where, of course, it might happen that $\zeta \in \mathbb{Q}_{\ell}$ when $p \mid \ell -1$. That's not quite what I was expecting in the $\ell = p$ case. Perhaps I've not translated my question from rep theory correctly in the $\ell$-adic case.
@OfirGorodetsky Thanks for the reference giving a description of $\mathbb{E}$. I was curious what that field would be.
Global question: If $p = 1 \bmod 4$, then the element $-1 \in F^\times$ is not a norm from $K^\times$, because $F^\times$ is totally real and $K^\times$ is totally complex; thus $-1$ is not a local norm at either of the infinite places of $F$, and hence cannot be a global norm either.
Local question, $\ell \ne p$: If $\ell \ne p$ then the fields $K$ and $F$ you define are both unramified extensions of $\mathbf{Q}_\ell$. Hence $N_{K/F}(O_K^\times) = O_F^\times$, and in particular contains $-1$. (This surjectivity result for the norm map on units follows from the analogous result for the residue fields, which is easy to prove by hand, see slide 6 of these lecture notes by Garrett.)
Local question, $\ell = p$: As pointed out by KConrad in the comments, if you take $\omega$ a $(p-1)$-st root of unity in $\mathbf{Q}_p^\times \subseteq F^\times$, then the composite $F^\times \hookrightarrow K^\times \xrightarrow{N_{K/F}} F^\times$ is raising to the $(p-1)/2$-th power; so its image contains $\omega^{(p-1)/2}$ for every $(p-1)$-st root of unity, i.e. it always contains $-1$.
(You were correct in deducing from Hasse's theorem that if $-1$ is not a norm globally then it had to fail to be a norm locally at some place. However, the bad place is not $p$ but $\infty$.)
Thanks a lot for your answer. Somehow I was not paying attention to the fact that $\mathbb{Q}(\sqrt{p})$ had two real primes. The argument for the $\ell \neq p$ case is also pleasingly close to the block theory argument.
|
2025-03-21T14:48:31.358287
| 2020-06-24T19:21:26 |
364052
|
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|
Stack Exchange
|
Average of product of matrix elements in the special orthogonal group
Given two lists $i$ and $j$ of $2n$ positive integers less than $N$, Collins and Sniady have computed, in Integration with respect to the Haar measure on unitary, orthogonal and symplectic group (see also On some properties of orthogonal Weingarten functions by Collins and Matsumoto), the integral over the orthogonal group,
$$ \int_{O(N)} \prod_{k=1}^{2n}u_{i_kj_k}du=\sum_{\sigma,\tau}\Delta_\sigma(i)\Delta_\tau(j) {\rm Wg}_N(\sigma^{-1}\tau),\qquad (1)$$ where the sum is over matchings, $\Delta_\sigma(i)=1$ if and only if the sequence $i$ satisfies the matching $\sigma$ and ${\rm Wg}_N$ is called the Weingarten function.
This implies for instance that $\int_{O(N)} u_{11}u_{22}du=0$ because the list $(1,2)$ does not match.
On the other hand, we know that a matrix from $SO(2)$ is of the form $u=\begin{pmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta \\
\end{pmatrix}$ so clearly we have $\int_{SO(2)} u_{11}u_{22}du=1/2$. This shows that the $SO(N)$ result can be quite different from the $O(N)$ one.
Is there a general theory of integrals like (1) over $SO(N)$?
you can of course multiply the integrand by the factor $1+{\rm det},u$ to constrain the average over $O(N)$ to $SO(N)$; I have used this approach in App.B of arXiv:1012.0932 for a particular problem in the context of topological superconductivity.
That's a good question. This is not quite an answer but a bit long for a comment.
A quick remark is that for $N$ odd a Haar-random element of $O(N)$ can be obtained as $\epsilon U$ where $\epsilon=\pm1$ with equal probability, and $U$ is a Haar-random element of $SO(N)$. So if your monomial has an even number of factors the integrals over $O(N)$ and $SO(N)$ coincide and thus Weingarten calculus is applicable. This is of course because for $N$ odd $-I$ has determinant $-1$ and is in the center of $O(N)$. I don't know if there is a similar trick for $N$ even.
I said it is a good question because, when looking at the vast probability/representation theory literature, I didn't see much as far an analogue of Weingarten calculus for special groups. Even the work of Chatterjee (and Basu and Ganguly,...) on $SO(N)$ lattice gauge theories does not seem to use Weingarten calculus. So for $SO(N)$, my answer to the question is: I don't know. However, for $SU(N)$ there is a combinatorial calculus. It is explained in my two answers to
How to constructively/combinatorially prove Schur-Weyl duality?
This technique was worked out explicitly by Creutz but it has its roots in the work of Clebsch and Hilbert in invariant theory. See for example, the averaging operator $[\cdot]$ used by Hilbert on p. 523 of "Ueber die Theorie der algebraischen Formen" is basically the same as Creutz's formula for $SU(2)$. Also note that if a combinatorial Weingarten-like calculus for $SO(N)$ is perhaps missing, there is at least an Euler angle parametrization due to Hurwitz (see this review by Diaconis and Forrester).
Thank you for your input. The idea of working with explicit Euler angles seems scary. Let me point out that it is easy to derive the Weingarten function for $O(N)$ using the theory of zonal polynomials, as I show here: https://arxiv.org/abs/1406.2182. Zonal polynomials appear because of the relation $\int_{O(N)}s_{2\lambda}(Au)du=Z_\lambda(A^TA)/Z_\lambda(1)$, where $s$ is a Schur function. So a Weingarten calculus for $SO(N)$ might follow from the integral $\int_{SO(N)}s_{\lambda}(Au)du$. Is this integral known?
|
2025-03-21T14:48:31.358543
| 2020-06-24T21:22:01 |
364058
|
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|
Stack Exchange
|
Polish spaces and isomorphisms
An isomorphism between two measurable spaces $(X_1,\mathcal{B}_1), (X_2,\mathcal{B}_2)$ is a measurable bijection $f:X_1\rightarrow X_2$ whose inverse is also measurable.
QUESTION. Can there be an isomorphism between an uncountable Polish space and a non-Hausdorff topological space, each endowed with its respective Borel $\sigma$-algebra?
Yes. Just consider the identity of $\mathbf{R}\cup{\infty}$. In the first case it's endowed with its topology of 1-point compactification. In the second it's endowed with the topology making ${\infty}$ dense (closed subsets are the whole subsets and closed subsets of $\mathbf{R}$). Borel subsets are the same for both topologies.
@YCor: thank you.
Here is a particularly simple way to see why the answer is yes, as YCor already pointed out in a comment. Uncountability is a red herring.
Ignore the uncountability condition for now. Consider the set $\{0,1\}$ once with the discrete topology and once with the non-Hausdorff topology whose only nontrivial open set is $\{1\}$, the Sierpiński space. They are clearly Borel isomorphic.
Now take your favorite uncountable Polish space $X$. Taking the disjoint union $X\sqcup\{0,1\}$ of both two point spaces will again give you two Borel isomorphic spaces, only one of which is Polish.
thank you for your answer.
|
2025-03-21T14:48:31.359025
| 2020-06-24T21:48:46 |
364062
|
{
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|
Stack Exchange
|
Can a continuous real-valued function on a large product space depend on uncountably many coordinates?
Is there a reasonably well-behaved topological space $X$ (ideally Polish), a set $\kappa$, and a continuous function $g: X^\kappa\to\mathbb{R}$ that depends on uncountable many coordinates?
If $X$ is a compact Hausdorff space, the answer is known to be no. To see this, note that the family of all continuous functions depending on only finitely many coordinates satisfies the conditions for the Stone-Weierstrass theorem and is, therefore, uniformly dense. The argument can be found in textbooks.
It seems like a negative answer follows pretty directly from the definition of the product topology and the fact that $\mathbb{R}$ is second countable. Is there a subtlety that I'm missing?
I don't think so. Thank you!
As I found by writing a wrong answer, it is not so simple as that. The argument I had in mind shows that for each given $x$, there are countably many coordinates such that if $y$ agrees with $x$ in those coordinates, then $g(y)=g(x)$. But that's not good enough.
Do you know whether $X^\kappa$ is Lindelöf, at least for small $\kappa$? That would give a negative answer as well.
Bockstein's theorem
Bockstein, M., Un théorème de séparabilité pour les produits topologiques, Fundam. Math. 35, 242-246 (1948). ZBL0032.19103.
This is the case of a product $\prod_{t \in T} X_t$ where all factors are second-countable. I that case any continuous function
$\prod_{t \in T} X_t \to \mathbb R$ depends on countably many coordinates.
PLUG... See Theorem 2.1 in
Edgar, G. A., Measurability in a Banach space, Indiana Univ. Math. J. 26, 663-677 (1977). ZBL0361.46017.
where the special case $X = \mathbb R$ is done. That is, a continuous function $\mathbb R^T \to \mathbb R$ depends on only countably many coordinates.
Thank you! An English language proof of Bockstein's theorem with an application to my question can be found in: Ross and Stone. Products of Separable Spaces. 1964.
They also show that second-countable can be weakened to separable.
"They" = Ross & Stone, not Bockstein?
I can't read French; I assumed from your description that Bockstein's argument required the space to be second countable.
Let $X$ be an uncountable discrete space with a distinguished element $0$. We view the product space $X^X$ as the space of maps $\phi: X \to X$. The set
$$ E := \{ \phi \in X^X: \phi(\phi(0)) = 0 \}$$
is easily seen to be clopen, hence the indicator function $1_E: X^X \to {\bf R}$ is continuous, but depends on all of the (uncountably many) coordinates of $X^X$.
The key point here (which was inspired by Nate's comment based on the earlier incorrect attempt at solving this problem) is that deciding whether a given map $\phi$ belongs to $E$ requires only a finite number of (adaptive) evaluations of $\phi$, but the set of (non-adaptive) locations where $\phi$ could potentially need to be evaluated is uncountable.
Note that a similar construction works for $X \times \{0,1\}^X$ using the set $E := \{ (x, \phi) \in X \times \{0,1\}^X: \phi(x)=0\}$; thus even a single highly non-compact factor is enough to generate a counterexample. (But I am not sure what happens if one insists that all of the factors be sigma-compact, in particular can one construct a continuous function $f: {\bf N}^{\bf R} \to {\bf R}$ that depends on uncountably many coordinates?)
Actually the case $X$ separable boils down to $X=\mathbf{N}$ which you suggest (just considering a map with dense image $\mathbf{N}\to X$ and composing with $\mathbf{N}^\kappa\to\mathbf{X}^\kappa$.
This is a beautiful example !!! (but I think there is a typo, the indicator function 1_E should be defined on X^X and not on X. I tried to change that, but for some reason edits need to change at least 6 characters and I did not know what to change in this wonderdul answer...
Typo corrected, thanks.
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2025-03-21T14:48:31.359355
| 2020-06-24T21:57:22 |
364064
|
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|
Stack Exchange
|
Reference request for linear matrix inequality with PSD matrices
In literature, people say a spectrahedron is the following set
$$\left\{x \in \mathbb{R}^d : x_1 A_1 + \cdots + x_d A_d \geq B \right\}$$
where $\geq$ is in the positive semidefinite sense. Is there a name for the set where one further restricts matrices $\{A_i\}_i,B$ to also be positive semidefinite?
This seems natural to me, but do not see people having looked at it, but also I am not sure what to google to understand this better.
Sure, I agree with that. I was more curious if they are just PSD but not diagonal, is there something known about such feasible sets (for example, do they have a name or are they interesting for other reasons?)
|
2025-03-21T14:48:31.359448
| 2020-06-24T23:29:19 |
364069
|
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|
Stack Exchange
|
Is there a general theory for Wilsonian renormalization?
I know that Wilson's renormalization group is not a theory per se and that there are many ways to implement it in a given system. Also, renormalization group techniques are applied in a large number of problems in e.g. QFT and Statistical mechanics.
However, I'd like to know if there is any general/abstract theory which rigorously formalizes all these incarnations or implementations in some unified way, so to give precise meaning to objects like renormalization group transformations, renormalization group flows etc, in which case these explicit implementations I mentioned become particular realizations of such a general theory. For instance, I know that we can define renormalization group maps on the space of measures and also on the space of Lagrangians. Thus, maybe these maps can be defined in abstract measure spaces or abstract functional spaces and some possible relations with, say, phase transitions and critica phenomena follow from theoretical considerations of this theory.
NOTE: I think "a general theory" accounting for all explicit models might be too much too ask, but I'd be happy with at least some general theory to some extent. It might be worth to mention that I'm mainly insterested in statistical mechanics.
EDIT 1: As pointed out in the comments, I should specify that the notion of renormalization I'm using here is Wilson's, as is now reflected in the title.
EDIT 2: I came across this beautiful article by Aernout van Enter, Roberto Fernández and Alan Sokal in which some abstract notion of renormalization group map on the space of measures and on space of Lagrangians is introduced. It seems to be possible to put Wilsonian renormalization group in a more abstract context, at least to some extent. This is the first paper (that I know) which treats renormalization in such general context (if there are other papers with similar approach, please let me know). However, I still have many questions to be answered. Some of them are listed below.
(1) The paper is pretty much self-contained but the section about renormalization group on Lagrangians is not very detailed in the sense that there are not many examples of explicit realizations of their definition of RG. Does the so-called renormalization group transformation or the renormalization group flow has anything to do with their definition, at least for finite volume systems?
(2) The authors seem to suggest that this general picture has, indeed, natural connections with the theory of Gibbs measures and, as a consequence, also with phase transitions. However, not much is said in the paper, so I'd like to know more about this. What kind of analyses can be made using this approach? What can physical phenomena can be explained by these general abstract approach?
(3) Is this approach just a particular point of view of the authors? Are there other aternative ways to define these objects? Is this the more general picture we have?
(4) Is it possible to relate this approach with dynamical systems? Is it possible to justify 'relevant directions' and fixed points and its consquences to physical theories?
NOTE 2: Although I posted the article and addressed some questions about it, it is not necessary to address the original question only in terms of this approach. My comments and questions on the paper only reflect the fact that I think this seems to be pretty much the most general picture one can come with, but I don't know much on the subject and I can be pretty much wrong, so any other approach is worthy to me.
Just a comment. "Renormalization" means different things to different people. It sounds like you mean "Wilsonian renormalization," unless there's some more specific label that you can attach to it. Perhaps that terminology should be reflected in your question.
@IgorKhavkine you are absolutely right! I will make and edit! Thank you!
Just a remark. At this point in time, a general rigorous theory might be too much to ask. One has a general philosophy/nonrigorous theory/loose set of methods and guidelines/...but within the scope of this philosophy there are examples treated rigorously. The general idea is very simple and surprisingly powerful when it can be implemented rigorously. See https://mathoverflow.net/questions/363119/every-mathematician-has-only-a-few-tricks/363383#363383
@AbdelmalekAbdesselam I really enjoyed your answer on the question linked. You said a general theory might be too much to ask, and I already expected that. But what about the abstract formulation I cited? Sounds pretty general to me.
Note that in his later years Wilson together with Głazek developed a different approach to renormalization, in which the "renormalization group" is actually a group, not semi-group. There, the operation of projecting out higher energies is replaced with a unitary transformation. A somewhat reduced version of this approach, widely used for non-relativistic systems, is known as Similarity Renormalization Group, while for relativistic systems it is known as Renormalization Group Procedure for Effective Particles.
|
2025-03-21T14:48:31.359802
| 2020-06-24T23:47:59 |
364070
|
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|
Stack Exchange
|
Books to develop a deep understanding of Algorithmic Information Theory?
I'm mathematical physicist working with hydrodynamics modelling. Recently, I had to turn to modelling of flows with particles and some questions I have I think are related to Algorithmic Information Theory. I have the general idea (like expected Kolmogorov complexity equals entropy, but this is more general physical intuition and I need more rigour), but I've never worked on AIT neither my colleagues, hence I'm here, asking for textbooks and advanced courses.
I think it is usual to start with Li and Vitanyi's textbook.
A comprehensive collection of resources is maintained by Marcus Hutter: Tutorials, Courses, Text Books, and more.
|
2025-03-21T14:48:31.359888
| 2020-06-25T00:22:04 |
364072
|
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|
Stack Exchange
|
Etale sites for stacks
Let $X$ be an algebraic stack, let $U\to X$ be a smooth cover by an algebraic space. In this setting, we have the big étale site of $X$ (if $X$ is a stack over a scheme $S$, this is the restriction of the big étale site of $S$ to $X$). We also have the Cech nerve $C(U/X)$, and the associated étale site. Are these two sites isomorphic? I.e. do abelian étale sheaves on the big étale site of $X$ correspond to abelian étale sheaves on $C(U/X)$, and do their cohomologies agree? Is there a reference where this is detailed?
So far, I only found Stacks Project Tag 06XJ, which gives a spectral sequence from the cohomology of the $C(U/X)_n$ to the cohomology of the big étale site of $X$, but this is not quite what I need.
I don't believe Cech cohomology computes etale cohomology, even in the case where $X$ is a scheme and $U\to X$ is an etale cover. I don't know an example or reference though.
I might be wrong, but I don't think I am considering Cech cohomology here. I am not saying "apply $F$ to $C(U/X)$, turn this into a Cech complex, and compute cohomology", but I am considering the sheaf on the étale site of $C(U/X)$ and computing sheaf cohomology there (so this involves choosing an injective resolution of $F$).
Ah, I see. Sorry, I think I misunderstood the question. I don't have anything substantial to say but in terms of modifying cech cohomology to compute derived cohomology, cohomological descent is interesting and useful (in case you haven't seen it before).
|
2025-03-21T14:48:31.360013
| 2020-06-25T01:54:38 |
364074
|
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|
Stack Exchange
|
How to combine two $d$-degenerate subgraphs into a $d$-degenerate graph?
A graph is $d$-degenerate if every its subgraph has a vertex of degree at most $d$. Suppose that $G$ is a graph and $G_1,G_2$ are two subgraphs of $G$ that are $d$-degenerate. Is there any way to efficiently decide whether the subgraph induced by $V(G_1)\cup V(G_2)$ is still $d$-degenerate? Furthermore, if $|V(G_1)\cap V(G_2)|$ is bounded by a constant, then is there any sufficient / necessary condition mainly relative to the structures of $G_1$ and $G_2$ so that $V(G_1)\cup V(G_2)$ induces a $d$-degenerate graph?
In order to make my quesion more precise, I expain it by an example when $d=1$.
Let $S=V(G_1)\cap V(G_2)$. If $|S|=1$, then it is trivial that $V(G_1)\cup V(G_2)$ induces a $1$-degenerate graph (forest)? If $|S|\geq 2$, we just need to test the following for each pair of non-adjacent vertices $u,v\in S$, and then can decide whether $V(G_1)\cup V(G_2)$ induces a forest.
($\star$) whether $u,v$ are in a same tree-componment of $G_1$, and meanwhile
in a same tree-componment of $G_2$?
if the answer is YES for some pair of $u$ and $v$, then there is a $(u,v)$-path in $G_1$, and another $(u,v)$-path in $G_2$, impying the existence of a cycle in $G$, and thus $V(G_1)\cup V(G_2)$ does not induce a forest;
if the answer is NO for every pair of $u$ and $v$, then $V(G_1)\cup V(G_2)$ induces a forest.
Therefore, while deciding whether $V(G_1)\cup V(G_2)$ induces a forest, we do not much care about vertices in $V(G_1)\setminus S$ and $V(G_2)\setminus S$, and just need to test the "local structure" of vertices in $S$.
So my previous question, in other words, is whether we can do something similar for $d\geq 2$, i.e, how to decide whether $V(G_1)\cup V(G_2)$ induces a $d$-degenerate graph just based on the "local structure" of vertices in $V(G_1)\cap V(G_2)$.
Can you precise what you mean by "efficiently decide"? Deciding if a graph is $d$-degenerate is easy, no? It suffices to remove any vertex of degree at most d and then check inductively if the remaining graph is d-degenerate (and if at any point the graph under consideration has minimum degree $>d$ then the answer is negative).
Thanks for your comments. I edited my question and described what I mean by "efficiently decide". Actually, I want to decide whether $V(G_1)\cup V(G_2)$ induces a $d$-degenerate graph only based on the "local structure" of vertices in $V(G_1)\cap V(G_2)$, without testing the existence of a $d$-degenerate ordering of $V(G_1)\cup V(G_2)$.
It's easy to write down examples where the union is or is not $d$-degenerate that have no edges between vertices in the intersection. Does that answer the question?
Many thanks but I don't quite undertand your answer. Actually, I want to have an algorithm to decide whether the union is or is not $d$-degenerate that just uses the information of the intersection. So I think examples are not enough for me...
|
2025-03-21T14:48:31.360211
| 2020-06-25T01:56:58 |
364075
|
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"Carlo Beenakker",
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|
Stack Exchange
|
Spectrum of the sum of two general matrices
Are there any restrictions on the possible spectrum of the sum of two arbitrary matrices with given spectra other than the trace identity?
In other words:
Let $\alpha, \beta, \gamma$ be $n$-tuples (nonordered) of complex numbers such that there exist matrices $A,B$ and $C$ respectively, such that $\alpha$ is the spectrum of $A$, $\beta$ is the spectrum of $B$ and $\gamma$ is the spectrum of $C$. Is the set of all possible triples of $\alpha, \beta, \gamma$ is a subset of $\mathbb{R}^{3n}/permutations$ defined by a single equation $\sum \alpha_i+\sum \beta_i=\sum \gamma_i$?
For Hermitian matrices (as well as some other special classes of matrices) the complete answer is well known by Klyachko-Knutson-Tao-...
I was wondering whether the structure of the set of triples of possible spectra is trivial or not known or out of reach for all matrices. I don't know the answer even for the case of $2\times 2$ matrices.
For $2\times 2$ matrices the answer is yes: the only constraint on the eigenvalues of $C$ is a constraint on the sum $\gamma_1+\gamma_2=\alpha_1+\alpha_2+\beta_1+\beta_2$; the product $\gamma_1\gamma_2=(\alpha_1+\beta_1)(\alpha_2+\beta_2)+\Delta$, where $\Delta$ can take any value independently of $\alpha_1,\alpha_2,\beta_1,\beta_2$, so the product is unconstrained.
For 2$\times$2 matrices, it suffices the consider the case
$$A=\left(\begin{matrix}\alpha_1&0\\ x&\alpha_2\end{matrix}\right),$$
$$B=\left(\begin{matrix}\beta_1&y\\ 0&\beta_2\end{matrix}\right).$$
|
2025-03-21T14:48:31.360455
| 2020-06-25T02:17:56 |
364078
|
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|
Stack Exchange
|
Heuristic for a density conjecture related to the Collatz $(3x+1)$-problem
First, some notation. Define $T(n)$ over $n\in \mathbb{N}$ as:
$$
T(n) = \left\{ \begin{array}{}
3n+1, & \text{if $n$ is odd}\ \\
n/2, & \text{if $n$ is even}
\end{array} \right.
$$
And let:
$$
\pi_a(x) = \#\{ n : n\leq x \mbox{ and } T^j(n) = a \mbox{ for some $j$ }\}
$$
Of course, the classic Collatz conjecture is that $\pi_1(x) = x$ for all $x$.
A succession of papers (see references below) have established bounds on $\pi_a(x)$ of the form: For $a \not\equiv 0 \; (\mbox{mod } 3)$,
$$
\pi_a(x) \geq x^\gamma, \; \forall x \geq x_0(a)
$$
The cases where $a \equiv 0\; (\mbox{mod } 3)$ are generally uninteresting for these questions since we know that the set $\{n : T^j(n) = a\}$ is precisely $\{2^k a : k=0,1,2,\ldots\}$.
In [1], the authors state the following conjecture (though I'm uncertain if they originated this conjecture or if there are earlier citations for it):
$$
\text{Conjecture A}:\mbox{For each $a\not\equiv 0 \;(\mbox{mod } 3)$, there is a positive constant $c_a$ such that}\\
\pi_a(x) \geq c_a x, \; \forall x \geq a
$$
There appears to be at least a tiny bit of empirical data to support this conjecture. Some arithmetic reveals that the last odd number (not including 1) in a Collatz trajectory must be of the form ${2^{2k}-1}\over{3}$. I took a pseudo-random sample of 10,000,000 numbers in the interval $[0, 2^m-1]$ for each $m \in \{64,128,256,512,1024,2048\}$ and recorded the last odd number in each trajectory. The results are summarized in the table below:
$$
\begin{array}{r|c|c|c|c|c|c}
\; & m=64 & 128 & 256 & 512 & 1024 & 2048 \\
\hline
k=2 & 9377219 & 9377605 & 9376867 & 9377728 & 9378367 & 9377611 \\
4 & 238003 & 237079 & 238246 & 237240 & 237296 & 237829 \\
5 & 378816 & 379476 & 379021 & 379013 & 378382 & 378506 \\
7 & 787 & 782 & 783 & 760 & 808 & 831 \\
8 & 4839 & 4736 & 4762 & 4916 & 4814 & 4896 \\
10 & 311 & 295 & 310 & 322 & 317 & 307 \\
11 & 21 & 24 & 9 & 19 & 15 & 17 \\
13 & 3 & 2 & 2 & 1 & 0 & 2 \\
14 & 1 & 1 & 0 & 1 & 1 & 1 \\
\end{array}
$$
Question: Is there any intution or heuristic that supports Conjecture A? Or, if not for the conjecture in full generality, at least for these set of "last" odd numbers of the form ${2^{2k}-1}\over {3}$, $ k\not\equiv0(\mbox{mod }3)$?
By "intuition or heuristic", I'm thinking of perhaps something in the same vein as this probabilistic heuristic for why the Collatz conjecture itself should be true.
Applegate, David; Lagarias, Jeffrey C., Density bounds for the (3x+1) problem. I: Tree-search method. II: Krasikov inequalities, Math. Comput. 64, No. 209, 411-426, 427-438 (1995). ZBL0820.11006.
Crandall, R. E., On the ”3x+1” problem, Math. Comput. 32, 1281-1292 (1978). ZBL0395.10013.
Krasikov, I., How many numbers satisfy the (3x+1) conjecture?, Int. J. Math. Math. Sci. 12, No. 4, 791-796 (1989). ZBL0685.10008.
Krasikov, Ilia; Lagarias, Jeffrey C., Bounds for the (3x+1) problem using difference inequalities, Acta Arith. 109, No. 3, 237-258 (2003). ZBL1069.11011.
Sander, J. W., On the ((3N+1))-conjecture, Acta Arith. 55, No. 3, 241-248 (1990). ZBL0707.11017.
Wirsching, Günther, An improved estimate concerning (3n+1) predecessor sets, Acta Arith. 63, No. 3, 205-210 (1993). ZBL0804.11022.
I'm sorry, I do not have the reference but if we think of some conventional integer $x$ as drawn from the 2-adic integers and suppose its Collatz sequence diverges then let $Q(x)$ be the function from $\Bbb Z_2\to\Bbb Z_2$ that sends any 2-adic integer to the sequence of parities $\pmod 2$ of its successors $T^n(x)$ then it turns out the density of ones in that sequence is greater than $\log_3(2)$ if and only if $x$ is positive. This appears perhaps similar or related to the heuristic you require.
|
2025-03-21T14:48:31.360726
| 2020-06-25T02:43:35 |
364080
|
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|
Stack Exchange
|
If $M$ and $N$ are closed and $M\times S^1$ is diffeomorphic to $N\times S^1$, is it true that $M$ and $N$ are diffeomorphic?
If $M$ and $N$ are closed smooth manifolds, and $M\times S^1$ is diffeomorphic to $N\times S^1$, is it true that $M$ and $N$ are diffeomorphic?
This has to be in the literature somewhere... Internet suggests it's true for simply-connected n-manifolds with n > 4.
Definitely not true if M is of dim 4.
@ Chris: I am sure it is. I don't know where to find it. I googled but did not find anything. I asked our topologists; no response yet. The one with S^2 instead of S^1 is (negative) pretty famous and important.
@ Mukherjee: Definitely means you know a result, please let me know.
There are closed flat manifolds that are not homotopy equivalent but become diffeomorphic after product with $S^1$, see https://math.stackexchange.com/questions/396608/does-x-times-s1-cong-y-times-s1-imply-that-x-times-mathbb-r-cong-y-times-ma. Also if $M$ is closed manifold of dimension $\ge 6$ (maybe $5$?) such that the Whitehead group of $\pi_1(M)$ is nonzero, then there is a $h$-cobordism $W$ from $M$ to a manifold $L$ not diffeomorphic to $M$, and $W\times S^1$ is trivial, so $M\times S^1$, $L\times S^1$ are diffeomorphic.
@IgorBelegradek: Sorry, I didn't see your comment when I posted my answer which is effectively your first sentence.
@MichaelAlbanese: no worries. I was too lazy to post an answer, and it is good that you did that, since the answers get more visibility.
If $M$ is of dimension $<4$ then the answer is YES because there are no exotic structures on $M$ and there are full classification results.[EDIT: In case of 3-manifolds this is true except some surface bundles over $S^1$ with fiber genus $>1$ and periodic monodromy, Stability of 3-manifolds.]
It is not true in dimension 4. For example, any closed simply-connected 4-manifold $M$ and an exotic copy $M'$ are h-cobordant by a theorem of Wall. Thus $M\times S^1$ is h-cobordant to $M'\times S^1$ (as one can extend the previus h-cobordsim trivially on the $S^1$ component). This is then a trivial cobordism by the high-dimensional s-cobordism theorem which says that such an h-cobordism is trivial if the Whitehead torsion of $\pi_1(M\times S^1)$ vanishes and indeed $Wh(\pi_1(M\times S^1))= Wh(\mathbb Z)=0$ by a result of Bass. So they are in fact diffeomorphic.
When the dimension is $>4$ the answer is YES if $M$ is simply-connected. To see this, notice that it is enough to show that $M$ and $M'$ are h-cobordant. Since $M\times S^1$ is diffeomorphic to $M'\times S^1$ there is a map $f:M \to M'\times S^1$. Since $M$ is simply-connected $f$ has a lift $\bar{f}$ to the universal cover $M'\times \mathbb R$. We claim that the image of $\bar{f}$ separates $M'\times \mathbb R$. Otherwise we can cut $M'\times \mathbb R$ along $Im(\bar{f})$ and connect the two boundary components by an arc $\gamma$. This arc in the original manifold $M'\times \mathbb R$ gives rise to a closed curve $\gamma'$ which transversally intersects $Im(\bar{f})$ at a single point. But $M'\times \mathbb R$ is simply-connected and thus $\gamma'$ is homotopic to a point disjoint from $Im(\bar{f})$, contradicting the homotopy-invariant count of transverse intersection points. Now, since $M$ is compact we can find a cobordism from $Im(\bar{f})\approx M$ to $M'\times \{t\}$ for some sufficiently large $t\in \mathbb R$. Since everything is simply-connected and the projection map induces isomorphisms on homologies, by Hurewitz' theorem we can conclude that this is an h-cobordism and so $M$ is diffeomorphic to $M'$.
Thanks, that was quick, though I am not familiar with some of the content. Probably you can expand the second and third sentences of your answer a little bit.
@MohammadFarajzadeh-Tehrani I tried to add a few references. Please let me know if it is helpful or not.
very good. thanks
@MohammadFarajzadeh-Tehrani I made a few changes in my ans.
Sorry if I am being daft, but why is the image of $\bar{f}$ separating?
@MichaelAlbanese I did an edit, please have a look.
The manifolds $M$ and $N$ may not even be homotopy equivalent!
In Compact Flat Riemannian Manifolds: I, Charlap showed that there are two closed flat manifolds $M$ and $N$ of the same dimension which are not homotopy equivalent (this is equivalent to $\pi_1(M) \not\cong \pi_1(N)$ as $M$ and $N$ are aspherical), such that $M\times S^1$ and $N\times S^1$ are diffeomorphic (which is equivalent to $\pi_1(M\times S^1) \cong \pi_1(N\times S^1)$ as closed flat manifolds are determined up to diffeomorphism by their fundamental group).
For a more explicit example, see this excellent answer by George Lowther.
Thanks for the answer. I was thinking about that too. In your response, $M$ and $N$ are compact but not closed, correct?
They are closed, I will edit to make this clear.
|
2025-03-21T14:48:31.361039
| 2020-06-25T02:45:50 |
364081
|
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|
Stack Exchange
|
"Oddity" in counting intervals of the Tamari lattice
Let $C_n=\frac1{n+1}\binom{2n}n$ be the Catalan numbers. Recall, in particular, that $C_n$ is odd iff $n=2^h-1$. A combinatorial proof is given by Deutsch and Sagan in Congruences for Catalan and Motzkin numbers and
related sequences and further extended by Postnikov and Sagan in What power of two divides a weighted Catalan number?.
Let's introduce the sequence $T_n=\frac2{n(n+1)}\binom{4n+1}{n-1}$ which enumerates intervals of the so-called Tamari lattice (see Bousquet-Mélou, Fusy, and Ratelle - The number of intervals in the m-Tamari lattices) (also counting triangular maps).
It is rather simple to prove the following using basic arithmetic means.
QUESTION. Can you provide a combinatorial justification that $T_n$ is odd if $n=2^h-1$?
More generally, odd values seem to appear for indices in https://oeis.org/A263133 .
There is a natural involution on Tamari intervals, namely reversal and exchange of bounds, with fixed points counted by 1, 1, 3, 4, 15, 22, 91, 140, 612, 969, 4389, 7084, 32890, 53820, ... One expects an instance of the q=-1 phenomenon for these numbers.
|
2025-03-21T14:48:31.361156
| 2020-06-25T06:18:50 |
364085
|
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|
Stack Exchange
|
A question of invertibility of matrices
Let $A$ and $B$ be self-adjoint $n \times n$ matrices. Let $A$ be diagonal. Suppose $A+tB$ and $tA+B$ are invertible for all $t \in \mathbb R$. What can we say about $A$ and $B$?
My guess is that $\mbox{Tr}(A) = \mbox{Tr}(B) = 0$. Definitely when $n$ is odd there exist no $A, B$ satisfying the hypothesis. Because in this case $\det(A+tB)$ is an odd-degree polynomial. I do not know what happens if $n$ is even.
If $B$ is also diagonal, then it cannot happen that $A+tB$ and $tA+B$ are invertible for all $t \in \mathbb R$. This is easy to see.
@People who are trying to close the question. Hi! The question looks very innocent and may be foolish. But look already it is leading to something nontrivial. I do not know why I am still getting negative votes!
There is a Jordan-like canonical form for symmetric matrix pairs $(A,B) = (A^*,B^*) \in \mathbb{C}^{n\times n} \times \mathbb{C}^{n\times n}$ under the transformation $(A,B) \to (M^*AM,M^*BM)$, with $M$ square invertible. You can find it stated, for instance, in Lemma~3 of Thompson's paper https://doi.org/10.1016/0024-3795(76)90021-5 . This canonical form is known today as even Kronecker canonical form (in a slightly different variant when $B$ is anti-Hermitian, but you can just multiply $B \gets iB$ to fix this).
Note that if you are interested in a canonical form under that transformation the requirement that $A$ is diagonal becomes superfluous, because you can always reduce to that case with another transformation of the same kind.
Each block in this canonical form determines a polynomial factor of $\det(A+tB)$, apart from type IV which is present only in pairs for which $\det(A+tB)\equiv 0$. So you just need to check which blocks correspond to factors that have no real zeros to find a complete characterization of these matrix pairs in terms of their blocks. If I am not mistaken, the allowed blocks are those of type II and III.
Thank you very much. Actually that is what was looking for. I think in your paper A and B are real matrices . Right? Also could not see E_4. I can see E_1,E_2 and E_3 conditions.
@Abeginnermathmatician Oops, I wrote this answer looking at the Arxiv preprint, where we had a variant with a fourth block type E4 (EKCF vs. EWCF). I have fixed it, as well as the theorem and table number according to the published version.
@Abeginnermathmatician You are correct that in my paper the theorem is phrased for real symmetric matrices only, but it holds without changes for complex Hermitian ones. I have now edited my answer and references to point to the original paper by Thompson that we cite; that reference treats directly the Hermitian-Hermitian case.
Thank you again. For the similarity can we take $X$ in the paper to be unitary? The matrix $X$ is referred to as "nonsingular constant matrix".
@Abeginnermathmatician no, $X$ is not unitary in general, unfortunately. Thompson's setting uses matrix pencils, i.e., linear expressions $\lambda A + \mu B$ in the variables $\lambda, \mu$, so he has to specify that $X$ does not depend on $\mu$ nor $\lambda$, i.e., it is "constant". However, this setting is equivalent to considering matrix pairs $(A,B)$, and in that case $X$ is just a general invertible complex matrix.
What about
$$A=\begin{pmatrix} 1 & 0 \\ 0 & -1 \end{pmatrix},\qquad B=\begin{pmatrix} 0 & 1 \\ 1 & 0 \end{pmatrix}\quad ?$$
I have edited the question. I guess under my hypothesis we must have $Tr(A)=Tr(B)=0$. Your example falls into this category.
@Abeginnermathmatician Replace $A \leftarrow A-\frac12I$ to get a counterexample without zero trace. More generally, $\det(A+xB) = -x^2-1$, so small perturbations to $A$ and $B$ will give a small perturbation of this polynomial, which still has no real zero. This is an open condition.
In complete generality, since $tA+B = t(A+(1/t)B)$, the second assumption is only adding the single requirement that $B$ is invertible. We can rephrase and say that we're looking at $A+tB$, $t\in\mathbb R_{\infty}$ (to interpret this rigorously, take advantage of the fact that we can always multiply by a non-zero number), which is a compact space, and the invertible matrices form an open set, so a small perturbation of an example always is an example too.
|
2025-03-21T14:48:31.361459
| 2020-06-25T06:21:45 |
364087
|
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|
Stack Exchange
|
On a reference for computing global spectrum of $A_n$-curve singularities, by H.Dao and E.Faber
This question is about chasing down a reference in a paper relating to non-commutative crepant resolutions and Cohen-Macaulay representation theory.
Allow me to first give a minor introduction.
Let $(R, \mathfrak m)$ be a commutative complete local Cohen-Macaulay ring. For a finitely generated $R$-module $M$, let $\text{End}_R(M):=\text{Hom}_R(M,M)$ be the endomorphism ring (this ring is Noetherian, albeit non-commutative) . A finitely generated $R$-module $M$ is called maximal Cohen-Macaulay iff $\text{depth}(M)=\dim R$. For any ring $S$, let $\text{gl}.\dim (S)$ denote the global dimension of $S$
The global spectrum of $R$ is the set $$gs(R):=\{n\ge 0 :n=\text{gl}.\dim \text{End}_R(M); M \text{ is a maximal Cohen-Macaulay module over } R\}$$.
In https://doi.org/10.1016/j.jalgebra.2016.03.020 , section 4.2, the authors calculate the global spectrum of $A_n : k[[x,y]]/(x^2+y^n), n\ge 2$ , curve singularities using ladders and state that in the reference [11] the same characterization has been carried out using a general fact about triangulated categories, and I'd really like to know this alternative proof. So I look at the reference [11] and see that it is a paper in preparation by H.Dao and E.Faber , On global dimension of endomorphism rings. Now when I searched Google and arxiv with this title, nothing showed up.
Does anyone know if this paper referenced [11] has been made available yet ?
Thank you.
Why don't you write to Dao at University of Kansas? He also may see your post here and answer it.
@Mohan: I am from a very small place and I really never have written to Professors at other Universities ever ... if Professor Dao is at Mathoverflow, then that's my best chance at getting a response ...
You could look him up on MO (Hailong Dao) and ping him.
@Mohan: how would I ping him ?
Exactly as you have pinged me here. Find his MO name prefix with the at symbol.
@Hailong Dao: Dear Professor Dao, would you be so kind enough to please provide an answer to this Question ? Thank you
@Mohan: I think I was able to do it. Thank you very much.
|
2025-03-21T14:48:31.361629
| 2020-06-25T08:15:25 |
364090
|
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|
Stack Exchange
|
Can an abelian group have multiple different actions of $\mathbb{Z}_p$?
This is perhaps a trivial question, but I've asked a few colleagues and they couldn't answer. For a given abelian group $M$, is it possible to have several different actions of the ring of $p$-adic numbers $\mathbb{Z}_p$? Or is it necessarily unique when it exists?
If the action were continuous of some sort then it would have to be unique: integers are dense in $\mathbb{Z}_p$, and for $n \in \mathbb{Z}$ one has $n \cdot x = x + \dots + x$ ($n$ times) so the action (if it exists) cannot be changed. But what if there are no assumptions about continuity?
(To be completely precise: by an action, I just mean a bilinear map $\mathbb{Z}_p \otimes_{\mathbb{Z}} M \to M$ satisfying $a \cdot (b \cdot m) = ab \cdot m$.)
Let $R$ be a commutative ring which admits two distinct ring homomorphisms $\mathbb Z_p\to R$, e.g. $R=\mathbb C$. Now take an abelian group $M$ such that $R$ embeds into the endomorphisms of $M$, e.g. $M=R$ viewed as a group.
@Wojowu Thanks! I am almost ashamed to ask this but is it possible to explicitly describe two distinct morphisms $\mathbb{Z}_p \to \mathbb{C}$? All I can find is about set theory (e.g. axiom of choice).
No you won't explicitly exhibit any such homomorphism at all. But for $x$ transcendental in $\mathbf{Z}_p$, for every transcendental $t$ in $\mathbf{C}$ there exists a ring homomorphism $\mathbf{Z}_p\to\mathbf{C}$ mapping $x$ to $t$ (it makes use of AC).
@YCor Thanks! (I knew I should have asked the logicians downstairs...)
An alternative is to consider the two obvious homomorphisms $\mathbf{Z}_p\to\mathbf{Z}p\otimes{\mathbf{Z}}\mathbf{Z}_p=:B$. While these homomorphisms are explicit, the fact that they are distinct (i.e., that $B$ is not reduced to $\mathbf{Z}_p$) makes use of some choice as far as I know.
@YCor Thanks again!
In fact I think YCor's example is in some sense "universal" (for lack of a better word), in that for a map of commutative rings $A\to B$, if the two maps $B\to B\otimes_A B$ agree, then the restriction $\mathrm{Mod}_B\to \mathrm{Mod}_A$ is fully faithful; so an $A$-module either is or isn't a $B$-module, but there is only one way to make it so.
Thinking twice, I tend to guess that "the two canonical maps $\mathbf{Z}_p\to \mathbf{Z}p\otimes\mathbf{Z} \mathbf{Z}_p$ are distinct", holds in ZF. Namely choose $x$ transcendental: the claim is that $x\otimes 1$ and $1\otimes x$ are distinct. Otherwise we have some "relation" saying that they are equal, and this relation holds in some finitely generated subring of $\mathbf{Z}_p$ containing $x$, and in the latter I guess that we distinguish by making an explicit pair of homomorphisms.
|
2025-03-21T14:48:31.361827
| 2020-06-25T09:13:35 |
364094
|
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|
Stack Exchange
|
Detect cycle in graph in logspace and linear time?
Let's consider graphs of bounded degree.
I know that it's possible to detect cycles in a graph in linear time -- essentially do a depth-first search, depositing a trail along the path you're currently exploring and checking if there's an edge back to the trail.
I also know that it's possible to detect cycles in log-space: think of the edges at each vertex as being cyclically ordered, so there is a “successor edge” function (defined on directed edges corresponding to the undirected edges of the original graph) which maps $(u,v)$ to $(v,u')$, where $u'$ is next after $u$ among the neighbours of $v$; then loop over all directed edges $e$, and check if the first edge in the sequence $e$, $\mathrm{succ}(e)$, $\mathrm{succ}^2(e)$, ... whose target is the source of $e$ is different from the opposite of $e$. Note that this may require at least quadratic time, e.g. if the graph is a cycle of length $n$ with spikes added at each vertex and you're unlucky enough to start your exploration at the spikes.
Now my question: is it possible to detect cycles both in logspace and linear time?
In fact, I would even be happy if there were an algorithm using sublinear space, say $O(\sqrt n)$. I know probabilistic algorithms achieving runtime $O(\sqrt n)$, but no deterministic one.
@MaxAlekseyev Sorry I wasn't being very precise. Let's assume the graphs have bounded degree, for simplicity. See e.g. https://www.geeksforgeeks.org/detect-cycle-undirected-graph/ for details.
This question might be more appropriate for cs.stackexchange or cstheory.stackexchange
|
2025-03-21T14:48:31.361965
| 2020-06-25T10:51:14 |
364097
|
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|
Stack Exchange
|
Continuity of the Lebesgue measure w.r.t the Hausdorff metric
I have a question linked to Interplay of Hausdorff metric and Lebesgue measure. Let us denote as $\mathcal K(\mathbb R^n)$ the space of compact subsets of $\mathbb R^n$ endowed with the Hausdorff metric $\rho$ and let $\lambda$ be the $n$-dimensional Lebesgue measure on $\mathbb R^n$. I want to know if there are (sufficient) conditions under which the measure $\lambda$ is continuous w.r.t. $\rho$, that is
$$
\lim_{k\rightarrow\infty}\rho(K, K_k)=0\qquad\Rightarrow\qquad \lim_{k\rightarrow\infty}\lambda(K_k)=\lambda(K).\qquad(\star)
$$
I tried to search it in the books Fractal geometry by Kenneth Falconer and Functions of Bounded Variation and Free Discontinuity Problems by Ambrosio, Fusco and Pallara but I did not find anything. In the second book it is written that, in the case $n=2$, the Hausdorff measure (which is a rescaling of the usual $\lambda$ on $\mathbb R^n$) is lower-semicontinuous w.r.t. the Hausdorff metric along sequences satisfying a suitable uniform concentration property, but this is not what I am looking for.
Some help? Do you have some references?
what do you want to put conditions on? the sequence, right?
@erz I don't know, maybe yes. But on the set $K$ too. I was asking if there are special cases in which the condition $(\star)$ holds.
$K$ has to be zero measure since finite sets are dense. Is there a zero measure compact that is a limit of $>\epsilon$ measure compacts?
If $K_k$ converges to $K$, then for any $\varepsilon>0$, $K_k$ is eventually included in $K+B(0,\varepsilon)$. Since $\lambda(K+B(0,\varepsilon))\to0$ as $\varepsilon\to0$ (Lebesgue dominated convergence), the Lebesgue measure is semicontinuous (I want to say upper?).
Proof by Socratic method
Conjecture (that I think is not difficult to prove with the above comments): for all neighbourhood $U$ for some fixed $K$, the set $\lambda_(U)=\lbrace \lambda(K'),\ K'\in U\rbrace$ contains $[0,\lambda(K)]$; more precisely, the intersection of all $\lambda_(U)$ where $U$ ranges over all neighbourhoods of $K$ is precisely $[0,\lambda(K)]$. I think I can write that down if it is of interest.
@PierrePC and is it actually an interval?
@VilleSalo Yes, precisely the interval $[0,\lambda(K)]$. I must say though, the answers on math.SE are rather detailed already (I just checked it), so I'm not sure what the OP has in mind.
@PierrePC I posted the question both here and in math.SE. You're right, the answers in math.SE are rather detailed but they were posted after my question on mathoverflow. Anyway, your comments were useful for me! Thank you!
|
2025-03-21T14:48:31.362162
| 2020-06-25T11:02:15 |
364098
|
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|
Stack Exchange
|
Proving equivalence of two definitions of a convex-type Hamming distance
Update: If somebody can answer my question there, then I will be able to fully answer my question here.
Consider $n\in\mathbb N$ and a non-empty set $M\subset\{0,1\}^n$. I have the following conjecture:
Conjecture. It is true that $$\sup_{\alpha\in[0,1]^n, \lVert \alpha\rVert_2=1}\min_{m\in M} \langle \alpha, m\rangle = \min_{\beta\in[0,1]^M, \sum_{m\in M} \beta_m = 1} \left\lVert\sum_{m\in M}\beta_m m\right\rVert_2.$$
Here, $\beta_m m$ is just the scalar multiplication of the number $\beta_m$ with $m\in M\subset\{0,1\}^n$. Also, $\lVert \cdot\rVert_2$ is the usual euclidean norm and $\langle\cdot,\cdot\rangle$ is the usual euclidean inner product. (And note, of course, that $[0,1]^M$ is the set of all functions $\beta: M\to[0,1]$ where I will write $\beta_m$ for $\beta(m)$.)
For instance, it is true if $M=\{m\}$, i.e. if $M$ only contains one element. In that case, the left-hand side equals $$\sup_{\alpha\in[0,1]^n, \lVert \alpha\rVert_2 =1}\langle \alpha,m\rangle.$$
By Cauchy-Schwarz, we know that $\langle\alpha, m\rangle\le\lVert \alpha\rVert_2\lVert m\rVert_2=\lVert m\rVert_2$ and we have equality if and only if $\alpha=\frac{m}{\lVert m\rVert_2}$. Hence the left-hand side equals $\lVert m\rVert_2$.
The right-hand side is, as we must have $\beta=1$, $\lVert m\rVert_2$.
If $M=\{m_1, m_2\}$, then we would have to prove
$$\sup_{\alpha\in[0,1]^n, \lVert \alpha\rVert_2 = 1} \min(\langle \alpha, m_1\rangle, \langle\alpha, m_2\rangle) = \min_{\beta\in[0,1]} \lVert \beta\, m_1+(1-\beta)\, m_2\rVert_2.$$
This is already not obvious to me. However, for example with $M=\{(1,0),(0,1)\}$, both sides can be computed to equal $\frac1{\sqrt 2}$.
Note: This conjecture is a Lemma that I would need to prove the equivalence of different definitions of convex distance that I found in the context of Talagrand's concentration inequality.
Another example: Consider $n=4$ and (with a slight abuse of notation) $M=\{m_1,m_2,m_3\}=\{(1,1,0,0),(0,1,1,0),(0,1,1,1)\}$.
The right-hand side is $$\min_{(\beta_1,\beta_2,\beta_3)\in[0,1]^3, \beta_1+\beta_2+\beta_3=1} \lVert (\beta_1,\beta_1+\beta_2+\beta_3,\beta_2+\beta_3,\beta_3)\rVert_2.$$
It is not too hard to see that the minimizer is $\beta=(1/2,1/2,0)$ for which we have $$\lVert (\beta_1,\beta_1+\beta_2+\beta_3,\beta_2+\beta_3,\beta_3)\rVert_2=\lVert (1/2,1,1/2,0)\rVert_2=\sqrt{\frac32}.$$
The left-hand side is $$\sup_{(\alpha_1,\alpha_2,\alpha_3,\alpha_4)\in[0,1]^4, \alpha_1^2+\alpha_2^2+\alpha_3^2+\alpha_4^2=1} \min(\alpha_1+\alpha_2,\alpha_2+\alpha_3,\alpha_2+\alpha_3)=\sup_{(\alpha_1,\alpha_2,\alpha_3,\alpha_4)\in[0,1]^4, \alpha_1^2+\alpha_2^2+\alpha_3^2+\alpha_4^2=1}\min(\alpha_1+\alpha_2,\alpha_2+\alpha_3).$$
The supremum occurs only if $\alpha_1+\alpha_2=\alpha_2+\alpha_3$, which happens for $\alpha=\left(\sqrt{\frac16},\sqrt{\frac23},\sqrt{\frac16},0\right)$.
For that $\alpha$, we have $$\alpha_1+\alpha_2=\alpha_2+\alpha_3=\sqrt{\frac32}$$ and so we indeed have equality of both sides.
Answer inspired by this great answer to a very related question by Paata Ivanishvili.
The right-hand side is equal to $\min_{m\in\operatorname{conv}(M)} \lVert m\rVert_2$, where $\operatorname{conv}(M)$ is the convex hull of $M$ in $\mathbb R^n$. Hence we have:
\begin{equation}
\begin{split}
\min_{m\in\mathrm{conv}(M)} \|m\|_{2}&=\min_{m \in \operatorname{conv} (M)} \max_{\|\alpha\|_{2}\leq 1} \langle \alpha, m\rangle
\\ &\overset{(*)}= \max_{\|\alpha\|_{2}\leq 1} \min_{m \in \operatorname{conv}(M)}\langle \alpha, m\rangle
\\ &\overset{(**)}= \max_{\substack{\alpha \in [0, \infty[^{n}\\\|\alpha\|_{2}\leq 1}} \min_{m \in M}\langle \alpha, m\rangle \\
&=\max_{\alpha\in[0,1]^n \\ \lVert\alpha\rVert_2=1}\min_{m \in M}\langle \alpha, m\rangle.
\end{split}
\end{equation}
(*): Here, a minimax Theorem was used.
(**): Here, the Bauer maximum principle was used (as the minimum of $m\mapsto\langle\alpha, m\rangle$ will be attained on the extremal points of $\operatorname{conv}(M)$, and they are contained in $M$.) Notice that it is also used that $M\subset\mathbb R_+^n$ in order to restrict the $\alpha$'s to $[0,\infty[^n$.
Note. This works for any compact $M\subset\mathbb R_+^n$, not just $M\subset\{0,1\}^n$.
|
2025-03-21T14:48:31.362522
| 2020-06-25T11:06:31 |
364099
|
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|
Stack Exchange
|
Approximation of smooth diffeomorphisms by polynomial diffeomorphisms?
Is it possible to (locally) approximate an arbitrary smooth diffeomorphism by a polynomial diffeomorphism?
More precisely: Let $f:\mathbb{R}^d\rightarrow\mathbb{R}^d$ be a smooth diffeomorphism for $d>1$. For $U\subset\mathbb{R}^d$ bounded and open and $\varepsilon>0$, is there a diffeomorphism $p=(p_1, \cdots, p_d) : U\rightarrow\mathbb{R}^d$ (with inverse $q:=p^{-1} : p(U)\rightarrow U$) such that both
$\|f - p\|_{\infty;\,U}:=\sup_{x\in U}|f(x) - p(x)| < \varepsilon$, $\ \textbf{and}$
each component of $p$ and of $q=(q_1,\cdots,q_d)$is a polynomial, i.e. $p_i, q_i\in\mathbb{R}[x_1, \ldots, x_d]$ for each $i=1, \ldots, d$?
Clearly, by Stone-Weierstrass there is a polynomial map $p : \mathbb{R}^d\rightarrow\mathbb{R}^d$ with $\|f - p\|_{\infty;\,U} < \varepsilon$ and such that $q:=(\left.p\right|_U)^{-1}$ exists; in general, however, this $q$ will not be a polynomial map.
Do you have any ideas/references under which conditions on $f$ an approximation of the above kind can be guaranteed nonetheless?
$\textbf{Note:}$ This is a crosspost from https://math.stackexchange.com/questions/3689873/approximation-of-smooth-diffeomorphisms-by-polynomial-diffeomorphisms
I know very little about this, but people have invested some time in finding some form of inversion procedure for a polynomial diffeomorphism: https://en.wikipedia.org/wiki/%C3%89tale_morphism#Inverse_function_theorem
Maybe this is already useful for you
Note that this is impossible when $d=1$, since, in that case, a polynomial map with polynomial inverse must be linear. I think it is very unlikely to be true for any $d>1$.
Thank you @GevaYashfe. (Not quite what I was looking for, but good to know either way.)
Thanks for your comment, @RobertBryant. The impossibility of the case $d=1$ was already pointed out in the stack-exchange version of the question (link above); I share your scepticism for $d>1$, but maybe there's some definite counterexample in the literature?
@qp10: In that case, you should have specified $d>1$ in your question.
@RobertBryant Yes you're right, I will edit that now.
The answer is 'no', because polynomial mappings with polynomial inverses preserve volumes up to a constant multiple.
To see why this property holds, suppose that $p:\mathbb{R}^d\to\mathbb{R}^d$ is a polynomial mapping with polynomial inverse $q:\mathbb{R}^d\to\mathbb{R}^d$. Then $p$ and $q$ extend to $\mathbb{C}^d$ as polynomial maps with polynomial inverses. This means that the Jacobian determinant of $p$ on $\mathbb{C}^d$ is a complex polynomial with no zeros and hence must be a (nonzero) constant.
Now, consider a diffeomorphism $f:\mathbb{R}^d\to\mathbb{R}^d$ that is radial, i.e., $f(x) = m(|x|^2)x$ for some smooth function $m>0$. One can easily choose $m$ in such a way that $m(4)=1/2$ and $m(9)=4/3$, so that $f$ maps the ball of radius $2$ about the origin diffeomorphically onto the ball of radius $1$ about the origin while it maps the ball of radius $3$ about the origin diffeomorphically onto the ball of radius $4$ about the origin.
Let $\epsilon>0$ be very small and suppose that $\|f-p\|_{\infty;U} <\epsilon$ for $U$ chosen to be some very large ball centered on the origin. Then $p$ maps the sphere of radius $2$ about the origin to within an $\epsilon$-neighborhood of the sphere of radius $1$, while it maps the sphere of radius $3$ about the origin to within an $\epsilon$-neighborhood of the sphere of radius $4$. It's easy to see from this that $p$ cannot have constant Jacobian determinant.
Added remark: The group $\mathrm{SDiff}(\mathbb{R}^d)$ consisting of volume-preserving diffeomorphisms of $\mathbb{R}^d$ is a 'Lie group' in Sophus Lie's original sense (i.e., a group of diffeomorphisms defined by the satisfaction of a system of differential equations; in this case, that the Jacobian determinant be equal to $1$).
The subgroup $\mathcal{SP}(\mathbb{R}^d)\subset \mathrm{SDiff}(\mathbb{R}^d)$ consisting of volume-preserving polynomial diffeomorphisms with polynomial inverses however, is not a 'Lie subgroup' in Lie's original sense when $d>1$, as it cannot be defined by the satisfaction of a system of differential equations: It contains all of the mappings of the form $p(x) = x + a\,(b{\cdot}x)^m$ where $a,b\in\mathbb{R}^d$ satisfy $a\cdot b = 0$ and $m>1$ is an integer (indeed, $p^{-1}(y) = y - a\,(b{\cdot}y)^m$), plus, it contains $\mathrm{SL}(d,\mathbb{R})$ and the subgroup consisting of the translations. Using this, it is easy to show that, for any $f\in\mathrm{SDiff}(\mathbb{R}^d)$ and for any integer $k$, there exists a $p\in \mathcal{SP}(\mathbb{R}^d)$ such that $f$ and $p$ have the same Taylor series at the origin up to and including order $k$. Thus, $\mathcal{SP}(\mathbb{R}^d)$ cannot be defined by a system of differential equations (in Lie's sense).
Using this Taylor approximation property, one can prove that $\mathcal{SP}(\mathbb{R}^d)$, like $\mathrm{SDiff}(\mathbb{R}^d)$, acts transitively on $n$-tuples of distinct points in $\mathbb{R}^d$ for any integer $n$. Whether one can prove that $\mathcal{SP}(\mathbb{R}^d)$ can 'uniformly approximate' $\mathrm{SDiff}(\mathbb{R}^d)$ on compact sets is an interesting question.
That's convincing. Thank you!
@qp10: It's still an interesting question if one assumes that $f$ is volume preserving. For volume preserving $f$, the statement is correct when $d=1$, because $f$ itself must be polynomial. However, for $d>1$, there are non-polynomial volume preserving diffeomorphisms, so the question becomes interesting again. I don't see any obvious reason that the volume-preserving diffeomorphisms that are polynomial with polynomial inverse (which is obviously a subgroup of the volume-preserving diffeomorphisms) does not approximate all volume-preserving diffeomorphisms uniformly on compact subsets.
Do you think there is any polynomial diffeomorphism of degree $\ge2$ with polynomial inverse? To me, this already seems implausible, let alone the approximation property.
@IosifPinelis : Such non-linear 'bi-polynomial' diffeomorphisms exist; e.g. for $d=2$, the Hénon map is an explicit example.
@qp10 : Thank you for your comment. Indeed, the Hénon map is is even volume-preserving (up to a constant factor).
@IosifPinelis: In fact, there are tons of such volume-preserving polynomial mappings with polynomial inverses as soon as $d>1$. Any mapping of the form $p(x,y) = \bigl(x,y+h(x)\bigr)$ where $h$ is a polynomial in one variable is such, and once you start composing these with maps of the form $p(x,y) = \bigl(x+g(y),y\bigr)$ where $g$ is a polynomial in one variable and toss in the invertible linear maps, you'll rapidly generate a huge number of such maps that do not fix any variable and are not obviously invertible.
@RobertBryant : Thank you for your comment. Indeed, the example of the Hénon map suggests similar ideas. Great answer, too!
An illustration for one of the examples in the answer by Robert Bryant.
It is supposed to convey the feeling of something extremely rigid, unyielding and inflexible.
Image of the square $[-1,1]\times[-1,1]$ under the map $(x,y)\mapsto(x-y^2-2x^2y-x^4,y+x^2)$ (composite of $(x,y)\mapsto(x-y^2,y)$ with $(x,y)\mapsto(x,y+x^2)$).
|
2025-03-21T14:48:31.362911
| 2020-06-25T11:14:18 |
364100
|
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}
|
Stack Exchange
|
Minimizing coefficients in a product related to the Rogers Ramanujan identity
Start with the product for partitions into parts congruent to $1$ or $4$ modulo $5$:
$(1 + x + x^2 + x^3 + ...)(1 + x^4 + x^8 + x^{12} +...)(1 + x^6 + x^{12} + x^{18} +...)$...
Now replace some of the plus signs with minus signs and expand the product into a series. Is it possible that the coefficients of the resulting series can all be in the set {$+1,0,-1$}?
This is a variant of the question "A Product Related to Unrestricted Partitions" asked by David Newman about six years ago.
Are you hoping that, with sparser factors (i.e., skipping $(1+x^2+x^4+ \cdots)$ and $(1+x^3+x^6+ \cdots)$ and their analogues modulo 5), this is more tractable than David's question?
Indeed, my father was hoping that this case would be more tractable. I suppose there is some chance that one could show this case to be impossible by a brute-force backtracking computation. Initial computations suggest this is unlikely though.
|
2025-03-21T14:48:31.363007
| 2020-06-25T12:00:28 |
364103
|
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"Chris",
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"url": "https://mathoverflow.net/questions/364103"
}
|
Stack Exchange
|
if $I$ is finitely presented nilpotent and $M/IM$ is finitely presented, then $M$ is finitely presented
Let $R$ be a commutative ring, and let $I \subseteq R$ be a nilpotent ideal. Let moreover $M$ be an $R$-module, and let $IM$ be the submodule generated by the products $xm$ with $x \in I$ and $m \in M$. It is quite easy to prove that, if $M/IM$ is finitely generated as an $R$-module, then $M$ is finitely generated. Indeed, let $\tilde{m}_1, \ldots , \tilde{m}_k$ be generators of $M/IM$ and lift them to elements $m_1, \ldots, m_k$ in $M$. Call $N$ the submodule generated by $m_1, \ldots, m_k$: then, we see that
$$
M= N + IM.
$$
Indeed, take $m \in M$. By definition of the $m_i$, we have $r_1,\ldots,r_k \in R$ such that $m - \sum_i r_im_i$ lies in $IM$. Now, since $I$ is nilpotent, we conclude (a variant of Nakayama's lemma if you want) that $M=N$, and we are done.
What about finite presentation? Namely:
Let us assume that $I$ is nilpotent and also finitely presented as an $R$-module. Can I conclude that if $M/IM$ is finitely presented as an $R$-module, then the same is true for $M$? (if necessary, I can also assume that $R$ is a coherent ring).
I'm not sure how you apply Nakayama in this case, even though I believe the conclusion is true, i.e. that $M$ is f.g. The way I see it is by multiplying the identity $M=N+IM$ by $I^j$ for $j=1,\dots, k-1$ where $k$ is such that $I^k=0$, and then working backwards.
It's a "variant" of Nakayama: more or less as you say, $M=N+IM$ is the same as $M/N= I(M/N)$, so if $I$ is nilpotent then $M/N=0$. This works for the finitely generated case. My question is about finite presentation, which I feel is quite more difficult...
By the way: if $R$ is Noetherian, then finitely generated and finitely presented modules coincide, so the result is true. Hence, I believe that the assumption of $R$ being coherent is kind of necessary... I'll perhaps modify the original post.
No. $R=k[X_n\mid n\in\mathbb{N}]/(X_1^2)$ is coherent (if $k$ is a field), and the ideal $I$ $=$ $X_1R$ is nilpotent and f.p. If $J$ $=$ $\Sigma,_{n>1},X_1X_nR$, then $M=R/J$ is not f.p., yet $M/IM$ $=$ $R/(I+J)$ $=$ $R/I$ is.
|
2025-03-21T14:48:31.363175
| 2020-06-25T12:36:57 |
364106
|
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|
Stack Exchange
|
A differential inequality involving gradient and laplacian
Let $V:\mathbb{R}^{n}\to\mathbb{R}$ smooth, such that $\lim_{|x|\to\infty}V(x)=+\infty$.
What are conditions on $V$ that guarantee the existence of a function $U:\mathbb{R}^{n}\to\mathbb{R}$ such that for all $\alpha>0$
$$\liminf_{|x|\to\infty} (\nabla V(x) \cdot \nabla U(x) - \alpha\, \Delta U(x)) > 0 \ ?$$
Of course choosing $U=V$ does the job if $\nabla V$ does not vanish at infinity and we assume $\limsup_{|x|\to\infty}\Delta V(x)\leq0$. This condition seems too restrictive to me, any other idea?
Edit: I've realized that what I actually need is the following: given any $\alpha>0$ prove the existence of a function $U$ (possibly $U=U(\alpha,x)$) satisfying the inequality (under suitable hypothesis on $V$). The fact that $U$ may depend on $\alpha$ should help.
Suppose you know that $\Delta V=o(|\nabla V|^2)$ as $|x|\to\infty$. This implies that for any $\theta\in(0,1)$ you have eventually the inequality $\Delta V\le \theta|\nabla V|^2$. Then for any eventually increasing function $\phi$, if you take $U=\phi(V)$ your expression is bounded below by
$$
|\nabla V|^2((1-\alpha\theta)\phi'-\alpha \phi'').
$$
Then you can play with the choice of $\phi$. For instance, if $\phi''=o(\phi')$ and $\phi'>c>0$ you are done.
Thank you for your answer. Taking $U=\phi(V)$ is great, but is it possible to weaken the hypothesis on $V$, for example $\Delta V \leq \theta |\nabla V|^2$ for some fixed $\theta$?
It seems difficult, since you want your inequality to hold for arbitrarily large $\alpha$, but I can not exclude it
If I weaken the request and ask the inequality to hold only for $0<\alpha<\alpha_0$, then $\Delta V\leq \theta |\nabla V|^2$ for some $\theta<\alpha_0^{-1}$ is a sufficient hypothesis. This can be seen by choosing $U=V$. Do you see any further possibility when only $0<\alpha<\alpha_0$ is considered?
You could try with a multiplier of the form $\phi(x,V)$. The derivatives w.r.to $x$ should be of lower order, that is, small compared to the main terms. For instance of polynomial growth in $x$
If $U$ is allowed to depend on $\alpha$, is everything easier?
Maybe, but I do not see how to use this extra freedom
|
2025-03-21T14:48:31.363336
| 2020-06-25T14:00:27 |
364108
|
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|
Stack Exchange
|
canonical divisor of a contraction
Let $X$ be a smooth quasi-projective variety and $Y$ be a positive dimensional subvariety. Let $Z$ be a variety obtained from $X$ by contracting $Y$.
My question is what is the relation between $K_X$ and $K_Z$, where $K_X$ and $K_Z$ are canonical divisors of $X$ and $Z$ respectively ?
The normal thing to do is to compare $K_X$ and the pullback of $K_Z$ (their difference being the discrepancy divisor), but if the codimension of $Y$ is greater than 1, then $K_Z$ fails to be ($\mathbb Q$)-Cartier, so you can't pull back $K_Z$ (at least not in the usual sense). If you're interested in the case where $Z$ is not a divisor, then, it's maybe unclear how one should compare the two canonical divisors.
is it true that if $Y$ is of codimension at least $2$ then $Z$ is not Cohen-Macualay ?
One has to be a little careful with terminology. It is possible to contract a subvariety of codimension 2 to get $Z$ with $\mathbf Q$-Cartier, even Cartier, canonical divisor. The basic example is the Atiyah flop: here the base of the contraction is an ordinary double point, which is Gorenstein (in particular CM). However, sometimes people use contraction to mean exclusively K-negative contraction: for a $K$-negative contraction (or any $K$-nontrival contraction), if the contracted locus has codimension 2 or more, it's true that the canonical divisor of the base is not $\mathbf Q$-Cartier.
By K-negative did you mean that anti canonical is ample ? How to prove that for K-negative contraction with co-dim at least $2$ contracted locus, the canonical divisor is not Q-Cartier ?
Related to the comment of @Pop: If $\mathrm{codim}(Y \subset X) > 1$ then $Z$ can't be $\mathbb{Q}$-factorial, but $K_Z$ could still be $\mathbb{Q}$-Cartier, for instance if $Z = C(\mathbb{P}^1 \times \mathbb{P}^1)$ (this is a hypersurface so $K_Z$ is Cartier) and $X = \mathrm{Bl}_{C(0 \times \mathbb{P}^1)} Z$ ($X$ is smooth, since the preimage of $C(0 \times \mathbb{P}^1)$ is a smooth Cartier divisor on X, and the exceptional set is a $\mathbb{P}^1$).
As mentioned in the comments, at least when $Y \subset X$ is a divisor and $K_Z$ is ($\mathbb{Q}$-)Cartier, $K_X, K_Z$ and $Y$ are related through the discrepancy $a(Y, X)$, defined by
$$
K_X = \pi^* K_Z + a(Y, Z)Y
$$
where $\pi: X \to Z$ is the contraction. Section 2.3 of Kollár-Mori's Birational geometry of algebraic varieties covers this in detail.
|
2025-03-21T14:48:31.363539
| 2020-06-25T14:02:08 |
364109
|
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}
|
Stack Exchange
|
Jordan decomposition on the dual Lie algebra
$\newcommand\fg{\mathfrak g}\newcommand\gl{\mathfrak{gl}}\DeclareMathOperator\Ad{Ad}\DeclareMathOperator\GL{GL}$Let $k$ be an algebraically closed field, and let $G$ be a smooth, affine algebraic group over $k$. In everything below, I intend to work on the level of algebraic groups, not just of rational points. I am also happy to restrict to the reductive case, and even, if it helps, to the adjoint case. However, I am specifically interested in the case of small, positive characteristic.
There are well defined notions of a Jordan decomposition in $G$ and in the Lie algebra $\fg$ of $G$, both of which involve identifying a semisimple part and a nilpotent part that commute.
On the dual Lie algebra $\fg^*$, I don't know any reasonable notion of when two elements commute, but one can still imagine defining at least the conjugacy class of the semisimple part, as follows: let $X^*$ be an element of $\fg^*$, and consider the coadjoint orbit $\mathscr O = \Ad^*(G)X^*$. Are the following two statements true?
The closure of $\mathscr O$ contains a unique closed coadjoint orbit $\mathscr S$.
For every closed coadjoint orbit $\mathscr S$, there are a maximal torus $T$ in $G$ and an element $S^* \in \mathscr S$ such that $S^*$ vanishes on all root subalgebras of $T$ in $\fg$.
If one or more is not true, then is there any other way of coming up with a reasonable notion at least of the conjugacy class of the semisimple part in the Jordan decomposition? I would prefer such a decomposition to satisfy two properties:
a. Every $G$-invariant isomorphism $\fg^* \to \fg$ (e.g., the Killing form, when it is non-degenerate), if any, should commute with the map "taking the orbit of the semisimple part".
b. For every faithful representation $G \to \GL(V)$, we obtain $\gl(V) \cong \gl(V)^* \to \fg^*$, where the first isomorphism is via the trace pairing. For every $X \in \gl(V)$, the semisimple part of $X$ is carried into the orbit of the semisimple part of its image.
For (a), a remark: the existence of such an isomorphism is a rare phenomenon, true for semisimple Lie algebras, but for some others including non-reductive such as including $\mathfrak{g}\ltimes\mathfrak{g}^*$ for arbitrary $\mathfrak{g}$ (I call it "quadrable Lie algebra"); has restrictive implications such as: dim(center)=codim(derived). It's called "metric/metrizable" Lie algebra, or "quadratic Lie algebra" depending on authors, some of these terminologies not being specific as to whether one fixes an invariant non-degenerate quadratic form, or if one assumes the existence thereof.
@YCor, I think that (a) can fail even for semisimple Lie algebras in small positive characteristic. (For example, I think that it doesn't hold for $\operatorname{SL}_2$ in characteristic $2$, although I think one can work around the problem for all $\operatorname{SL}_n$'s.) As I mentioned, but should probably make more emphatic, I am perfectly happy to restrict to the reductive, and even the adjoint, case. Anyway, I agree that (a) is a weak condition, but it is one of the basic sanity checks I want to impose.
Oh, sure, I had char 0 in mind.
With a few restrictions ($p \ne 2$ and $G$ has no components of special orthogonal type), my (1) and (2) are Theorem 4, parts (iv) and (ii), of Kac and Weisfeiler - Coadjoint action of a semi-simple algebraic group and the center of the enveloping algebra in characteristic $p$.
By the way, the theorem of Kac and Weisfeiler assumes either $p\not=2$ or $G$ is not $\mathrm{SO}_{2n+1}$ when $G$ is almost simple. In particular $p\not=2$ is always enough. I think it doesn't want any coroot to be divisible by $p$, but I am not sure.
@Cheng-ChiangTsai, re, thanks! Of course you are right that it should be ‘or’, not ‘and’. If you are interested in some more remarks on the dual Jordan decomposition, you might be interested in Section 3 of my Explicit asymptotic expansions II, which has been souped up a bit in v3.
Wow! Yes yes those are very beautiful remarks. Thank you!
|
2025-03-21T14:48:31.363833
| 2020-06-25T14:56:06 |
364113
|
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|
Stack Exchange
|
How to write computer-assisted mathematics well?
Much has been said about writting good papers in mathematics. A short google query yields countless sources of advice. This skill also appears to be quite transferrable between basic branches of mathematics: a well-written paper in analysis follows the same basic principles as a well-written paper in algebra, etc.
Recently, I found my first (hopefully) publishable result that uses the help of a computer, and - to my embarassment - realised that I have very little idea what a well-written computer-assisted paper looks like. Surprisingly, I also had no success finding general writting advice on this subject.
It would probably be too broad to just ask "How to write computer-assisted mathematics well?", although to be honest that's the question to which I'm trying to find an answer. Let me try to be (marginally) more specific.
Are there any style manuals that specifically address the question of writting computer-assisted mathematics? Barring that, what are some well-known and well-written papers that one can try and emulate?
In my particular case, I have around 20 pages explaining how to reduce a certain problem in number theory / combinatorics to a finite computation and around 100 lines of Mathematica code that perform the computatio. Is it fair game to simply say, once the explanation is completed, "I took the argument above and I coded it up, and the computer produced 42 as output, so that's the solution"? If not, what are reasonable steps to take in order to ensure my findings are verifiable?
What steps, if any, should I take to make the code I used to perform computations accessible? In principle, the reader could recreate it themselves, the same way as they can retrace any technical computation that is routinely omitted, but that hardly seems polite.
My opinion is that writing the computer-assisted mathematics should now include sharing the code, and to share the code in a universally accessible way, ideally requires sharing it on an open-source language. In any case it's IMHO an excellent and important question.
It is also courteous to provide checksums. For instance, if one needs to evaluate the output $f(N)$ some computable function $f$ at some large number $N = 10^{10}$, also provide evaluations of $f$ at other values (e.g., $10^j$ for $j=1,\dots,9$), or also report the values of a related function $g(N)$. In particular any reported computations that are consistent with the theory or heuristics of what you are computing are reassuring.
In terms of sharing your code, I made some comments in my answer to a related MO question.
If you use a closed-source and proprietary CAS then I think providing the source is less important than a clear description of the formulation.
This earlier MO question may be helpful: Computer calculations in a paper. There I suggest that Thomas Hales' work on the Kepler Conjecture can serve as a (high-bar) model.
Another high-bar model, from the field of geometric topology, is the paper by D. Gabai, R. Meyerhoff, and N. Thurston, Homotopy hyperbolic 3-manifolds are hyperbolic.
Follow the golden rule: do unto readers as you would have them do unto you.
Include technical computations if doing so makes readers' life easier; omit them if doing so only obscures the exposition; move them to other sections or appendices if they are not useful on a first reading, but are still of potential value to (some) readers.
The code is no different. Include it in the text itself if doing so communicates something important about mathematics itself. Delegate it to appendix/external page otherwise. Include any comments on implementation, if these are not obvious.
Special considerations for code:
When uploading a paper on arXiv, you can upload ancillary files. That is a good way to include code with your paper.
Do not forget to comment your code. Code is meant to be read by people first, and computers second.
In writing code, exercise the same care you would that you would in relying on someone else's result in your proof. Try to make sure that the other people's code that you use actually works correctly. This is especially difficult with closed proprietary systems such as Mathematica. You can still perform tests, and if possible write more than one implementation. I have ran into a non-insignificant number of bugs this way.
There are so many things to say here. "The concept of 'well' depends on the (position of the) observer." "Don't just involve the reader, recruit them!" "Reading code is more boring than running it." "Is it really proved if it takes a computer?" Tempting as it is to address these and other points, I will elaborate on just one: recruitment.
A good paper informs, but a well written paper inspires. Imagine that you want someone to not only verify your result but extend it. You should not only explain your work with utmost clarity, you should also indicate ways in which your work could be verified or extended. Ideally you have done some of this verification or extension yourself, and left some of the fun of (re-) discovery to the interested reader.
If you can encapsulate the ideas of the Mathematica code into a paragraph, that makes the paper more approachable than presenting a block of well commented code. My opinion (as a reader, not as a professional writer) is that code listings are best left to an appendix or the end of the paper. Only if you are writing an extremely literate program, where every subroutine teaches some mathematics to the reader, do you include it in the paper. Describing how the program performs in a run without explaining well why each computation branch was chosen has all the thrill of watching paint dry. Instead, try to challenge the reader to code with you, by describing the relevant portion of the computation, and then you presenting your solution and subtly asking the reader to come up with code that is good or better.
I have no examples that apply directly to your situation. For inspiration, I recommend the New Turing Omnibus by A.K. Dewdney, which is a collection of short articles in computer science. If you can write the core of your paper in the style of one of these articles, you can at least get people to read and understand the core, and leave the less exciting stuff for a series of appendices.
Gerhard "Doesn't Always Remove An Appendix" Paseman, 2020.06.25.
In theory, good writing is good writing. Writing (or for that matter, doing) computer-assisted mathematics well is, at bottom, not different from writing (or doing) any kind of mathematics well.
In practice, computer-assisted mathematics does pose special challenges. Mathematical writing serves (at least) two purposes: (1) you want to convey your ideas clearly so that people understand them; (2) you want to present your arguments in a way that allows others to confirm their correctness. It would be wonderful if you could accomplish both of these goals simultaneously, and sometimes you can, but computer-assisted mathematics tends to pose special challenges. In particular, one often has to address these two goals separately.
It sounds like you've already done a pretty good job of identifying these two distinct goals, and addressing them separately. Your 20 pages sounds like it's geared toward conveying your ideas clearly, and motivating the computational part. All the usual guidelines for writing mathematics apply here. If you've done your job well, the reader will understand what the computer-assisted part is supposed to do, and how it does it. All that's left is part (2), ensuring that the reader can confirm correctness without too much pain.
To do this, you should first convince yourself that the computation is correct. It is good scholarly practice to have some healthy skepticism of the correctness of computer code, whether it's someone else's code or your own. Victor Miller likes to tell the story of how, historically, several published papers on the computation of the number of primes less than $n$ suffered from the "curse" that all the entries in their table were correct, except for the last and largest value. For any kind of nontrivial computation where your stated theorems actually depend on the computation being correct, you should try to code up the computation in two completely different ways (or at least use two completely different computer algebra packages). If the computation is too large or complicated for this to be practical, then try to think of ways of generating corroborating evidence that your computation is being carried out correctly (e.g., checksums, as Terry Tao commented). For an example of "best practices," I recommend this StackOverflow answer to the question of how to verify the correctness of a computation of the digits of $\pi$. Of course, in your writeup, you should at least summarize the cross-checks that you performed.
Finally, there is the question of how to make your code available to others. This is an important question, but I think that this has already been addressed by the MO questions that commenters have linked to.
|
2025-03-21T14:48:31.364855
| 2020-06-25T14:59:22 |
364114
|
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"Joel Villatoro",
"LSpice",
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}
|
Stack Exchange
|
Regarding extensions of finite groups by Tori
I know how to prove the following result. However, my proof is a little bit long and complicated and only uses fairly low tech results in group cohomology. It would be nice if I could find a citation in the literature that directly implies this claim but I am not experienced in group theory or homological algebra and to be honest I'm not sure where to start looking for such a result.
Lemma: Suppose $\mathbb{T}^n = \mathbb{R}^n/ \mathbb{Z}^n $ is the standard $n$-dimensional torus and $\Gamma$ is a finite group of order $k$. Consider a short exact sequence:
$$ 0 \to \mathbb{T}^n \to E \to \Gamma \to 1. $$
Suppose $ R := \{ t \in \mathbb{T}^n \mathrel| kt = 0 \} $. Then the following short exact sequence splits:
$$ 0 \to \mathbb{T}^n/R \to E/R \to \Gamma \to 1. $$
The proof basically boils down to the fact that $ H^2( \Gamma, \mathbb{T}^n) $ is annihilated by $k$ and using that to show that the induced map $ H^2(\Gamma, \mathbb{T}^n) \to H^2(\Gamma, \mathbb{T}^n/R) $ is the zero map and therefore the curvature class of the top short exact sequence projects to zero.
I guess the things I would like to know are:
Is this 'obvious' to those experienced with finite group extensions or Lie theory?
Is there a textbook/paper where I can find a result or exercise that directly implies this lemma?
Using \text inside math mode for formatting results in funny-looking text. Use stars starsrather than $\textbf{textbf}$$\textbf{textbf}$`. I have edited accordingly. (Also, I think that this is an interesting question!)
Yes, it is almost obvious.
@MikhailBorovoi, as a sorta-Lie theorist for whom it is not obvious, would you be willing to explain why it should be obvious?
It is almost obvious, see my one-page answer.
(I write an answer rather than a comment in order to accommodate exact sequences.)
Let
$$0\to T\to E\to\Gamma\to 1\tag{$E_1$}$$
be your first group extension, where $T$ is the torus ${\Bbb R}^n/{\Bbb Z}^n$.
Write $R_k\subset T$ for the kernel of multiplication by $k$ in $T$
and consider your second exact sequence
$$0\to T/R_k\to E/R_k\to\Gamma\to 1.\tag{$E_2$}$$
To the extension $(E_1)$ we associate its cohomology class $\eta_1\in H^2(\Gamma,T)$, and to extension $(E_2)$ we associate its class $\eta_2\in H^2(\Gamma,T/R_k)$.
Then it follows from the constructions of $\eta_1$ and $\eta_2$ that $\eta_2$ is the image of $\eta_1$ under the homomorphism
$$\phi_*\colon H^2(\Gamma,T) \to H^2(\Gamma,T/R_k)$$
induced by the canonical homomorphism
$$\phi\colon T\to T/R_k.$$
We have not yet used the assumption that $T$ is a torus and that $\#\Gamma=k$.
Now consider the surjective homomorphism
$$\alpha\colon T\to T,\quad x\mapsto kx.$$
Its kernel is $R_k$, and so it induces an isomorphism
$$\alpha_*\colon T/R_k\to T.$$
Identifying $T/R_k$ with $T$ using $\alpha_*$, we obtain that our
$$\phi\colon T\to T$$
is multiplication by $k$.
It follows that
$$\phi_*\colon H^2(\Gamma,T) \to H^2(\Gamma,T)$$
is multiplication by $k$ as well.
Since $\Gamma$ is a group of order $k$, multiplication by $k$ annihilates $H^2(\Gamma,T)$.
See Corollary 1 of Proposition 8 in Section 6, page 105, of:
Atiyah and Wall, Cohomology of groups, in: Cassels and Fröhlich (eds.), Algebraic Number Theory, Acad. Press 1967, pp. 94-115. It follows that $\eta_2=0$ and the sequence $(E_2)$ splits.
Without details from @JoelVillatoro it's hard to know for sure, but this seems similar to the proof that they very briefly sketched in the post.
The argument uses only the following property of $T$: the map $$T\to T, \quad x\mapsto kx$$ is surjective.
@LSpice: OP complained that his proof is low tech. Mine is not.
@MikhailBorovoi, thank you very much for your reply. Your proof here is certainly much better than the one I wrote originally. Certainly more efficient.
|
2025-03-21T14:48:31.365142
| 2020-06-25T15:17:52 |
364116
|
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"François Brunault",
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}
|
Stack Exchange
|
Toward a cyclotomic Riemann hypothesis
For an integer $n \ge 3$, consider the function $$u(n) = \frac{\sigma(n)}{n \log \log n}$$ with $\sigma$ the divisor function. Now consider the sequence (bounded below and decreasing) $$v_n = \sup_{m>n} u(m).$$
Then Robin's theorem states that the Riemann hypothesis (RH) is true if and only if the sequence $(v_n)$ reaches its limit (which means the existence of $m \in \mathbb{N}$ such that $v_m = \lim_{n \to \infty} v_n$).
To illustrate the complexity of the ring $\mathbb{Z}[\zeta_r]$ of cyclotomic integers of degree $r$ (with $\zeta_r=e^{2i\pi/r}$), here is the picture of the set $$\{\sum_{i=0}^4 a_i\zeta_5^i \ | \ a_i \in \mathbb{Z}, |a_i| \le 3 \} \subset \mathbb{Z}[\zeta_5].$$
Consider the quotient map $$i: \mathbb{Z}[\zeta_r] \to \mathbb{Z}[\zeta_r]/\sim$$ with $x \sim y$ if and only if there is a unit $u \in \mathbb{Z}[\zeta_r]$ such that $x=uy$. Then consider $$\mathcal{P}_r = i(\mathbb{Z}[\zeta_r] \cap \mathbb{R}),$$ where $\mathbb{Z}[\zeta_r] \cap \mathbb{R}$ is the ring of real cyclotomic integers of degree $r$.
For $x \in \mathcal{P}_r$, and $a \in i^{-1}(\{ x\})$, consider the set of divisors $$D(x)= i(\{ b \in \mathbb{Z}[\zeta_r] \cap \mathbb{R} \ | \ a/b \in \mathbb{Z}[\zeta_r] \cap \mathbb{R} \}).$$
Question 1: Is the set $D(x)$ finite (for $x$ nonzero)?
As pointed out by Wojowu in comment (before revision), the group of units can be huge, and so without quotient, the set of divisors could be infinite (for example, let $\phi$ be the golden ratio, then $\phi^k, \phi^{-k} \in \mathbb{Z}[\zeta_5]$, and $1 = \phi^k \cdot \phi^{-k}$ for all $k$). Then Question 1 asks whether the quotient is sufficient to make the set finite.
Answer to Question 1 (after the comments of François Brunault): Yes.
Proof: the ring of algebraic integers in a finite extension of $\mathbb{Q}$ is a Dedekind domain. So any nonzero proper ideal of $\mathbb{Z}[\zeta_r] \cap \mathbb{R}$ decomposes uniquely (up to permutation) into a product of (finitely many) prime ideals. Then a nonzero ideal admits only finitely many dividing ideals. But $$x \sim y \Leftrightarrow x(\mathbb{Z}[\zeta_r] \cap \mathbb{R}) = y(\mathbb{Z}[\zeta_r] \cap \mathbb{R}),$$ so the set $D(x)$ is in bijection with the set of singly generated dividing ideals of $x(\mathbb{Z}[\zeta_r] \cap \mathbb{R})$. $\square$
To define a divisor function $\sigma$, we first need to define a natural non-negative representative for each element of $\mathcal{P}_r$, by a map
$$\tau: \mathcal{P}_r \to \mathbb{R}_{\ge 0}.$$ For so, first consider the surjective map: $$ s: \mathbb{Z}^{r} \to \mathbb{Z}[\zeta_r]$$ such that $s(n_0,n_1, \dots, n_{r-1}) = \sum_i n_i \zeta_r^i$.
Let $P(\mathbb{Z}^{r})$ be the set of subsets of $\mathbb{Z}^{r}$. Consider the minimizers map $$m: P(\mathbb{Z}^{r}) \to P(\mathbb{Z}^{r})$$
such that $$m(S) = \{ v \in S \ | \ \Vert v \Vert_2 \le \Vert w \Vert_2, \ \forall w \in S \},$$
next consider the lexicographical order on $\mathbb{Z}^{r}$ defined by $$(n_0, n_1, \dots, n_{r-1}) \le (n'_0, n'_1, \dots, n'_{r-1}) \Leftrightarrow $$ $$n_{r-1} < n'_{r-1} \text{ or } (n_{r-1} = n'_{r-1} \text{ and } n_{r-2} < n'_{r-2}) \text{ or } \dots$$ Finally consider the lexicographical first map $l: P(\mathbb{Z}^{r}) \to \mathbb{Z}^{r}$ such that $l(S) = min(S,\le)$.
Now we can defined the teased map $\tau: \mathcal{P}_r \to \mathbb{R}_{\ge 0}$ as follows: $$ \tau(x) = s \circ l \circ m \circ s^{-1} (i^{-1} (\{x \})\cap \mathbb{R}_{\ge 0}).$$
Note that $\tau$ is unital, i.e. for every unit $u \in \mathbb{Z}[\zeta_r] \cap \mathbb{R}$ then $\tau(i(u))=1$.
For $x \in \mathcal{P}_r$, consider the divisor function $$\sigma(x) = \sum_{y \in D(x)} \tau(y),$$
next, for $\tau(x)>e$, consider the function $$u(x) = \frac{\sigma(x)}{\tau(x) \log \log \tau(x)},$$
and finally consider the sequence $$v_n = \sup_{\tau(x)>n} u(x).$$
Question 2: Is $v_n$ finite?
Assume that Question 2 admits a positive answer (otherwise the definition of $u$ should be modified using a more appropriate asymptotic). Then a degree $r$ cyclotomic Riemann hypothesis (CRH$_r$) can be stated as follows:
The decreasing sequence $v_n$ (defined above with $\mathcal{P}_r$)
reaches its limit.
Note that for $r=1,2,3,4,6$, (CRH$_r$) is obviously (RH) because then $\mathbb{Z}[\zeta_r] \cap \mathbb{R} = \mathbb{Z} $.
Question 3.1: Is (CRH$_r$) equivalent to (RH) for all $r$?
Question 3.2: Is there $r$ such that (CRH$_r$) is false?
Note that we can get alternative versions of the post replacing $\mathbb{Z}[\zeta_r] $ by any Noetherian subring of the algebraic integers.
I'm fairly certain the cyclotomic units are dense in $\mathbb R$ for all $r$ a prime larger than $5$, and hence $D(x)$ is going to be very much infinite. This should follow from Dirichlet Unit Theorem and the classical fact that all units are a root of unity times a real number.
@Wojowu: I just did some modifications according to your comment. Now Q1 asks whether the quotient by the unit group makes the divisor set finite. Next to define the divisor function, and I introduced a natural representative for each quotient class.
You are (essentially) looking at the ring of integers of the maximal real subfield $\mathbb{Q}(\zeta_r)^+$ of the cyclotomic field. There are only finitely many ideals dividing a given nonzero ideal.
@FrançoisBrunault What (or where) is the proof of that? A chain of dividing ideals (from a nonzero one) is finite because the ring is Noetherian, but how to prove that a nonzero ideal has only finitely many minimal dividing ideals?
@SebastienPalcoux This is true because of uniqueness of factorisation into prime ideals. More generally, this holds in Dedekind rings.
@FrançoisBrunault: Thanks! I improved the post according to your comments. I realize that historically the notion of Dedeking ring was specifically created to study the cyclotomic ring $\mathbb{Z}[\zeta_r]$, which is factorial for $1 \le r \le 22$ but not in general (in particular $r=23$).
Your definition of $\tau$ is extraordinarily complicated. Furthermore, if it is not stable under multiplication by units, then your divisor function sum is not well-defined. Is there a reason to not just use the norm?
@WillSawin: No specific reason, this is what came in my mind when I wanted to define a non-negative representative. But I am not number theorist and I am open to improvement. What is the definition of the norm on $\mathcal{P}_r$?
@WillSawin: Got it, you certainly mean the absolute norm after identifying $\mathcal{P}_r$ with the set of principal ideals of $\mathbb{Z}[\zeta_r] \cap \mathbb{R}$. This norm maps to $\mathbb{N}$, so does not provide a representative of an element of $\mathcal{P}_r$ (seen as an equivalence class), but seems easier to deal with. Then the mean question is whether the cyclotomic RH defined in this post is equivalent to the extended RH for the corresponding Dedekind zeta function (i.e. does Robin's theorem extended to ERH?).
(1) No map that provides a representative can possibly be invariant under multiplication by a unit (easy to check) so certainly that is not a good idea. (2) Yes, I think it's almost certain that it is equivalent to the RH for the appropriate Dedekind zeta function - just go through the same proof.
@WillSawin:This post only deals with the principal ideals, whereas the Dedekind zeta function is a sum over all the nonzero (possibly non-principal) ideals. So perhaps it is equivalent to an ERH for the principal part of the Dedekind zeta function. I don't know if such a "principal" ERH is possibly inequivalent to ERH, or if it is even relevant.
You would probably count principal divisors using characters of the ideal class group. The contribution of the trivial character would dominate. If not, it's possibly the RH for Hecke L-functions associated to characters of the ideal class group would come into play.
@WillSawin: I disagree with your sentence stating that $\tau$ (as defined in the post) is not invariant by multiplication by a unit (and then that no invariant representative map exists), because an element of $x \in \mathcal{P}_r$ is precisely an equivalence class of the form $x = a \mathcal{O}$ with $\mathcal{O}$ the group of units, so $x$ itself is already invariant by mutliplication by units: $\forall u \in \mathcal{O}$, $ux = ua\mathcal{O} = au\mathcal{O} = a \mathcal{O} = x$.
Yes, I misunderstood what you meant by representative. Your map is well-defined, but I believe the expected asymptotic will be different as soon as the degree is bigger than $1$ (so $n$ not $1,2,3,4,6$.) This is because your quantity is essentially a small power of the norm. You'll get something like $e^{ log n^{ 1-1/d} / \log \log n}$ if I did the math right, where $d$ is the degree. It may not be as sensitive to the Riemann hypothesis in this case.
|
2025-03-21T14:48:31.365699
| 2020-06-25T15:28:13 |
364118
|
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"authors": [
"Daniil Rudenko",
"Will Sawin",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/364118"
}
|
Stack Exchange
|
Cohomology of the moduli space of rational curves with $n$ marked points with spin structure
Consider $\mathcal{M}_{0,n}$- the moduli space of rational curves with $n$ marked points. A map
$$
p\colon \pi_1(\mathcal{M}_{0,n})\longrightarrow H_1(\mathcal{M}_{0,n},\mathbb{Z}/2\mathbb{Z})
$$
defines a finite cover $\mathcal{M}_{0,n}^s.$ One can think of it as a space, on which square roots of cross ratios are regular functions. If I understand correctly, this is the underlying manifold of the super-moduli space of rational curves with $n$ marked points of NS type.
Question 1 (easy): Points of $\mathcal{M}_{0,n}^s$ should parametrize marked rational curves with spin structure, but I am not sure that I understand, what that means. I would be grateful for an explanation or a reference.
Question 2 (more interesting): What is known about cohomology of $\mathcal{M}_{0,n}^s?$ For instance, are they mixed Tate?
This definition doesn't define a finite cover of $\mathcal M_{0,n}$ over $\mathbb Q$ - you need to pick a base point. However, it's possible to check, regardless of the base point, that the cohomology of $\mathcal M_{0,4}^s$ is Artin-Tate, by just bashing it out.
Thank you! I guess that I was implicitly working over C.
You want the Hodge structure to be mixed Tate type then? It is for $n=4$, because $\mathcal M_{0,4}^s $ is $ \mathbb P^1$ minus six points. Probably is for $n=5$ as well. I guess this should stop at some point though, but I don't know.
I see, thank you.
The cohomology is not mixed Tate for $n\geq 12$, and this possibly can be improved.
We can view $\mathcal M_{0,n}$ as the locus of $n$-tuples of points $x_1,\dots,x_n$ in $\mathbb P^1$ which are all distinct, and where $x_1=0, x_2=1, x_3=\infty$.
Consider the $2^{n-4}$-fold cover of $\mathcal M_{0,n}$ defined by adjoining square-roots of $x_j (x_j-1) (x_j- x_4)$ for $j$ from $5$ to $n$. This cover is an open subset of the $n-4$-fold self-product of the Legendre family of elliptic curves $y^2 =x (x-1) (x-x_4)$.
This $n-4$-fold self-product includes in its cohomology the space of cusp forms of weight $(n-4)+2$ and level $2$. The associated Hodge structures are pure of weight $(n-4)+1$, lie in degree $(n-4)+1$, and are not mixed Tate (unless they are trivial). Because of this purity they are preserved when restricting to an open subset, and thus these non-mixed-Tate cohomology classes show up in this $2^{n-4}$-fold covering.
Because the spin moduli space is the universal covering with monodromy group $\mathbb F_2^k$, it admits a finite map to this $2^{n-4}$-fold covering, and so the non-mixed-Tate classes will also show up there.
There is a cusp form of level $2$ and weight $8$ so this will show up for $n\geq 12$.
Thank you, this is very clarifying!
Also, I would be really glad for a reference, in what sense it is a spin moduli space? I found a lot of contradicting definitions in the literature. I guess that it should be the underlying variety for the corresponding super-moduli space, but even that is not completely clear to me.
@DaniilRudenko I know nothing about the spin terminology unfortunately.
Thank you, I will try to find out.
|
2025-03-21T14:48:31.365940
| 2020-06-25T15:32:54 |
364120
|
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"Enrico",
"Evgeny Shinder",
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"sort": "votes",
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|
Stack Exchange
|
Clemens-Griffiths component birational invariant
Let $X$ be a smooth variety over complex numbers $\mathbb{C}$, say a threefold for sake of better intuition. Is there any geometrical intuition behind the fact that
the Clemens-Griffiths component of the intermediate
Jacobian
$$J(X)= H^1(\Omega^2,X)/H^3(X, \mathbb{Z})$$
is a birational invariant. Recall the
Clemens-Griffiths component is called the associated
polarized torus $(J(X), \theta)$ with non-degenerated divisor
$\theta$.
I took some time to look in the proof and
understood the single steps but haven't still any
geometric intuition how to think about it. For example
we know that $H^1(\Omega^2,X)$ is a priori not
a birational invariant, but quotient out $H^3(X, \mathbb{Z})$
seems somehow to play important role in "clearing"
the obstuctions that could occure.
Thinking of blowing up
as archetypical examples for birational transformations,
is there any picture one can have in mind how
$H^3(X, \mathbb{Z})$ "cleans" the defect that in general
prevents $H^1(\Omega^2,X)$ from beeing birational invariant.
Is there any "geometry" behind?
The geometry is that when you blow up a curve in a threefold, you modify the intermediate Jacobian by adding the Jacobian of the curve. This seems pretty geometric to me.
At the level of intuition --so a very sloppy comment (I assume that we are speaking about threefolds here).
Suppose you take a (say) smooth threefold and you want to blow it up in a (say) smooth curve. $H^{2,1}(X)$ will not tell you much about the curve you are blowing up (just its genus). However, if you quotient out by $H^3(X, \mathbb{Z})$ then you want to recover the Jacobian of curve, which by Torelli determines the curve up to isomorphism (and smooth projective birational curves are isomorphic). Do not take it as a complete answer, but I found this picture helpful back in the days.
@abx: where can I find a proof of this fact you have quoted.
In Clemens-Griffiths of course.
An excellent modern survey by Beauville: https://arxiv.org/pdf/1507.02476.pdf; see section 3 for the intermediate Jacobian.
|
2025-03-21T14:48:31.366115
| 2020-06-25T15:50:03 |
364121
|
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|
Stack Exchange
|
Interpretation of probability statements in Nina Zubrilina's paper
I asked this question on Math.stackexchange but got no answer.
In the paper Zubrilina - Asymptotic behavior of the edge metric dimension of the random graph (MR, the main result is
$$\operatorname{edim}(G(n,p)) \leq (1+o(1))\frac{4 \log n}{\log(1/q)},$$ where $q= 1-2p(1-p)^2(2-p)$.
My first question is how should I interpret the result, what is $\operatorname{edim}$ of random graph. Should I interpret it as
$$\mathbb{P}\left[\operatorname{edim}(G(n,p)) \leq (1+o(1))\frac{4 \log n}{\log(1/q)}\right] \rightarrow 1 \text{ as $n \rightarrow \infty$?} $$
My second question is concerned with how to interpret lemma 2.2, it is stated that
Let $G=G(n,p)$ be the random graph. Let $V,E$ denote the vertex and edge sets. Let $\omega \in \{1,\dotsc,n\}$ be such that for any two distinct edges $e_1$, $e_2$ of $E$, a uniformly random subset $W \subset V$ of size $\omega$ satisfies
$$\mathbb{P}( \text{$W$ does not distinguish $e_1$, $e_2$}) \leq 1/n^4p^2. $$
Then
$$\operatorname{edim}(G) \leq \omega.$$
So, firstly how should I understand $E$ as subset of a random graph, and how can I fix two edges of this seemingly random set by saying "for any two distinct edges $e_1,e_2 \in E$". I am confused about how I interpret such statement. Can any one clarify them?
You should define terms in your question (like $\mathrm{edim}$, which is surely not a ubiquitous notion).
Also, you should link to your MSE question to avoid duplication of effort. Anyways, it seems your concerns are about basic shorthands when dealing with random objects, which are standard but I agree can be confusing when not spelled out.
The original link went to a page with only generic content for me ("0 Articles Found"). I pasted in the DOI link, which takes me to an articles-to-appear page which doesn't seem to list Zubrilina's. So I pasted in the name of the article and a link to the MR. I hope that was all all right.
I just skimmed the paper. When she writes
$\mathrm{edim}(G(n,p)) \leq (1+o(1))\frac{4 \log n}{\log(1/q)}$, where $q= 1-2p(1-p)^2(2-p)$
she means this inequality holds "asymptotically almost surely", i.e., the probability that this inequality holds goes to 1 as $n\to \infty$ (note: this is for a fixed $p$). This matches what you thought, and it's clear from her use of "a.a.s" in Lemma 2.3 and in Section 3. I previously misread your interpretation to think you were suggesting the result was a bound on $P(edim(G(n,p)))$ which of course makes no sense.
Similarly, in Lemma 2.2, edim$(G)$ refers to the family of graphs, and it's enough to prove the inequality holds a.a.s. as $n\to \infty$. So, when she says "Let $G = G(n,p)$ be the random graph" she is saying $G$ represents the family of graphs drawn from this random variable. That's confirmed in the proof, where she talks about the expected size of the edge set, $E$ -- it's a random variable. I think the definition of $\omega$ is fine as stated. She's choosing a number to guarantee a probabilistic inequality. It would be like if I said, "Consider an $n$-sided die, and let $n$ be a number such that the probability of rolling two consecutive 1s is less than 0.01." That statement does define a number. Of course, there is a probability on a given draw of $G = G(n,p)$ that you get a graph with no edges, or only 1 edge, and maybe she could have spelled out how to handle those cases. But I think they are vacuous, because the definition of "edge metric dimension" says "for any distinct $e_1,e_2\in E$". Also, the probability of this situation occurring goes to zero as $n\to \infty$.
I don't understand what you mean by "the edge metric dimension of the family of random graphs." This notion is just defined for a graph. I think that mahmoud's interpretation of the inequality holding a.a.s. is correct. For instance, the paper cites https://arxiv.org/abs/1208.3801 which proves a similar thing but for metric dimension instead of edge metric dimension (and states the a.a.s. bit in their theorem more clearly).
@SamHopkins: I think you're right. I'll edit.
The formal interpretation of Lemma 2.2 should also involve some kind of a.a.s. statement, but there's a little more unwinding to do there.
I edited a second time. I just jumped the gun on the first edit.
For the first case: I think that for many readers, it would still be unclear what it means to say that "the probability that $X_n\leq (1+o(1))f(n)$ goes to $1$ as $n\to\infty$". The "$o()$" notation already includes an $n\to\infty$ limit, which makes it hard to understand that statement. To interpret it really clearly one could write: for any $\epsilon>0$, the probability that $X_n\leq (1+\epsilon)f(n)$ goes to $1$ as $n\to\infty$.
|
2025-03-21T14:48:31.366477
| 2020-06-25T16:40:16 |
364128
|
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|
Stack Exchange
|
The sign of the tail of Fourier transform of a positive function/ characteristic function
I am interested in a specific density (positive function) and would like to prove that the tail of its characteristic function (Fourier transform) is positive ($>0$). Here is the density $f(x)=c_\alpha \exp(-|x|^{\alpha})(|x|^\alpha\log(|x|)+1)$, $\alpha \in(1,2)$, $x\in R$, $c_\alpha>0$ is a normalizing constant. I can show that the tail of the characteristic function is $O(\log(\omega)/\omega^{\alpha+1})$. I just cannot find any theorems for the sign of the limit. From the numerical analysis I see it is positive.
$x\in R$ or $x>0$? Or did you want $|x|$ in place of $x$, to make sure the characteristic function is real?
Looks like regular variation and Karamata's theory is what you are looking for. I would start by looking up the desired result in [K. Soni, R. P. Soni, Slowly varying functions and asymptotic behavior of a class of integral transforms: I, II, III. J. Anal. Appl. 49 (1975), 166–179, 477–495, 612–628].
@IosifPinelis, yes, thank you, edited
@MateuszKwaśnicki, yes, thank you, that is how I got the rates in limit, however I need to show it is positive. The Fourier transform itself is posttive for some values of $\omega$ and negative for others
I do not remember well (and I have no time to check it now), but I thought Soni–Soni give explicit constants? I mean, an explicit expression for the limit of $x \hat{f}(x) / f(1/x)$ as $x \to \infty$.
@MateuszKwaśnicki In Soni/Soni and bunch of others the framework is
$F(x)=\int_{0}^{\infty}k(xt)f(t) dt$ and behavior of $f(t)$ at $\infty$ relates to $F(x)$ at $0$.
In my case $F(x)$ is density and $f(t)$ is a characteristic function.
Basically I have to invert Fourier, which is fine. I have a problem with the assumption about $f(t)$. $f(t)$ is bounded and decreases to zero as $t \to \infty$. I know $f(t)$ is bounded as positive definite function and it decreases in absolute value to 0 when $t$ is large, so I cannot quite use these results
I see. Then, how about the following approach: Write $G(z) = c_1 \log(1 + z^2) / (1 + z^2)^{(\alpha + 1)/2}$. Notice that $g(x)$, the inverse Fourier transform of $G(z)$, is $c_2 + c_3 |x|^\alpha \log (1/|x|) (1 + o(1))$ as $x \to 0$ (see Soni–Soni), and smooth otherwise. Therefore, $f(x) - c_4 g(x)$ is $c_5 + c_6 |x|^\alpha (1 + o(1))$ as $|x| \to 0$ and smoth otherwise, and consequently $F(z) - c_4 G(z)$ is $O(|z|^{-\alpha-1})$ as $|z| \to \infty$. (This is of course just a sketch, and I do not claim it will work. I rather write this to ask whether you tried the above approach).
@MateuszKwaśnicki well the rate itself is not a problem, the sign of the coefficients is, I need something like $\sim$ not $O(\cdot)$
This is an extended version of my comment (and now heavily edited). Your function $f$ is
$$
f(x) = c_\alpha e^{-|x|^\alpha} (1 + |x|^\alpha \log |x|) = c_\alpha (1 + |x|^\alpha \log |x| - |x|^\alpha) + O(x^2)
$$
as $x \to 0$. Similarly,
$$
f'(x) \operatorname{sign} x = c_\alpha (\alpha |x|^{\alpha - 1} \log |x| - (\alpha - 1) |x|^{\alpha - 1}) + O(|x|) ,
$$
and
$$
f''(x) = c_\alpha (\alpha (\alpha - 1) |x|^{\alpha - 2} \log |x| - (\alpha (\alpha - 1) - (2 \alpha + 1)) |x|^{\alpha - 1}) + O(1) .
$$
Define
$$
G(z) = \frac{1}{(1 + z^2)^{(\alpha + 1)/2}}
$$
and
$$
H(z) = \frac{\log(1 + |z|^2)}{(1 + |z|^2)^{(\alpha + 1)/2}} = -\frac{d}{d\alpha} \frac{2}{(1 + |z|^2)^{(\alpha + 1)/2}} \, .
$$
Then the inverse Fourier transforms of $G(z)$ and $H(z)$ are
$$
g(x) = \frac{1}{2^{\alpha/2} \sqrt{\pi} \, \Gamma(\tfrac{\alpha + 1}{2})} \, |x|^{\alpha/2} K_{\alpha/2}(|x|)
$$
and (by the dominated convergence theorem)
$$
h(x) = -\frac{d}{d\alpha} \biggl( \frac{1}{2^{\alpha/2} \sqrt{\pi} \, \Gamma(\tfrac{\alpha + 1}{2})} \, |x|^{\alpha/2} K_{\alpha/2}(|x|) \biggr) .
$$
Note that
$$
g(x) = \frac{A_\alpha}{2} - \frac{B_\alpha}{2} \, |x|^\alpha + O(x^2) ,
$$
where
$$
A_\alpha = \frac{\Gamma(\tfrac{\alpha}{2})}{\sqrt{\pi} \, \Gamma(\tfrac{1 + \alpha}{2})} , \qquad B_\alpha = \frac{-2^{-\alpha} \Gamma(-\tfrac{\alpha}{2})}{\sqrt{\pi} \, \Gamma(\tfrac{1 + \alpha}{2})} .
$$
Similarly,
$$
g'(x) = -\frac{B_\alpha}{2} \, \alpha |x|^{\alpha - 1} + O(|x|) ,
$$
and
$$
g''(x) = -\frac{B_\alpha}{2} \, \alpha (\alpha - 1) |x|^{\alpha - 2} + O(1) .
$$
The derivative of the Bessel function $K_\nu$ with respect to the parameter $\nu$ can in fact be expressed in terms of other special functions, but the formula is quite complicated and I will not copy it here and only refer to the paper Higher derivatives of the Bessel functions with respect to the order by Yu.A. Brychkov, DOI:10.1080/10652469.2016.1164156. An integral expression for this derivative follows directly from the integral expression for $K_\nu(x)$; see formula 10.32.9 in DLMF, for example. What is important here is that the series expansion of $K_\nu$ near $0$ can be differentiated with respect to $\nu$ term-by-term (and either the explicit expression or the integral formula, both mentioned above, provide a way to make this claim rigorous; I did not verify this carefully, though). It follows that
$$
h(x) = -A_\alpha' + B_\alpha |x|^\alpha \log |x| - B_\alpha' |x|^\alpha + O(x^2 \log |x|) ,
$$
where primes stand for derivatives with respect to $\alpha$. Similarly,
$$
h'(x) \operatorname{sign} x = \alpha B_\alpha |x|^{\alpha - 1} \log |x| - (\alpha B_\alpha' - B_\alpha) |x|^{\alpha - 1} + O(|x| \log |x|) ,
$$
and
$$
h''(x) = \alpha (\alpha - 1) B_\alpha |x|^{\alpha - 2} \log |x| - (\alpha (\alpha - 1) B_\alpha' - (2 \alpha - 1) B_\alpha) |x|^{\alpha - 1} + O(\log |x|) .
$$
Consider an auxiliary function
$$
\phi(x) = f(x) - \frac{c_\alpha}{B_\alpha} h(x) - \frac{2 c_\alpha (B_\alpha - B_\alpha')}{B_\alpha^2} g(x) .
$$
The constants are chosen in such a way that
$$\begin{gathered}
\phi(x) = C_\alpha + O(x^2 \log |x|) , \\
\phi'(x) = O(|x| \log |x|) , \\
\phi''(x) = O(\log |x|) .
\end{gathered}$$
In order to estimate the Fourier transform of $\phi$, decompose $\phi$ into $\phi_1(x) = \phi(x) u(x)$ and $\phi_2(x) = \phi(x) (1 - u(x))$, where $u$ is smooth, $u = 1$ in $[-1, 1]$ and $u = 0$ outside $(-2, 2)$. Since $\phi_2$ is infinitely smooth and all derivatives of $\phi_2$ are integrable, the Fourier transform of $\phi_2$ decays faster than $|z|^{-p}$ for every $p > 0$, so we only need to handle $\phi_1$.
Observe that for an arbitrary $\varepsilon \in (0, 1)$, we have $|\phi_1''(x)| \leqslant a_\alpha \log (e + \tfrac{1}{|x|}) \leqslant b_{\alpha, \varepsilon} (1 + |x|^{-\varepsilon})$ for some constants $a_\alpha$ and $b_{\alpha, \varepsilon}$. Thus, $\phi_1'$ is Hölder continuous with exponent $\beta = 1 - \varepsilon$. As a consequence, the Fourier transform of $\phi_1'$ is $O(|z|^{-\beta})$ as $|z| \to \infty$. It follows that the Fourier transform of $\phi_1$ is $O(|z|^{-1 - \beta})$ for any $\beta \in (0, 1)$, and by choosing $\beta > \alpha - 1$ we find that, in particular, it is $o(|z|^{-\alpha})$. This implies that the Fourier transform of $\phi$ is $o(|z|^{-\alpha})$, and thus, finally
$$
F(z) = \frac{c_\alpha}{B_\alpha} H(z) + \frac{2 c_\alpha (B_\alpha - B_\alpha')}{B_\alpha^2} G(z) + o(|z|^{-\alpha}) .
$$
This, of course, implies that $F(z) > 0$ when $|z|$ is large enough, as desired.
Note: Apparently it can be proved that the Fourier transform of $\phi$ is $O(|z|^{-3} \log |z|)$ as $|z| \to \infty$, and consequently the Fourier transform of $f(x)$ satisfies
$$
F(z) = \frac{c_\alpha}{B_\alpha} H(z) + \frac{2 c_\alpha (B_\alpha - B_\alpha')}{B_\alpha^2} G(z) + O(|z|^{-3} \log |z|)
$$
as $|z| \to \infty$. The proof of the above bound, however, more effort: it involves monotonicity of $\phi''(x)$ for $x > 0$ small enough, not just Hölder continuity of $\phi'(x)$. The idea is the same as in the proof of $F(z) = O(|z|^{-1-\alpha} \log|z|)$ result mentioned in your post, I suppose. It is not really needed in this answer, and I am short of time, so I will skip the details if you do not mind.
I do not see why "It follows that the Fourier transform of is (||−3log||) as ||→∞" you need an assumption on $F(z)$ , it has to be decreasing according to the Soni/Soni for Tauberian theorems, am i missing soemthing?
By the way, $f(x)\sim c_\alpha(1+x^{\alpha}\log(x))$. the sign is $+$
@TanyaVladi: Thanks for spotting the typo. Regarding your first comment, I edited the answer, hopefully it is more clear now.
I guess I still have that same concern, in order for the rates to work out, the assumptions on the functions need tp be checked. Tauberian theorems do not work for ANY functions. I know how to apply the theorem. I am not sure the assumptions on functions are met. The monotonicity for $\phi()$, for example is not there
@TanyaVladi: I'm afraid I do not understand your concerns: no Tauberian theorem is involved here.
" Tauberian theorems give the asymptotic behaviour of the original function based on properties of the transform but usually require some restrictions on the original function " ..in my notation as fas as I se it "original function " is a characteristic function as I care about tis asymptotic behaviour at $\infty$
big O notation does not give ma a sign as it refers to an absolute value. $\sim$ is a better choice. The characteristic function I am interested in not monotone, I need to show that it becomes positive and converges to $0$ from above, aka decreases, I am sorry if I was not clear. I can get a rate from differencing Fourier of $\exp(-|x|^{\alpha})$
@TanyaVladi: Doesn't the last display in the answer show that $F(z) \sim (c_\alpha/B_\alpha) H(z)$ as $|z| \to \infty$?
I will write a separate comment about it. It seems the framework Soni/Soni is confusing and the devil in the details. What they have is $F(x)=\int_{0}^{\infty}k(xt)f(t) dt$, where $F(x)\to 0$ as $x\to 0$ with some rate which is related to the rate of $f(t)\to 0$ as $t\to\infty$. Let $\psi(\cdot)$ be a density, symmetric around $0$. So it seems $F(x)=1-\psi(x)/\psi(0)$, $k(tx)=1-\cos(xt)$ and $f(t)=\phi(t)/\psi(0)$
Basically, under which assumptions the following is true? If $F(x)\sim G(x)$ as $x\to 0$ means their transforms $f(t)\sim g(t)$ as $ t\to\infty$
@TanyaVladi: This answer (as well as my last comment to the question) does not refer to any Tauberian theorem. In particular, it does not rely on the results of Soni–Soni.
well then why bother with Bessel functions? $\exp(-|x|^{\alpha})\sim 1-|x|^{\alpha}$, as $x\to 0$ and Fourier of $\exp(-|x|^{\alpha})$ $\sim {\Gamma{(\alpha+1)/2}\Gamma(1/\alpha)\over 2 t^{\alpha+1}}$, taking the derivative with respect to $\alpha$ with give the coefficient. I was not sure whether it is a legitimate operation. May be it is - I just cannot find any referees to support it
@TanyaVladi: Sorry, only now I noticed your last comment. Why bother with Bessel functions? Because everything can be made rigorous here. Since the question was bumped anyway, I edited the answer to provide further details. Does this work for you now?
thank you for getting back to me ."Thus, $\phi_1'$ is compactly supported and Hölder continuous with exponent $\beta$ for any $\beta < 1$. As a consequence, the Fourier transform of $\phi_1'$ is $O(|z|^{-\beta})$ as $|z| \to \infty$. It follows that the Fourier transform of $\phi_1$ is $O(|z|^{-1 - \beta})$ for any $\beta < 1$, and hence, in particular, it is $o(|z|^{-\alpha})$." I do not understand that part. Specifically, why the rate is $o(|z|^{-\alpha})$. This is the critical park of the proof and I was struggling with it when you posted your original proof. Thank you
@TanyaVladi: If we choose any $\beta \in (\alpha-1, 1)$, then $O(|z|^{-1-\beta})$ is $o(|z|^{-\alpha})$ (both as $|z| \to \infty$), right?
OK, I guess I do not understand, how you can choose $\beta $ and why $\phi_1^{\prime}$ is Hölder continuous with exponent $\beta $for any $\beta<1$"
sorry the whole trick with $\phi_1$ and $\phi_2$ is new to me, why did you ignore $f^{\prime}$ in your argument about $\phi_1^{\prime}$
@TanyaVladi: I am sorry, I do not quite understand your last two comments. I added again some details to my answer, but I am not sure if they address your concerns. Note, however, that I am no expert in asymptotic analysis, there are certainly better ways to write this argument.
Thank you for your details. It is more clear now. I have seen that trick with multiplying by $u$ before, here, on this site, I have not seen it in the textbooks or papers though. I really appreciate your attention to my problem.
|
2025-03-21T14:48:31.367317
| 2020-06-25T17:08:57 |
364129
|
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|
Stack Exchange
|
Simultaneous vanishing $\mathbb{Q}$-linear relations between $N$-th roots of unity
Let $\zeta$ be a primitive $N$-th root of unity and $\Gamma \subset (\mathbb{Z}/N)^\times$ a subgroup. Let $|\Gamma|$ be the cardinality of $\Gamma$ and consider the linear map $M_\Gamma\colon \mathbb{Q}^{N-1}\rightarrow \mathbb{Q}(\zeta)^{(N-1)\cdot |\Gamma|}$ given by the matrix
\begin{align*}
M_\Gamma = (\zeta^{nir}-\zeta^{ni})_{i=1,...,N-1 \atop (n,r)\in \{1,...,N-1\}\times \Gamma}.
\end{align*}
My question(s): What is the kernel of $M_\Gamma$? Is there any known result about this that I am missing?
Remark: There is Theorem 2.2 in the paper of Lam and Leung “On vanishing sums of roots of unity”, which gives a description of all $\mathbb{Z}$-linear relations among $N$-th roots of unity. However, I'm interested in those $\mathbb{Z}$-linear combinations given by the set of equations encoded in $M_\Gamma$ for a subgroup $\Gamma\subset (\mathbb{Z}/N\mathbb{Z})^\times$.
|
2025-03-21T14:48:31.367415
| 2020-06-25T17:32:29 |
364131
|
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"Connor Malin",
"https://mathoverflow.net/users/134512",
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"user168706"
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"sort": "votes",
"url": "https://mathoverflow.net/questions/364131"
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|
Stack Exchange
|
Groupoid completion of a topological category vs its homotopy category?
Given a category $\mathcal{C}$ enriched in spaces, we can take the nerve (a simplicial space) and then geometric realization to get a space $B\mathcal{C}$. If we view spaces as $\infty$-groupoids, then this process should be thought of as a ($\infty$-)groupoidification.
We can also consider the homotopy category $h\mathcal{C}$, which has the same objects as $\mathcal{C}$ but where the morphisms from x to y are given by $\pi_0 \mathcal{C}(x,y)$. This is an ordinary category and we can take the nerve and geometrically realize it to get the classifying space $Bh\mathcal{C}$.
In general the spaces $B\mathcal{C}$ and $Bh\mathcal{C}$ will be very different, but they might agree on some low dimensional homotopy groups.
Fix an object $x \in \mathcal{C}$. Is it true that $\pi_1(B\mathcal{C}, x)$ is isomorphic to $\pi_1( Bh\mathcal{C}, x)$? If not, what is a good counter example? Are there conditions under which these will be isomorphic? For example I am interested in the case where $\mathcal{C}$ is symmetric monoidal and $x$ is the unit object.
Note that we can view a set as a discrete topological space and so $h\mathcal{C}$ is also a (discrete) topological category. There is a functor $\mathcal{C} \to h\mathcal{C}$, and so there is a natural comparison map $\pi_1(B\mathcal{C}, x)\to\pi_1( Bh\mathcal{C}, x)$.
Alternative argument: The induced map $N_p(\mathcal{C}) \to N_p(h\mathcal{C})$ is $(2-p)$-connected for all $p$, which implies the map of (thick) geometric realizations is 2-connected (e.g. [arXiv:1403.2334, Prop 2.7]).
The $\pi_0$ and $\pi_1$ are the same. The former is obvious since, taking homotopy categories and groupoidifying do not affect connected components.
The fundamental group of an infinity category $S$ by van Kampen has a generator and relation description in terms of the 1 and 2 simplices. In particular, it has generators given by strings of 1-simplices and formal inverses that start and end at $*$ subject to the relation that we can exchange homotopic simplices, and that $ee^{-1}=e^{-1}e=Id$.
This group is the same as the group where we pick a single representative 1-simplex in each homotopy class and add in all the relations involving only these representatives.
Again by van Kampen, this group is exactly the fundamental group of the realization of $Ho(S)$, since we have just named each path component of the morphism space.
Then we simply transfer this back to topologically enriched categories and we are done.
This is the best one can hope for in general, since if $S$ is a Kan complex, its homotopy category is a groupoid and the realization of this is a 1-type (in particular the map you describe is the Postnikov approximation map).
This does generalize to the higher homotopy groups in the sense that the homotopy category is a model for the infinity category where you replace the mapping space with its n-type for $n=0$. I believe in this context, we have isomorphisms on homotopy groups up to $n+1$, since these should be possible to achieve just by attaching $(n+3)$-simplices.
|
2025-03-21T14:48:31.367670
| 2020-06-25T17:53:28 |
364133
|
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|
Stack Exchange
|
Which stable homotopy groups are represented by parallelizable manifolds?
The Pontryagin-Thom construction allows one to identify the stable homotopy groups of spheres with bordism classes of stably normally framed manifolds. A stable framing of the stable normal bundle induces a stable framing of the stable tangent bundle.
This means that a framed manifold (one whose tangent bundle is trivial, e.g. a Lie group) represents an element of the stable homotopy groups of spheres.
So some elements are represented by honestly framed manifolds (as opposed to stably framed).
What is known about such elements? Is every element of the stable homotopy groups of spheres represented by an honestly framed manifold (i.e. with a trivial tangent bundle)?
At least for Lie group framings there are some articles: https://core.ac.uk/download/pdf/82048719.pdf, https://projecteuclid.org/euclid.jmsj/1468956166
Am I confused or doesn't every tangential framing on $S^1$ give the trivial stable homotopy class?
No, if you take the Lie group framing you obtain the generator of $\Omega_1^{\rm{fr}}$. If you would like to obtain the trivial element in this framed bordism group, then you need to trivialize the stable tangent bundle accordingly. With other words: Not every stable framing is induced by a framing of the tangent bundle.
Suppose $S^1 \subset \mathbb R^n$ and suppose furthermore you have the Lie group framing on $S^1$ than you obtain a framing on the normal bundle of $S^1$. But this framing has to be in such way that if you compare it to the standard background framing of $\mathbb R^n$ you obtain a map $S^1 \to \mathrm{SO}(n)$ which has to be the trivial homotopy class (since you want to obtain the standard framing of $\mathbb R^n$). Pontryagin showed that the stable homotopy class is given by this homotopy class $S^1 \to \mathrm{SO}(n)$ in $\pi_1(\mathrm{SO}(n))$ plus the generator of that group
But if you specify a normal framing instead of a tangential framing than you obtain (using the standard backgroundframing) a stable framing of the tangent bundle which can lead to the trivial element. E.g. take the normal vector pointing outwards of $S^1 \subset \mathbb R^2$ and suspend it to a framing of the normal bundle of $S^1$ in $\mathbb R^n$. The homotopy class of the corresponding map $S^1 \to \mathrm{SO}(n)$ is a generator of $\pi_1(\mathrm{SO}(n))$.
Doesn't this show that the answer for the last question of Chris is negative? (Sorry for the spamming the comment section)
@PanagiotisKonstantis I don't think so. As you observe the Lie group framing on $S^1$ gives the non-trivial element of $\Omega^{fr}_1$. The disjoint union of two circles, each with the Lie group framing gives a tangentially framed manifold representing the trivial element of $\Omega^{fr}_1$. Also for the sake of this question the empty manifold should also probably be considered a tangentially framed manifold.
Okay, I somehow assumed falsely the manifold should be connected. Very interesting question!
Is $\sigma\in\pi_7^s$ represented by a unstably framed manifold? The $e$-invariant can be computed as (half of the) relative $\hat{A}$-genus of a bounding spin-manifold.
@archipelago Yes. The 7-sphere has a "Lie group" framing coming from viewing it as the unit octionians. It is not really a Lie group since it is not associative, but there is enough structure to talk about "left-invariant vector fields" and this induces a trivialization of the tangent bundle like for Lie groups. I believe that with this framing it represents $\sigma$.
All the useful comments were deleted. Maybe they were wrong, but at least they were trying. You have to take risks! Did Ryan's comment lead to the Kervaire invariant 1 problem? Is that what @archipelago was talking about?
For trivial reasons, it can in odd dimensions at most fail at the prime $2$ : The obstruction to destabilising a stable framing is $\mathbb{Z}/2$-valued.
@archipelago That is a nice idea. I think you are suggesting something along the following lines. Suppose that X is a stably framed manifold where the obstruction to destabilizing is non-trivial. Suppose for concreteness that [X] generates a $\mathbb{Z}/3\mathbb{Z}$ in $\pi^s_*$. Then the obstruction will vanish on 2[X], which also generates this group. But 2[X] is just two disjoint copies of X, so how could it all of a sudden become framable? Really the obstruction lives in a $\mathbb{Z}/2$ for each component of the manifold in question.
Your definition of a framed manifold is not standard. Surely the 2 sphere is framed, but its tangent bundle isn't trivial.
@NicholasKuhn It depends. In algebraic topology we usually (but not always) mean stably framed and shorten it to "framed". But, for example, in the cobordism hypothesis one talks of the framed bordism category and in this case one means unstable tangential framings. So it all depends on context. Perhaps you would prefer the term "parallelizable"? I will change the title.
I think all elements are representable by honestly framed manifolds.
Let $M$ be a closed $d$-manifold with a stable framing, and consider the obstructions to destabilising a stable framing. Asumng $M$ is connected, which we can arrange by stably-framed surgery, there is a single obstruction, lying in $H^d(M ; \pi_d(SO/SO(d)))$.
If $d$ is even then $\pi_d(SO/SO(d)) = \mathbb{Z}$ and this obstruction may be identified with half the Euler characteristic of $M$. (As $M$ is stably framed, its top Stiefel--Whitney class vanishes and so its Euler characteristic is even.) We can change $M$ to $M \# S^p \times S^{2n-p}$ by doing a trivial surgery in a ball, and the stable framing extends over the trace of such a surgery. By taking $p$ to be 1 or 2 we can therefore change the Euler characteristic by $\mp 2$: thus we can change $M$ by stably framed cobordism until its Euler characteristic is 0, whence the stable framing destabilises to an actual framing.
If $d$ is odd then then $\pi_d(SO/SO(d)) = \mathbb{Z}/2$ and the obstruction is obscure to me (it is realised by the stable framing induced by $S^d \subset \mathbb{R}^{d+1}$, and is non-trivial even in Hopf invariant 1 dimensions where $S^d$ does admit a framing). I can't see an elementary argument for $d$ odd, but I think it is nontheless true by the following.
Let $d=2n+1$ with $d \geq 7$ (lower dimensions can be handled manually). Consider the manifold
$$W_g^{2n} = \#g S^n \times S^n.$$
This has a stable framing by viewing it as the boundary of a handlebody in $\mathbb{R}^{2n+1}$. By doing some trivial stably-framed surgeries as above (with $p=2,3$ say, to keep it simply-connected), we can change it by a cobordism to a manifold $X$ having an honest framing $\xi$. I wish to apply [Corollary 1.8 of Galatius, Randal-Williams, ``Homological stability for moduli spaces of high dimensional manifolds. II"], to $(X, \xi)$. There is a map
$$B\mathrm{Diff}^{fr}(X, \xi) \to \Omega^{\infty+2n} \mathbf{S}$$
given by a parameterised Pontrjagin--Thom construction. Now there is a step that I would have to think about carefully, but I think that the choices made can be arranged so that $(X,\xi)$ has genus $g$ in the sense of that paper, and so taking $g$ large enough the map above is an isomorphism on first homology. But this has the following consequence: any element $x \in \pi_{2n+1}(\mathbf{S})$ is represented by the total space of a fibre bundle
$$X \to E^{2n+1} \overset{\pi}\to S^1$$
with a framing of the vertical tangent bundle (and the Lie framing of $S^1$).
(Again, I'm sure there must be a more elementary way of seeing this.)
It seems to me that the obstruction in odd dimension could be the semi-characteristic of the manifold. Bredon and Kosinski showed in (https://www.jstor.org/stable/1970531) that a stably framed manifold of dimension $n$ admits $n$ linear independent vector fields if only if the ($\mathbb Z_2$-valued) semi-characteristic vanishes.
Forgive me if I am just stating the obvious here, but wouldn't the GMTW-theorem for cobordisms with honest framings of the tangent bundle immediately show that all are represented by manifolds which have tangent bundle trivial after adding a single trivial bundle. Obviously not as strong as the result you prove, but at least a sanity check.
Ah, after reading @archipelago's answer I realize that since tangent bundles are the same rank as the dimension of the manifold, you are stably trivial if and only if adding a single trivial bundle makes you trivial.
Repeating the first part of Oscar's answer and elaborating on comments by Chris and Panagiotis, here is a down-to-earth argument in all cases:
The cases $n=1,3,7$ are fine, since the stable stems are in these degrees generated by $S^1$, $S^3$, $S^7$ with the unstable framing induced by the multiplication in the unit complex numbers, quaternions, or octonions.
In the other cases, we use that the obstruction to destabilising a given stable framing $F$ of an oriented closed manifold $M^n$ lies in $H^n(M,\pi_n(SO/SO(d))$, which is isomorphic (in a preferred way) to $\mathbb{Z}$ if $n$ is even and to $\mathbb{Z}/2$ if $n$ is odd. It is not too hard to see that, with respect to this isomorphism, the obstruction is given by the semi-characterstic: half the Euler characteristic for $n=2d$ and $\sum_{i=0}^d\mathrm{dim}(H_i(M,\mathbb{Z}/2))\text{ mod }(2)$ for $n=2d+1$ and $n\neq1,3,7$. In particular, the obstruction to destabilising is independent of $F$ which is somewhat surprising.
Originally this was proved by to Bredon and Kosinksi [1] who used a more geometric description of this obstruction: it is the degree (mod $2$ if $n$ is odd) of the Gauss map $M\rightarrow{S^n}$ induced by the stable framing $TM\oplus \varepsilon\cong \varepsilon^{n+1}$ (take the image of the canonical vector field in the trivial line bundle and normalize).
Now observe that, as Oscar explained, by doing a couple of trivial surgeries in a ball corresponding to taking connected sums with $S^1\times S^{n-1}$ or $S^2\times S^{n-2}$ and extending the stable framing, any stably framed bordism class in even dimensions contains a representative with trivial Euler-characteristic. The same works with the semi-characteristic in odd dimensions (here at most one surgery is necessary), so by the discussion above every stably framed bordism class has a representative whose stable framing can be destabilised.
[1] G.E. Bredon and A. Kosinski, Vector fields on $\pi$-manifolds. Annals of Math. 84, 85– 90 (1960).
Could you explain why your last paragraph is true?
Oscar explained it in his answer in the even-dimensional case and the same works in odd dimensions. I added it to the answer.
Thanks for this summary and for an easier argument in the odd dimensional case.
$k \cdot[\mathrm{point}]\in \pi_0^s$ is represented by an honestly framed 0-manifold if and only if $k \geq 0$.
In terms of homotopy theory, the questions is that when an element of $\pi_n^s$ pulls back to an element of $\pi_{n+i}S^i$ for some $i$ and the answer to this question is positive by Freudenthal's suspension theorem as there is an epimorphism
$$\pi_{2n+1}S^{n+1}\longrightarrow \pi_n^s$$
unless you put more restrictions, e.g. ask for some specific $i$. The point is that there could be two different pull backs whose normal bundles are not isomorphic, only stably isomorphic. An example is given by $\eta_3\in\pi_8^3$ which equals to $\eta\sigma=\sigma\eta$ in $\pi_*^s$. However, one can do some unstable computations to show that one pulls back one step further than the other. Hence, as unstable elements they are not the same really, but map to the same element. I think it could be interesting to sort this out in terms of bordism theory and I don't know if such specific examples are considered somewhere in the literature.
ADDED I think the answer still is positive. I think manifolds with tangential structures are understood in terms of Madsen-Tillmann spectra using the Madsen-Tillmann-Weiss map; experts can comment more on this and correct me if this is wrong or vague. In the case of trivialisation of the tangent bundle of $m$ dimensional manifolds, the relative spectrum is $\mathbb{S}^{-m}=\Sigma^{-m}S^0$. The general result of Galatius-Madsen-Tillmann-Weiss provides an interpretation of $\pi_i\Omega^\infty\mathbb{S}^{-m}$ in terms of specific submersions (I guess). Now, the point is that $\pi_i^s\simeq\pi_{i-m}\mathbb{S}^{-m}$ for any $m>0$ and I think again using Freudenthal's theorem one can see the answer is positive.
No, the question is whether the class is represented by a manifold with trivial tangent bundle, not trivial normal bundle.
Your edit does not answer the question: roughly speaking, MT-spectra encode bordism groups of manifolds equipped with a vector bundle with some tangential structure that is stably equivalent to the tangent bundle, not tangential structures on the tangent bundle itself.
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2025-03-21T14:48:31.368490
| 2020-06-25T18:01:28 |
364134
|
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|
Stack Exchange
|
On reflection properties of convex regions
It is well known that any ray of light passing thru a focus of an ellipse will pass thru the other focus after a single reflection from the ellipse boundary. If $A$ and $B$ are the foci of an ellipse, this property of rays holds both ways (those passing thru $A$ meet at $B$ and vice versa).
Is there a closed convex region $C$ with the property: there exists a pair of points $A$ and $B$ within $C$ such that all rays thru $A$ will reflect once on $C$ and pass thru $B$ but not all rays thru $B$ will pass thru $A$ after one reflection from $C$?
Is there a closed convex region $C$ such that: there is a pair of points $A$ and $B$ in the interior such that all rays thru $A$ pass thru $B$ after exactly 2 reflections from $C$? It is sufficient for the ray going directly from $A$ to $B$ to get reflected exactly twice from somewhere on $C$ and then pass thru $B$.
Note 1: Question 2 can have 'one-way' (convergence only of rays thru $A$ at $B$) and 'both-ways' variants.
Note 2: If not explicit constructions, even existence/non-existence arguments could be sought as answers to these questions.
One can also ask if relaxing convexity has any implications.
you need $C$ to be smooth, right?
The guess is that a candidate C needs to be smooth for it to have a chance of achieving the above properties.
but we cannot even define reflection at the non-smooth points
I understand. Yes. The ellipse is smooth. So, one is looking for other curves in the same spirit.
I will assume your regions are closed and have differentiable boundary, otherwsie reflections are not defined for certain directions.
Question 1: a no-go result.
The answer to Question 1 is no: the reflection property you are interested in is symmetric in $A$ and $B$.
For every ray $v$ originating at $A$, write $f_A(v)$ for the (unique, by convexity) point where $v$ hits the boundary of the convex region. Because of convexity, the map $f$ is a bijection between the unit circle and the boundary of the region. If $v$ is now a ray originating at $B$, similarly define $f_B(v)$ to be the unique point of the boundary hit by a ray from $B$ in direction $v$.
If $x$ is a point on the boundary, write $n_x$ for the normal to the boundary at that point.
By hypothesis, every ray through $A$ reflects through $B$ after hitting the boundary once. This is the same as saying that for every $x$ on the boundary the directions $f_A^{-1}(x)$ and $f_B^{-1}(x)$ form the same angle to $n_x$. The statement is symmetric in $A$ and $B$, so your condition holds for all reflections from $A$ to $B$ if and only if it holds for all reflections from $B$ to $A$.
Question 2: a no-go result.
The answer to Question 2 is also no: at some point the two reflections must collapse to one.
Indeed, consider any convex region and let $f_A(v)$, $f_B(v)$ and $n_x$ as before. Because the boundary is closed and differentiable and the region is convex, there must exist an $x$ on the boundary such that $n_x = -f_A^{-1}(x)$, i.e. a ray from $A$ towards $x$ reflects back through $A$. Either $B$ is on the way of the reflection, in which case we're done, or the ray must pass through $A$, hit the boundary again at another point $y$ and from there reflect through $B$. But then the ray from $A$ in direction $f_A^{-1}(y)$ reflects at $y$ and then goes through $B$ after a single reflection.
Question 2: relaxing assumptions.
The answer to Question 2 can become yes if some assumptions are relaxed. In the example(s) below, we must allow a finite number of directions in which the second focus is hit only after a single reflection (because of the the no-go result above).
Cofocal parabolas
For this first example we have to allow the points A and B to coincide (must as the circle is a degenerate case of an ellipse where the foci coincide). Then the answer is yes: joining two parabolic arcs with the same focus and axis of symmetry (but opposite vertices wrt the focus) always yields a convex shape with the desired property. Indeed, every ray from the focus (point A) will reflect exactly twice before reaching the focus again (point B), with the exception of two rays parallel to the axis of symmetry of the parabolas. Note that the two rays orthogonal to the axis of symmetry of the parabolas hit the boundary in a non-differentiable point, but the reflection is nevertheless well-defined by continuation.
Half-ellipse and two half-circles
For this second example we have to allow non-convex shapes and we have to relax the requirement of two reflections: let's allow the ray from A to travel through B after the first reflection, as long as it always goes through B after the second reflection.
Consider two distinct points A and B in the plane and let $a > 0$ be their distance. Without loss of generality, let the points be at $(\pm a/2, 0)$ in the Cartesian plane. Draw the positive y-coordinate half of an ellipse with foci A and B and having semimajor axis $a$ (so that the semiminor axis $b$ satisfies $\frac{a}{2} = \sqrt{a^2-b^2}$, i.e. $b = \frac{\sqrt{3}a}{2}$):
$$
y = \frac{\sqrt{3}}{2}\sqrt{a^2-x^2}
$$
Draw the negative y-coordinate half of a circle around $A$ with diameter $a$:
$$
y = -\sqrt{\frac{a^2}{4}-\left(x+\frac{a}{2}\right)^2}
$$
Draw the negative y-coordinate half of a circle around $B$ with diameter $a$:
$$
y = -\sqrt{\frac{a^2}{4}-\left(x-\frac{a}{2}\right)^2}
$$
A ray from A with positive y direction will first reflect against the ellipse boundary, then reflect against the circle boundary below B and finally go through B (but it will have passed through B on its way from the ellipse boundary to the circle boundary).
A ray from A with negative y direction will first reflect against the circle boundary below A, then reflect against the ellipse boundary and finally go through B.
Bonus: additional "reflections" of a physical nature.
The answer to Question 1 is related to a property of classical geometric optics sometimes known as reciprocity: the path described by light is independent of the direction in which light travels through the path. Specifically, if a ray starting at $A$ in direction $v$ follows a path which ends at $B$ in direction $w$, then light starting at $B$ in direction $-w$ follows the same path (travelling in the opposite direction) and arrives at $A$ in direction $-v$.
It may interest you to know that this property fails in curved spacetimes: a ray starting at $A$ in direction $v$ may follow a path which ends at $B$ in direction $w$, but light starting at $B$ in direction $-w$ might not even reach $A$ at all. As a pictoresque (and extreme) example of this, consider light falling radially into a Schwarzschild black hole: rays starting a point $A$ outside directed radially towards a point $B$ inside will reach $B$, but rays starting from $B$ inside directed radially towards the point $A$ outside will never reach $A$.
Hi Stefano, I like your argument for question 2) but the criteria does not strictly exclude the ray passing through the other focus after one reflection as long as it hits again after 2 reflections. I wonder, if we allow this, is there an argument that still works? As you can see I managed to find a curve that works for only 2 rays excluded for each focus.
Indeed, as your (clever!) example shows a surface can be constructed with the required property as long as some singular directions are allowed where the two reflections collapse to one. Also, your remark made me notice that the question does not implicitly require all reflection directions to be defined, but only those from the two points, which makes one of my assumptions (differentiability) perhaps too strong.
Thanks Stefano, note that I actually have to exclude some rays but an arbitrarily small percentage from each focus. I wonder therefore if there is an argument that shows that you have to exclude at least some measurable set of rays from any convex example?
I see: if you could take the limit you'd get to a zero fraction, but for every finite shape you have a non-zero fraction of the rays that don't actually end up reflecting through B.
I added a comment to your answer to this effect: it would be interesting to see if such a deformation is possible. After all, my proof only excludes finitely many points, so it doesn't rule out such a possibility.
Thanks to both Ivan Meir and Stefano Gogioso for the nice and illuminating answers. I understand the only remaining bit is to confirm this statement made in the above discussion reg question 2: "for every finite shape you have a non-zero fraction of the rays (not just a few) thru A that don't actually end up passing through B (after 2 hits on C)"
@NandakumarR Yes it needs to be proved that any convex shape will have a set of rays from A of non-zero measure that do not pass through B after 2 reflections.
I have edited my answer to provide examples of how the question could be relaxed to allow for interesting solutions.
The answer to question 1) is no:
Take any ray R passing through B. Since C is convex and B lies in the interior of C, R will intersect C at a point O. Now the ray $\overrightarrow {AO}$ lies in the interior of C and its reflected ray passes through B. But this means we have a path of light $\overrightarrow {AO},\overrightarrow {OB}$. Since C is smooth the tangent at O is unique. Hence we can reverse this to give a light path $\overrightarrow {BO},\overrightarrow {OA}$ and hence the ray R from B will reflect off C at O and pass through A.
Since R was arbitrary, any ray passing through B will reflect off C and pass through A.
For question 2) a good example is to take two parabolas facing each other with a common axis, where the parabolas intersect at points A and A':
Rays through the focus of either parabola, not passing through A or A' or hitting the other parabola first will be reflected twice and then pass through the other focus. If we move the foci apart we can get an arbitrarily large percentage of directions to work.
Thus the answer to 2) is yes if we allow an arbitrarily small percentage of rays from each focus to be excluded. This is the 2-way version.
Note that we can prove that such any example has to be 2-way in the same way as we did for question 1).
I wonder whether there is a way to take the limit of this construction where the two parabolas are taken to infinity and then apply some transformation to the space which brings the shape back to a finite but preserving the horizontal directions followed by the rays between the two reflections. The resulting shape would still have two directions where a single, degenerate reflection happens, at the two points where the parabolas join, but those would be the only places where the two-reflection property fails.
@StefanoGogioso I don't think this can be done unfortunately as any time you have such a parallel set of rays you have a parabolic curve and hence the curve can never become parallel to the parabola axis. This axis must join the two points A and B hence you could never achieve a complete curve without a point where the tangent is discontinuous.
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2025-03-21T14:48:31.369339
| 2020-06-25T18:09:20 |
364135
|
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|
Stack Exchange
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Definition of signs of isomorphisms $c_u : o(x_1) \to o(x_0)$ in the definition of Floer cohomology via Seidel's book
I'm reading Paul Seidel's book "Fukaya Categories and Picard-Lefschetz Theory", chapter 12, and I'm currently trying to understand the differential on Floer cohomology in terms of orientation lines $o_y$ for $y \in \mathcal{C}(L_0, L_1)$, in Seidel's notation. $L_0$ and $L_1$ are exact Lagrangians in $(M, \omega = d\theta)$ equipped with brane structures, and $\mathcal{C}(L_0, L_1)$ are time-1 Hamiltonian chords which start at $L_0$ and end at $L_1$. $o_y$ is a certain one-dimensional real vector space associated to $y$, which can either be seen as a tensor product of certain one-dimensional spaces obtained as intersections of Lagrangian subspaces at $y(1)$ and their duals, or as the determinant line of a certain Fredholm operator on the upper half-plane (the precise definitions are in equations $(11.20)$ and $(11.25)$ in the book, as well as lemma $11.11$).
For a field $\mathbb{K}$, $|o_y|$ is defined by taking the direct sum of the rank-1 vector spaces generated by the two orientations of $o_y$, and then setting their sum to be $0$. The Floer cochain space is defined as $$CF^* (L_0, L_1) = \bigoplus_{y} |o_y|_{\mathbb{K}}.$$ The grading on each Lagrangian induces a grading on $CF^*(L_0, L_1)$.
Now suppose I have $y_0$, $y_1 \in \mathcal{C}(L_0, L_1)$ with $y_0$ of degree $1$ greater than $y_1$. From what I understood, Proposition $11.13$ implies that there is a canonical isomorphism $$\lambda^{top}(T_u\mathcal{M}(y_0, y_1)) \otimes o_{y_1} \cong o_{y_0},$$
where by $\mathcal{M}(y_0, y_1)$ I mean the "unquotiented" moduli space (although it can easily be replaced by the quotiented one). If $u$ is a regular point, then the first term on the left vanishes and we obtain an isomorphism $$c_u: o_{y_1} \to o_{y_0}.$$ This induces an isomorphism of $\mathbb{K}$-normalizations $$|c_u|_{\mathbb{K}}: |o_{y_1}|_{\mathbb{K}} \to |o_{y_0}|_{\mathbb{K}}.$$ Now when defining the differential, $u$ contributes a factor of $\pm 1$ depending on the "sign" of $c_u$.
My question: what is the sign of $c_u$? If there were a canonical orientation of the orientation lines $o_y$ for each $y$ I would understand this, but I was not able to find something like that.
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2025-03-21T14:48:31.369528
| 2020-06-25T19:04:08 |
364141
|
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|
Stack Exchange
|
A best constant in an integral inequality
Assume that $f$ is a real continuous function on $[0,1]$.
I need the smallest constant $C$ in the following inequality
$$4\int_0^1\left(\int_0^\rho r f(r)dr\right)^2\frac{d\rho}{\rho}+\left(\int_0^1 r f(r)dr\right)^2\le C \int_0^1 r f^2(r)dr.$$
Is there some context for where this comes form? Do you need the precise best constant and do you have reason to suppose it will have a nice closed form?
It has to do with the norm of some operator, but complicated to explain...
$C=1.0986702957852257$ is the solution. It can be proved by using the Langrange multipliers.
If you are satisfied with this answer, then you should accept it.
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2025-03-21T14:48:31.369617
| 2020-06-25T19:04:09 |
364142
|
{
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"Ali Taghavi",
"B K",
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}
|
Stack Exchange
|
When is the exterior derivation $d$ a Lie algebra morphism?
In this question we search for some conditions under which the exterior derivation $d:\Omega^i(M)\to \Omega^{i+1}(M)$ on a differentiable manifold $M$ is a Lie algebra morphism in a certain sense. We consider $2$ different cases:
For our first question we consider $(M,\omega)$ a symplectic manifold. Then $\Omega^0(M)$ has a natural Lie algebra structure via Poisson bracket. On the other hand for every Riemannian metric on $M$ we get a Lie algebra structure on $\Omega^1(M)$ since the metric gives us a linear isomorphism between $\Omega^1(M)$ and $\chi^{\infty}(M)$, the Lie algebra of smooth vector fields on $M$. In the simplest case, $M=\mathbb{R}^2$ with its standard symplectic and Riemannian structure, we observe that the differential operator $d:\Omega^0(M) \to \Omega^1(M)$ does not preserve the corresponding Lie brackets. This motivates us to ask the following question:
Question 1:
Let $(M,\omega)$ be a symplectic manifold. Does there exist a Riemannian metric on $M$ such that $d:\Omega^0(M) \to \Omega^1(M)$ is a Lie algebra morphism?
In our next question we search for possible Lie algebra structures on higher order differential forms $\Omega^i(M)$, $i>1$, of a Riemannian manifold such that the exterior derivation $d$ would be a Lie algebra morphism for all dimensions $i$. More precisely:
Question 2:
Let $(M,g)$ be a Riemannian manifold. Can we equip each $\Omega^i(M)$ with a Lie algebra structure such that $\forall i>0$, $d:\Omega^i(M)\to \Omega^{i+1}(M)$ preserves the corresponding Lie brackets?
On Question 1: The Poisson bracket (or its negative, depending on the sign convention) has the property that $d:\Omega^0(M)\to \Omega^1(M)$ becomes a Lie algebra morphism if $\Omega^1(M)$ is identified with the space of smooth vector fields $\chi^\infty(M)$ using the symplectic form. No need for a Riemannian metric here. I even doubt that there can exist one as in the question, as such a metric would probably have to induce the same isomorphism from $\Omega^1(M)$ to $\chi^\infty(M)$ as the symplectic form, which would make it an antisymmetric tensor, a contradiction.
I feel like it's more natural to use the Schouten–Nijenhuis bracket on higher degree forms (along with the symplectic form vector-covector identification), rather than a plain Lie bracket.
@BK Thank you very much for your interesting comment. Could you please more explain about its last part?(About the contradiction you pointed out)
@PaulReynolds Thank you very much for informing me of this graded bracket.
On question 1, to expand on what @BK said: If you have a symplectic structure $\omega$ on a manifold $M$, you get a natural Lie bracket on $\Omega^1(M)$ by the following rule:
$$ [\alpha, \beta ] = \omega^\flat([\omega^\sharp (\alpha), \omega^\sharp(\beta)]) $$
Where:
$$ \omega^\sharp \colon \Omega^1(M) \to \mathfrak{X}(M) \quad \omega^\sharp(\alpha) = X \ \Leftrightarrow \iota_X \omega = \alpha $$
$$ \omega^\flat \colon \mathfrak{X}(M) \to \Omega^1(M) \quad \omega^\flat(X) = \alpha \ \Leftrightarrow \iota_X \omega = \alpha $$
If you have a metric $ g $ on $M$ then you can define similar sharp and flat maps by pairing with the metric tensor instead. The problem is that the sharp and flat maps completely characterize the tensor. Therefore if $g$ induces the same isomorphism between $\Omega^1(M)$ and $\mathfrak{X}(M)$ as $\omega$ it follows that they are equal. The problem with this is that a metric cannot ever be equal to $\omega$.
In fact, symplectic structures are much more closely related to Lie algebra structures than metrics. So I think if you're trying to construct some Lie theoretic object, my recommendation is that you look more into the symplectic universe.
That said, regarding Question 2: There is no canonical way to make the $\Omega^i(M)$ into Lie algebras by using a symplectic structure. Is it possible? Sure, why not? By picking a basis for each vector space and splitting them along the images, kernels and cokernels of the differentials you can construct a variety of split Lie algebra structures on the resulting infinite dimensional vector spaces. But there is nothing geometrically interesting about this.
If you want a more natural Lie-theoretic structure that has some real geometric meaning, you can continue the suggestion of @PaulReynolds to look at graded brackets. I'll refer you to wikipedia for the definition of the Schouten-Nijenhuis bracket:
https://en.wikipedia.org/wiki/Schouten%E2%80%93Nijenhuis_bracket
Since the symplectic structure on $M $ produces a bunch of isomorphisms $ \omega^\flat \colon \Omega^i(M) \to \wedge^i \mathfrak{X}(M) $, you can transport the Schouten-Nijenhuis bracket to the complex of differential forms.
A closely related, but different option is to weaken the symplectic structure to a Poisson structure. Basically, this is just a Lie bracket:
$$ \{ \cdot , \cdot \} \colon C^\infty(M) \times C^\infty(M) \to C^\infty(M) $$
which satisfies:
$$ \{ f, gh \} = g \{ f, h \} + h \{ f, g \} $$
By using this bracket, you can actually construct a Lie bracket on $\Omega^1(M)$. By using the exact same formulas for the Schouten-Nijehuis bracket, except using 1-forms instead, you can get a graded bracket on the whole complex of forms. Depending on the Poisson bracket you started with, these carry a lot of geometric meaning regarding symplectic foliations and other cool stuff.
To be clear, these structures I just described do not make the differential into a Lie algebra homomorphism. Rather, I believe you get that the differential is a derivation of the graded bracket. That is:
$$ \forall \alpha \in \Omega^i(M) , \ \beta \in \Omega^j(M) \quad d[\alpha,\beta] = [d \alpha , \beta] + (-1)^{i}[\alpha , d \beta] $$
Maybe try looking at Gerstenhaber algebras and BV-algebras in Poisson
geometry by Ping Xu for some more advanced reading on the topic.
|
2025-03-21T14:48:31.370023
| 2020-06-25T19:56:01 |
364146
|
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"Anthony Quas",
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|
Stack Exchange
|
Confusion on the assumption when discussing the kneading invariants for unimodal maps
A unimodal map is a continuous map $f:[0,1]\longrightarrow [0,1]$ such that there is only one turning point (critical point), denoted by $c$, and $f(0)=f(1)=0$.
Unimodal map is related to kneading invariant and kneading map, constructed by Hofbauer and Keller.
One assumption is always being made by several books in dynamics. They all assume $$f^{2}(c)<c<f(c),$$ where $f^{n}(x)$ is the $n-$times iteration of $f$ at $x$.
They all use an one-line explanation:
If this condition does not hold, then $f$ has no interesting dynamics.
I have drew several pictures but I cannot see why. Even though we have $c<f^{2}(c)<f(c)$, the dynamics is still complicated, since I cannot see a clear rule about where $f^{n}(c)$ will go for $n\geq 3$.
So what does it mean by "interesting" here? and why does this assumption make the dynamics "interesting"?
Thank you!
In this case, the interval $I=[c,f(c)]$ is mapped into itself and the restriction of $f$ to $I$ is decreasing. This means that every orbit in $I$ converges either to a fixed point or a period two orbit. For points in $[0,c]$, the orbit is either monotonic (and so converges to a fixed point), or it eventually leaves $[0,c]$, whereupon it enters $I$. For points in $[f(c),1]$, the image is in $[0,f^2(c)]\subset [0,c]\cup I$, so that we have shown that every orbit approaches a fixed point or period 2 orbit. Pretty dull...
@AnthonyQuas brilliant! Post an answer so that I can accept and vote?
@AnthonyQuas could you please elaborate a little bit why every orbit in $I$ converges either to a fixed point or a period two orbit?
The point is $f$ is a decreasing map from $I$ to $I$, so that $f^2$ is an increasing map from $I$ to $I$. It's pretty easy to see that if you have an increasing map, everything converges to a fixed point. Probably math.stackexchange.com would be a better place if you want to ask about that.
|
2025-03-21T14:48:31.370185
| 2020-06-25T20:24:12 |
364148
|
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"Carlo Beenakker",
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"sort": "votes",
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|
Stack Exchange
|
Simultaneous diagonalization in Matlab
Crossposted from StackOverflow. The generalised diagonalization of two matrices $A$ and $B$ can be done in Matlab via
[V,D] = eig(A,B);
where the columns of $V$ are are the generalised eigenvectors of the pair ($A$, $B$), while $D$ is a diagonal matrix containing the corresponding generalised eigenvalues.
First, is it correct to assume that the generalised eigenvectors $V$ are also simultaneous eigenvectors of both $A$ and $B$, under the assumption that $A$ and $B$ commute?
If so, I could compute the respective eigenvalues of $A$ and $B$ (in the same order as the eigenvectors in $V$) as follows
eigA = V\A*V;
eigB = V\B*V;
However, is there a simpler way to achieve this, i.e. without having to compute the inverse of $V$ as in the snippet above?
if you know an eigenvector $v$ of $A$, can't you just find the corresponding eigenvalue by calculating $w=Av$ and comparing the first component of $w$ and $v$?
Thank you. This is OK except that $v(1)$ might be zero or close to zero, which makes the computation unstable. One might fix that by comparing the whole $v$ and $w$.
I think your premise is wrong; [V,D] = diag(A, B) does not, in general, simultaneously diagonalize matrices. It works only in sufficiently simple cases without repeated eigenvalues (try rng(0); V = randn(5); A = V*blkdiag(1,1,3,3,3)/V; B = V * blkdiag(1,1,1,3,3)/V;). As a workaround with possibly bad stability, try [V, D] = eig(randn*A + randn*B).
@FedericoPoloni Thank you very much, excellent counterexample. In my case, $A$ and $B$ were symmetric, positive definite, tridiagonal Toeplitz matrices, so the generalised diagonalisation $AV = BVD$ happened to be a simultaneous diagonalization as well: $AV = VD_1$ and $BV = VD_2$.
@MassimoFrittelli Uh, that looks like an oddly specific problem, since in that case $B=\alpha A + \beta I$ for suitable $\alpha$ and $\beta$. Then you know that there is a closed-form solution to that problem? https://en.wikipedia.org/wiki/Tridiagonal_matrix#Eigenvalues ? The eigenvectors have a closed form as well.
@FedericoPoloni Thank you, this is a life changer. Yes, as you noted, my eigenvector problem oddly recalls discrete Laplacians...
Happy to be of help @MassimoFrittelli !
From comments: OP mentions that his matrices are Toeplitz tridiagonal; in particular, they can all be written as $\alpha L + \beta I$, where $L$ is the finite-difference matrix tridiag(-1,2,-1).
So the problem has a closed-form solution: the eigenvalues of $L$ are given by $\lambda_k = 2(1-\cos \frac{k\pi}{n+1})$, and its eigenvectors are $(v_k)_j = \sin \frac{jk\pi}{n+1}$. (From https://en.wikipedia.org/wiki/Tridiagonal_matrix#Eigenvalues and https://en.wikipedia.org/wiki/Eigenvalues_and_eigenvectors_of_the_second_derivative#Pure_Dirichlet_boundary_conditions_2 ).
Thank you. A remarkable consequence of your answer is that all symmetric Toeplitz triadiagonal matrices of given dimension share the same eigenvectors. After all, the set of such matrices is a vector space of dimension just two.
is there a similar result for the case of higher order discrete Laplacians? I am happy to open a separate question if that is the case.
|
2025-03-21T14:48:31.370430
| 2020-06-25T21:18:30 |
364151
|
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"Andrew",
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|
Stack Exchange
|
Existence of classical solution for a parabolic equation without Hölder continuity in time for its coefficients
Consider equation
$$\partial_t u = \partial_x u + \partial_{xx} u - c u + f, \hbox{ on } (t, x) \in (0, \infty) \times \mathbb R$$
with initial condition $u(0, x) = g(x).$
Suppose that $c(t, x)$ and $f(t,x)$ are continuous in $(t, x)$
and $\phi (\cdot) = c(t, \cdot), f(t, \cdot), g(\cdot)$ satisfy
$$|\phi|_0 + |\partial_x \phi|_0 + |\partial_{xx} \phi|_0 <K$$
for some $K>0$.
[Question.] Is there unique classical solution for the equation with the above conditions?
Remark: I have seen that some conditions for the existence requires at least Holder continuity in $t$ for $c$ and $f$. I want to know if it is still true by dropping Holder $t$-continuity?
Solution is not unique if not to require some growth bounds for solutions at infinity. https://mathoverflow.net/questions/89120/maximum-principle-for-heat-equation-on-infinite-domain/89123#89123
Yes it's true if to drop Holder $t$-continuity.
@Andrew Yes, true. I concentrated on existence which is where regularity enters.
In your case the domain of the operator is $UCB^2(R)$ (uniformly continuous and bounded functions up to the second derivative) and $f$ is continuous with values in the domain of the operator. Semigroup theory yields that the mild solution is a classical one, that one you are looking for.
In the case of $D^2$, you can also write down the explicit formula for the solution $u$ and check that $u_{xx}$ exists, by differentiating $f$ under the integral. Then one needs an argument for $u_t$...this is the point where I prefer semigroup theory. Another possibility is to approximate $f$ with $f_n$, better in $t$, consider $u_n$ the corresponding solutions and let $n \to \infty$. Then $u_n \to u$, $(u_{n)_{xx}} \to u_{xx}$, using the fundamental solution, and then, by difference, $(u_n)_t$ also converges.
It is helpful, I will look for details as you suggested.
The first attempt with method of continuity does not go through. This is second attempt based on the hints of other replies. The answer seems to me Yes if the state domain is changed to 1-torus $\mathbb T$ from $\mathbb R$.
[Conclsion] There exists $u\in C^{1,2}([0, T]\times \mathbb T)$.
Let
$$F(c, u) = F[c] u = \partial_t u - \partial_{xx} u - \partial_x u + cu.$$
We use the following claims. (C1) and (C2) are standard, and (C3) will be proved later.
(C1) The operator $F: C^{0,0} \times C^{1,2} \mapsto C^{0,0}.$ is a continuous mapping
(C2) If $c, f$ are Holder continuous, then exists unique $u\in C^{1,2}$, s.t. $F(c, u) = f$, i.e. $u = F^{-1}[c] f$ is well defined.
(C3)For $c_i, f_i \in C^{1,3}$ satisfying $|c_i|_{0,2} + |f_i|_{0,2} < K$ with $i = 1, 2$, we have
$$|F^{-1}[c_1] f_1 - F^{-1} [c_2] f_2 |_{1,2} \le\Psi(K, T)( |f_1 - f_2|_{0,2} + |c_1 - c_2|_{0,2})$$
for some strictly increasing continuous function $\Psi(K, T)$ with $\Psi(0, T) = \Psi(K, 0) = 0.$
(C4) if $\phi \in C^{0,2}$, then there exists $\phi_n \in C^{1,3}$ s.t.
$|\phi_n - \phi|_{0,2} \to 0$. This is the place why the domain is changed to $\mathbb T$. Indeed, one can use polynomial $p_n$ approximate $\partial_{xx} \phi$ by stone-weirstauss and take integral twice to get $\phi_n$.
Now, let $c, f\in C^{0,2}$ be given. Then, since $C^{1,3}$ is dense in $C^{0,2}$, there exists $c_n, f_n \in C^{1,2}$, s.t.
$$|c_n -c|_{0,2} + |f_n -f|_{0,2} \to 0$$
and
$$|c_n|_0 + |f_n|_0 \le 2(|c|_0 + |f|_0) := K.$$
We denote $u_n = F^{-1} [c_n] f_n$. Then,
$u_n$ is Cauchy in $C^{1,2}$ since by (C3)
$$|u_n - u_m|_{1,2} \le \Psi(K, T) (|f_n - f_m|_{0,2} + |c_n - c_m|_{0,2}).$$
So there exists $u\in C^{1,2}$ s.t. $|u_n - u|_{1,2} \to 0$. Now we can verify $u$ is the solution by checking
$$F(c, u) = \lim_n F (c_n, u_n) = \lim_n f_n = f.$$
In the above, we used (C1).
The remaining part is the proof of (C3). For $c_i, f_i \in C^{1,3}$ satisfying $|c_i|_{0,2} + |f_i|_{0,2}< K$ with $i = 1, 2$, $u_n = F^{-1} [c_n] f_n$ for $n=1 , 2$ is a classical solution by (C2) and $v_n (t, x) = u_n(T - t, x)$ has probability representation of the form
$$v_n(t, x) =\mathbb E \Big[ \int_t^{T} \exp\{- \int_t^{s} c_n(r, X^{t,x}(r)) dr\} f_n(s, X^{t,x}(s) )ds\Big] $$
where
$$X^{t, x} (s)= x + (t-s) + W(s) -W(t).$$
By direct approximation, one can have
$$|v_1 - v_2|_0 \le KT^2 e^{KT}(|f_1 - f_2|_0 + |c_1 - c_2|_0) .$$
This also holds for
\begin{equation}
\label{eq:01}
|u_1 - u_2|_0 \le KT^2 e^{KT}(|f_1 - f_2|_0 + |c_1 - c_2|_0) := \Psi(K, T)(|f_1 - f_2|_0 + |c_1 - c_2|_0) .
\end{equation}
Next, we can check that, by (C2)
$\bar u_n = \partial_x u_n$ is the classical solution of
$$\partial_t \bar u_n = \partial_x \bar u_n + \partial_{xx} \bar u_n - c_n\bar u_n - \partial_x c_n \cdot u_n + \partial_x f_n$$
with $\bar u_n(0, x) = 0.$ Similarly, we have
$$|\partial_x (u_1 - u_2)|_0 \le \Psi(K, T)(|(- u_1\partial_x c_1 + \partial_x f_1) - (- u_2\partial_x c_2 + \partial_x f_2)|_0 + |c_1 - c_2|_0).$$
Combined with the earlier estimation on $|u_1 - u_2|_0$, we have
$$|u_1 - u_2|_{0, 1} \le \Psi(K, T) (|f_1 - f_2|_{0, 1} + |c_1 - c_2|_{0,1}).$$
Using exactly the same approach, we have
$$|u_1 - u_2|_{0, 2} \le \Psi(K, T) (|f_1 - f_2|_{0, 2} + |c_1 - c_2|_{0,2}).$$
Together with original equation $\partial_t u = ...$, we have final estimation
$$|u_1 - u_2|_{1, 2} \le \Psi(K, T) (|f_1 - f_2|_{0, 2} + |c_1 - c_2|_{0,2}).$$
This completes the proof of (C1).
I am not convinced since $L_0$ maps $C^{1,3}$ to $C^{0,1}$.
agreed, my proof is wrong.
|
2025-03-21T14:48:31.371139
| 2020-06-25T22:10:54 |
364154
|
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|
Stack Exchange
|
Definition of Iwahori subgroup independently of the Bruhat-Tits building
Let $G$ be the points of a connected, semisimple algebraic group over a $p$-adic field $k$. To make life easy, let's assume the underlying group scheme is simply connected. The Bruhat-Tits building $\mathscr B$ of $G$ can be constructed from a valued root datum of $G$. An Iwahori subgroup $B$ of $G$ is defined to be a stabilizer of a chamber in $\mathscr B$.
I know that if the Iwahori subgroup $B$ of $G$ is already given, as well as the normalizer $N$ of a maximal split torus of $G$ in good position relative to $B$, then $\mathscr B$ can be constructed from the BN-pair $(G,B,N)$. This is explained in the first part of Bruhat-Tits article Groupes reductifs sur un corps locale I.
Is it possible to construct the Iwahori subgroup independently of the Bruhat-Tits building? I would like to approach Bruhat-Tits theory from this point of view, first defining the Iwahori subgroup, then the building.
I know that if (and only if) $G$ is quasi-split to split over an unramified extension of $k$, there is a smooth group scheme $\mathscr G$ over $\mathcal O_k$ such that $G = \mathscr G(k)$ and $\mathscr G(\mathcal O_k)$ is a maximal compact subgroup of $G$, and then the Iwahori subgroup $B$ of $G$ can be defined as the preimage of a Borel subgroup of $\mathscr G(\mathcal O_k/\mathfrak p_k)$ under the projection $\mathscr G(\mathcal O_k) \rightarrow \mathscr G(\mathcal O_k/\mathfrak p_k)$. But I think proving this requires Bruhat-Tits theory.
I think that one can saw something like: the Iwahori is a maximal pro-$p$ subgroup. This might not be literally true if one has non-trivial $p$-torsion in the $\Omega$-group, though.
@LSpice probably you want the normalizer of the maximal pro-$p$ subgroup.
Schneider and Stuhler - Representation theory and sheaves on the Bruhat–Tits building do something like what you propose, but they define the building in terms of all the parahoric subgroups, not just the Iwahori.
@WillSawin, you are right; I was describing something closer to the unipotent radical of the Iwahori.
I don't think the last claim requires Bruhat-Tits theory. You could instead check that quasi-split groups are classified by maps from the Galois group into the automorphisms of the Dynkin diagram (= automorphisms of the pair of a group and the Borel, modulo the Borel), and then observe that each unramified such map corresponds to a map from $\pi_1(\mathcal O_k)$ to the automorphisms and therefore to a group scheme.
By the way, do think carefully about whether your order of definition is the right one. If you're doing it because you find the definition of the building complicated and would like to avoid it, (1) you're in good company, but (2) it might be a good idea to bite that bullet and just learn what you need about the building, while knowing that (3) you can do an awful lot understanding how the building behaves without going into every detail of how it's defined.
One introduction that is much friendlier than BT itself is Yu - Bruhat–Tits theory and buildings.
I understand very well the desire to re-order and simplify the logic in this situation. After trying and failing for a few years, I went so far as to write a book about classical groups and buildings... In fact, after doing so, I realized that some things could be simplified. But I think not the relation between affine and spherical buildings.
|
2025-03-21T14:48:31.371402
| 2020-06-26T00:09:02 |
364158
|
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|
Stack Exchange
|
Problem about two elastic ropes in equilibrium
I have an elementary geometric problem that has thus far resisted all efforts from my end. The problem concerns "elastic ropes" which I model as a sequence of points $\gamma=(x_1,x_2,\dots,x_m)$ in $\mathbb{R}^d$ ($d\geq 2$). The "rope" has unit tension and is supported on the union of line segments connecting this path:
$$
S(\gamma) := \bigcup_{j=1}^{m-1} \overline{x_jx_{j+1}} \subset\mathbb{R}^d.
$$
The $j$-th segment of the path $\gamma$ points in the direction $\tau_j$,
$$
\tau_j = \frac{x_{j+1}-x_j}{|x_{j+1}-x_j|}.
$$
The rope is at unit tension and exerts a force $f_\gamma$ represented by the measure
$$
f_\gamma = \sum_{j=1}^{m-1} \tau_j (\delta_{x_j} - \delta_{x_{j+1}}).
$$
Here $\delta_x$ is the Dirac delta supported at $x$, and $f_\gamma$ is a vector-valued measure.
Notice that the support of $f_\gamma$ might be a strict subset of $\{x_1,\dots,x_m\}$, as it is possible for the rope to intersect itself in such a way as to balance the forces at a particular intersection.
Question:
Let $\gamma$ and $\gamma'$ be two ropes satisfying $f_\gamma=f_{\gamma'}$. Does it follow
that $S(\gamma)=S(\gamma')$?
As an example, consider a path $\gamma$ with a loop, so $\gamma=(x_1,x_2,x_3,x_4,x_2, x_5)$. One can check that the path which traces the loop in the opposite direction,
$\gamma'=(x_1,x_2,x_4,x_3,x_2,x_5)$ has the same forces ($f_\gamma=f_{\gamma'}$), but they share the same support ($S(\gamma)=S(\gamma')$).
By case work I can show that this holds for $m\leq 4$, and I have not found any counterexamples by computer search (but I do not have any sense what an effective algorithm for finding counterexamples would look like anyway...), so I would appreciate any insight whatsoever into this question. There are several equivalent formulations and non-examples I can provide if there is interest.
We can reformulate the problem to allow arbitrary rational tensions in each rope by scaling everything by twice the common denominator to get even integer tensions, then finding an Eulerian path.
This is an example of two arrangements with the same $f_\gamma$ and different supports. All angles are $\pi/3$ or $2\pi/3$. In order to make each one be a single rope, we specify that each segment drawn is actually two lengths of rope, and find an Eulerian cycle.
Thanks! I like the idea of doubling up the edges on the rope, I hadn't considered that before.
|
2025-03-21T14:48:31.371590
| 2020-06-26T00:19:42 |
364161
|
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|
Stack Exchange
|
Exterior algebra of normed spaces
This question is related to my prior question, but this one is aimed, even though it's more general. If $V$ is a vector space, we define the exterior algebra of $V$ do be:
$$\bigwedge V := \bigoplus_{n=0}^{\infty}\bigwedge^{n}V $$
where $\bigwedge^{n}V$ is the $n$-fold exterior power of $V$ and I used the identification $\bigwedge^{0}V = \mathbb{C}$ and $\bigwedge V = V$. Thus, an element $v \in \bigwedge V$ is a sequence $v=(v_{0},v_{1},...)$, with $v_{n}\in \bigwedge^{n}V$, with all but finitely many nonzero entries.. $\bigwedge^{n}V$ can be realized as the subspace of all skew-symmetric tensors of $\overbrace{V\otimes \cdots \otimes V}^{\text{n times}}$.
My question is: If $V$ is replaced by a normed vector space $U$, is $\bigwedge^{n}U$ defined in the same algebraic way as before? I know that I can induce a norm on $\bigwedge^{n}U$ from $U$ but I realy don't know anything about the construction of exterior powers of normed vector spaces. Is that any different from the algebraic one? Also, if $U$ is Banach, is $\bigwedge U$ also Banach or we need to complete it?
If $H$ is a Hilbert space, then probably the right definition for $\bigwedge H$ is the antisymmetric Fock space over $H$.
In my paper with Philippe Thieullen and Mohammed Zarrabi, we construct in a somewhat Bourbaki style exterior powers of a Banach space. See Appendix A4 of http://www.math.uvic.ca/faculty/aquas/papers/paper53.pdf
Both of these seem like great answers (possibly with a little expansion)!
@AnthonyQuas I've not had time to read this appendix in detail, but can't one just take the usual n-fold projective tensor power of a Banach space E (much studied) and then mod out by the (closure of) the usual subspace of degenerate tensors? This is how one gets e.g. the symmetric (projective) tensor powers in BSp world
@YemonChoi: I am really not too expert in this area. Our goal was to have a framework which makes it easy to calculate growth rates of finite-dimensional volumes for random matrix products. I suspect the quotient formulation you’re proposing might be hard to calculate with.
If $V$ is a Hilbert space there is a standard notion of alternating tensor. First, we have a definition of full tensor products of Hilbert spaces such that if $\{e_i\}$ is an orthonormal basis of $V$ then $\{e_i \otimes e_j\}$ is an orthonormal basis of $V \otimes V$ (and similarly for more than two factors). Then we have a notion of symmetric and antisymmetric parts of $V^{\otimes n}$ coming from the natural action of $S_n$ on this space. The antisymmetric part is considered as a space of alternating tensor over $V$, and the exterior algebra is taken to be the direct sum of the alternating tensor powers of $V$. This is the antisymmetric or "bosonic" Fock space.
I don't know about other Banach spaces. Anthony Quas gives a reference for this in the comments. But in general there are many reasonable ways to norm tensor products of Banach spaces, so I don't think it's fair to expect there to be a really canonical answer.
|
2025-03-21T14:48:31.371809
| 2020-06-26T00:22:39 |
364162
|
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|
Stack Exchange
|
The least distance of $f\in\ell_\infty(K,X)$ to $C_b(K,X)$
Let $K$ be a paracompact space and consider a bounded function $f:K\to\mathbb R$ not necessarily continuous, that is, $f\in\ell_\infty(K,\mathbb R)$. It's a well-known fact that the least distance of $f$ to some continuous $g:K\to \mathbb R$ is half of the maximum oscilation of $f$. More precisely, we define the oscilation of $f$ at $k$ by
$$
osc_f(k) = \inf_{U\in\mathcal V_k} \sup_{u,v\in U} |f(u)-f(v)|,
$$
where $\mathcal V_k$ is the set of open neighborhoods of $k$, and we have the following result, which is the Proposition 1.18.(ii) of the book Geometric Nonlinear Functional Analysis, by Y. Benyaminni and J. Lindenstrauss:
Let $f:K\to\mathbb R$ and put $\delta =\|osc_f\|_\infty$, where $\|\cdot\|_\infty$ is the supremum norm. Then, there exists a continuous function $g:K\to\mathbb R$ such that
$$
\|f-g\|_\infty=\frac{\delta}{2},
$$
and this is the least distance of a continuous function from $f$.
This proposition is an application of Michael's Selection Theorem, which is valid for any selection with domain in a paracompact space and counter-domain in the set of non-empty closed convex subsets of a Banach space $E$. Recently, I've proved a generalization of the proposition above for functions with counterdomain in $\mathbb C$ (which I posted here). Of course the measure of the oscilation changed, but we still get the least distance of $f$ to a continuous function.
I can see, by my proof, that it can be extended to Hilbert spaces in general, but, I wonder if this result can be extended to a much more general class, I would say the uniformly convex Banach space, or something close to that.
My question is if there exists a result analogous to the proposition for a more general class of Banach spaces, or if someone could guide me on an argument that could be valid to prove such a result.
|
2025-03-21T14:48:31.371958
| 2020-06-26T02:47:54 |
364165
|
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|
Stack Exchange
|
Some computational results and goals of stable motivic homotopy theory of schemes
I am trying to learn ($\mathbb{P}^1$-)stable motivic ($\mathbb{A}^1$-)homotopy theory of schemes from the Cisinski-Deglise book, Triangulated Categories of Mixed Motives. In order to keep myself going during its long abstract development as well as to help see parallels with the topological stable homotopy theory I'd like to know in advance some of the concrete computational results and goals the theory is reaching for.
For example, in the topological setting calculating the stable homotopy groups of spheres has been a long-running motivating theme. Are there similar computational goals animating the development in the algebraic-geometric context (as opposed to those concerned with having a satisfactory theory with certain features, e.g., the six-functor formalism, spectra for cohomology theories etc. which are wonderfully explained in the preface to the book?) I'm specially interested in those that are of arithmetical nature.
An obvious one that comes to mind is the Bloch–Kato conjecture (now a theorem of Voevodsky), which drove the development of $\mathbf A^1$-homotopy theory. Or are you asking specifically about the stable variant?
I think independence of $\ell$ is a rather good topic to have in mind: a goal of the six operations in the motivic world is to promote many cohomological features of $\ell$-adic sheaves to the motivic world. Anything which is trace-like (Euler characteristics, zeta functions) falls in this category. A nice paper to read on these kind of features is Olsson's https://link.springer.com/article/10.1007/s00229-015-0765-3 (or https://math.berkeley.edu/~molsson/Chernandmotives4.0.pdf for free access).
I also wrote a survey which revists this kind of things here: http://www.mathematik.uni-regensburg.de/cisinski/traces.pdf (but many of this topic can be found in various remarks in my joint papers with Déglise).
Such $\ell$-independence of traces is still a subject of research: this paper of Hiroki Kato https://arxiv.org/pdf/1904.02324.pdf for instance. However, the mere existence of motivic sheaves and fo their $\ell$-adic realizations make the independence of $\ell$ part of the aforementioned paper trivial.
|
2025-03-21T14:48:31.372170
| 2020-06-26T06:53:26 |
364174
|
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"Ivan Meir",
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|
Stack Exchange
|
Maximizing distance between points on the positive surface of the unit hyper-sphere
Suppose we want to place $k$ ($k \geq 3$) points on the positive surface of a unit hyper-sphere in $\mathbb{R}^n$ ($n \geq 3$), where all coordinates of a point are positive, such that the minimum distance, $x$, between any pair of points is maximized. Is there a general formula to determine the value $x$?
Why do you want all coordinates positives? This would be only 1/8 of a standard sphere in $\mathbb {R}^3$ say.
Equidistribution over the entire sphere is more standard and if you have a lower bound for the min distance over the whole surface it will also hold for the positive quadrant as well since deleting points only increases min distances between points.
For the sphere in $\mathbb {R}^3$ you may want to look at this post
I have no hard reasons, but given that there are no such formulas for spheres, balls (solid spheres), cubes, ... etc, I highly doubt that there is one for the positive part of a sphere.
|
2025-03-21T14:48:31.372274
| 2020-06-26T06:56:51 |
364175
|
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"Beta",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/364175"
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|
Stack Exchange
|
How to solve equation $e^x \log x=2$
I want to know how to solve the equation
$$e^x\log x=2.$$
We can get a numerical solution but it seems difficult to get an exact solution. I know the Lambert W function but unable to use it for the above equation.
$x=1.5372017025783550472\cdots $ --- I'm not sure what you mean by an "exact" solution --- like in terms of some known special function? that is not likely to be forthcoming.
Is there a closed form solution not only the numerical solution?
"closed form" in the sense of a known special function: unlikely; however, you could define a new special function $U(a)$ by $e^x \log x=a$, and then the number you seek is $U(2)$.
Maple does not find a closed form, and Maple is pretty good with the Lambert W, so it seems that W will not help with closed form for this.
I do not think that there is any known special function which solves your equation. However, a solution can be given in a sort of 'infinite exponent tower'. More precisely, rewrite the equation as:
$$ \ln x = 2e^{-x} \Rightarrow x = e^{2e^{-x}} $$
Then, the solution is:
$$ x=e^{2e^{-e^{2e^{...}}}} $$
Following a similar idea to the one given by @Carlo Beenakker in the comments, you could then define a function $U$ as follows:
$$ U(a,b,c):= a^{ba^{ ca^{ ba^{ ...}}}} $$
(alternating $b$ and $c$)
Then, you can study when this new function converges. It is clear that the solution to your equation is:
$$ x=U(e,2,-1) \approx 1.537201702578...$$
Notice that the function $U$ is a generalisation of the operation of infinite tetration. Indeed:
$$ U(a,1,1) = a^{a^{a^{ ...}}} = \frac{W(-\ln a)}{- \ln a} $$
EDIT: as noticed by @LSpice, we can even simplify this and consider the function (with only two variables):
$$ Z(a,b) := a^{-2b a^{ba^{ ...}}}$$
(again, alternating) and get:
$$ x=Z(e,-1) $$
Of course $U$ could be boiled down to a function of $2$ variables; you can just put $x = e^{2U(e, -2)}$, in the obvious notation.
Yes: this is even better, because you simplify the study of convergence of such new function
Thanks for your discussion and help!
It seems to me, from $x=\exp(2\exp(-x))$ we get not what you said, but instead
$$
x = \exp\Bigg(2\exp\bigg(-\exp\Big(2\exp(-\dots)\big)\Big)\bigg)\Bigg)\
$$
where we alternate coefficnents $2$ and $-1$.
Thanks for noting this; I have now corrected the answer.
|
2025-03-21T14:48:31.372457
| 2020-06-26T07:21:28 |
364176
|
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|
Stack Exchange
|
Gaussian integral $\int_X \|x\|_X^2 \mu(dx)$ in Banach space
For a centered Gaussian measure $\mu$ on a Hilbert space $X$, it is known that
$$\int_X \|x\|^2 \mu(dx) = tr(Q)$$ where $Q$ is the covariance operator. Is there a similar version for Gaussian measures in Banach spaces? I know that I can bound it from above by Fernique's theorem but I'm interested in calculating this explicitly.
A minor comment. The equation in the title gives a Math Processing error when viewed from the page where the titles are listed.
There is no explicit expression for $\int_X \|x\|^2 \mu(dx)$ in general, even if $X=\mathbb R^d$ with $d\in\{2,3,\dots\}$ and $\mu$ is the standard normal distribution on $\mathbb R^d$.
Indeed, any norm on $X=\mathbb R^d$ is characterized by the corresponding unit ball $B:=\{x\in X\colon\|x\|\le1\}$. More specifically, $\|\cdot\|=M_B$, where $M_B$ is the Minkowski functional for $B$, defined by the formula
$$M_B(x):=\inf\{t\in(0,\infty)\colon x/t\in B\}$$
for $x\in X$. Moreover, a set $C\subset X=\mathbb R^d$ is the unit ball for some norm on $X$ iff $C$ is compact, convex, symmetric (i.e., $-C=C$), and absorbing (i.e., $\bigcup_{n>0}nC=X$); let us call such a set a Minkowski set. Equivalently, a set $C\subset X=\mathbb R^d$ is the unit ball for some norm on $X$ iff $C=D-D$ for some compact convex set $D$ with nonempty interior. Thus, the set of all norms on $X=\mathbb R^d$ is almost as big as the set of all convex bodies in $X$.
So, your integral is
$$\int_X (\inf\{t\in(0,\infty)\colon x/t\in C\})^2 \mu(dx)$$
for a general Minkowski set $C$.
Another way to write this integral is as follows:
\begin{aligned}\int_X \|x\|^2 \mu(dx)
&=\int_X \mu(dx)\int_0^\infty ds\,1\{s<\|x\|^2\} \\
&=\int_0^\infty ds\int_X \mu(dx)\,1\{s<\|x\|^2\} \\
&=\int_0^\infty ds\int_X \mu(dx)(1-\,1\{\|x\|^2\le s\}) \\
&=\int_0^\infty ds\,(1-\mu(\sqrt s B)).
\end{aligned}
We see that to compute your integral, you need to be able to compute the $\mu$-measure of an arbitrary Minkowski set, which is impossible to write explicitly even when $\mu$ is the standard normal distribution on $\mathbb R^d$.
|
2025-03-21T14:48:31.372603
| 2020-06-26T09:17:28 |
364179
|
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|
Stack Exchange
|
Interpreting the Bockstein lemma?
I am reading through "Cohomology Operations and Applications in Homotopy Theory" by Mosher and Tangora and I had a little bit of confusion with the Bockstein lemma. All cohomology will be with $\mathbb{Z}/2$-coefficients. The map $d_i$ is the $i$-th Bockstein differential and the map $\tau$ is the transgression in the Serre spectral sequence. The statement of the lemma is as follows, suppose that $F \to E \to B$ is a fibration with maps $j: F \to E$ and $p: E \to B$ and let $i \geq 1$ be an integer. Let $u \in H^n(F)$ is transgressive and $d_i v = \tau (u)$. Then the Bockstein lemma says that $d_{i+1} p^\ast(v)$ is defined and $j^\ast d_{i+1} p^\ast(v) = d_1(u)$.
The authors say that "The equalities should be suitably interpreted." and I wanted to see if I am doing that correctly, namely the last equality. In particular, when they write $j^\ast d_{i+1} p^\ast(v)$ does that mean that I should take a lift of $d_{i+1} p^\ast(v)$ to $H^{n+1}(E)$, apply $j^\ast$ to that lift, and then interpret the result in $H^{n+1}(F)$? So the equality $j^\ast d_{i+1} p^\ast(v) = d_1(u)$ is an equality of elements in $H^{n+1}(F)$?
Why does $j^\ast d_{i+1} p^\ast(v)$ not depend on the choice of lift of $d_{i+1} p^\ast(v)$ to $H^{n+1}(E)$?
|
2025-03-21T14:48:31.372719
| 2020-06-26T11:54:59 |
364185
|
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|
Stack Exchange
|
Why the name `Lipschitz-Free Banach spaces'?
There are many names for the same objects that is known as the Arens--Eells spaces, transportation cost spaces, free Banach spaces over a (pointed) metric space, and Lipschitz-free Banach spaces.
The last one bothers me a bit. Doesn't the construction something-free mean free of that something? Think of alcohol-free drinks. The name free in Mathematics means free of further relations, so the name means free of Mr Lipschitz, who's been quite dead for some time already.
Is it just slopiness or is there are good reason to call these spaces Lipschitz-free?
After Isaiah Berlin, are these spaces free from Lipschitz, or free to Lipschitz?
I think 'slopiness' is a bit harsh. Godefroy and Kalton presumably wished to say: free with respect to Lipschitz maps. I think they were the originators of the terminology "Lipschitz-free spaces" and some later authors have followed them; every time the terminology gets repeated, there is less incentive to change it :-/
I suppose that this is the standard use of “free” as in free group, free topological group, free vector space, free locally convex space, etc. The technical definition is that it can be regarded as a functor which is adjoint to a forgetful one. In this case, the one from the category of Banach spaces to that of pointed metric spaces (both with the condition that the morphisms are contractive) where one maps the Banach space to its unit ball and forgets the linear structure. By the way do you mean “slopiness” or “sloppiness”?
Add the phrase “with radius at most 1” above after “pointed metric space”.
@user131781 Undoubtedly Yemon Choi was mimicking the OPs misspelling, the standard is "sloppiness" ("slopiness" is a conceivable way of referring to the quality of being sloped). I think you are right that it is "free" in the sense of left adjoint. Of course, $\ell^1(X)$ is the left adjoint to the forgetful closed unit ball functor to $\mathbf{Set}$, so some extra thing is needed to specify that it is not this forgetful functor that is meant, but the one to the category with Lipschitz maps, as you mentioned.
@user131781 Thank you for making precise what I was alluding to (I typed the original comment in a rush). Yes, G+K's paper, and some talks by G that I happened to attend, made it clear that the "freeness" property they had in mind was what the more category-theoretic inclined would call: left adjoint to the forgetful functor between the two categories described in your comment.
|
2025-03-21T14:48:31.372952
| 2020-06-26T12:01:47 |
364186
|
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|
Stack Exchange
|
When does this limiting ratio give a real root $x$ to the equation of the form $\sum\limits_{k=0}^d \frac{x^k a_{k+1}}{k!}=0$?
By searching the Inverse Symbolic Calculator, we appear to be able to make the following conjecture about a real root to the equation:
$$\sum\limits_{k=0}^d \frac{x^k a_{k+1}}{k!}=0 \tag{1}$$
Let the lower triangular matrix $A$ be:
$$A=\binom{n-1}{k-1} a_{n-k+1} \tag{2}$$
where $n=1,2,3,4,5,...N$ and $k=1,2,3,4,5,...N$, with $N>>d$, and
where the parentheses is the binomial function.
Calculate the matrix inverse
$$B=A^{-1} \tag{3}$$,
consider the first column of matrix $B$:
$$b_n=B(n,1) \tag{4}$$
and take the limiting ratio:
$$x=\lim_{n\to \infty } \, \frac{(n-1) b_{n-1}}{b_n} \tag{5}$$
Under what conditions for the coefficients:
$$a_1,...,a_{d} \tag{6}$$
is the limiting ratio $x$ in $(5)$ a real root solution to $(1)$
$$\sum\limits_{k=0}^d \frac{x^k a_{k+1}}{k!}=0$$ ?
Is the conjecture true at all?
I apologize for not letting the index of $a$ begin with $0$ instead of $1$.
Here is the Mathematica program for the conjecture:
Clear[A, B, a, b, x];
a = {1, 3, 5, 8, 5, 41, 39, 57, 53, 47, 13, 19, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
0, 0, 0, 0, 0, 0, 0, 0, 0};
nn = Length[a];
d = Max[Flatten[Position[Sign[Abs[a]], 1]]]
A = Table[
Table[If[n >= k, Binomial[n - 1, k - 1]*a[[n - k + 1]], 0], {k, 1,
Length[a]}], {n, 1, Length[a]}];
b = Inverse[A][[All, 1]];
x = N[Table[(n - 1)*b[[n - 1]]/b[[n]], {n, nn - 8, nn - 1}], 30]
Sum[a[[k + 1]]/k!*x^k, {k, 0, d}]
Clear[x];
Sum[a[[k + 1]]/k!*x^k, {k, 0, d}]
The limiting ratio output from the program is:
$$x=-0.474390307209018254579812222047$$
and this appears to be a solution to:
$$1+3 x+\frac{5 x^2}{2}+\frac{4 x^3}{3}+\frac{5 x^4}{24}+\frac{41 x^5}{120}+\frac{13 x^6}{240}+\frac{19 x^7}{1680}+\frac{53 x^8}{40320}+\frac{47 x^9}{362880}+\frac{13 x^{10}}{3628800}+\frac{19 x^{11}}{39916800}=0$$
In the program one needs to have $N$ much bigger than $d$ in order to see the conjecture in the output. Therefore there are a lot of trailing zeros in the coefficients $a$. You can add more zeros manually to the vector $a$ if you want to.
I don't know how to tag this question properly.
A related proof by joriki:
https://math.stackexchange.com/a/60385/8530
OEIS entry:
https://oeis.org/A167196
Related limiting ratio:
https://oeis.org/A132049
OEIS searches:
https://oeis.org/A322262
https://oeis.org/A006153
A much more conventional formulation of what I am doing in Mathematica:
Clear[x, b];
polynomial = (1 + 2*x + x^2/2! + x^3/3! + x^4/4! + x^5/5!);
digits = 100;
b = With[{nn = 200},
CoefficientList[Series[1/polynomial, {x, 0, nn}],
x] Range[0, nn]!] ;
nn = Length[b] - 10;
x = N[Table[(n - 1)*b[[n - 1]]/b[[n]], {n, nn - 8, nn - 1}], digits]
polynomial
Thanks to Harvey Dale in the OEIS for how to write the part:
b = With[{nn = 200},
CoefficientList[Series[1/polynomial, {x, 0, nn}],
x] Range[0, nn]!] ;
And thereby the polynomial need not have factorials:
Clear[x, b];
polynomial = (1 - 2 x + 3*x^2 - 5 x^3 + 7 x^4 - 11 x^5);
digits = 100;
b = With[{nn = 4000},
CoefficientList[Series[1/polynomial, {x, 0, nn}],
x] Range[0, nn]!] ;
nn = Length[b] - 10;
x = N[Table[(n - 1)*b[[n - 1]]/b[[n]], {n, nn - 8, nn - 1}], digits]
polynomial
The following program uses the method in the question by first Taylor expanding the Riemann zeta function at $zi$ and real part equal to $0$, and then adding trailing zeros to the vector $a$. Expanding at real part equal to $1$ gives a similar plot.
(*start*)
start = 10;
end = 30;
Monitor[list = Table[zz = 10;
d = 20;
a = Flatten[{CoefficientList[
Normal[Series[Zeta[x + z*N[I, d]], {x, 0, zz}]], x]*
Range[0, zz]!, Range[d]*0}];
nn = Length[a];
A = Table[
Table[If[n >= k, Binomial[n - 1, k - 1]*a[[n - k + 1]], 0], {k,
1, Length[a]}], {n, 1, Length[a]}];
Quiet[b = Inverse[A][[All, 1]]];
z*I + N[(nn - 1)*b[[nn - 1]]/b[[nn]], 40], {z, start, end, 1/10}],
z*10]
ListLinePlot[Re[list], PlotRange -> {-1, 3}, DataRange -> {start, end}]
ListLinePlot[Im[list], DataRange -> {start, end}]
(*end*)
The result is an approximation to the Riemann zeta zeros, where the plot of the real part stays around $\frac{1}{2}$ except at singularities at Gram points:
The heights of the steps in the staircase in the second plot are at imaginary parts of the Riemann zeta zeros.
The plots below are the same as above but from $z=10$ to $z=60$:
I now later posted a similar simpler question at Mathematics Stack Exchange: https://math.stackexchange.com/q/3735496/8530
If you define $c_n=a_{n+1}/a_1$ and $d_n=b_{n+1}$, then $b_n$ is obtained also by $e^{d.t}=1/e^{c.t}$, with the e.g.f.s in umbral notation, and they satisfy an intertwined recursion relation also (see https://oeis.org/A133314).
This looks related: https://arxiv.org/ftp/math/papers/0408/0408264.pdf
Also related: https://math.stackexchange.com/a/60385/8530
Perturbing x in the polynomial allows for solving those polynomials that are otherwise unsolvable: https://pastebin.com/YQD9rHyp
(Mathematica start)
x = -1.1184521423567664364899882962729454658601429873285997501246`30.
97507448583345;
1 + x + x^2 - 2 x^3 + x^4 - 5 x^5 + x^6 + 2 x^7 + 3 x^8 + 7 x^9
(end)
https://pastebin.com/q0mu1Fj3
Update 8/25/2021:
"On Schröder’s families of root-finding methods" by M. Petković, L. Petković, and Ð. Herceg presents the method you used and attributes it to Schröder. See eqn. (9) on and Thm. 1 on pg. 1758. Reference is made to a translation by Stewart of the 1869/70 paper by Schröder "On infinitely many algorithms for solving equations" (Ueber unendlich viele hlgorithmen zur hufliisnng der Gleichungen) with the set of polynomials (I) on pg. 31 of Stewart being the first few refined Euler characteristic polynomials for the permutahedra (OEIS A1333147, mod signs) discussed in the comments below, which encode multiplicative inversion of formal Taylor series, or e.g.f.s (neither Schröder nor the later authors noted any relation to the polytopes). An approach related to compositional inversion is also presented in Schröder's paper as an extension of Newton's method.
Extended comments made on Aug. 1, 2020, in reply to Matt's comment to me:
The answer to your MO question was provided in the MSE question couched in terms of polynomials expressed as truncated power series, or ordinary generating functions (o.g.f.s) and their reciprocals. Here you use truncated Taylor series, or exponential generating functions (e.g.f.s). The series for the reciprocals can be found several parallel ways. Three are 1) by recursion relations (as in the MSE-Q) related to regular or binomial convolution of the coefficients of the generating function with those of its reciprocal, generated by the product of the pair of g.f.s, which by construction is unity; 2) for e.g.f.s, through evaluation of the signed face partition polynomials of the of permutahedra (A133314), and, for o.g.f.s, by the signed, refined Pascal partition polynomials (A263633), encoding the Newton identity relating the symmetric elementary polynomials to complete symmetric homogeneous polynomials; and 3) by inverting a Pascal matrix diagonally multiplied by the Taylor coefficients as you do here. All three methods are explained in A133314 (see also my blog post "Skipping over Dimensions, Juggling Zeros in the Matrix," and others, in the companion OEIS entry.
Whenever you are looking at multiplicative inversion of g.f.s, Appell sequences are lurking nearby.
Interesting paper in your comment. if you throw in some binomial coefficients and normalize, the polynomial in eqn. 1 in the paper (any polynomial) can be re-expressed in umbral notation as $(e.+x)^n=E_n(x)=s(x)$ with lower order polynomials in the related Appell sequence obtained by differentiation or by noting $E_k(x)=(x+e.)^{k}$ for $k \leq n$. The paper seeks to construct zeros of this polynomial equation.
Eqn. 2 in the paper can be couched in terms of the umbral operational calculus. If you let $\frac{m_{n-k+1}}{m_n}=\frac{b_{k-1}}{(k-1)!}$ for $1 \leq k \leq n$ and $b_k=0$ for $k \geq n-1$ , then the eqn. becomes
$$e^{b.D_s}x(s)=x(b.+s)=x(B.(s)) = \frac{-m_{n+1}}{m_n}$$
where $D_s=d/ds$ and $B.(s)^n=B_n(s)= (b.+s)$. Clearly, $e^{b.D_x}$ is the diff op that generates the Appell polynomials when acting on $x^n$.
So, the analysis in the paper also seems closely related to Appell sequences, so perhaps there is something to gain from this perspective, but I don't have much time to pursue it at the moment.
In addition, a solution about $x=0$ to eqn. 1 in the paper can be approached through compositional inversion of the o.g.f. $s(x) = a./(1-a.x)$ through A133437 or A134264, but I'm not sure yet how this might relate to your results.
As I believe you already know, this Appell approach ties in with your interest in the Riemann zeta function through investigations of Jensen and later Polya (probably generating the initial interest in what became known as Appell polynomials). The Riemann hypothesis is true if and only if all the Appell polynomials formed from the Taylor series coefficients of the shifted, rotated, entire Landau-Riemann xi function $\xi(1/2+it)$ are hyberbolic, i.e., have only real zeros.
(Titchmarsh also investigated the zeros of the partial sums of Taylor series, no doubt due to the RH.)
log(2) = A052882/A000670, Pi/4 = n*A001586(n-1)/A001586(n) as n-->infinity, and this https://oeis.org/A167196 were my starting points. I noticed that the coefficients of your list partition transform turned up in one of the series expansions. Also I have a typo in the oeis in the limiting ratio for square roots.
I posted a new question: https://mathoverflow.net/a/368534/25104
|
2025-03-21T14:48:31.373636
| 2020-06-26T12:28:23 |
364188
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"authors": [
"Darth Vader",
"Yemon Choi",
"https://mathoverflow.net/users/149852",
"https://mathoverflow.net/users/763"
],
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/364188"
}
|
Stack Exchange
|
On the automorphisms of the unitary group in the strong operator topology
Let $H$ be an infinite dimensional complex (or real) Hilbert space, and let $U(H)$ be the unitary (or orthogonal) group. We equip $U(H)$ with the strong topology.
Now, suppose that $\phi: U(H) \rightarrow U(H)$ is a continuous group automorphism.
Is it true that $\phi$ is automatically continuous with respect to the norm topology on $U(H)$?
I tried proving this directly, but I did not get very far.
Next, I tried characterizing the automorphisms of $U(H)$ more abstractly.
I vaguely recall reading somewhere that any automorphism $\phi$ of $U(H)$ can be realized by conjugation by either an element of $U(H)$ itself, or by conjugation by a conjugate linear isometry of $H$, which would thus automatically imply that $\phi$ is continuous with respect to the norm topology, but I haven't been able to find this result or prove it myself.
This result is true, and is surprisingly (to me) nontrivial. One reference I could come up with is the paper " Transformations of the unitary group on a Hilbert space" by L. Molnar and P. Semrl. Please look at Theorem 2.5 in this paper : https://pdfs.semanticscholar.org/386f/31da52d66592ccee13ed046cac4d9ed2444f.pdf
The authors credit the result to M. Broise "Commutateurs dans le groupe unitaire d’un facteur"(Cor 3).
(I didn't check Broise's paper though, as it's in a language that I can't read :)).
I find your lack of faith ... disturbing
@YemonChoi: :D I read that in the voice of Vader himself!
|
2025-03-21T14:48:31.373768
| 2020-06-26T13:09:44 |
364189
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/364189"
}
|
Stack Exchange
|
When is the intersection of two determinantal ideals equal to the product?
Let $S = k[x_{i,j}\mid 1\leq i\leq n, 1\leq j\leq m]$ be a polynomial ring over an arbitrary field $k$. Let $M$ be a generic $n\times m$ matrix of indeterminates in the ring $S$ where $n\leq m$. For any $k\leq n$, let $M'$ denote the matrix $M$ with column $k$ removed and let $M_k$ denote the matrix $M$ restricted to columns $1, \ldots, k$. Consider the ideals $J_1 = I_n(M')$, the ideal of maximal minors of $M'$, and $J_2 = I_k(M_k)$, the ideal of maximal minors of $M_k$. I need to show that
$$
J_1\cap J_2 = J_1\cdot J_2
$$
or, equivalently,
$$
Tor_1(S/J_1,S/J_2) = 0.
$$
I feel like this must follow somehow from the fact that NO determinants in $J_1$ contain variables from column $k$, but EVERY determinant in $J_2$ contains variables from column $k$, but I am not sure how to show it.
After some thought, I came up with the following answer to my own question.
First, observe that
$$
\mathbf{x} = (x_{1,k}, \ldots, x_{n,k})
$$
is a regular sequence over $R/I_n(M')$. Take an element $f\in I_n(M')\cap I_k(M_k)$. Then
$$
f = \sum_{I\in I_n(M')} c_I [i_1, \ldots, i_n] = \sum_{J\in I_k(M_k)} d_J [j_1, \ldots, j_k].
$$
Expanding the minor $[j_1,\ldots, j_k]$ along column $k$ using the Laplace expansion, one obtains the equality
\begin{align*}
\sum_{I\in I_n(M')} c_I [i_1, \ldots, i_n] &= \sum_{J\in I_k(M_k)} d_J \left(\sum_{\ell\in \{1,\ldots, n\}} (-1)^\ell \, x_{\ell, k} \, [1, \ldots, \hat \ell, \ldots, n\mid j_1, \ldots, j_{k-1}]\right)\\
&= \sum_{J\in I_k(M_k)} \sum_{\ell\in \{1,\ldots, n\}} (-1)^\ell d_J\cdot x_{\ell,k}\, [1, \ldots, \hat \ell, \ldots, n\mid j_1, \ldots, j_{k-1}].
\end{align*}
Observe that since $\mathbf{x}$ is regular over $R/I_n(M')$, each $d_J\cdot [1, \ldots, \hat \ell, \ldots, n\mid j_1, \ldots, j_{k-1}]$ must be in $I_n(M')$. But $I_n(M')$ is a prime ideal, so either $d_J$ or $[1, \ldots, \hat \ell, \ldots, n\mid j_1, \ldots, j_{k-1}]$ must be in $M'$. Clearly $[1, \ldots, \hat \ell, \ldots, n\mid j_1, \ldots, j_{k-1}]$ is not in $M'$, so we have that $d_J$ must be in $I_n(M')$, so $I_n(M')\cap I_k(M_k) = I_n(M')\cdot I_k(M_k)$.
|
2025-03-21T14:48:31.373912
| 2020-06-26T13:11:29 |
364190
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"0xbadf00d",
"Jochen Glueck",
"Nate Eldredge",
"R W",
"https://mathoverflow.net/users/102946",
"https://mathoverflow.net/users/4832",
"https://mathoverflow.net/users/8588",
"https://mathoverflow.net/users/91890"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/364190"
}
|
Stack Exchange
|
If $(κ_t)$ is a semigroup with invariant measure $\mu$ and $ν$ is singular to $\mu$, then $νκ_t$ might not converge to $\mu$ in total variation norm
Let $E$ be a Polish space, $(\kappa_t)_{t\ge0}$ be a Markov semigroup on $(E,\mathcal B(E))$, $\mu$ be a probability measure on $(E,\mathcal B(E))$ invariant with respect to $(\kappa_t)_{t\ge0}$ and $\nu$ be a probability measure $(E,\mathcal B(E))$ singular to $\mu$.
I've read that if $(\kappa_t)_{t\ge0}$ is not strongly Feller$^1$, then we cannot expact that the composition $\nu\kappa_t$ tends to $\mu$ in total variation distance. How do we see this?
And does "cannot expect" mean that there are instances of the setting such that the convergence does not hold or can we even show that it's impossible to hold in general?
$^1$ $(\kappa_t)_{t\ge0}$ is called strongly Feller at time $t\ge0$ if $\kappa_t f$ is continuous for all bounded $\mathcal E$-measurable $f:E\to\mathbb R$.
"We cannot expect" would ordinarily mean "it is not generally true that". In other words, there exists at least one non-strongly Feller example in which $\nu \kappa_t$ does not converge to $\mu$ in total variation. Probably there are many such examples, but that's not necessarily implied by the wording. It is certainly not claiming that every non-strongly-Feller semigroup fails to converge.
In the general situation that you describe here, the wording that you quote seems a bit odd: In fact, even if the semigroup is strongly Feller we cannot expect convergence in total variation norm of $\nu \kappa_t$ in case that $\mu$ and $\nu$ are singular. And conversely, if the semigroup is not strongly Feller, we cannot expect convergence even if $\mu$ and $\nu$ are not singular. So in the situation you describe there are actually two different reasons why we cannot expect convergence, in general.
@JochenGlueck Yes, I know that even for a strongly Feller semigroup it is not guaranteed that convergence takes place. And that's why (@Nate) why I questioned what precisely is meant by "cannot expect", since it sounded like $\nu$ being singular to $\mu$ would somehow rule out that this is possible. So, maybe I didn't get your comment correctly, but can we generally infer something from knowing that $\nu$ is singular (or not singular) to $\mu$ at all?
@NateEldredge Please take note of my comment.
Well, what is true is the following: If $\nu$ is absolutely continuous with respect to $\mu$ and the semigroup is strongly Feller, then we always have convergence of $\nu \kappa_t$ as $t \to \infty$ with respect to the total variation norm (but not necessarily to $\mu$, if we do not assume uniqueness of the invariant measure). If one of the assumptions (strongly Feller or absolute continuity of $\nu$ with respect to $\mu$) fails, convergence can fail (but does not necessarily fail).
Maybe one additional comment about the case where the semigroup is strongly Feller. In a way, the question whether $\mu$ and $\nu$ are singular is a bit besides the point (as I said: if $\mu$ and $\nu$ are singular, convergence can fail, but there are also many situations where convergence still holds). What is more important is the dual structure of the fixed spaces of the semigroup: for strong Feller semigroups, we have convergence of $\nu \kappa_t$ for each $\nu$ if and only if the fixed measures of the semigroup separate the fixed functions.
As a simple example, consider $E = [0,1]$, $\mu$ Lebesgue measure, and $\kappa_t f = e^{-t} f + (1-e^{-t}) \int f,d\mu \cdot 1$, a process which jumps at rate 1 to a uniformly chosen point. This is not strongly Feller but we have $\nu \kappa_t = e^{-t} \nu + (1-e^{-t}) \mu$ which does indeed converge to $\mu$ in total variation, even for $\nu$ which are singular to $\mu$.
@NateEldredge So, can you make any sense of the quote? (it's taken from this paper(p. 4, second paragraph)
I don't think it's anything more than that their semigroup is not strong Feller, and therefore they cannot conclude from standard theory alone that $P_t^* \mu \to \mu$ in total variation in the singular case. It may be possible to get that conclusion, or convergence in some other mode, under additional assumptions, and that's what the paper is about.
You are asking for a condition that, in particular, guarantees that your semigroup has a unique invariant (stationary) measure, which is quite strong. There are lots of examples without this property. For a very simple (if somewhat degenerate) example notice that if you do not impose any continuity assumptions on the transition probabilities at all, then there is no way the behaviours of $\kappa_t\mu$ and $\kappa_t\nu$ can be related if $\mu$ and $\nu$ are singular. Take a partition of $E$ into two measurable sets $A$ and $B$, take $a\in A, b\in B$, and let $\kappa_t$ send any point from $A$ to $a$, and any point from $B$ to $b$. Then the semigroup $\kappa_t$ has two invariant measures $\delta_a,\delta_b$.
Thank you for your answer. (a) I guess your intention is that $\mu(A)=\nu(B)=0$? (b) You wrote that "there is no way" when $\mu$ and $\nu$ are singular, but as long as I got you right you've only provided an example that there is a particular semigroup such that there is more than one invariant measure, which is a different claim.
(c) What I want to know is how this is particularly related to the convergence in total variation norm. The quoted claim in the question is taken from this paper (p. 4, second paragraph) and it's indicated that convergence might hold with respect to a coarser topology (like an appropriate Waserstein distance).
@RW: I'm not sure whether I understand your reference to "continuity assumptions on the transition probabilities" correctly. It seems that continuity could only help to avoid the scenario that you describe if we assume in addition that $E$ is connected.
@Jochen Glueck Any Markov kernel can be considered as a map assigning to any point $x$ from the state space the corresponding transition probability measure $\pi_x$, and what I meant is the continuity of this map in an appropriate topology (for instance, the standard Feller condition is precisely the continuity of this map in the weak topology). It has nothing to do with the connectedness of the state space.
@0xbadf00d Look up the description of various Feller type properties for the Markov kernel determined by a family of transition probabilities $\pi_x$ in terms of the continuity of the map $x\mapsto\pi_x$ in the corresponding topologies. For instance, see Chapter IX Definition 8 of the book by Dellacherie - Meyer quoted in the article you mention.
|
2025-03-21T14:48:31.374321
| 2020-06-26T14:10:57 |
364196
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"authors": [
"Iosif Pinelis",
"Statguy",
"https://mathoverflow.net/users/160245",
"https://mathoverflow.net/users/36721"
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/364196"
}
|
Stack Exchange
|
Lower bound on $L^2$ norm of a strongly convex function
Let $f\colon[0, 1] \to \mathbb R$ be an $m$-strongly convex function and $\mu$ be a probability measure on $[0,1].$ For any $t<1$, the goal is to find a lower bound on $\int_{0}^t f^2(x) d\mu(x)$ in terms of $t$, $m$, and $\mu$ (and nothing else). We currently have the following bound
$$\int_{0}^t f^2(x) d\mu(x) \ge \frac{ m^2 t^4}{36} \mu[0,t].$$
We do not know if our bound is tight. Moreover, our proof is really long and messy. A clean/simple proof of such an elementary result would be helpful.
The left-hand side, $\int_{0}^t f^2(x)\, dx$, of your inequality does not contain $\mu$. Since you wanted "to find lowerbound on $\int_{0}^t f^2(x) d\mu(x)$", it appears that your desired inequality is
$$\int_0^t f^2(x)\,\mu(dx)\ge cm^2t^4 \mu[0,t] \tag{1}$$
for some real $c>0$.
However, (1) is obviously false in general. E.g., let $t=1/2$, $m=1$, $f(x)\equiv x^2/2$, and let $\mu$ be the Dirac measure supported on $\{0\}$. Then $f$ is $m$-strongly convex and $\int_0^t f^2(x)\,\mu(dx)=0$, so that (1) fails to hold for any $c>0$.
Sorry about that @Iosif. An assumption I forgot to mention, we assume that $\mu$ has a density with respect to the Lebesgue measure, and the density is bounded on $[0,1]$ (but not bounded away from zero) Thank you for your answer and time, however.
@Statguy : Your question was fully answered. If you actually wanted to ask something else, you can do that in a separate post.
|
2025-03-21T14:48:31.374438
| 2020-06-26T14:25:47 |
364197
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
],
"authors": [
"Maximilian Janisch",
"https://mathoverflow.net/users/129831"
],
"include_comments": true,
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/364197"
}
|
Stack Exchange
|
How to apply Hahn-Banach to the convex hull?
I am trying to understand the proof of Lemma 4.1.2 in Michel Talagrand's publication from 1995 on concentration inequalities (see below for the precise question statement):
A bit of context: Talagrand fixes a point $x\in X$ (he uses the notation $X=\Omega$) and a subset $A\subset X$, where $X=X_1\times X_2\times\dots X_n$ is the product space of arbitrary non-empty sets $X_1,\dots, X_n$. The $\alpha_i$ and $t$ are all supposed to be positive real numbers. He defines $A_t^c$ as follows:
My question. I understand why (4.1.4) implies (4.1.5). However, Talagrand says that "the converse follows from the Hahn-Banach theorem". How does it follow from the Hahn-Banach theorem?
Note: The problem can be slightly reformulated by saying that we want to prove that for $t>0$ and all $M\subset\{0,1\}^n$, we have that whenever $$\text{for all }\alpha\in]0,\infty[^n, \text{ there exists a } m\in M \text{ such that } \langle \alpha, m \rangle \le t\lVert \alpha\rVert_2,$$ then
$$\min_{m \in \text{Convex hull of } M} \lVert m \rVert_2 \le t.$$
In fact, if somebody can show this, then I will be able to prove the conjecture formulated by me yesterday.
To solve the problem you mention at the end you can argue in this way:
$$
\min_{m \in \mathrm{conv} (M)} \|m\|_{2}=\min_{m \in \mathrm{conv} (M)} \max_{\|\alpha\|_{2}\leq 1} \langle \alpha, m\rangle = \max_{\|\alpha\|_{2}\leq 1} \min_{m \in \mathrm{Conv}(M)}\langle \alpha, m\rangle \leq \max_{\|\alpha\|_{2}\leq 1, \alpha \in [0, \infty)^{n}} \min_{m \in M}\langle \alpha, m\rangle \leq t
$$
The only nontrivial observation was used is min-max theorem, which says that if $X, Y$ are convex compact sets, $f(x,y)$ continuous, convex in $x$ and concave in $y$ then $\min_{x \in X} \max_{y \in Y} f(x,y) = \max_{y\in Y} \min_{x\in X} f(x,y)$. Choose $f(x,y)=\langle x, y\rangle$
Amazing! Can't this be used to also directly prove my conjecture here as the last inequality seems like an equality to me if you consider the Bauer maximum principle ? (The right-hand side is just the minimum of norms over the convex hull of $M$.)
Turns out the answer is "yes, it can!". Please have a look at my answer here which is a very small extension of your answer.
|
2025-03-21T14:48:31.374616
| 2020-06-26T15:13:58 |
364203
|
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|
Stack Exchange
|
Probability in Chromatic number upper bound of induced subgraph
Let $G=(V, E)$ be a graph with chromatic number $\chi(G)=1000 .$ Let $U \subset V$ be a random subset of $V$ chosen uniformly from among all $2^{|V|}$ subsets of $V$. Let $H=G[U]$ be the induced subgraph of $G$ on $U$. Prove that
$$
\operatorname{Pr}[\chi(H) \leq 400]<1 / 100
$$
This is an exercise on The Probabilistic Method, authors are Noga Alon, Joel H. Spencer.
My copy of the book being in my locked-down office makes it easy to avoid checking which section this question is from for a hint of the expected method, so here's a sledge hammer.
Fix a partition $V = V_1 \cup V_2 \cup \cdots \cup V_{1000}$ into independent sets and let $U_i = U \cap V_i$. A uniformly random subset of $V$ includes each element of $V$ independently with probability $1/2$, so the $U_i$ are independent.
Apply the Hoeffding-Azuma inequality with the martingale that reveals each $U_i$ one at a time. We always have the option to give the vertices of $U_i$ in $H$ their own colour, so each $U_i$ can affect $\chi(H)$ by at most $1$. It follows that $\chi(H)$ is exponentially concentrated in some interval of length on the order of $\sqrt{1000}$, but we don't know where.
Now note that (i) $\chi(G[V\setminus U])$ has the same distribution as $\chi(H)$ and (ii) $\chi(H) + \chi(G[V\setminus U]) \geq 1000$ (else we can combine their colourings to obtain a cheaper colouring of $G$). So if $\Pr(\chi(H) \leq 400) > 1/100$, then we also have $\Pr(\chi(H) \geq 600) > 1/100$, but this isn't consistent with exponential concentration in a short interval.
I haven't checked that the numbers work out in this case, but this method should at least solve some problems of this basic shape.
Right. Your argument shows the expected chromatic number is at least 500, and Azuma then gives us Prob(\chi \leq 400) \leq Prob(\chi - \mu \leq -100) \leq exp(-(100)^2/(2*1000)) = exp(-5) = 0.00673... < 1/100.
|
2025-03-21T14:48:31.374772
| 2020-06-26T15:15:12 |
364204
|
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|
Stack Exchange
|
Constructing functorial homotopies in derived infinity-category
I'm interested in the following problem : let $\mathcal{C}$ be an $\infty$-category and $\mathcal{D}:=D_\infty(\mathbb{Z})$ the derived $\infty$-category of abelian groups. Consider functors $A, B, C ,D :\mathcal{C}\to \mathcal{D}$, and natural transformations $B\to C$, $A\to C$ such that $D=Cone(B\to C)$. To construct a natural transformation $A\to B$, it suffices to construct a functorial homotopy from the composite $\tau : A\to C \to D$ to $0$. Are there some known methods to tackle this problem ?
Let me describe some of the approaches I tried. In the specific case I'm studying, we have some information about the homotopy groups $\pi_n \mathrm{Map}(Ax,Dy)$ for $x,y\in \mathcal{C}$, in particular they are trivial except for $n=1,2,3$.
Since we have $\pi_0 \mathrm{Map}(Ax,Dy)=0$ for all $x,y\in\mathcal{C}$, a result of the type "a natural transformation is (homotopic to) $0$ if and only if it is at each point" would suffice, but I don't expect it to hold.
Since the mapping space $\mathrm{Map}(A,D)$ is the end $\int_{c\in\mathcal{C}}\mathrm{Map}(Ac,Dc)$ which is the totalization of a certain cosimplicial object in $\mathcal{D}$, we might be able to use the spectral sequence of a totalization in a stable $\infty$-category from Remark <IP_ADDRESS> in J. Lurie's "Higher Algebra" ; its first page has rows given by the normalized Moore complex of the cosimplicial abelian groupe formed by the homotopy groups of the cosimplicial object. I don't know how to progress from here since 1) I do not know the exat form of the cosimplicial object and 2) I expect to obtain a fourth quadrant spectral sequence which I don't know how to analyze.
Having learned derived categories from Gelfand & Manin's book, I noticed that in proofs where they construct an homotopy, often the homotopy seems to "write itself". Maybe a brute-force approach, using the specifities of quasicategories, could enable one to construct a functorial homotopy that similarly "writes itself". Are there known examples of a construction of this type ?
Is there maybe a spectral sequence approach for the homotopy groups of the space $\mathrm{Map}_{\mathrm{Map}(A,D)}(\tau, 0)$ ? What I would really like is to have an essentially unique choice, i.e. show that the above space is contractible.
|
2025-03-21T14:48:31.374934
| 2020-06-26T15:15:51 |
364205
|
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"sort": "votes",
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|
Stack Exchange
|
Why doesn't the Manin obstruction work for quadratic forms?
The Manin obstruction explains why the Hasse principle doesn't work for non-quadratic forms. Let's write the notation first;
$V(\mathbb{Q})$ is variety for rational numbers.
$V(A_\mathbb{Q})$ is variety for adèlic points (i.e., it contains real solution and p-adic solution for all $p$).
It's clear that $V(\mathbb{Q}) \subset V(A_\mathbb{Q})$.
Manin obsruction says, we can find new variety set $V(A_\mathbb{Q})^{\text{Br}}$ such that $V(\mathbb{Q}) \subset V(A_\mathbb{Q})^{\text{Br}} \subset V(A_\mathbb{Q})$. If $V(A_\mathbb{Q})$ is not empty but $V(A_\mathbb{Q})^{\text{Br}}$ is empty, then $V(\mathbb{Q})$ must be empty. So we can conclude the defining equations have no rational solution.
But why doesn't it work for quadratic forms?
I estimate that, for quadratic forms, $V(A_\mathbb{Q})^{\text{Br}}$ must be equal to $V(A_\mathbb{Q})$ or $V(\mathbb{Q})$. Is this approach correct?
I could not find an explanation of this in the article and books I have examined. I think, I can use this for my thesis. Therefore I have to explain why it doesn't work or I must have a reference for this.
What are the article(s) and books you examined?
Introduction to Modern Number Theory- Yuri Ivanovic Manin, Alexei A. Panchishkin
The Brauer-Manin Obstruction to the
Hasse Principle - Boris Lerner
Survey of Diophantine Geometry- Serge Lang
I examine this source but I can't find.
|
2025-03-21T14:48:31.375170
| 2020-06-26T16:04:54 |
364210
|
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|
Stack Exchange
|
Fixed locus of a Kahler $S^1$-action
Given a compact Kahler manifold $M$ with an $S^1$-action by Kahler isometries, we know that
Its fixed loci $F=M^{S^1}$is a smooth Kahler submanifold.
It splits $F=\sqcup_{\alpha \in A} F_{\alpha}$ into connected components
The tangent space of M at fixed point splits as a complex $S^1$-representation into weight spaces
$$T_x M =\oplus_{k\in\mathbb{Z}} H_{k}.$$
These are all classical results, e.g. claim 1. follows from the work of Atiyah-Bott and F. Kirwan (the moment map of the $S^1$-action is a Morse-Bott function). Claims 2. and 3. are immediate.
The question is:
Does the tangent space decomposition from claim 3. induces a split of $T_{F_\alpha} M$ into bundles that correspond to weight-spaces?
It doesn't give a trivialization (think of the trivial action...), but indeed a splitting. For $y\in F_{\alpha }$, $S^1$ acts on $T_y(M)$; denote by $t_y$ the action of an element $t\in S^{1}$. The coefficients of the characteristic polynomial $\det(X\cdot \mathrm{I}- t_{y})$ are holomorphic functions on $F_{\alpha }$, hence constant; therefore the eigenvalues of $t_y$ and their multiplicities are independent of $y$. It follows that the eigenspaces $E^\chi_y $ of the action of $S^1$ associated to a character $\chi$ of $S^1$ form a subbundle $E^{\chi }$ of $T(M)_{|F_{\alpha }}$, and that $T(M)_{|F_{\alpha }}$ is the direct sum of these subbundles.
Does this decomposition of the tangent bundle generalises to a compact Lie group action on a smooth manifold? The decomposition of a tangent space according to characters is just representation theory so generalises easily, but the continuation argument from the answer above uses the holomorphicity...
Yes, the argument definitely uses the holomorphicity, and also the compactness. I would guess that the result does not hold in the smooth category, though I don't see an obvious counter-example.
Hmm, maybe it still works: given a non-trivial element $t \in S^1,$ the spectrum of $t_y$ consists of a discrete set ${t^{w_i} \mid w_i \text{ are weights on }y}$ so one cannot "jump" from one set of weights to an another, while on a path from $y$ to $y'.$ Thus I would still think that the weight decomposition is uniform on a path component $F_\alpha,$ even in the case of smooth actions. And possibly even compactness of $F_\alpha$ is not important. Then the eigenspaces $E_k=ker(t^k \cdot I - t_y)$ are smooth in $y$ hence constitute bundles, for each $k$ that appears as a weight.
|
2025-03-21T14:48:31.375369
| 2020-06-26T18:03:18 |
364216
|
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|
Stack Exchange
|
Elliptic curves: about a passage in J. Silverman's "Advanced topics of elliptic curves"
Reading the proof of the main theorem of complex multiplication for elliptic curves over number fields in J. Silverman's book "Advanced topics of elliptic curves" I got stuck at a passage which looks quite innocuous but which I can not provide a reasonable proof about. On page 126, proving Proposition 4.2 the following statement is given. Let $p$ be a prime ideal in $\mathbb{Z}$ which splits completely in the quadratic field extension $K$, so that $pO_{K}=\mathfrak{p}\mathfrak{q}$. Then it is always possible to find an ideal $\mathfrak{a}$ in $O_{K}$, coprime with $pO_{K}$ and such that $\mathfrak{a}\mathfrak{p}$ is principal. I do not see any immediate reason why this should be true? I have found a paper by K. Conrad stating that for any $\mathfrak{a}$ in $O_{K}$ then if $\overline{\mathfrak{a}}$ is the conjugate of $\mathfrak{a}$ (the ideal of all conjugates of the elements of $\mathfrak{a}$), we have that $\mathfrak{a}\overline{\mathfrak{a}}$ is principal. This seems weaker than the statement in J. Silverman's book and still very complicated to prove. Is there any simple proof I am missing?
Are you sure you've got the question right? $pO_K$ itself is principal, so you could take $\mathfrak a=O_K$.
Thank you I was not meaning $pO_{K}$ but $\mathfrak{p}$ above $p$, I have edited the question now
Quickest, although maybe overkill, consider the ideal class of $\mathfrak p^{-1}$, Every ideal class contains infinitely many prime ideals, take any of them (other than $\mathfrak p$ itself if $\mathfrak p^2$ is principal) for $\mathfrak a$.
It looks even easier because there was no requirement that be prime.
@NoamD.Elkies I agree, there probably is an easier way. But for a book at this level, I think it's fair to use results from a first course in algbraic number theory. Also, what I wrote isn't quite right, one should take a prime ideal in the ideal class of $\mathfrak p^{-1}$ other than $\mathfrak p$ and $\mathfrak q$.
Let $a\in O_K$ be such that $a\in\mathfrak p,a\not\in\mathfrak p^2,a\not\in\mathfrak q$ (CRT). Then $(a)\subseteq\mathfrak p$, so $(a)=\mathfrak a\mathfrak p$ for some $\mathfrak a$. The choice of $a$ guarantees $\mathfrak a$ is not divisible by $\mathfrak p$ or $\mathfrak q$.
|
2025-03-21T14:48:31.375560
| 2020-06-26T18:07:34 |
364217
|
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|
Stack Exchange
|
Upper bounds on the sectional curvature of the real Grassmannian
Consider the real Grassmannian as the symmetric space $\operatorname{Gr}(n,k) \cong \operatorname{O}(n)/(\operatorname{O}(k) \times \operatorname{O}(n-k))$ for $n \geq 3$, $k \geq 2$, where the metric is that induced from the bi-invariant metric on $\operatorname{O}(n)$, $\langle X,Y\rangle =\frac{1}{2}\operatorname{tr}(X^\intercal Y)$. The sectional curvature on $\operatorname{O}(n)$ with this bi-invariant metric is given by
$$
\operatorname{sec}_{\operatorname{O}(n)}(X, Y) = \frac{1}{4}\lVert[X,Y]\rVert^2.
$$
where the norm is that induced by the scalar product.
Writing $\mathfrak{h} = \mathfrak{o}(n-k) \oplus \mathfrak{o}(k)$ and $\mathfrak{m} = \mathfrak{h}^\perp \subset \mathfrak{o}(n)$, by the O'Neill's formula and identifying a tangent space of the Grassmannian with a subspace of the Lie algebra of $\operatorname{O}(n)$, we have that the sectional curvature of $\operatorname{Gr}(n,k)$ for a pair of orthonormal vectors $X, Y \in \mathfrak{m}$
$$
\operatorname{sec}_{\operatorname{Gr}(n,k)}(X, Y) = \frac{1}{4}\lVert[X,Y]\rVert^2 + \frac{3}{4}\lVert [X,Y]_{\mathfrak{h}}\rVert^2 = \lVert[X,Y]\rVert^2
$$
since $[\mathfrak{m}, \mathfrak{m}] \subset \mathfrak{h}$.
Using now the bounds for the Lie bracket in $\operatorname{O}(n)$ (see this other MO answer)
$$
\operatorname{sec}_{\operatorname{Gr}(n,k)}(X, Y) \leq 4.
$$
This bound is not tight, as it can be seen by looking at the equality cases for the inequality used, as per Lemma 2.5 in this article.
On the other hand, in this paper the author announces (without proof) in Theorem 3a that
$$
\operatorname{sec}_{\operatorname{Gr}(n,k)}(X, Y) \leq 2.
$$
and even gives examples in Theorem 5a of a submanifold where this bound is achieved.
Is there any reference in which the tighter bound of $2$ is computed? Is there a reference where the tightness of the bound $2$ is also derived?
A proof can be found in this article by Hildebrandt, Jost, and Widman. I reproduce here the proof for completeness.
Consider the usual representation of $\mathfrak{m}$ as matrices of the form
$$
\mathfrak{m} = \left\{
\begin{pmatrix}
0 & A \\
-A^\intercal & 0
\end{pmatrix}
\bigm\vert A \in \mathbb{R}^{(n-k)\times k} \right\}
$$
We can write the sectional curvature of the Grassmannian at $A, B \in \mathbb{R}^{(n-k)\times k}$ for two matrices such that $\operatorname{tr}(A^\intercal B)$ in terms of their Frobenius norm as
$$
\operatorname{sec}_{\operatorname{Gr}(n,k)}(X, Y) =
\frac{
\lVert AB^\intercal - BA^\intercal \rVert_F^2 +
\lVert A^\intercal B - B^\intercal A \rVert_F^2
}{
2\lVert A \rVert_F^2 \lVert B \rVert_F^2
}
$$
In the rest, we will be deliberately imprecise with the limits of the indices to not clutter innecessarily the proof.
Considering the SVD of A, we may assume that $A$ just has non-zero elements in its main diagonal. We can then write the numerator of the sectional curvature as
$$
N = \sum_{i\neq j} (a_{ii}b_{ji}-a_{jj}b_{ij})^2 +
\sum_{i\neq j} (a_{ii}b_{ij}-a_{jj}b_{ji})^2
$$
$$
D = 2(\sum_i a_{ii}^2)(\sum_{j,k}b_{jk}^2)
$$
We can bound $N$ using $(a+b)^2 \leq 2(a^2 + b^2)$ on the summation terms so that
$$
N \leq
4\sum_{i\neq j} a_{ii}^2b_{ji}^2 + 4\sum_{i\neq j} a_{ii}^2b_{ij}^2 \leq
4\sum_{i} a_{ii}^2(\sum_{j \neq i}b_{ji}^2 + \sum_{j \neq i}b_{ij}^2)
$$
and we can bound the denominator as
$$
D = 2(\sum_i a_{ii}^2)(\sum_j b^2_{ji} + \sum_k\sum_{j \neq i}b_{kj}^2)
\geq \frac{1}{2}N.
$$
In the paper they also show the tightness of this bound considering $A = \mathrm{Id}$ and $B =\begin{pmatrix}
0 & -1 \\
1 & 0
\end{pmatrix}$.
|
2025-03-21T14:48:31.375779
| 2020-06-26T19:07:19 |
364225
|
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|
Stack Exchange
|
On Glaeser's result for the square-root of a smooth non-negative function
One of the results due to Georges Glaeser is the following: there exists a non-negative $C^\infty$ function $f$ on the real line, flat at its zeroes, such that $\sqrt{f}$ is not $C^2$. On the other hand, $\sqrt{f}$ is $C^1$ for any such $f$.
Question 1: is it possible to find $f$ as above such that $\sqrt{f}$ is not twice differentiable at a point? In Glaeser's counterexample mentioned above, $\sqrt{f}$ is twice differentiable with an unbounded second derivative.
Question 2: is it possible to find $f$ as above such that there is no function $g$, $C^2$ on the real line such that $f=g^2$. Here $g$ is allowed to take negative values, which is not the case of $\sqrt f$.
Question 3: is it possible to find $f$ as above such that there is no $C^\infty$ function $g:\mathbb R\longrightarrow \mathbb C$ such that $f=\vert g\vert^2$.
Question 1: In
Alekseevsky, Dmitri; Kriegl, Andreas; Michor, Peter W.; Losik, Mark
Choosing roots of polynomials smoothly. (English summary)
Israel J. Math. 105 (1998), 203–233.
it is shown that such an $f$ always has a twice differentiable square root. But this square root is not necessarily positive.
Question 2:
$$
f(t) = \sin^2(1/t)e^{-1/t} + e^{-2/t} \text{ for }t>0,\quad f(t) =
0\text{ for }t\le0.
$$
This is a sum of two non-negative $C^\infty$ functions each of which
has a $C^\infty$ square root.
But the second derivative of the square
root of $f$ is not continuous at the origin. This is also a counter example to question 3.
J.-M. Bony, F. Broglia, F. Colombini and L. Pernazza, Nonnegative. functions as squares or sums of squares, J. Funct. Anal. 232(2006), 137–147.
have shown that twice differentible is best possible; it cannot be improved to $C^{1,\alpha}$ for any continuity module $\alpha$.
See also
Bony, Jean-Michel; Colombini, Ferruccio; Pernazza, Ludovico
On the differentiability class of the admissible square roots of regular nonnegative functions. (English summary) Phase space analysis of partial differential equations, 45–53,
Progr. Nonlinear Differential Equations Appl., 69, Birkhäuser Boston, Boston, MA, 2006.
Thank you. Why do you write that your answer to Question 2 gives a counterexample to Question 3? I claim (in fact Paul Cohen claimed) that there exist non-negative smooth functions which are not finite sum of squares of smooth functions. Question 3 was limited to the case of sum of two squares.
|
2025-03-21T14:48:31.375953
| 2020-06-26T19:30:10 |
364228
|
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|
Stack Exchange
|
Are categories special, foundationally?
Some folks over at nLab want to use categories as a foundation for all of mathematics, I'm guessing as an alternative to sets. Sets work fine, and so do categories, so I have started wondering what other kinds of objects everything could be founded on. The Wikipedia page for categories has a nice table of different group-like structures. In addition to the group-like structures, there are also a lot of other structures that come up in abstract algebra, like rings, vector spaces and modules. Could there be an v-Lab where everyone gets together to phrase everything in math in terms of vector spaces, or is there something special about categories? Is there some kind of property or requirement that an abstract object must have in order to work as a foundational concept?
There is a lot of theory about how powerful each different kind of logic is (for example, some theorems can be proven in ZFC but not in constructive mathematics), so I am curious if there is a similar multiverse for foundational definitions.
Here is a closely related question, but for groups. It is encouraging to see that for groups the answer is "yes." This makes it even more interesting to ask, what are the requirements for foundational objects in general?
Is there any evidence that “the folks at nLab want to use categories as a foundation for all of mathematics”? I have the impression that most of them (if they think about these issues) like (homotopy) types.
@JakobWerner Actually you're right, I have been hearing a lot more about types than categories lately. However, there are a few people that like categories enough to want to found mathematics on them. Maybe after this one I should open a similar question about alternatives to types... (I edited my question to say "some," to hedge.)
This must be referring to ETCC, but I do not know anybody on the nLab who wants to take categories with noninvertible morphisms as a foundation of mathematics. Both sets (e.g., ZFC, elementary toposes, and ETCS) and ∞-groupoids (e.g., HoTT) can be successfully used as a foundation, but I don't think much has been done with ETCC, and we don't even have a reasonably precise working definition of ETCC.
By the way, using ETCS as foundation is not "using categories as foundation"
Yes, any more than using ZFC is "using first-order logic as foundations".
Related: Category theory and set theory: just a different language, or different foundation of mathematics?
The term "foundations of mathematics" is all well and good when one has fixed a foundation to work with. But when you start trying to compare different foundations of mathematics, you quickly realize that the term "foundations of mathematics" requires a great deal of unpacking. I recall once reading a perspicuous article by Penelope Maddy [1] identifying on the order of a dozen different roles we implicitly expect a "foundations of mathematics" to play. This is a testament to the resounding successes of the most widely-used foundations of math, namely ZFC layered on top of (classical, finitary) first-order logic, and variations thereof.
Anyway, if the question is "What are the requirements for foundational objects in general?", then I'm not sure I can do better than pointing you to Maddy. I don't think the question has a simple answer.
Maybe I'll also point out that there are many different ways one might "use category theory as a foundations of mathematics".
One way would be to use Lawvere's ETCC (plus additional axioms as needed) in a similar way to the way ZFC (plus additional axioms as needed) is used by most mathematicians. That is, we write down a particular first-order theory (either ZFC or ETCC) and then just encode everything we want to do inside of it. This is perfectly doable, but I think most would agree that swapping out ZFC in favor of ETCC in this way is awkward and doesn't really have many tangible benefits.
Another way would be to use ETCS, or perhaps the theory of an elementary topos with natural number object, or something like this, instead of ZFC. Again, this is perfectly doable. I think there are some who feel that this approach has merit.
Another way would be to back up, and decide that one isn't interested in just one foundational theory, but in comparing many different ones (note that even the usual foundations are open-ended in this way -- we always allow ourselves to extend ZFC by large cardinal hypotheses and whatnot when necessary). We might find a privileged role for category theory in deciding what is and isn't a "foundations of math", or in comparing different foundations of math. For instance, we might want to compare two different foundations by comparing what each of them says about "the category of sets".
We might also just use category theory as a tool in the foundations of mathematics. For instance, type theoretic foundations are often compared to classical foundations by building models of the former in the latter, and what a "model" is is often expressed in terms of some kind of category.
One might stick with ZFC or whatever, but emphasize that category theory plays a privileged role in comparing different types of mathematical objects, based on the empirical fact that mathematical objects often organize themselves naturally into categories, and such categories can be compared via functors, etc.
As an example of (3), different notions of constructivism can be fruitfully compared in terms of what properties they say the category of sets has (things like being an "exact category", a "pretopos", or whatnot). It's probably due to my ignorance, but I'm not aware of similarly enlightening concepts being more naturally formulated in terms of membership-based set theory. Moreover, new constructivist theories can be formulated and motivated starting from such categorical considerations.
In another direction, support for certain large cardinal principles, such as measurable cardinals or Vopenka's principle, can be given based on categorical formulations of these principles, and categorical considerations about these formulations. Conceivably, one might imagine formulating new large cardinal principles based primarily on categorical considerations, although I'm not sure this has been done much in practice, unless you count Vopenka's Principle or Weak Vopenka's Principle. I actually find it very suggestive that the strongest large cardinal principles (Reinhardt and Berkeley cardinals) are inconsistent with ZFC, so even classically one most naturally retreats to ZF to study them. Although I don't have any evidence for the following scenario, I wouldn't be surprised if even stronger large cardinal principles might one day be found which are inconsistent with ZF, and instead most naturally studied in the context of topos theory or something like that.
And so forth.
[1] I'm not sure, but it may have been What do we want a Foundation to do? (non-paywalled link). Full citation: Maddy P. (2019) What Do We Want a Foundation to Do?. In: Centrone S., Kant D., Sarikaya D. (eds) Reflections on the Foundations of Mathematics. Synthese Library (Studies in Epistemology, Logic, Methodology, and Philosophy of Science), vol 407. Springer, Cham
Excellent answer! I’d add a suggestion that your 5 points fall into two main groups, and this distinction is worth emphasising. Points 1 & 2 use “foundation” in a comparable sense to how ZFC is a foundation: a basic logical formal system for encoding. Points 3–5 use “foundation” in a quite different sense: a set of particularly fundamental organising tools (with 3 & 4 emphasising the application of these tools to “foundations” in the first sense).
Regarding (1): it is common nowadays to assume that of course all the axioms we use in math are unproblematic, and of course all of math can be brought under a single umbrella. Given this, there is indeed "no tangible benefit" to showing these obvious facts. But such conclusions were not always so obvious, and foundational systems such as ZFC were in large part motivated by them. As Maddy says, it's important to be explicit about what you want a foundation to do. You'll get very different answers depending on whether or not you care about Metamathematical Corral or Risk Assessment.
@TimothyChow I see the confusion -- when I wrote "I think most would agree it is awkward and doesn't really have any tangible benefits", I should have clarified that "it" refers to encoding all of mathematics in terms of ETCC. By contrast, I think most would agree that encoding all of mathematics in terms of ZFC is less awkward than encoding all of mathematics in terms of ETCC, and does have the tangible benefits you allude to. What I'm driving at is that to the mathematician interested in "replacing ZFC", ETCS or a topos or type-theoretic approach is probably more attractive than ETCC.
|
2025-03-21T14:48:31.376791
| 2020-06-26T20:36:40 |
364230
|
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|
Stack Exchange
|
Riemann hypothesis for the motivic zeta function
To repair the failure of rationality in general (as shown by Larsen and Lunts for products of two curves of genus > 1) of M. Kapranov's zeta function defined for a variety over a field $k$ and valued in the power series ring over the Grothendieck ring of varieties over $k$, Y. André instead defines the following motivic zeta function for a pure motive $M$ over $k$:
$$ Z_{mot} (M, t) = 1 + \sum_{n=1}^\infty [Sym^nM] t^n \in K_0(Mot_k)[[t]]$$
where $K_0(Mot_k)$ is the $K$-ring of pure motives over $k$.
He shows that $ Z_{mot} (M, t)$ is a rational function of $t$ for any finite-dimensional motive with rational coefficients. B. Kahn has established a functional equation for $ Z_{mot} (M, t)$ relating it to $ Z_{mot} (M^\vee, t)$.
There is no mention of a Riemann hypothesis anywhere, however, that I have been able to find. In analogy with the Weil conjectures and by the requirement of compatibility with them under the point-counting homomorphism when $k$ is a finite field, the motivic Riemann hypothesis should take the form: if $\alpha \in K_0(Mot_k)$ is a root of a polynomial appearing in the numerator or denominator of the rational function $ Z_{mot} (M, t)$, then $\alpha \otimes \alpha^\vee = \mathbb{L}^i$, for an appropriate power $i$ of the class of the affine line.
Is there an obvious reason why it or something similar would not hold true and which is why it is not discussed?
|
2025-03-21T14:48:31.376924
| 2020-06-26T23:12:07 |
364236
|
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|
Stack Exchange
|
Are "strongly finite dimensional" homotopy invariant sheaves with transfers (locally) constant?
Let $k$ be an algebraically closed field. Let $S$ be a homotopy invariant $\mathbb{Q}$-linear sheaf with transfers in the sense of Voevodsky–Suslin, and assume that the dimension of $S(U)$ (over $\mathbb{Q}$) is bounded for a constant $c$ if $U$ is a smooth (connected) $k$-variety. Is it known that $S$ is constant?
I probably have a somewhat clumsy proof of this fact using Suslin rigidity-type arguments; yet I wonder which facts related to this one are already known. Moreover, I am actually interested in the extension of $S$ to pro-smooth (say, affine) $k$-schemes; and my finite dimensionality assumption corresponds to the finite dimensionality of $S(\operatorname{Spec} K)$, where $K$ is an algebraically closed field extension of $k$ of infinite transcendence degree. Consequently, I would also like to know which statements should I cite to deal with "colimit extensions" of this sort; is this section 8.13 of EGA 4?
I have proved this statement as Lemma 5.1.3 at Bondarko and Sosnilo - On Chow-weight homology of geometric motives. Comments are very welcome!
Why not on the arXiv, or at least your web-page? (I always find the maybe-available-maybe-not interface of ResearchGate confusing and slightly scammy seeming.)
I should create my webpage, but I didn't find time to do this yet.(: As about arxiv: there is an alternative version of our paper there. Neither of the versions is currently "strictly better" than another one; thus I would prefer to keep them at distinct places. Maybe, I will merge the versions eventually.
I noticed on the arXiv that there is a paper with different authors: Bondarko and Zumallagov - On Chow-weight homology of motivic complexes and its relation to motivic homology. Is that the one to which you refer?
No; this is the next paper on the subject; it is very much different from both versions of the text that I refer above (the other one is https://arxiv.org/abs/1411.6354). My new co-author is Kumallagov.:)
Oops, sorry for misspelling Kumallagov's name!
|
2025-03-21T14:48:31.377104
| 2020-06-27T00:22:32 |
364238
|
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|
Stack Exchange
|
On a quasi-separated assumption in a lemma for the homotopy exact sequence of the etale fundamental group
Background:
I've seen two versions of the homotopy exact sequence for etale fundamental groups. One from Stacks:
Stacks 0BTX: Let $k$ be a field with algebraic closure $\overline{k}$. Let $X$ be a quasi-compact and quasi-separated scheme over $k$. If the base change $X_{\overline{k}}$ is connected, then there is a short exact sequence $$ 1\to \pi_1(X_{\overline{k}}) \to \pi_1(X)\to \pi_1(\operatorname{Spec} k) \to 1$$ of profinite topological groups.
And one from SGA:
SGA I, Theoreme 6.1 of chapter IX: Suppose $S$ is the spectrum of an Artinian ring $A$ with residue field $k$, $\overline{k}$ an algebraic closure of $k$, $X$ a $S$-scheme, $X_0=X\times_A k$, $\overline{X}_0=X\times_A \overline{k}$, $\overline{a}$ a geometric point of $\overline{X}$, $a$ the image in $X$, and $b$ the image in $S$. We suppose that $X_0$ is quasi-compact and geometrically connected over $k$ (N.B. if $X$ is proper over $S$, this means that $H^0(X_0,\mathcal{O}_{X_0})$ is an artinian local ring whose residue field is radicial over $k$). Then the canonical sequence of homomorphisms $$1\to \pi_1(\overline{X}_0,\overline{a})\to \pi_1(X,a)\to \pi_1(S,b) \to 1$$ is exact, and we have $$\pi_1(S,b)\stackrel{\sim}{\leftarrow}\pi_1(k,\overline{k})= Gal(\overline{k},k).$$
A key step in both of these proofs is the following lemma:
Lemma: Let $X$ be quasi-compact and geometrically connected. If we have a finite etale cover $\overline{Y}$ of $\overline{X}=X\times_k \operatorname{Spec}\overline{k}$, then it comes from a finite etale cover of $X\times_k \operatorname{Spec} K$, where $k\subset K$ is a finite extension.
Stacks adds the assumption that $X$ is quasi-separated, and SGA omits this assumption. In the case when $X$ is assumed quasi-separated, I think I understand how to show this lemma and I can even write down a recipe for producing the extension we need: using the fact that $X$ is quasi-compact and quasi-separated in combination with the definition of etale morphisms as locally of finite presentation, we can pick a finite affine open cover $U_i=\operatorname{Spec} A_i$ of $X$ and get a finite affine open cover $\overline{U_i}=U_i\times_k \operatorname{Spec} \overline{k}$ of $\overline{Y}$ by spectra of rings of the form $(A_i\otimes_k \overline{k})[x_1,\cdots,x_n]/(f_1,\cdots,f_m)$ which gives us a finite list of coefficients from $\overline{k}$ needed to define the $\overline{U_i}$. Covering $U_i\cap U_j$ with a finite number of open affines $U_{ijk}$ since $X$ is quasi-separated, we see that there's a finite number of coefficients from $\overline{k}$ necessary to define the maps $\overline{U_{ijk}}\to \overline{U_i}$ and $\overline{U_{ijk}}\to \overline{U_j}$, and we can apply the same trick to get a finite list of coefficients of $\overline{k}$ needed to define the data we use to patch together the $\overline{U_i}$ into $\overline{Y}$. We end up with a finite list of elements of $\overline{k}$ which are enough to specify all the data needed to put together $\overline{Y}$, and we can define our cover over a finite extension of $k$ containing all these elements.
Question: How can I prove the lemma when $X$ is not quasi-separated? SGA leaves the proof of the lemma to the reader, and it appears to me that my strategy fails without the quasi-separated hypothesis (some $U_i\cap U_j$ could fail to be quasi-compact and then I would have to deal with a potentially infinite list of elements of $\overline{k}$). Stacks' copy of the lemma seems to rely on quasi-separatedness in an essential way, and I don't see how to remove it.
This is more a comment than an answer: a few years back, in 2011, while working with some friends on SGA1, we also found out that we could not prove this statement without the hypothesis that $X$ is quasi-separated. Our question: Is this hypothesis simply missing in SGA1 ? reached Michel Raynaud and his answer was reported to be something like: Probably, but this is not very interesting.
Thank you - I think this more or less resolves my issue (I don't need to prove this in the case that $X$ fails to be quasi-separated, and it does seem reasonable to me that it's a minor error in SGA), but I'm going to leave the question open for a bit longer just in case a more definitive answer should happen to come along.
|
2025-03-21T14:48:31.377502
| 2020-06-27T00:37:06 |
364240
|
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"Nate Eldredge",
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|
Stack Exchange
|
Seminorm which is zero on dense subset
Let $X$ be a Banach space and let $\hat{X}$ be a dense subset of $X$. If $p$ is a seminorm on $X$ such that $p(x) =0 $ for all $x \in \hat{X}$, does $p(x) =0$ for all $x\in X$ (is $p$ the trivial seminorm)? In other words, if $p$ is uniformly zero on a dense subset of $X$, is it zero for all $X$?
For instance, as a special case, are there any seminorms $p$ on $\mathbb{R}$ such that $p(x)=0$ for all $x\in \mathbb{Q}$ but there exists $r$ such that $p(r) \ne 0$?
If the seminorm is continuous this follows immediately, but this is now always the case.
It depends on whether $\hat{X}$ spans $X$ (in the algebraic sense, i.e. finite linear combinations).
If it does, then for every $x \in X$, we can write $x = a_1 x_1 + \dots + a_n x_n$ for some $x_1, \dots, x_n \in \hat{X}$ and some $a_1, \dots, a_n \in \mathbb{R}$. Then if $p$ is any seminorm that vanishes on $\hat{X}$ we have
$$p(x) \le |a_1| p(x_1) + \dots + |a_n| p(x_n) = 0.$$
If $\hat{X}$ does not span $X$, by Zorn's lemma we can find a (necessarily discontinuous) nontrivial linear functional $f : X \to \mathbb{R}$ that vanishes on $\hat{X}$. Set $p(x) = |f(x)|$.
Note that if $X$ is finite dimensional, only the first case can happen, since all proper subspaces of a finite dimensional normed space are closed. In particular your example with $X=\mathbb{R}$, $\hat{X}=\mathbb{Q}$ is impossible.
Thanks so much, I had the spanning case also proved but wasn’t sure if it didn’t span. Is it possible to construct such a functional if a Schauder basis exists for X? If every vector in the basis is in $\hat{X}$ then any finite linear combination will have be zero but in the limit this is necessarily not the case.
@Shash: Yes, the Zorn argument works in any vector space; no other structure matters, and in particular the topology plays no role. It will work as you say: since the functional need not be continuous, it does not have to vanish on infinite linear combinations of elements of $\hat{X}$.
Sorry, should have been more clear, I was wondering if there was an explicit method of constructing one of these. The mention of a Schauder basis was because I'd imagine its impossible for certain uncountable spaces
@Shash: I think that an explicit example will be impossible - that every "example" must use the axiom of choice in an essential way. There are models of set theory in which the axiom of choice fails (but dependent choice still holds) and in which every linear functional on every Banach space is continuous. In such a model I suspect one could also show that every seminorm on a Banach space is continuous. I'd have to think about the details, though.
@Shash: Indeed, in the models I mentioned, every linear map from a Banach space to a normed space is continuous. So if $p$ is any seminorm on $X$, let $E = \ker p$ and let $Y = X / E$, where $Y$ is equipped with the norm $|x+E|_Y = p(x)$. We check this is well defined and is a norm. Then the projection map $\pi : X \to Y$ is continuous, so $p(x) = |\pi(x)|_Y$ is continuous as well. Therefore, in such a model, the second case cannot occur.
|
2025-03-21T14:48:31.377736
| 2020-06-27T02:30:03 |
364244
|
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|
Stack Exchange
|
Run-away functions
Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a continuous function. We say that f has the run-away property if for every compact subset $K\subseteq \mathbb{R}$ there is some positive integer N such that for every $n \geq N$
$$
f^n(K) \cap K = \emptyset.
$$
Some toy examples include:
$f(x)=x+b$ for non-zero b.
$f(x)=\exp(x)$.
Some non-examples are:
$f(x)=x^2 - b$ for positive b (since iterates of any compact neighborhood of the root of $x^2 -x -b$ always contain that fixed point).
$f(x)=\sin(x)$ (since the compact $[-1,1]$ is never escaped).
In general is there a known sufficient condition on f for it to be run-away?
Thoughts:
It seems that unbounded range (not necessarily surjective), and no periodic points are necessary...
Why is monotone necessary? Doesn't $f(x)=|x|+1$ work?
No fixed points (i.e. either $f(x) > x$ for all $x$ or $f(x) < x$ for all $x$) should be necessary and sufficient.
@SamHopkins I guess you're thinking that since $\sum \frac{1}{n^2}$ is convergent we will travel only a finite distance from the initial point. But it is actually wrong! Because $\frac{1}{|x_n|^2}$ is not $\frac{1}{n^2}$, but something much smaller this would give us divergence.
@AlekseiKulikov: yes, I see, you are right!
If we relax the continuity assumption then I think $|f(x)|>x$ works (since then we can take something like $f(x)= -|x|-1$ on intervals $[n,n+1]$ with n even and $f(x)= |x| +1$ otherwise?
You seem to refer to $x = \sqrt b$ as a fixed point of $f(x) = x^2 - b$, but it isn't.
Ah you're right!
As noted in the question's comments by Aleksei Kulikov, a necessary and sufficient condition is given by the following:
Theorem 1
A real continuous function f is a runaway function iff $f(x)=x$ has no solution for $x\in \mathbb{R}$.
To prove this we need the following lemma:
Lemma
Let $f$ be continuous on $\mathbb{R}$ and $f(x)>x$ for all $x \in \mathbb{R}$.
Then for any real values $x$ and $u$ with $x<u$ $$\exists_{N \in \mathbb{N}} \mid \forall_{n\geq N} f^{(n)}(x) > u,$$ and $N$ can be chosen to be less than $1+(u-x)/G$ where $G=\min_{s \in [x,u]} (f(s)-s)$.
Proof
Since $g(x)=f(x)-x$ is continuous, by the extreme value theorem it attains its bounds on $[x,u]$ and in particular there exists $\theta \in [x,u]$ s.t. $g(\theta)\leq g(t)$ for all $t \in [x,u]$. Since $f(x)>x$ for all $x$, $g(\theta)>0$. Hence there exists real $G=g(\theta)>0$ s.t. $f(s)\geq s+G$, for all $s \in [x,u].$
We know that $f(x)\geq x+G$. If $f(x)>u$ we are done so assume that $f(x)\in [x,u]$. Then by the above we have $f^{(2)}(x)=f(f(x))\geq f(x)+G\geq x+2G$. Clearly by induction we can prove $$\forall_{n \in \mathbb{N}}f^{(n)}(x)\leq u \implies \forall_{n \in \mathbb{N}}f^{(n)}(x)\geq x+nG.$$
However if we choose $n>(u-x)/G$ then $f^{(n)}(x)\geq x+nG > x+(u-x)=u$.
This is a contradiction. Hence there must exist an $N \in \mathbb{N}$ s.t. $f^{(N)}(x)>u$ and then clearly since $f(t)>t$ for all $t \in \mathbb{R}$, $f^{(n)}(x)>f^{(n-1)}(x)>\dotsb>f^{(N)}(x)>u$ for all $n\geq N$. Clearly $N$ can be chosen to not exceed $1+(u-x)/G$ and we are done. $\blacksquare$
Proof of Theorem 1
If $f(x)=x$ for some $x\in \mathbb{R}$ then the non-empty compact set $X=\{x\}$ is fixed by $f$ and hence $f^{(n)}(X)\cap X = X \cap X = X\neq \emptyset$ for all $n \in \mathbb{N}$. Thus $f$ is not a runaway function.
If $f(x)\neq x$ for any $x\in \mathbb{R}$ then since $f$ is continuous either $f(x)>x$ or $f(x)<x$ for all $x\in \mathbb{R}$. This is because if the continuous function $g(x)=f(x)-x$ takes both strictly positive and strictly negative values then by the intermediate value theorem it has a real root $a$ which satisfies $f(a)=a$.
Assume wlog $f(x)>x$ for all $x$ (for the other case take continuous $f_1(x)=-f(-x)>x$).
Take any interval $[a,b]$. Lemma 1 then says that for any $x\in[a,b]$ we can find $h(x) \in \mathbb{Z}_{>0}$ s.t. $\forall_{n\geq h(x)} f^{(n)}(x) > b$ and $h(x)$ can be chosen to not exceed $1+(b-x)/G$ where $G=\min_{s \in [a,b]} (f(s)-s)$.
However $1+(b-x)/G<1+(b-a)/G$ and $G'=\min_{s \in [a,b]} (f(s)-s)\leq G = \min_{s \in [x,b]} (f(s)-s)$, and since $f(x)-x$ is continuous, by the extreme value theorem, the minimum is attained at some point $\mu \in [a,b]$. Thus $G'=f(\mu)-\mu>0$.
Hence $h(x)$ can be chosen not to exceed $h=1+(b-a)/G'$ where $G'=\min_{s \in [a,b]} (f(s)-s)>0$. Thus for any $x\in[a,b]$ we can find $h \in \mathbb{Z}_{>0}$ s.t. $\forall_{n\geq h} f^{(n)}(x) > b$. Clearly this implies that $\forall_{n\geq h} f^{(n)}([a,b]) \cap [a,b] = \emptyset$.
Now any compact set $S$ in $\mathbb{R}$ is bounded hence we can find a closed interval $[a,b]$ which contains it. By the above we can find $h \in \mathbb{Z}_{>0}$ s.t. $\forall_{n\geq h} f^{(n)}([a,b]) \cap [a,b] = \emptyset $ which implies that $$\forall_{n\geq h}( f^{(n)}(S)\cap S \subset f^{(n)}([a,b]) \cap [a,b] = \emptyset ).$$
Hence we have proved that $f$ is a runaway function if there does not exist $x\in \mathbb{R}$ s.t. $f(x)=x$. This, combined with the first implication, proves the result. $\blacksquare$
this seems like it should be a known fact...do you have a reference by chance?
@AIM_BLB Sorry I did search but didn't find anything relevant so I just worked this out from scratch. I am reasonably sure there must be something more general you can deduce this from if you look hard enough.
You assume that WLOG $f(x)>0$ but the function $x \mapsto x + 1$ is runnaway, continuous, and fails this.... Why is this not a problem?
Thanks for your comment - I meant to say wlog $g(x)>0$ so $f(x)>x$. I've corrected this now.
That's what I assumed you meant but I just wanted to be sure :)
In the statement of your lemma, you take $G$ to be the minimum of $f(s) - s$ for $s \in [x, u]$; but, in the proof, you require $G$ to be strictly smaller than the minimum. (This doesn't much matter, since it only makes the upper bound $1 + (u - x)/G$ bigger, but I found the notational mismatch confusing.)
@LSpice Thank you for your comment which I definitely agree with, I will fix it now.
|
2025-03-21T14:48:31.378135
| 2020-06-27T03:09:07 |
364245
|
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|
Stack Exchange
|
Bound for multinomial expansion involving Poisson random variables
Let $x_i, i=1, \ldots n$ be Poisson random variables with parameters $\lambda_i$ correspondingly with condition that $\sum_{i=1}^nx_i=T$. Due to linearity of the expectation one can write:
$$
E\left(\left|\sum_{i=1}^n a_ix_i\right|^{2k} \big| \sum_{i=1}^nx_i=T\right)\\
=\sum_{k_1+\ldots k_n=2k}\frac{(2k)!}{k_1!\ldots k_n!}a_1^{k_1} \ldots a_n^{k_n}E\left(x_1^{k_1}\ldots x_n^{k_n}\big | \sum_{i=1}^nx_i=T\right)
$$
I would like to bound this expression from above. Ideally, I would like to get something like $C\times E\left(x_1^{k_1}\ldots x_n^{k_n}\big | \sum_{i=1}^nx_i=T\right)\times \|a\|_1$ in the right hand side. Or, at least to understand in which cases this bound would hold.
But I am not sure on how to take into account all the possible cases for $k_i \in \{0, \ldots, 2k\}$?
Do you assume independence of the random variables $x_1,\ldots,x_n$?
The poisson variables are slightly dependent by the condition that their sum is equal to $T$.
@user124297: The term "slightly dependent" is unknown to me. I suppose that $x_1,\ldots,x_n$ are independent and that you then work with conditional probabilities. Right?
Yes you can think about it in this way. In the beginning $x_i$ are independent, but when you put conditional probability they are not independent anymore. In any case, you are working with conditional probability.
Have you tried the case $n = 2$? In this case with respect to $P(.|x_1+x_2 = T)$ $x_1$ has the binomial distribution $Bin(T,\lambda_1/(\lambda_1+\lambda_2))$. (N.B.: Without independence of $x_1,x_2$ the distribution of $x_1$ can be rather arbitrary.)
Yes, I am ok with this case. But when we are getting to n variables, when the distribution is multinational.
For the following we only need that $X_1,\ldots,X_n$ are arbitrary random variables with values in $\mathbb{N}_0$ such that $\mathbb{P}(X_1+\ldots+X_n = T) > 0$. (In particular the original situation is included.) Let $Q(A) := \mathbb{P}(A | X_1+\ldots+X_n = T)$ for measurable $A$. Let $E_Q$ be the expectation w.r.t. $Q$. Then the question reduces to: Does $C > 0$ exist with
$$E_Q\left( |\sum_{i=1}^n a_iX_i|^{2k} \right) \leq C \cdot E_Q\left( |\sum_{i=1}^n X_i|^{2k} \right) \cdot \|a\|_1$$
First, such $C$ cannot exist if $2k > 1$. Let $a = (t,\ldots,t)$ with $t > 0$. Then $E_Q\left( |\sum_{i=1}^n a_iX_i|^{2k} \right) = \|a\|_1^{2k}/n^{2k} \cdot E_Q\left( |\sum_{i=1}^n X_i|^{2k} \right)$. Thus we have to replace at least $\|a\|_1$, f.i. with $\|a\|_1^{2k}$. But then the inequality it trivial. Given $a$, let $t := \max\{|a_1|,\ldots,|a_n|\} = \|a\|_\infty$. Then
$$E_Q\left( |\sum_{i=1}^n a_iX_i|^{2k} \right) \leq E_Q\left( |\sum_{i=1}^n X_i|^{2k} \cdot \|a\|_\infty^{2k} \right) \leq E_Q\left( |\sum_{i=1}^n X_i|^{2k} \right) \cdot \|a\|_1^{2k},$$
since $\|a\|_\infty \leq \|a\|_1$. As simple examples show there is no better $C$.
Thank you, is it possible to actually find this constant $C$? Also, does it mean that the bound would hold for negative moments instead? And is it possible to find an upper bound for the original problem?
First, $C = 1$, since $|a|_\infty \leq |a|1$ and there is no better $C$, since $|(1,0,\ldots,0)|\infty = |(1,0,\ldots,0)|_1$. Further, as shown in the first part, there actually is no $C$ with the required properties. I've edited my answer.
Thank you. How about situation when $C$ is dependent on $N$? Would it be possible to get such a $C$? I guess one would have to go through combination of all the possible $k_i$?
Hello, actually $C$ does not depend on $n$!
|
2025-03-21T14:48:31.378671
| 2020-06-27T07:56:28 |
364251
|
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"Noah",
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"sort": "votes",
"url": "https://mathoverflow.net/questions/364251"
}
|
Stack Exchange
|
pro-commutative group schemes
When $k$ is field, Demazure and Gabriel defined and worked with the category of commutative pro-algebraic groups over $k$. In their book, they proved that $Ext^n(\varprojlim G_i, H)= \varinjlim Ext^n(G_i,H)$ (Chap.5, cor. 3.9), the Ext is taken within the category of commutative pro-algebraic groups. When working with general base $S$, I am aware that the category of commuatative group schemes over $S$ needs not be abelian. But do we have similar construction of the pro-category here? More importantly (this is really the question I concern) do we have the isomorphism $Ext^n(\varprojlim G_i, H)= \varinjlim Ext^n(G_i,H)$ in general?
@მამუკაჯიბლაძე Oh no.. sorry for the typos. edited.
|
2025-03-21T14:48:31.378752
| 2020-06-27T09:06:15 |
364254
|
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/364254"
}
|
Stack Exchange
|
Hermite polynomial after rotation
When we consider the $n$-dimensional standard normal distribution, the orthogonal basis is $\{H_S(x)\}_{S}$ where $H_S(x) = \prod_{k=1}^n H_{s_k}(x_k)$. Here $H_*(x)$ is the normalized probabilist's hermite polynomial. Suppose $U$ is any real orthogonal matrix. How to express $H_S(Ux)$ in terms of $\{H_S(x)\}_{S}$?
The expansion coefficients of a function $f(x_1,x_2,\ldots x_n)$ in the rotated basis of Hermite polynomials are related to the original expansion coefficients by an orthogonal matrix of "steering coefficients". Explicit expressions for $n=1,2,3$ are given in section 3.6 of K.L. Reynolds, Convolution, Rotation, and Data Fusion with Orthogonal Expansions. See also Accurate Image Rotation using Hermite Expansions.
|
2025-03-21T14:48:31.378824
| 2020-06-27T09:27:16 |
364256
|
{
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"Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/"
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"authors": [
"Derek Holt",
"HIMANSHU",
"https://mathoverflow.net/users/101534",
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/364256"
}
|
Stack Exchange
|
Presentations of groups of order $p^4$
In the (xi) group of the classification of groups of order $p^4$ given by W.Burnside in his book," Theory of Groups Of Finite Order". The group ($\mathbb{Z_{p^{2}}}\rtimes \mathbb{Z_{p^{}}}) \rtimes_{\phi}\mathbb{Z_{p^{}}} $, have presentation
$$<a,b,c : a^{p^{2}}=b^p=c^p=e, ab=ba^{1+p},ac=cab,bc=cb>$$
From the above relations, I can produce following relation
$$a^ib^j=b^ja^{{(1+p)}^ji},$$ Now I am trying to obtain similar relation for the generators $a$ and $c$. (Here $i$ and $j$ are natural numbers). Kindly help me out as I am not able to proceed further to write $a^ic^j= ? ?$ in simple manner.
Cross posted to https://math.stackexchange.com/questions/3736295
$$a^ic^j = c^j(ab^j)^i.$$
I expect you could use your existing formula for $a^ib^j$ to write $(ab^j)^i$ in the form $b^ka^l$, but I will leave that to you!
I got my final answer as $$a^ic^j=c^jb^{ij}a^{i+\frac{ji(i+1)}{2}p.}$$
Yes I checked a few cases, and that seems to be correct!
|
2025-03-21T14:48:31.378912
| 2020-06-27T12:07:11 |
364263
|
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/364263"
}
|
Stack Exchange
|
Boundary regularity of rectifiable multiplicity 1 hypercurrents
Background. I have just recently started studying this aspect of geometric measure theory (and I am also by no means well versed in the latter) and I really can not seem to get the slightest hang of it and the existing older publications. Since I furthermore cannot always bother specific smarter people with me not understanding and the problem is sadly quite important for me, I'm now passing to posting a question here.
Problem setting. My question deals with sufficient conditions for boundary regularity and under which circumstances they can be achieved (hence implying the corresponding regularity near a given regular boundary support). Moreover, this is for me only interesting in codimension 1, ie. we are dealing only with hypercurrents. I am aware of two particular papers about boundary regularity, namely one by Hardt (Comm. PDE) and one by Hardt-Simon (Ann. Math.), and also of some miscellaneous related works (eg. by Brothers about tangent cones). I also tried to consult the book of Francesco Maggi, which unfortunately performs the stunt of relating in its last sentence of the corresponding Part III to the more general works above that I do not get.
Terminology. We consider a rectifiable current $T$ in $ \mathbb{R}^n $ of codimension $1$. We assume moreover that $T$ has multiplicity $1$ almost everywhere. (We may also as well for my problem assume that
$$
T=\partial [E] \llcorner \Omega,
$$
ie. $T$ is generated as the restriction of an oriented boundary to some differentiable domain $\Omega$.) Let also $\psi$ denote an elliptic parametric integrand of sufficient differentiability on $\mathbb{R}^n$ and assume that $T$ is absolutely minimizing for $\psi$ on $\mathbb{R}^n$. Then let $B$ denote a sufficiently differentiable orientable submanifold of $\mathbb{R}^n$ of codimension 2 without boundary and assume
$$
\partial T = [B]
$$
with $B$ sufficiently oriented. Assuming that $0\in B$, the above mentioned work by Hardt (Thm. 3.6) assures that $T$ is actually a manifold with boundary near $0$ given
$$
\Theta^{n-1}_*(T,a) \leq \frac12,
$$
ie. we may bound at least the lower density by what one would expect for a manifold with boundary.
Question. Can one always verify the density-estimate in this case of a multiplicity 1 current? Or can one do this under the assumption of dealing with an oriented boundary? Are there other sufficient conditions on $T$ implying such an estimate in this case (ie. codimension and multiplicity one)?
From what I have read and heard, it should be possible (and intuitively makes sense but that is always very dangerous), but I am kind of blindsided on how to get the tech right (and even what to use to get it right). Thanks to everyone in advance.
Edit: I just realised that I misread your question; let me correct my answer accordingly.
When $\psi$ is the area-functional the paper by Hardt-Simon that you cite seems to answer your question, by giving a complete boundary regularity theorem in codimension one. To paraphrase this result here, it states that if $U$ is an open subset of $\mathbf{R}^{n+1}$, $0 < \alpha < 1$, $T$ is an $n$-dimensional area-minimising current, and $\partial T$ is a connected oriented embedded $C^{1,\alpha}$ submanifold of $U$, then $V \cap \mathrm{spt} \, T$ is a connected embedded $C^{1,\alpha}$ hypersurface with boundary. In particular, near a boundary point $X \in U \cap \mathrm{spt} \partial T$ the Allard-type condition $\Theta(T,X) = 1/2$ is a posteriori satisfied.
Moreover Hardt-Simon remark in the introduction that it when $\psi$ is a general elliptic integrand then the boundary regularity is not known even in dimension $n = 2$, although this might well have changed since then.
I am bit confused about the part of your question relating to the orientability of the boundary, as working with currents imposes this hypothesis. If you are interested in the 'unoriented' setting, you should consult the literature on flat chains mod two.
Although you claim to only be interested in the codimension one case, let me finally point out that Hardt-Simon's boundary regularity result does not extend to higher codimension, even with multiplicity one. White gives the example of $\{ (z^3,z^4) \mid \mathrm{Im} \, z \geq 0 \} \subset \mathbf{C}^2 = \mathbf{R}^4$, which is area-minimising but has a boundary singularity at the origin.
|
2025-03-21T14:48:31.379314
| 2020-06-27T12:33:06 |
364265
|
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"authors": [
"Andreas Blass",
"Andrés E. Caicedo",
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"url": "https://mathoverflow.net/questions/364265"
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|
Stack Exchange
|
Is it consistent to have a function that is sensitive to subset relation from the power set of a set to that set?
Is it consistent with $ZF$ to have a set $S$ and a function $F: P(S) \to S$ such that:
$\forall X,Y \in P(S): X \subsetneq Y \implies F(X) \neq F(Y)$
No, this is not consistent. Todorčević has shown in ZF that, in fact, there is no function $F\!:\mathcal W(S)\to S$ with the property you require. Here, $\mathcal W(S)$ is the collection of subsets of $S$ that are well-orderable.
This is corollary 6 in
MR0793235 (87d:03126). Todorčević, Stevo. Partition relations for partially ordered sets. Acta Math. 155 (1985), no. 1-2, 1–25.
The corollary is a consequence of theorem 5, stating that for any structure $\mathcal M$ with one binary relation, $\sigma \mathcal M$ is not "$\mathcal M$-embeddable". Here, if $\mathcal M=(M,R)$, then $\sigma\mathcal M=(\sigma M,\subseteq)$, where $\sigma M$ is the set of all one-to-one maps $s$ with domain an ordinal $\alpha$ such that whenever $\beta<\gamma<\alpha$, we have $s(\beta)\mathrel R s(\gamma)$. In Stevo's notation, that $(A,S)$ is $(B, T)$-embeddable means that there is a function $f\!:A\to B$ such that $f(a) \mathrel T f(b)$ and $f(a)\ne f(b)$ whenever $a,b\in A$, $a\mathrel S b$, and $a\ne b$.
An alternative proof of the result about $F:\mathcal W(S)\to S$ without introducing $\sigma\mathcal M$: If there were such an $F$, then define, by recursion on ordinals $\alpha$, $G(\alpha)=F({G(\beta):\beta<\alpha})$ and note that $G$ maps all the ordinals one-to-one into $S$. (To avoid mentioning the proper class of all ordinals, just restrict to $\alpha<$ Hartogs number of $S$.)
@Andreas Yes, this is in essence the argument.
Here's an argument that doesn't use ordinals, as an alternative to the nice proof described by Andrés and Andreas.
Take $F: P(S) \to S$ satisfying your hypothesis. Define a function $\Phi: P(S) \to P(S)$ by
$$
\Phi(X) = \{F(Y): Y \subseteq X\}.
$$
Then $\Phi$ is an order-preserving map from the complete lattice $P(S)$ to itself, so there is a least $X \in P(S)$ such that $\Phi(X) \subseteq X$. (That such an $X$ exists is part of the proof of the Knaster-Tarski fixed point theorem, and is in any case easy: put $X = \bigcap\{ Y \in P(S): \Phi(Y) \subseteq Y\}$, then use the fact that $\Phi$ is order-preserving to show that $\Phi(X) \subseteq X$.)
Now:
$F(X) \in \Phi(X)$ by definition of $\Phi$, so $F(X) \in X$, so if we write $X' = X \setminus\{F(X)\}$ then $X' \subsetneqq X$.
$\Phi(X') \subseteq \Phi(X) \subseteq X$, so $\Phi(X') \subseteq X$.
Any subset $Y$ of $X'$ is a proper subset of $X$, so your hypothesis on $F$ gives $F(Y) \neq F(X)$. Hence $F(X) \not\in \Phi(X')$.
Combining the three bullet points, we have a proper subset $X'$ of $X$ satisfying $\Phi(X') \subseteq X'$. This contradicts the minimality of $X$.
Very nice! $\ \ $
|
2025-03-21T14:48:31.379532
| 2020-06-27T13:04:53 |
364267
|
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"sort": "votes",
"url": "https://mathoverflow.net/questions/364267"
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|
Stack Exchange
|
Rings or algebras with many nilpotent elements and efficient computation
Crossposted from quantum.SE
where comment appears to suggest that solving modulo 2 might
be possible.
Searching the web for '"quantum computer" nilpotent'
returns many results, so maybe the question is ontopic
for this site.
Can a quantum computer solve the following mathematical
problem:
This is related to an open problem, so likely the answer is negative.
The problem is Cycle Enumeration using Nilpotent Adjacency
Matrices with Algorithm Runtime Comparisons pp 2-3
Is there commutative ring or commutative algebra $R$ with the following properties:
There are $n$ nilpotent elements $a_i$ satisfying $a_i^2=0$
$a_1 a_2 \cdots a_n \ne 0$.
Computation in $R$ is efficient: for an $n$ by $n$ matrix $M$
with entries zero and $a_i$, for natural $m$ we can compute
$M^m$ in time polynomial in $nm$.
If we omit the efficiency constraint, the answer is easy:
Take $R=K[a_1,a_2,...a_n]/(a_1^2,a_2^2,...a_n^2)$ for any ring $K$.
If we omit commutativity, there are solutions with matrices.
Let's consider a commutative $R$-algebra satisfying assumptions 1 and 2 in the question, where $R$ is a commutative ring. Let's start with a few observations:
For every $U \subseteq \{1,...,n\}$, we have $a_U := \prod_{u \in U}a_u \neq 0$. Otherwise, $\prod_{u=1}^n a_u = 0$, which contradicts assumption 2 in the question. (We set $a_\emptyset = 1$ by convention).
For every $U, V \subseteq \{1,...,n\}$, if $a_U = a_V$ then necessarily $U = V$. To see this, assume that $a_U = a_V$ and that $U \neq V$. Without loss of generality assume $U \not\subseteq V$ and let $x \in U \backslash V$. Then $0 = a_x^2a_{U \backslash\{x\}}= a_{V \sqcup \{x\}}\neq 0$. Conclude by contradiction with point 1 above.
For each $U \subseteq V$ write $q_U^V$ for the quotient map of $R$-modules $q_U^V: R a_U \rightarrow R a_V$ defined by $x \mapsto x a_{V \backslash U}$. Write $K := R a_{\{1,...,n\}})$ and $q_U := q_U^{\{1,...,n\}}: R a_U \rightarrow K$.
We can now prove an interesting fact about $R$-linear combinations of the $a_U$ elements, namely that if for some coefficients $x_U \in R$ we have $\sum_{U \subseteq \{1,...,n\}} x_U a_U = 0$ then for every $V \subseteq \{1,...,n\}$ we also have:
$$
\sum_{U \subseteq V}
q_U^{U\sqcup \overline{V}}\!\!\left(x_U\right) a_{U \sqcup \overline{V}}
=
\left(\sum_{U \subseteq \{1,...,n\}} x_U a_U\right)a_{\overline{V}}
= 0
$$
where we have written $\overline{V} := \{1,...,n\} \backslash V$. In particular, if $\mathcal{U}\subset \mathcal{P}$ is an antichain (i.e. if for all $U, V \in \mathcal{U}$ we have that $U \subseteq V$ implies $U=V$), then the elements $(a_U)_{U \in \mathcal{U}}$ are linearly independent over some suitable (non-trivial) quotient of $R$.
Because there are antichains $\mathcal{U}$ with size exponential in $n$, e.g. those formed by subsets $U$ with size $\frac{n}{2}$ for even $n$, this is an indication that computation in the $R$-algebra is not going to be efficient over some suitable (non-trivial) quotient of $R$.
PS: I think this last argument can be made formal by defining a surjective ring homomorphism $f: S \rightarrow Q$ from the sub-ring $S$ spanned by $R$-linear combinations of the elements $a_U$ for all $U \subseteq \{1,...,n\}$ to a suitable quotient ring $Q$ of $K[a_1,...,a_n]/(a_1^2,...,a_n^2)$ as follows:
$$
f\left(\sum_{U \subseteq \{1,...,n\}} x_U a_U\right)
:=
\sum_{U \subseteq \{1,...,n\}} q_U(x_U) a_U
$$
I have not worked out the details yet, I might do so in the future if of interest.
|
2025-03-21T14:48:31.379764
| 2020-06-27T13:05:00 |
364268
|
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|
Stack Exchange
|
Scheme-theoretic image of the inverse image of a morphism of schemes
Let $f:X \to Y$ be a finite, surjective morphisms between noetherian, integral varieties (over $\mathbb{C}$). I am looking for conditions on $f$ under which I can say that for any closed subscheme $Z \subset Y$, the scheme-theoretic image (under $f$) of the scheme-theoretic inverse image $f^{-1}(Z)$ (obtained by taking fiber product with $f$) is isomorphic to $Z$? Do I need to assume that $f$ is smooth or a normalization map or ...?
$f$ flat is certainly enough.
@abx Thanks, how about if $f$ is a normalization map? Are there counterexamples in this case?
In fact, in your situation this property is equivalent to $f$ flat — see Bourbaki's Commutative Algebra, ch. I, §3.5, Proposition 9. So it is never true when $f$ is a normalization map, except if it is an isomorphism.
|
2025-03-21T14:48:31.379843
| 2020-06-27T13:26:28 |
364269
|
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"site": "mathoverflow.net",
"sort": "votes",
"url": "https://mathoverflow.net/questions/364269"
}
|
Stack Exchange
|
On maximum principle of spectral fractional Laplacian
Suppose $(-\Delta)^s u=g \geq 0$ in $\Omega$ and $u=0$ in $ \partial \Omega.$ Also suppose $u$ is $C^{2}$ non-negative and $(-\Delta)^s u=0$ in $\Omega \setminus B$ and $u\leq a$ on $\partial B $ where $B\subset \Omega$ is a unit ball and $a$ is a positive constant. Is it true that $u\leq a$ in $\Omega \setminus B$ where $s\in (0,1)$ and $\Omega$ is a smooth bounded domain.
|
2025-03-21T14:48:31.379907
| 2020-06-27T13:37:13 |
364270
|
{
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"sort": "votes",
"url": "https://mathoverflow.net/questions/364270"
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|
Stack Exchange
|
Existence of entire function that yields periodicity
I have the following question:
Does there exist an entire function $f(z)$ where $z=x+iy$ such that
$$g(x,y) =e^{-2\pi y^2}f(z)$$
is periodic in both $x$ and $y$ direction, i.e. $$\forall x,y: g(1,y)=g(0,y) \text{ and }g(x,1)=g(x,0).$$
$g(1, y) = g(0, y)$ doesn't sound like periodicity. May be you've meant $g(x+1, y) = g(x, y)$? If so, then I believe the following should be a promising approach: from your periodicity prove an a priory bound that $f$ is entire of order $2$ with periodic zeroes. This means that $f$ is a finite product of shifts of Weierstrass sigma functions times $\exp(az^2+bz+c)$, which then should be easy to check whether it works or not.
If you correct your definition to the correct definition of periodicity, $g(x,y+1)=g(x,y)$, for all $x,y$, then the answer is no (except when $f=0$). Indeed, let $z=x+iy$, and assuming $g$ is periodic with respect to $y$, we obtain
$$f(z+i)=g(x,y+1)e^{-2\pi(y+1)^2}=g(x,y)e^{-2\pi y^2}e^{-4\pi y-2\pi}=f(z)e^{-4\pi y-2\pi},$$
and this holds identically in $z$. Therefore $f(z+i)/f(z)=\exp(-4\pi y-2\pi)$ is not an analytic function anywhere, since it is real, contradiction.
On the other hand, if one takes your conditions literally as you wrote them (and don't call them "periodicity"), then such a function exists:
namely $f(z)=\exp(-2\pi iz)$. It satisfies $f(z+1)=f(z)$ and $f(z+i)=e^{2\pi}f(z)$, therefore it satisfies your conditions.
Moreover, this $f$ is unique, up to a constant factor.
Indeed, your first conditon is equivalent to $f(z+1)=f(z)$ therefore $f$ has period one, and has Fourier expansion
$$f(z)=\sum_{-\infty}^\infty c_ne^{2\pi inz}=g(e^{2\pi i z}),$$
where $g$ is analytic in $C^*$. And your second condition means that $g(e^{-2\pi} w)=e^{2\pi}g(w)$ which easily implies that all $c_j$ except $c_{-1}$ are zero. This gives $f(z)=ce^{-2\pi iz}$.
Technically one needs to exclude the degenerate case $f=0$ in your analysis of interpretation 1.
@Terry Tao: Thanks. I corrected.
|
2025-03-21T14:48:31.380047
| 2020-06-27T13:40:01 |
364271
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|
Stack Exchange
|
Marcinkiewicz-Mihlin-Hormander Fourier multiplier theorem
I'm trying to understand the hypothesis of the Marcinkiewicz-Mihlin-Hörmander multiplier theorem. See for instance Theorem A in this paper of Elias Stein.
Theorem A: Assume that $m: (0, \infty)\to \mathbb R$ satisfies the following equation
$$
|m^{(j)}(x)| \leq C x^{-j}\quad0 \leq j \leq k,\; k>\frac{d}{2},\label{1}\tag{1}
$$
rr more generally
$$
\sup_{t>0} \|\chi m(t \cdot)\|_{L^2_{\alpha}}< \infty \label{2}\tag{2}
$$
where $\chi$ is a non-zero smooth cut-off function of compact support that vanishes near the origin.
Then Fourier-multiplier operator $\widehat{Tf}= m(|\xi|^2) \hat{f}$ is bounded on $L^p (\mathbb R^d)$ for $1<p< \infty.$
My questions are:
(A) How to define $\|\cdot\|_{L^2_{\alpha}}$? Is it standard notation? (It seems that Stein has not defined this notation in his paper. Thus I guess it must be standard)
(B) Why condition \eqref{1} implies \eqref{2} in Theorem A?
(C) Does symbol $m(\xi)=e^{i|\xi|^2}$ satisfy condition \eqref{2} of Theorem A? (I am interested in this problem since, by using this symbol, we can solve Schrodinger's equation)
(D) How this theorem has been developed historically? I mean is it correct to say that the first version of this result goes back to Marcinkiewicz in 1939, and later versions were Mihlin's in 1957 and finally Hormander's in 1960?
About question A above, my heuristic guess is that we can define $\|f\|_{L^2_{\alpha}}^2$ as
$$
\|f\|_{L^2_{\alpha}}^2= \int_{\mathbb R^d} |f(x)|^2 (1+|x|^2)^{\alpha} \mathrm{d}x
$$
or
$$
\|f\|_{L^2_{\alpha}}^2= \int_{\mathbb R^d} |\widehat{f}(\xi)|^2 (1+|\xi|^2)^{\alpha} \mathrm{d}x
$$
Is my guess correct or am I missing something?
Your second definition of $L^2_\alpha$ is right. Notice that $|\partial^j(\chi m(tx))|\le\sum_l c_lt^l|\partial^{j-l}\chi,(\partial^l m)(tx)|\le C\sum_l|\partial^{j-l}\chi||x|^{-l}$ and the $L^2$-norm is uniformly bounded. The symbol $e^{i|\xi|^2}$ doesn't satisfy the hypotheses; furthermore, the $L^p\to L^p$ bound only holds for $p=2$; use examples $e^{i|x|^2}f$ for suitable functions $f$, say $f=1/(1+|x|^2)^\alpha$ for $\alpha<n/2$.
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